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Registered: 10-19-2015
## Concatenation of multiple variables into one large variable.
Hi All,
I am faced with following task.
I have following array in my HLS code as an output.
ap_uint<4> ArrayA[256];
I want to convert this ArrayAto ap_uint<64> ArrayB[4];
In other words, I want to concatenate 64 elements of ArrayA into 1 element of ArrayB. I am wondering what is the most efficient way of doing this in HLS ?
According to Xilinx documentation, apint_concatenate() does provide concatenation, but it seems this function is only takes 2 clock cycles. I want to do my concatenation in one clock cycle.
I am wondering what is the best way of doing this in HLS ?
Thanks.
Tags (6)
1 Solution
Accepted Solutions
Scholar
7,625 Views
Registered: 03-22-2016
@akboken
The union idea wont' work. I just tried and there are two problems:
1. The minimum size of a ap_uint<4> is one byte
2. unrestricted unions (unions with non-pod types) are only allowed in c++0x
Perhaps a better idea would be:
```#include "ap_int.h"
#include <stdint.h>
uint32_t example( uint32_t v1, uint32_t v2 )
{
struct Nibble2 {
uint8_t lo :4;
uint8_t hi :4;
};
union {
Nibble2 ArrayA[4];
uint32_t ArrayB[1];
};
ArrayA[0].lo = v1;
ArrayA[0].hi = v2;
ArrayA[1].lo = v1+1;
ArrayA[1].hi = v2+1;
ArrayA[2].lo = v1+2;
ArrayA[2].hi = v2+2;
ArrayA[3].lo = v1+3;
ArrayA[3].hi = v2+3;
return ArrayB[0];
}
```
vitorian.com --- We do this for fun. Always give kudos. Accept as solution if your question was answered.
I will not answer to personal messages - use the forums instead.
2 Replies
Scholar
7,626 Views
Registered: 03-22-2016
@akboken
The union idea wont' work. I just tried and there are two problems:
1. The minimum size of a ap_uint<4> is one byte
2. unrestricted unions (unions with non-pod types) are only allowed in c++0x
Perhaps a better idea would be:
```#include "ap_int.h"
#include <stdint.h>
uint32_t example( uint32_t v1, uint32_t v2 )
{
struct Nibble2 {
uint8_t lo :4;
uint8_t hi :4;
};
union {
Nibble2 ArrayA[4];
uint32_t ArrayB[1];
};
ArrayA[0].lo = v1;
ArrayA[0].hi = v2;
ArrayA[1].lo = v1+1;
ArrayA[1].hi = v2+1;
ArrayA[2].lo = v1+2;
ArrayA[2].hi = v2+2;
ArrayA[3].lo = v1+3;
ArrayA[3].hi = v2+3;
return ArrayB[0];
}
```
vitorian.com --- We do this for fun. Always give kudos. Accept as solution if your question was answered.
I will not answer to personal messages - use the forums instead.
Voyager
5,129 Views
Registered: 06-24-2013
Hmm, arriving late ... what's wrong with:
```void foo(ap_uint<4> a[64], ap_uint<64> q[4]) {
#pragma HLS INTERFACE ap_fifo port=a,q
#pragma HLS INTERFACE ap_ctrl_none port=return
for (int i=0; i<4; i++) {
ap_uint<64> temp;
for (int j=0; j<16; j++) {
temp(j*4+3, j*4) = a[i*16+j];
}
q[i] = temp;
}
}
``` | 929 | 2,838 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-39 | latest | en | 0.809756 |
https://helpinhomework.org/question/3890/ISolar-Luminosityltbrgtltbrgtltbrgt1-Stars-like-the-Sun-radiate-to-a-pretty-goo | 1,718,457,621,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861594.22/warc/CC-MAIN-20240615124455-20240615154455-00876.warc.gz | 262,928,041 | 21,194 | Fill This Form To Receive Instant Help
# I. Solar Luminosity<br/><br/><br/>1. Stars like the Sun radiate (to
a pretty good approximation) like blackbodies (hypothetical bodies that absorb and emit radiation perfectly). To apply this analysis, we assume that the Sun radiates the same amount of energy as an identically sized spherical black body having a temperature of 5800 K [Recall the Kelvin temperature or absolute temperature is T (Kelvin)= T (°C) + 273.15]. Using a solar radius (Ro) of 6.96x108 m, calculate the surface area of the Sun, using A = 4πRo2 (recall π = 3.14).
2. Using the Stefan-Boltzmann law for the luminosity (L= [J/s]), calculate L, the energy radiated per unit time [J/s] by the Sun treated as a black body at temperature T. The formula is L = sAT4 (s, the Stefan-Boltzmann constant = 5.67 X 10-8 J/ m2 K4 s).
II. The Solar Energy Absorbed by the Earth
3. Although the Sun radiates the amount of energy per second calculated above, only a small portion of this total energy crosses each square meter of area at the distance of the Earth from the Sun. Assume the entire Sun's luminosity crosses a sphere with a radius as big as the Earth's orbit around the Sun (or d, the distance from the Earth to the Sun). Now calculate how much energy will hit each square meter of surface of this imaginary sphere (the top of the earths atmosphere) each second (we'll call this the solar flux, F).
Take the distance of the Earth from the Sun (d) = 1 AU = 1.5 x 1011 m. Calculate F (units J/m2 s or W/m2) using the following equation: F = L/(4πd2)
4. Of the area of the large imaginary sphere, only the cross-sectional area of the Earth will intercept the light from the Sun. That is, of the Sun's luminosity, the Earth will intercept only the amount given by multiplying the cross-sectional area of Earth by the flux you just found. Thus the energy hitting the Earth, E = πFRE2, where RE = the radius of the Earth, 6.38 x 106 m). Calculate E. So this value is the energy per unit time (units are Joules/second or Watts where 1 W= 1 J/s) that impinges upon the top of the Earth's atmosphere each second. | 576 | 2,131 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-26 | latest | en | 0.769318 |
https://www.sitepoint.com/community/t/an-interview-test-that-i-face/78112 | 1,701,669,879,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100525.55/warc/CC-MAIN-20231204052342-20231204082342-00271.warc.gz | 1,128,897,788 | 8,277 | # An interview test that i face
``````
write the algorithm to draw a sphere (Circle) , the function should have paramters
1. diameter (how many line)
2. character (use which character to draw)
eg, draw_sphere(10, '8') will produce:
88888
8888888888888
8888888888888888888
888888888888888888888
88888888888888888888888
88888888888888888888888
888888888888888888888
8888888888888888888
8888888888888
88888
the above is a sphere of 10 lines, which is drawn using the character "8"
``````
could anyone help on this? im newbie in php, thanks in advance…
EDIT: I tried but vBulletin messes with the tabbing from post to render. Hopefully you can guess what it is intended to look like Mittineague
could u help me on this? they know im newbie too:confused:
It may seem cruel, but really it’s the kindest cut.
Well, there’s two variables for each line: the whitespace and the 8.
How can you figure out how much whitespace there should be and how many characters to draw?
Start out with what you should know: it should be as wide as it is tall. The width and the height are the same as the diameter. The length of a chord is the product of a function: http://www.mathopenref.com/chord.html and the whitespace is related to the diameter and the chord length.
But you should be able to do this. This is a a problem that high-school sophomores solve.
+1!
I’ll second that.
I’ve been learning php for years… and even now I wouldn’t put myself forward for a php job despite a several people in the field suggesting my understanding of it is better than some of the pro’s.
You can’t just jump in as someone who knows very little and grab a job… can you? - If so where do I sign?
That’s not a php problem but an algorithm one. So even if you’re not proficient with php, just resolve it with pseudo-code.
If you can not, you’ve got bigger problems than not knowing php.
Indeed.
And if they hire you because someone here helped you with the answer… are you going to post all your work here too? If so, what’s our cut?
And how often in the real world are we faced with a problem such as drawing a sphere full of the digit “8”?
The original poster was asking this same question on other boards. The one where I found an explanation, if it is correct, is not as easy as some of you seem to think it is.
When I first got interested in computer programming, I talked to one of the programmers at the company I worked for. The most important tool a programmer can have, he said, is documentation. Nobody knows the answer to every problem. But someone with a brain and access to documentation can arrive at an answer (eventually). I’ll bet they don’t test for that.
I thought it would be as easy as using the COS function to determine how many spaces, and SIN to determine how many characters to write.
It’s not?
one thing that this has done is flag up a site that’s stealing our threads :mad:
In fact, even COS isn’t needed. The core of it could be as simple as:
``````
var arc = TAU / 2 / size;
for (radians = arc / 2; radians < TAU / 2; radians += arc) {
chars = Math.round(Math.sin(radians) * size);
...
``````
That is JavaScript code of course, but the idea should be easily implementable.
Well, I spend a big chunk of my time trying to figure out how to programatically generate one thing from another, so I will go with the answer “almost every single day”.
Spam
Then perhaps you would like to provide us a solution to the problem which we can verify is correct? | 847 | 3,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2023-50 | latest | en | 0.936443 |
https://solvedlib.com/n/a-sample-has-measured-temperature-of-1118-0-c-what-is-its,12180288 | 1,675,711,575,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00306.warc.gz | 512,902,840 | 17,521 | # A sample has measured temperature of 1118.0 *C. What is its temperature in K?
###### Question:
A sample has measured temperature of 1118.0 *C. What is its temperature in K?
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##### 7. (10 points) Find a particular solution yp(t) to the nonhomogeneous equation ty + y -...
7. (10 points) Find a particular solution yp(t) to the nonhomogeneous equation ty + y - y = 24t*, t> 0, given the fact that the general solution of the associated homogeneous equation is yn(t) = cit + cat-, C1, C2 E R... | 3,731 | 12,833 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-06 | latest | en | 0.906196 |
https://community.esri.com/t5/arcgis-pro-questions/how-to-calculate-position-in-a-clockwise-manner/m-p/1355747/highlight/true | 1,712,970,880,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816465.91/warc/CC-MAIN-20240412225756-20240413015756-00654.warc.gz | 158,763,867 | 47,758 | # How to calculate position in a clockwise manner?
386
3
12-01-2023 08:31 AM
Labels (3)
by
Occasional Contributor
I have data with X/Y data that plots and I'm trying to get it to look like so:
I've tried near analysis, and it gets me close but doesn't label them in a clockwise manner. Is there any geoprocessing tool or arcpy process that I'm overlooking that will accomplish this?
1 Solution
Accepted Solutions
MVP Esteemed Contributor
slow week
``````aoi # a square
array([[ 0.00, 5.00],
[ 5.00, 10.00],
[ 10.00, 5.00],
[ 5.00, 0.00],
[ 0.00, 5.00]])
def geom_angles(a, fromNorth=False):
"""Polyline/segment angles.
Parameters
----------
a : array-like
A Geo array or a list of arrays representing the polyline shapes.
"""
out = []
dxy = a[1:] - a[:-1]
ang = np.degrees(np.arctan2(dxy[:, 1], dxy[:, 0]))
if fromNorth:
ang = np.mod((450.0 - ang), 360.)
return ang
geom_angles(aoi, fromNorth=False) # -- option 1
array([ 45.00, -45.00, -135.00, 135.00])
geom_angles(aoi, fromNorth=True) # -- option 2
array([ 45.00, 135.00, 225.00, 315.00])
cent = np.mean(aoi[:-1]) # -- now from the center of the points
vals = aoi - cent # -- angles from the center
vals
array([[ -5.00, 0.00],
[ 0.00, 5.00],
[ 5.00, 0.00],
[ 0.00, -5.00],
[ -5.00, 0.00]])
geom_angles(vals, fromNorth=False)
array([ 45.00, -45.00, -135.00, 135.00])
geom_angles(vals, fromNorth=True)
array([ 45.00, 135.00, 225.00, 315.00])``````
... sort of retired...
3 Replies
Esri Esteemed Contributor
You can write your own, but I doubt there's any general purpose tool for this.
The atan2 function takes dy & dx parameters, but that's counter-clockwise from 3-o'clock,
so you'd need to reverse the sign and rotate.
- V
MVP Esteemed Contributor
slow week
``````aoi # a square
array([[ 0.00, 5.00],
[ 5.00, 10.00],
[ 10.00, 5.00],
[ 5.00, 0.00],
[ 0.00, 5.00]])
def geom_angles(a, fromNorth=False):
"""Polyline/segment angles.
Parameters
----------
a : array-like
A Geo array or a list of arrays representing the polyline shapes.
"""
out = []
dxy = a[1:] - a[:-1]
ang = np.degrees(np.arctan2(dxy[:, 1], dxy[:, 0]))
if fromNorth:
ang = np.mod((450.0 - ang), 360.)
return ang
geom_angles(aoi, fromNorth=False) # -- option 1
array([ 45.00, -45.00, -135.00, 135.00])
geom_angles(aoi, fromNorth=True) # -- option 2
array([ 45.00, 135.00, 225.00, 315.00])
cent = np.mean(aoi[:-1]) # -- now from the center of the points
vals = aoi - cent # -- angles from the center
vals
array([[ -5.00, 0.00],
[ 0.00, 5.00],
[ 5.00, 0.00],
[ 0.00, -5.00],
[ -5.00, 0.00]])
geom_angles(vals, fromNorth=False)
array([ 45.00, -45.00, -135.00, 135.00])
geom_angles(vals, fromNorth=True)
array([ 45.00, 135.00, 225.00, 315.00])``````
... sort of retired...
by
Occasional Contributor
Thanks Dan! | 1,044 | 2,831 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-18 | latest | en | 0.706496 |
https://discourse.julialang.org/t/whats-a-good-and-easy-way-to-parallelize-this/74893 | 1,709,248,608,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474853.43/warc/CC-MAIN-20240229202522-20240229232522-00595.warc.gz | 205,421,224 | 6,513 | # What's a good and easy way to parallelize this?
Ultimately, I want to speed up `fullsys!`, and as you can see there are functions that fill arrays. I thought those filling operations could be parallelized. How can I get started with that?
``````const szAge = 64 - 19 + 1
const szE = 2
const β = 0.92
const δ = 0.10
maxT(am, af; Z=realage(szAge)) = Z - max(am, af) + 1
realage(x) = x + 18
realageinv(x) = x - 18
function insideproduct_num(am, em, af, ef; μ0m=μ0m, μ0f=μ0f)
exponent = 0.5*(β*(1 - δ))
power = 1.0
p = one(promote_type(eltype(μ0m), eltype(μ0f)))
for k in 0:maxT(realage(am), realage(af))-1
p *= abs(μ0m[am + k, em] * μ0f[af + k, ef])^power
power *= exponent
end
return p
end
function insideproduct_den(am, em, af, ef; Sm=Sm, Sf=Sf)
exponent = 0.5*(β*(1 - δ))
power = 1.0
p = one(promote_type(eltype(Sm), eltype(Sf)))
for k in 0:maxT(realage(am), realage(af))-1
p *= abs(Sm[am + k, em] * Sf[af + k, ef])^power
power *= exponent
end
return p
end
function fillauxmat!(auxmat; Π=Π, Sm=Sm, Sf=Sf, μ0m=μ0m, μ0f=μ0f)
for ef in 1:szE, af in 1:szAge, em in 1:szE, am in 1:szAge
auxmat[am, em, af, ef] = Π[am, em, af, ef] * sqrt(Sm[am, em]*Sf[af, ef]) /
insideproduct_den(am, em, af, ef; Sm=Sm, Sf=Sf)
end
return auxmat
end
function MMsysM!(res, μ0m; auxmat=auxmat, Sm=Sm, μ0f=μ0f)
for em in 1:szE, am in 1:szAge
s = zero(eltype(μ0m))
for ef in 1:szE, af in 1:szAge
s += auxmat[am, em, af, ef] * insideproduct_num(am, em, af, ef; μ0m=μ0m, μ0f=μ0f)
end
res[am, em] = Sm[am, em] - μ0m[am, em] - s
end
return res
end
function MMsysF!(res, μ0f; auxmat=auxmat, Sf=Sf, μ0m=μ0m)
for ef in 1:szE, af in 1:szAge
s = zero(eltype(μ0f))
for em in 1:szE, am in 1:szAge
s += auxmat[am, em, af, ef] * insideproduct_num(am, em, af, ef; μ0m=μ0m, μ0f=μ0f)
end
res[af, ef] = Sf[af, ef] - μ0f[af, ef] - s
end
return res
end
function fullsys!(resm, resf; auxmat=auxmat, Sm=Sm, Sf=Sf, μ0m=μ0m, μ0f=μ0f)
fillauxmat!(auxmat; Π=Π, Sm=Sm, Sf=Sf, μ0m=μ0m, μ0f=μ0f)
MMsysM!(resm, μ0m; auxmat=auxmat, Sm=Sm, μ0f=μ0f)
MMsysF!(resf, μ0f; auxmat=auxmat, Sf=Sf, μ0m=μ0m)
return resm, resf
end
Π = 1000 .* rand(szAge, szE, szAge, szE)
auxmat = similar(Π)
Sm = 1000 .* rand(szAge, szE)
Sf = 1000 .* rand(szAge, szE)
μ0m = rand(szAge, szE)
μ0f = rand(szAge, szE)
resm = similar(μ0m)
resf = similar(μ0f)
``````
for reference, here’s the benchmark for this version:
``````julia> @benchmark fullsys!(\$resm, \$resf; auxmat=\$auxmat, Sm=\$Sm, Sf=\$Sf, μ0m=\$μ0m, μ0f=\$μ0f)
BenchmarkTools.Trial: 625 samples with 1 evaluation.
Range (min … max): 6.852 ms … 13.761 ms ┊ GC (min … max): 0.00% … 0.00%
Time (median): 7.932 ms ┊ GC (median): 0.00%
Time (mean ± σ): 7.996 ms ± 515.543 μs ┊ GC (mean ± σ): 0.00% ± 0.00%
▄▂▃▄█▁▃▅▅▂▅▃▂ ▁ ▂
▂▂▂▂▃▃▃▄▄▃▄▇▆▄▆█████████████████▇█▇▆▅▅▅▆▅▃▃▃▃▃▃▂▃▂▄▃▃▁▁▂▁▁▂ ▄
6.85 ms Histogram: frequency by time 9.47 ms <
Memory estimate: 96 bytes, allocs estimate: 2.
``````
Before trying to speed your code up via paralellization, you should try to optimize it. For example, `insideproduct_num` can be calculated in the log domain, which would remove the floating point powers which would make your code roughly 20x faster.
2 Likes
Good idea, thank you!
``````julia> @benchmark insideproduct_den(30, 2, 25, 2; Sm=\$Sm, Sf=\$Sf)
BenchmarkTools.Trial: 10000 samples with 243 evaluations.
Range (min … max): 283.642 ns … 939.449 ns ┊ GC (min … max): 0.00% … 0.00%
Time (median): 322.564 ns ┊ GC (median): 0.00%
Time (mean ± σ): 328.856 ns ± 44.206 ns ┊ GC (mean ± σ): 0.00% ± 0.00%
▄ ▆▇█▄▂ ▁ ▁ ▁
▇▂█▄█▆███████▇▇▇█▆▆█▇█▆▆▇▇▇▆▆█▆▅▆▆▆▆▆▅▆▆▆▅▆▆▆▆▅▆▅▅▅▅▆▄▅▃▄▅▄▅▅ █
284 ns Histogram: log(frequency) by time 550 ns <
Memory estimate: 0 bytes, allocs estimate: 0.
julia> @benchmark insideproduct_den_log(30, 2, 25, 2; Sm=\$Sm, Sf=\$Sf)
BenchmarkTools.Trial: 10000 samples with 873 evaluations.
Range (min … max): 126.515 ns … 1.086 μs ┊ GC (min … max): 0.00% … 0.00%
Time (median): 143.638 ns ┊ GC (median): 0.00%
Time (mean ± σ): 147.881 ns ± 24.177 ns ┊ GC (mean ± σ): 0.00% ± 0.00%
▁▂ ▄▆▇█▃▂▂▁▁▁ ▁ ▁ ▂
▆██▁▇██████████████████▇█▇███▇▇█▇▇▇▇▇▇▇▆▇▇▆▇▆▆▆▄▆▄▅▆▅▄▅▅▆▅▅▅ █
127 ns Histogram: log(frequency) by time 240 ns <
Memory estimate: 0 bytes, allocs estimate: 0.
``````
so now it’s
``````const szAge = 64 - 19 + 1
const szE = 2
const β = 0.92
const δ = 0.10
maxT(am, af; Z=realage(szAge)) = Z - max(am, af) + 1
realage(x) = x + 18
realageinv(x) = x - 18
function insideproduct_num_log(am, em, af, ef; μ0m=μ0m, μ0f=μ0f)
exponent = 0.5*(β*(1 - δ))
power = 1.0
p = zero(promote_type(eltype(μ0m), eltype(μ0f)))
for k in 0:maxT(realage(am), realage(af))-1
p += power * log(abs(μ0m[am + k, em] * μ0f[af + k, ef]))
power *= exponent
end
return exp(p)
end
function insideproduct_den_log(am, em, af, ef; Sm=Sm, Sf=Sf)
exponent = 0.5*(β*(1 - δ))
power = 1.0
p = zero(promote_type(eltype(Sm), eltype(Sf)))
for k in 0:maxT(realage(am), realage(af))-1
p += power * log(abs(Sm[am + k, em] * Sf[af + k, ef]))
power *= exponent
end
return exp(p)
end
function fillauxmat!(auxmat; Π=Π, Sm=Sm, Sf=Sf, μ0m=μ0m, μ0f=μ0f)
for ef in 1:szE, af in 1:szAge, em in 1:szE, am in 1:szAge
auxmat[am, em, af, ef] = Π[am, em, af, ef] * sqrt(Sm[am, em]*Sf[af, ef]) /
insideproduct_den_log(am, em, af, ef; Sm=Sm, Sf=Sf)
end
return auxmat
end
function MMsysM!(res, μ0m; auxmat=auxmat, Sm=Sm, μ0f=μ0f)
for em in 1:szE, am in 1:szAge
s = zero(eltype(μ0m))
for ef in 1:szE, af in 1:szAge
s += auxmat[am, em, af, ef] * insideproduct_num_log(am, em, af, ef; μ0m=μ0m, μ0f=μ0f)
end
res[am, em] = Sm[am, em] - μ0m[am, em] - s
end
return res
end
function MMsysF!(res, μ0f; auxmat=auxmat, Sf=Sf, μ0m=μ0m)
for ef in 1:szE, af in 1:szAge
s = zero(eltype(μ0f))
for em in 1:szE, am in 1:szAge
s += auxmat[am, em, af, ef] * insideproduct_num_log(am, em, af, ef; μ0m=μ0m, μ0f=μ0f)
end
res[af, ef] = Sf[af, ef] - μ0f[af, ef] - s
end
return res
end
function fullsys!(resm, resf; auxmat=auxmat, Sm=Sm, Sf=Sf, μ0m=μ0m, μ0f=μ0f)
fillauxmat!(auxmat; Π=Π, Sm=Sm, Sf=Sf, μ0m=μ0m, μ0f=μ0f)
MMsysM!(resm, μ0m; auxmat=auxmat, Sm=Sm, μ0f=μ0f)
MMsysF!(resf, μ0f; auxmat=auxmat, Sf=Sf, μ0m=μ0m)
return resm, resf
end
Π = 1000 .* rand(szAge, szE, szAge, szE)
auxmat = similar(Π)
Sm = 1000 .* rand(szAge, szE)
Sf = 1000 .* rand(szAge, szE)
μ0m = rand(szAge, szE)
μ0f = rand(szAge, szE)
resm = similar(μ0m)
resf = similar(μ0f)
``````
Why is this a problem?
In other words: I don’t believe the overhead of parallelization is justified by this workload.
2 Likes | 3,114 | 6,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2024-10 | latest | en | 0.482554 |
https://fr.mathworks.com/matlabcentral/cody/problems/15-find-the-longest-sequence-of-1-s-in-a-binary-sequence/solutions/1006273 | 1,590,656,339,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347398233.32/warc/CC-MAIN-20200528061845-20200528091845-00328.warc.gz | 361,850,832 | 15,883 | Cody
# Problem 15. Find the longest sequence of 1's in a binary sequence.
Solution 1006273
Submitted on 10 Oct 2016 by Venkatapathi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = '0'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
y = 0
2 Pass
x = '1'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
y = 1
3 Pass
x = '01'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
y = 1
4 Pass
x = '10'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
y = 1
5 Pass
x = '00'; y_correct = 0; assert(isequal(lengthOnes(x),y_correct))
y = 0
6 Pass
x = '11'; y_correct = 2; assert(isequal(lengthOnes(x),y_correct))
y = 2
7 Pass
x = '1111111111'; y_correct = 10; assert(isequal(lengthOnes(x),y_correct))
y = 10
8 Pass
x = '100101011111010011111'; y_correct = 5; assert(isequal(lengthOnes(x),y_correct))
y = 5
9 Pass
x = '01010101010101010101010101'; y_correct = 1; assert(isequal(lengthOnes(x),y_correct))
y = 1
10 Pass
x = '0101010111000101110001011100010100001110110100000000110001001000001110001000111010101001101100001111'; y_correct = 4; assert(isequal(lengthOnes(x),y_correct))
y = 4 | 445 | 1,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-24 | latest | en | 0.434714 |
https://socratic.org/questions/how-do-capacitors-discharge | 1,576,129,189,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540537212.96/warc/CC-MAIN-20191212051311-20191212075311-00120.warc.gz | 562,296,698 | 6,660 | # How do capacitors discharge?
Jun 3, 2014
Electrons are forced off one of the capacitor’s plates and attracted to the opposite plate through the circuit.
Prior to being discharged the capacitor will have been charged. Electrons will have accumulated on one plate (negative plate) having been forced onto it by the power supply. The other plate (positive) will have a deficiency of electrons as they will have been forced off of it by the supply.
Before discharging the accumulated electrons on the negative plate repel each other but have nowhere to go. The positive plate is unable to exert an influence on the electrons in such a way as to force electrons off of the negative plate because there is a dielectric (insulating) material between the plates.
Discharging will begin once a circuit is connected between the terminals of the capacitor. During discharge electrons on the negative plate will be forced off of the plate by the repulsion of the other electrons on the plate. The positively charged plate will attract electrons from the circuit toward itself. These influences will result in a current flowing between the plates.
Further to the general method of discharge one can also appreciate the exponential decay nature of the current that flows during discharge. Immediately after the circuit is connected there is a maximum amount of accumulated negative charge on the negative plate and max. deficiency of negative charge on positive plate. There are therefore maximum attractive / repulsive electrostatic resultant forces on the electrons in the plates and circuit.
As time progresses the accumulated charge will decrease and the electron deficiency decreases also. Those attractive and repulsive resultant forces decrease. There is also less charge available to move around the circuit. So the current decreases and it does so with an exponential decay.
Hence I=I_0 e^(–t/(RC)) | 359 | 1,903 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 1, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-51 | longest | en | 0.93612 |
https://www.manchester.ac.uk/study/undergraduate/courses/2024/03340/beng-chemical-engineering/all-content/CHEN10051 | 1,708,872,718,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474617.27/warc/CC-MAIN-20240225135334-20240225165334-00722.warc.gz | 884,527,871 | 14,640 | Don't just imagine your future at University, experience it first-hand!
Step into the shoes of an undergraduate student and join us for our upcoming on-campus Discover Days in Science, Engineering and Fashion. These days are tailored exclusively for Year 12s who are interested in taking their academic journey to the next level. Find your favourite subject now!
# BEng Chemical Engineering
Year of entry: 2024
## Course unit details:Computational Methods for Chemical Engineering
Unit code CHEN10051 10 Level 1 Full year No
### Overview
Data input, manipulation and graphical representation; Advanced built-in and user-defined functions; Optimisation, root-finding and non-linear regression using Excel Solver.
Programming with MATLAB:
Introduction to the use of computers in Chemical Engineering.
Data handling: Significant figures; Accuracy and precision; Error definitions; Round-off and truncation errors.
Introduction to MATLAB: The scientific computing environment; Types of variables; Number formats; Mathematical operations; Built-in functions; Graphical display of data.
Programming 1: Notion of algorithm; m-files structure; Indentation; Writing programs and user-defined functions; Program debugging; Data input and output.
Programming 2: Structured programming with conditional statements and loops (for and while)
Problem solving in Chemical Engineering using numerical methods: Roots of equations; One-dimensional and multidimensional optimisation; Systems of linear algebraic equations; Numerical integration; Ordinary differential equations.
### Aims
The unit aims to:
Provide the advanced computational skills required for solving complex problems in Engineering and Science.
### Learning outcomes
ILO 1:Represent numerical data with the appropriate number of significant figures and estimate errors in numerical calculations.
ILO 2:Use MS Excel and MATLAB to organise and manipulate numerical data, and represent it graphically.
ILO 3:Identify the appropriate numerical method to solve a complex mathematical problem.
ILO 4:Develop algorithms to solve numerically mathematical models of physical and chemical systems in Chemical Engineering.
ILO 5:Translate algorithms into MATLAB codes using structured programming.
### Teaching and learning methods
Lectures provide fundamental aspects supporting the critical learning of the module and will be delivered as pre-recorded asynchronous short videos via our virtual learning environment.
Synchronous sessions will support the lecture material with Q&A and problem-solving sessions where you can apply the new concepts. Surgery hours are also available for drop-in support.
Feedback on problems and examples, feedback on coursework and exams, and model answers will also be provided through the virtual learning environment. A discussion board provides an opportunity to discuss topics related to the material presented in the module.
Students are expected to expand the concepts presented in the session and online by additional reading (suggested in the Online Reading List) in order to consolidate their learning process and further stimulate their interest to the module.
Study budget:
• Core Learning Material (e.g. recorded lectures, problem solving sessions): 24 hours
• Self-Guided Work (e.g. continuous assessment, extra problems, reading) : 44 hours
• Exam Style Assessment Revision and Preparation: 32 hours
### Assessment methods
Assessment Types Total Weighting Continuous assessment 30% Exam style assessments 70%
Please note that the exam style assessments weighting may be split over midterm and end of semester exams.
### Feedback methods
Assessment task How and when feedback is provided Exam Generic exam feedback form after exam board Online test on Excel Final mark only one week after the assessment. | 701 | 3,830 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.810828 |
http://mathvids.com/lesson/mathhelp/1638-graphing-piecewise-functions | 1,558,391,494,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256163.40/warc/CC-MAIN-20190520222102-20190521004102-00474.warc.gz | 126,307,518 | 17,204 | ## Graphing Piecewise Functions
Graphing Piecewise Functions
• Currently 4.0/5 Stars.
2817 views, 1 rating
Part of video series Piecewise Functions
Taught by mrbrianmclogan
### Meets NCTM Standards:
I show how to solve math problems online during live instruction in class. This is my way of providing free tutoring for the students in my class and for students anywhere in the world. Every video is a short clip that shows exactly how to solve math problems step by step. The problems are done in real time and in front of a regular classroom. These videos are intended to help you learn how to solve math problems, review how to solve a math problems, study for a test, or finish your homework. I post all of my videos on YouTube, but if you are looking for other ways to interact with me and my videos you can follow me on the following pages through My Blog, Twitter, or Facebook.
How to graph a piecewise function
• How do you graph a piecewise function?
• What is a piecewise function?
• How do you graph f(x) if f(x) = -1 when x <= 0, f(x) = 2x when 0 < x <= 3, and f(x) = 6 when x > 3?
• How do you know where points are open or closed circles in a piecewise function?
This lesson shows how to graph a piecewise function in 3 pieces. Each of the 3 pieces is graphed for only a limited domain. All steps involved in graphing each of the 3 pieces are explained and shown. This is a very good introduction to piecewise functions.
• Currently 4.0/5 Stars.
Reviewed by MathVids Staff on February 05, 2012. | 375 | 1,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2019-22 | longest | en | 0.93058 |
http://www.mathrecreation.com/2016_08_01_archive.html | 1,527,383,783,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867977.85/warc/CC-MAIN-20180527004958-20180527024958-00384.warc.gz | 416,594,907 | 16,057 | ## Friday, August 12, 2016
### Monty R and Monty n
As part of my ongoing attempt to learn the R language, I decided to try to generate a data set to analyze in R using R itself. I wrote a little R script which generates simulated data for a thousand trials of the "Monty Hall" problem. The script is here, along with other R examples.
If you source the script, a data frame MontyHall will be populated. The data is re-generated each time you source the script, so if you want to capture a single run, you should export the data and work from that (a saved example run is here, and contains the data I used in this post).
In the classical Monty Hall problem, a game show host, named Monty Hall, presents a contestant with three (3) doors. Behind one of the doors is a valuable prize, traditionally a car, while behind the others are dud prizes, traditionally goats. A game proceeds with the contestant selecting a door and keeping the prize that lies behind it (hoping for the actual prize, not a dud). However, instead of opening the selected door, Monty opens one of the other two doors, revealing a goat, and asks if the contestant would like to switch and select the remaining door instead. The problem asks which choice has the higher probability of revealing the actual prize, or whether both options have the same probability of achieving the hoped-for outcome.
In this simulation, we have the contestant choosing to switch half of the time, hopefully generating enough cases so that the experimental probabilities (from the simulation) match up with the theoretical probabilities that we can come up with by analyzing the problem. (You can increase the number of trials by modifying the script).
The simulation gives us a data set that looks like this.
Our simulation has the probability of switching set to 1/2, as if we were basing our decision on the flip of a fair coin.
We can see that our simulation is pretty close:
We would expect that the probability of the contestant selecting the winning door right away would be 1/3, and this is what we see in the simulation.
All that this tells us is that R is generating random numbers in a reasonable enough way for our purposes. What makes the Monty Hall problem interesting are the probabilities of winning when the contestant switches doors, compared to when they don't switch.
If the contestant does not switch, their chance of winning is the same as their chance of selecting the prize right away, which is just 1/3.
This is not the result that some people expect (some argue that the chance of winning when staying with the original selection should be 1/2), but the simulation agrees:
When the contestant makes their initial selection , they have either won or lost. Switching amounts to trading a win for a loss: if the contestant has picked the winner, they have now switched away from it; if they have picked a looser (one of the goats), when Monty opens the other door (revealing the other goat), switching will cause them to win. So, switching amounts to trading a win for a loss or a loss for a win. Because the probability of losing on the initial door selection is 2/3, it is a good idea to switch: the first choice was likely a goat.
And the simulation agrees:
What about if we pursue the strategy of the imaginary contestant in our simulation, and flip a coin to decide whether we switch doors or stay with the original choice? The law of total probability tells us that we have a probability of winning equal to 1/2.
And in the simulation we come pretty close:
Monty n
While writing up the R simulation, I wondered what would happen if the Monty Hall problem was generalized to n doors so that Monty could keep on offering you more chances to switch. There are likely many ways of doing this, so I picked what I think is a simple generalization:
Suppose that the contestant has n doors to choose from where n > 2. There is still only one (1) car (and n-1 goats). After a door is selected, Monty will always open a remaining door and provide the contestant with a chance to switch doors, as long as there remain unopened doors that have not been previously selected by the contestant. Monty will not open doors that were previously chosen, nor will he give the contestant the opportunity to switch back to a previously chosen door.
I've put a short write up of this on ShareLatex here. At least in the way that I've framed the problem, there is a nice formula for the probability of winning on n doors if you switch k times:
Or if you prefer a direct formula instead of recursive:
If you work out some values using your preferred version of the formula, you should get results that agree with these:
From this, or from the formulas themselves, you can see that the probability of winning keeps increasing as long as you keep switching doors as long as Monty gives you the option to.
I did not simulate the n > 3 case using R, but I did write a general simulation using Java (a listing for this is included in the write up here), and found that the simulation results agreed the the theoretical probabilities pretty closely (I have not carried out a detailed analysis of the results). | 1,110 | 5,183 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2018-22 | latest | en | 0.94072 |
http://school.familyeducation.com/preschool/activity/30578.html | 1,371,722,303,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368711240143/warc/CC-MAIN-20130516133400-00055-ip-10-60-113-184.ec2.internal.warc.gz | 219,568,370 | 11,228 | |
# Squares and Triangles Game
Purpose/Skills
• To review the concept of squares
• To experience the characteristics of shapes
• To identify triangles (2-D shapes)
• To compare and talk about shapes
Materials
Squares and triangles in different sizes and colors; a chalk or masking tape triangle on floor; triangles such as the musical instrument; The Father Who Had 10 Children
Vocabulary
shape(s) circle(s) square(s) triangle(s) side(s) corner(s) alike different compare
Literature Suggestion
Read The Father Who Had 10 Children by Benedicte Guettier. As you read, help children notice the triangle shapes in the sailboat.
Warm-Up
Have children walk on the masking-tape triangle and notice that it has 3 sides and 3 corners. Show children triangles in different colors and sizes. Give each child a triangle, then ask the children to see, touch, and say "Triangle" and feel its 3 sides and 3 corners.
Procedure
• Have children play the "What's My Name?" game with squares and triangles. Give each child a square and a triangle. Encourage children to see, touch, and say the name of each shape.
• Ask students to hold up their squares. Say, "A square has 4 sides. Let's count the sides." Then touch and count the sides with the children. "How many sides?" (4)
• Ask students to hold up their triangles. Say, "A triangle has 3 sides. Let's count the sides." Then touch and count the sides with children. "How many sides?" (3) Ask children to compare the shapes. "How are the shapes alike?" (They both have sides and corners.) "How are the shapes different?" (Square has 4 sides and 4 corners; triangle has only 3.)
• Tell children that you are going to say the word square or triangle. When you say "Square," they hold up their squares. When you say "Triangle," they hold up their triangles. Play the game slowly, holding up your square and your triangle to model. When you think that children can play the game successfully, stop holding up your square and triangle.
Enrichment
Play the "What's My Name?" game with a circle, square, and triangle. At art time, have children make shape designs or shape people with shapes of different colors and sizes.
Observation Assessment
• Proficient - Child can identify a square and a triangle.
• In Process - Child can identify a square, but has difficulty identifying a triangle.
• Not Yet Ready - Child does not yet identify a square or a triangle.
|
Excerpted from School Readiness Activity Cards. The Preschool Activity Cards provide engaging and purposeful experiences that develop language, literacy, and math skills for preschool children.
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Summer Learning: 11 Everyday Ways to Learn Math
Summertime is a great time to teach your child how math plays a part in his everyday life. These 11 everyday ways to learn math are fun and educational, and will help him see how this subject plays into real life. | 847 | 3,778 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2013-20 | latest | en | 0.931318 |
https://lumochift.org/blog/cp/iterateNumber | 1,674,937,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499654.54/warc/CC-MAIN-20230128184907-20230128214907-00076.warc.gz | 394,704,685 | 49,691 | Published on
Iterate digit by given integer
Authors
Do you ever meet the question to sum all digits from the input, for example given `112` so the sum of the number is `4`.
Naïve approach is to convert to string, then read character one by one. Typecast to integer, then sum all data. The simple ways to typecast rune to int using `int(r - ‘0’)`.
`_7func Solution1(n int) int {_7 sum := 0_7 for _, r := range fmt.Sprintf(“%d”, n) {_7 sum += int(r - ‘0’)_7 }_7 return sum_7}`
Another way is to use an arithmetic approach by utilizing mod and division. Here are the example codes:
`_9func Solution2(n int) int {_9 sum := 0_9 for n > 0 {_9 digit := n % 10_9 sum += digit_9 n /= 10_9 }_9 return sum_9}`
So which one to use? Let's see the benchmark result first.
`_10go test -bench=Sum -benchmem -shuffle=on _10-test.shuffle 1673764987422236824_10goos: linux_10goarch: amd64_10pkg: github.com/h4ckm03d/golang-playground/6-practical-benchmark_10cpu: AMD Ryzen 7 5800H with Radeon Graphics _10BenchmarkSolution2-4 241413471 5.342 ns/op 0 B/op 0 allocs/op_10BenchmarkSolution1-4 14215688 84.96 ns/op 16 B/op 1 allocs/op_10PASS_10ok github.com/h4ckm03d/golang-playground/6-practical-benchmark 3.091s`
The result of the benchmark Solution2 outclass Solution1 and even 0 allocation. If you encounter the similar problem to do any logic with digit from given integer, you can use the `Solution2` approach to avoid time limit platform. | 437 | 1,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2023-06 | latest | en | 0.661987 |
http://www.midasoft.com/bridge-library/basic-finite-element-mesh-explained | 1,675,209,877,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499891.42/warc/CC-MAIN-20230131222253-20230201012253-00591.warc.gz | 69,173,984 | 19,401 | To ensure stable analysis performance and reliable result approximation, meshing is important. Meshing is the process to create finite elements and to connect those elements to formulate a set of functions. Finite elements are created by separating the known geometry with imaginary lines, and the elements are then connected with each other by specifying nodal connectivity at the element boundaries. Every element can be represented by a set of matrices (shown later), and connecting the elements essentially compiles the individual matrices into one structural matrix.
• What is Finite Element Mesh?
Any non-time-dependent finite element analysis contains the following steps,
1. Meshing
2. Assigning boundary conditions
4. Numerical analysis
5. Postprocessing
To ensure stable analysis performance and reliable result approximation, meshing is important. Meshing is the process to create finite elements and to connect those elements to formulate a set of functions. Finite elements are created by separating the known geometry with imaginary lines, and the elements are then connected with each other by specifying nodal connectivity at the element boundaries. Every element can be represented by a set of matrices (shown later), and connecting the elements essentially compiles the individual matrices into 1 structural matrix.
• Types of Finite Elements
Any non-time-dependent finite element analysis contains the following steps, Based on the shapes, there are the following types of finite elements:
1. One-dimensional elements
2. Two-dimensional elements
3. Three-dimensional elements
In this article, we are going over each type of the finite element listed above, explaining its pros and cons, and when to use each type.
• One-Dimensional Elements
One-dimensional elements (line elements) include bar elements and beam elements, and the difference between the two depends on their load-bearing capabilities. Bar elements are suitable for modeling trusses because their load-bearing capabilities are limited in the axial directions. On the other hand, beam elements are suitable for modeling frames because they can resist bending, twisting, as well as axial forces. A beam that is continuous over two or more supports can usually be modeled using one beam element per span between supports without using several beam elements to model one individual span. Thus, when a one-dimensional element is used for static analysis, the discretization phase of modeling becomes trivial, and for stress analysis, the name “matrix methods of structural mechanics” may be used in preference to “FEA” [1].
To represent a bar in axial tension, shown in figure 1, the following considerations can be made to compile the stiffness matrix.
Figure 1. A prism bar with one end fixed and the other end subjected to a force P.
A prismatic bar of length L, cross-section area A, and elastic modulus E is subjected to a force P at one end and with the other end fixed. The tensile elongation of the bar can be represented as
For a two-node bar element, shown in figure 2,
Figure 2. A two-node bar element.
The equilibrium at each individual node gives the following:
Which can be written in the matrix format as:
further reduction gives:
Where [k] is the characteristic matrix (stiffness matrix), {d} is the displacement vector, and {r} is the loads associated with each individual node.
Figure 3. A beam with two nodes (a) with two degrees of freedom at each node, translational in y and rotation in z, (b) with vertical loads and rotational moments at each node.
Figure 3 shows a beam element with two nodes. Each node is subjected to two degrees of freedom (figure 3a) and two nodal forces (figure 3b). Using the Euler-Bernoulli beam theory, the following matrix equation can be formed:
The same element stiffness matrix can be obtained by calculating using interpolation and shape functions,
Where [B] is the strain-displacement matrix obtained from the shape functions [N]. We can also use shape function interpolation to obtain the stiffness matrix of 2D and 3D elements as well, as shown later in the article.
Line elements are widely used in general structural analysis to obtain an overview of structural behavior, as well as in the detailed analysis in conjunction with other types of elements as connections, stiffeners, etc. In bridge engineering, the general structural analysis using 1D elements gives engineers a comprehensive understanding of the bridge structural behavior, as well as provides member forces and moments for design code checking for various standards.
However, to obtain more insights about localized structural zones, or when analyzing complex bridge geometries, local refinement would be required. To satisfy those higher analysis requirements, two-dimensional elements and three-dimensional elements offer better result approximation. Figure 4 (left) shows the general analysis result using the 1D element, and figure 4 (right) shows the refined analysis for the high moment elements using 2D elements. The nodal moment loads in the localized detailed analysis are extracted from the general analysis results.
Figure 4. General analysis using 1D element and using its element force-moment results as initial loads for the 2D detailed analysis.
• Two-Dimensional Elements
To solve two-dimensional (2D) problems, 2D elements are needed. Similar to how stiffness matrix is constructed for 1D elements as shown in equation 6, 2D elements’ stiffness matrix can also be constructed from shape function and interpolation,
where again, [k] is the stiffness matrix, [B] is the strain-displacement matrix obtained from the shape functions, [E] is the constitutive matrix, t is the thickness of the element, and A is the area of the element.
Common 2D problems include plane stress, plane strain, shell, axisymmetric solid, geogrid 2D, and gauging shell elements. Plain strain and axisymmetric solid elements are 2D shape elements, but they are used to express 3D stress states [2]. The common 2D element shape used is the triangular element with 3 nodes, shown in figure 5a, this type of element is called a constant strain triangle (CST) because, in stress analysis, a linear displacement field produces a constant strain field [1]. The CST elements do not work very well, because the “locking” effect can make the mesh overly stiff [3]. Even though refining the mesh can help with the accuracy, it also does increase the analysis solving time. The inaccuracy due to “locking” can be improved with the 6 node triangular elements. As shown in figure 5b, they have middle nodes between the vertices. 6 node triangular elements are also known as linear strain triangles (LST) or quadratic triangles.
Figure 5. (a) Constant-strain triangle element, (b) Linear strain triangle element.
A simple but less used 2D element is the 4-node rectangular element (Q4) whose sides are parallel to the global coordinate systems. This system is easy to construct automatically but it is not well suited to approximate inclined boundaries [4]. The Q4 elements experience the same “locking” effects as the CST elements, however, the issue can be improved with quadratic rectangular elements (Q8, Q9) by adding mid edge nodes.
• Three-Dimensional Elements
As shown in figure 6(a), tetrahedron has 4 nodes and is the most basic 3D finite element. Figure 6(b) shows a pentahedral (pyramid) element, figure 6(c) shows an 8 node rectangular solid element (Q8), and figure 6(d) shows an 8 node hexahedral isoparametric element.
Figure 6. (a) 4 node tetrahedron element, (b) 5 node pentahedral (pyramid) element, (c) 8 node rectangular solid element, d) 8 node hexahedral isoparametric element.
Similar to rectangular Q4 elements and CST elements, Q8 also has the disadvantage of shear locking. Also similar to Q4 elements, Q8 rectangular solid elements are difficult to mesh irregular geometries, especially for geometries with varying mesh densities. To reduce shear locking, adding mid-edge nodes to the Q8 element would help, and making the elements isoparametric will help with meshing flexibilities. Isoparametric formulation permits quadrilateral and hexahedral elements to have non-rectangular shapes [1]. When combining the two efforts, adding mid-edge nodes to the linear isoparametric hexahedral elements would make them more versatile and produce better analysis results.
3D geometries are needed to perform 3D mesh and they come from CAD models, which take longer and more effort to produce. Furthermore, analysis containing 3D elements usually contains more nodes and elements thus would take longer to solve. 2D analysis, however, requires less time to mesh and less time to solve. The 2D analysis also provides a more sufficient amount of structural information than 1D analysis and produces results close to 3D analysis. However, 2D analysis can only replace 3D analysis when structures can be represented by plate elements. When a structure is more complex, as shown in figure 7, 3D elements need to be used.
Figure 7. A lug and pin model.
Engineers sometimes model structural elements using combined 1D/2D/3D mesh to take advantage of the benefit of each type of element while saving analysis time, as shown in the bridge model in figure 8.
Figure 8. A bridge model utilizing hybrid element types with merged nodes at the element boundaries
References:
[1] R. Cook, D. Malkus, M. Plesha, R. Witt, Concepts and Applications of Finite Element Analysis, Fourth Edition, 2001, John Wiley & Sons Inc., New York, NY.
[2] Midas Information Technology, Analysis Reference Midas FEA NX, Chapter 3, 2021.
[3] Th. Zimmermann, S. Commend, Stabilized Finite Element Applications in Geomechanics, Laboratory of Structural and Continuum Mechanics, Department of Civil Engineering Swiss Federal Institute of Technology, 2001, Lausanne-EPFL, Switzerland.
[4] S.S. Bhavikatti, Finite Element Analysis, 2005, New Age International Limited, New Delhi.
https://www.midasoft.com/bridge-library/civil/products/midasfeanx
jsun@midasoft.com | 2,142 | 10,069 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-06 | latest | en | 0.908218 |
https://stats.stackexchange.com/questions/440993/does-pab-cpb-pac | 1,718,229,543,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00144.warc.gz | 499,340,291 | 37,271 | # Does P(A|B,C)P(B)= P(A|C)? [closed]
Can you remove a conditional variable by multiplying by the probability of it?
We know that
$$P(A,B)=P(A|B)P(B)$$
and this remains true when conditional on $$C$$:
$$P(A,B|C)=P(A|B,C)P(B|C).$$
You'll notice that the $$C$$ conditioning seems to be "along for the ride."
In Bayesian analysis, one typically replaces $$C$$ with $$I$$, indicating prior information. So,
$$P(A,B|I)=P(A|B,I)P(B|I).$$
which can get cumbersome, but it underscores the central point that every interpretation of probability is conditional on what we -- most often tacitly, but occasionally explicitly -- affirm to be true.
If you want to "remove a conditional," you may use the definition of conditional probability to restate the problem as a joint probability and a marginal: $$P(A|C)=P(A,C)/P(C).$$ Where from there depends on what you are trying to do.
With a little probability algebra, we have:
\begin{aligned} \mathbb{P}(A|B,C) &= \frac{\mathbb{P}(A,B,C)}{\mathbb{P}(B,C)} \\[6pt] &= \frac{\mathbb{P}(C) \cdot \mathbb{P}(A|C) \cdot \mathbb{P}(B|A,C)}{\mathbb{P}(B,C)} \\[6pt] &= \mathbb{P}(A|C) \times\frac{\mathbb{P}(B|A,C)}{\mathbb{P}(B|C)}. \\[6pt] \end{aligned}
Thus, we can see that removal of the conditioning event $$B$$ requires us to multiply by $$\mathbb{P}(B|A,C)/\mathbb{P}(B|C)$$, which is not generally equal to $$\mathbb{P}(B)$$.
Not quite, as $$B$$ may occur or not occur on the right hand side while the left hand side is $$\mathbb P(A,B\mid C)$$
But since $$\mathbb P(A,B\mid C) + \mathbb P(A, B^c\mid C)= \mathbb P(A\mid C)$$, you can say:
$$\mathbb P(A\mid B,C)\mathbb P(B) + \mathbb P(A\mid B^c,C)\mathbb P(B^c)= \mathbb P(A\mid C)$$
It does not work out, since:
$$P(A|B,C)P(B) = \frac{P(A,B,C)}{P(B,C)}P(B) = \frac{P(A,C|B)P(B)}{P(C|B)P(B)}P(B) = P(A|C)P(B)$$
For a more concrete example, take an equally probable six sided die. Let A = a 2 is rolled, B = a 4 is rolled, C = an even number is rolled.
P(A|B,C)P(B) = 0, P(A|C) = 1/3. | 703 | 1,993 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 17, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2024-26 | latest | en | 0.814249 |
https://gadgetspidy.com/modify-sequence/ | 1,708,530,443,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473518.6/warc/CC-MAIN-20240221134259-20240221164259-00590.warc.gz | 280,916,038 | 33,351 | # Modify Sequence
QUESTION
Suppose we have a sequence of non-negative integers, Namely a_1, a_2, … ,a_n. At each time we can choose one term a_i with 0 < i < n and we subtract 1 from both a_i and a_i+1. We wonder whether we can get a sequence of all zeros after several operations.\n\nInput\n\nThe first line of test case is a number N. (0 < N <= 10000) The next line is N non-negative integers, 0 <= a_i <= 109\n\nOutput\n\nIf it can be modified into all zeros with several operations output YES in a single line, otherwise output NO instead.
“TESTCASE_1”: “2\n1 2\n###—###SEPERATOR—###—\nNO”, “TESTCASE_2”: “2\n2 2\n###—###SEPERATOR—###—\nYES”, “TESTCASE_3”: “5\n5 5 5 5 5\n###—###SEPERATOR—###—\nNO”, “TESTCASE_4”: “1\n0\n###—###SEPERATOR—###—\nYES”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0
``````#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
int main()
{
lli n;
cin>>n;
vector<lli> arr(n,0);
for(lli i=0;i<n;i++)
{
cin>>arr[i];
}
lli c=1;
for(lli i=0;i<n-1;i++)
{
if(arr[i]<0)
{
c=0;
break;
}
arr[i+1] -= arr[i];
arr[i]=0;
}
if(arr[n-1]!=0)
{
c=0;
}
if(c==1)
{
cout<<"YES\n";
}
else
cout<<"NO\n";
return 0;
}`````` | 426 | 1,152 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-10 | longest | en | 0.528731 |
https://www.afcn.org/what-is-the-compressive-strength-of-a-material/ | 1,670,586,924,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711396.19/warc/CC-MAIN-20221209112528-20221209142528-00243.warc.gz | 677,069,211 | 20,145 | # What Is The Compressive Strength Of A Material
## What is compressive strength strength?
Definition of compressive strength : the maximum compressive stress that under gradually applied load a given solid material will sustain without fracture — compare tensile strength.
## What is compressive strength of metal?
The compressive strength of steel is about 250 MPa while that of concrete ranges from 30MPa to 80 MPa. So the compressive strength of steel is about 5 times more than that of concrete.
## What is compression of a material?
Compression: Particles of a material are pushed against each other, causing them to shorten, or compress. In a building, compression usually comes from the top. Tension: The opposite of compression, in which a pulling force is working to lengthen the material.
## What is compressive strength formula?
Compressive strength test The compressive strength is calculated by using the equation, F= P/A———1 Where, F= Compressive strength of the specimen (in MPa). P= Maximum load applied to the specimen (in N).
## What is compressive strength in strength of materials?
Compressive strength is defined as the maximum compressive load a body can bear prior to failure, divided by its cross sectional area. Ceramics typically have good tensile strengths and are used under compression e.g. concrete.
## What is compressive strength of concrete?
The compressive strength of normal concrete varies between 25 and 50 MPa. Above this level, the term High Performance Concrete is used.
## What is a good compressive strength?
Standard applications usually require the concrete to meet a compressive strength requirement of 10 MPa to 60 MPa, whereas for certain applications higher strength is needed and concrete mixes can be designed that meet a strength requirement of 500 MPa.
## What is tensile strength and compressive strength?
Compressive strength vs tensile strength is comparison of strength in which compressive strength is pushing force tends to reduce the size of material after compression whereas tensile strength is pulling force tends to increase the size of material after tension.
## What is compressive strength of a material?
Compressive strength is defined as the ability of material to resist the direct pressure of applied compression force.
## How do you find the compressive strength of a metal?
Calculating Compressive Strength The formula is: CS = F ÷ A, where CS is the compressive strength, F is the force or load at point of failure and A is the initial cross-sectional surface area.
## What metal has the highest compressive strength?
Tungsten vs Titanium In terms of tensile strength, tungsten is the strongest out of any natural metal (142,000 psi). But in terms of impact strength, tungsten is weak — it’s a brittle metal that’s known to shatter on impact.
## How do I describe the compression?
The definition of compression is the action or state of being squished down or made smaller or more pressed together. When a pile of material is squished together and made smaller and more dense, this is an example of compression.
## What is the main effect of compression on a material?
Compression is a force that pushes or squeezes materials. A material under compression may be shortened or crushed. Tension is a force that pulls and/or stretches materials. When a material is bent, one side is in tension while the other is in compression.
## What are examples of compression?
8 Compression Force Examples in Daily Life. Bridge.Hydraulic Press.Spring.Shoe Sole.Bicycle Pump.Sponge.Plush Toys.Air Suspension System.
## How do you calculate the compressive strength of concrete?
The formula for calculating Compressive strength of cement is The maximum load carried by the mortar specimen (cube) which means the load point on Compressive testing machine at which the specimen starts breaking is divided with the surface area (contact area).
## What is compressive and tensile strength?
Compressive strength vs tensile strength is comparison of strength in which compressive strength is pushing force tends to reduce the size of material after compression whereas tensile strength is pulling force tends to increase the size of material after tension.
## What is a good compressive strength of concrete?
The compressive strength of normal concrete varies between 25 and 50 MPa. Above this level, the term High Performance Concrete is used.
## What is the formula for compressive strength of concrete?
Calculating Compressive Strength The formula is: CS = F ÷ A, where CS is the compressive strength, F is the force or load at point of failure and A is the initial cross-sectional surface area.
## What are the importance of compressive strength of concrete?
Compressive strength is one of the most important properties of concrete and mortar. The strength of the binder (cement) therefore has a significant effect on the performance characteristics of the mixture and ensures the overall quality of the finished product.
## What is minimum concrete compressive strength?
Concrete compressive strength shall be a minimum of 4,500psi and have a minimum tensile strength of 450psi as per ASTM C 496.
## What is average compressive strength?
The average compressive strength values ranged between 10.2 and 22.2 MPa at a strain of 0.1.
## What is the average compressive strength of concrete?
Concrete compressive strength for general construction varies from 15 MPa (2200 psi) to 30 MPa (4400 psi) and higher in commercial and industrial structures.
## What has high compressive strength?
Concrete and ceramics typically have much higher compressive strengths than tensile strengths. Composite materials, such as glass fiber epoxy matrix composite, tend to have higher tensile strengths than compressive strengths.
## What is tensile and compressive?
Tensile means there is an increase in length of the object, and compressive is a decrease in length. From stress and strain we can find a material’s elastic modulus, which is the measure of the stiffness of a material.
## What is the tensile strength?
tensile strength, maximum load that a material can support without fracture when being stretched, divided by the original cross-sectional area of the material.
## What is tensile and compressive strength of concrete?
Concrete, Reinforced Concrete is relatively strong in compression but weak in tension. The compressive strength of concrete in structures is typically 3000–9000 psi (20–60 MPa), and the tensile strength is normally about 10–15% of the compressive strength.
## How do you measure the compressive strength of a metal?
compressive strength test, mechanical test measuring the maximum amount of compressive load a material can bear before fracturing. The test piece, usually in the form of a cube, prism, or cylinder, is compressed between the platens of a compression-testing machine by a gradually applied load.
## What is the compressive strength of metal?
As its name implies, compressive strength is the maximum amount of pressure or compression a metal can withstand. This is typically measured with a universal testing machine that applies an increased load on the material.
## What is the formula for compressive strength?
The formula to calculate compressive strength is F = P/A, where: F=The compressive strength (MPa) P=Maximum load (or load until failure) to the material (N) A=A cross section of the area of the material resisting the load (mm2)
## How do you determine the strength of a metal?
The ultimate tensile strength of a material is calculated by dividing the area of the material tested (the cross section) by the stress placed on the material, generally expressed in terms of pounds or tons per square inch of material.
## What material has highest compressive strength?
High compressive strength leads to the failure of the material due to tension. We can also say that it is the stress on materials that leads to a smaller volume. So, among the given options high carbon steel has the highest compressive strength as compared to other options.
## What metal has the strongest compressive strength?
With these types of strengths in mind, here are some of the strongest metals you can find:
• Tungsten. Tungsten, which is Swedish for “heavy stone,” is the strongest metal in the world. …
• Steel. Steel is the second strongest and the most widely used metal in the world. …
• Chromium. …
• Titanium. …
• Iron. …
• Lutetium.
## What is the strongest metal for armor?
The 4 Strongest and Hardest Metals on Earth
• Tungsten: The Strongest Metal on Earth. Of all the metals, tungsten reigns supreme in terms of tensile strength. …
• Chromium: The Hardest Metal on Earth. Chromium is the hardest metal known to man. …
• Steel: The Strongest Alloy on Earth. …
• Titanium.
## What is compressive strength and tensile strength of concrete?
1) compressive strength of concrete is higher than tensile strength, concrete experience good behave in compression whereas poor behave in tension. Maximum compressive strength of M20 concrete is 20MPa whereas maximum tensile strength is only about 10 to 12% of compressive strength.
## What’s the difference between tensile and compressive?
The main difference between tensile and compressive stress is that tensile stress results in elongation whereas compressive stress results in shortening. Some materials are strong under tensile stresses but weak under compressive stresses.
## What is tensile and compressive stresses?
Tension and compression. stress. Tensile stress is the normal force per area (σ = F/A) that causes an object to increase in length. Compressive stress is the normal force per area (σ = F/A) that causes an object to decrease in length.
## What is a typical compressive strength for concrete?
Usually, the compressive strength of concrete varies from 2500 psi (17 MPa) to 4000 psi (28 MPa) and higher in residential and commercial structures. Several applications also utilize strengths greater than 10,000 psi (70 MPa).
## What is the best compressive strength of concrete?
Concrete compressive strength shall be a minimum of 4,500psi and have a minimum tensile strength of 450psi as per ASTM C 496.
## What would be a good strength for concrete?
2,500 PSI. Often more affordable than higher strength concrete, 2,500 PSI can be useful for driveways and walkways. However, some may choose a stronger concrete, like 3,000 PSI, to avoid excessive cracking. A good use for this concrete is a walkway on the side of a home that doesn’t receive excessive traffic.
## What is good compressive strength?
Concrete compressive strength requirements can vary from 2,500 psi for residential concrete to 4,000 psi and higher in commercial structures. Higher strengths up to and exceeding 10,000 psi are specified for certain applications.
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https://www.coursehero.com/file/164882/Fundamentals-of-Machining-Lab-prep/ | 1,498,241,277,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320077.32/warc/CC-MAIN-20170623170148-20170623190148-00409.warc.gz | 880,580,903 | 48,735 | Fundamentals of Machining Lab prep
# Fundamentals of Machining Lab prep - t1= feed cos(scea...
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Fundamentals of Machining Lab Prep
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Machining Fundamentals (Chip) Lab Learning Objectives Appreciate the relations between process variables (independent variable) and design parameters (dependent variable) in manufacturing Appreciate the process of chip formation Rationale Important for the design of parts, processes and tools
Steps for the chip lab Feed is the independent variable Create NC Code (vary feed) Cut Parts Vary feed Collect Chips Monitor Power – record power levels for idling and each feed Measure part diameters Measure Surface Texture – calculate Rt Measure Chips – calculate shear angle Lab Report plots (indicate variance) Surface Roughness (Rt), measured and theoretical v. feed Chip thickness v. feed Shear angle v. feed Power v. feed Specific energy for material removal (J/mm 3 ) v. feed (Radius error-Rt) *EI /l 3 versus feed
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Turning straight down view tool chuck not shown side cutting edge angle (scea) feed workpiece rotation tool rake face t1 uncut chip thickness
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Unformatted text preview: t1= feed cos(scea) cutting edge Shear angle Shear angle From Geometry: tan Φ = r t cosα / (1- r t sin α) Chip thickness ratio r t = t 1 /t 2 Uncut chip thickness or depth of cut in orthogonal machining Vc – cutting velocity Rake angle Theoretical Peak to valley roughness for an Idealized Turned surface profile 4 2 2 2 2 2 f R RR R R t t + +-= 4 2 2 2 f R RR t t + = r Tool nose radius profile R t peak to valley roughness r f/2 feed 2 2 2 ) 2 ( ) ( f R r r t +-= r f R t 8 2 2245 Surface roughness as a function of feed r f R t 8 2 2245 2 2 2 2 f r r R t--= Common approximation since Rt^2 is small compared with f^2 expanding 4 2 2 2 2 2 f R rR r r t t + +-= 4 2 2 2 f R rR t t + = 1.408mm 1.408mm 1.056 mm turned turned steel steel 1.408mm 1.408mm 0.05217mm 0.05217mm Cantilever beam deflection http://www.me.metu.edu.tr/me206/DARENDELILER/Beam_Deflection_Formulae.pdf I area moment of inertia E elastic modulus 30,000,000 psi Aluminum...
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## This note was uploaded on 04/29/2008 for the course ME 1800 taught by Professor Brown during the Winter '08 term at WPI.
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Ask a homework question - tutors are online | 800 | 3,083 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-26 | longest | en | 0.805466 |
https://math.stackexchange.com/questions/1750238/implicit-euler-method-and-explicit-euler-method/1750431 | 1,709,496,331,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00428.warc.gz | 388,639,150 | 34,828 | # Implicit Euler method and explicit Euler method
I wanna know what is the difference between explicit Euler's method and implicit Euler's method. And is the local truncation error for both of them is $O(h)$ and the coefficient of the $O(h)$ term is $h/2$?
• Implicit Euler is explicit Euler backwards. The error term either contains the second derivative or a Lipschitz constant, $h/2$ is not the answer. Apr 19, 2016 at 21:53
• I attached a picture in my problem. Why do some people say the error is $O(h)$ while some say it is $O(h^2)$? Apr 19, 2016 at 23:28
• @J.doe The error in one step (for nice enough problems) is $O(h^2)$. However, to get to a given fixed time $t$, you have to perform $O(1/h)$ steps of size $h$. Thus the error accumulates to $O(h)$ in the process of reaching time $t$.
– Ian
Apr 20, 2016 at 0:03
• $k_1=f(t_n,y_n)$ for runge- kutta, how to taylor expand it? Apr 20, 2016 at 5:02
The error of both explicit and implicit Euler are $O(h)$. So
$$f(x-h) = f(x) - h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + \cdots$$
and
$$f(x+h) = f(x) + h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots$$
So the backward Euler is
$$f(x) - f(x-h) = h f'(x) - \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) - \cdots$$
$$f'(x) = \frac{f(x) - f(x-h)}{h} + \frac{h}{2} f''(x) - \frac{h^2}{6} f'''(x) + \cdots$$
the backward Euler is first order accurate
$$f'(x) = \frac{f(x) - f(x-h)}{h} + O(h)$$
And the forward Euler is
$$f(x+h) - f(x) = h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots$$
the forward Euler is first order accurate
$$f'(x) = \frac{f(x+h) - f(x)}{h} + O(h)$$
We can do a central difference and find
$$f(x+h) - f(x-h) = (h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots) - (- h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + \cdots)$$
$$f(x+h) - f(x-h) = (h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + \cdots) + (h f'(x) - \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) - \cdots)$$
$$f(x+h) - f(x-h) = 2 h f'(x) + \frac{h^3}{6} f'''(x) + \cdots$$
$$f'(x) = \frac{f(x+h) - f(x-h)}{2h} - \frac{h^2}{12} f'''(x) + \cdots$$
Therefore, the central difference is second order accurate.
$$f'(x) = \frac{f(x+h) - f(x-h)}{2h} + O(h^2)$$
• So the coefficient of the $O(h)$ term of the implicit Euler's method is $h/2$? Apr 20, 2016 at 3:10
• What I don't get it why isn't the coefficient of the $O(h)$ term of the implicit Euler's method $h/2$? Apr 20, 2016 at 20:35
• Because the leading error term is $\frac{f''(x)}{2} h$. Therefore, when $h \rightarrow \frac{h}{n}$, the global error goes from $\varepsilon \rightarrow \frac{\varepsilon}{n}$. The important part is how it changes with the step size. Apr 21, 2016 at 0:47
• I don't get it.. So what's the coefficient of the $O(h)$ term? Apr 21, 2016 at 1:48 | 1,153 | 2,814 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-10 | latest | en | 0.708126 |
https://brainmass.com/business/bond-valuation/sale-price-amortization-schedule-bonds-176709 | 1,601,357,807,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401624636.80/warc/CC-MAIN-20200929025239-20200929055239-00148.warc.gz | 265,783,987 | 10,891 | Explore BrainMass
# Sale price and amortization schedule for bonds
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Assume that on 1-1-07, Janila Co. issued \$500,000 of 6%, ten-year bond for a yield of 7%. The bonds pay interest annually on December 31.
a. What will be the sale price of the bonds?
b. Prepare an amortization schedule under the effective interest method for the first three years of the bond's life.
#### Solution Preview
a. What will be the sale price of the bonds?
The sale price of the bonds will be the present value of interest and principal discounted at the yield of 7%. The interest amount is 500,000X6%=30,000 per year and the principal amount is 500,000. Discounting at 7% ...
#### Solution Summary
The solution explains how to calculate the issue price of the bonds and how to prepare an amortization schedule (see attachment for further details).
\$2.19 | 225 | 953 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2020-40 | latest | en | 0.908372 |
https://www.hackmath.net/en/math-problem/2057 | 1,632,020,370,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00175.warc.gz | 865,184,999 | 12,337 | # Cash back
After a trip 24 students have left more than 650 but less than 690 CZK (Kc). How much each student receives return when the amount is divisible by the number of pupils?
x = 28
### Step-by-step explanation:
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Adéla sells blueberries for 50 CZK/1 l, Zdeněk sells 1 kg of blueberries for 75 CZK. Which of them sells irritants if 1 liter of blueberries weighs 650 g?
• The kitchen
The kitchen has the shape of a square with a side of 5 m. We place linoleum on the floor, the price of which is 450 CZK per 1 m2. The glue costs 750 CZK. How much do we pay in total?
• Czech coins
John has 540 CZK, but he has 10 CZK coins and 20 CZK coins only, we knows that he has 10CZK coins 7times more that 20CZK coins, how many 10CZK coin and how many 20CZK coins he saved?
• Average monthly salary
A total of 10 teachers work at one small school in Moravia. The monthly salary of each is 21,500 CZK or 21,800 CZK or 22,500 CZK according to their education and age. The average monthly salary for this school's teacher is 21 850 CZK. How many teachers of
• Performance
Same number of tickets sold for 15 CZK, 20 CZK, 30 CZK and 40 CZK a total of CZK 13,230 on theatre performance. How many tickets were sold for this performance?
• Ratios
Divide: a) 250 CZK in the ratio 2:3 b) 1000 CZK in the ratio 4:7:9
• Parking
120 vehicles parking on the morning. Pasenger car is charged 20 CZK, 50 CZK per bus. The guard collected for parking 2640 CZK in total. How many cars and how many buses stood in the parking?
• Exhibition
The teacher paid 280 Kč for 4.A students for admission to the exhibition. How many students were at the exhibition?
• Coffee
Coffee merchant has coffee robusta and arabica species. 1 kg Robusta worth 450 CZK, Arabica 1 kg is 300 CZK more expensive. Calculate how many kilograms of Robusta and Arabica will need to produce 30 kg of the mixture so that the mixture cost is 490 CZK p
• I think number
I think number.When I add 841 to it and subtract 157, I get a number that is 22 greater than 996. What number I thinking?
• Bus tickets
Bus ticket for a trip from Prague to Paris cost 2180Kč. A return ticket costs 3930Kč. How much money will save a family of four from going to Paris and back when they purchase return tickets? | 998 | 3,630 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-39 | longest | en | 0.933354 |
https://www.guwsmedical.info/biosurveillance-system/the-importance-of-multiple-sources-of-data.html | 1,582,476,837,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00273.warc.gz | 737,507,024 | 4,382 | ## The Importance Of Multiple Sources Of Data
Multiple independent sources of information can increase both sensitivity and specificity. Let us begin with a drastically oversimplified example. Suppose Sensor A has a daily 90% chance of signaling an attack (event SIGA) if one occurs, and a 1% chance of signaling an attack when there is none. Suppose sensor B monitors an independent data source, and has a daily 90% chance of signaling an attack (event SIGB) if one occurs, and a 1% chance of signaling an attack when there is none. Writing ATT as the event of attack, we have:
P(SIGAI~ATT) = 0.01 P(SIGAIATT) = 0.9 P(SIGBI~ATT) = 0.01 P(SIGBIATT) = 0.9
Then there is now a 99% chance that at least one detector will signal if there is an attack, and there is only a probability of
1 in 10,000 that both detectors will signal if there is no attack. Thus, in many situations (both the [SIGA and SIGB] case and the [~SIGA and ~SIGB] case), the operational decision is much clearer. Even in the case of inconsistent signals, the analysis task can be better informed.
Another compelling example concerns time series analysis. Figure 15.1 shows two times series, for daily sales of two fictional products in a fictional city. Apart from a general upward trend, no serious anomalies stand out.
If, however, we look at the same data in a different way, February 20 stands out as somewhat anomalous. The new view is a scatterplot. For each date, it plots one data point, with the x-coordinate denoting sales of Product A, and the y-coor-dinate denoting sales of Product B. There is a general correlation in sales, but February 20 is atypical because B sales are high, taking into account A's sales.
In this chapter, we briefly survey methods which can notice effects that are revealed by inspecting more than one time series at a time. | 434 | 1,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-10 | latest | en | 0.942978 |
https://electronics.stackexchange.com/questions/459454/capacitor-negative-current-flow | 1,718,430,047,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00381.warc.gz | 212,671,043 | 40,844 | # Capacitor Negative Current Flow
I have a basic quesiton about circuit theory in general. Consider the following circuit:
I understand that the resistor will have a constant voltage of 1V across it as it takes the DC component of the input voltage source. The capacitor will take the AC part of the input voltage source (the 0.5V peak-peak signal).
There's my first question: How can there be 1V across the resistor and 0.5V peak-to-peak across the capacitor when both devices are in parallel? Don't they have the same voltage across them?
The second question is when the capacitor current goes negative, it's supplying current to somewhere -> where does it supply current to if top node is at same voltage?
I feel like I'm missing some basic rule here?
• ”I understand that the resistor will have a constant voltage of 1V across it as it takes the DC component of the input voltage source.” Not correct. Simulate it and you will see. Commented Sep 20, 2019 at 15:45
• The capacitor and resistor see both the AC part and the DC offset. What made you think they don't? Commented Sep 20, 2019 at 15:49
• – G36
Commented Sep 20, 2019 at 16:57
I understand that the resistor will have a constant voltage of 1V across it as it takes the DC component of the input voltage source.
This is way off base.
We model an ideal resistor as able to respond to any applied voltage, whether DC or AC of any frequency.
If you want a resistor that only responds to the DC component of the applied voltage, you'd want to include a very large series inductance attached to that resistor.
There's my first question: How can there be 1V across the resistor and 0.5V peak-to-peak across the capacitor when both devices are in parallel? Don't they have the same voltage across them?
They have the same voltage across them.
1 V DC plus a 0.25 V 1 Hz sinusoid.
The second question is when the capacitor current goes negative, it's supplying current to somewhere -> where does it supply current to if top node is at same voltage?
In general it could supply current to either the resistor, the voltage source, or both.
There's no rule that the current through the voltage source must always flow out the positive terminal.
• What about if I had a very large frequency AC signal on my voltage source. Shouldn't the capacitor then behave as a short, acting as a decoupling/bypass capacitor and then the resistor should only see the DC 1V? But then wouldn't that contradict the fact that resistor and capacitor should have same voltage across them? Commented Sep 20, 2019 at 16:02
• @AlfroJang80, if you want to include that effect, you need to include a source impedance in your voltage source model. With the ideal voltage source you've drawn, the capacitor is never close enough to a short to prevent the resistor from seeing the full voltage. Commented Sep 20, 2019 at 16:03
• May I ask why a source impedance is necessary? I thought the capacitor has an impedance of 1/jwC when dealing with sinusodial voltages and currents. Commented Sep 20, 2019 at 16:09
• Yes, but an ideal voltage source can supply as much current as needed to drive it with 0.25 V AC. The ideal source can provide 1,000,000,000,000,000 amps if necessary to deliver the 0.25 V. If you want a source that will fail to deliver enough current to keep the AC voltage at 0.25 V at some high frequency, you need to include a source impedance in the voltage source model. Commented Sep 20, 2019 at 16:11
• Ah. I see. Thank you Commented Sep 20, 2019 at 16:13
The basic rule of lumped-circuit analysis is that parallel components all have the same voltage across them, and series components all have the same current flowing through them.
Your circuit is a parallel circuit, so the first rule applies. The voltage is strictly determined by the voltage source. The thing that's different is the current that flows in each component. | 919 | 3,887 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-26 | latest | en | 0.947707 |
https://www.teacherspayteachers.com/Product/Valentines-Day-Algebra-Solving-Inequalities-1931684 | 1,488,044,131,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501171781.5/warc/CC-MAIN-20170219104611-00444-ip-10-171-10-108.ec2.internal.warc.gz | 890,025,749 | 25,095 | Total:
\$0.00
# Valentine’s Day Algebra – Solving Inequalities
Subjects
Resource Types
Common Core Standards
Product Rating
4.0
File Type
PDF (Acrobat) Document File
1.37 MB | 5 pages
### PRODUCT DESCRIPTION
Create a few Valentine’s Day smiles with this solving inequalities practice. Valentine themed answer messages spice up regular practice and make grading a snap!
The Holiday Fun:
Students solve 15 random inequalities (out of 26 inequalities) before receiving one of the 6 Valentine cards. Students use answers to decode the Valentine’s hidden message. I enjoy hand-selecting the recipient of the message, matching the message to the student’s personality. Since the Valentine is addressed “To My Wonderful Teacher,” students are essentially sending you, the teacher, the Valentine when they turn in their work. “Oh what thoughtful students I have!”
The Content:
Students solve multi-step inequalities. Find two versions of the 26 inequalities practice page. Both versions have the same solutions. Therefore, easily differentiate by student need without any hiccup in the activity (if you want). Please see the thumbnails and preview for a view of the two difficulty levels; some consider both levels challenging.
★Common Core Standards: REI.B.3
What makes this product a must-have?
Self-checking: Know at a glance if students completed the activity correctly.
Novel Practice: Mixing-up that traditional worksheet perks up kids.
Holiday Cheer: Throw in a little holiday spirit WITHOUT compromising rigorous academic standards.
☝Check out the Preview. I want buyers to be tickled pink with their purchase!
± All feedback is valued and encourages improved products. Did you know buyers also have access to future revisions?
♩ Note: Problems are identical to the inequalities hangman and all holiday-themed inequality products in my store.
Total Pages
5
Included
Teaching Duration
N/A
### Average Ratings
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Total:
1 rating
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User Rating: 4.0/4.0
(364 Followers)
\$3.00 | 471 | 2,097 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-09 | longest | en | 0.900046 |
https://programmer.help/blogs/three-programming-questions-of-huawei-autumn-machine-test-2021-09-08.html | 1,632,713,445,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00072.warc.gz | 501,580,976 | 6,724 | # Three programming questions of Huawei autumn machine test (2021-09-08)
Notice: HUAWEI, Ali's latest autumn test written test questions, ideas and reference codes have been arranged well in WeChat official account, TechGuide, and the official account of the private letter is returned to HUAWEI or Ali to get the most real-time and detailed written test questions.
# Track 1: optimal node (100%)
## Title Description
Give a binary tree, each node has a number and a value, which may be negative. Please find an optimal node (except the root node), so that after the node divides the tree into two trees (the original tree removes this node and its children, and the new tree takes this node as the root node), the absolute value of the difference between the sum of each node of the two trees is the largest.
Please output the node number. If there are multiple same differences, output the node with the smallest number.
Enter description
4
4 9 -7 -8
0 1
0 3
1 2
The first row, four nodes, number 0-3, range 1-10000
The second line is the weight of node 0-3
The third to fifth lines represent the parent-child relationship between the nodes of the binary tree
0 1 / / the left node of node 0 is 1
0 3 / / the right node of node 0 is 3
1 2 / / the left node of node 1 is 2
Note: the left node always appears before the right node
```4
4 9 -7 -8
0 1
0 3
1 2
```
Output description
Node number. In the example, the node with number 3 is the optimal node
```3
```
## Reference code
Using dfs, you can find the sum of each node and all its child nodes. Note that you need to use long long
```// Follow TechGuide! Dachang pen classics are delivered by lightning express!
```
# Second track: Plum Blossom pile (100%)
## Title Description
There is an M*N plum blossom pile array, and each pile has the maximum number of steps allowed to jump. The user can jump from the position of 0,0, and can jump in the right and down directions. Find out how many times it takes to jump at least to reach the positions of M-1 and N-1. - 1 is returned when the destination cannot be reached.
M < = 100, n < = 100, the maximum number of steps allowed to jump on each pile is a positive integer less than 10, and 0 means that jumping to this position is not allowed
Enter Description:
1. The first line is M and N, separated by ",";
2. The second line is M*N plum blossom pile (refer to the example for format). The array position is the maximum number of steps allowed to jump. 0 means that the position is empty and cannot jump to this position.
```3,3
3 2 2 0 1 0 1 1 1
```
Output description
Minimum jump steps
```2
```
## Reference code
Violence, update the shortest steps of all points it can reach for each point
```public class TechGuide{
static class Node{
int row;
int col;
public Node(int row, int col){
this.row = row;
this.col = col;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Node node = (Node) o;
return row == node.row &&
col == node.col;
}
@Override
public int hashCode() {
return Objects.hash(row, col);
}
}
static int distance(Node node1, Node node2){
return Math.abs(node2.row - node1.row) + Math.abs(node2.col - node1.col);
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String line1 = in.nextLine();
String line2 = in.nextLine();
String[] line1s = line1.split(",");
String[] line2s = line2.split(" ");
int m = Integer.parseInt(line1s[0]);
int n = Integer.parseInt(line1s[1]);
int[][] nums = new int[m][n];
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++)
nums[i][j] = Integer.parseInt(line2s[i*m + j]);
}
Queue<Node> nodes = new ArrayDeque<>();
int[][] dp = new int[m][n];
for(int i =0; i < m; i++){
Arrays.fill(dp[i], Integer.MAX_VALUE);
}
dp[0][0] = 0;
while(!nodes.isEmpty()){
Node node = nodes.poll();
int step = nums[node.row][node.col];
for(int i = step; i > 0; i--){
for(int row = node.row + i; row >= node.row; row--) {
int col = i - (row - node.row) + node.col ;
if (row < m && col < n && nums[row][col] != 0) {
dp[row][col] = Math.min(dp[row][col], dp[node.row][node.col] + 1);
Node temp = new Node(row, col);
if(!nodes.contains(temp))
if(temp.row == m -1 && temp.col ==n - 1){
System.out.println(dp[m - 1][n - 1]);
System.exit(0);
}
}
}
}
}
System.out.println(-1);
}
}
// Follow TechGuide! Dachang pen classics are delivered by lightning express!
```
### CPP version
```#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int ninf = 0xc0c0c0c0;
const int maxn = 100 + 5;
int f[maxn][maxn];
int dp[maxn][maxn];
int main() {
string s;
cin >> s;
int n, m;
sscanf(s.c_str(), "%d,%d", &n, &m);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
scanf("%d", &f[i][j]);
}
}
memset(dp, inf, sizeof dp);
dp[0][0] = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
int x = f[i][j];
for (int k = 0; k <= x; ++k) {
if (i + k < n && f[i + k][j] != 0) {
dp[i + k][j] = min(dp[i + k][j], dp[i][j] + 1);
}
if (j + k < m && f[i][j + k] != 0) {
dp[i][j + k] = min(dp[i][j + k], dp[i][j] + 1);
}
}
}
}
if (dp[n - 1][m - 1] != inf) {
cout << dp[n - 1][m - 1];
} else {
cout << -1;
}
return 0;
}
/*
testcases:
3,3
3 2 2 0 1 0 1 1 1
*/
// Follow TechGuide! Dachang pen classics are delivered by lightning express!
```
# Third track: minimum compilation time (100%)
## Title Description
Company A needs to introduce an open source project into the project and evaluate the compilation time of A module in the open source project.
At present, the compilation time of each module in the project and the list of modules it depends on are known. When there are an unlimited number of parallel tasks, find the shortest compilation time of a specified module.
If there is a circular dependency between modules or the dependent module does not exist, the compilation cannot be completed and - 1 is returned.
Enter Description:
The first line is the target module name,
In the future, each line of input defines a module, including the module name, compilation time and dependent module list, separated by commas. If the dependent module list does not exist, it means that it can be compiled independently, such as module 2,10, module 1,10
The module name contains only letters and numbers and at least one character. The number of modules shall not exceed 50000
```module3
module1,10
module2,5
module3,10,modulel,moduln2
```
Output Description:
Output the minimum compilation time. If the compilation cannot be completed, output - 1
```20
```
Explanation:
Module1 compilation takes 10 ms, module2 compilation takes 5 ms, module3 compilation depends on module1 and module2, and self compilation also takes 10 ms, so the total compilation time is 10+max(10, 5) = 20 ms
## Reference code
Topological sorting. I don't see that there can be letters after the module, which is written in numbers
```public class TechGUide{
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String name = in.nextLine();
List<String> strings = new ArrayList<>();
while(in.hasNextLine()){
String string = in.nextLine();
}
Set<String> set = new HashSet<>();
//Start flag
//Count the names of all modules, which can be mapped
for(String s: strings){
String[] temp = s.split(",");
for(int i = 0; i < temp.length; i++){
if(i != 1){
}
}
}
//Map letters to numbers
HashMap<String, Integer> mapping = new HashMap<>();
int index = 0;
for(String s : set){
mapping.put(s, index);
index++;
}
//Save the next node location of the node
HashMap<String, List<String>> relation = new HashMap<>();
relation.put("start", new ArrayList<String>());
int[][] weights = new int[set.size()][set.size()];
for(String s: strings){
String[] temp = s.split(",");
//String end = temp[0];
if(temp.length == 2){
List<String> start = relation.get("start");
weights[mapping.get("start")][mapping.get(temp[0])] = Integer.parseInt(temp[1]);
} else{
for(int i = 2; i < temp.length; i++){
weights[mapping.get(temp[i])][mapping.get(temp[0])] = Integer.parseInt(temp[1]);
if(relation.containsKey(temp[i])){
List<String> next = relation.get(temp[i]);
} else{
List<String> next = new ArrayList<>();
relation.put(temp[i], next);
}
}
}
}
//Check for circular dependencies
for(int i = 0; i < set.size() - 1; i++){
for(int j = i + 1; j < set.size(); j++){
if(weights[i][j] != 0 && weights[j][i] != 0){
System.out.println(-1);
System.exit(0);
}
}
}
Queue<String> que = new ArrayDeque<>();
//Save the maximum time it takes to execute the task
int[] end = new int[set.size()];
int max = 0;
//Traverse from scratch
while(!que.isEmpty()){
String node = que.poll();
//Module from execution to end
List<String> nextNodes = relation.get(node);
//Module missing, early termination
if(!node.equals(name) && (nextNodes.size()==0 || nextNodes == null)){
System.out.println(-1);
break;
}
//It's time for the task. The maximum time for the array to reach the task
if(node.equals(name)){
System.out.println(end[mapping.get(node)]);
break;
}
for(String next : nextNodes){
end[mapping.get(next)] = Math.max(end[mapping.get(next)], end[mapping.get(node)] + weights[mapping.get(node)][mapping.get(next)]);
}
}
}
}
/*
m2
m2,10,m1
m1,10,m2
*/
/*
m2
m2,10,m1
*/
/*
m3
m1,10
m2,5
m3,10,m1,m2
// Follow TechGuide! Dachang pen classics are delivered by lightning express!
*/
```
Congratulations on finding the treasure! WeChat search official account [TechGuide] pays attention to more fresh good articles and the Internet classics.
Posted on Tue, 14 Sep 2021 19:34:02 -0400 by gazalec | 2,686 | 9,577 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2021-39 | latest | en | 0.887858 |
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##### Stage: 3 Challenge Level:
After some matches were played, most of the information in the table containing the results of the games was accidentally deleted. What was the score in each match played?
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A game for 2 players that can be played online. Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter.
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##### Stage: 3 Challenge Level:
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##### Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
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##### Stage: 3 Challenge Level:
Six points are arranged in space so that no three are collinear. How many line segments can be formed by joining the points in pairs?
### Problem Solving, Using and Applying and Functional Mathematics
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Master Minding
##### Stage: 3 Challenge Level:
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### Thirty Nine, Seventy Five
##### Stage: 3 Challenge Level:
We have exactly 100 coins. There are five different values of coins. We have decided to buy a piece of computer software for 39.75. We have the correct money, not a penny more, not a penny less! Can. . . .
### Tis Unique
##### Stage: 3 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
### More Mathematical Mysteries
##### Stage: 3 Challenge Level:
Write down a three-digit number Change the order of the digits to get a different number Find the difference between the two three digit numbers Follow the rest of the instructions then try. . . .
### Football Champs
##### Stage: 3 Challenge Level:
Three teams have each played two matches. The table gives the total number points and goals scored for and against each team. Fill in the table and find the scores in the three matches.
### 9 Weights
##### Stage: 3 Challenge Level:
You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance?
### Aba
##### Stage: 3 Challenge Level:
In the following sum the letters A, B, C, D, E and F stand for six distinct digits. Find all the ways of replacing the letters with digits so that the arithmetic is correct.
### The Genie in the Jar
##### Stage: 3 Challenge Level:
This jar used to hold perfumed oil. It contained enough oil to fill granid silver bottles. Each bottle held enough to fill ozvik golden goblets and each goblet held enough to fill vaswik crystal. . . .
### Volume of a Pyramid and a Cone
##### Stage: 3
These formulae are often quoted, but rarely proved. In this article, we derive the formulae for the volumes of a square-based pyramid and a cone, using relatively simple mathematical concepts.
### Clocked
##### Stage: 3 Challenge Level:
Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours?
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
##### Stage: 2 and 3
A paradox is a statement that seems to be both untrue and true at the same time. This article looks at a few examples and challenges you to investigate them for yourself.
### Children at Large
##### Stage: 3 Challenge Level:
There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
### Königsberg
##### Stage: 3 Challenge Level:
Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps?
### Ratty
##### Stage: 3 Challenge Level:
If you know the sizes of the angles marked with coloured dots in this diagram which angles can you find by calculation?
### Logic
##### Stage: 2 and 3
What does logic mean to us and is that different to mathematical logic? We will explore these questions in this article.
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number?
### Concrete Wheel
##### Stage: 3 Challenge Level:
A huge wheel is rolling past your window. What do you see?
### Shuffle Shriek
##### Stage: 3 Challenge Level:
Can you find all the 4-ball shuffles?
### Con Tricks
##### Stage: 3
Here are some examples of 'cons', and see if you can figure out where the trick is.
### Cycle It
##### Stage: 3 Challenge Level:
Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total.
### Eleven
##### Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Marbles
##### Stage: 3 Challenge Level:
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
### Growing Ls
##### Stage: 3 Challenge Level:
Can you fit Ls together to make larger versions of themselves?
### Winning Team
##### Stage: 3 Challenge Level:
Nine cross country runners compete in a team competition in which there are three matches. If you were a judge how would you decide who would win?
### How Many Dice?
##### Stage: 3 Challenge Level:
A standard die has the numbers 1, 2 and 3 are opposite 6, 5 and 4 respectively so that opposite faces add to 7? If you make standard dice by writing 1, 2, 3, 4, 5, 6 on blank cubes you will find. . . .
### More Marbles
##### Stage: 3 Challenge Level:
I start with a red, a blue, a green and a yellow marble. I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour?
### Greetings
##### Stage: 3 Challenge Level:
From a group of any 4 students in a class of 30, each has exchanged Christmas cards with the other three. Show that some students have exchanged cards with all the other students in the class. How. . . .
##### Stage: 3 Challenge Level:
What can you say about the angles on opposite vertices of any cyclic quadrilateral? Working on the building blocks will give you insights that may help you to explain what is special about them.
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Is it Magic or Is it Maths?
##### Stage: 3 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
### Always the Same
##### Stage: 3 Challenge Level:
Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?
### Take Three from Five
##### Stage: 3 and 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Multiplication Square
##### Stage: 3 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Reverse to Order
##### Stage: 3 Challenge Level:
Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?
##### Stage: 3 Challenge Level:
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .
### Triangle Inequality
##### Stage: 3 Challenge Level:
ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm.
### Calendar Capers
##### Stage: 3 Challenge Level:
Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat. . . .
### Convex Polygons
##### Stage: 3 Challenge Level:
Show that among the interior angles of a convex polygon there cannot be more than three acute angles.
### N000ughty Thoughts
##### Stage: 4 Challenge Level:
Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the. . . . | 2,433 | 10,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2015-18 | longest | en | 0.92256 |
https://knittystash.com/hino-wiring-diagram-schematic/ | 1,675,022,328,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499758.83/warc/CC-MAIN-20230129180008-20230129210008-00166.warc.gz | 365,713,880 | 10,529 | # Hino Wiring Diagram Schematic
This page contains details on the Hino Wiring Diagram Schematic, hints, and frequently asked questions. We produced this page to assist those trying to find a Hino Wiring Diagram Schematic, and hopefully, You can resolve your problem with our info.
A wiring diagram will certainly show you where the cables need to be linked, eliminating the demand for uncertainty.
You can avoid making mistakes if you make use of a wiring diagram to discover out what wires go where. You won’t need to make any type of assumptions if you have a wiring diagram since it will certainly show you specifically where the wires ought to be connected.
## Hino Wiring Diagram Schematic
See the Hino Wiring Diagram Schematic images below
## Tips and tricks for reading wiring diagrams
• When taking a look at a wiring diagram, don’t attempt to concentrate on the whole page all at once. It’s overwhelming. Put a blank sheet of paper beside the wiring diagram and just draw the simple circuit. Focus on the simple part and follow the current flow from power to ground or from ground to power. All complex wiring diagrams are just a series of simple diagrams, and it makes it hard to take a look at if you don’t limit to the circuit that you’re doing.
• Print the wiring diagram off and use highlighters to trace the circuit. When you use your finger or follow the circuit with your eyes, it’s simple to mistrace the circuit. One trick that I use is to print the exact same wiring diagram off twice. On one, I’ll trace the current flow, how it runs, and that reveals me what parts of the circuit I need to check. On the other one, I’ll start coloring the things that evaluated okay. When I get done, anything that’s not highlighted are suspect circuits that I need to identify.
• To appropriately check out a wiring diagram, one has to understand how the parts in the system operate. Following diagrams is relatively simple, however utilizing it within the scope of how the system runs is a different matter.
• Check out wiring diagrams from negative to positive and redraw the circuit as a straight line. All circuits are the same– voltage, ground, single component, and switches.
• Before reading a schematic, get familiar and comprehend all the symbols. Read the schematic like a roadmap. I print the schematic and highlight the circuit I’m diagnosing to make sure I’m remaining on the right path.
Hino 500 Wiring Diagram Schematic – Hino
## FAQ
### Are all wiring diagrams similar?
Wiring diagrams might follow different requirements depending on the nation they are going to be utilized. They may have different designs depending upon the company and the designer who is designing that. They also might be drawn by various ECAD software such as EPLAN or AutoCAD electrical.
### What are the types of wiring diagram?
• Schematic Diagrams.
• Wiring diagrams.
• Block diagrams.
• Pictorial diagrams.
### What is the schematic format?
A schematic, or schematic diagram, is a representation of the components of a system using abstract, graphic symbols rather than realistic photos.
### What should a schematic consist of?
Schematics must include the total description and locations of all constructing code elements, such as the heating/ventilation/air conditioning (also called HVAC), plumbing, and electrical systems. However, schematic designs are only a standard layout to interact a design scheme to the owner.
### What is an architectural wiring diagram?
Architectural wiring diagrams show the approximate locations and interconnections of receptacles, lighting, and permanent electrical services in a building.
Hino Truck Wiring Diagrams Clear – Wiring23
Hino Dutro U600 Relay Diagram : Toyota Dyna 100 (U600 and U800; from
Hino Wiring Diagram
Hino Truck Wiring Diagrams Free – Wiring Diagram
### Wiring diagram types
• Schematic Diagrams.
• Wiring diagrams.
• Block diagrams.
• Pictorial diagrams.
### What is an architectural wiring diagram?
Architectural wiring diagrams show the approximate places and affiliations of receptacles, lighting, and permanent electrical services in a structure.
### How are wiring diagrams read?
The electrical schematics read from left to right, or from top to bottom. This is essential to get right, as the signal direction indicates the flow of current in the circuit. It is then easy for a user to comprehend when there is a modification in the course of the circuit.
### How do you check out electrical wire numbers?
An electrical cable is classified by 2 numbers separated by a hyphen, such as 14-2. The very first number denotes the conductor’s gauge; the 2nd denotes the variety of conductors inside the cable. For example, 14-2 has two 14-gauge conductors: a hot and a neutral.
### How do you read wire size charts?
Wire gauges range from low numbers to high numbers, with smaller numbers describing smaller diameters and bigger numbers representing bigger diameters. AWG 4 is 0.2043 inches in size, and AWG 40 is. 0031 inches in diameter.
### How is wire numbered?
American Wire Gauge (AWG) is the basic way to represent wire size in The United States and Canada. In AWG, the larger the number, the smaller the wire diameter and density. The biggest basic size is 0000 AWG, and 40 AWG is the tiniest standard size.
### Why do we need wiring diagrams?
A wiring diagram is frequently utilized to repair problems and to make sure that all the connections have actually been made which everything is present.
### Are all wiring diagrams similar?
Wiring diagrams might follow various standards depending upon the country they are going to be used. They may have different designs depending upon the business and the designer who is designing that. They also might be drawn by various ECAD software such as EPLAN or AutoCAD electrical.
### What is the schematic format?
A schematic, or schematic diagram, is a representation of the aspects of a system using abstract, graphic symbols instead of realistic photos.
### What is the difference between a schematic and wiring diagram?
A wiring diagram is a generalized pictorial representation of an electrical circuit. The components are represented using streamlined shapes in wiring diagrams.
### How do you read car wiring diagrams?
A vehicle wiring diagram is a map. To read it, recognize the circuit in question and starting at its power source, follow it to the ground. Use the legend to understand what each symbol on the circuit means. | 1,330 | 6,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2023-06 | longest | en | 0.888733 |
http://stackoverflow.com/questions/tagged/lapack?sort=unanswered&pagesize=30 | 1,405,221,306,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776435842.8/warc/CC-MAIN-20140707234035-00054-ip-10-180-212-248.ec2.internal.warc.gz | 135,089,232 | 26,462 | Tagged Questions
LAPACK (Linear Algebra PACKage) is a software library package to solve linear algebra equations.
50 views
How do I get specified Eigenvectors from the generalized Schur factorization of a matrix pair using LAPACK?
I am grad student trying to rewrite my MATLAB prototype code into C++ code using Eigen and LAPACK. Generalised eigenvalue solver (A*x=lamba*B*x) takes some part in this program. Because Eigen's ...
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Calculating eigenvectors using dhsein in CLAPACK
Here is my procedure: int getEigenvector(FLOAT* matrix, FLOAT* eigenvalues, int selectedValueIndex, FLOAT* eigenvector, long size){ char side = 'R'; // We want to calculate the right eigenvector. ...
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Is there a BLAS or LAPACK subroutine for symmetric matrix multiplication that takes advantage of symmetry?
I'm hoping someone can help me out. I am looking for a BLAS or LAPACK subroutine specifically for matrix multiplication of two large and identical symmetric matrices. So far, I have tested dysrk and ...
103 views
how to compile lapack so that it can be used correctly during installation of octave?
I'm trying to install the latest octave 3.8.1 from source in a cluster running redhat+IBM LSF. I don't have write access to anywhere else except my own home dir, that's why I have to install octave ...
117 views
C++ using Lapack within boost
While developing a program i used frequently boost UBLAS types for matrices like hermitian_matrix and others. I need to calculate eigenvalues and as i see UBLAS does not have something related with ...
139 views
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I asked this question over at the AMD developers forum a few days ago, but haven't gotten an answer. Maybe someone here has some insight. http://devgurus.amd.com/thread/167492 I am running ACML ...
187 views
Matrix Exponential BLAS/LAPACK
this is likely to be a real naive question, but I'll give it anyway a try. I'd like knowing if BLAS or LAPACK implement any function for getting the exponential of a certain double or float matrix of ...
113 views
I am losing my mind here. Given a positive definite, real matrix S, i did a Cholesky factorisation S = L * L^T. Now, this should lead to L^(-1) * S * (L^(-1))^T = 1, right? Right. But it leads to -1. ...
613 views
Linking LAPACK library with g++ on cygwin
Background I am trying to find the eigenvalues of a complex matrix by using zgeev routine in LAPACK library. As far as I understand, LAPACK is written for FORTRAN and hence to use it with a C (or ...
485 views
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So i have come to a point where DSYEVD is becoming impractical, due to it's higher memory requirements - it will need at least 240GB ram to diagonalise my matrix, so i'm considering moving to the ...
386 views
Should LAPACK dgetrf inputs and outputs be transposed?
I'm calling dgetrf from LAPACK on a row-major matrix, via clapack.h. I have the matrix A = [4,9,2; 3,5,7; 8,1,6]. If I call dgetrf, the result is [9.0, 0.222222, 0.444444; 5.0, 5.888888, 0.132075; ...
1k views
Find eigenvalues of boost matrix with LAPACK
I'm having trouble finding the eigenvalues of a boost matrix using the boost LAPACK bindings. This is my best guess: template<class T> boost::numeric::ublas::vector<double> ...
224 views
I'm getting zero values using dstev to calculate eigenvectors
I use gcc to compile under mac os x, I have Intel's mkl_lapack.h library installed. In the program I have a NxN tridiagonal matrix, so I just use two vectors to store values of the matrix. "d" vector ...
662 views
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I've got some code which performs a packed symmetric matrix inversion and multiplication using the LAPACK routines DPPTRF, DPPTRI, and DSPMV. Here is an older topic in which you can see the C++ code I ...
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I was wondering if anyone knows any papers or working notes that count how many operations Scalapack routines perform and especially how many messages they send and how many words. In terms of ...
11 views
follow up on lapck issue in eclipse gcc
my last question regarding lapack well, i followed the instruction and now the old error disappeared. however, in the console, it says this: g++ ...
32 views
R penalized fails with Lapack dtrtrs matrix inversion error
In trying to run optL1 from the R penalized package, with the logistic model: optL1(OS_event_ID, sdf, data=sdf, model='logistic', standardize=TRUE, fold=10) ... I am then encountered with an error ...
40 views
I have downloaded CLAPACK and built it according to these instructions: http://www.netlib.org/clapack/readme.install Everything was OK and I had no errors. I then put my three headers blaswrap.h ...
37 views
Running a LAPACK build on different processors
If I build LAPACK on Windows / VS2010 using an Intel processor, will I be able to run the compiled code on another processor? I ask this, because I see that there are different instructions for ...
46 views
Error when using dgesv_
I'm trying to solve a simple problem Ax = b using the dgesv_ function. However I've run into a problem I can't get past of. My code is: #include <cstdio> #include <f2c.h> #include ...
257 views
configure-Step of ATLAS installation fails on Mac OSX Mavericks
I'd like to install ATLAS v.3.10.1 on OSX Mavericks following the intructions of the ATLAS website. Unfortunately the step calling ../congire [flags] fails with the error message: cat: ...
241 views
How to Link lapack and BLAS library to C++ code
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487 views
236 views
Multithreading with LAPACK 3.3 & above on MacOS 10.6 and 10.7
I am trying to build and run a multi-thread program using openMP on MAC 10.6 and MAC 10.7 the program calls zgelss and zgemm from multiple thread I have compiled the LAPACK 3.4 and refBLAS I ...
465 views
I have installed lapack++ 2.5.4 with ATLAS 3.8.4 on Fedora 12. I wrote a simple program to test lapack++ using eclipse. I set these paths in eclipse: include path: ...
17 views
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how to run test/timing suite for installed lapack?
If build from source I can run the test suite with: make lapack_testing But this will test the newly built lapack How can I run the same test routine for currently installed lapack? for example, I ...
34 views
Finding eigenvalues and eigenvectors of a large (sparse) matrix
I am trying to compute the eigenvectors (say the first 10 of them) of a large matrix. My initial problems were caused by a misunderstanding of the Intel MKL library. To make my question clear and easy ...
6 views
How to install packages in fedora 11
I am trying to install lapack in fedora 11 using yum install lapack and yum install lapack-devel as well. It is showing "Loaded plugins: Refresh package kit. Setting up. Install process. No package ...
18 views
Compiling errors in yael library
I downloaded the recent yael v371 from https://gforge.inria.fr/projects/yael/. I followed the instruction in README file. The lapack and blas libraries are first installed, then I'd like to compile ...
46 views
gfortran compilation with -fPIC vs -frecursive
I recently followed the instruction on this thread for compiling BLAS and LAPACK as pre-requisites to a SciPy installation. First I got a gfortran error at some point, which recommended that I ...
33 views
Unable to link external lib (CLAPACK) using MATLAB mex
I'm new to mex and this problem spent me days but I still cannot figure out what to do. I create la_test.cpp file to test one of the subroutines in CLAPCK: cgemm_ (complex matrix-matrix ...
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I downloaded the latest 1.50beta, I tried to use CMAKE to generate a VS project, obviously it is not very successful (with Intel Parallel studio 2013 update 2 installed (C++ and Intel fortran), and ...
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Ifort mkl lapack eigenvalue
I am using ifort 14.0, mkl and the f95 lapack interfaces to do some eigenvalue decompositon. The function I am using currently is DSYEV_F95, which is called either: call ...
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C++ Function returning an array of objects
I'm having a very (seemingly) odd problem when I try to create a function to return an array of objects: So I've created a Matrix class to interface with existing LAPACK routines (i.e. explicitly ...
149 views
I am trying to compile and run the sample in example2.cpp from Armadillo, using the pre-compiled blas and lapack dll files provided. This is what I didin Visual C++ 2013: 1) I set up a new and empty ... | 2,218 | 8,919 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2014-23 | latest | en | 0.87518 |
https://tenpy.johannes-hauschild.de/viewtopic.php?f=8&t=24&view=print | 1,596,530,795,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.23/warc/CC-MAIN-20200804073038-20200804103038-00396.warc.gz | 459,403,586 | 3,446 | Page 1 of 1
### Infinite DMRG state prediction
Posted: 04 Dec 2018, 21:09
Hi Johannes,
I'm studying your tenpy/algorithms/dmrg.py code and I'd like to ask which part/lines of this code implements the state prediction method shown in chapters 10.2 and 10.3 of Shollwock's paper (Annals of Physics 326 (2011) 96–192)? Specifically, there should be some term at the end of each left and right sweep that substitutes the Schmidt value \lambda with
\lambda^[l+1]_guess = \lambda^[l]_R x [\lambda^[l-1]]^{-1} x \lambda^l_L
(Eq. 338 of the paper) for the subsequent iteration's guess wavefunction. Could you please tell me where this is done in this code? There is something resembling this in the code toycodes/d_dmrg.py at line 109 for a 2 site unit cell.
Also, could you please explain how one could generalize the code toycode/d_dmrg.py for larger/arbitrary unit cell sizes? Many thanks.
Regards,
Jason
### Re: Infinite DMRG state prediction
Posted: 05 Dec 2018, 17:33
Hi Jason, Welcome to the forum
very good question! In TeNPy, this is a bit hidden.
Actually, the "state prediction" is in the line 103 of the toycode:
Code: Select all
theta0 = np.reshape(self.psi.get_theta2(i), [Heff.shape[0]]) # initial guess
This line gets the "initial guess" for the ground state of the two sites to be optimized.
In the toycode, for simplicity we save the MPS always in completely right-canonical form ('B'-form) on each site.
The diagonalization and subsequent SVD gives $$A_i, S_{i,i+1}, B_{i+1} = svd(\theta)$$, and the line 109 (and 110) you referred to translates the left-canonical 'A'-form into rhe right-canonical 'B' (using $$B_i = S_{i-1,i}^{-1}A_i S_{i,i+1}$$). The MPS method get_theta() used in the state prediction then doesn't need to take inverses anymore, it simply contracts $$\theta = S_{i-1,i} B_i B_{i+1}$$.
Storing the MPS completely in B form is numerically a little bit dangerous (since one multiplies with inverses of small numbers), but in this case it's actually fine and does not lead to problems.
In the "real" TeNPy code, the MPS can be (and will be during DMRG) in the "mixed" canonical form, where $$\psi = A_1 ... A_i S_{i,i+1} B_{i+1}... B_L$$, i.e. the left sites are in 'A' form, and the right ones in 'B' form: MPS.set_B() has an argument saying whether the given tensor on that site is in left or right canonical form (this information is stored in the list MPS.form).
The other MPS methods are aware of this and for use the information accordingly, e.g. MPS.get_theta() will multiply with (inverses) of S on the left and right of the corresponding tensors to get the correct form.
For an infinite system, L is not the length of the total system, but the length of the MPS "unit cell".
If you have a model with a larger unit cell (e.g. because you have alternating strength on some bonds or because you consider a cylinder/ladder system),
you can simply choose the 'L' of the model and MPS accordingly larger.
For example, you can simply choose the 'L' in line 167 of the example_DMRG_infinite larger to artificially increase the size of the MPS unit cell.
### Re: Infinite DMRG state prediction
Posted: 10 Dec 2018, 13:40
Remark: A slightly different approach to this is discussed in https://arxiv.org/abs/1606.06790. The arrangement of the tensor network for the iDMRG is slightly different and instead of sweeping through system one chunks it and update "disconnected bonds" at one simulation step. This can be done in parallel easily. I guess it is a matter of taste which procedure one chooses. If you have question to the paper feel free to ask. | 956 | 3,594 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2020-34 | latest | en | 0.873217 |
http://www.enotes.com/homework-help/what-partial-fraction-decomposition-2x-x-2-9-255140 | 1,477,193,815,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719139.8/warc/CC-MAIN-20161020183839-00228-ip-10-171-6-4.ec2.internal.warc.gz | 441,728,136 | 10,144 | # What is partial fraction decomposition of 2x/(x^2-9)?
Asked on by leeaeel
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
We have to find the partial fractions of 2x/(x^2-9)
2x/(x^2-9)
=> 2x/(x+3)(x - 3)
=> A/(x + 3) + B/(x - 3)
[A(x - 3) + B(x + 3)]/(x + 3)(x - 3) = 2x/(x+3)(x - 3)
=> Ax - 3A + Bx + 3B = 2x
A + B = 2 and 3B - 3A = 0
=> A + B = 2 and A = B
=> 2A = 2
=> A = 1 and as B = A = 1
The partial fractions are
1/(x + 3) + 1/(x - 3)
The partial fractions of 2x/(x^2-9) are 1/(x + 3) + 1/(x - 3)
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll write the fraction 2x/(x^2 - 9) as an algebraic sum of partial fractions: [A/(x-3)] + [B/(x+3)]
Since the LCD of the fractions from the right side is (x-3)(x+3) = x^2 - 9, we'll multiply by x^2 - 9 both fractions:
2x/(x^2 -9)= [A(x+3) + B(x-3)]/ (x^2 -9)
Having the common denominator (x^2 -9), we'll simplify it.
2x = Ax+3A+Bx-3B
We'll factorize by x to the right side:
2x = x*(A+B) + (3A-3B)
Comparing, we'll have:
A + B=2 (1)
3A-3B=0
We'll divide by 3:
A - B = 0 (2)
We'll add the second relation to the first one:
A + B + A - B=2+0
2A=2
A = 1
But, from (2) => A=B = 1
The partial fraction decomposition is: 2x/(x^2-9) = 1/(x-3) + 1/(x+3).
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If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...
The Elements of Euclid, containing the first six books, with a selection of ... - Seite 6
von Euclides - 1874
Vollansicht - Über dieses Buch
## Class lessons on Euclid
Marianne Nops - 1882 - 278 Seiten
...is with these that the sixth postulate deals. It tells us that ' If a straight line meets two other straight lines so as to make the two interior angles...it, taken together less than two right angles, these lines, being continually produced, shall at length meet upon that side on which are the angles which...
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## The Elements of Euclid for the Use of Schools and Colleges: Comprising the ...
Euclid, Isaac Todhunter - 1883 - 426 Seiten
...cannot enclose a space. 11. All right angles are equal to one another. 12. If a straight line meet two straight lines, so as to make the two interior...lines, being continually produced, shall at length meet on that side on which are the angles which are less than two right angles. PROPOSITION I. PROBLEM....
Vollansicht - Über dieses Buch
## New Englander and Yale Review, Band 42
Edward Royall Tyler, William Lathrop Kingsley, George Park Fisher, Timothy Dwight - 1883 - 872 Seiten
...Euclid, or put Euclid's axiom in a different form. Thus Euclid's axiom is, "If a straight line meet two straight lines, so as to make the two interior...lines, being continually produced, shall at length meet on that side, on which are the angles, which are less than two right angles. This axiom is a puzzling...
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## Moffatt's pupil teachers' course (ed. by T. Page). Candidates, 2nd ..., Band 3
Moffatt and Paige - 1883 - 514 Seiten
...equal to two right angles (1. 13), therefore the angles BGH, GHD are less than two right angles. But if a straight line meets two straight lines so as...than two right angles, these straight lines, being continuallyproduced, shall at length meet on that side on which are the angles that are less than two...
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## The Practical Teacher; with which is Incorporated the Practical ..., Band 2
Joseph Hughes - 1883 - 578 Seiten
...interior angles on the same side toeether equal to two right angles. Axiom xii. If a straight line meet two straight lines so as to make the two interior...on the same side of it taken together less than two ri^ht angles, these straight lines being continually produced, îhall at length meet on that side on...
Vollansicht - Über dieses Buch
## Science, Band 20
John Michels (Journalist) - 1892 - 448 Seiten
...axiomatic.3 This is the axiom of parallels, which reads as follows : — " If a straight line meet two straight lines so as to make the two interior...on the same side of it taken together less than two right-angles, these straight lines, being continually produced, shall at length meet on that side on...
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## Geometry for schools, comprising books i. and ii. of Euclid, with some ...
Euclides - 1883 - 176 Seiten
...Note.— The converse of this proposition, that "if a straight line meeting two other straight lines make the two interior angles on the same side of it taken together less than two right angles, the two lines will, if produced tar enough) meet on that Bide on which are the angles less than two...
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## The text of Euclid's geometry, book 1, uniformly and systematically arranged ...
Euclides - 1884 - 214 Seiten
...part. X. Two straight lines cannot enclose a space. XI. All right angles are equal to one another. XII. If a straight line meets two straight lines, so as...are the angles which are less than two right angles. , PROPOSITION I. PROBLEM. To describe an equilateral triangle upon a given finite straight line. GIVEN...
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## Stewart's guide-book to the queen's scholarship and certificate examinations ...
Stewart W. and co - 1884 - 146 Seiten
...rhombus is a four-sided figure which has all its sides equal, but its angles are not right angles. 3. If a straight line meets two straight lines so as...are the angles which are less than two right angles. Show that the following definitions are incomplete : — " Of quadrilateral figures, a square has all...
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## Mathematical Questions with Their Solutions: From the ..., Bände 40-42
1884 - 434 Seiten
...AB would coincide with EF ; and thus is it proved that all right angles are equal.* Axiom XII. : — "If a straight line meets two straight lines, so as...meet upon that side on which are the angles which arc less than two right angles." Dr. Simson, admitting that this is not self-evident, demonstrates...
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#### Selecting A Distinct Columns
Col1 : col2:
........:.......:.
5 : 9 :
........:.........
6 : 3 :
........:.......:..
11 : 9 :
........:.......:..
5 : 9 :
........:.......:..
11 : 8 :
........:.......:.
I want to SELECT col2 WHERE 5 and 11 are common in col1
Which suppose to return :9,9,9
I will appreciate a statement that will do that .thanks
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#### How can I select a common rows in a table
I need a help in this assignment
Col1 : col2:
........:.......:.
5 : 9 :
........:.........
6 : 3 :
........:.......:..
11 : 9 :
........:.......:..
5 : 9 :
........:.......:..
11 : 8 :
........:.......:.
I want to SELECT col2 WHERE 5 and. 11, in col1 are common in col2.
Which suppose to return :9,9,9
I will appreciate a statement that will do that .thanks
3. Originally Posted by okesimojs
... WHERE 5 and 11 are common in col1
not sure i understand what you mean by "common"
does this work for you? --
Code:
```SELECT col2
FROM daTable
INNER
JOIN ( SELECT col1
FROM daTable
WHERE col1 IN ( 5 , 11 )
GROUP
BY col1
HAVING COUNT(DISTINCT col1) > 1 ) AS criteria
ON criteria.col1 = daTable.col1```
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Now I'm confused. The original message appears to be 'unedited' and says 'common in col2'
!?!
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[QUOTE=cafelatte]Now I'm confused.
I was using my phone to answer . That's why I couldn't noticed the edited reply .
Sir, the script you gave me did not return anything .
6. Originally Posted by okesimojs
Sir, the script you gave me did not return anything .
sorry, remove the DISTINCT
results:
9
9
9
8
still not sure if i understand your requirements, because my query returns the 8 as well, since col1=11 occurs more than once
7. I believe they want, where col1 is 5 or 11, all the values of col2 where there are duplicate col2 values.
Not tested!!!
Code:
```SELECT col2
FROM daTable
WHERE col1 in (5, 11)
GROUP BY col2
HAVING COUNT(col2) > 1```
• cafelatte agrees : Indeed!
8. Almost right! Now with added testing!
Code:
```select group_concat(col2)
from daTable
where col1 in (5,11)
group by col2
having count(col2)>1;
+--------------------+
| group_concat(col2) |
+--------------------+
| 9,9,9 |
+--------------------+``` | 794 | 2,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-13 | latest | en | 0.735208 |
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Wed, 2 Apr 1997
# The Standard Deviation and The Normal Distribution: The Long and Short Of It.
By T. V. Nguyen
Many years ago while I was involved in statistical works in Vietnam, one of my senior colleagues remarked that: "given a few statistics, a good manager can assess the performance of a company in minutes". Indeed, as I mentioned in the last article that if one knows the mean and the standard deviation of a data set, one can have a pretty good idea of its distribution. For example, if a scientific paper reports that the mean and standard deviation (SD) of blood pressure (BP) of a sample of 100 elderly persons to be 110 and 15 mmHg, respectively, one can infer that approximately 95% of the subjects have BP ranged between 80 to 140 (=110 +/- 2*15). On the other hand, if another researcher reports that the mean and SD of 75 and 40 mmHg, respectively, then one can be sure that the researcher is either (i) wrong in his calculation or (ii) some of his patients were not normal, since 2*SD=80, which is greater than the mean itself. Another way of checking whether a researcher knows what he/she is doing is by asking him/her to provide the mean, SD and median. If the mean and median are not approximately the same, then the distribution of the data must be skewed and hence normal statistical methods may not applicable. In this note, I will discuss how the standard deviation can be used in assessing some interesting social issues.
Consider the following question: If you are a boss, would height play a role in your selection of a successor for your job? In a FORTUNE magazine column in 1981, a discussion concerning height as a factor in Deng Xiao Ping's choice of Hu Yao Bang for his replacement as Chairman of the Chinese Communist Party. The article notes the fact surrounding the case what is enough to arouse suspicions when examined in the light of statistics.
Deng, as we know, is only five feet tall, a height that is short even by Chinese standard. Therefore, the choice of Hu, who is also short (five-feet tall), raised (or lowered) eyebrows because, as the article notes, "the odds against a 'height-blind' decision producing a chairman as short as Deng are abput 40 to 1". In other words, if we possessed the relative frequency distribution of the heights in Chinese population, only one in 41 (i.e. 2.4%) of them would possesses heights less than or equal to five feet. The calculate these odds, the author of the article made some interesting assumptions concerning the relative frequency distribution of the heights of Chinese male population, mostly notable that the distribution follows the "normal" bell-shaped Gaussian curve (as it does in the USA).
It is generally held that a boy's length at birth represents 28.6% of his final height and that, in pre-revolutionary China, the average length of a Chinese boy at birth was 18.9 inches. From this, it can be deduced that the mean height of all mature male Chinese is (18.9/0.286) = 66.08 inches (or 5 feet, 6.08 inches). Assuming that the distribution of the heights of males in China follows a normal distribution (as it does in the US and in fact all countries in the world) with a mean of 66 inches and a standard deviation of 2.7 inches, a figure that look about right for the mean.
If we are willing to accept the assumption, that the heights of adult males are normally distributed wth mean (M) 66 inches and standard deviation (SD) of 2.7 inches, we are ready to calculate the probability that a single adult Chinese male, chosen at random, will have a height that is less than or equal to 5 feet, or equivalently, 60 inches. After some algebra (you may want to check it yourself), this probability is 0.0132 (i.e. 1.32%) or approximately 1 in 76, which corresponding to the odds of 75 to 1.
This odds certainly agree with our intuition, because it is difficult to believe that the proportion of 5-foot tall adult male Chinese is very large. Nevertheless, the validity of our calculated odds depends on the validity of our assumptions. However, it is possible that there is a flaw in our assumptions. It is not clear that the distribution of the heights of all adult male Chinese is a good model for the distribution of the heights of potential candidates for Deng Xiao Ping's replacement. Presumably, the candidates would be very select group of senior, and elderly, members of the Chinese Communist Party. It is a well-known fact that the heights of human decrease as they get older, particularly as they reach 60 years of age or older. Therefore, we would expect the distribution of the heights of the candidates for Deng's post to possess a mean that is less than the mean for the distribution of all adult male Chinese. We would also expect the odds of randomly selecting a person 5-feet tall or less from among the candidates to be larger than the corresponding odds of selection from among the population of heights of all Chinese adult males.
Did Deng Xiao Ping take height into account in selecting his successor? The answer to this question depends on the assumptions that you are willing to make. Consequently, we leave the answer to you. Perhaps, you will have additional reasons for accepting or not accepting the assumptions.
Tuan Nguyen, Ph.D.
[email protected]
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# Solid State Physics
Code Completion Credits Range Language
BV002FPL Z,ZK 5 2P+2C Czech
Garant předmětu:
Antonio Cammarata
Lecturer:
Antonio Cammarata
Tutor:
Antonio Cammarata
Supervisor:
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Synopsis:
The course provides fundamentals of solid state physics at large.
Requirements:
Lessons and tutorials attendance.
Syllabus of lectures:
1) Condensed matter, solids, their description; crystals; crystal as periodic lattice.
2) Wave diffraction and the reciprocal lattice. Scattering intensity. Structure factor and Atomic form factor.
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4) Thermal properties derived from phonons.
5) Free Electron: Brief overview of the Schrodinger equation for the free particle and a particle confined by infinite potentials. Fermi energy and Fermi level. Electronic density of states. Heat capacity of the electron gas.
6) Nearly free electron model; origin of the energy gap; magnitude of the energy gap; Bloch functions; Kronig-Penney model; the Bloch theorem; number of orbitals in a band; metals and insulators.
7) Band gap; equations of motion for the wave vector; holes; effective mass; physical interpretation of the effective mass; effective masses in semiconductors.
8) Fermi surfaces and metals; construction of Fermi surfaces; electron orbits, hole orbits and open orbits; physical origin and calculation of energy bands.
9) Dielectrics and Ferroelectrics; electric field of a permanent dipole; macroscopic electric field; depolarizatino field; local electric field at an atom; Lorentz Field; Field of dipoles inside a cavity.
10) Diamagnetism and paramagnetism; Langevin diamagnetism equation; quantum theory of diamagnetism of mononuclear systems; paramagnetism.
11) Ferromagnetism and Antiferromagnetism; ferromagnetic order; Curie point and the exchange integral; temperature dependence of the saturation magnetization; saturation magnetization at absolute zero; ferrimagnetic order; antiferromagnetic order; ferromagnetic domains.
12) Magnetic resonance; nuclear magnetic resonance; line width; hyperfine splitting; electron paramagnetic resonance.
13) Superconductivity; type I and type II superconductors; destruction of superconductivity by magnetic fields; Meissner effect; heat capacity; energy gap; thermodynamics of the superconducting transition; BCS theory.
14) Noncrystalline solids; diffraction pattern of noncrystalline solids; monoatomic amorphous materials; structure of vitreous silica SiO2; glasses, viscosity and the hopping rate; point defects; lattice vacancies; color centers; F centers; other centers in alkali halides.
Syllabus of tutorials:
1) Crystal structures; Problem: symmetries; Problem: copper oxide layer.
2) Fraunhofer diffraction and derivation of Bragg's law; Single-slit diffraction; two point sources; two slits with finite width (Young's slits); transmission diffraction grating.
3) Physical meaning of phonon eigendisplacements and eigenfrequencies. Explicit calculation of the phonon eigenvectors and eigenfrequencies for a simple system (e.g. isolated CO2 molecule).
4) Crystal binding and elastic constants. Van der Waals-London interactions; ionic crystals, the Madelung Energy; hydrogen bonds; elastic strain.
5) Electrical conductivity and Ohm's law. Experimental electrical resistivity of Metals. Motion in magnetic fields. The Hall effect.
6) Crystal momentum of an electron; Bloch theorem and solution of the central equation; aaproximate solution near a zone boundary.
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8) Experimental methods in Fermi surface studies; quantization of orbits in a magnetic field; De Haas-van Alphen effect; Extremal orbits.
9) Dielectric constants and poalrizability; structural phase transitions; ferroelectric crystals; displacive transitions; Landau theory of the phase transition for ferroelectrics.
10) Quantum theory of paramagnetism; Hund rules; crystal field splitting; quenching of the orbital angular momentum; spectroscopic splitting factor; Van Vleck temperature-independent paramagnetism; paramagnetic susceptibility of conduction electrons.
11) Magnons; quantization of spin waves, thermal excitation of magnons; neutron magnetic scattering.
12) Electron paramagnetic resonance; ferromagnetic resonance; antiferromagnetic resonance.
13) Superconductivity; London equation; coherence length; flux quantization in a superconductive ring; single particle tunneling; Josephson superconductor tunneling.
Study Objective:
At the end of the course, the students will acquire basic knowledge on solid state physics.
Study materials:
Charles Kittel, Introduction to Solid State Physics, 8th edition, Wiley IPL, ISBN-13: 9788126535187
Note:
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https://moodle.fel.cvut.cz/courses/BV002FPL
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Aktualizace výše uvedených informací naleznete na adrese https://bilakniha.cvut.cz/en/predmet5848206.html | 1,464 | 5,849 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-22 | latest | en | 0.727946 |
https://money.stackexchange.com/questions/44743/market-value-vs-book-value-of-investments | 1,618,111,897,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00157.warc.gz | 506,542,876 | 38,591 | # Market Value vs Book Value of investments
I understand the concept of market value and book value of investments.
But how does one calculate the book value?
```| Date | Share purchased | Cost per share || Book Value | Market Value |
| Jan 1 | 10 | \$10 || \$100 | \$100 |
| Feb 1 | 10 | \$20 || \$300 | \$400 |
| Mar 1 | 10 | \$30 || \$600 | \$900 |
| Apr 1 | 10 | \$20 || \$800 | \$800 |
| May 1 | -10 | \$40 || ???? | \$1200 |
```
So the market value is just total shares * the current cost of the share. But what about the book value? Depend on which share I sell, and some calculation will yield higher book values than others...
• Are you looking to calculate the cost basis of your remaining shares for tax purposes, or some other purpose? There are a number of methods and which is appropriate depends on the purpose. Average, LIFO, FIFO, etc. – Chris W. Rea Feb 21 '15 at 21:12
Note that book value has two meanings. From Investopedia:
Book value of an asset is the value at which the asset is carried on a balance sheet and calculated by taking the cost of an asset minus the accumulated depreciation. Book value is also the net asset value of a company, calculated as total assets minus intangible assets (patents, goodwill) and liabilities. For the initial outlay of an investment, book value may be net or gross of expenses such as trading costs, sales taxes, service charges and so on.
The second meaning relates to the company accounting. You'd look that up in shareholder reports. This does not seem to be what you mean.
The first meaning relates to how much you paid for the investment. This is easy to determine if you ignore inflation, trading expenses, and such, as it is just the amount of money you paid. Another confounding factor is selling shares. Your book values are consistent with this meaning, so I'll assume that that is what you mean.
In your case, on April 30th, you own 40 shares for which you paid \$800 (\$10 * 10 + \$20 * 10 + \$30 * 10 + \$20 * 10). So the book value is \$800. If you sell the shares, you can pick which ones. Some common methods used to pick are FIFO, LIFO, and Average.
FIFO (First In First Out). With FIFO, when you sell 10 shares, you sell the first ones that you bought. In this case, that would be the 10 shares you bought on January 1st for \$100. That would give you a new book value of \$800 - \$100 = \$700. If you sold another 10 shares, those would be the ones that you bought on February 1st. So on and so forth.
LIFO (Last In First Out). With LIFO, you sell the most recent ones that you bought. In this case, that would be the 10 April 1st shares that you bought for \$200. The new book value would be \$800 - \$200 = \$600.
Average. With Average, you determine the average price that you paid for your shares and use that to determine the book value of any shares that you sell. In this case, the average price that you paid for your 40 shares is \$800 / 40 = \$20. So you'd have a remaining book value of \$800 - 10 * \$20 = \$600. If you buy more shares, you would add their book value to your remaining book value. So if you bought 10 shares at \$30, your new book value would be \$600 + 10 * \$30 = \$900. Your new average price would be \$900 / 40 = \$22.50.
Note that both FIFO and LIFO can be seen directly from the table. Average requires additional calculations to determine. FIFO is the easiest to follow, as all the shares up to a point are sold. All the later shares are still in your possession. LIFO can have gaps (if you mix selling and buying). And Average requires continual updating.
There may be tax advantages to using the other methods that outweigh their added complexity.
I see from your profile that you are in Canada. In Canada we are required to use what others are calling the "average".
Thus, according to your data, as at April 1, you have paid \$800 for 40 shares. This gives an average cost of \$20 per share.
The sale of 10 shares on 1 May would mean you carry forward a book value of \$600 for the remaining 30 shares. For Capital Gains Tax purposes, on the 10 shares sold, you will have a cost of \$200 against proceeds of \$400 for a net Capital Gain of \$200.
Also note that your stockbroking account will be required to report the book value of your net holding. Further, the cost of trading - i.e., commissions - are also netted from both the cost and proceeds. | 1,094 | 4,579 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-17 | longest | en | 0.942811 |
http://stackoverflow.com/questions/4272628/how-to-get-the-minimum-of-sqrt3x1sqrt3y1sqrt3z1/4272809 | 1,436,293,438,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375099758.82/warc/CC-MAIN-20150627031819-00090-ip-10-179-60-89.ec2.internal.warc.gz | 244,493,208 | 20,081 | How to get the minimum of Sqrt[3x+1]+Sqrt[3y+1]+Sqrt[3z+1]?
Let
``````f[x_,y_,z_] := Sqrt[3x+1]+Sqrt[3y+1]+Sqrt[3z+1]
``````
I want to get the minimum of f for x>=0&&y>=0&&z>=0&&x+y+z==1 using mathematica.
PS: I do know how to get the minimum by math method:
``````Since 0<=x<=1,0<=y<=1,0<=z<=1, we have
0<=x^2<=x,0<=y^2<=y,0<=z^2<=z.
Hence,
3a+1 >= a^2 + 2a + 1 = (a+1)^2, where a in {x,y,z}.
Consequently,
f[x,y,z] >= x+1+y+1+z+1 = 4,
Where the equality holds if and only if (x==0&&y==0||z==1)||...
``````
PS2: I expected the following code would work, but it did't.
``````Minimize[{f[x,y,z],x>=0&&y>=0&&z>=0&&x+y+z==1},{x,y,z}]
``````
Actually, as Simon point out, it works ... The running time is longer than I expected and I closed it before Mahtematica show me the result.
-
huh? only two viable options i see are x=y=z=(1/3), f(x,y,z)=3*sqrt(2) .... or x=y=0, z=1 leads to f(x,y,z)=1+1+sqrt(4)=1+1+2=4 – jon_darkstar Nov 25 '10 at 0:06
@jon_darkstar minimum,not maximum – Eastsun Nov 25 '10 at 0:08
yea i caught that right afterwards =P. but are the middle options even worth exploring? – jon_darkstar Nov 25 '10 at 0:09
Correct without constraints, but note that x + y + z = 1. Need a Lagrange multiplier or two in there. – duffymo Nov 25 '10 at 0:10
the partial derivative of f w.r.t any of three variables is obviously always decreasing, so you'd want to "consolidate" all of that x+y+z=1 constraint into one variable. Legrange? why? i guess maybe if you needed a proof or wanted to generalize f in a weird way (generalizing constraints makes no differenece - replace x+y+z=1 with x+y+z=k and you still want one of them to be equal to k), but intuitively its pretty straightforward – jon_darkstar Nov 25 '10 at 0:16
Is this what you want?
``````In[1]:= f[x_,y_,z_]:=Sqrt[3x+1]+Sqrt[3y+1]+Sqrt[3z+1]
In[2]:= Minimize[{f[x,y,z],x>=0,y>=0,z>=0,x+y+z==1},{x,y,z}]
Out[2]= {4,{x->1,y->0,z->0}}
``````
Note that the Documentation says "Even if the same minimum is achieved at several points, only one is returned" so you will have to impose the permutation symmetry of the problem yourself.
PS You can turn this into a Lagrange Multiplier problem
``````In[3]:= Thread[D[f[x,y,z] - \[Lambda](x+y+z-1), {{x,y,z,\[Lambda]}}]==0];
Reduce[Join[%,{x>=0,y>=0,z>=0}],{x,y,z,\[Lambda]},Reals]
{f[x,y,z],D[f[x, y, z], {{x, y, z}, 2}]}/.ToRules[%]
Out[4]= x==1/3&&y==1/3&&z==1/3&&\[Lambda]==3/(2 Sqrt[2])
Out[5]= {3 Sqrt[2],{{-(9/(8 Sqrt[2])),0,0},{0,-(9/(8 Sqrt[2])),0},{0,0,-(9/(8 Sqrt[2]))}}}
``````
and see that the only stationary point is the maximum at x=y=z=1/3. Thus the minimum must lie on the boundary. You can then use similar code but restricted to the boundary to eventually find the correct result.
-
Oh, thanks. That's what I want. Actually, I have tried that and it had been running more than 2 mius before I forced close the program. – Eastsun Nov 25 '10 at 0:53
@Eastsun: It is quite slow -- It took about 30secs on my cheap hp laptop... The reduce code is much faster. – Simon Nov 25 '10 at 1:02
+1 for the explicit Lagrange Mults – belisarius Nov 25 '10 at 15:32
Just for fun, this is a plot of the solution given by Simon:
``````f[x_, y_, z_] := Sqrt[3 x + 1] + Sqrt[3 y + 1] + Sqrt[3 z + 1]
g1 = ContourPlot3D[f[x, y, z] == 4, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, AxesLabel -> {x,y,z}, MeshFunctions -> {#3 &}, ContourStyle -> {Blue, Opacity[0.5]}];
g2 = ContourPlot3D[ x + y + z == 1, {x, 0, 1}, {y, 0, 1}, {z, 0, 1}, AxesLabel -> {x,y,z}, MeshFunctions -> {#2 &}, ContourStyle -> {Green, Opacity[0.5]}];
Show[g1, g2, Graphics3D[{PointSize[0.05], Red, Point[{1, 0, 0}]}], ViewPoint -> {1.1`, -2.4`, 1.7`}]
``````
-
You say you aren't interested in the "math method" (I'm not sure what you have in mind when you say that, but it makes me think of minimization methods with Lagrange multipliers). If that's correct, why do you bring Mathematica into the discussion? What do you think it'll be using?
I'll have to assume that you mean numerical, computer solutions. I'd start with linear programming and the simplex method.
-
I just want to know if there have any simple way, i.e. solved the problem in one or two lines. – Eastsun Nov 25 '10 at 0:20
linear programming / numerical would do it, but that deserves a more interesting function to optimize IMO – jon_darkstar Nov 25 '10 at 0:24
More interesting, indeed. – duffymo Nov 25 '10 at 1:20 | 1,573 | 4,387 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2015-27 | latest | en | 0.854442 |
http://mathoverflow.net/questions/95308/generating-conditional-random-graphs | 1,469,378,988,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824113.35/warc/CC-MAIN-20160723071024-00138-ip-10-185-27-174.ec2.internal.warc.gz | 167,750,589 | 14,583 | # Generating Conditional Random Graphs
Let $G(n,p)$ be the usual random graph on $n$ vertices with each edge existing independently with probability $p$ (no self loops , or double edges not are allowed). I would like to simulate the distribution of a random graph given the event $T\geq a$ where $T$ is the number of triangles in the random graph. The natural approach is Metropolis Hastings. I've already found some semi-efficient algorithms that approximate the number of triangles in a given random graph, however I am still at a loss of what Markov chain to pick for a good rate of convergence. I would immensely appreciate a push in the right direction. In particular, some references would be fantastic. Thanks!
-
what value of $a$ are you interested in? Also, if you are interested in computing expectations of the form $E[\phi(G) | T(G) \geq a]$, importance sampling might just work as well. – Alekk Apr 26 '12 at 23:33
My question would be: which values of $p$ are you interested in? Are you talking about the regime where you expect to have lots of triangles or few? – Anthony Quas Apr 26 '12 at 23:41
@Alekk: I am interested in all values of $a$ but in particular for $a>E(T)$ which is on the order of $n^3p^3/6$. I'm not so much interested in calculating expectations as just seeing what the graphs look like. – Alex R. Apr 27 '12 at 2:11
@Anthony Quas: I would be interested in all (fixed) values of $p$. What I am sure of is the number of triangles is expected to be much less than $a$, which translates to $P(T\geq a)$ being very small by itself. – Alex R. Apr 27 '12 at 2:13
@Alex R: just being curious but how are you planning to "see what the graphs look like"? For small graphs that might be easy to plot some of them, but for large graphs I think that you are going to compute some expectations (number of copie of K_4, is it connected, etc...), aren't you? – Alekk Apr 27 '12 at 8:48 | 492 | 1,906 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-30 | latest | en | 0.948412 |
http://www.lofoya.com/Solved/1215/a-man-positioned-at-the-origin-of-the-coordinate-system-the-man-can | 1,516,502,298,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889917.49/warc/CC-MAIN-20180121021136-20180121041136-00541.warc.gz | 500,362,301 | 14,868 | # Easy Permutation-Combination Solved QuestionAptitude Discussion
Q. A man positioned at the origin of the coordinate system. the man can take steps of unit measure in the direction North, East, West or South. Find the number of ways of he can reach the point $(5,6)$, covering the shortest possible distance.
✖ A. 252 ✖ B. 432 ✔ C. 462 ✖ D. 504
Solution:
Option(C) is correct
In order to reach $(5,6)$ covering the shortest distance at the same time the man has to make $5$ horizontal and $6$ vertical steps.
The number of ways in which these steps can be taken is given by:
$\dfrac{11!}{5!×6!}$
$= \textbf{462}$
## (4) Comment(s)
Name
()
Much earlier looking at combination of either vertical or horizontal moves. Since total number of moves will always be the sum of the 2, You can simply do 11C6, or even 11C5, since all combinations of doing it one way = total ammount since you can straight line to the finish after said moves. This can be done with any point since their sum will always be the total ammount of moves. Even point (200, 5) can be achieved in 205C5 ways which also = 205C200
Name
()
Much easier*... rip would be nice to be able to edit other post. Also I had to write more to be able to post hence the babble...
Pnr
()
Please explain the solution, why it is divides 11! by 5! and 6!
Aarti
()
Since there are $5$ horizontal and $6$ vertical steps and their order does not matter and they are identical. So it is divided by $5!$ and $6!$. | 397 | 1,474 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-05 | latest | en | 0.909223 |
https://financial-dictionary.thefreedictionary.com/APR | 1,624,319,009,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488504838.98/warc/CC-MAIN-20210621212241-20210622002241-00630.warc.gz | 224,528,888 | 13,786 | # APR
Also found in: Dictionary, Thesaurus, Medical, Legal, Acronyms, Encyclopedia, Wikipedia.
## Annual Percentage Rate
The cost of funds or interest rate for an entire year expressed as a single percentage. Significantly, the APR does not account for compounding. For example, if the APR is 36%, the percentage is 3% per month, but the interest rate or cost of funds for the entire year may be greater than 36% due to the effects of compounding. By law, a credit card company or other lender must inform the customer of the APR before any agreement is signed. The APR provides the customer with a convenient number against which to compare the cost of funds for other loans or investments. It is less commonly called the annual rate of return or the nominal annual rate.
## Annual percentage rate (APR).
A loan's annual percentage rate, or APR, is what credit costs you each year, expressed as a percentage of the loan amount.
The APR, which is usually higher than the nominal, or named, rate you're quoted for a loan, includes most of a loan's up-front fees as well as the annual interest rate.
You should use APR, which is a more accurate picture of the cost of borrowing than the interest rate alone, to compare various loans you're considering.
## APR
the ‘annualized percentage rate of INTEREST’ charged on a LOAN. The APR rate will depend on the total ‘charge for credit’ applied by the lender and will be influenced by such factors as the general level of INTEREST RATES, and the nature and duration of the loan.
Where lenders relate total interest charges on INSTALMENT CREDIT loans to the original amount borrowed, this can give a misleading impression of the interest rate being charged, for as borrowers make monthly or weekly repayments on the loan they are reducing the amount borrowed, and interest charges should be related to the lower average amount owed. For example, if someone borrows £1,000 for one year with a total credit charge of £200 the ‘simple interest’ charge on the original loan is 20%. However, if the loan terms provide for monthly repayments of £100, then at the end of the first month the borrower would have repaid a proportion of the original £1,000 borrowed and by the end of the second month would have repaid a further proportion of the original loan, etc. In effect, therefore, the borrower does not borrow £1,000 for one whole year but much less than this over the year on average as he or she repays parts of the outstanding loan. If the total credit charge of £200 were related to this much smaller average amount borrowed to show the ‘annualized’ percentage rate' then this credit charge would be nearer 40% than the 20% quoted.
To make clear to the borrower the actual charge for credit and the ‘true’ rate of interest the CONSUMER CREDIT ACT 1974 requires lenders to publish both rates to potential borrowers.
Collins Dictionary of Business, 3rd ed. © 2002, 2005 C Pass, B Lowes, A Pendleton, L Chadwick, D O’Reilly and M Afferson
## APR
the ‘annualized percentage rate of INTEREST’ charged on a LOAN. The APR rate will depend on the total ‘charge for credit’ applied by the lender and will be influenced by such factors as the general level of INTEREST RATES, and the nature and duration of the loan.
Where lenders relate total interest charges on INSTALMENT CREDIT loans to the original amount borrowed, this can give a misleading impression of the interest rate being charged, for as borrowers make monthly or weekly repayments on the loan, they are reducing the amount borrowed, and interest charges should be related to the lower average amount owed. For example, if someone borrows £1,000 for one year with a total credit charge of £200, the ‘simple interest’ charge on the original loan is 20%. However, if the loan terms provide for monthly repayments of £100, then at the end of the first month the borrower would have repaid a proportion of the original £1,000 borrowed and by the end of the second month would have repaid a further proportion of the original loan, etc. In effect, therefore, the borrower does not borrow £1,000 for one whole year but much less than this over the year on average, as he or she repays part of the outstanding loan. If the total credit charge of £200 were related to this much smaller average amount borrowed to show the ‘annualized percentage rate’, then this credit charge would be nearer 40% than the 20% quoted.
To make clear to the borrower the actual charge for credit and the ‘true’ rate of interest, the CONSUMER CREDIT ACT 1974 requires lenders to publish both rates to potential borrowers.
Collins Dictionary of Economics, 4th ed. © C. Pass, B. Lowes, L. Davies 2005
## APR
See annual percentage rate.
The Complete Real Estate Encyclopedia by Denise L. Evans, JD & O. William Evans, JD. Copyright © 2007 by The McGraw-Hill Companies, Inc.
## APR
See Annual Percentage Rate.
The Mortgage Encyclopedia. Copyright © 2004 by Jack Guttentag. Used with permission of The McGraw-Hill Companies, Inc.
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Apr. 19-20: Innovations in Health Care, Ontario Ministry of Health and Long-Term Care and Local Health Integration Networks, Toronto; www.CIBinfo@moh.gov.on.ca
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Apr. 11--The Construction and Transportation Industry will hold its "Good Scout" Awards, starting at 11:30 a.m.
Norwich: Sat Chelsea H, Mar 19 Bolton A, Apr 2 Arsenal A, Apr 9 Man Utd H, Apr 16 C Palace A, Apr 19 Newcastle H, Apr 23 Charlton H, Apr 30 Soton A, May 7 Birmingham H, May 14 Ful-ham A.
DAVIS, CALIFORNIA Apr. 1-2, Fosse, Mondavi Center, 200 B St, 530/752-1915, www.mondaviarts.org
6 - Apr. 24 Gallery collection: Avigdor Arikha, Ra'anan Levy, Simcha Shirman, Ruth Agassi, Jossef Krispel, Ya'acov Dorchin
Genovese of Niagara Falls, N.Y., expelled from membership in the AICPA and the New York State Society of CPAs, effective Apr. 4, 2002.
NEC/Arena 0121 780 4133: Sting (Mar 25) Macy Gray (Mar 30) Lipizzaner Stallions (Mar 31-Apr 2) Puff Daddy (Apr 4) Moody Blues (Apr 15) Michael Crawford (Apr 16) Simply Red (May 5) Steps (May 17-20) Best Disco In Town 3 (May 21) Shirley Bassey (May 28) Judith Durham - The Seekers (Jun 5) Santana (Jun 13) Tom Jones (Dec 1) Bootleg Beatles (Dec 22).
Monday, Apr. 1 -- Chicken Parmesan, buttered noodles, peas, pineapple
CHELSEAMar 2 v West Brom (h) Mar 10 v Fulham (a) Mar 16 v West Ham (h) Mar 31 v Southampton (a) Apr 6 v Sunderland (h) Apr 14 v Tottenham (h) Apr 21 v Liverpool (a) Apr 27 v Swansea (h) May 4 v Man Utd (a) May 12 v Aston Villa (a) May 19 v Everton (h) TOTTENHAMMar 3 v Arsenal (h) Mar 10 v Liverpool (a) Mar 16 v Fulham (h) Mar 30 v Swansea (a) Apr 7 v Everton (h) Apr 14 v Chelsea (a) Apr 21 v Man City (h) Apr 27 v Wigan (a) May 4 v Southampton (h) May 12 v Stoke (a) May 19 v Sunderland (h) ARSENALMar 3 v Tottenham (a) Mar 9 v Everton (h) Mar 16 v Swansea (a) Mar 30 v Reading (h) Apr 6 v West Brom (a) Apr 13 v Norwich (h) Apr 20 v Fulham (a) Apr 28 v Man Utd (h) May 4 v QPR (a) May 12 v Wigan (h) May 19 v Newcastle (a) By WILL BAKER-GRIFFIN
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Open / Close | 2,058 | 7,792 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-25 | latest | en | 0.958606 |
https://www.mathworks.com/matlabcentral/cody/problems/55-counting-sequence/solutions/1154066 | 1,579,707,582,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250607118.51/warc/CC-MAIN-20200122131612-20200122160612-00056.warc.gz | 990,733,181 | 15,759 | Cody
# Problem 55. Counting Sequence
Solution 1154066
Submitted on 5 Apr 2017 by Chong Zhang
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = [5 5 2 1 1 1 1 3]; correct = [2 5 1 2 4 1 1 3]; assert(isequal(correct, CountSeq(x)));
y = 2 5 0 0 0 0 0 0 y = 2 5 1 2 0 0 0 0 y = 2 5 1 2 4 1 0 0 y = 2 5 1 2 4 1 1 3
2 Pass
x = [9]; correct = [1 9]; assert(isequal(correct, CountSeq(x)));
y = 1 9
3 Pass
x = ones(1,9); correct = [9 1]; assert(isequal(correct, CountSeq(x)));
y = 9 1
4 Pass
x = 1:9; correct = [1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9]; assert(isequal(correct, CountSeq(x)));
y = 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 y = 1 1 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 y = 1 1 1 2 1 3 0 0 0 0 0 0 0 0 0 0 0 0 y = 1 1 1 2 1 3 1 4 0 0 0 0 0 0 0 0 0 0 y = 1 1 1 2 1 3 1 4 1 5 0 0 0 0 0 0 0 0 y = 1 1 1 2 1 3 1 4 1 5 1 6 0 0 0 0 0 0 y = 1 1 1 2 1 3 1 4 1 5 1 6 1 7 0 0 0 0 y = 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 0 0 y = 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9
5 Pass
x = [1 2 2 1]; correct = [1 1 2 2 1 1]; assert(isequal(correct, CountSeq(x))); | 703 | 1,166 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-05 | latest | en | 0.618503 |
https://www.tes.com/news/tes-archive/tes-publication/everyone-can-share-successmathematicssubject-week | 1,519,540,876,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891816138.91/warc/CC-MAIN-20180225051024-20180225071024-00180.warc.gz | 975,034,775 | 18,250 | # Everyone can share in success;Mathematics;Subject of the week
21st May 1999 at 01:00
Victoria Neumark describes how one inner-city authority is raising maths standards by involving parents and setting targets
Anna Lodge beams at her 22 Year 1 children. It is Thursday morning, coming to the end of the second numeracy lesson of the week, but the circle of children is still filled with eagerness. They are concentrating on the cards laid out on the carpet. What's going to happen next?
Mrs Lodge, maths co-ordinator at St Francis de Sales infant school, Haringey, north London, reminds them that they have played this game before and is encouraging.
"Before Christmas you only knew numbers up to 30, but now I hope you'll know them up to 60 and by the end of the year we expect you'll know all the numbers up to 100."
There is a pleased wriggle on the carpet. "So, today we are going to look at the numbers between 30 and 60. We are going to use some words - can you say 'Between'?" "Between," chorus the children. As Mrs Lodge runs through some key mathematical words and phrases such as "above" "below" "more than" "less than", hands shoot up to have a go.
"Christopher, can you say what is 31 add 2?" Christopher can, and does. He turns over card 33. Can Lauren say what is between 31 and 33? She is pretty sure but there's a bit of bating of breath until she has turned over the card 32, and a sigh of triumph when Mrs Lodge says, "It is 32, isn't it?" So, for about 10 minutes, the game goes on, with Erin finding four more than 40, Jessica counting on her fingers to get to three more than 43 and Sharleen needing a bit of help to establish the number before 37.
Then we get to the hard bit. Can Jenna tell the group what is 2 less than 60? She shakes her head. There are wildly divergent views on offer, with Larissa's confident 68 leading the field but other contenders such as 62 and 67 pointing to a fundamental misconception. Mrs Lodge hears the group's opinions before closing firmly.
"When we say 'less than', it means a smaller number. Is 68 a smaller number than 60?" Some squinting at the grid of number cards here, but a definite consensus: "No."
"So which way are we going to go to get 2 less than 60?" Fifty-eight being correctly identified, Mrs Lodge prepares to wind the session up with counting in 10s, to claps. She points out how 60 always comes after 50, so that 68 could not be less than 50 and adds cheerfully, "We can't move on past 60 yet, can we, so we'll play this game again."
As the children stream out to play, Mai trots off to her daily stint on the SuccessMaker software and is soon totally absorbed in picking out ordinal numbers. Jessica hangs back. She has finished her go on SuccessMaker and longs for more. "When I go home," she says proudly, "my mum has tidied up my room and my dad cleans the computer screen and I do maths on my big computer and I write poems and I like it."
"You see," says Mrs Lodge, "parental involvement is the key to what we do here. It is what really makes the difference."
For the past three years, St Francis de Sales, an inner-city school whose excellence was recognised in the "Ofsted Oscars", has been running phased maths teaching in Years 1 and 2, as well as phased literacy. For one hour, twice a week, the whole year group is split into four groups. Phase 1 is the ablest children who have reached or exceeded the expected key stage standard; phase 2 those who are near that level; phase 3a those not yet there and phase 3b those with special needs.
It has been, says Margarita Mooney, headteacher of St Francis for 12 years, "really motivating, really challenging". Separating groups with different learning speeds means, says Mrs Lodge, that whether a child is in the 18 high-fliers of phase 1 or the six statemented children in phase 3b, he or she is "with like-minded pupils - they breathe the same air".
For the past two years, Mrs Lodge has also been using the ideas pioneered by Margaret McKeever, maths adviser on haringey's STARS (Setting Targets and Raising Standards) project, funded by the Single Regeneration Budget.
The key to success, believes Ms McKeever, is "sharing everything with the parents". The numeracy objectives are established using her own Number Recovery scheme, a detailed breakdown of basic number facts for the primary years; their implementation is tied to a structured interaction with families. Also, each child has near-daily use of the expensive SuccessMaker software, an integrated learning system from Research Machines. The pound;2,000 package electronically teaches, assesses, prints out homework sheets and gives reports to teachers.
Target-setting is intimately bound in to dealings with parents. Each child is allocated two specific targets for each half-term - one achievable and one challenging - agreed at meetings with parents; each child has short, regular homework in books read by teacher and parents; each child gains recognition for reaching targets. Leaflets on each of the national curriculum targets make it easy for the teachers; photocopies of these sheets are stuck in to the child's target books.
For many families in inner-city Haringey, language is a problem. English is frequently an additional language and mathematical terminology can be hard to grasp. "Language here is our whole problem," says Mrs Lodge. "If I had said 'take away' instead of 'less than' just now, the children would have not been able to give the answer." Bringing the children's numeracy along means bringing the parents' language along too - but the rewards are immense. "Parents are so thrilled to be part of the child's learning," says Ms McKeever.
So thrilled that, says Mrs Lodge, they will regularly bring in their children half an hour early in the morning to ensure they get their go on SuccessMaker, a vital part of the STARS project. Even more vital, everyone agrees, is the weekly assembly in which success certificates are given out. "They are just so delighted to get them," says Ms Mooney.
Praise and the knowledge of success go hand in hand with raising teacher expectations, says Ms McKeever. As the project progresses, success in turn is fed back into refining of the targets and in-service training on numeracy. It works, and, says Ms McKeever, "It's certainly raised our self-esteem as well as the children's."
Every one of the 32 schools in Tottenham has had some involvement in STARS, if only by using SuccessMaker. Six have used the Number Recovery scheme and 12 IMPACT, the parent involvement plan. Those using IMPACT have registered an 18 per cent increase in performance in key stage 1 SATs, comparing 1995 with 1998. The two schools which have implemented all three strands of the project - IMPACT, target-setting through Number Recovery, and SuccessMaker - registered a 14 per cent improvement, well above the other schools in the borough.
There are lessons here for the implementation of the numeracy strategy. Lessons about gearing learning to pupil level of knowledge, about staff commitment. As Ms Mooney says, "We're no different from any other school. What makes the difference is staff commitment. It could never work without the dedication of our teachers working together as a team."
But, above all, the lesson is that parents are the key to children's numeracy.
For more information contact Alan Boyle, Chief Inspector, Haringey, tel: 0181 829 5050
Tips
* Involve the parents at a level they can relate to * Focus on mathematical language * Use target-setting in home- work as well as in class * Give frequent, real rewards and public acknowledgement * Use SuccessMaker, but make sure it is in constant use * Review in teams regularly, being flexible to the children's progress
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Enter subscription number | 1,788 | 8,056 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-09 | latest | en | 0.972036 |
https://glasp.co/youtube/1lA1hWdw62Q | 1,713,800,176,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818312.80/warc/CC-MAIN-20240422144517-20240422174517-00074.warc.gz | 240,106,453 | 75,544 | # Mistakes when finding inflection points: not checking candidates | AP Calculus AB | Khan Academy | Summary and Q&A
7.6K views
September 7, 2017
by
Mistakes when finding inflection points: not checking candidates | AP Calculus AB | Khan Academy
## TL;DR
Olga's solution for finding inflection points is incorrect because she incorrectly concludes that there is an inflection point at x equals two based on the fact that the second derivative is zero at that point.
## Install to Summarize YouTube Videos and Get Transcripts
### Q: What did Olga do correctly in her solution for finding inflection points?
Olga correctly found the first and second derivatives of the function and correctly determined that x equals two is a critical point.
### Q: What is Olga's mistake in her approach to finding inflection points?
Olga's mistake is concluding that there is an inflection point at x equals two solely based on the fact that the second derivative is zero at that point. She fails to test for concavity change by analyzing the sign of the second derivative on both sides of x equals two.
### Q: How does one determine if there is an inflection point at a given point?
To determine if there is an inflection point at a given point, one must check if the sign of the second derivative changes as you move from values less than the given point to values greater than the given point. If the signs do not change, there is no inflection point.
### Q: Can inflection points occur when the second derivative is zero?
Inflection points can occur when the second derivative is zero, but the second derivative being zero at a point does not guarantee the presence of an inflection point. The change in concavity must be checked for to confirm the existence of an inflection point.
## Summary & Key Takeaways
• Olga attempts to find inflection points by finding the first and second derivatives of the given function.
• She correctly finds the first and second derivatives, and determines that x equals two is a critical point.
• However, she incorrectly concludes that there is an inflection point at x equals two based solely on the fact that the second derivative is zero at that point. | 461 | 2,193 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-18 | latest | en | 0.921378 |
http://www.icoachmath.com/math_dictionary/cylinder.html | 1,501,108,613,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426639.7/warc/CC-MAIN-20170726222036-20170727002036-00150.warc.gz | 433,793,096 | 6,610 | Definition of Cylinder | Define Cylinder - Geometry - Free Math Dictionary Online
# Cylinder
## Definition of Cylinder
A Cylinder is a three-dimensional geometric figure that has two congruent and parallel bases.
• Right Cylinder: A right cylinder is a cylinder in which the centers of the bases are aligned directly one above the other.
• Right Circular Cylinder: When the base of a right cylinder is a circle, it is called a right circular cylinder.
• Oblique Cylinder: When the centers of the bases of a cylinder are not aligned directly one above the other, it is called an oblique cylinder.
### Solved Example on Cylinder
#### Ques: Which of the following coincides with the axis of rotation of a right cylinder?
##### Choices:
A. its altitude
B. its base
C. its surface area
D. none of these | 172 | 806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-30 | longest | en | 0.877539 |
http://www.java-forums.org/jcreator/45871-help-arrays-magicsquare.html | 1,394,974,922,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678702690/warc/CC-MAIN-20140313024502-00011-ip-10-183-142-35.ec2.internal.warc.gz | 365,754,887 | 15,012 | # Thread: [Help] Arrays and MagicSquare
1. ## [Help] Arrays and MagicSquare
Hi.. I need your help in jcreator with Arrays...
ive been working on these programs but still cant get it..
Activity 1:
Create class ArrayApp with 20 elements. The program must ask the user to input an element to search for and prints message " Found[search key] " Then prints the data eliminating the searched element. Otherwise, print "[search key] not Found".
Below is the sample output you should acquire.
Output:
77 99 44 55 22 88 11 0 66 33
Found 66.
77 99 44 55 22 88 11 0 33
************************************************** ***********
Activity 2:
Create a MagicSquare with [7][7] dimension
The magic Square should display a magic square below
Output:
28 19 10 1 48 39 30
29 27 18 9 7 47 38
37 35 26 17 8 6 46
45 36 34 25 16 14 5
4 44 42 33 24 15 13
12 3 43 41 32 23 21
20 11 2 49 40 31 22
2. Is this question posted elsewhere?
Help in Multidimensional Arrays :(
3. Member
Join Date
Nov 2011
Posts
1
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0
public class ArrayApp {
long[] arrayElems;
int numElements;
public ArrayApp() {
this(10);
}
public ArrayApp(int nElems) {
arrayElems = new long[nElems];
numElements = 0;
}
arrayElems[0] = 77;
arrayElems[1] = 99;
arrayElems[2] = 44;
arrayElems[3] = 55;
arrayElems[4] = 22;
arrayElems[5] = 88;
arrayElems[6] = 11;
arrayElems[7] = 00;
arrayElems[8] = 66;
arrayElems[9] = 33;
numElements = 10;
}
public void displayElements() {
System.out.println();
System.out.print ("");
for (int j=0; j<numElements; j++) {
System.out.print(arrayElems[j] + " ");
}
System.out.println("");
}
public void searchElement(long sKey) {
System.out.println();
int j;
for (j=0; j<numElements; j++) {
if (arrayElems[j] == sKey) {
break;
}
}
if (j == numElements) {
} else {
System.out.println("Found " + sKey);
}
}
public void deleteElement(long dKey) {
System.out.println();
int j;
for(j=0; j<numElements; j++) {
if(arrayElems[j] == dKey) {
break;
}
}
if (j == numElements) {
} else {
for (int k=j; k<numElements; k++) {
arrayElems[k] = arrayElems[k+1];
}
numElements--;
}
}
public static void main(String[] args) {
ArrayApp arrayApp = new ArrayApp(20);
arrayApp.displayElements();
arrayApp.searchElement(66);
arrayApp.searchElement(34);
arrayApp.deleteElement(55);
arrayApp.deleteElement(34);
arrayApp.displayElements();
}
}
4. ## Re: [Help] Arrays and MagicSquare
What is the purpose of your post? What is this code supposed to do?
If you have a question you should start your own thread and not hijack someone else's thread. | 789 | 2,516 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2014-10 | longest | en | 0.419224 |
http://www.numbersaplenty.com/9901 | 1,590,841,671,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347409171.27/warc/CC-MAIN-20200530102741-20200530132741-00380.warc.gz | 189,455,929 | 3,312 | Search a number
9901 is a prime number
BaseRepresentation
bin10011010101101
3111120201
42122231
5304101
6113501
740603
oct23255
914521
109901
117491
125891
134678
143873
152e01
9901 has 2 divisors, whose sum is σ = 9902. Its totient is φ = 9900.
The previous prime is 9887. The next prime is 9907. The reversal of 9901 is 1099.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 9801 + 100 = 99^2 + 10^2 .
It is a cyclic number.
It is not a de Polignac number, because 9901 - 211 = 7853 is a prime.
It is a Chen prime.
It is the 100-th Hogben number.
It is a plaindrome in base 13 and base 16.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (9907) by changing a digit.
It is a nontrivial repunit in base 99.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 4950 + 4951.
It is an arithmetic number, because the mean of its divisors is an integer number (4951).
29901 is an apocalyptic number.
9901 is the 45-th centered decagonal number.
It is an amenable number.
9901 is a deficient number, since it is larger than the sum of its proper divisors (1).
9901 is an equidigital number, since it uses as much as digits as its factorization.
9901 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 81, while the sum is 19.
The square root of 9901 is about 99.5037687728. The cubic root of 9901 is about 21.4730146386.
The spelling of 9901 in words is "nine thousand, nine hundred one". | 478 | 1,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-24 | latest | en | 0.919133 |
http://nattierosewrites.com/yljlo9jy/3a11c4-how-to-calculate-normal-force-with-friction | 1,627,746,827,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154089.68/warc/CC-MAIN-20210731141123-20210731171123-00623.warc.gz | 29,567,747 | 9,538 | The resistance force to the motion of the object over the other surfaces is called as the friction. Generally, the coefficient of sliding friction is smaller than the coefficient of static friction. … That's why, in the normal force equations listed below, angles are included. The gravitational force of the object is not opposite and equal to the normal force, but one of the force of gravity's vector's components is. μ is the coefficient of friction. If the box was on a soft surface, because of your additional force, it might collapse. Friction calculator solving for normal force given maximum static friction and coefficient Check out 44 similar classical mechanics calculators ⚙️, Normal force formula with an external force, How to use normal force equation - an example. Copyright 2021 Leaf Group Ltd. / Leaf Group Media, All Rights Reserved. Calculation of a normal force with an angle. For sliding friction, the frictional force is given by: F k = μ k F n. If the coefficient of the sliding friction between the floor and sculpture is 0.57, Calculate the mass, normal force, force of friction, and the force with which the curator pushes the sculpture. Example 2 A force F a is now applied horizontally to the floor to pull the box from left to right . This law can be represented in the form of an equation. These factors make it harder to move them past each other. This simply matches the “weight” of the object. For a still object on a flat surface, the force must exactly oppose the force due to gravity, otherwise the object would move, according to Newton’s laws of motion. Where, the normal force is F N For an object lying on a flat surface, the formula is. Do not ignore tension in the string when calculating the net torque (including friction) on the cylinder. Physicists sometimes write Fmax to make this point clear. It always acts perpendicular to the surface. With the given details, we can calculate the normal force as N = 2 kg × 9.8 N/kg = 19.6 N Now that we have the values of normal force and static friction co-efficient, we can calculate the frictional force as follows: If the force is directed straight upwards and equal to the gravitational force, the normal force is zero. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Ask Question Asked 5 years, 8 months ago. Calculate the static friction force using equation (5). Solution. A typical MCD for zero rake angle is given below. How to calculate static friction? The force resists motion, and in most cases the force acts in the opposite direction to the motion. Once the block is moving, you use μslide = 0.2, in this case: Lee Johnson is a freelance writer and science enthusiast, with a passion for distilling complex concepts into simple, digestible language. To find out how to calculate normal force with friction, use the friction calculator. You don’t work at this level when you calculate the force of friction, though. Draw a quick sketch of the object. $F_{fs} = \mu_{s} \cdot N = \mu_{s} \cdot G = \mu_{s} \cdot m \cdot g = 0.61 \cdot 50 \cdot 9.81 = 298.9 \text{ N}$ Step 3. Calculate the net force acting on a stationary or moving object. This ratio is dependent on material properties and most materials have a value between 0 and 1. If one object exerts a force on a second object, the second object exerts a force of equal magnitude and opposite direction on the first object (action equals reaction). Normal force is the perpendicular force that the surface exerts on an object. Why? The formula for the force of friction states: For the example, consider a wood block of 2-kg mass on a wooden table, being pushed from stationary. Friction describes the force between two surfaces when you try to move one across the other. where μ s is the coefficient of static friction and N is the force normal to the surface acting on the box. $\begin{split} \sum F_{y} &= 0 \\ N – G &= 0\\ N &= G \end{split}$ Step 2. The kinetic friction of a moving object is equal to the co-efficient of friction multiplied by object's normal force. Homework Equations For everyday situations, physicists group all of these factors together in the “coefficient” μ. Note that the normal force is perpendicular to the surface the object sits on. Unit of frictional force: Newton or N. Dimensions of frictional force: MLT-2. Suppose a force is acting on an object of mass m that displaces it on a surface. (Don't confuse it with mass! It displays shear force (P S), normal shear force (P N), main cutting force (P Z), resultant thrust force (P XY), friction force (F), normal force (N) and a resultant force (R). The friction force is best defined using an equation. Weight is the same as gravitational force.) Even though they act in the direction opposite to that in which an object moves, the normal force (F N) produces these forces, which acts perpendicular to the … According to Newton’s third law of motion, the surface will apply an equal and opposite force on the object, which is the force of sliding friction F K. According to friction … To find out how to calculate normal force with friction, use the friction calculator. 2. On an inclined surface (assuming that the object doesn't slide down), the weight of the object is supported by both the normal force and friction. The normal force calculator helps you find the force that a surface exerts to prevent an object from falling through it. N = 100 * 9.807 + 250 * sin(45o) = 980.7 + 250 * √2 / 2 = 1,157.4 N. The ground exerts 1,157.4 N of force on the box. Force is a vector quantity, which means you must consider the direction in which it acts. Have you enjoyed our normal force calculator? F NET = net force . F ∝ F N. Or, F = μF N. Symbol. In other words, it’s easier to slide something that’s already sliding than to slide something that’s still. The force of friction depends on the friction coefficient for the type of friction under consideration and the magnitude of the normal force that the surface exerts on the object. It's the opposite case for an external force directed upward. Its direction is opposite to the applied force. On an inclined surface (assuming that the object doesn't slide down), the weight of the object is supported by both the normal force and friction. Since you're here, you might also enjoy our Newton's second law calculator. If the object isn’t already moving across the surface, you use the coefficient of static friction μstatic, but if it is moving you use the coefficient of sliding friction μslide. Scroll down to find out how to calculate normal force. In this calculator, calculate the normal force, friction force and coefficient with the other known values. The sliding friction equation can be calculated by applying physics principles and Newton’s laws of motion. Coefficient of Friction calculator uses Coefficient of Friction=Limiting Force/Normal reaction to calculate the Coefficient of Friction, Coefficient of Friction is the ratio defining the force that resists the motion of one body in relation to another body in contact with it. When we roll a ball on the ground, friction plays a major role to bring the ball to halt. For inclined surfaces, the strength of the normal force is reduced the more the surface is inclined, so the formula becomes: With θ standing for the angle the surface is inclined to. He studied physics at the Open University and graduated in 2018. The normal force can be calculated using physics principles and balancing the forces using Newton’s laws of motion. It also displays three relevant angles – orthogonal rake angle (γ O), friction angle (η) and shear angle (β O). Check out the work and power calculator too! Draw a free body diagram. How to find normal force on your own? Draw an arrow showing every force acting on the object. About the Book Author For most purposes, you can use the formula: to calculate friction, with N standing for the “normal” force and “μ” incorporating the characteristics of the surface. Look these up for your surface using an online table (see Resources). That's way, to save strength, it's best to push things by their side, straight toward the destination. For most purposes, you can use the formula: F=\mu N F = μN to calculate friction, with N standing for the “normal” force and “ μ ” incorporating the characteristics of the surface. F N = normal force . Static Friction Formula Questions: 1) A 5500 N force is applied to a sled full of firewood in a snow-covered forest. Start studying Everyday Forces (Calculating normal force, friction..etc). F f = force due to friction . So, a normal force is equal to the force exerted by the object on the surface. How to Calculate Friction Force. Replace the μ and N values in the above formula. In this case, you use the static coefficient, with μstatic = 0.25 to 0.5 for wood. This increases the normal force, the outside force is pushing the object into the ground. The normal reaction R on the body is equal to its weight. Assume the coefficient of friction between all surfaces is μ= 0.0350. To counteract this force, the table exerts a force on the book, preventing it from falling. Surfaces exert a frictional force that resists sliding motions, and you need to calculate the size of this force as part of many physics problems. Calculate the normal reaction force on the crate. Two main types of frictional forces exist: the static force (F st) and the sliding force (F sl). In the case of the figure, because the ingot slides along the horizontal ground and is pulled by a force that is parallel to the ground, the normal force has the same magnitude as the weight of the ingot, so Fnormal = mg. You have the normal force, which is the force pressing the ingot and the ground together. The normal force would point directly upwards (to support the weight of the block), and you would calculate: The coefficient depends on the object and the specific situation you’re working with. i.e., weight = mass x acceleration due to gravity (10m/s 2) R = W Solution. Friction calculator solving for normal force given kinetic friction and coefficient ... Equations Calculator Constant Acceleration Motion Physics Adiabatic Lapse Rate Calculator Net Income Multiplier Calculator Force Equations Physics Calculator Wind Power Generator Calculator Euler Number Formulas Calculator . For a simple example calculation, consider a flat surface with a 2-kg block of wood sitting on it. For example, if you put a book on a table, there is a gravitational force that is pulling it toward the ground. For ice on ice, the static coefficient is 0.1. This is because a … You push on it at a 45 degrees angle with 250 N of force. ” force ( F st ) and the sliding coefficient reduces this even more, to for... You put a book on a surface Question, along with all the forces that acting. It pulls an object of mass m is lying on the cylinder, ignore in. To calculate normal force the opposite direction to the force between two surfaces when you the... 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Might collapse 2 a force F a is now applied horizontally to the motion of the normal force kinetic... A block of mass m that displaces it on a surface which it acts into the ground surface... For several websites including eHow UK and WiseGeek, mainly covering physics and astronomy in existing... At a 45 degrees angle with 250 N of force t work at this level when you the. Of kinetic friction calculations the effect of the normal force with friction, the! Open University and graduated in 2018 terms, and more with flashcards, games, and more with flashcards games..., a normal force and the friction calculator its vector component the surface it! On material properties and most materials have a coefficient of static friction laws motion. Directed downwards, you might also enjoy our Newton 's second law calculator formula is ignore in! Friction equation can be represented in the string when calculating the normal force will apply a force F a now. Offers all types of frictional force 5 years, 8 months ago University and graduated in.! Friction force using equation ( 5 ) other forces are acting on an object on... A surface table, there is an outside force is counter to the weight so the normal force is coefficient! Equation ( 5 ) also affect the coefficient of static friction and normal force friction! Enjoy our Newton 's second law calculator which means you must consider the direction in it! Be represented in the string when calculating the net torque ( including friction on! Exist: the static coefficient is 0.1 F N on the ground, friction plays major... 19.6 = 3.92\text { N } you know the normal force most materials a... … this engineering statics tutorial introduces how to calculate normal force is counter to the weight,... Friction calculator to move F N. Or, F = μF N..! Μ * N. where, N is the name for the force of friction F = μF N... Into consideration the parallel vector component to the motion of the surface exerts on an object second law calculator coefficient... Major role to bring the ball to halt sits on ( including friction ) on the box also enjoy Newton... “ coefficient ” μ how to calculate normal force with friction for wood on wood don ’ t work at this when! For your surface using an equation object over the other known values friction calculations 's best push. By applying physics principles and Newton ’ s easier to slide something that ’ laws... He studied physics at the Open University and graduated in 2018 to bring the ball to halt to. Means you must consider the direction in which it acts and other study tools ” force ( F )... Forces acting on the object in its existing form 1 Index 2 Index 3 4... Terms, and more with flashcards, games, and no other forces are on... A ball on the book, preventing it from falling slide } =\mu_ { slide } =\mu_ slide., N is the product of coefficient of static friction and N is perpendicular! This level when you calculate the friction coefficient if you know the force... Object into the ground, friction plays a major role to bring the ball to halt add its vector to! The string when calculating the normal force and the friction calculator studied physics at the Open University and graduated 2018... The perpendicular force that does this do not ignore tension in the normal force equations below! Is n't concurrent to the objects weight a science blogger for Elements Behavioral Health 's how to calculate normal force with friction network for five.. ( F sl ) add its vector component flat surface, the normal force, static... N'T concurrent to the floor to pull the box was on a horizontal surface μ=.. Motion, and more with flashcards, games, and more with flashcards games... Given below ( 5 ) re considering also affect the coefficient of static friction μ s. a. Is equal to the co-efficient of friction calculator it acts including friction ) on the book, preventing it falling! Means you must consider the direction in which it acts every force acting on it directed upward smaller than coefficient! Forces acting on it this engineering statics tutorial introduces how to calculate static friction is smaller than the coefficient static! Angles are included wood sitting on it for your surface using an online table ( see Resources.... Toward the ground from falling replace the μ and N is the force acts in the “ weight of... Harder to move imagine there is an outside force is counter to the surface on! Of sliding friction equation can be calculated by applying physics principles and balancing the acting!, … this engineering statics tutorial introduces how to calculate normal force.. A is how to calculate normal force with friction applied horizontally to the co-efficient of friction calculator objects weight to the the! Other words, it 's best to push things by their side straight. Index 3 Index 4 Infant Chart … how to calculate as static friction μ s. draw a how to calculate normal force with friction body is... To maintain an object in its existing form at this level when you calculate the force in... Coefficient of sliding friction is smaller that the force acts in the string calculating. ( N ) is the perpendicular force that does this straight toward the destination formulas vary with snow. Smaller than the coefficient of static friction and N is the normal force is directed downwards you... Up for your surface using an online table ( see Resources how to calculate normal force with friction as static friction so, normal. Kinetic friction because more forces are acting to maintain an object lying on a surface. Ice and 0.2 for wood on wood known values friction increases with it Question Asked 5 years, months! Weight ” of the object on the surface that are acting to maintain object! Or, F = μ * N. where, N is the normal force expressed in Newtons plays major... Things by their side, straight toward the ground object on the ground, plays! | 3,836 | 17,864 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-31 | latest | en | 0.913048 |
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All Right Reserved. Copyright 2005-2018 | 1,452 | 6,352 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-09 | latest | en | 0.831328 |
https://www.kylesconverter.com/mass-flow/kilograms-per-week-to-ounces-per-day | 1,670,093,264,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00588.warc.gz | 913,826,741 | 5,747 | # Convert Kilograms Per Week to Ounces Per Day
### Kyle's Converter > Mass Flow > Kilograms Per Week > Kilograms Per Week to Ounces Per Day
Kilograms Per Week (kg/wk) Ounces Per Day (oz/d) Precision: 0 1 2 3 4 5 6 7 8 9 12 15 18
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Ounces Per Day to Kilograms Per Week
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Unit Descriptions
1 Kilogram per Week:
Mass flow of kilograms across a threshold per unit time of a week. A week containing 604800 seconds. 1 Kilogram per week = 1/604800 kilograms per second (SI base unit). 1 kg/wk ≈ 0.000 001 653 439 153 439 kg/s.
1 Ounce per Day:
Mass flow of ounces across a threshold per unit time of a day. 1 Ounce per day = 0.028349523125/86400 kilograms per second (SI base unit). 1 oz/d ≈ 3.281 194 806 13 x 10-7 kg/s.
Conversions Table
1 Kilograms Per Week to Ounces Per Day = 5.039170 Kilograms Per Week to Ounces Per Day = 352.7396
2 Kilograms Per Week to Ounces Per Day = 10.078380 Kilograms Per Week to Ounces Per Day = 403.131
3 Kilograms Per Week to Ounces Per Day = 15.117490 Kilograms Per Week to Ounces Per Day = 453.5224
4 Kilograms Per Week to Ounces Per Day = 20.1565100 Kilograms Per Week to Ounces Per Day = 503.9137
5 Kilograms Per Week to Ounces Per Day = 25.1957200 Kilograms Per Week to Ounces Per Day = 1007.8275
6 Kilograms Per Week to Ounces Per Day = 30.2348300 Kilograms Per Week to Ounces Per Day = 1511.7412
7 Kilograms Per Week to Ounces Per Day = 35.274400 Kilograms Per Week to Ounces Per Day = 2015.655
8 Kilograms Per Week to Ounces Per Day = 40.3131500 Kilograms Per Week to Ounces Per Day = 2519.5687
9 Kilograms Per Week to Ounces Per Day = 45.3522600 Kilograms Per Week to Ounces Per Day = 3023.4825
10 Kilograms Per Week to Ounces Per Day = 50.3914800 Kilograms Per Week to Ounces Per Day = 4031.3099
20 Kilograms Per Week to Ounces Per Day = 100.7827900 Kilograms Per Week to Ounces Per Day = 4535.2237
30 Kilograms Per Week to Ounces Per Day = 151.17411,000 Kilograms Per Week to Ounces Per Day = 5039.1374
40 Kilograms Per Week to Ounces Per Day = 201.565510,000 Kilograms Per Week to Ounces Per Day = 50391.3742
50 Kilograms Per Week to Ounces Per Day = 251.9569100,000 Kilograms Per Week to Ounces Per Day = 503913.7421
60 Kilograms Per Week to Ounces Per Day = 302.34821,000,000 Kilograms Per Week to Ounces Per Day = 5039137.4214 | 805 | 2,381 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2022-49 | latest | en | 0.631912 |
http://docplayer.net/16418804-3-6-the-factor-theorem.html | 1,603,810,340,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107894203.73/warc/CC-MAIN-20201027140911-20201027170911-00127.warc.gz | 28,608,836 | 27,894 | # 3.6. The factor theorem
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1 3.6. The factor theorem Example 1. At the right we have drawn the graph of the polynomial y = x 4 9x 3 + 8x 36x Your problem is to write the polynomial in factored form. Does the geometry of the graph give you any help here? Solution If this graph had appeared in the previous section, we'd have expected it to belong to an equation which was a product of linear factors corresponding to the zeros x=1, and 4. That is, we'd expect to have the factors (x 1), (x ) and (x 4). Actually, since the graph does not cross the axis at x=, we'd expect (x ) to occur an even number of times, perhaps twice. Well, if we have one copy of (x 1), two copies of (x ) and one copy of (x 4), that's already a product of degree 4 and the whole polynomial has degree 4, so that must be it. So we propose: y = (x 1) (x ) (x 4). Is this the answer? Let's check it out. What we do is expand the factored form and see if we get the original. y = (x 1)(x 4) (x ) = (x 5x + 4) (x 4x + 4) = x 4 + ( 4 5)x 3 + (4+4+0)x + ( 0 16)x + 16 x 4 9x 3 + 8x 36x which is the original equation. So we did it! We used the graph to factor the polynomial. The graph gave us the zeros and the zeros gave us the factors. Though you might not have noticed, our argument actually made use of a wonderful theorem. It s called the factor theorem, and it isn t obviously true, but the proof is a nice argument, and we give a version of it below. What the theorem says, roughly speaking, is that if you have a zero of a polynomial, then you have a factor y The previous section was really about going from the factored form of the equation to the geometric form of the graph. Now we go the other way, and get information about the factors from the picture. To be precise, the intersections of the graph with the x-axis tell us about the factors of the polynomial. In the second line of this calculation we have the product of two terms each with 3 entries. To expand this we multiply each entry in the first term by each entry in the second. That gives us 3 3=9 terms and these are displayed on the third line. There's 1 x 4 -term, x 3 -terms, 3 x - terms, x-terms, and 1 constant term. For example, the three ways to get an x term are diagrammed below: 4 4 ( x - 5 x + 4) ( x - 4 x+ 4) 0 x 36factor theorem 9/4/007 1
2 The factor theorem. Geometric version. If f(x) is a polynomial whose graph crosses the x-axis at x=a, then (x a) is a factor of f(x). Algebraic version. If f(x) is a polynomial and f(a) = 0, then (x a) is a factor of f(x). Proof of the factor theorem Let's start with an example. Consider 4 3 ( x) = x 9x + 8x 36x + 18 f. [This is the polynomial of Example 1 with last term 18 instead of 16.] Our question is whether or not it has x-4 as a factor. Well, one way to find that out is to divide it by x 4 and see what we get. Either it will go evenly or there will be a remainder. If we do this (perhaps by long division of polynomials) we find: x 4 9x 3 + 8x 36x + 18 = (x 4)(x 3 5x + 8x 4) + which has the form: f(x) = (x 4)Q(x) + r for a quotient Q(x) and a remainder r. The remainder here is, which is not zero, and thus we can conclude that x 4 does not divide evenly into f(x). Now here's the interesting result that remainder is the value of f(x) at x=4. You can verify that by plugging x= into the expression for the polynomial f(x), but an easier and more insightful argument is obtained by plugging x=4 into the above equation. We get: f(4) = (4 4)Q(4) + r and the stuff in the middle is zero, giving us r = f(4). So we can write the above equation as: f(x) = (x 4)Q(x) + f(4). The reason this fast argument is so nice is that it works in general. We can do this for any polynomial f(x) and for any factor (x a). When we perform the above routine, we get: f(x) = (x a)q(x) + f(a). And just as above, the remainder when f(x) is divided by (x a) is f(a). It follows from this that (x a) will be a factor of f(x) exactly when f(a)=0. And that s the factor theorem. x 4 3 x 5x + 8x x 9x + 8x 36x x 4x 3 5x + 8x 3 5x + 0x 8x 36x 8x 3x 4x x factor theorem 9/4/007
3 Example. Consider the polynomial f(x) = x 3 + 4x 17x 60. Verify that f( 3) = 0, and use this observation to find the other two zeros of the polynomial. Solution First we show that 3 is a zero: f( 3) = ( 3) 3 + 4( 3) 17( 3) 60 = = 0. Now how might we find the other zeros? Well, that s where the factor theorem comes in. The zero 3 gives us the factor: (x 3) = (x+3). If we divide this factor into f(x), we ll get a quotient of degree. That s a quadratic polynomial and we can find its zeros either by factoring it or using the quadratic formula. Let s do it. To divide (x+3) into f(x), we can use long division, but in simple cases, trial and error is just as easy. We write: x 3 + 4x 17x 60 = (x+3)(x + x 0) We have filled in the first and last terms of the quotient. Both of these are easy to obtain. We get the x by seeing what has to go with the x to make x 3, and we get the 0 by seeing what has to go with the 3 to make 60. So all that remains is to fill in the x-term, and we have written its coefficient as an underlined blank. We can do this in two way, choosing it so as to get the 4x right or choosing it so as to get the 17x right. Let s go for the 4x. We already have 3x from the 3 and the x so we need one more, and that means that the should be 1. Then that will go with the x in the (x+3) term to give 1x. Thus the factorization is x 3 + 4x 17x 60 = (x+3)(x + x 0). The remaining two zeros will be zeros of the quadratic factor. We can use the quadratic formula, or factor it. In fact it factors: x + x 0 = (x+5)(x 4). Thus the entire factorization is: x 3 + 4x 17x 60 = (x+3) (x+5) (x 4) and the zeros are 3, 5, and 4. This is a bit of a drawn out presentation, but here s the scoop. 1. Since 3 is a zero of f(x), x+3 is a factor.. Divide x+3 into f(x) to get the factorization: (x+3)(x +x 0) 3. Factor the second term to get (x+3)(x+5)(x 4). 4. The zeros are 3, 5, and 4. As a check on our work, we see if the x-term is right. We need 17 of these. The above factorization gives us = 17, as expected. 36factor theorem 9/4/007 3
4 Example 3. Find all solutions of the equation: x 3 x 6x 4 = 0. Solution This time we are not given a zero of the polynomial. One idea is to plug a few numbers in, hoping that by some stroke of luck, a small integer (like ±1 or ±) might work. We try: f(1) = (1) 3 (1) 6(1) 4 = 10 f( 1) = ( 1) 3 ( 1) 6( 1) 4 = 0 and 1 works. [I guess we're lucky after all.] It follows from the factor theorem that (x+1) is a factor. We divide this into f(x) and get x 3 x 6x 4 = (x+1)(x x 4). The quadratic quotient x x 4 will not factor nicely, so we use the quadratic formula to find its zeros: x = ± = 1 ± 5. Thus the zeros of f(x) are 1, and 1 5. Cubic equations It s worth pointing out that cubic equations are not so easy to solve. If the equation in Example 3 were quadratic, we could use the quadratic formula, but it's cubic. Is there an analogue to the quadratic formula that works for cubics? Well, yes and no. There is a routine which will give you the solutions of a cubic equation, but it isn't nearly as friendly as the quadratic formula, and I can never remember how it goes. So the best way for us to solve a cubic right now is to find that first zero. For an excellent application of this method, see 3.1 #11. Finding that first zero Quite often in my work as an applied mathematician, I come across an equation that I have to solve. That is, I need to find the zeros of some function. And sometimes what I need is a first zero, something that will let me start taking the function apart. But how do I find it? Do I make a guess? Am I lucky enough to find a small integer that works? Is the universe kind to me? Well that's an interesting question. It turns out that the universe is kind. In fact, to mathematicians, it's incredibly, unexpectedly, awesomely kind. And that's because it's so "together," so connected, so coherent. You see the equation I have to solve isn't just an abstract mathematical equation it comes out of a physical problem, it carries with it a rich context. And that context gives a physical meaning to the polynomial, and to all its various pieces, and that meaning often gives me a hook that let s me find a zero. For example, last month I had to find the roots of the quartic equation x 4 sx 3 + (s 1)x s(s 1)x = 0. This is actually a whole family of equations, one for every value of the parameter s. And my job was to find the roots in terms of s. Well, x=0 is clearly a root, so I can factor that out. But then what? I am left with a cubic equation. Well the context of the problem told me to expect that x=s would always be a root. So I could factor out (x s). And that left me a quadratic equation I could solve. (Problem 6). 36factor theorem 9/4/007 4
5 Problems 1. Find all solutions of the following equations. The strategy here is to guess at a zero and then factor it out. (a) x 3 9x 1x + 0 = 0 (b) x 3 3x x + 4 = 0 Look at the coefficients in (a): = 0. What does that tell you about a zero?. Write the polynomial f(x) = 5x 4 in the form f(x) = (x )Q(x) + f() for some polynomial Q(x). 3. Write the polynomial f(x) = x 4 11x x 55x + 5 in the form f(x) = (x a)q(x) + f(a) for some polynomial Q(x) for the following values of a. (a) a=0 (b) a=1 (c) a= (d) a=3 4. At the right, we have drawn the graph of the polynomial y = x 4 11x x 55x + 5. It intersects the x-axis at x=1 and x=5 and at two other places. Find these other two intersections y 5. Verify that if we set x equal to a, we get a solution of the equation: x 3 + (a )x a(1+a)x + 4a = 0 Find two other solutions x 6. Verify that x=s is always a solution of the equation: x 4 sx 3 + (s 1)x s(s 1)x = 0 and use this fact to find all the roots of the equation in terms of s. 36factor theorem 9/4/007 5
6 7. A feast of a problem. {Notices of the AMS 47, March 000, p. 360.] Mark Saul a mathematician who teaches in the Bronxville school district was in Taiwan observing an after school class of 400 grade 10 students working on the following problem. Find the remainder when the polynomial f(x) = x + x + x + x + is divided by x + x Now of course if we had lots of time (and were kinda bored) we could do this by brute force long division, but the challenge is to find a clever way, and the factor theorem (or the ideas behind it) provide the lever. We know that if we did the long division we d get an equation of the form: f(x) = ( x + x + 1)Q(x) + rx+s where the remainder rx+s will be a polynomial of degree 1. Now the factor theorem trick was to plug something into the equation which would make the Q(x) term zero. In the case of the factor theorem, the divisor was x a and we could make it zero by setting x=a. But here the divisor is quadratic, and so to make it zero we need to find one of its roots. If we had such a root α. then we could write: f(α) = ( α +α + 1 )Q(α) + rα+s = rα+s where the last equality follows from the fact that α was chosen to make α +α + 1 = 0. Now recall what the original problem is to find the remainder rx+s, and that means finding r and s. Well what we have above is a single equation in r and s. To solve for them we will need another equation (two unknowns usually requires two equations). Where should we get it from? Well the divisor is quadratic, so it should have two roots, α and a second root β. Using them both, we would get two equations: f(α) = rα+s f(β) = rβ+s which we should be able to solve for r and s. Well that all sounds just fine, but the fly in the ointment (as they say) is that the divisor x + x + 1 does not have any real roots. The quadratic formula gives us: 1± 1 4 x = and we get a negative number under the root sign. But that s not such a problem after all. One of the triumphs of mathematics is the discovery that we can handle such situations with standard algebra by simply treating 1 as a number whose square is 1. Thus we write the two roots as α 1± 3i = β where i = 1. So there we are. Believe it or not there s still a fascinating joker in the deck. When we come to calculate f(α) and f(β) we will be faced with the problem of calculating numbers like α 33 and b 33. How on earth do we do that? Well the divisor belongs to an interesting family. For any n, the zeros of the polynomials n n 1 x + x x + 1 have a wonderful algebraic property. Can you discover it? [Hint: factor x n 1.] 11 36factor theorem 9/4/007 6
7 7. A little logic (goes a long way). Note that the converse of the Factor Theorem is immediately true. If I have a polynomial which has (x a) as a factor, then when I set x equal to a I will certainly get zero. And that also means the graph will cross the axis at x=a. But this fact does not give us a proof of the Factor Theorem. If we know that the converse of a theorem is true, that does not allow us to deduce that the theorem itself is true. Here we investigate that just a bit. An implication is a statement of the form: if A then B, where A and B are statements which might or might not be true. The converse of the implication if A then B is the implication if B then A. As an example, consider the implication: 1) If no one is at home, then the machine answers the phone. The converse of this is: ) If the machine answers the phone, then no one is at home. The implication if A then B is said to hold if whenever A is true, B will also be true. For example, in a certain home the implication 1) above might well hold. It is possible for an implication to hold but its converse to fail to hold. For example this might the case for 1) and ) above we might have 1) holding, but ) not holding. (a) Give an example of a mathematical implication that holds but whose converse fails to hold. (b) Give an example of a mathematical implication that holds and whose converse holds. (c) Give an example of a mathematical implication that fails to hold and whose converse fails to hold. Often an implication is not stated in the nice if A then B form, and sometimes care must be taken to see what it is really saying. Here are some examples. 3) Employees of the school board are not eligible. 4) You'll only find it on sale at the end of the season. 5) I never drink coffee before 10. 6) First come, first served. Formulate each of these in standard form (if A then B) and then write the converse in as elegant as form as you can. Note to avoid confusion. For some of the above questions there will be two correct answers, one implication being the contrapositive of the other. By definition, the contrapositive of the implication if A then B is the implication if not-a then not-b. For example, the contrapositive of 1) above is the implication 1*) If the machine does not answer the phone then someone is at home. You can see that this really says the same thing as 1). We say that they are equivalent. In fact, an implication and its contrapositive are always equivalent. 36factor theorem 9/4/007 7
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MATH 21 College Algebra 1 Lecture Notes MATH 21 3.6 Factoring Review College Algebra 1 Factoring and Foiling 1. (a + b) 2 = a 2 + 2ab + b 2. 2. (a b) 2 = a 2 2ab + b 2. 3. (a + b)(a b) = a 2 b 2. 4. (a | 11,122 | 41,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-45 | latest | en | 0.961869 |
https://scicomp.stackexchange.com/questions/tagged/openmodelica | 1,685,494,141,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646181.29/warc/CC-MAIN-20230530230622-20230531020622-00418.warc.gz | 579,928,195 | 33,068 | # Questions tagged [openmodelica]
OpenModelica is an open-source implementation for the Modelica modeling language.
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### Recommended language/environment for large scale semi-continuous biological models
We have a fairly large (maybe 1000 equations) differential-algebraic equation model written in ACSLX, an obsolete modelling environment similar to Modelica. The model represents the evolution of a ...
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### Modelica and SIMULINK yield completely different results
I'm trying to simulate a 1D model of two bodies sliding on each other with a Coulomb friction in between. As I have explained here I modeled the friction as : The force required to keep two bodies ...
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### Finite difference method not working for advection PDE with negative coefficient
I'm trying to solve a very simple advection PDE $\frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x}=0$ where $c<0$. I have been able to implement a simple Modelica code to solve the ...
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### Object-oriented non-linear solving in python
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### Alternatives to Comsol Multiphysics
This might be a question better suited for the Software Recommendations side of S.E., however I do believe that people who frequent this part of S.E. are more likely to be able to answer this question....
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### Modelica "derivative" annotation problem
I have questions about the use of "derivative" annotation I wanted to ask you. Lets put a sample. The equation is this: z_level = LevelZ(time); where LevelZ(...
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### How to do something in Modelica every solver iteration?
How is it possible to do something once every solver iteration? The expressions when change(time) or when time > pre(time) ...
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### How to do co-simulation with two FMU models
I have two FMU developed with different simulation tools. I would like to connect them through some variables and run a co-simulation with an open source tool as OpenModelica. This co-simulation has ...
• 95 | 565 | 2,407 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-23 | latest | en | 0.923916 |
https://www.studypool.com/discuss/531012/algebraic-expressions-using-angles-help?free | 1,480,766,267,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540928.63/warc/CC-MAIN-20161202170900-00487-ip-10-31-129-80.ec2.internal.warc.gz | 1,029,949,297 | 13,953 | Algebraic expressions using angles...help!!
Mathematics Tutor: None Selected Time limit: 1 Day
The measure of the angles supplement is 20 degrees more than three times that of its complement. Find the measure of the angle described.
May 12th, 2015
180 - x - 20 = 3(90 - x)
160 - x = 270 - 3x
2x = 90
x = 45
May 12th, 2015
...
May 12th, 2015
...
May 12th, 2015
Dec 3rd, 2016
check_circle | 134 | 393 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.328125 | 3 | CC-MAIN-2016-50 | longest | en | 0.916854 |
https://se.mathworks.com/matlabcentral/profile/authors/18681596 | 1,717,074,684,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00380.warc.gz | 430,499,260 | 25,258 | # Andreas Apostolatos
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Andreas holds a Diploma in Applied Mathematics with specialization in Applied Mechanics and Statistics from the National Technical University of Athens. Currently, Andreas applies his expertise as a Senior Application Engineer in Education at MathWorks, where he supports academics using state-of-the-art MathWorks solutions for their teaching and research activities, while also engaging in the preparation of high-quality courseware for science and engineering disciplines. Moreover, Andreas is also an active lecturer at the Technical University of Munich where he applies modern teaching methods, such as Inverted Classroom and Project-Based Learning (PBL).
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# Map Basics - Scale, Directions, Labeling, & Coloring
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Scale, directions, labeling, and coloring instructions are the focal point of this map essentials packet. You'll receive a four page handout of notes that cover these basics and then one worksheet each to practice directions and scale.
Before purchasing this item, you may be interested in my more in depth Map Skills Packet, which includes everything seen here, plus another 15 pages of worksheets, tasks, and a mapping project.
The notes are contained in a four page document that is a handout to give your students or a lesson plan for your entire class on the basics of map labeling. It explains map scales, directions, map labeling, and coloring.
Ratio and linear scale are explained with words and diagrams. Three examples to practice calculating distance using scale as well as step by step procedures are provided.
Cardinal and intermediate directions are explained with words and diagrams, as well as a mnemonic device to help students remember the correct order of the directions around the compass rose.
Directions for properly labeling maps are broken down into four steps including handwriting, orientation, and when to use arrows versus using a legend.
Finally, there is a page on properly coloring maps correctly including only using blue for water, not allowing the same color to touch, etc. This includes a practice diagram for students to color.
Next is a beginner's worksheet on directions called "Here's Looking at Directions" where students can practice naming directions and locating places based on directional cues.
Finally, is a basic scale worksheet entitled "A Basic Look at Scale" on which students can practice measuring and determining distances using the posted scale and easily marked locations.
Other history and geography based items can also be found in my store, Brittany Naujok - The Colorado Classroom. I hope you will take a look!
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Coordinates of the United States: Latitude & Longitude Practice
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\$2.50 | 686 | 3,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-51 | longest | en | 0.933638 |
https://www.quickpickdeal.com/energy-calculator/microwave-oven-power-consumption-calculator-what-is-the-power-consumption-of-a | 1,716,604,286,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00668.warc.gz | 853,567,184 | 17,714 | A microwave oven is a kitchen appliance. Microwave ovens are used to cook food in the kitchen
as well as to heat food. If you cook in a microwave oven, it takes very little time. Microwave ovens are used in most homes nowadays.
A Microwave will use from 500 to 2000 watts depending on the capacity of the Microwave, a domestic small Microwave consumes about 500-850 watts While larger, commercial Microwave models may use 2,000 watts or higher.
We recently measured the energy consumption of our Microwave using an energy monitoring device
and we conclude that our Compact Microwave used 550 watts for the most optimal and energy-efficient mode available on the Microwave.
You can find the wattage on the compliance label in the Microwave, using below tool Calculate How much electricity does a Microwave consume using our Microwave cost calculator.
Choose Country: Australia Canada Denmark France Germany India Philippines Singapore United Kingdom United States Other country Wattage of Microwave : watts (W) kilowatts (kW) Average Usage Per Day: minutes/day hours/day 1 kWh(1 unit)Cost: cent pence rupee peso kr other Daily cost : Monthly Cost: Yearly Cost Daily Energy Usage: kWh/day Monthly Energy Usage: kWh/month Yearly Energy Usage: kWh/year
At what power setting do you use your microwave? If you always use your microwave at 100 percent power, it will consume 100% energy(you can find on the back of microwave.
most microwaves provide settings that allow you to use in energy efficiency mode. In these settings, your energy consumption is slightly less.
## How many kwh does a microwave oven use?
Let's take an example. Suppose the power consumption of your microwave oven is 500W. You use the microwave oven one-hour every day.
So Daily Energy consumption of microwave
500W × 1 hour= 500 watts.
Monthly Energy consumption of microwave
So thus 500W × 1 hour × 30 days = 15,000 watts.
Divide by 1000 to convert watts into kilowatt.
500-watt/1000= .5 kWh
15000-watt / 1000 = 15 Kwh.
Let assume the electricity rate in your city is 12 cents per kilowatt-hour.
So thus
Daily cost is:- .5*.12=\$.06
Monthly cost is:- 15*.12=\$1.8
### Microwave Power Consumption Chart
@Assumes electricity cost per unit .12c
### The cost of microwaving our favourite foods
@Assumes electricity cost per unit .12c for 800 watts
## Does a microwave oven use more energy than a convection oven?
Microwave consumes less power as compared to a conventional oven. The reason is that microwave ovens work for a short time, much less than conventional ovens. Since you are billed for kilowatt-hours, the shorter the time, the lower your electricity bill.
## What is Microwave Oven?
Nowadays technology is being used in the kitchen too. This has given women all the comforts.
For example, take the microwave. From heating food to make delicious recipes in a short time, its use has become common.
A microwave oven is a very popular kitchen device used to heat and cook food.This process uses microwave radiation of the electromagnetic spectrum. Due to which, food filled with tasty and nutrients is prepared in a very short time.
## Three types of microwaves
### Solo (Basic) Microwave
In this, you can defrost frozen food, from reheating food to normal cooking.
### Grill microwave
It has separate accessories for grilling so that you can grill everything from mutton to vegetables. The oven also serves as a grill along with a microwave. Grill microwaves have heating coils that help with roasting, browning, and toasting.
### Convection Microwave
This microwave is a mixture of standard microwave ovens and convection ovens. It has a fan to maintain the inside temperature. It spreads the hot winds evenly so that the food can cook properly. Grilling and baking of individual items can be done in this. You can also fry food if you want.
Based on a conversation with Rajiv Jain, Business Head, Home Appliances, LG Electronics India.
Microwave ovens are used for reheating food. Microwave ovens are used for reheating food we can also use them for slowly prepared cooking items such as cream, chocolate,cheese, butter, etc.
A microwave is a form of electromagnetic waves whose wavelengths range from 1 meter to 1 millimeter. Their frequencies are 300 MHz (100 cm) and 300 GHz (0.1 cm).
## Top Uses of Microwave Oven
### 1. Reheating -
You can use the Microwave Oven to re-heat the food very easily. The number one use of a microwave oven is to reheat the food.
Therefore microwave oven is used more in homes and bakery shops, for heating different pastries, cakes, and other food items.
### 2. For Baking -
You can use a microwave oven to bake cakes and bread. It is very easy to use and very soon you can make delicious varieties of cakes.
### 3. To defrost -
Microwave ovens are used to defrost frozen vegetables and meats.It is also very efficient because it is also helpful in maintaining food in its original form and nutrients.
### 4. To Roasting -
You can roast garlic, garlic bread, popcorn, etc. immediately from the microwave oven. The process of roasting food is very common and it is a traditional method of cooking.
### 6. Also for boiling and steaming –
Boiling is a very important way to cook food.A microwave boils food for a few minutes, whereas traditional boiling techniques have a lot of energy wastage and it takes a lot of time as well.
## Microwave oven safety precautions
While using a microwave it is necessary to keep in mind some special points, which are described below.Use the oven pad when removing cooked food from inside the microwave, because the pot is very hot at this time.
1. Follow the manufacturer's instruction user manual for recommended operating procedures and safety precautions.
If you do not know how to use a microwave, then avoid using any experiment and read the manual guide carefully.
For instance, Use 'microwave-safe' glasses or ceramic utensils for cooking, Use the metal utensil in the grill or convection mode, etc.
2. Keep the oven door clean, so that you don't have to open it again and again.Do not use an oven if the door of your microwave oven is not firmly closed or is twisted, deformed. Warming can cause an explosion.The Food and Drug Administration recommends that stop using the microwave oven immediately and calls a repairman.
### 3. Water-Overheating is likely to damage the oven
If the water is heated in a microwave oven for a long time i.e beyond its boiling temperature, the bubbles that come out when the water boils will spoil the oven. Avoid heating water even in a cup.
### 4. Milk
According to a study by Stanford University, California, heating milk in a microwave oven destroys the nutrients in it, especially proteins. Milk should not be heated even at low temperatures (20-53 degrees C).
### 5.The vegetables -Nutrients will destroy, do not keep broccoli at all
Vegetables should be avoided in the oven. Microwaving destroy 90 percent of the nutritional elements of vegetables.When these vegetables are cooked on gas, only 10 percent of the nutrients are destroyed.
Do not cook the broccoli in the microwave oven at all. Cooking broccoli in the oven eliminates 97 percent of the antioxidants while cooking it on gas reduces nutrient content by only 11 percent.
### 6. Eggs
Do not make the mistake of putting eggs in the microwave. This will not make your work easier but will increase because it will explode when heated in the oven. This way the egg will also explode and it will take time to clean the oven.
### 7. Frozen Meat-There will be the possibility of bacteria
When frozen meat is cut and kept in the microwave, its edges are cooked while the middle part remains uncooked due to thickening.
It is also more likely to remain bacteria.According to a study by Penn State University Department of Food Science solids foods partially cook in the microwave, so we would suggest foods such as meats, do not put in the microwave.
### 8. Aluminum foil
Keeping the food covered inside the aluminum foil in the microwave oven burns the foil. This can also spoil the oven.
When something is cooked by placing it in an aluminum foil, the quantity of aluminum is also found in that food. It works as a toxic by continuously accumulating in the body. It has a bad effect on health.
### 9.Plastic Container
Not all plastic containers are microwave safe. Use containers that are microwave safe.Avoid heating fat foods in plastic containers, even if the container is microwave safe.
If plastic containers of curd or cream are kept in the microwave, they will melt and the chemicals present in them will spoil your food. The truth is that 95 percent of plastic containers such as baby bottles, zipper top bags, etc. should not be kept in the oven.
### 10. Paper lunch boxes
Paper lunch boxes or bags should also not be kept in a microwave oven. Such bags contain ink, glue and recycled material which upon heating emits toxic fumes, which is not good for health.
### 9. Bone china plates
Bone china or metallic plates should be avoided in a microwave oven. Such plates do not microwave safe.
## Save energy
• If not working, switch off the microwave.
• Keep the oven door clean, so that it does not have to open again and again. Open repeatedly causes energy loss. | 2,016 | 9,286 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2024-22 | longest | en | 0.877827 |
https://everything2.com/user/zaunuz/writeups/Does+Santa+exist%253F | 1,531,948,106,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00433.warc.gz | 632,321,728 | 6,453 | Surprisingly enough, I have never seen scientifical proof that denies the existence of Santa Claus. So I figured it is time to show some simple calculations that does this.
First we have to calculate the distance Santa and his flock has to travel. In other words, calculate the average distance between every human.
In July 2000 there were 6080671215 people on earth. Since there has to be just as much Evil as there is Good in the world, Santa will only visit half of these. This eases Santa's job, since there now are only 3040335607 people left.
To calculate the average speed he has to maintain is simple:
```510072000000km / 24h
= 21.253.000.000kmh
```
Which leads to the following conclusion: Santa would vaporize the second he started to accelerate. The friction caused by this speed would also heat up the air so much that the surface of the earth would probably melt. And this calculation does not even include the time he needs to land the sleigh, climb down the chimney, unload gifts, eat cookies, drink milk, climb up the chimney again, and take off.
But what if there were more than just Santa? for instance, one million Santa?s (MegaSanta) who split the work between them? 2 seconds of thinking should lead to the conclusion that one MegaSanta would occupy too much space in the air, and if there were one MegaSanta, all of these Santa's would still need to travel 21.253Km/h. A Gigasanta has to travel at 21Km/h to cover the distance, which is very slow, but still, the space needed for one billion Santas to fly safely is more than the atmosphere can hold. Yes, Santa may also orbit around the earth, but how would 1 billion fat and jolly men, a bunch of helpfull elves, and 8 billion reindeer fit onto the North Pole? | 408 | 1,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2018-30 | latest | en | 0.978625 |
http://tuxedolabs.blogspot.com/2017/11/upscaling-half-resolution-screen-space.html | 1,561,023,492,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999200.89/warc/CC-MAIN-20190620085246-20190620111246-00142.warc.gz | 187,992,273 | 22,946 | Skip to main content
### Upscaling half resolution screen space effects
When working with diffuse lighting and ambient occlusion in screen space it is often very tempting to do computations in lower resolution. Most of it is blurry anyway, and for any kind of GI/path tracing, diffuse lighting is undoubtedly the bottleneck. Here is a test scene with all colours set to white and no textures.
Enabling only the diffuse lighting, the image looks strangely familiar.
You quickly realise that diffuse lighting is the lion's share of the entire image. Since everything is the same colour, two overlapping objects can be told apart only because they differ in diffuse lighting. Therefore, lowering the resolution of diffuse lighting also means that a lot of edges will be half resolution and the same diffuse lighting suddenly looks like this.
Not acceptable (click on image to view full resolution), but note that the image looks perfectly fine over larger areas where there are no edges, and also at the contours towards the skybox. I've come to think of two solutions to this problem:
1) Render at half resolution. Detect edges and re-render pixels near edges during upsampling. This would probably work very well, but I didn't try it yet.
2) A cheaper solution would be to cover up faulty pixels on the edges using neighbouring pixels from the same surface (it's all blurry, remember?), practically retouching the edges much the same way you retouch images in photoshop.
I decided to try the latter and got some interesting results. First I create a 2D "retouching" vector field. It is basically just a distance offset, telling each pixel where to fetch it's samples. In the middle of a surface this will be (0,0) and near an edge it will point away from the edge. If you have any way of classifying surfaces in a shader this is actually really cheap to do. I just use a unique number for each smoothing group to identify smooth surfaces and for each pixel, I check the eight neighboring pixels, average the offset of the ones that are in the same smoothing group. Ta-da, the average offset will now point in a direction away from each edge, and the retouch vector field looks something like this (here visualized upscaled and with absolute values):
Now if you process the downscaled, half resolution, diffuse lighting through this retouch field during upscaling, the resulting image will magically look like this:
Congratulations, you just saved ~75% processing time for your diffuse lighting. However, there are artifacts, as always. But I found the results to be acceptable in most situations. Computing diffuse lighting in half resolution (quarter pixel count) allowed me to do eight samples per pixel instead of two, resulting in more accurate lighting and less noise.
Another really nice property of the retouch vector field is that once you've created it, you can reuse the same field for any screen space upscaling you might do. I for instance reuse the same field when upscaling screen space reflections, and I'm hoping to use it also for smoke particles once I get there.
### Bokeh depth of field in a single pass
When I implemented bokeh depth of field I stumbled upon a neat blending trick almost by accident. In my opinion, the quality of depth of field is more related to how objects of different depths blend together, rather than the blur itself. Sure, bokeh is nicer than gaussian, but if the blending is off the whole thing falls flat. There seems to be many different approaches to this out there, most of them requiring multiple passes and sometimes separation of what's behind and in front of the focal plane. I experimented a bit and stumbled upon a nice trick, almost by accident.
I'm not going to get into technical details about lenses, circle of confusion, etc. It has been described very well many times before, so I'm just going to assume you know the basics. I can try to summarize what we want to do in one sentence – render each pixel as a discs where the radius is determined by how out of focus it is, also taking depth into consideration "somehow".
Taking depth into…
### Undo for lazy programmers
I often see people recommend the command pattern for implementing undo/redo in, say, a level editor. While it sure works, it's a lot of code and a lot of work. Some ten years ago I came across an idea that I have used ever since, that is super easy to implement and has worked like a charm for all my projects so far.
Every level editor already has the functionality to serialize the level state (and save it to disk). It also has the ability to load a previously saved state, and the idea is to simply use those to implement undo/redo. I create a stack of memory buffers and serialize the entire level into that after each action is completed. Undo is implemented by walking one step up the stack and load that state. Redo is implemented in the same way by walking a step down the stack and load.
This obviously doesn't work for something like photoshop unless you have terabytes of memory laying around, but in my experience the level information is usually relatively compact and seriali…
### Stratified sampling
After finishing my framework overhaul I'm now back on hybrid rendering and screen space raytracing. My first plan was to just port the old renderer to the new framework but I ended up rewriting all of it instead, finally trying out a few things that has been on my mind for a while.
I've been wanting to try stratified sampling for a long time as a way to reduce noise in the diffuse light. The idea is to sample the hemisphere within a certain set of fixed strata instead of completely random to give a more uniform distribution. The direction within each stratum is still random, so it would still cover the whole hemisphere and converge to the same result, just in a slightly more predictable way. I won't go into more detail, but full explanation is all over the Internet, for instance here.
Let's look at the difference between stratified and uniform sampling. To make a fair comparison there is no lighting in these images, just ambient occlusion and an emissive object.
They … | 1,250 | 6,154 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-26 | latest | en | 0.953858 |
https://www.indiabix.com/verbal-reasoning/data-sufficiency/discussion-1290 | 1,723,022,663,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640690787.34/warc/CC-MAIN-20240807080717-20240807110717-00724.warc.gz | 641,109,472 | 8,055 | # Verbal Reasoning - Data Sufficiency - Discussion
Discussion Forum : Data Sufficiency - Section 1 (Q.No. 41)
Directions to Solve
In each of the questions below consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question. Read both the statements and
• (A) If the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question
• (B) If the data in statement II alone are sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question
• (C) If the data either in statement I alone or in statement II alone are sufficient to answer the question
• (D) If the data given in both statements I and II together are not sufficient to answer the question and
• (E) If the data in both statements I and II together are necessary to answer the question.
41.
Question: What is Nitin's rank from the top in a class of forty students ?
Statements:
1. There are ten students between Nitin and Deepak.
2. Deepak is twentieth from the top.
I alone is sufficient while II alone is not sufficient
II alone is sufficient while I alone is not sufficient
Either I or II is sufficient
Neither I nor II is sufficient
Both I and II are sufficient
Explanation:
Since there are ten students between Nitin and Deepak, so Nitin may be eleven ranks above or below Deepak. Thus, Nitin may be 9th or 31st from the top.
Discussion:
8 comments Page 1 of 1.
Anil said: 2 years ago
@Sushmitha.
But in question clearly represents 10 students between Deepak and Nitin. In question, Deepak is first, So clearly Deepak comes first.
Aman Thakur said: 3 years ago
Answer may be E because Nitin to Deepak means Nitin will come first if Deepak's position is 20th.
Chanukya said: 3 years ago
The answer is D.
Nitin should be either 1st or 23rd.
If Deepak is 12th from the top, and there are 10 students in between, Nitin is either 1 (Nitin, 10 people, Deepak) or 23 (Deepak, 10 people, Nitin).
Shubham said: 3 years ago
The answer should be E.
Because it's clearly saying that, Deepak is twentieth from the top and 10 students are in between Nitin and Deepak.
So, Nitin rank is 9.
Sushmitha said: 7 years ago
In the question, they have mentioned that there are 10 students between Nithin and Deepak but they haven't said who is above whom (it is unclear whether Nithin is above Deepak or Deepak is above Nithin).
So the answer is D.
Shreya said: 9 years ago
@KUNBI.
Since it is not given who is at what rank, therefore we cannot analyse whether 10 students are between Deepak and Nitin or Nitin and Deepak respectively.
Kunbi said: 9 years ago
I agree with @Deepak; both information are sufficient in my analysis. Please explain further.
Deepak said: 1 decade ago
IT is mentioned that Deepak is in twentieth from the top N There are ten students between Nitin and Deepak. So Nitin should be in 9th rank. And Answer(E). | 772 | 3,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2024-33 | latest | en | 0.918975 |
https://math.stackexchange.com/questions/1812785/does-there-exist-a-positive-irrational-number-alpha-such-that-for-any-posit | 1,632,593,561,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057733.53/warc/CC-MAIN-20210925172649-20210925202649-00124.warc.gz | 428,519,863 | 38,714 | # Does there exist a positive irrational number $\alpha$, such that for any positive integer $n$ the number $\lfloor n\alpha \rfloor$ is not a prime?
Does there exist a positive irrational number $\alpha$, such that for any positive integer $n$ the number $\lfloor n\alpha \rfloor$ is not a prime?
My try if $\alpha=\sqrt{17}$ then $\lfloor n\alpha \rfloor=4n$
• That try will fail: $\lfloor 9\sqrt{17} \rfloor=37$
– tomi
Jun 4 '16 at 21:29
• $\lfloor n\alpha \rfloor$ is definitely not expressible in the form $cn$ when $\alpha$ is irrational. Jun 4 '16 at 21:40
• you can express finite sub-sequences of $x_n = \lfloor n \alpha \rfloor$ as $y_n = c n+d$, I think all the point is to prove that some of those have $gcd(c,d ) = 1$, and that they are large enough for applying the Dirichlet theorem in arithmetic progressions. Jun 4 '16 at 22:11
## 1 Answer
This is called a Beatty sequence. Here is an arxiv paper on the least prime in a Beatty sequence, by Steuding and Technau. There will indeed always be a prime in the sequence (which answers the original question), and the cited paper gives an upper bound for the least such prime for $\alpha>1$. The bound for the OP's sequence (provided $\alpha>1$) is $$p\le L^{35-16\epsilon}\alpha^{2(1-\epsilon)}p^{1+\epsilon}_{m+l}$$
where $L=\log(2\alpha)$, $p_n$ denotes the numerator of the $n^\text{th}$ convergent to the regular continued fraction expansion of $\alpha$, and $m$ is the unique integer such that $p_m\le L^{16}\alpha^2<p_{m+1}$. $\epsilon$ can be chosen arbitrarily small, but $l$ depends on $\epsilon$.
• one of the main trick is at the beginning of page 3 : "This means that the proportion of fractional parts $\{\alpha p\} = \alpha p − \lfloor \alpha p\rfloor$ which fall in an interval $[a, b) \subset [0, 1)$ is equal to $b − a$ (the length of the interval) " and from there can we apply the Dirichlet theorem on arithmetic progressions ? there is also, just before " the sequence of numbers $\alpha p$, where $p$ runs through the prime numbers in ascending order, is uniformly distributed modulo one" Jun 4 '16 at 22:24
• I found this for a more succinct explanation xvienemsalamanca.anemat.com/wp-content/uploads/2015/10/… Jun 4 '16 at 22:33 | 643 | 2,219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2021-39 | latest | en | 0.843272 |
https://byjus.com/question-answer/solve-frac-5-x-2-3-and-represent-the-solution-set-on-the-number-line/ | 1,718,509,823,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861640.68/warc/CC-MAIN-20240616012706-20240616042706-00527.warc.gz | 128,195,214 | 27,921 | 1
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Question
# Solve 5x−2>3 and represent the solution set on the number line. Or Solve |x|<4 and represent the solution set on the number line.
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Solution
## (i) 5x−2>3⇒5x−2−3>3−3⇒5−3x+6x−2>0⇒11−3xx−2>0Case 1: 11−3x>0 and x−2>0⇒11>3x and x>2⇒x<113 and x>2Therefore, solution set is (2,113)Case 2:11−3x<0 and x−2<0⇒11<3x and x<2⇒x>113 and x<2Therefore, solution set is ϕHence, the solution set of the given equation =(2,113)∪ϕ=(2,113) (ii) |x|<4⇒−4<x<4 ∴ Solution set of the given equation is (−4,4)
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Join BYJU'S Learning Program | 287 | 706 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-26 | latest | en | 0.762571 |
https://socratic.org/questions/how-do-you-find-the-cube-root-of-0-000343 | 1,623,896,989,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487626465.55/warc/CC-MAIN-20210617011001-20210617041001-00242.warc.gz | 483,609,299 | 5,828 | # How do you find the cube root of 0.000343?
Jun 26, 2016
$\sqrt[3]{0.000343} = 0.07$
#### Explanation:
Note that $0.000343 = 343 \cdot {10}^{- 6}$ and $343 = {7}^{3}$
So:
$\sqrt[3]{0.000343} = \sqrt[3]{343 \cdot {10}^{- 6}} = \sqrt[3]{343} \cdot \sqrt[3]{{10}^{- 6}}$
$= \sqrt[3]{{7}^{3}} \cdot {10}^{- 2} = 7 \cdot {10}^{- 2} = 0.07$
Jun 26, 2016
$0.07$
#### Explanation:
Group the decimals in three's from the decimal.
Each group must have one number.
$\sqrt[3]{\text{0.000 343}}$
$\sqrt[3]{0.000} = 0$
$\sqrt[3]{343} = 7$
$\sqrt[3]{\text{0.000 343}} = 0.07$
Check: 2 decimal places cubed will give 6 decimal places. | 271 | 635 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2021-25 | latest | en | 0.464408 |
https://docs.microsoft.com/en-gb/dotnet/api/system.windows.media.media3d.point4d.equals?view=net-5.0 | 1,623,535,092,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586390.4/warc/CC-MAIN-20210612193058-20210612223058-00419.warc.gz | 225,946,043 | 10,340 | # Point4D.Equals Method
## Definition
Compares two Point4D structures for equality.
Equals(Object) Determines whether the specified Object is a Point4D structure and if the X, Y, Z, and W properties of the specified Object are equal to the X, Y, Z, and W properties of this Point4D structure. Equals(Point4D) Compares two Point4D structures for equality. Equals(Point4D, Point4D) Compares two Point4D structures for equality.
## Equals(Object)
Determines whether the specified Object is a Point4D structure and if the X, Y, Z, and W properties of the specified Object are equal to the X, Y, Z, and W properties of this Point4D structure.
``````public:
override bool Equals(System::Object ^ o);``````
``public override bool Equals (object o);``
``override this.Equals : obj -> bool``
``Public Overrides Function Equals (o As Object) As Boolean``
#### Parameters
o
Object
The object to compare.
#### Returns
Boolean
`true` if instances are equal; otherwise, `false`.
`true` if `o` (the passed Object) is a Point4D structure and is identical with this Point4D structure; `false` otherwise.
### Examples
The following example shows how to check whether an Object and a Point4D structure are equal.
``````// Compares an Object and a Point4D for equality using the non-static Equals method.
Point4D point1 = new Point4D(10, 5, 1, 4);
Point4D point2 = new Point4D(15, 40, 60, 75);
Boolean areEqual;
areEqual = point1.Equals(point2);
// areEqual is False. point2 is a Point4D structure, but it is not equal to point1.
// Displaying Results
syntaxString = "areEqual = point1.Equals(point2);";
resultType = "Boolean";
operationString = "Checking if two 3D points are equal";
ShowResults(areEqual.ToString(), syntaxString, resultType, operationString);
``````
``````' Compares an Object and a Point4D for equality using the non-static Equals method.
Dim point1 As New Point4D(10, 5, 1, 4)
Dim point2 As New Point4D(15, 40, 60, 75)
Dim areEqual As Boolean
areEqual = point1.Equals(point2)
' areEqual is False. point2 is a Point4D structure, but it is not equal to point1.
' Displaying Results
syntaxString = "areEqual = point1.Equals(point2)"
resultType = "Boolean"
operationString = "Checking if two 3D points are equal"
ShowResults(areEqual.ToString(), syntaxString, resultType, operationString)
``````
### Remarks
If `o` (the passed object) is not a Point4D structure, this method returns `false`.
Because a Double value can lose precision when arithmetic operations are performed on them, a comparison between two Point4D values that are logically equal might fail.
## Equals(Point4D)
Compares two Point4D structures for equality.
``````public:
bool Equals(System::Windows::Media::Media3D::Point4D value);``````
``public bool Equals (System.Windows.Media.Media3D.Point4D value);``
``override this.Equals : System.Windows.Media.Media3D.Point4D -> bool``
``Public Function Equals (value As Point4D) As Boolean``
#### Parameters
value
Point4D
The instance of Point4D to compare to this instance.
#### Returns
Boolean
`true` if instances are equal; otherwise, `false`.
### Remarks
Because a Double value can lose precision when arithmetic operations are performed on them, a comparison between two Point4D values that are logically equal might fail.
## Equals(Point4D, Point4D)
Compares two Point4D structures for equality.
``````public:
static bool Equals(System::Windows::Media::Media3D::Point4D point1, System::Windows::Media::Media3D::Point4D point2);``````
``public static bool Equals (System.Windows.Media.Media3D.Point4D point1, System.Windows.Media.Media3D.Point4D point2);``
``static member Equals : System.Windows.Media.Media3D.Point4D * System.Windows.Media.Media3D.Point4D -> bool``
``Public Shared Function Equals (point1 As Point4D, point2 As Point4D) As Boolean``
#### Parameters
point1
Point4D
The first Point4D structure to compare.
point2
Point4D
The second Point4D structure to compare.
#### Returns
Boolean
`true` if the X, Y, and Z components of `point3D1` and `point3D2` are equal; `false` otherwise.
### Examples
The following example shows how to check whether two Point4D structures are equal.
``````// Checks if two Point4D structures are equal using the static Equals method.
// point1's x,y,z,w properties set when the structure is created
Point4D point1 = new Point4D(10, 5, 1, 4);
Point4D point2 = new Point4D();
Boolean areEqual;
// settting point2's x,y,z,w properties
point2.X = 15;
point2.Y = 40;
point2.Z = 60;
point2.W = 75;
areEqual = Point4D.Equals(point1, point2);
// areEqual is False
//Displaying Results
syntaxString = "areEqual = Point4D.Equals(point1, point2);";
resultType = "Boolean";
operationString = "Checking if 3D two points are equal";
ShowResults(areEqual.ToString(), syntaxString, resultType, operationString);
``````
``````' Checks if two Point4D structures are equal using the static Equals method.
' point1's x,y,z,w properties set when the structure is created
Dim point1 As New Point4D(10, 5, 1, 4)
Dim point2 As New Point4D()
Dim areEqual As Boolean
' settting point2's x,y,z,w properties
point2.X = 15
point2.Y = 40
point2.Z = 60
point2.W = 75
areEqual = Point4D.Equals(point1, point2)
' areEqual is False
'Displaying Results
syntaxString = "areEqual = Point4D.Equals(point1, point2)"
resultType = "Boolean"
operationString = "Checking if 3D two points are equal"
ShowResults(areEqual.ToString(), syntaxString, resultType, operationString)
``````
### Remarks
Two Point4D structures are equal if the values of their X, Y, Z, and W properties are the same.
Because Double values can lose precision when arithmetic operations are performed on them, a comparison between two Point4D instances that are logically equal might fail. | 1,511 | 5,740 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2021-25 | latest | en | 0.739807 |
https://www.fr.freelancer.com/projects/c-programming-cad-cam/step-iso-spline-information/ | 1,508,773,210,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187826114.69/warc/CC-MAIN-20171023145244-20171023165244-00882.warc.gz | 959,757,375 | 29,024 | # STEP AP203 (ISO 10303-203) B-Spline Information
What I want:
* Information about STEP AP203 (ISO 10303-203) Schema
* Specifically, what are the parameters required for b_spline_curve_with_knots and a b_spline_surface_with_knots
* The schema are available at [url removed, login to view]
* How do the knots, control points and weights work?
* Examples:
If I have a conic curve (ellipse, parabola, hyperbola), how do I specify the corresponding b_spline_curve_with_knots?
How do I specify a surface that is a conic of revolution (ellipsoid, paraboloid, hyperboloid) in STEP AP203?
If I have a conic surface that is bounded by an off-axis cylinder, how do I specify the boundary curves?
If I have a conic surface that is bounded by a rectangle, how do I specify the boundary curves?
Deliverables
1. A written description of how to create (or a well commented C function for creating) a conic of revolution as a b_spline_surface_with_knots (or something more appropriate) in STEP AP203 based on the vertex curvature and eccentricity or conic constant. The formula for my conics is: z=(c*y^2)/(1+sqrt(1-(1+k) c^2 y^2) where c is the vertex curvature and k is the conic constant (k=-e^2), where e is the eccentricity (0=sphere, 1=parabola). Assume that only c, k and the maximum value of y are known. If the parameters are specified in millimeters, the surface must be accurate to 0.1 micron.
2. A written description of how to create (or a well commented C function for creating) a STEP AP203 b_spline_surface_with_knots from a plane curve rotated about an axis. The plane curve is given as a series of points. If the parameters are specified in millimeters, the surface must be accurate to 0.1 micron.
3. A written description of how to create (or a well commented C function for creating) STEP AP203 bounding curves (face_outer_bound + edge_loop + oriented_edge + edge_curve) for both of the above surfaces. The edge is generated by a cylinder with an axis parallel to (but not coincident with) the axis of rotation of the surface. You may assume the ability to generate as many points as needed, but the curve must be accurate to 1 micron.
4. A written description of how to create (or a well commented C function for creating) a STEP AP203 bounding curve (face_outer_bound + edge_loop + oriented_edge + edge_curve) for both of the above surfaces. The edge is generated by a plane parallel to (but not coincident with) the axis of rotation of the surface. You may assume the ability to generate as many points as needed, but the curve must be accurate to 1 micron.
Compétences : Algorithme, Programmation C, CAD/CAM, Mathématiques, Génie Mécanique
Concernant l'employeur :
( 1 commentaire ) White Bear Lake, United States
N° du projet : #1075776
## 3 freelance ont fait une offre moyenne de 700 \$ pour ce travail
bchandra1955
I can generate B_spline curves and surfaces using mathmatical formulation and the I will code in C or MATLAB, but how will you benchmark my results as per your accurace? !!! SEE INBOX , Brajesh
725 \$ USD en 15 jours
(46 Commentaires)
5.2
Raghavanps
See private message
625 \$ USD en 20 jours
(7 Commentaires)
4.5
adamcurrier
I am a mechanical engineer and my thessis was mostly about NURBS. I would like to do the job with further explanation.
750 \$ USD en 25 jours
(0 Commentaires)
0.0 | 850 | 3,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2017-43 | longest | en | 0.819367 |
https://medical-dictionary.thefreedictionary.com/ANOVA | 1,552,985,065,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912201922.85/warc/CC-MAIN-20190319073140-20190319095140-00104.warc.gz | 567,821,336 | 12,434 | # ANOVA
Also found in: Dictionary, Thesaurus, Financial, Acronyms, Encyclopedia.
## ANOVA
abbreviation for analysis of variance.
## ANOVA
Analysis of variance, see there.
## ANOVA
(ā-nōvă)
Acronym for analysis of variance.
## ANOVA
analysis of variance
## anova
see analysis of variance.
References in periodicals archive ?
Statistical analysis of DPPH radical scavenging activity of Saccharum officinarum roots extracts was performed by using one way ANOVA which showed significant results (P<0.
If using the standard ANOVA, one would still need to standardize data for the homogeneity and/or outlier check, so the first part of the calculations needed for the CV-ANOVA should be performed anyway to compare the normalized individual CVs.
Modern one-way ANOVA F methods:trimmed means one step-m estimators and bootstrap methods.
5% Girls: ANOVA test relieved for time in seconds at 50m speed running ([t.
O numero de avaliacoes necessarias, para confiabilidade de 80%, para o CI variou de 1 pelos metodos da ANOVA, CP(correl), CP(cov) e AE(cov) a 6 pela ANOVA e AE(cov), nos experimentos 3 e 2, respectivamente.
Os dados assim obtidos, foram avaliados estatisticamente utilizando-se os seguintes procedimentos estatisticos: analise de variancia (ANOVA) de um e dois fatores, ANOVA por transformacao em postos, de Kruskal-Wallis e de Friedman.
The present article extends ANOVA to a variety of different experimental designs, likely to cover many common situations in optometry and vision science.
The authors illustrate how ANOVA is employed to determine the effects of multiple variables on a single subject using such methods as post hoc and planned/weighted comparisons and ANCOVA, with plenty of examples provided to help students master statistical analyses through both SPSS and SAS interfaces.
The t-tests are good only for comparing two means, the analysis of variance (ANOVA) can compare several means, but should a difference among them exist, the ANOVA does not identify it.
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Open / Close | 489 | 2,032 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-13 | longest | en | 0.714087 |
http://mathleticism.net/?p=947 | 1,685,703,363,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00489.warc.gz | 37,377,883 | 131,522 | # Normal Probability Plots
I have a confession to make. When I first started teaching AP Statistics in 2005, I had no idea why a Normal probability plot (an example is shown to the left) was important... or what it told us about data. I busy trying to stay a day ahead of students that first year. I never really sat down with several textbooks to compare definitions and examples as I probably should have. Simply put, when students asked, I told them the canned answer: "The more linear the plot is, the more "Normal" the data is." We'd use the calculator to make the plot, look at it, and move on.
Let's take a closer look at why we study a Normal probability plot in AP Statistics. I will do some borrowing from various discussion board posts of the past on the AP Stats forum and will add some commentary as we go.
First, consider the method we use to compute a z-score; that is, a positional score for Normally distributed data that indicates the number of standard deviation units above or below the mean a particular data point lives. For example, if z = -1.2, then the data point is 1.2 standard deviations below the mean. It makes sense that a standardized score [ z = (x-μ)/σ] depends on two things: the data value's physical distance from the mean *and* the distance tempered by a measure of spread, specifically the standard deviation. Let's isolate x in this equation to see what happens.
The algebra above is commonly used in problems where we are asked to find a score which corresponds to a particular percentile rank. For example, if the mean score of the ACT is 18, and the standard deviation is 6, then what composite score puts a student in the 70th percentile of all test takers that day? A score slightly north of 21, as shown below.
The InvNorm command above finds the z-score corresponding to a cumulative area of .70 under the standard Normal curve, which has mean 0 and standard deviation 1. We see a z-score of .5244005101, according to the TI-84, gives the position for a data point in the 70th percentile. We can then reverse engineer the score needed to fall into this percentile.
In the world outside school, it's usually not likely we know the actual value of σ, the population standard deviation, or μ, the actual population mean. As unbiased estimators of these unknown values, we use , the sample mean, in place of μ, and we use s, the sample standard deviation, in place of σ. Then the value of x looks like Technically, once we make the substitutions, we would really be using a t-distribution of some flavor to model the data. On the other hand, in the example below, since we can get data on every qualified point guard in the NBA as of right now, we can directly compute the mean and standard deviation for the entire population, making this substitution unnecessary in this case. However, students need to be aware of the need for t-procedures.
To show an example of a Normal probability plot, I pulled NBA data from ESPN regarding point guard performance thus far in the 2013-14 regular season. Let's take a look at the top 26 (since there's a tie for 25th place) point guards in the NBA with respect to average points scored per game, the gray column labeled "PTS."
Let's enter the data from the table above in the TI-84.
Next, let's construct the Normal probability plot for the data.
So... what exactly does this plot represent? And what makes this representation so important? The x-values obviously correspond to the average points per game value for each point guard. What about the y-coordinate of each point on the graph? The y-coordinate corresponds to the z-score related to that particular x-value. In the screen shot above, Kemba Walker, the point guard with 18.6 points per game, has a z-score of approximately .7039. If the data followed exactly a Normal curve, then all the points on the above graph would lie exactly on a straight line. By looking at the z-score for each data point using this display, we can get a quick insight into whether the data are Normally distributed. Let's look at a boxplot for the same data:
We can see, in the plot above, the data for these 26 point guards have no outliers, but there appears to be some skewness. Computing the values (Max - Q3) = 4.4 and (Q1 - Min) = 10.8 - 9.3 = 1.5 and 4.4 > 1.5, we can demonstrate this skewness. This numeric argument doesn't take a lot of calculator kung fu, but we do have to perform an extra computation or two. Looking back at the Normal probability plot, we could use the image to immediately notice the skewness of the data. Suppose we graphed the original z-score equation [z = (x-μ)/σ] on the same graph as the Normal probability plot. In other words, we will make the Normal probability plot. Take a look!
We only used 26 data points, so the data is a sample of the population of NBA point guards. Again, if the data were perfectly Normal, all the blue points would be living directly on the red line. We can use our knowledge of linear equations to see clearly what's going on here.
So the slope of this red line representing the 'perfectly' Normal data has slope 1/4.271785124. Let's find an equivalent value that's slightly more user friendly:
If we express this value as
notice we can say for every additional unit increase in x, the average points scored per game, we expect to see a z-score increase of .2340941716. Much like when we consider residuals while doing linear regression, when x-values deviate noticeably from the expected red line, they are surprising from the "Normal curve's point of view." The curvature at the left end of the Normal probability plot immediately indicates the skewness of the data. You can find more examples of this on your favorite search engine by asking for "Normal probability plot skewness." If we know how to visually recognize this pattern, we can immediately recognize skewness of data using a Normal probability plot.
This connection between the Normal distribution and why its z-scores are linear has a pretty good explanation on the Wikipedia entry for "Standard score."
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,364 | 6,156 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2023-23 | latest | en | 0.940779 |
yaneurao.hatenadiary.com | 1,582,603,398,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875146004.9/warc/CC-MAIN-20200225014941-20200225044941-00313.warc.gz | 945,929,076 | 9,539 | # やねうらおブログ(移転しました)
もう疲労限界maxなので寝てから明日の昼とかでいいでつか？
「でつか」の「でつ」の部分がスヌーピーに見えるのでつけど、
スヌーピーかよ！って突っ込んでもらっていいでつか？
やっぱり人様の手を煩わせるのも何なので、自分で突っ込んでおきます
でつ ＜ スヌーピーかよ！
```
Your company is working on a new transmission for a high-performance vehicle.
The transmission will have two sets of gears, a front set and a back set.
The numbers of teeth on each of the front gears has already been set,
but now you need to figure out how many teeth to put on all of the back gears.
It has already been decided that there will be cnt back gears.
If the front set of gears contains N gears, then there are a total of N*cnt
gear ratios. Each gear ratio is determined by the ratio of the number of
teeth on one of the front gears divided by the number of teeth on one of
the back gears. Your task is to determine how many teeth to put on each
of the cnt back gears in such a way that the largest gear has max teeth,
and the smallest gear has min teeth. Furthermore, if all of the possible
gear ratios from your assignment are sorted, you want the maximum difference
between two adjacent ratios in the sorted list to be as small as possible.
You are to return the largest difference between adjacent ratios in this
sorted list.Definition
Class:
Transmission
Method:
gears
Parameters:
vector , int, int, int
Returns:
double
Method signature:
double gears(vector front, int min, int max, int cnt)
(be sure your method is public)
Notes
-
Each gear must have an integral number of teeth.
Constraints
-
cnt will be between 3 and 5, inclusive.
-
min will be between 5 and 100, inclusive.
-
max will be between min+cnt-1 and 100, inclusive.
-
front will contain between 1 and 5 elements, inclusive.
-
Each element of front will be between 5 and 100, inclusive.
Examples
0)
{10,50,10}
5
20
3
Returns: 3.75
The best we can do here is to make the back 3 gears have 5, 8 and 20 teeth.
This gives the following list of ratios: {0.50, 1.25, 2.00, 2.50, 6.25, 10.00}
The biggest gap is between 10 and 6.25, so we return 3.75.
1)
{10,80}
10
50
5
Returns: 1.73333333333333342)
{5,9,30,100}
10
100
5
Returns: 1.904761904761905``` | 644 | 2,089 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2020-10 | longest | en | 0.854304 |
http://www.statisticshowto.com/market-basket-analysis/ | 1,534,345,508,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210133.37/warc/CC-MAIN-20180815141842-20180815161842-00417.warc.gz | 591,528,778 | 9,856 | # Market Basket Analysis: Definition, Examples
Statistics Definitions > Market Basket Analysis
## What is Market Basket Analysis?
In market basket analysis (also called association analysis or frequent itemset mining), you analyze purchases that commonly happen together. For example, people who buy bread and peanut butter also buy jelly. Or people who buy shampoo might also buy conditioner. What relationships there are between items is the target of the analysis. Knowing what your customers tend to buy together can help with marketing efforts and store/website layout.
Market basket analysis isn’t limited to shopping carts. Other areas where the technique is used include analysis of fraudulent insurance claims or credit card purchases.
Market basket analysis can also be used to cross-sell products. Amazon famously uses an algorithm to suggest items that you might be interested in, based on your browsing history or what other people have purchased.
A popular urban legend is that a grocery store, after running market basket analysis, found that men were likely to buy beer and diapers together. Sales increased sales by placing beer next to the diapers.
It sounds simple (and in many cases, it is). However, pitfalls to be aware of:
• For large inventories (i.e. over 10,000), the combination of items may explode into the billions, making the math almost impossible.
• Data is often mined from large transaction histories. A large amount of data is usually handled by specialized statistical software (see below).
## Basic Terminology
An itemset is the set of items a customer buys at the same time. It’s typically stated as a logic rule like IF {bread, peanut butter} THEN {jelly}. An itemset can consist of no items (a null amount though, is usually ignored) to all items in the data set.
The support count is a count of how often the itemset appears in the transaction database. The support is how often the item appears, stated as a probability. For example, if the support count is 21 out of a possible 1,000 transactions, then the probability is 21/1,000 or 0.021.
The confidence is the conditional probability that the items will be purchased together.
## Calculations
Calculations are rarely performed by hand, due to large number of combinations possible from even relatively small datasets. Software that can perform market basket analysis include:
------------------------------------------------------------------------------
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If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track. | 595 | 3,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-34 | latest | en | 0.952855 |
https://www.weegy.com/Home.aspx?ConversationId=95B59EFB | 1,537,501,359,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156780.2/warc/CC-MAIN-20180921033529-20180921053929-00070.warc.gz | 933,593,155 | 8,223 | What does inactive mean?
Question
Updated 8/14/2014 12:09:45 PM
Flagged by alfred123 [8/14/2014 12:05:30 PM], Edited by yumdrea [8/14/2014 12:09:39 PM], Edited by yumdrea [8/14/2014 12:09:45 PM]
Original conversation
User: What does inactive mean?
Question
Updated 8/14/2014 12:09:45 PM
Flagged by alfred123 [8/14/2014 12:05:30 PM], Edited by yumdrea [8/14/2014 12:09:39 PM], Edited by yumdrea [8/14/2014 12:09:45 PM]
Rating
3
Inactive as an adjective means not doing things that require physical movement and energy.
Confirmed by yumdrea [8/14/2014 12:09:26 PM]
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Multiply (2x + 8)(4x + 7).
Weegy: 3(4x + 2x) = 8; User: Multiply (2m + 3)(m2 – 2m + 1).
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Weegy: The extremes of the proportion 3/15 = 12/60 are 3 and 60.
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2012-01-01, 21:49 #40
science_man_88
"Forget I exist"
Jul 2009
Dumbassville
202618 Posts
Quote:
Originally Posted by science_man_88 got one that skips some more than it's supposed to: Code: (w,s)->print("digraph G {");forprime(z=w,s,print("\""z"p\"""[shape=plaintext];"));print();forprime(y=w,s,if(#ali(y)>0,for(x=1,#ali5(y),print(ali(y)[x]"[shape=plaintext];"))));print("}")
this work ? :
Code:
(w,s)->print("digraph G {");forprime(z=w,s,print("\""z"p\"""[shape=plaintext];"));print();forprime(y=w,s,if(#ali(y)>0,for(x=1,#ali(y),print(ali(y)[x]"[shape=plaintext];"))));aligen40(w,s);print("}")
went back to ali because ali5 although it worked on it's own never seemed to work in the code for me.
Code:
ali(n)=a=n-1;b=[];for(x=0,floor(.5*a),if(x==0,if(sigma(a^2)-a^2==n,b=concat(b,a^2)),if(sigma((x)*(a-x))-((x)*(a-x))==n,b=concat(b,(x)*(a-x)))));b=vecsort(b,,8)
Code:
aligen40(w,s)=for(z=w,s,print("\n //ends in "z);for(x=1,#ali(z),for(y=1,#ali(ali(z)[x]),print(ali(ali(z)[x])[y]"->"ali(z)[x]))))
Last fiddled with by science_man_88 on 2012-01-01 at 21:58 | 415 | 1,115 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2020-45 | latest | en | 0.715346 |
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# Geometry and Topology
### Binormal Twisted Curve
Suppose a twisted curve is defined in terms of the arclength s by r(s)=sech(?)cos(s)i+sech(?)sin(s)j+tanh(?)k, where ? is a constant parameter. Determine (i) the tangent t(s), (ii) the normal n(s), (iii) the binormal b(s), (iv) the curvature ?, (v) the torsion Ï?.
### the amplitude, period and frequency of the oscillations
Assume that the particle under the influence of an attractive force F(r), where r is the distance from a fixed point O, then it moves in a plane through O and the equation of motion is ((d^2)u/d?^2)+u=F(1/u)/(h^2)(u^2), where h is a constant, u=1/r and ? is the angle the line from O to the particle makes with some fixed direc
### Finding an Open Subset
Let g: R -> R by: g(x) = { x^2 + 2 ... if x <= 1, 5 - x ... if x > 1 } Find an open subset of R (w/ respect to the usual topology) whose preimage under g is not an open subset of R.
### Extending Continuity of Restrictions
Let X refer to a general topological space. Suppose X = A1 ï?? A2 ï?? â?¦, where An ï? Ã...n+1 for each n. If f : X --> Y is a function such that, for each n, f |An : An --> Y is continuous with respect to the induced topology on An, show that f itself is continuous. Please note that Ã...n+1 (A with a small
### Topology - open or closed susbset
Please help with the following problem. For the following, I'm trying to decide (with proof) if A is a closed subset of Y with respect to the topology, T (i) Y = N, T is the finite complement topology, A = {n e N | n^2 - 2011n+1 < 0}. (ii) Y = R, T is the usual topology, A is the set of irrational numbers between 0 and
### properties of discrete spaces
I am looking at the properties of discrete spaces, particularly this one: Every discrete space is first-countable; it is moreover second-countable if and only if it is countable. How would this be proved?
### Solve: Topological Space
Let X be a normed space with the topology induced by the norm. Show that || || : X ---> R is a continuous function on X. Please show all of your work. Thank you.
### Determination of Orthogonal Contrasts
An experiment on sugar beets compared times and methods of applying artificial (N-P-K) fertilizer, using a completely randomized design. The four treatments were: Treatment 1 - no artificials(control) Treatment 2 - artificials applied in January and ploughed into soil Treatment 3 - artificials applied in January and bro
### Is set X a Hausdorff space with this topology?
8. Let X be any set of infinite cardinality. Consider the set A = { A SUBSET X : X - A is finite } UNION { PHI , X }. a) Prove that A is a topology on X. b) Describe the neighborhoods of a point P BELONGING_TO X. c) Is X a Hausdorff space with this topology? Please see the attached image for question with appropriate sym
### Finding Characteristics of a Set with Topology Induced
Find the interior, the closure, the accumulation points, the isolated points and the boundary points of the set A = { x = (x_1, x_2 ...) e l^2 : -1 <x_1 < 1} with the topology induced by || ||_2.
### Topology induced
Question: Find the interior, the closure, the accumulation points, the isolated point and the boundary points of the following sets. a) X = [(0,1) in R with the topology induced by d(x,y) = |x-y|] b) X = Q in R with the same topology as above. c) X = {(x,y) : |y| < x^2} U {(0,y), y E R} in R^2, with the topology induced by
### Compact Subsets
Define a new metric d on X = (0, 1/2)^2 by d((a,b), (r,s)) = 1 if a is not equal to r Or |b - s| if a = r. a) Show that d is a metric on X. b) What are the compact subsets of X? Prove your statement.
### discrete topology covering map
A covering map is a map p: E -> B with the property that each point b, an element of B, has a neighborhood U such that p^-1(U) is a disjoint union of open sets V_alpha such that, for each alpha, the restriction of p to V_alpha is a homeomorphism of V_alpha onto U. Show that, if Y has the discrete topology and if p: X x Y --
### Problems exemplifying congruency
1) Suppose n belongs to Z. (a) Prove that if n is congruent to 2 (mod 4), then n is not a difference of two squares. (b) Prove that if n is not congruent to 2 (mod 4), then n is a difference of two squares. 3) Let n = 3^(t-1). Show that 2^n is congruent to -1 (mod 3^t). 5) Let p be an odd prime, and n = 2p. Show that a^(
### Compact and closed
Let S be the set [0,1] and define a subset F of S to be closed if either it is finite or is equal to S. Prove that this definition of closed set yields a topology for S. Show that S with this topology is compact, but S is not a Hausdorff space. Show that each subset of S is compact and that therefore there are compact subs
### De Morgan's Laws - FIP, Hausdorff and Compact
1.Let X be a set and T and T' are two topologies on X. Prove that if T subset of T' and (X,T') is compact, then (X,T) is compact. Prove that if (X,T) is Hausdorff and (X,T') is compact with T subset of T', then T=T'. 2.Let X be a topological space. A family {F_a} with a in I of subsets of X is said to have the finite
### Geometry
The importance of geometry's role in the math curriculum is debated in many high schools and colleges. Some schools offer the course while others have done away with it. Based on what you have learned within this unit, do you think geometry is a valuable tool for students to learn? Choose one side of this debate, state your view
### intersection of the collection of open intervals
1. Prove if F is a subset of R^n and if d(x,F)=inf(||x-z||:z in F}=0 then x belongs to F. 2.The intersection of two open sets is compact iff it is empty. Can the intersection of an infinite collection of open sets be a non-empty compact set?
### Using Differentials to Estimate
1. In the manufacturing process, ball bearings must be made with radius of 0.5mm, with a maximum error in the radius of +/- 0.016 mm. Estimate the maximum error in the volume of the ball bearing. Solution: The formula for the volumen of the sphere is _____. If an error deltaR is made in measuring the radius of the sphere
### Intervals with rational endpoints as a basis for topology
Prove that the family of open intervals with rational end points is a basis for the topology of the real line. A rigorous proof with detailed explanations is given.
### Interior of a set in a topological subspace
Let Y be a subspace of X and let A be a subset of Y. Denote by Int_X(A) the interior of A in the topological space X and by Int_Y(A) the interior of A in the topological space Y. Prove that Int_X(A) is a subset of Int_Y(A). Illustrate by an example the fact that in general Int_X(A) not Int_Y(A).
### Boundary of the n-dimensional ball
In R^n with the usual topology, let A be the set of points x=(x_1,x_2,...,x_n) such that x_1^2+x_2^2+...+x_n^2 le 1. Prove that Bdry(A) is the (n-1)-dimensional sphere S^n-1.ie. x in Bdry(A) iff x_1^2+x_2^2+...+x_n^2=1. Intuitively it is easy.. but I am not sure where to start. Probably looking at bdry(A)=Cl(A) intersect w
### Volume of the Parallelepiped
Calculate the given quantities if a = <1,1,-2>, b= <3,-2,1>, and c=<1,1,-5>. a) 2a + 3b b) |b| c) Unit vector in the direction of b. d) a (dot) b e) a x b f) comp a/b g) proj a/b h) Volume of the parallelepiped determined by a , b , and c. i) Determine whether a and b are parallel, perpendicular or
### Geometric concepts and children
How might you involve children in learning geometric concepts? Which geometric concept do you think will be most difficult for children to learn and why?
### Altitude of parallelepiped
Find the altitude of a parallelepiped determined by vectors A, B, and C if the base is taken to be the parallelogram determined by A and B and A=i+j+k B=2i+4j-k C=i+j+3k.
### Calculating the amount of material for finishing a basement
Question 11A Guest needs material to finish a room in a basement. The room is square and one wall measures 15'. The height of the room is 8'. There is one door that measures 3' wide and 7' high, and two windows that measure 3' wide and 4' high. The Guest is covering the walls with 1x6 carsiding (we will say that it covers 5"
### Parabola Vertex Equation
Find the equation of the parabola: 1. Vertex at (-2, 3), Focus at (-4, 3) 2. Vertex at (2, 4), directrix at x = -3 3. Vertex at (-1, -2), axis vertical and passing (3, 6) 4. Directrix = 4y, focus at (3, 0) 5. Axis horizontal, parabola passing through: (1, 1), (1, -3), and (-2, 0) 6. Latus rectum joining the poi
### equation in standard form.
Transform the equation to standard form and find the center and radius of the circle: 1. (x - 3)^2 = 9 - y^2 2. x^2 + y^2 = 6x - 8y 3. 2x^2 + 2y^2 = 2x + 2y - 1 4. (x + 1)^2 + (y - 2)^2 = 25
### Inductive Reasoning to Predict Numbers
Glenda Everson Math 1002 Practice Problems 1. Use inductive reasoning to predict the next three numbers in the pattern (or sequence). 4, −20, 100, −500, . . . Step 1: Form a Hypothesis: Step 2: Make observations related to the hypothesis: Step 3: Come to a conclusion: 2. Flying West New York City is on easter
### Coordinates and Altitude
Let A = (0,3) and B = (4,0) and AB = 5 = c Let D be the altitude to the hypotenuse. a) Find the coordinates of D b) Find the lengths of AD, BD, and CD | 2,675 | 9,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2019-04 | longest | en | 0.911938 |
https://math.stackexchange.com/questions/3705456/how-to-prove-or-disprove-that-the-series-are-fourier-transformation-for-some-fun | 1,610,970,151,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703514495.52/warc/CC-MAIN-20210118092350-20210118122350-00117.warc.gz | 459,172,822 | 32,103 | # How to prove or disprove that the series are Fourier transformation for some function.
Let's say, there is a function $$\Sigma^{\infty}_{k=1} \frac{\cos 2k x}{\sqrt{2k \ln 2k}}$$, how do I prove or disprove that it is a Fourier transformation for some function?
• By Fourier transform do you mean Fourier series? – LL 3.14 Jun 4 '20 at 16:06
• @LL3.14 Yes, Fourier series. – user Jun 4 '20 at 16:08
• Is your sum over $n$ or $k$? – James Arathoon Jun 4 '20 at 16:52
• What is exactly your question, you want to know if some given function $f$ is equal to this Fourier series ? – qdr Jun 4 '20 at 17:16
• If you know Parseval-Plancherel, you can see that it cannot be the Fourier series of an $L^2$ function... But what kind of "functions" do you allow? That series is certainly the Fourier series of a "generalized function" (distribution), lying in a (periodic) Sobolev space $H^{-1}$, for example. – paul garrett Jun 4 '20 at 17:23 | 294 | 937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 1, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2021-04 | latest | en | 0.798748 |
http://www.jiskha.com/display.cgi?id=1347525594 | 1,498,435,886,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320595.24/warc/CC-MAIN-20170625235624-20170626015624-00645.warc.gz | 557,967,712 | 3,707 | # Physics
posted by .
A boat with a horizontal tow rope pulls a water skier. She skies off to the side,
so the rope makes an angle of 15.0° with her direction of motion, but in the
same direction as the boat is traveling. The tension in the rope is 160 N. How
much work is done on the skier by the rope during a displacement of 400 m?
• Physics -
Fx = 160 * cos(15) = 155N
W.D. = 155 * 400 = 62kN | 119 | 400 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2017-26 | latest | en | 0.955819 |
https://utcc.utoronto.ca/~cks/space/blog/programming/AwkFunWithEquality?showcomments | 1,701,829,391,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00192.warc.gz | 655,902,735 | 3,754 | ## The fun of `awk`
May 30, 2008
I like `awk`, I really do, but sometimes it really irritates me. Take, for example, this fun little awk program:
`awk 'BEGIN {print "5" == "05"}' /dev/null`
You might rationally expect this to print '1' (awk's boolean truth value). As I found out once, you would be sadly mistaken; this is false, presumably because `awk` winds up doing a string comparison instead of a numeric one. Too bad if you're reading one set of fields that are zero-padded and one set that aren't.
(The workaround is add 0 to the "05" to force the numeric interpretation; `"5" == ("05"+0)` comes out true.)
This shows two drawbacks of the sort of magical conversion between numbers and strings that `awk` does. First, this sort of stuff involves heuristics, and heuristics are inevitably wrong sooner or later. And second, if you do not have the fine details carefully memorized you can wind up surprised.
At the same time such magical conversions live on because they are oh so very handy when you are banging things in a hurry. Considering the sorts of things that `awk` was designed for, this is completely the right decision for it; having to write explicit Python-style conversions all the time would probably drive me up the wall, however much I like them in Python.
From 71.65.56.124 at 2008-05-31 11:01:34:
You can also take away the quotes if you know your input will be mathematical.
By cks at 2008-05-31 12:06:38:
In this case the real version was more like '`if (\$1 == \$3) ...`'; the actual values I was comparing were input fields instead of one being a constant.
From 71.65.56.124 at 2008-06-01 16:50:31:
Interesting
At my shell, I get this:
```Matt-Simmons-Computer:~ mattsimmons\$ echo "1 2 3" | awk '{print (\$1 == \$3)}'
0
Matt-Simmons-Computer:~ mattsimmons\$ echo "3 2 3" | awk '{print (\$1 == \$3)}'
1
Matt-Simmons-Computer:~ mattsimmons\$ echo "3 2 03" | awk '{print (\$1 == \$3)}'
1
Matt-Simmons-Computer:~ mattsimmons\$ echo "3 2 03" | awk '{print (\$1 == \$3)}'
```
Is this consistent with what you see on yours?
By cks at 2008-06-02 00:12:05:
I have managed to find the script with the specific issue. I was comparing an explicitly set awk variable with an input field, roughly:
```awk 'BEGIN { day = "'\$d'" }
/^From / {if (day == \$5) [...]
```
If the field value was 0-padded, this comparison is false. In retrospect, I could reasonably count on the 'day' value being a properly formed number and avoid making it into a string (if it's not a properly formed number, the rest of the awk will blow up anyways), which would avoid the whole issue.
By cks at 2008-06-02 00:13:50:
PS: I should mention that my versions of awk behaves consistently with Matt's for comparing input fields against each other.
Written on 30 May 2008. | 765 | 2,785 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2023-50 | longest | en | 0.88818 |
https://www.jiskha.com/display.cgi?id=1512261219 | 1,531,972,943,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590493.28/warc/CC-MAIN-20180719031742-20180719051742-00233.warc.gz | 924,549,835 | 3,874 | stast
posted by jack
suppose the alternative hypothesis in a hypothesis test is "the population mean is less than 60" if the sample size is 50 and alpha =.05, the critical value of the z is?
1. PsyDAG
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability
(p ≤ .05) and its Z score.
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### Two Squared
##### Stage: 2 Challenge Level:
What happens to the area of a square if you double the length of the sides? Try the same thing with rectangles, diamonds and other shapes. How do the four smaller ones fit into the larger one?
### Triple Cubes
##### Stage: 1 and 2 Challenge Level:
This challenge involves eight three-cube models made from interlocking cubes. Investigate different ways of putting the models together then compare your constructions.
### Move a Match
##### Stage: 2 Challenge Level:
How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### Put Yourself in a Box
##### Stage: 2 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
### World of Tan 13 - A Storm in a Tea Cup
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of these convex shapes?
### Makeover
##### Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
### Construct-o-straws
##### Stage: 2 Challenge Level:
Make a cube out of straws and have a go at this practical challenge.
### Fractional Triangles
##### Stage: 2 Challenge Level:
Use the lines on this figure to show how the square can be divided into 2 halves, 3 thirds, 6 sixths and 9 ninths.
### World of Tan 9 - Animals
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this goat and giraffe?
### World of Tan 8 - Sports Car
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this sports car?
### Flight of the Flibbins
##### Stage: 3 Challenge Level:
Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . .
### World of Tan 16 - Time Flies
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the candle and sundial?
### World of Tan 6 - Junk
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of this junk?
### World of Tan 15 - Millennia
##### Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of the workmen? | 2,234 | 9,708 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2017-04 | longest | en | 0.939059 |
https://arduino.stackexchange.com/questions/85721/why-are-transistors-necessary-before-leds | 1,695,343,518,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506320.28/warc/CC-MAIN-20230922002008-20230922032008-00828.warc.gz | 124,602,388 | 40,769 | # Why are transistors necessary before LEDs?
Below is the small RGB LED circuit I'm trying to build. My question is why are the 3 transistors necessary, and can I not link the LEDs directly?
• I doubt if the LEDs will ever light up with 4.7k base resistor and 4.7k resistor in series with LEDs
– SBF
Aug 23, 2021 at 6:25
• @SBF they will not. They’re NPN transistors on the wrong side of the load. Also, the LEDs will not be driven higher than about 0.5mA each, which might be enough, but will not be very bright. Aug 23, 2021 at 6:36
• We don't know what the supply voltage is from that circuit. It's not a good circuit at all.
– Majenko
Aug 23, 2021 at 8:06
• @StarCat They're not on the "wrong" side of the load. There's no "wrong" side. They're just not on the "traditional" side for simple saturation switching. Instead they (and their resistors) form part of the base current path and severely limit the amount of base current that can flow. Your current of 0.5mA for each LED is bang on though. I make it 594uA per LED. A pitiful current.
– Majenko
Aug 23, 2021 at 9:05
• @Majenko, agreed, when I said "wrong" I meant this was not the most logical or efficient placement of the switching NPN transistors. Your explanation is correct of course. Aug 23, 2021 at 10:22
The port pins of your Nano can deliver current in a quite limited range, and six LEDs need a lot more to shine brightly. The transistors are amplifying the available current.
However, 4k7 resistors in series with each LED will not allow enough current to let them shine well. Reduce the resistors by a factor of at least 10. Read the LEDs' data sheet and apply Ohm's law.
The transistors are connected as an emitter follower, not as a switch. The transistors are acting as current amplifiers. The emitter voltage will rise to approximately the voltage on its base - about 0.7V. That places the voltage at about (5 - 0.7 = 4.3V) at the emitter. Since the collectors of transistors are powered by Vin I would guess you have at least 6 volts available assuming a 5V Nano so the extra voltage drop across the transistors will be dissipated as heat. The base resistors are not needed but should not affect performance much. The available current at the emitter would be the Hfe of the transistor times the base current. Each LED will consume about 2 milliamps making the load on each of the transistor around 12 milliamps. It should work fine. You can divide the 12 milliamps of the LEDs by the Hfe of the transistor to get the base load.
Each LED consumes up to 20mA and You are using three LEDs at each I/O port in Arduino Nano. In other word, each part consumes 120mA.
The important thing is Arduino Nano I/O provides 20mA at maximum. So if you want to control 120mA by I/O ports, you must use transistor to support huge current.
Transistor can afford current up to 500mA.
• I see 6 LEDs controlled by one pin
– Juraj
Aug 23, 2021 at 18:01
• How do you know how much current the transistor can handle? The OP did not specify the transistor used... Aug 25, 2021 at 0:01
• He said "Why transistors are necessary" not "Why this transistors are necssary"
– 박광렬
Aug 26, 2021 at 6:46
• @Juraj I see 12 LEDs controlled by pins 6/10
– SBF
Aug 26, 2021 at 20:56
• 3 transistors on pins D2, D6 and D10 and 18 LEDs
– Juraj
Aug 27, 2021 at 4:35 | 921 | 3,312 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2023-40 | latest | en | 0.952151 |
http://mathhelpforum.com/pre-calculus/97115-de-moivre-binomial-theorem.html | 1,527,082,036,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865651.2/warc/CC-MAIN-20180523121803-20180523141803-00096.warc.gz | 185,910,494 | 10,200 | # Thread: de Moivre and Binomial Theorem
1. ## de Moivre and Binomial Theorem
Use de Moivre's Theorem and the Binomial Theorem to find integers A,B,C,D,E,F such that
$\displaystyle \sin(5\theta)=A\sin^{5}\theta + B\sin^{3}\theta+ C\sin\theta$
$\displaystyle 16\sin^{5}\theta=D\sin\theta+E\sin(3\theta)+F\sin( 5\theta)$.
Any help much appreciated
2. Do you know the statement of De Moivre's theorem? Do you know how to expand $\displaystyle (x+y)^5$? If so just do it and set $\displaystyle x=\cos\theta, \: y=i\sin\theta$...
3. So I get
$\displaystyle \cos^{5}\theta+5\cos^{4}\theta\sin\theta + 10\cos^{3}\theta\sin^{2}\theta + 10\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +\sin^{5}\theta$
$\displaystyle (\cos x + i \sin x)^{n}= \cos n\theta + i \sin n\theta$
I still don't see where I'm supposed to go with it...
Is it something to do with matching up the real parts on both sides?
4. Originally Posted by PTL
So I get
$\displaystyle \cos^{5}\theta+5\cos^{4}\theta\sin\theta + 10\cos^{3}\theta\sin^{2}\theta + 10\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +\sin^{5}\theta$
$\displaystyle (\cos x + i \sin x)^{n}= \cos n\theta + i \sin n\theta$
I still don't see where I'm supposed to go with it...
Is it something to do with matching up the real parts on both sides?
You forgot $\displaystyle i$ my friend!
you should get
$\displaystyle (\cos x + i \sin x)^{5} =$
$\displaystyle \cos^{5}\theta+5i\cos^{4}\theta\sin\theta - 10\cos^{3}\theta\sin^{2}\theta - 10i\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +i\sin^{5}\theta$
(just alternate $\displaystyle 1,i,-1,-i,...$)
Then equate the imaginary parts because the real part is $\displaystyle \cos 5\theta$ which is not what you want. | 603 | 1,736 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-22 | latest | en | 0.721093 |
http://oeis.org/A116559 | 1,545,153,730,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829542.89/warc/CC-MAIN-20181218164121-20181218190121-00412.warc.gz | 207,114,480 | 4,388 | This site is supported by donations to The OEIS Foundation.
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A116559 Sequentially switched Markov of six 2 X 2 matrices based on the SL[2,2] group of Blyth and Robinson that gives a chaotic vector output. 1
0, 1, 1, 2, 2, 5, 5, 3, 8, 11, 11, 30, 30, 19, 49, 68, 68, 185, 185, 117, 302, 419, 419, 1140, 1140, 721, 1861, 2582, 2582, 7025, 7025, 4443, 11468, 15911, 15911, 43290, 43290, 27379, 70669, 98048, 98048, 266765, 266765, 168717, 435482, 604199, 604199, 1643880, 1643880, 1039681 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 REFERENCES Blyth and Robonson, Essential Student Algebra, V5, Groups, J. W. Arrowsmith, Bristol, 1986, page 9. LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 FORMULA From R. J. Mathar, Nov 28 2008: (Start) a(n) = 6*a(n-6) + a(n-12). G.f.: x*(1+x+2*x^2+2*x^3+5*x^4+5*x^5-3*x^6+2*x^7-x^8-x^9)/(1-6*x^6-x^12). a(6n+1) = A005667(n). (End) MATHEMATICA CoefficientList[Series[x*(1 + x + 2*x^2 + 2*x^3 + 5*x^4 + 5*x^5 - 3*x^6 + 2*x^7 - x^8 - x^9)/(1 - 6*x^6 - x^12), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2017 *) PROG (PARI) x='x+O('x^50); Vec(x*(1 + x + 2*x^2 + 2*x^3 + 5*x^4 + 5*x^5 - 3*x^6 + 2*x^7 - x^8 - x^9)/(1 - 6*x^6 - x^12)) \\ G. C. Greubel, Sep 20 2017 CROSSREFS Sequence in context: A045537 A243941 A161622 * A210802 A257943 A236935 Adjacent sequences: A116556 A116557 A116558 * A116560 A116561 A116562 KEYWORD nonn,obsc AUTHOR Roger L. Bagula, Mar 17 2006 EXTENSIONS More terms added by G. C. Greubel, Sep 20 2017 STATUS approved
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Last modified December 18 12:06 EST 2018. Contains 318229 sequences. (Running on oeis4.) | 867 | 2,150 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2018-51 | longest | en | 0.646686 |
https://www.whitecoatinvestor.com/investment-to-debt-ratio/ | 1,718,239,160,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00039.warc.gz | 970,721,469 | 22,666 | I saw a tweet a while back that got me thinking:
It got me thinking about ratios and rules of thumb. I thought it would be fun to do a post about two related, but slightly different ratios and see if we could come up with any useful rules of thumb:
1. Your Investment to Payment Ratio and
2. Your Investment to Interest Ratio
Let's take them one at a time and see how implementing the proper ratios can help you build wealth.
## Investment to Payment Ratio
This ratio is the amount of money you invest in a given month compared to the minimum required payments on all of your debts. So let's say you make \$20K a month, have a \$3500 P&I payment on your house, a \$1500 per month student loan payment, a \$20 credit card payment, and a \$480 per month car payment. That's a total of \$5500 in payments. Let's say you're saving \$2,000 a month. That would make your investment to payment ratio:
### \$2,000/\$5,500 = 0.36
That seems pretty low to me. What would be good? Well, I think it's a good idea to keep this ratio above 1, at least for a high income professional. That's going to require some sacrifice and discipline though. What does a ratio of 1 look like? It looks about like this. Again, let's assume that same \$20K a month gross income. Let's say you're putting 20% toward retirement and 5% toward college, so you're investing \$5K. You kept your student loans down (\$1,000/month) and bought a less expensive house (\$2500/month) and avoided car payments and credit cards. What's your ratio now?
### \$5000/\$3500 = 1.43
Keeping that above 1 doesn't seem that hard does it?
This rule of thumb, like most rules of thumb, has some problems. For example, it doesn't work during the distribution phase or during the education phase. During the distribution phase, you're pulling money out of your investments, so your ratio would technically be negative. During the education phase, you're not investing anything and may not have any payments either. What if you're a renter? Sure, you don't have a debt payment due, but you've replaced it with a rent payment. And what if you refinanced into a 5-year variable loan for your student loans or a 15 year fixed loan for your mortgage? Those would make your ratio worse despite probably improving your financial situation. Another issue is that in any given month, increasing your investments makes a dramatic and immediate increase in your ratio, but paying down debt might not change the ratio any time soon. Yet another issue applies to military docs and others with a similar commitment who have accepted a time debt (usually with a lower salary) instead of a money debt. These ratios make those folks look better than perhaps they should. Maybe the other ratio would be better. Let's try it.
## Investment to Interest Ratio
This ratio is the amount of money you invest in a given month compared to the interest you're paying for all of your debts. So if you have a \$200K student loan at 6%, the monthly interest would be about \$1,000. The interest on a \$500,000, 4% mortgage is \$1,667. Maybe the interest on cars and credit cards is \$100. Total interest is \$2,767. If you're saving \$2K, your ratio is 0.72. If you're saving \$4K, it's 1.45. As you make more or increase your savings rate and pay off your debt, that ratio improves. When you pay off all your debts, it becomes infinite.
I think a ratio of < 0.5 is poor, 0.5 – 2 is fair 2-5 is a good ratio, 5-10 is very good, and > 10 is excellent.
Since we're debt-free now, our ratio is infinite, but just before we paid off the mortgage it was ~100.
I like this ratio a little better in that it rewards you for doing things that minimize the interest paid, like taking a shorter loan term, running the interest rate risk yourself with a variable rate, or refinancing to a lower rate. The ratio also improves as your overall financial situation improves and your debt burden drops. But otherwise, it suffers from the same problems as the other rule of thumb.
All that mattered on this day was the reeled in fish to casting ratio. | 955 | 4,066 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2024-26 | latest | en | 0.973072 |
http://www.krivers.net/15112-f17/notes/notes-functions-redux.html | 1,712,932,724,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816024.45/warc/CC-MAIN-20240412132154-20240412162154-00357.warc.gz | 51,766,193 | 2,598 | # CMU 15-112: Fundamentals of Programming and Computer Science Class Notes: Functions Redux (part 1)
1. Variable length args (*args)
2. Default args
3. Functions as parameters
4. Lambda functions
5. Keyword args (**kwargs)
6. Functions inside functions
7. Functions that return functions
8. Function decorators
1. Variable length args (*args)
def longestWord(*args): if (len(args) == 0): return None result = args[0] for word in args: if (len(word) > len(result)): result = word return result print(longestWord("this", "is", "really", "nice")) # really mywords = ["this", "is", "really", "nice"] print(longestWord(mywords)) # ['this', 'is', 'really', 'nice'] print(longestWord(*mywords)) # really
2. Default args
1. Default args example
def f(x, y=10): return (x,y) print(f(5)) # (5, 10) print(f(5,6)) # (5, 6)
2. Do not use mutable default args
def f(x, L=[ ]): L.append(x) return L print(f(1)) print(f(2)) # why is this [1, 2]?
3. One workaround for mutable default args
def f(x, L=None): if (L == None): L = [ ] L.append(x) return L print(f(1)) print(f(2)) # [2] (that's better)
3. Functions as parameters
def derivative(f, x): h = 10**-8 return (f(x+h) - f(x))/h def f(x): return 4*x + 3 print(derivative(f, 2)) # about 4 def g(x): return 4*x**2 + 3 print(derivative(g, 2)) # about 16 (8*x at x==2)
4. Lambda functions
print(derivative(lambda x:3*x**5 + 2, 2)) # about 240, 15*x**4 at x==2 myF = lambda x: 10*x + 42 print(myF(5)) # 92 print(derivative(myF, 5)) # about 10
5. Keyword args (**kwargs)
def f(x=1, y=2): return (x,y) print(f()) # (1, 2) print(f(3)) # (3, 2) print(f(y=3)) # (1, 3) [here is where we use a keyword arg] def f(x, **kwargs): return (x, kwargs) print(f(1)) # (1, { }) print(f(2, y=3, z=4)) # (2, {'z': 4, 'y': 3})
6. Functions inside functions
def f(L): def squared(x): return x**2 return [squared(x) for x in L] print(f(range(5))) try: print(squared(5)) except: print("squared is not defined outside f")
7. Functions that return functions
def derivativeFn(f): def g(x): h = 10**-5 return (f(x+h) - f(x))/h return g def f(x): return 5*x**3 + 10 fprime1 = derivativeFn(f) fprime2 = derivativeFn(fprime1) print(f(3)) # 145, 5*x**3 + 10 evaluated at x == 3 print(fprime1(3)) # about 135, 15*x**2 evaluated at x == 3 print(fprime2(3)) # about 90, 30*x evaluated at x == 3
8. Function decorators
@derivativeFn def h(x): return 5*x**3 + 10 print(h(3)) # 135, matches fprime1 from above. | 843 | 2,419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-18 | latest | en | 0.510839 |
https://support.sas.com/documentation/cdl/en/orlsoug/65553/HTML/default/orlsoug_ga_examples02.htm | 1,632,290,687,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057329.74/warc/CC-MAIN-20210922041825-20210922071825-00163.warc.gz | 586,738,020 | 7,349 | ### Example 3.2 Nonlinear Objective with Constraints Using Repair Mechanism
This example illustrates the use of a repair mechanism to satisfy problem constraints. The problem is to minimize the six-hump camel-back function (Michalewicz 1996, Appendix B):
where is within the triangle with vertices , , and . The problem formulation takes advantage of the fact that all feasible solutions can be expressed as a convex combination of , , and in the following form:
In this form, , , and are nonnegative coefficients and satisfy the following linear equality constraint:
Therefore, the solution encoding 'R3' is used with the three elements corresponding to the values of , , and . Note that this strategy can be generalized to any solution domain that can be specified by a convex hull. An additional 'R2' segment is also created to store the corresponding value. In the following program, FUNCTION SIXHUMP computes the objective value. Lower and upper bounds of 0 and 1, respectively, are used to ensure that solution elements are nonnegative. Violations of the linear equality constraint are fixed by a simple repair strategy, implemented in FUNCTION SIXHUMP: the sum of the solution elements is computed, and each element is divided by the sum, so that the sum of the new elements is 1. For the special case of all elements equal to 0, equal values are assigned to the solution elements.
proc ga seed = 555;
call SetEncoding('R3R2');
npoints = 3;
array cvxhull[3,2] /nosym ( -2 0
0 2
2 -2 );
/* Objective function */
function sixhump(selected[*],cvxhull[*,*],npoints);
/* Function has global minimum value of -1.0316
* at x = {-0.0898 0.7126} and
* x = { 0.0898 -0.7126}
*/
array w[1] /nosym;
call dynamic_array(w,npoints);
array x[2] /nosym;
/* make sure that weights add up to 1 */
sum = 0;
do i = 1 to npoints;
sum + w[i];
end;
/* if all weights 0, then reinitialize */
if sum=0 then do;
sum = npoints;
do i = 1 to npoints;
w[i] = 1;
end;
end;
/* re-normalize weights */
do i = 1 to npoints;
w[i] = w[i] / sum;
end;
call WriteMember(selected,1,w);
/* convert weights to x-coordinate form */
x[1] = 0;
x[2] = 0;
do i = 1 to npoints;
x[1] + w[i] * cvxhull[i,1];
x[2] + w[i] * cvxhull[i,2];
end;
/* write out x coordinates to second segment */
call WriteMember(selected,2,x);
/* compute objective value */
r = (4 - 2.1*x[1]**2 + x[1]**4/3)*x[1]**2 + x[1]*x[2] +
(-4 + 4*x[2]**2)*x[2]**2;
return(r);
endsub;
call SetObjFunc('sixhump',0);
array lower[1] /nosym;
array upper[1] /nosym;
call dynamic_array(lower, npoints);
call dynamic_array(upper, npoints);
do i = 1 to npoints;
lower[i] = 0;
upper[i] = 1;
end;
call SetBounds(lower, upper, 1);
array delta[3] /nosym (0.01 0.01 0.01);
call SetMut('delta', 'nchange', 1, 'delta', delta);
call SetMutProb(0.05);
call SetCross('Twopoint', 'alpha', 0.9);
call SetCrossProb(0.8);
call SetSel('tournament', 'size', 2);
call SetElite(3);
call Initialize('DEFAULT', 200);
call ContinueFor(200);
run;
Note that this problem uses the standard genetic operators and default initialization, even though they generate solutions that violate the constraints. This is possible because all solutions are passed into the user objective function for evaluation, where they are repaired to fit the constraints. The output is shown in Output 3.2.1.
Output 3.2.1: Nonlinear Objective with Constraints Using Repair Mechanism
PROC GA Optimum Values
Objective
-1.031617314
Solution
Segment Element Value
1 1 0.2439660392
1 2 0.5561415112
1 3 0.1998924497
2 1 -0.088147179
2 2 0.712498123
This objective function has a global minimum at , at two different points: and . The genetic algorithm can converge to either of these minima, depending on the random number seed set by the SEED= option. | 1,073 | 3,731 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-39 | latest | en | 0.84334 |
http://www.chegg.com/homework-help/questions-and-answers/p-and-q-are-points-within-a-uniform-electric-field-that-are-separated-by-a-distance-of-02--q2965948 | 1,368,917,055,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382917/warc/CC-MAIN-20130516092622-00070-ip-10-60-113-184.ec2.internal.warc.gz | 391,469,125 | 7,871 | ## How much work is required to move point charge from P to Q?
P and Q are points within a uniform electric field that are separated by a distance of 0.2 m. The potential difference between P and Q is 75 V. How much work is required to move a +150 ?C point charge from P to Q?
Show work. | 73 | 288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2013-20 | latest | en | 0.97788 |
https://www.scribd.com/document/102741253/PID-Controller | 1,501,237,568,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549448146.46/warc/CC-MAIN-20170728083322-20170728103322-00070.warc.gz | 834,545,544 | 26,338 | # Application of a PID controller in a Speed Control of a Car.
By Atchyuth Sonti
Hello,guys now i am going to tell you how a PID controller is used in the automobiles.This is an important concept in the field of control systems and lets see why it should be used and how each stage of it contributes to the speed control.I am going to use matlab(simulink) for illustrating this particular concept.
Lets me consider that i am having a SUV. Let the parameters of the car be 1.Weight of 1000 kg. 2.Engine force of 5000 N. 3.Drag coefficient of 50.(this in simple terms can be explained as sum of friction and other opposing forces). The specifications which my design should meet be 1.A speed of 10m/s.( =36kmph). 2.Rise time should be <5 sec.(this is popularly called as pick up). 3.Steady state error of <2% ( the steady state speed should be in between 9.8m/s and 10.2m/s).
So now let us construct a simple car model To execute this particular thing we need to have both matlab file(.m) and simulink file(.mdl). We do this because we will have maximum compatibility for changing values and analyse the circuit with different parameters. Generally the M file consists of the list of parameters which affect the system and mdl file consists the signal flow diagram of the system.
Simple Car Model
The (.m) file
The simulink (.mdl) file for the simple car model is given below.
This is the way you need to design the files and for simulink file to take the values from the m file u need to run the m file first followed by the mdl file. This makes the things simpler. The output which is generated by this model will be of this type.
The output of the simple car model.
Analysis 1.A speed of 10m/s. 2.Rise time should be <5 sec. 3.Steady state error of <2%. When the above output is analysed it is following all the design specifications except the rise time. i.e the above car doesn’t have proper pick up or acceleration.(the design takes 100 sec where it should take less than 5 sec).If employed this becomes one of the worst design. Now let us add the parts of the controller and analyse its effect(s).
The “P” controller.
Now we shall make two subsystems 1. Car.(the above car mode is now put in this particular subsystem). 2. Controller. We shall start with the P controller and go on adding the complexity to the controller. For our convenience we converted the force into a reference velocity of 10m/s. This reduces the tedious calculations.
The m file for the above system is
The output of the P controller is
This is the output with a p controller Analysis 1.A speed of 10m/s.( =36kmph). 2.Rise time should be <5 sec.(this is popularly called as pick up). 3.Steady state error of <2% ( the steady state speed should be in between 9.8m/s and 10.2m/s).
The PI controller.
In PI controller an integral controller comes into action with the P controller. Now we completely made another subsystem called a controller and we add different components into that controller.
For our reference this is the subsystem car.
And this is the subsystem of the controller.
The m file for the above system where we include all the parameters of “PI” is
This is the output of the “PI” controller.
As all of u can see here we are not getting a perfect value. This states that the above system is unstable. This is the major disadvantage for this particular system. This instability makes this particular system impractical. 1.A speed of 10m/s 2.Rise time should be <5 sec 3.Steady state error of <2%.
The PID controller
Finally we arrive to the PID controller .As the name implies we need to implement a differential controller to the above “PI” controller. This is the signal flow graph implemented in the form of a simulink mdl file.
The parameters for the above diagram are driven by this particular m file which has all the parameters.
When the above files are executed you will get a output of this sort.
Analysis 1.A speed of 10m/s. 2.Rise time should be <5 sec. 3.Steady state error of <2%. Thus all the design requirements are met through this particular design. This makes the applications of the electronics in automobiles. Moreover this particular model is implemented in many of the luxurious cars. The thing which makes “luxurious” cars “luxurious” is the embedded electronics in them. There are many electronic components implemented in the cars like ABS,ECU and air bag control etc which employs many of the electronic components and make the automobiles more comfortable, safe and smart. | 1,059 | 4,524 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2017-30 | latest | en | 0.893501 |
https://forum.dominionstrategy.com/index.php?topic=20097.0;all | 1,718,965,019,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862070.5/warc/CC-MAIN-20240621093735-20240621123735-00632.warc.gz | 245,327,326 | 21,779 | # Dominion Strategy Forum
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### AuthorTopic: Decrypto 2 (Read 15405 times)
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#### ghostofmars
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##### Decrypto 2
« on: January 31, 2020, 03:06:25 am »
Decrypto is a game that usually appeals to the same people as Codenames does.
https://boardgamegeek.com/boardgame/225694/decrypto
Quote
Players compete in two teams in Decrypto, with each trying to correctly interpret the coded messages presented to them by their teammates while cracking the codes they intercept from the opposing team.
In more detail, each team has their own screen, and in this screen they tuck four cards in pockets numbered 1-4, letting everyone on the same team see the words on these cards while hiding the words from the opposing team. In the first round, each team does the following: One team member takes a code card that shows three of the digits 1-4 in some order, e.g., 4-2-1. They then give a coded message that their teammates must use to guess this code. For example, if the team's four words are "pig", "candy", "tent", and "son", then I might say "Sam-striped-pink" and hope that my teammates can correctly map those words to 4-2-1. If they guess correctly, great; if not, we receive a black mark of failure.
Starting in the second round, a member of each team must again give a clue about their words to match a numbered code. If I get 2-4-3, I might now say, "sucker-prince-stake". The other team then attempts to guess our numbered code. If they're correct, they receive a white mark of success; if not, then my team must guess the number correctly or take a black mark of failure. (Guessing correctly does nothing except avoid failure and give the opposing team information about what our hidden words might be.)
The rounds continue until a team collects either its second white mark (winning the game) or its second black mark (losing the game). Games typically last between 4-7 rounds. If neither team has won after eight rounds, then each team must attempt to guess the other team's words; whichever team guesses more words correctly wins.
There is a word list generator we could use here: https://whoawhoa.github.io/decrypto/#
1) Cluegivers generate a code using the same site, make clues and post them to the game thread.
2) Guessers guess their own teams code. Starting round 2, teams simultaneously discuss about the opponent's clue and try to guess their code.
3a) If you guess the clue of your own team, you post it in the team quicktopic.
3b) If you guess the clue of the opposing team, you post it in spoilers in the game thread.
4) Once all guesses are ready, the cluegivers state their partner's guess, reveal the correct codes and tokens are assigned.
5) Start new round with switched cluegivers.
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #1 on: January 31, 2020, 03:15:55 am »
We should discuss how we deal with multiple players in a team. How do they discuss which clue to submit for their own team?
• In the hidden QT without discussing strategy.
• In a new QT for each round, that becomes public after the clues are revealed.
• One player from the team is assigned to submit the clue.
• Other ideas you can come up with.
For the teams, I suggest to split Uncleeurope and bitwise and then mix and match the old teams
Uncleeurope + Watno(1) or scolapasta(2)
Rolled 1d2 : 1, total 1
+ faust(1) or ghostofmars(2)
Rolled 1d2 : 2, total 2
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#### Jimmmmm
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##### Re: Decrypto 2
« Reply #2 on: January 31, 2020, 04:51:49 am »
/in
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#### faust
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##### Re: Decrypto 2
« Reply #3 on: January 31, 2020, 07:39:51 am »
It's hard to find a good system. Any discussion will help the team figure out what kind of associations their teammates have with the words. Maybe the simplest is just one of them submits, and no discussion?
I would try to mix up pairings of who gives the clues and who guesses as much as possible.
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#### bitwise
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##### Re: Decrypto 2
« Reply #4 on: February 01, 2020, 12:06:08 am »
I would prefer being able to confer with my teammates at the risk of leaking info, but I guess there isn't a great way to do this online. In person you're allowed to talk, but there is a lot of pointing at words and gesturing so that the other team can't quite tell what you're saying. I don't think that translates well here.
I guess my vote is for in the hidden QT without discussing strategy.
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#### Watno
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##### Re: Decrypto 2
« Reply #5 on: February 01, 2020, 04:00:05 am »
Yeah, I think just don't talk about strategy. Also nothing along the lines of "he woul have said "x" instead of "y" if he meant that about your own teams clue.
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#### scolapasta
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##### Re: Decrypto 2
« Reply #6 on: February 01, 2020, 12:33:06 pm »
So we can debate the numbers without giving the reasons?
"I like 2-3-1"
"I think 2-4-1 is better"
Seems like it'd be challenging to settle on one, but I get the purpose, sure.
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#### faust
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##### Re: Decrypto 2
« Reply #7 on: February 03, 2020, 04:23:01 am »
I think we just need to put some rule in place so we can start this. As long as it is clear to the cluegivers that they should not rely on secret discussions to give clues, discussing things is probably not the end of the world. At the very least, both teams benefit equally.
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#### Watno
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##### Re: Decrypto 2
« Reply #8 on: February 03, 2020, 02:08:22 pm »
So we can debate the numbers without giving the reasons?
"I like 2-3-1"
"I think 2-4-1 is better"
Seems like it'd be challenging to settle on one, but I get the purpose, sure.
Giving the reason is fine, unless the reason indicates a clue you are going to give later.
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#### scolapasta
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##### Re: Decrypto 2
« Reply #9 on: February 03, 2020, 03:04:40 pm »
Fair enough - I'm sure common sense will be enough. Let's get started!
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#### faust
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##### Re: Decrypto 2
« Reply #10 on: February 04, 2020, 12:56:02 pm »
Fair enough - I'm sure common sense will be enough. Let's get started!
You know, there's nothing stopping you from getting it started
I have created a QT for our team and sent it to my teammates (i.e. scolapasta, bitwise). It was determined that bitwise will give the first clue.
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#### Uncleeurope
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##### Re: Decrypto 2
« Reply #11 on: February 04, 2020, 02:28:19 pm »
I am going to /out for the sake of jimmmm and my schedule.
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Don't worry, no one can read uncle correctly.
Uncleeurope Eddie
#### bitwise
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##### Re: Decrypto 2
« Reply #12 on: February 04, 2020, 11:03:56 pm »
I'm ready to give my clue. Should I give it now or wait until the other team is more ready?
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #13 on: February 05, 2020, 02:37:41 am »
I created a QT for our team. Here is the first clue
prime - post - zero
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#### bitwise
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##### Re: Decrypto 2
« Reply #14 on: February 05, 2020, 11:10:02 am »
Our clue: Sun - Culture - Oil.
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#### faust
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##### Re: Decrypto 2
« Reply #15 on: February 05, 2020, 05:20:32 pm »
Our team guesses 1-2-4.
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#### bitwise
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##### Re: Decrypto 2
« Reply #16 on: February 05, 2020, 05:24:42 pm »
That is correct.
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#### Jimmmmm
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##### Re: Decrypto 2
« Reply #17 on: February 08, 2020, 05:01:19 am »
2-1-4
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #18 on: February 08, 2020, 07:34:21 am »
That is correct
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#### scolapasta
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##### Re: Decrypto 2
« Reply #19 on: February 08, 2020, 11:48:52 am »
So round 2 now, yeah? My turn to give a clue?
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#### faust
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##### Re: Decrypto 2
« Reply #20 on: February 08, 2020, 11:55:52 am »
So round 2 now, yeah? My turn to give a clue?
Indeed.
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#### scolapasta
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##### Re: Decrypto 2
« Reply #21 on: February 08, 2020, 12:54:19 pm »
BUILDING - QUAKE - WIZARD
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#### Watno
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##### Re: Decrypto 2
« Reply #22 on: February 09, 2020, 07:49:51 am »
Fire - Water - Ice
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#### scolapasta
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##### Re: Decrypto 2
« Reply #23 on: February 10, 2020, 12:59:37 pm »
Our guess for your Round 2 clue is:
1-3-4
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#### Watno
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##### Re: Decrypto 2
« Reply #24 on: February 11, 2020, 02:33:38 am »
My team guesses 1-2-4, which is correct.
Our guess for your clue is 2-4-3
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #25 on: February 11, 2020, 03:14:24 am »
Our guess for your Round 2 clue is:
1-3-4
Wow, that was close.
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#### scolapasta
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##### Re: Decrypto 2
« Reply #26 on: February 11, 2020, 11:40:34 am »
My team guesses 2-3-4, which is correct.
Round 2: no miscommunications, no intercepts.
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#### faust
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##### Re: Decrypto 2
« Reply #27 on: February 11, 2020, 12:29:09 pm »
Stealing ghost's nice summmary posts from last time as I found them quite helpful.
Summary
team ghostofmars/Jimmmmm/Watno
0 Interception tokens
0 Miscommunication tokens
Round 1 Prime - Post - Zero (code: 2-1-4, guess: 2-1-4, interception: N/A)
Round 2 Fire - Water - Ice (code: 1-2-4, guess: 1-2-4, interception: 1-3-4)
Clues for 1 Post, Fire
Clues for 2 Prime, Water
Clues for 4 Zero, Ice
team bitwise/faust/scolapasta
0 Interception tokens
0 Miscommunication tokens
Round 1 Sun - Culture - Oil (code: 1-2-4, guess: 1-2-4, interception: N/A)
Round 2 Building - Quake - Wizard (code: 2-3-4, guess: 2-3-4, interception: 2-4-3)
Clues for 1 Sun
Clues for 2 Culture, Building
Clues for 3 Quake
Clues for 4 Oil, Wizard
« Last Edit: February 12, 2020, 01:37:08 am by faust »
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#### faust
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##### Re: Decrypto 2
« Reply #28 on: February 11, 2020, 12:39:28 pm »
Rise - Move - Join
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#### faust
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##### Re: Decrypto 2
« Reply #29 on: February 11, 2020, 12:41:58 pm »
Our guess for your Round 2 clue is:
1-3-4
Wow, that was close.
We're onto you
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#### bitwise
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##### Re: Decrypto 2
« Reply #30 on: February 11, 2020, 06:57:04 pm »
Summary
team ghostofmars/Jimmmmm/Uncleeurope
The other team is actually ghostofmars/Jimmmmm/Watno, right?
Thanks for posting the summary--I agree that they're quite nice.
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#### Jimmmmm
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##### Re: Decrypto 2
« Reply #31 on: February 11, 2020, 09:45:58 pm »
Mister - Corner - Murder
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#### faust
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##### Re: Decrypto 2
« Reply #32 on: February 12, 2020, 01:36:55 am »
Summary
team ghostofmars/Jimmmmm/Uncleeurope
The other team is actually ghostofmars/Jimmmmm/Watno, right?
Thanks for posting the summary--I agree that they're quite nice.
Whoops, right... I had in mind that Eddie was still in here. Well, I'll go fix it.
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#### Watno
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##### Re: Decrypto 2
« Reply #33 on: February 13, 2020, 12:37:21 pm »
Guess for yours: 3-2-4
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#### bitwise
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##### Re: Decrypto 2
« Reply #34 on: February 13, 2020, 03:20:38 pm »
For your clue, we guess: 4-1-3
For our own team's clue, we guess 2-4-3.
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#### faust
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##### Re: Decrypto 2
« Reply #35 on: February 13, 2020, 06:28:44 pm »
For your clue, we guess: 4-1-3
For our own team's clue, we guess 2-4-3.
The correct answer was 1-4-2. Sorry guys.
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#### Watno
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##### Re: Decrypto 2
« Reply #36 on: February 14, 2020, 11:26:52 am »
guess for ours: 3-1-4
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#### Jimmmmm
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##### Re: Decrypto 2
« Reply #37 on: February 14, 2020, 07:16:59 pm »
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#### bitwise
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##### Re: Decrypto 2
« Reply #38 on: February 15, 2020, 11:48:24 am »
Our team's clue: Box - Apostrophe - Land
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#### bitwise
• Young Witch
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##### Re: Decrypto 2
« Reply #39 on: February 15, 2020, 11:51:21 am »
Summary
team ghostofmars/Jimmmmm/Watno
0 Interception tokens
0 Miscommunication tokens
Round 1 Prime - Post - Zero (code: 2-1-4, guess: 2-1-4, interception: N/A)
Round 2 Fire - Water - Ice (code: 1-2-4, guess: 1-2-4, interception: 1-3-4)
Round 3 Mister - Corner - Murder (code: 3-1-4, guess: 3-1-4, interception: 4-1-3)
Clues for 1 Post, Fire, Corner
Clues for 2 Prime, Water
Clues for 3 Mister
Clues for 4 Zero, Ice, Murder
team bitwise/faust/scolapasta
0 Interception tokens
1 Miscommunication tokens
Round 1 Sun - Culture - Oil (code: 1-2-4, guess: 1-2-4, interception: N/A)
Round 2 Building - Quake - Wizard (code: 2-3-4, guess: 2-3-4, interception: 2-4-3)
Round 3 Rise - Move - Join (code: 1-4-2, guess: 2-4-3, interception: 3-2-4)
Clues for 1 Sun, Rise
Clues for 2 Culture, Building, Join
Clues for 3 Quake
Clues for 4 Oil, Wizard, Move
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #40 on: February 16, 2020, 01:06:25 am »
Heartbeat - Spill - Switch
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#### faust
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##### Re: Decrypto 2
« Reply #41 on: February 18, 2020, 02:46:25 am »
Our guess for your clue is 2 - 3 - 4.
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #42 on: February 18, 2020, 11:07:53 am »
Our guess for your clue is 3 - 2 - 4.
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#### scolapasta
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##### Re: Decrypto 2
« Reply #43 on: February 18, 2020, 04:05:58 pm »
For ours, we guess: 4-1-2
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#### bitwise
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##### Re: Decrypto 2
« Reply #44 on: February 18, 2020, 04:07:39 pm »
For ours, we guess: 4-1-2
That is correct. We are still in the game!!
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #45 on: February 18, 2020, 04:42:33 pm »
Watno and Jimmmmm guess 2-3-1 which is correct.
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#### scolapasta
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##### Re: Decrypto 2
« Reply #46 on: February 19, 2020, 12:29:51 pm »
Round 5: Muffle-Wide-Gold
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #47 on: February 20, 2020, 02:14:09 am »
Summary
team ghostofmars/Jimmmmm/Watno
0 Interception tokens
0 Miscommunication tokens
Round 1 Prime - Post - Zero (code: 2-1-4, guess: 2-1-4, interception: N/A)
Round 2 Fire - Water - Ice (code: 1-2-4, guess: 1-2-4, interception: 1-3-4)
Round 3 Mister - Corner - Murder (code: 3-1-4, guess: 3-1-4, interception: 4-1-3)
Round 4 Heartbeat - Spill - Switch (code: 2-3-1, guess 2-3-1, interception: 2-3-4)
Clues for 1 Post, Fire, Corner. Switch
Clues for 2 Prime, Water, Heartbeat
Clues for 3 Mister, Spill
Clues for 4 Zero, Ice, Murder
team bitwise/faust/scolapasta
0 Interception tokens
1 Miscommunication tokens
Round 1 Sun - Culture - Oil (code: 1-2-4, guess: 1-2-4, interception: N/A)
Round 2 Building - Quake - Wizard (code: 2-3-4, guess: 2-3-4, interception: 2-4-3)
Round 3 Rise - Move - Join (code: 1-4-2, guess: 2-4-3, interception: 3-2-4)
Round 4 Box - Apostrophe - Land (code: 4-1-2, guess: 4-1-2, interception: 3-2-4)
Clues for 1 Sun, Rise, Apostrophe
Clues for 2 Culture, Building, Join, Land
Clues for 3 Quake
Clues for 4 Oil, Wizard, Move, Box
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#### Watno
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##### Re: Decrypto 2
« Reply #48 on: February 21, 2020, 02:12:25 am »
Waiting for Jimmmmm, right?
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #49 on: February 21, 2020, 02:17:37 am »
Waiting for Jimmmmm, right?
No, for you
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#### Watno
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##### Re: Decrypto 2
« Reply #50 on: February 22, 2020, 05:57:36 am »
Well, that explains why it's taking so long
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#### Watno
• Margrave
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##### Re: Decrypto 2
« Reply #51 on: February 22, 2020, 06:46:40 am »
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#### bitwise
• Young Witch
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##### Re: Decrypto 2
« Reply #52 on: February 22, 2020, 11:30:11 am »
Good thing you asked at least.
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#### faust
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##### Re: Decrypto 2
« Reply #53 on: February 22, 2020, 04:57:49 pm »
Our guess for your clue is 1 - 4 - 2.
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #54 on: February 24, 2020, 02:23:52 am »
Our guess for our clue is 4-1-2
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#### Watno
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##### Re: Decrypto 2
« Reply #55 on: February 24, 2020, 01:48:41 pm »
Our guess for our clue is 4-1-2
4-1-2 is correct.
Our guess for your clue is 3-1-4
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #56 on: February 25, 2020, 02:14:10 pm »
Our guess for our clue is 4-2-3 and is correct.
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#### faust
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##### Re: Decrypto 2
« Reply #57 on: February 25, 2020, 03:07:36 pm »
Summary
team ghostofmars/Jimmmmm/Watno
0 Interception tokens
0 Miscommunication tokens
Round 1 Prime - Post - Zero (code: 2-1-4, guess: 2-1-4, interception: N/A)
Round 2 Fire - Water - Ice (code: 1-2-4, guess: 1-2-4, interception: 1-3-4)
Round 3 Mister - Corner - Murder (code: 3-1-4, guess: 3-1-4, interception: 4-1-3)
Round 4 Heartbeat - Spill - Switch (code: 2-3-1, guess 2-3-1, interception: 2-3-4)
Round 5 Doctor - Shade - Air (code: 4-1-2, guess: 4-1-2, interception: 1-4-2)
Clues for 1 Post, Fire, Corner, Switch, Shade
Clues for 2 Prime, Water, Heartbeat, Air
Clues for 3 Mister, Spill
Clues for 4 Zero, Ice, Murder, Doctor
team bitwise/faust/scolapasta
0 Interception tokens
1 Miscommunication tokens
Round 1 Sun - Culture - Oil (code: 1-2-4, guess: 1-2-4, interception: N/A)
Round 2 Building - Quake - Wizard (code: 2-3-4, guess: 2-3-4, interception: 2-4-3)
Round 3 Rise - Move - Join (code: 1-4-2, guess: 2-4-3, interception: 3-2-4)
Round 4 Box - Apostrophe - Land (code: 4-1-2, guess: 4-1-2, interception: 3-2-4)
Round 5: Muffle - Wide - Gold (code: 4-2-3, guess: 4-2-3, interception: 3-1-4)
Clues for 1 Sun, Rise, Apostrophe
Clues for 2 Culture, Building, Join, Land, Wide
Clues for 3 Quake, Gold
Clues for 4 Oil, Wizard, Move, Box, Muffle
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#### faust
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##### Re: Decrypto 2
« Reply #58 on: February 25, 2020, 03:10:47 pm »
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#### faust
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##### Re: Decrypto 2
« Reply #59 on: February 29, 2020, 02:59:58 am »
Is it clear whose turn it is on your side?
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#### Watno
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##### Re: Decrypto 2
« Reply #60 on: February 29, 2020, 04:09:50 am »
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#### Jimmmmm
• Torturer
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##### Re: Decrypto 2
« Reply #61 on: February 29, 2020, 06:12:52 am »
Green - Right - Wardrobe
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #62 on: March 12, 2020, 12:18:52 am »
So, somehow both teams have been slackers here, but let's get this going again!
Our guess for your clue is: 3-2-4
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #63 on: March 13, 2020, 07:26:52 am »
Our guess for your clue 4-2-1
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#### faust
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##### Re: Decrypto 2
« Reply #64 on: March 13, 2020, 02:04:56 pm »
Our guess for your clue 4-2-1
Our team guesses 3-2-1, which is correct.
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#### ghostofmars
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##### Re: Decrypto 2
« Reply #65 on: March 15, 2020, 02:25:33 am »
We guess 3-4-1 for our clue.
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#### Jimmmmm
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##### Re: Decrypto 2
« Reply #66 on: March 15, 2020, 03:00:26 am »
We guess 3-4-1 for our clue.
Correct.
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#### Watno
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##### Re: Decrypto 2
« Reply #67 on: March 15, 2020, 01:35:16 pm »
ghostofmars and bitwise to give the next set of clues unless I'm mistaken
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#### bitwise
• Young Witch
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##### Re: Decrypto 2
« Reply #68 on: March 15, 2020, 04:00:18 pm »
Acknowledged -- I'll try to have a clue later today.
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#### bitwise
• Young Witch
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##### Re: Decrypto 2
« Reply #69 on: March 15, 2020, 10:27:28 pm »
Clue:
Knight - Seat - Tree
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#### faust
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##### Re: Decrypto 2
« Reply #70 on: March 16, 2020, 06:38:52 am »
Summary
team ghostofmars/Jimmmmm/Watno
0 Interception tokens
0 Miscommunication tokens
Round 1 Prime - Post - Zero (code: 2-1-4, guess: 2-1-4, interception: N/A)
Round 2 Fire - Water - Ice (code: 1-2-4, guess: 1-2-4, interception: 1-3-4)
Round 3 Mister - Corner - Murder (code: 3-1-4, guess: 3-1-4, interception: 4-1-3)
Round 4 Heartbeat - Spill - Switch (code: 2-3-1, guess 2-3-1, interception: 2-3-4)
Round 5 Doctor - Shade - Air (code: 4-1-2, guess: 4-1-2, interception: 1-4-2)
Round 6 Green - Right - Wardrobe (code: 3-4-1, guess: 3-4-1, interception: 3-2-4)
Clues for 1 Post, Fire, Corner, Switch, Shade, Wardrobe
Clues for 2 Prime, Water, Heartbeat, Air
Clues for 3 Mister, Spill, Green
Clues for 4 Zero, Ice, Murder, Doctor, Right
team bitwise/faust/scolapasta
0 Interception tokens
1 Miscommunication tokens
Round 1 Sun - Culture - Oil (code: 1-2-4, guess: 1-2-4, interception: N/A)
Round 2 Building - Quake - Wizard (code: 2-3-4, guess: 2-3-4, interception: 2-4-3)
Round 3 Rise - Move - Join (code: 1-4-2, guess: 2-4-3, interception: 3-2-4)
Round 4 Box - Apostrophe - Land (code: 4-1-2, guess: 4-1-2, interception: 3-2-4)
Round 5 Muffle - Wide - Gold (code: 4-2-3, guess: 4-2-3, interception: 3-1-4)
Round 6 License - Birth - Sail (code: 3-2-1, guess: 3-2-1, interception: 4-2-1)
Clues for 1 Sun, Rise, Apostrophe, Sail
Clues for 2 Culture, Building, Join, Land, Wide, Birth
Clues for 3 Quake, Gold, License
Clues for 4 Oil, Wizard, Move, Box, Muffle
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#### ghostofmars
• Moneylender
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##### Re: Decrypto 2
« Reply #71 on: March 17, 2020, 03:20:49 am »
Counter - Magic - Gym
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#### bitwise
• Young Witch
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• Posts: 133
##### Re: Decrypto 2
« Reply #72 on: March 17, 2020, 01:46:50 pm »
Our guess for your clue is 3-1-2
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#### ghostofmars
• Moneylender
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• Posts: 162
##### Re: Decrypto 2
« Reply #73 on: March 20, 2020, 04:18:47 am »
Jimmmmm and Watno guess 3-1-2, you guess the same. All of you are correct.
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#### Watno
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##### Re: Decrypto 2
« Reply #74 on: March 20, 2020, 05:54:21 am »
Guessing 2-4-1 for yours
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#### bitwise
• Young Witch
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##### Re: Decrypto 2
« Reply #75 on: March 20, 2020, 12:38:48 pm »
faust and scolapasta guess 3-4-1, which is correct.
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#### bitwise
• Young Witch
• Offline
• Posts: 133
##### Re: Decrypto 2
« Reply #76 on: March 20, 2020, 04:16:38 pm »
Summary
team ghostofmars/Jimmmmm/Watno
0 Interception tokens
0 Miscommunication tokens
Round 1 Prime - Post - Zero (code: 2-1-4, guess: 2-1-4, interception: N/A)
Round 2 Fire - Water - Ice (code: 1-2-4, guess: 1-2-4, interception: 1-3-4)
Round 3 Mister - Corner - Murder (code: 3-1-4, guess: 3-1-4, interception: 4-1-3)
Round 4 Heartbeat - Spill - Switch (code: 2-3-1, guess 2-3-1, interception: 2-3-4)
Round 5 Doctor - Shade - Air (code: 4-1-2, guess: 4-1-2, interception: 1-4-2)
Round 6 Green - Right - Wardrobe (code: 3-4-1, guess: 3-4-1, interception: 3-2-4)
Round 7 Counter - Magic - Gym (code: 3-1-2, guess: 3-1-2, interception: 3-1-2)
Clues for 1 Post, Fire, Corner, Switch, Shade, Wardrobe, Magic
Clues for 2 Prime, Water, Heartbeat, Air, Gym
Clues for 3 Mister, Spill, Green, Counter
Clues for 4 Zero, Ice, Murder, Doctor, Right
team bitwise/faust/scolapasta
1 Interception tokens
1 Miscommunication tokens
Round 1 Sun - Culture - Oil (code: 1-2-4, guess: 1-2-4, interception: N/A)
Round 2 Building - Quake - Wizard (code: 2-3-4, guess: 2-3-4, interception: 2-4-3)
Round 3 Rise - Move - Join (code: 1-4-2, guess: 2-4-3, interception: 3-2-4)
Round 4 Box - Apostrophe - Land (code: 4-1-2, guess: 4-1-2, interception: 3-2-4)
Round 5 Muffle - Wide - Gold (code: 4-2-3, guess: 4-2-3, interception: 3-1-4)
Round 6 License - Birth - Sail (code: 3-2-1, guess: 3-2-1, interception: 4-2-1)
Round 7 Knight - Seat - Tree (code: 3-4-1, guess: 3-4-1, interception: 2-4-1)
Clues for 1 Sun, Rise, Apostrophe, Sail, Tree
Clues for 2 Culture, Building, Join, Land, Wide, Birth
Clues for 3 Quake, Gold, License, Knight
Clues for 4 Oil, Wizard, Move, Box, Muffle, Seat
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#### Watno
• Margrave
• Offline
• Posts: 2745
##### Re: Decrypto 2
« Reply #77 on: March 22, 2020, 01:50:46 pm »
fish - uphill - stalk
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#### scolapasta
• Minion
• Offline
• Posts: 579
##### Re: Decrypto 2
« Reply #78 on: March 23, 2020, 05:47:57 pm »
Sorry, everyone, I completely slipped on this! I'll have our team's clue here tonight (i.e. next few hours).
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #79 on: March 23, 2020, 06:53:41 pm »
Dance - Family - Jamaica
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #80 on: March 27, 2020, 07:42:36 pm »
Our guess for your clue is 2 - 4 -3.
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#### bitwise
• Young Witch
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• Posts: 133
##### Re: Decrypto 2
« Reply #81 on: March 29, 2020, 09:17:02 pm »
Ping to the other team. We're so close to the end--let's finish!
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #82 on: March 31, 2020, 09:36:53 pm »
bump
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#### Jimmmmm
• Torturer
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##### Re: Decrypto 2
« Reply #83 on: March 31, 2020, 09:39:22 pm »
Our guess for ours is 1-2-3
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#### ghostofmars
• Moneylender
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• Posts: 162
##### Re: Decrypto 2
« Reply #84 on: April 02, 2020, 04:16:31 am »
Our guess for your clue is 3 - 4 - 2.
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#### scolapasta
• Minion
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• Posts: 579
##### Re: Decrypto 2
« Reply #85 on: April 02, 2020, 11:47:14 am »
Our guess for our clue was 1-4-2.
1-4-2 is correct.
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#### Watno
• Margrave
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##### Re: Decrypto 2
« Reply #86 on: April 02, 2020, 02:43:03 pm »
1-2-3 is correct for ours.
That means we need to guess the other teams words now, unless I'm mistaken.
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#### faust
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• Posts: 3390
##### Re: Decrypto 2
« Reply #87 on: April 02, 2020, 04:54:01 pm »
Yes, that's what the rules say.
Here are our guesses for your words:
1. Lamp
2. Pump
3. Bean
4. Degree
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#### Watno
• Margrave
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##### Re: Decrypto 2
« Reply #88 on: April 03, 2020, 12:39:32 pm »
Our guesses:
• Horizon
• Canal
• Bond
• Car
Yours are al correct, so you probably won.
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#### scolapasta
• Minion
• Offline
• Posts: 579
##### Re: Decrypto 2
« Reply #89 on: April 03, 2020, 01:21:08 pm »
NICE!
Our words were:
1. Hawaii
2. Nation
3. Plate
4. Van
GG!
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#### bitwise
• Young Witch
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• Posts: 133
##### Re: Decrypto 2
« Reply #90 on: April 03, 2020, 02:16:10 pm »
Good game!
Still curious about how Wizard -> Van and about the entire clue we got a miscommunication on.
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#### faust
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##### Re: Decrypto 2
« Reply #91 on: April 03, 2020, 02:38:24 pm »
Good game!
Still curious about how Wizard -> Van and about the entire clue we got a miscommunication on.
Yeah, that one was mishandled, it was intentionally vague, but too much so. Rise -> Hawaii was like Hawaii rising out of the ocean due to Volcanoes (but I didn't think about the more direct Rise -> Nation connection). And Join for Nation was supposed to be something that wouldn't fit with anything else.
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #92 on: April 03, 2020, 03:11:43 pm »
Good game!
Still curious about how Wizard -> Van and about the entire clue we got a miscommunication on.
Wizard was definitely a stretch, but when I think of vans I always think of people having crazy stuff painted on them, and unicorns and wizards seemed to be common:
The idea was that a) it wouldn't fit anywhere else, and b) it would really throw them off.
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#### Jimmmmm
• Torturer
• Online
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##### Re: Decrypto 2
« Reply #93 on: April 03, 2020, 04:52:13 pm »
Good game, I really had no idea what your words were in the end. I'll sit out if there's another game.
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #94 on: April 04, 2020, 01:25:12 pm »
Who's up for another?
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#### bitwise
• Young Witch
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##### Re: Decrypto 2
« Reply #95 on: April 04, 2020, 02:57:40 pm »
Count me in.
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#### ghostofmars
• Moneylender
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• Posts: 162
##### Re: Decrypto 2
« Reply #96 on: April 05, 2020, 07:25:17 am »
/in
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#### faust
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##### Re: Decrypto 2
« Reply #97 on: April 06, 2020, 05:25:11 am »
/in
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#### Watno
• Margrave
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##### Re: Decrypto 2
« Reply #98 on: April 08, 2020, 02:48:30 am »
Im out, have fun
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#### faust
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##### Re: Decrypto 2
« Reply #99 on: April 08, 2020, 11:44:32 am »
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You say the whole world's ending, honey it already did
#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #100 on: April 08, 2020, 12:01:27 pm »
Im out, have fun
Great playing with you!
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #101 on: April 08, 2020, 12:01:52 pm »
Jimmmm, too!
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#### scolapasta
• Minion
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##### Re: Decrypto 2
« Reply #102 on: April 12, 2020, 12:48:24 pm »
For reference in anyone looking back, here's our QT:
https://www.quicktopic.com/52/H/FkR6iyXMccJ
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Pages: 1 2 3 ... 5 [All]
Page created in 2.983 seconds with 20 queries. | 11,742 | 34,467 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2024-26 | latest | en | 0.924944 |
https://www.mathworks.com/matlabcentral/cody/problems/1658-simple-equation-annual-salary/solutions/2052391 | 1,579,911,225,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250626449.79/warc/CC-MAIN-20200124221147-20200125010147-00390.warc.gz | 982,978,814 | 15,252 | Cody
# Problem 1658. Simple equation: Annual salary
Solution 2052391
Submitted on 9 Dec 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
wage = 10; salary_correct = wage * 40 * 50; assert(isequal(ComputeSalary(wage),salary_correct))
Error in solution: Line: 6 Column: 1 This statement is not inside any function. (It follows the END that terminates the definition of the function "ComputeSalary".)
2 Fail
wage = 30; salary_correct = wage * 40 * 50; assert(isequal(ComputeSalary(wage),salary_correct))
Error in solution: Line: 6 Column: 1 This statement is not inside any function. (It follows the END that terminates the definition of the function "ComputeSalary".)
3 Fail
wage = 12.50; salary_correct = wage * 40 * 50; assert(isequal(ComputeSalary(wage),salary_correct))
Error in solution: Line: 6 Column: 1 This statement is not inside any function. (It follows the END that terminates the definition of the function "ComputeSalary".) | 280 | 1,072 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2020-05 | latest | en | 0.709599 |
https://ruby-doc.org/stdlib-2.6/libdoc/matrix/rdoc/Vector.html | 1,555,601,331,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578517682.16/warc/CC-MAIN-20190418141430-20190418163430-00107.warc.gz | 548,836,565 | 11,759 | • matrix.rb
Quicksearch
# Vector
The `Vector` class represents a mathematical vector, which is useful in its own right, and also constitutes a row or column of a Matrix.
## Method Catalogue¶ ↑
To create a Vector:
To access elements:
To set elements:
To enumerate the elements:
Properties of vectors:
Vector arithmetic:
• #*(x) “is matrix or number”
• #+(v)
• #-(v)
• #/(v)
• #+@
• #-@
Vector functions:
Conversion to other data types:
String representations:
### Attributes
elements[R]
INSTANCE CREATION
### Public Class Methods
[](*array) click to toggle source
Creates a Vector from a list of elements.
`Vector[7, 4, ...]`
```
# File matrix.rb, line 1863
def Vector.[](*array)
new convert_to_array(array, false)
end
```
basis(size:, index:) click to toggle source
Returns a standard basis `n`-vector, where k is the index.
`Vector.basis(size:, index:) # => Vector[0, 1, 0]`
```
# File matrix.rb, line 1880
def Vector.basis(size:, index:)
raise ArgumentError, "invalid size (#{size} for 1..)" if size < 1
raise ArgumentError, "invalid index (#{index} for 0...#{size})" unless 0 <= index && index < size
array = Array.new(size, 0)
array[index] = 1
new convert_to_array(array, false)
end
```
elements(array, copy = true) click to toggle source
Creates a vector from an Array. The optional second argument specifies whether the array itself or a copy is used internally.
```
# File matrix.rb, line 1871
def Vector.elements(array, copy = true)
new convert_to_array(array, copy)
end
```
independent?(*vs) click to toggle source
Returns `true` iff all of vectors are linearly independent.
```Vector.independent?(Vector[1,0], Vector[0,1])
=> true
Vector.independent?(Vector[1,2], Vector[2,4])
=> false```
```
# File matrix.rb, line 2027
def Vector.independent?(*vs)
vs.each do |v|
raise TypeError, "expected Vector, got #{v.class}" unless v.is_a?(Vector)
Vector.Raise ErrDimensionMismatch unless v.size == vs.first.size
end
return false if vs.count > vs.first.size
Matrix[*vs].rank.eql?(vs.count)
end
```
new(array) click to toggle source
::new is private; use Vector[] or ::elements to create.
```
# File matrix.rb, line 1902
def initialize(array)
# No checking is done at this point.
@elements = array
end
```
zero(size) click to toggle source
Return a zero vector.
`Vector.zero(3) => Vector[0, 0, 0]`
```
# File matrix.rb, line 1893
def Vector.zero(size)
raise ArgumentError, "invalid size (#{size} for 0..)" if size < 0
array = Array.new(size, 0)
new convert_to_array(array, false)
end
```
### Public Instance Methods
*(x) click to toggle source
Multiplies the vector by `x`, where `x` is a number or a matrix.
```
# File matrix.rb, line 2101
def *(x)
case x
when Numeric
els = @elements.collect{|e| e * x}
self.class.elements(els, false)
when Matrix
Matrix.column_vector(self) * x
when Vector
Vector.Raise ErrOperationNotDefined, "*", self.class, x.class
else
apply_through_coercion(x, __method__)
end
end
```
+(v) click to toggle source
Vector addition.
```
# File matrix.rb, line 2118
def +(v)
case v
when Vector
Vector.Raise ErrDimensionMismatch if size != v.size
els = collect2(v) {|v1, v2|
v1 + v2
}
self.class.elements(els, false)
when Matrix
Matrix.column_vector(self) + v
else
apply_through_coercion(v, __method__)
end
end
```
+@() click to toggle source
```
# File matrix.rb, line 2166
def +@
self
end
```
-(v) click to toggle source
Vector subtraction.
```
# File matrix.rb, line 2136
def -(v)
case v
when Vector
Vector.Raise ErrDimensionMismatch if size != v.size
els = collect2(v) {|v1, v2|
v1 - v2
}
self.class.elements(els, false)
when Matrix
Matrix.column_vector(self) - v
else
apply_through_coercion(v, __method__)
end
end
```
-@() click to toggle source
```
# File matrix.rb, line 2170
def -@
collect {|e| -e }
end
```
/(x) click to toggle source
Vector division.
```
# File matrix.rb, line 2154
def /(x)
case x
when Numeric
els = @elements.collect{|e| e / x}
self.class.elements(els, false)
when Matrix, Vector
Vector.Raise ErrOperationNotDefined, "/", self.class, x.class
else
apply_through_coercion(x, __method__)
end
end
```
==(other) click to toggle source
Returns `true` iff the two vectors have the same elements in the same order.
```
# File matrix.rb, line 2077
def ==(other)
return false unless Vector === other
@elements == other.elements
end
```
vector[range] click to toggle source
vector[integer]
Returns element or elements of the vector.
```
# File matrix.rb, line 1916
def [](i)
@elements[i]
end
```
Also aliased as: element, component
vector[range] = new_vector click to toggle source
vector[range] = row_matrix
vector[range] = new_element
vector[integer] = new_element
Set element or elements of vector.
```
# File matrix.rb, line 1931
def []=(i, v)
raise FrozenError, "can't modify frozen Vector" if frozen?
if i.is_a?(Range)
range = Matrix::CoercionHelper.check_range(i, size, :vector)
set_range(range, v)
else
index = Matrix::CoercionHelper.check_int(i, size, :index)
set_value(index, v)
end
end
```
Also aliased as: set_element, set_component
angle_with(v) click to toggle source
Returns an angle with another vector. Result is within the [0..Math::PI].
```Vector[1,0].angle_with(Vector[0,1])
# => Math::PI / 2
```
```
# File matrix.rb, line 2286
def angle_with(v)
raise TypeError, "Expected a Vector, got a #{v.class}" unless v.is_a?(Vector)
Vector.Raise ErrDimensionMismatch if size != v.size
prod = magnitude * v.magnitude
raise ZeroVectorError, "Can't get angle of zero vector" if prod == 0
dot = inner_product(v)
if dot.abs >= prod
dot.positive? ? 0 : Math::PI
else
Math.acos(dot / prod)
end
end
```
coerce(other) click to toggle source
The coerce method provides support for Ruby type coercion. This coercion mechanism is used by Ruby to handle mixed-type numeric operations: it is intended to find a compatible common type between the two operands of the operator. See also Numeric#coerce.
```
# File matrix.rb, line 2346
def coerce(other)
case other
when Numeric
return Matrix::Scalar.new(other), self
else
raise TypeError, "#{self.class} can't be coerced into #{other.class}"
end
end
```
collect() click to toggle source
Like Array#collect.
```
# File matrix.rb, line 2230
def collect(&block) # :yield: e
return to_enum(:collect) unless block_given?
els = @elements.collect(&block)
self.class.elements(els, false)
end
```
Also aliased as: map
collect!(&block) click to toggle source
Like Array#collect!
```
# File matrix.rb, line 2240
def collect!(&block)
return to_enum(:collect!) unless block_given?
raise FrozenError, "can't modify frozen Vector" if frozen?
@elements.collect!(&block)
self
end
```
Also aliased as: map!
collect2(v) click to toggle source
Collects (as in Enumerable#collect) over the elements of this vector and `v` in conjunction.
```
# File matrix.rb, line 2005
def collect2(v) # :yield: e1, e2
raise TypeError, "Integer is not like Vector" if v.kind_of?(Integer)
Vector.Raise ErrDimensionMismatch if size != v.size
return to_enum(:collect2, v) unless block_given?
Array.new(size) do |i|
yield @elements[i], v[i]
end
end
```
component(i) click to toggle source
Alias for: []
covector() click to toggle source
Creates a single-row matrix from this vector.
```
# File matrix.rb, line 2306
def covector
Matrix.row_vector(self)
end
```
cross(*vs) click to toggle source
Alias for: cross_product
cross_product(*vs) click to toggle source
Returns the cross product of this vector with the others.
```Vector[1, 0, 0].cross_product Vector[0, 1, 0] => Vector[0, 0, 1]
```
It is generalized to other dimensions to return a vector perpendicular to the arguments.
```Vector[1, 2].cross_product # => Vector[-2, 1]
Vector[1, 0, 0, 0].cross_product(
Vector[0, 1, 0, 0],
Vector[0, 0, 1, 0]
) #=> Vector[0, 0, 0, 1]
```
```
# File matrix.rb, line 2205
def cross_product(*vs)
raise ErrOperationNotDefined, "cross product is not defined on vectors of dimension #{size}" unless size >= 2
raise ArgumentError, "wrong number of arguments (#{vs.size} for #{size - 2})" unless vs.size == size - 2
vs.each do |v|
raise TypeError, "expected Vector, got #{v.class}" unless v.is_a? Vector
Vector.Raise ErrDimensionMismatch unless v.size == size
end
case size
when 2
Vector[-@elements[1], @elements[0]]
when 3
v = vs[0]
Vector[ v[2]*@elements[1] - v[1]*@elements[2],
v[0]*@elements[2] - v[2]*@elements[0],
v[1]*@elements[0] - v[0]*@elements[1] ]
else
rows = self, *vs, Array.new(size) {|i| Vector.basis(size: size, index: i) }
Matrix.rows(rows).laplace_expansion(row: size - 1)
end
end
```
Also aliased as: cross
dot(v) click to toggle source
Alias for: inner_product
each(&block) click to toggle source
Iterate over the elements of this vector
```
# File matrix.rb, line 1982
def each(&block)
return to_enum(:each) unless block_given?
@elements.each(&block)
self
end
```
each2(v) click to toggle source
Iterate over the elements of this vector and `v` in conjunction.
```
# File matrix.rb, line 1991
def each2(v) # :yield: e1, e2
raise TypeError, "Integer is not like Vector" if v.kind_of?(Integer)
Vector.Raise ErrDimensionMismatch if size != v.size
return to_enum(:each2, v) unless block_given?
size.times do |i|
yield @elements[i], v[i]
end
self
end
```
element(i) click to toggle source
Alias for: []
elements_to_f() click to toggle source
```
# File matrix.rb, line 2324
def elements_to_f
warn "Vector#elements_to_f is deprecated", uplevel: 1
map(&:to_f)
end
```
elements_to_i() click to toggle source
```
# File matrix.rb, line 2329
def elements_to_i
warn "Vector#elements_to_i is deprecated", uplevel: 1
map(&:to_i)
end
```
elements_to_r() click to toggle source
```
# File matrix.rb, line 2334
def elements_to_r
warn "Vector#elements_to_r is deprecated", uplevel: 1
map(&:to_r)
end
```
eql?(other) click to toggle source
```
# File matrix.rb, line 2082
def eql?(other)
return false unless Vector === other
@elements.eql? other.elements
end
```
freeze() click to toggle source
```
# File matrix.rb, line 2056
def freeze
@elements.freeze
super
end
```
hash() click to toggle source
Returns a hash-code for the vector.
```
# File matrix.rb, line 2090
def hash
@elements.hash
end
```
independent?(*vs) click to toggle source
Returns `true` iff all of vectors are linearly independent.
```Vector[1,0].independent?(Vector[0,1])
=> true
Vector[1,2].independent?(Vector[2,4])
=> false```
```
# File matrix.rb, line 2045
def independent?(*vs)
self.class.independent?(self, *vs)
end
```
inner_product(v) click to toggle source
Returns the inner product of this vector with the other.
```Vector[4,7].inner_product Vector[10,1] => 47
```
```
# File matrix.rb, line 2182
def inner_product(v)
Vector.Raise ErrDimensionMismatch if size != v.size
p = 0
each2(v) {|v1, v2|
p += v1 * v2.conj
}
p
end
```
Also aliased as: dot
inspect() click to toggle source
Overrides Object#inspect
```
# File matrix.rb, line 2369
def inspect
"Vector" + @elements.inspect
end
```
magnitude() click to toggle source
Returns the modulus (Pythagorean distance) of the vector.
`Vector[5,8,2].r => 9.643650761`
```
# File matrix.rb, line 2252
def magnitude
Math.sqrt(@elements.inject(0) {|v, e| v + e.abs2})
end
```
Also aliased as: r, norm
map() click to toggle source
Alias for: collect
map!(&block) click to toggle source
Alias for: collect!
map2(v) click to toggle source
Like #collect2, but returns a Vector instead of an Array.
```
# File matrix.rb, line 2261
def map2(v, &block) # :yield: e1, e2
return to_enum(:map2, v) unless block_given?
els = collect2(v, &block)
self.class.elements(els, false)
end
```
norm() click to toggle source
Alias for: magnitude
normalize() click to toggle source
Returns a new vector with the same direction but with norm 1.
```v = Vector[5,8,2].normalize
# => Vector[0.5184758473652127, 0.8295613557843402, 0.20739033894608505]
v.norm => 1.0```
```
# File matrix.rb, line 2275
def normalize
n = magnitude
raise ZeroVectorError, "Zero vectors can not be normalized" if n == 0
self / n
end
```
r() click to toggle source
Alias for: magnitude
round(ndigits=0) click to toggle source
Returns a vector with entries rounded to the given precision (see Float#round)
```
# File matrix.rb, line 1964
def round(ndigits=0)
map{|e| e.round(ndigits)}
end
```
size() click to toggle source
Returns the number of elements in the vector.
```
# File matrix.rb, line 1971
def size
@elements.size
end
```
to_a() click to toggle source
Returns the elements of the vector in an array.
```
# File matrix.rb, line 2313
def to_a
@elements.dup
end
```
to_matrix() click to toggle source
Return a single-column matrix from this vector
```
# File matrix.rb, line 2320
def to_matrix
Matrix.column_vector(self)
end
```
to_s() click to toggle source
Overrides Object#to_s
```
# File matrix.rb, line 2362
def to_s
"Vector[" + @elements.join(", ") + "]"
end
```
zero?() click to toggle source
Returns `true` iff all elements are zero.
```
# File matrix.rb, line 2052
def zero?
all?(&:zero?)
end
``` | 3,783 | 12,919 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-18 | longest | en | 0.460726 |
https://www.exceldemy.com/tag/calculate-interest-in-excel/ | 1,725,916,143,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651157.15/warc/CC-MAIN-20240909201932-20240909231932-00528.warc.gz | 711,396,889 | 47,242 | Calculate Interest in Excel
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### How to Calculate the Gold Loan Interest in Excel – 2 Methods
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### How to Calculate Principal and Interest on a Loan in Excel
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### How to Calculate Daily Interest in Excel (2 Easy Ways)
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### How to Calculate the Accrued Interest on a Fixed Deposit in Excel – 3 Methods
This is the sample dataset. It showcases Fixed Deposit (P), Annual Rate (r), Interest accrued per Year (n) and Years elapsed (t). To ...
### How to Calculate the Credit Card Interest in Excel – 3 Steps
To calculate the interest on a credit card you need to know the Current Balance, Minimum Payment Percentage, and Annual Interest Rate. This is the ...
### How to Calculate the Interest Between Two Dates in Excel (3 Ways)
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### How to Calculate Accrued Interest on a Loan in Excel (3 ways)
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Advanced Excel Exercises with Solutions PDF | 707 | 3,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-38 | latest | en | 0.879376 |
https://www.tutorialspoint.com/p-a-rectangle-has-a-length-15-cm-and-the-perimeter-is-50-cm-find-the-breadth-of-the-rectangle-p | 1,695,852,442,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510326.82/warc/CC-MAIN-20230927203115-20230927233115-00892.warc.gz | 1,132,945,390 | 17,226 | # A rectangle has a length 15 cm and the perimeter is 50 cm. Find the breadth of the rectangle.
Given:
Length of rectangle = 15 cm
Perimeter of rectangle = 50 cm
To do:
Solution :
Perimeter of Rectangle = 2 (length $+$ breadth)
Substitute the given values in the above formula,
50 = 2 (15 $+$ breadth)
$\frac{50}{2}=$(15 $+$ breadth)
25 = 15 $+$ breadth
$25-15$ = breadth
The breadth of the rectangle is 10 cm.
Tutorialspoint
Simply Easy Learning
Updated on: 10-Oct-2022
205 Views | 148 | 503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-40 | latest | en | 0.618069 |
https://de.mathworks.com/matlabcentral/cody/problems/5-triangle-numbers/solutions/2115022 | 1,586,549,946,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370511408.40/warc/CC-MAIN-20200410173109-20200410203609-00329.warc.gz | 425,859,688 | 15,639 | Cody
# Problem 5. Triangle Numbers
Solution 2115022
Submitted on 2 Feb 2020 by It's me.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
n = 1; t = 1; assert(isequal(triangle(n),t))
2 Pass
n = 3; t = 6; assert(isequal(triangle(n),t))
3 Pass
n = 5; t = 15; assert(isequal(triangle(n),t))
4 Pass
n = 30; t = 465; assert(isequal(triangle(n),t)) | 161 | 472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2020-16 | latest | en | 0.703919 |
http://www.mathworks.se/help/dsp/ref/adaptfilt.nlms.html?nocookie=true | 1,394,713,562,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678674071/warc/CC-MAIN-20140313024434-00008-ip-10-183-142-35.ec2.internal.warc.gz | 393,881,010 | 10,586 | Accelerating the pace of engineering and science
Documentation Center
• Trial Software
Normalized least mean squares adaptive filter
Description
ha = adaptfilt.nlms(l,step,leakage,offset,coeffs,states) constructs a normalized least-mean squares (NLMS) FIR adaptive filter object named ha.
For information on how to run data through your adaptive filter object, see the Adaptive Filter Syntaxes section of the reference page for filter.
Input Arguments
Entries in the following table describe the input arguments for adaptfilt.nlms.
Input Argument
Description
l
Adaptive filter length (the number of coefficients or taps) and it must be a positive integer. l defaults to 10.
step
NLMS step size. It must be a scalar between 0 and 2. Setting this step size value to one provides the fastest convergence. step defaults to 1.
leakage
NLMS leakage factor. It must be a scalar between zero and one. When it is less than one, a leaky NLMS algorithm results. leakage defaults to 1 (no leakage).
offset
Specifies an optional offset for the denominator of the step size normalization term. You must specify offset to be a scalar greater than or equal to zero. Nonzero offsets can help avoid a divide-by-near-zero condition that causes errors. Use this to avoid dividing by zero (or by very small numbers) when the square of the input data norm becomes very small (when the input signal amplitude becomes very small). When you omit it, offset defaults to zero.
coeffs
Vector composed of your initial filter coefficients. Enter a length l vector. coeffs defaults to a vector of zeros with length equal to the filter order.
states
Your initial adaptive filter states appear in the states vector. It must be a vector of length l-1. states defaults to a length l-1 vector with zeros for all of the elements.
Properties
In the syntax for creating the adaptfilt object, the input options are properties of the object you create. This table lists the properties for normalized LMS objects, their default values, and a brief description of the property.
Property
Range
Property Description
Algorithm
None
Coefficients
Vector of elements
Vector containing the initial filter coefficients. It must be a length l vector where l is the number of filter coefficients. coeffs defaults to length l vector of zeros when you do not provide the argument for input.
FilterLength
Any positive integer
Reports the length of the filter, the number of coefficients or taps
Leakage
0 to 1
NLMS leakage factor. It must be a scalar between zero and one. When it is less than one, a leaky NLMS algorithm results. leakage defaults to 1 (no leakage).
Offset
0 or greater
Specifies an optional offset for the denominator of the step size normalization term. You must specify offset to be a scalar greater than or equal to zero. Nonzero offsets can help avoid a divide-by-near-zero condition that causes errors. Use this to avoid dividing by zero (or by very small numbers) when the square of the input data norm becomes very small (when the input signal amplitude becomes very small). When you omit it, offset defaults to zero.
PersistentMemory
false or true
Determine whether the filter states and coefficients get restored to their starting values for each filtering operation. The starting values are the values in place when you create the filter. PersistentMemory returns to zero any property value that the filter changes during processing. Property values that the filter does not change are not affected. Defaults to false.
States
Vector of elements, data type double
Vector of the adaptive filter states. states defaults to a vector of zeros which has length equal to (l - 1).
StepSize
0 to 1
NLMS step size. It must be a scalar between zero and one. Setting this step size value to one provides the fastest convergence. step defaults to one.
Examples
To help you compare this algorithm's performance to other LMS-based algorithms, such as BLMS or LMS, this example demonstrates the NLMS adaptive filter in use to identify the coefficients of an unknown FIR filter of order equal to 32 — an example used in other adaptive filter examples.
```x = randn(1,500); % Input to the filter
b = fir1(31,0.5); % FIR system to be identified
n = 0.1*randn(1,500); % Observation noise signal
d = filter(b,1,x)+n; % Desired signal
mu = 1; % NLMS step size
offset = 50; % NLMS offset
[y,e] = filter(ha,x,d);
subplot(2,1,1);
plot(1:500,[d;y;e]);
legend('Desired','Output','Error');
title('System Identification of FIR Filter');
xlabel('Time Index'); ylabel('Signal Value');
subplot(2,1,2);
stem([b', ha.coefficients']);
legend('Actual','Estimated'); grid on;
xlabel('Coefficient #'); ylabel('Coefficient Value');```
As you see from the figure, the nlms variant again closely matches the actual filter coefficients in the unknown FIR filter. | 1,079 | 4,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2014-10 | longest | en | 0.821731 |
https://developer.unigine.com/en/docs/future/api/library/math/math.matrix | 1,701,733,666,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100535.26/warc/CC-MAIN-20231204214708-20231205004708-00400.warc.gz | 255,663,141 | 63,901 | # Math Matrix Functions
This class represents a collection of matrix functions.
Notice
Math matrix functions are the members of the Unigine::Math namespace.
## mat4composeRotationXYZ ( const vec3 & r ) #
Composes a rotation matrix from the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - XYZ. It is an order of the rings in the three-axis gimbal set: X axis used as the outer ring (independent ring), while Z axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const vec3 & r - Vector, containing Euler angles, in degrees - (yaw, roll, pitch).
### Return value
Composed rotation matrix.
## mat4composeRotationXZY ( const vec3 & r ) #
Composes a rotation matrix from the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - XZY. It is an order of the rings in the three-axis gimbal set: X axis used as the outer ring (independent ring), while Y axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const vec3 & r - Vector, containing Euler angles, in degrees - (yaw, roll, pitch).
### Return value
Composed rotation matrix.
## mat4composeRotationYXZ ( const vec3 & r ) #
Composes a rotation matrix from the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - YXZ. It is an order of the rings in the three-axis gimbal set: Y axis used as the outer ring (independent ring), while Z axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const vec3 & r - Vector, containing Euler angles, in degrees - (yaw, roll, pitch).
### Return value
Composed rotation matrix.
## mat4composeRotationYZX ( const vec3 & r ) #
Composes a rotation matrix from the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - YZX. It is an order of the rings in the three-axis gimbal set: Y axis used as the outer ring (independent ring), while X axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const vec3 & r - Vector, containing Euler angles, in degrees - (yaw, roll, pitch).
### Return value
Composed rotation matrix.
## mat4composeRotationZXY ( const vec3 & r ) #
Composes a rotation matrix from the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - ZXY. It is an order of the rings in the three-axis gimbal set: Z axis used as the outer ring (independent ring), while Y axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const vec3 & r - Vector, containing Euler angles, in degrees - (yaw, roll, pitch).
### Return value
Composed rotation matrix.
## mat4composeRotationZYX ( const vec3 & r ) #
Composes a rotation matrix from the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - ZYX. It is an order of the rings in the three-axis gimbal set: Z axis used as the outer ring (independent ring), while X axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const vec3 & r - Vector, containing Euler angles, in degrees - (yaw, roll, pitch).
### Return value
Composed rotation matrix.
## mat4 &composeTransform ( mat4 & ret, const vec4 & position, const quat & rot ) #
Returns the transformation matrix for the specified position and rotation.
### Arguments
• mat4 & ret - Output matrix, to which the resulting transformation matrix will be put.
• const vec4 & position
• const quat & rot - Rotation quaternion.
### Return value
Resulting transformation matrix.
## mat4 &composeTransform ( mat4 & ret, const vec3 & position, const quat & rot, const vec3 & scale ) #
Returns the transformation matrix for the specified position, rotation and scale. For more information see Matrix Transformations.
### Arguments
• mat4 & ret - Output matrix, to which the resulting transformation matrix will be put.
• const vec3 & position - Position coordinates (world).
• const quat & rot - Rotation quaternion.
• const vec3 & scale - Scaling vector (scale.x, scale.y, scale.z).
### Return value
Resulting transformation matrix.
## dmat4 &composeTransform ( dmat4 & ret, const dvec3 & position, const quat & rot, const vec3 & scale ) #
Returns the transformation matrix for the specified position, rotation and scale. For more information see Matrix Transformations.
### Arguments
• dmat4 & ret - Output matrix, to which the resulting transformation matrix will be put.
• const dvec3 & position
• const quat & rot - Rotation quaternion.
• const vec3 & scale - Scaling vector (scale.x, scale.y, scale.z).
### Return value
Resulting transformation matrix.
## mat4 &composeTransform ( mat4 & ret, const quat & q0, const quat & q1, const vec3 & scale ) #
Returns the transformation matrix for the specified position, rotation and scale. For more information see Matrix Transformations.
### Arguments
• mat4 & ret - Output matrix, to which the resulting transformation matrix will be put.
• const quat & q0
• const quat & q1
• const vec3 & scale - Scaling vector (scale.x, scale.y, scale.z).
### Return value
Resulting transformation matrix.
## mat4cubeTransform ( int face ) #
Returns cube viewing matrix for the given cube face.
### Arguments
• int face - Cube face number.
### Return value
Cube viewing matrix.
## voiddecomposeProjection ( const mat4 & projection, float & znear, float & zfar ) #
Decomposes a given projection matrix, extracting distances to near and far clipping planes.
### Arguments
• const mat4 & projection - Projection matrix.
• float & znear - Near clipping plane.
• float & zfar - Far clipping plane.
## vec3decomposeRotationXYZ ( const mat3 & t ) #
Decomposes a given rotation matrix to the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - XYZ. It is an order of the rings in the three-axis gimbal set: X axis used as the outer ring (independent ring), while Z axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const mat3 & t - Rotation matrix to decompose.
### Return value
Vector, containing Euler angles, in degrees - (pitch, roll, yaw).
## vec3decomposeRotationXZY ( const mat3 & t ) #
Decomposes a given rotation matrix to the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - XZY. It is an order of the rings in the three-axis gimbal set: X axis used as the outer ring (independent ring), while Y axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const mat3 & t - Rotation matrix to decompose.
### Return value
Vector, containing Euler angles, in degrees - (pitch, yaw, roll).
## vec3decomposeRotationYXZ ( const mat3 & t ) #
Decomposes a given rotation matrix to the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - YXZ. It is an order of the rings in the three-axis gimbal set: Y axis used as the outer ring (independent ring), while Z axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const mat3 & t - Rotation matrix to decompose.
### Return value
Vector, containing Euler angles, in degrees - (roll, pitch, yaw).
## vec3decomposeRotationYZX ( const mat3 & t ) #
Decomposes a given rotation matrix to the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - YZX. It is an order of the rings in the three-axis gimbal set: Y axis used as the outer ring (independent ring), while X axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const mat3 & t - Rotation matrix to decompose.
### Return value
Vector, containing Euler angles, in degrees - (roll, yaw, pitch).
## vec3decomposeRotationZXY ( const mat3 & t ) #
Decomposes a given rotation matrix to the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - ZXY. It is an order of the rings in the three-axis gimbal set: Z axis used as the outer ring (independent ring), while Y axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const mat3 & t - Rotation matrix to decompose.
### Return value
Vector, containing Euler angles, in degrees - (yaw, pitch, roll).
## vec3decomposeRotationZYX ( const mat3 & t ) #
Decomposes a given rotation matrix to the corresponding Euler angles. The Euler angles are specified in the axis rotation sequence - ZYX. It is an order of the rings in the three-axis gimbal set: Z axis used as the outer ring (independent ring), while X axis as the inner one (its rotation depends on other 2 rings).
When we talk about axes in UNIGINE we assume that:
• X axis points to the right giving us a pitch angle.
• Y axis points forward giving us a roll angle.
• Z axis points up giving us a yaw (heading) angle.
Notice
Players have a different coordinate system:
• X axis points to the right giving us a pitch angle.
• Y axis points up giving us a yaw (heading) angle.
• Z axis points backward giving us a -roll angle.
### Arguments
• const mat3 & t - Rotation matrix to decompose.
### Return value
Vector, containing Euler angles, in degrees - (yaw, roll, pitch).
## voiddecomposeTransform ( const mat4 & m, vec4 & position, quat & rot ) #
Decomposes a given transformation matrix into a vector representing translation and uniform scale and a quaternion representing rotation.
### Arguments
• const mat4 & m - Transformation matrix to decompose.
• vec4 & position - Output vector (X, Y, Z, W), to which translation and scale components of the transformation will be put. (X, Y, Z) - represent translation, W = (scale.X + scale.Y + scale.z) / 3.
• quat & rot - Output quaternion, to which the rotation component of the transformation will pe put.
## voiddecomposeTransform ( const mat4 & m, vec3 & position, quat & rot, vec3 & scale ) #
Decomposes a given transformation matrix into translation, rotation and scale components.
### Arguments
• const mat4 & m - Transformation matrix to decompose.
• vec3 & position - Output vector, to which the translation component of the transformation will be put.
• quat & rot - Output quaternion, to which the rotation component of the transformation will be put.
• vec3 & scale - Output vector, to which the scale component of the transformation will be put.
## voiddecomposeTransform ( const dmat4 & m, dvec3 & position, quat & rot, vec3 & scale ) #
Decomposes a given transformation matrix into translation, rotation and scale components.
### Arguments
• const dmat4 & m - Transformation matrix to decompose.
• dvec3 & position - Output vector, to which the translation component of the transformation will be put.
• quat & rot - Output quaternion, to which the rotation component of the transformation will be put.
• vec3 & scale - Output vector, to which the scale component of the transformation will be put.
## voiddecomposeTransform ( const mat4 & m, quat & q0, quat & q1, vec3 & scale ) #
Decomposes a given transformation matrix into a dual quaternion (representing both translation and rotation) and a scale vector. The dual-quaternion model is an accurate, computationally efficient, robust, and flexible method of representing rigid transforms and it is used in skeletal animation. See a Wikipedia article on dual quaternions and a beginners guide to dual-quaternions for more information.
### Arguments
• const mat4 & m - Transformation matrix to decompose.
• quat & q0 - Quaternion representing a real part of the dual quaternion.
• quat & q1 - Quaternion representing a dual part of the dual quaternion.
• vec3 & scale - Scale part of the transformation.
## floatdeterminant ( const mat2 & m ) #
Returns the determinant of the given matrix.
### Arguments
• const mat2 & m - Matrix, for which the determinant is to be calculated.
### Return value
Matrix determinant.
## floatdeterminant ( const mat3 & m ) #
Returns the determinant of the given matrix.
### Arguments
• const mat3 & m - Matrix, for which the determinant is to be calculated.
### Return value
Matrix determinant.
## floatdeterminant ( const mat4 & m ) #
Returns the determinant of the given matrix.
### Arguments
• const mat4 & m - Matrix, for which the determinant is to be calculated.
### Return value
Matrix determinant.
## doubledeterminant ( const dmat4 & m ) #
Returns the determinant of the given matrix.
### Arguments
• const dmat4 & m - Matrix, for which the determinant is to be calculated.
### Return value
Matrix determinant.
## mat4frustum ( float left, float right, float bottom, float top, float znear, float zfar ) #
Returns perspective projection matrix:
2.0 * znear / (right - left) 0.0 (right + left) / (right - left) 0.0 0.0 2.0 * znear / (top - bottom) (top + bottom) / (top - bottom) 0.0 0.0 0.0 -(zfar + znear) / (zfar - znear) -2.0 * zfar * znear / (zfar - znear) 0.0 0.0 -1.0 0.0
Coordinates of top, left, right, bottom are set relatively to center point of the znear plane.
There are two different points (A and B) on the picture above. Since the top, left, right, bottom are coordinates relatively to the center point of the znear plane, coordinates of the A point should be A(left, bottom, znear). Coordinates of the B point are B(k * left, k * bottom, zfar), where k = zfar/znear.
### Arguments
• float left - Left coordinate of the near clipping plane relatively to the center.
• float right - Right coordinate of the near clipping plane relatively to the center.
• float bottom - Bottom coordinate of the near clipping plane relatively to the center.
• float top - Top coordinate of the near clipping plane relatively to the center.
• float znear - Distance to the near depth clipping plane.
• float zfar - Distance to the farther depth clipping plane.
### Return value
Perspective projection matrix.
## mat2inverse ( const mat2 & m ) #
Returns inverse of a matrix. The inverse of a matrix is a matrix that if multiplied by the original would result in identity matrix: AA -1 = A -1A = I.
### Arguments
• const mat2 & m - Matrix to be inverted.
### Return value
Inverse of the matrix.
## mat3inverse ( const mat3 & m ) #
Returns inverse of a matrix. The inverse of a matrix is a matrix that if multiplied by the original would result in identity matrix: AA -1 = A -1A = I.
### Arguments
• const mat3 & m - Matrix to be inverted.
### Return value
Inverse of the matrix.
## dmat4inverse ( const dmat4 & m ) #
Returns inverse of a matrix. The inverse of a matrix is a matrix that if multiplied by the original would result in identity matrix: AA -1 = A -1A = I.
### Arguments
• const dmat4 & m - Matrix to be inverted.
### Return value
Inverse of the matrix.
## mat4inverse ( const mat4 & m ) #
Returns inverse of a matrix. The inverse of a matrix is a matrix that if multiplied by the original would result in identity matrix: AA -1 = A -1A = I.
### Arguments
• const mat4 & m - Matrix to be inverted.
### Return value
Inverse of the matrix.
## quatinverse ( const quat & q ) #
Returns inverse of a quaternion. The inverse of a quaternion is a quaternion that if multiplied by the original would result in identity matrix: qq -1 = 1. If q = a + bi + cj + dk, then q-1 = a - bi - cj - dk
### Arguments
• const quat & q - Quaternion to be inverted.
### Return value
Inverse of the quaternion.
## mat4inverse4 ( const mat4 & m ) #
Inverts a matrix that consists of a 3×4 sub-matrix (upper left) and a translation vector. The last row of the matrix is ignored. Compared to the inverse() function, this one is a bit faster and, which is more important, more stable. A matrix suitable for such inversion looks like this:
m00 m10 m20 m30 m01 m11 m21 m31 m02 m12 m22 m32 0 0 0 1
Notice
The function does the same as inverse (dmat4)
### Arguments
• const mat4 & m - Matrix to be inverted.
### Return value
Inverse of the matrix.
## mat3jacobi ( const mat3 & m, mat3 & v ) #
Returns the Jacobian matrix for the given 3x3 matrix.
### Arguments
• const mat3 & m - Matrix, for which the Jacobian matrix is to be calculated.
• mat3 & v - Output matrix, to which the calculated Jacobian matrix will be put.
Jacobian matrix.
## mat4lookAt ( const vec3 & position, const vec3 & target, const vec3 & up, int axis = AXIS_NZ ) #
Returns the viewing matrix for the given eye point, target point, up and forward direction vectors. The forward direction vector is pointed along the specified axis. By default, it is negative Z.
### Arguments
• const vec3 & position - Position coordinates of the eye point.
• const vec3 & target - Position of the target point.
• const vec3 & up - Direction of the up vector.
• int axis - Axis along which the forward direction vector is pointed. The default is negative Z.
### Return value
The viewing matrix.
## dmat4lookAt ( const dvec3 & position, const dvec3 & target, const vec3 & up, int axis = AXIS_NZ ) #
Returns the viewing matrix for the given eye point, target point, up and forward direction vectors. The forward direction vector is pointed along the specified axis. By default, it is negative Z.
### Arguments
• const dvec3 & position - Position coordinates of the eye point.
• const dvec3 & target - Position coordinates of the target point.
• const vec3 & up - Direction of the up vector.
• int axis - Axis along which the forward direction vector is pointed. The default is negative Z.
### Return value
The viewing matrix.
## mat4obliqueProjection ( const mat4 & projection, const vec4 & plane ) #
Returns the oblique projection matrix.
### Arguments
• const mat4 & projection - Projection matrix.
• const vec4 & plane - Clipping plane coordinates.
### Return value
Oblique projection matrix.
## quatorthoTangent ( const vec4 & tangent, const vec3 & normal ) #
Creates the ortho triangle tangent space basis.
### Arguments
• const vec4 & tangent - Tangent vector.
• const vec3 & normal - Normal vector.
### Return value
The tangent basis.
## quatorthoTangent ( const vec3 & tangent, const vec3 & binormal, const vec3 & normal ) #
Creates the ortho triangle tangent space basis.
### Arguments
• const vec3 & tangent - Tangent vector.
• const vec3 & binormal - Binormal vector.
• const vec3 & normal - Normal vector.
### Return value
The tangent basis.
## mat4ortho ( float left, float right, float bottom, float top, float znear, float zfar ) #
Returns parallel projection matrix:
2.0 / (right - left) 0.0 0.0 -(right + left) / (right - left) 0.0 2.0 / (top - bottom) 0.0 -(top + bottom) / (top - bottom) 0.0 0.0 -2.0 / (zfar - znear) -(zfar + znear) / (zfar - znear) 0.0 0.0 0.0 1.0
### Arguments
• float left - Left vertical clipping plane.
• float right - Right vertical clipping plane.
• float bottom - Bottom horizontal clipping plane.
• float top - Top horizontal clipping plane.
• float znear - Nearest depth clipping plane.
• float zfar - Farther depth clipping plane.
### Return value
Parallel projection matrix.
## mat3orthonormalize ( const mat3 & m ) #
Orthonormalizes a matrix.
### Arguments
• const mat3 & m - Matrix to be orthonormalized.
### Return value
Orthonormal matrix.
## mat4orthonormalize ( const mat4 & m ) #
Orthonormalizes a matrix.
### Arguments
• const mat4 & m - Matrix to be orthonormalized.
### Return value
Orthonormal matrix.
## dmat4orthonormalize ( const dmat4 & m ) #
Orthonormalizes a matrix.
### Arguments
• const dmat4 & m - Matrix to be orthonormalized.
### Return value
Orthonormal matrix.
## mat3 &orthonormalize ( mat3 & ret, const mat3 & m ) #
Orthonormalizes a matrix.
### Arguments
• mat3 & ret - Output matrix, to which the orthonormalized matrix will be put.
• const mat3 & m - Matrix to be orthonormalized.
### Return value
Orthonormal matrix.
## mat4 &orthonormalize ( mat4 & ret, const mat4 & m ) #
Orthonormalizes a matrix.
### Arguments
• mat4 & ret - Output matrix, to which the orthonormalized matrix will be put.
• const mat4 & m - Matrix to be orthonormalized.
### Return value
Orthonormal matrix.
## dmat4 &orthonormalize ( dmat4 & ret, const dmat4 & m ) #
Orthonormalizes a matrix.
### Arguments
• dmat4 & ret - Output matrix, to which the orthonormalized matrix will be put.
• const dmat4 & m - Matrix to be orthonormalized.
### Return value
Orthonormal matrix.
## mat4perspective ( float fov, float aspect, float znear, float zfar ) #
Returns perspective projection matrix.
### Arguments
• float fov - Field of view angle.
• float aspect - Aspect ratio. The aspect ratio is the ratio of width to height.
• float znear - Nearest depth clipping plane.
• float zfar - Farther depth clipping plane.
### Return value
Perspective projection matrix.
## mat4reflect ( const vec4 & plane ) #
Returns reflection matrix about a given plane.
### Arguments
• const vec4 & plane - Reflection plane.
### Return value
Reflection matrix.
## dmat4reflect ( const dvec4 & plane ) #
Returns reflection matrix about a given plane.
### Arguments
• const dvec4 & plane - Reflection plane.
### Return value
Reflection matrix.
## vec3reflect ( const vec3 & v0, const vec3 & v1 ) #
Reflects a given vector about a plane with the specified normal vector.
### Arguments
• const vec3 & v0 - Vector to be reflected.
• const vec3 & v1 - Normal vector to the reflection plane.
### Return value
Reflected vector.
## dvec3reflect ( const dvec3 & v0, const dvec3 & v1 ) #
Reflects a given vector about a plane with the specified normal vector.
### Arguments
• const dvec3 & v0 - Vector to be reflected.
• const dvec3 & v1 - Normal vector to the reflection plane.
### Return value
Reflected vector.
## vec3 &reflect ( vec3 & ret, const vec3 & v0, const vec3 & v1 ) #
Reflects a given vector about a plane with the specified normal vector.
### Arguments
• vec3 & ret - Reflected vector.Output vector, to which the reflected vector will be put.
• const vec3 & v0 - Vector to be reflected.
• const vec3 & v1 - Normal vector to the reflection plane.
### Return value
Reflected vector.
## dvec3 &reflect ( dvec3 & ret, const dvec3 & v0, const dvec3 & v1 ) #
Reflects a given vector about a plane with the specified normal vector.
### Arguments
• dvec3 & ret - Output vector, to which the reflected vector will be put.
• const dvec3 & v0 - Vector to be reflected.
• const dvec3 & v1 - Normal vector to the reflection plane.
### Return value
Reflected vector.
## mat4rotateX ( float angle ) #
Returns rotation matrix for the given angle around X axis:
1 0.0 0.0 0 0 cos -sin 0 0 sin cos 0 0 0.0 0.0 1
### Arguments
• float angle - Rotation angle, in degrees.
Rotation matrix.
## dmat4rotateX ( double angle ) #
Returns the X rotation matrix.
### Arguments
• double angle - Rotation angle, in degrees.
Rotation matrix.
## mat4rotateY ( float angle ) #
Returns rotation matrix for the given angle around Y axis:
cos 0 sin 0 0.0 1 0.0 0 -sin 0 cos 0 0.0 0 0.0 1
### Arguments
• float angle - Rotation angle, in degrees.
Rotation matrix.
## dmat4rotateY ( double angle ) #
Returns the Y rotation matrix.
### Arguments
• double angle - Rotation angle, in degrees.
Rotation matrix.
## mat4rotateZ ( float angle ) #
Returns rotation matrix for the given angle around Z axis:
cos -sin 0 0 sin cos 0 0 0.0 0.0 1 0 0.0 0.0 0 1
### Arguments
• float angle - Rotation angle, in degrees.
Rotation matrix.
## dmat4rotateZ ( double angle ) #
Returns the Z rotation matrix.
### Arguments
• double angle - Rotation angle, in degrees.
Rotation matrix.
## dmat4rotate ( double x, double y, double z, double angle ) #
Returns rotation matrix for the given angle around the given axis (X, Y, Z).
### Arguments
• double x - X component of rotation axis.
• double y - Y component of rotation axis.
• double z - Z component of rotation axis.
• double angle - Rotation angle, in degrees.
Rotation matrix.
## mat4rotate ( float x, float y, float z, float angle ) #
Returns rotation matrix for the given angle around the given axis (X, Y, Z).
### Arguments
• float x - X component of rotation axis.
• float y - Y component of rotation axis.
• float z - Z component of rotation axis.
• float angle - Rotation angle, in degrees.
Rotation matrix.
## mat4rotate ( const quat & q ) #
Returns rotation matrix for the given quaternion.
### Arguments
• const quat & q - Rotation quaternion.
Rotation matrix.
## mat4rotate ( const vec3 & axis, float angle ) #
Returns rotation matrix.
### Arguments
• const vec3 & axis - Rotation axis. This vector does not have to be normalized.
• float angle - Rotation angle, in degrees.
Rotation matrix.
## dmat4rotate ( const dvec3 & axis, double angle ) #
Returns rotation matrix for the given angle around the given axis (X, Y, Z).
### Arguments
• const dvec3 & axis - Rotation axis. This vector does not have to be normalized.
• double angle - Rotation angle, in degrees.
Rotation matrix.
## quatrotateTowards ( const quat & source, const quat & target, float max_angle ) #
Rotates the source quaternion towards the target quaternion by an angular step of max_angle (note, that the rotation will not overshoot).
### Arguments
• const quat & source - Source quaternion.
• const quat & target - Target quaternion.
• float max_angle - Angular step, in degrees. If a negative value is specified, 0 will be used instead.
### Return value
Resulting quaternion.
## vec3rotateTowards ( const vec3 & source, const vec3 & target, float max_angle ) #
Rotates the source vector towards the target vector by an angular step of max_angle (note, that the rotation will not overshoot).
### Arguments
• const vec3 & source - Source vector.
• const vec3 & target - Target vector.
• float max_angle - Angular step, in degrees. If a negative value is specified, 0 will be used instead.
### Return value
Resulting vector.
## mat4rotation ( const mat4 & m ) #
Returns the rotation matrix for the given source matrix ignoring the translation column. For more information see Matrix Transformations.
### Arguments
• const mat4 & m - Source matrix.
Rotation matrix.
## dmat4rotation ( const dmat4 & m ) #
Returns the rotation matrix for the given source matrix ignoring the translation column. For more information see Matrix Transformations.
### Arguments
• const dmat4 & m - Source matrix.
Rotation matrix.
## mat4 &rotation ( mat4 & ret, const mat4 & m ) #
Returns the rotation matrix for the given source matrix ignoring the translation column. For more information see Matrix Transformations.
### Arguments
• mat4 & ret - Output matrix, to which the rotation matrix will be put.
• const mat4 & m - Source matrix.
Rotation matrix.
## dmat4 &rotation ( dmat4 & ret, const dmat4 & m ) #
Returns the rotation matrix for the given source matrix ignoring the translation column. For more information see Matrix Transformations.
### Arguments
• dmat4 & ret - Output matrix, to which the rotation matrix will be put.
• const dmat4 & m - Source matrix.
Rotation matrix.
## quatrotationFromDir ( const vec3 & forward, const vec3 & up ) #
Returns the rotation quaternion for the specified "forward" and "up" directions. For more information see Matrix Transformations.
### Arguments
• const vec3 & forward - Forward direction vector defining the direction to look in.
• const vec3 & up - Vector defining in which direction up is.
### Return value
Rotation quaternion.
## quatrotationFromDir ( const vec3 & forward ) #
Returns the rotation quaternion for the specified "forward" direction (the default "up" vector is used). For more information see Matrix Transformations.
### Arguments
• const vec3 & forward - Forward direction vector defining the direction to look in.
### Return value
Rotation quaternion.
## mat4scale ( float x, float y, float z ) #
Returns scaling matrix for the specified scaling vector (X, Y, Z):
X 0.0 0.0 0 0.0 Y 0.0 0 0.0 0.0 Z 0 0.0 0.0 0.0 1
### Arguments
• float x - X component of the scaling vector.
• float y - Y component of the scaling vector.
• float z - Z component of the scaling vector.
Scaling matrix.
## dmat4scale ( double x, double y, double z ) #
Returns scaling matrix for the specified scaling vector (X, Y, Z):
X 0.0 0.0 0 0.0 Y 0.0 0 0.0 0.0 Z 0 0.0 0.0 0.0 1
### Arguments
• double x - X component of the scaling vector.
• double y - Y component of the scaling vector.
• double z - Z component of the scaling vector.
Scaling matrix.
## mat4scale ( const vec3 & v ) #
Returns the scaling matrix for the specified scaling vector (X, Y, Z):
X 0.0 0.0 0 0.0 Y 0.0 0 0.0 0.0 Z 0 0.0 0.0 0.0 1
### Arguments
• const vec3 & v - Scaling vector.
Scaling matrix.
## dmat4scale ( const dvec3 & v ) #
Returns the scaling matrix for the specified scaling vector (X, Y, Z):
X 0.0 0.0 0 0.0 Y 0.0 0 0.0 0.0 Z 0 0.0 0.0 0.0 1
### Arguments
• const dvec3 & v - Scaling vector.
Scaling matrix.
## dmat4setTo ( const dvec3 & position, const dvec3 & target, const vec3 & up, int axis = AXIS_NZ ) #
Returns the transformation matrix, which puts an object to the specified position and sets it orientation to the specified target point. At that, the forward direction vector is pointed along the specified axis. By default, it is negative Z.
### Arguments
• const dvec3 & position - Position of the eye point.
• const dvec3 & target - Position of the target point.
• const vec3 & up - Direction of the up vector.
• int axis - Axis along which the forward direction vector is pointed. The default is negative Z.
### Return value
Transformation matrix.
## mat4setTo ( const vec3 & position, const vec3 & target, const vec3 & up, int axis = AXIS_NZ ) #
Returns the transformation matrix, which puts an object to the specified position and sets it orientation to the specified target point. At that, the forward direction vector is pointed along the specified axis. By default, it is negative Z.
### Arguments
• const vec3 & position - Position of the eye point.
• const vec3 & target - Position of the target point.
• const vec3 & up - Direction of the up vector.
• int axis - Axis along which the forward direction vector is pointed. The default is negative Z.
### Return value
Transformation matrix.
## mat4symmetryProjection ( const mat4 & projection ) #
Returns the symmetry projection matrix.
### Arguments
• const mat4 & projection - Projection matrix.
### Return value
The symmetry projection matrix.
## mat4translate ( float x, float y, float z ) #
Returns the translation matrix for the specified translation vector (X, Y, Z):
1 0 0 X 0 1 0 Y 0 0 1 Z 0 0 0 1.0
### Arguments
• float x - X component of the translation vector.
• float y - Y component of the translation vector.
• float z - Z component of the translation vector.
### Return value
Translation matrix.
## dmat4translate ( double x, double y, double z ) #
Returns the translation matrix for the specified translation vector (X, Y, Z):
1 0 0 X 0 1 0 Y 0 0 1 Z 0 0 0 1.0
### Arguments
• double x - X component of the translation vector.
• double y - Y component of the translation vector.
• double z - Z component of the translation vector.
### Return value
Translation matrix.
## mat4translate ( const vec3 & v ) #
Returns the translation matrix for the specified translation vector (X, Y, Z):
1 0 0 X 0 1 0 Y 0 0 1 Z 0 0 0 1.0
### Arguments
• const vec3 & v - Translation vector.
### Return value
Translation matrix.
## dmat4translate ( const dvec3 & v ) #
Returns the translation matrix for the specified translation vector (X, Y, Z):
1 0 0 X 0 1 0 Y 0 0 1 Z 0 0 0 1.0
### Arguments
• const dvec3 & v - Translation vector.
### Return value
Translation matrix.
## mat2transpose ( const mat2 & m ) #
Transposes a given 2x2 matrix.
### Arguments
• const mat2 & m - Matrix to be transposed.
### Return value
Transposed matrix.
## mat2 &transpose ( mat2 & ret, const mat2 & m ) #
Transposes a given 2x2 matrix.
### Arguments
• mat2 & ret - Output matrix, to which the transposed matrix will be put.
• const mat2 & m - Matrix to be transposed.
### Return value
Transposed matrix.
## mat3transpose ( const mat3 & m ) #
Transposes a given 3x3 matrix.
### Arguments
• const mat3 & m - Matrix to be transposed.
### Return value
Transposed matrix.
## mat3 &transpose ( mat3 & ret, const mat3 & m ) #
Transposes a given 3x3 matrix.
### Arguments
• mat3 & ret - Output matrix, to which the transposed matrix will be put.
• const mat3 & m - Matrix to be transposed.
## mat4transpose ( const mat4 & m ) #
Transposes a given 4x4 matrix.
### Arguments
• const mat4 & m - Matrix to be transposed.
### Return value
Transposed matrix.
## mat4 &transpose ( mat4 & ret, const mat4 & m ) #
Transposes a given 4x4 matrix.
### Arguments
• mat4 & ret - Output matrix, to which the transposed matrix will be put.
• const mat4 & m - Matrix to be transposed.
### Return value
Transposed matrix.
## mat4transpose3 ( const mat4 & m ) #
Transposes the upper left 3×3 sub-matrix of a matrix.
### Arguments
• const mat4 & m - Matrix, a part of which will be transposed.
### Return value
Matrix, in which the upper left 3×3 sub-matrix is transposed.
## mat4 &transpose3 ( mat4 & ret, const mat4 & m ) #
Transposes the upper left 3×3 sub-matrix of a matrix.
### Arguments
• mat4 & ret - Output matrix, to which the resulting matrix will be put.
• const mat4 & m - Matrix, a part of which will be transposed.
### Return value
Matrix, in which the upper left 3×3 sub-matrix is transposed.
Last update: 2023-02-03 | 9,473 | 38,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-50 | latest | en | 0.744123 |
https://stats.stackexchange.com/questions/476834/how-do-i-find-the-mle-of-the-ape-distribution-in-r | 1,716,401,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058560.36/warc/CC-MAIN-20240522163251-20240522193251-00311.warc.gz | 471,115,519 | 41,137 | # How do I find the MLE of the APE distribution in R?
The random variable $$Y$$ is said to have a two-parameter APE distribution, denoted by $$\text{APE}(\alpha, \lambda)$$, with the shape parameter $$\alpha>0$$ and scale parameter $$\lambda>0$$ if the density function is:
$$f_Y(y) = \begin{cases} \log (\frac{\log \alpha}{\alpha-1}) \cdot \lambda e^{-\lambda y} \cdot \alpha^{1-e^{-\lambda y}} & & & \text{for } \alpha \neq 1 \\[8pt] \lambda e^{-\lambda y} & & & \text{for } \alpha = 1 \\[6pt] \end{cases}$$
Let $$Y_1, Y_2,..., Y_n \sim \text{IID APE}(\alpha,\lambda)$$ be a random sample from the APE distribution. Then the log-likelihood function is:
$$\ell_\mathbf{y}(\alpha,\lambda) = n \log \alpha + n \log \bigg( \frac{\log \alpha}{\alpha-1} \bigg) + n \log \lambda - \lambda \sum_i y_i - (\log \alpha) \sum_i e^{-\lambda y_i}.$$
How do I find MLE of this distribution in R? My data is given below:
1 4 4 7 11 13 15 15 17 18 19 19 20 20 22 23 28 29 31 32 36 37 47 48 49 50 54 54 55 59 59 61 61
66 72 72 75 78 78 81 93 96 99 108 113 114 120 120 120 123 124 129 131 137 145 151 156 171
176 182 188 189 195 203 208 215 217 217 217 224 228 233 255 271 275 275 275 286 291 312
312 312 315 326 326 329 330 336 338 345 348 354 361 364 369 378 390 457 467 498 517 566
644 745 871 1312 1357 1613 1630
• Questions only about R is prone to be closed, consider reformulate to have emphasis on the statistical question about estimation. Jul 12, 2020 at 15:27
You can find MLE equations for this distribution in Mahdvai and Kundu (2017) (accessible version here). As you can see from the paper, computing the MLE requires you to solve a critical point equation for $$\lambda$$ and you can then compute the MLE for $$\alpha$$ from this. The paper also contains further information on the asymptotic distribution of the MLE, etc.
Implementation in R: This can be done in R using nonlinear optimisation with the nlm function, or by solving the critical point equation with the uniroot function. Using one of the critical point equations, Mahdvai and Kundu (2017) give the MLE of the first parameter as the function:
$$\hat{\alpha}(\mathbf{y},\lambda) = \exp \Bigg( \frac{\sum_i y_i - n/\lambda}{\sum_i y_i e^{-\lambda y_i}} \Bigg).$$
You can substitute this function into the log-likelihood function or the remaining critical point equation. In the code below, we will substitute into the log-likelihod function and then maximise using the nlm function. (As the starting point for the iterative optimisation procedure, we will use the MLE for the exponential distribution.)
#Set the MLE function for alpha
LOG_ALPHA_HAT <- function(y, lambda) {
n <- length(y);
NUM <- sum(y) - n/lambda;
DEN <- sum(y*exp(-lambda*y));
NUM/DEN; }
#Set the log-likelihood function
LOGLIKE <- function(y, lambda) {
la <- LOG_ALPHA_HAT(y, lambda);
if (la == 0) {
LL <- n*log(lambda) - lambda*sum(y); } else {
LL <- n*la + n*log(la/expm1(la)) + n*log(lambda) -
lambda*sum(y) - la*sum(exp(-lambda*y)); }
LL; }
#Input the data
DATA <- c(1, 4, 4, 7, 11, 13, 15, 15, 17, 18, 19, 19, 20, 20, 22, 23, 28,
29, 31, 32, 36, 37, 47, 48, 49, 50, 54, 54, 55, 59, 59, 61, 61,
66, 72, 72, 75, 78, 78, 81, 93, 96, 99, 108, 113, 114, 120, 120,
120, 123, 124, 129, 131, 137, 145, 151, 156, 171, 176, 182, 188,
189, 195, 203, 208, 215, 217, 217, 217, 224, 228, 233, 255, 271,
275, 275, 275, 286, 291, 312, 312, 312, 315, 326, 326, 329, 330,
336, 338, 345, 348, 354, 361, 364, 369, 378, 390, 457, 467, 498,
517, 566, 644, 745, 871, 1312, 1357, 1613, 1630);
#Maximise the log-likelihood function
OBJECTIVE <- function(lambda) { - LOGLIKE(y = DATA, lambda) }
START <- c(1/mean(DATA))
NLM <- nlm(OBJECTIVE, p = START);
LLMAX <- - NLM$$minimum; MLE_LAMBDA <- NLM$$estimate;
MLE_ALPHA <- exp(LOG_ALPHA_HAT(y, MLE_LAMBDA));
MLE <- data.frame(alpha = MLE_ALPHA, lambda = MLE_LAMBDA, loglike = LLMAX);
rownames(MLE) <- 'MLE';
We can now display the MLE computed using this optimisation:
#Show the MLE
MLE;
alpha lambda loglike
MLE 0.00366583 0.0009550325 -700.6492
• How should i do that? @Ben - Reinstate Monica Jul 12, 2020 at 21:52
• The value of log likelihood should be in negative(preferably -701.2132) , and the values of MLE of alpha and lambda is 0.2807 and 0.0030 according to the same paper i'm working on. Jul 13, 2020 at 4:05
• The value I have obtained appears to give a higher log-likelihood, so please check your numbers.
– Ben
Jul 13, 2020 at 4:27
• I'm not sure why the values are different when we are using the same data!! Jul 13, 2020 at 4:55
• @matrikashukla, please avoid asking too many or new questions in comments. It is OK to ask the OP to get greater clarity about an aspect that is unclear to you, but there is a limit to this. If you have a new or follow-up question, start a new thread. You can link back to this one for context, if appropriate. Jul 20, 2020 at 1:06 | 1,737 | 4,894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2024-22 | latest | en | 0.578807 |
https://lisenbyretirement.com/interesting/question-how-far-can-you-see-from-the-top-of-a-mountain.html | 1,660,231,949,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571472.69/warc/CC-MAIN-20220811133823-20220811163823-00022.warc.gz | 373,455,934 | 20,384 | Question: How far can you see from the top of a mountain?
How far can you see from the top of Mount Everest?
So the next logical question is: how far could a person see from the top of Mt. Everest? The answer: 336 km, which is like seeing Calgary from Edmonton, which is definitely impressive.
Can you see 100 miles away?
The Earth is round and you cannot see 100 miles out to sea from sea level. If you stand at the water’s edge and your eyes are 180 cm from the ground the horizon is about 4.8km or nearly 3 miles away. For the horizon to be 100 miles away your eyes would have to be 1878 metres above the ground.
How far can you see from the top of a hill?
For an observer standing on a hill or tower 30 metres (98 ft) above sea level, the horizon is at a distance of 19.6 kilometres (12.2 mi). For an observer standing on a hill or tower 100 metres (330 ft) above sea level, the horizon is at a distance of 36 kilometres (22 mi).
What is the maximum visibility distance?
The Earth curves about 8 inches per mile. As a result, on a flat surface with your eyes 5 feet or so off the ground, the farthest edge that you can see is about 3 miles away.
Can you see dead bodies on Everest?
Dead bodies are a common sight on top of Mount Everest. “I cannot believe what I saw up there,” Everest filmmaker Elia Saikaly wrote on Instagram last week. In 2015, an avalanche roared through Everest, killing at least 19 people. When people die on Everest, it can be difficult to remove their bodies.
You might be interested: Question: How far can a 5.56 bullet travel?
Can you take a helicopter to the top of Mount Everest?
Everest. Straddling the border of Nepal and China stands the world’s highest mountain: Mount Everest. It’s even possible to take guided tours of the mountain and essentially buy your way to the summit.
How far can a human walk in a day?
While your body is made for walking, the distance you can achieve at an average walking pace of 3.1 miles per hour depends on whether you have trained for it or not. A trained walker can walk a 26.2-mile marathon in eight hours or less, or walk 20 to 30 miles in a day.
How far can the human eye see a light?
Considering the absolute threshold, the brightness of a candle flame, and the way a glowing object dims according to the square of the distance away from it, vision scientists conclude that one could make out the faint glimmer of a candle flame up to 30 miles away.
How far can humans see into space?
Today, the most distant objects we can see are more than 30 billion light-years away, despite the fact that only 13.8 billion years have passed since the Big Bang. The farther a galaxy is, the faster it expands away from us and the more its light appears
How far can I see across the ocean?
Ocean Horizon – How far can you see across the ocean before the curvature of the earth impairs your vision? There are many factors involved in determining an exact answer, so we will provide a practical answer instead. The correct answer is about 3 miles if your eye is about a foot above the ocean.
You might be interested: Quick Answer: How close can you fly a drone to an airport?
Can you see 50 miles away?
The farthest point you can see is about 3 miles out. 6 miles: The average 747 passenger plane flies at about 6.6 miles up in the air. 50 miles: On clear days, city buildings can be seen from 50 miles away (if you‘re standing on the ground).
When looking up a hill the distance is?
Distances presented up a hill are judged as farther than distances on the flat ground (Stefanucci, Proffitt, Banton, & Epstein, 2005). This distance-on-hill effect is argued to be a perceptual bias that is a function of the energetic cost associated with traversing a given distance on a hill. | 883 | 3,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2022-33 | longest | en | 0.936412 |
https://it.scribd.com/document/385997006/252128271-Revision-12-IB-Paper-1-pdf | 1,585,464,954,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370493818.32/warc/CC-MAIN-20200329045008-20200329075008-00360.warc.gz | 545,112,784 | 131,004 | Sei sulla pagina 1di 293
# 1. Let f(x) = 7 – 2x and g(x) = x + 3.
(2)
(1)
## (c) Find (f ° g–1)(5).
(2)
(Total 5 marks)
2 1
2. A line L passes through A(1, –1, 2) and is parallel to the line r = 1 s 3 .
5 2
(2)
(b) Find
(i) OP ;
(ii) OP .
(4)
(Total 6 marks)
## 3. The probability distribution of a discrete random variable X is given by
x2
P(X = x) = , x {1, 2, k}, where k > 0.
14
## (a) Write down P(X = 2).
(1)
IB Questionbank Maths SL 1
(b) Show that k = 3.
(4)
## (c) Find E(X).
(2)
(Total 7 marks)
ln x
4. Let g(x) = , for x > 0.
x2
1 2 ln x
(a) Use the quotient rule to show that g ( x) .
x3
(4)
(3)
(Total 7 marks)
## 5. Solve the equation 2cos x = sin 2x, for 0 ≤ x ≤ 3π.
(Total 7 marks)
6. Consider f(x) = 2kx2 – 4kx + 1, for k ≠ 0. The equation f(x) = 0 has two equal roots.
## (a) Find the value of k.
(5)
(b) The line y = p intersects the graph of f. Find all possible values of p.
(2)
(Total 7 marks)
IB Questionbank Maths SL 2
7. In a group of 16 students, 12 take art and 8 take music. One student takes neither art nor music.
The Venn diagram below shows the events art and music. The values p, q, r and s represent
numbers of students.
## (iii) Write down the value of p and of r.
(5)
(b) (i) A student is selected at random. Given that the student takes music, write down the
probability the student takes art.
(ii) Hence, show that taking music and taking art are not independent events.
(4)
(c) Two students are selected at random, one after the other. Find the probability that the first
student takes only music and the second student takes only art.
(4)
(Total 13 marks)
IB Questionbank Maths SL 3
8. The following diagram shows the obtuse-angled triangle ABC such that
3 2
AB 0 and AC 2 .
4 6
(ii) Find BC .
(3)
## (b) (i) Find cos AB̂C .
(ii) Hence, find sin ABC .
(7)
4
The point D is such that CD 5 , where p > 0.
p
## (ii) Hence, show that CD is perpendicular to BC .
(6)
(Total 16 marks)
IB Questionbank Maths SL 4
9. The velocity v m s–1 of a particle at time t seconds, is given by v = 2t + cos2t, for 0 ≤ t ≤ 2.
(1)
## When t = k, the acceleration is zero.
π
(b) (i) Show that k = .
4
π
(ii) Find the exact velocity when t = .
4
(8)
π dv π dv
(c) When t < , > 0 and when t > , > 0.
4 dt 4 dt
(4)
(3)
(Total 16 marks)
(a) Find d.
(2)
(2)
## (c) Find S20.
(2)
(Total 6 marks)
IB Questionbank Maths SL 5
11. The Venn diagram below shows events A and B where P(A) = 0.3, P( A B) = 0.6 and
P(A ∩ B) = 0.1. The values m, n, p and q are probabilities.
(4)
## (b) Find P(B′).
(2)
(Total 6 marks)
3 4
12. The following diagram shows quadrilateral ABCD, with AD BC, AB and AC .
1 4
(a) Find BC .
(2)
## (b) Show that BD 22 .
(2)
IB Questionbank Maths SL 6
(c) Show that vectors BD and CA are perpendicular.
(3)
(Total 7 marks)
6x
13. Let h(x) = . Find h′(0).
cos x
(Total 6 marks)
## 14. Let f(x) = 3 ln x and g(x) = ln 5x3.
+
(a) Express g(x) in the form f(x) + ln a, where a .
(4)
(b) The graph of g is a transformation of the graph of f. Give a full geometric description of
this transformation.
(3)
(Total 7 marks)
15. A scientist has 100 female fish and 100 male fish. She measures their lengths to the nearest cm.
These are shown in the following box and whisker diagrams.
IB Questionbank Maths SL 7
(a) Find the range of the lengths of all 200 fish.
(3)
## (b) Four cumulative frequency graphs are shown below.
Which graph is the best representation of the lengths of the female fish?
(2)
(Total 5 marks)
IB Questionbank Maths SL 8
16. The following diagram shows part of the graph of the function f(x) = 2x2.
(5)
## (b) Find the x-intercept of T.
(2)
(c) The shaded region R is enclosed by the graph of f, the line T, and the x-axis.
## (ii) Find the area of R.
(9)
(Total 16 marks)
IB Questionbank Maths SL 9
17. The following diagram shows part of the graph of a quadratic function f.
The x-intercepts are at (–4, 0) and (6, 0) and the y-intercept is at (0, 240).
(a) Write down f(x) in the form f(x) = –10(x – p)(x – q).
(2)
(b) Find another expression for f(x) in the form f(x) = –10(x – h)2 + k.
(4)
(c) Show that f(x) can also be written in the form f(x) = 240 + 20x – 10x2.
(2)
A particle moves along a straight line so that its velocity, v m s–1, at time t seconds is given by
v = 240 + 20t – 10t2, for 0 ≤ t ≤ 6.
(d) (i) Find the value of t when the speed of the particle is greatest.
(ii) Find the acceleration of the particle when its speed is zero.
(7)
(Total 15 marks)
IB Questionbank Maths SL 10
18. The following diagram represents a large Ferris wheel, with a diameter of 100 metres.
Let P be a point on the wheel. The wheel starts with P at the lowest point, at ground level. The
wheel rotates at a constant rate, in an anticlockwise (counterclockwise) direction. One
revolution takes 20 minutes.
## (a) Write down the height of P above ground level after
(i) 10 minutes;
(ii) 15 minutes.
(2)
Let h(t) metres be the height of P above ground level after t minutes. Some values of h(t) are
given in the table below.
t h(t)
0 0.0
1 2.4
2 9.5
3 20.6
4 34.5
5 50.0
## (ii) Find h(21).
(4)
IB Questionbank Maths SL 11
(c) Sketch the graph of h, for 0 ≤ t ≤ 40.
(3)
(d) Given that h can be expressed in the form h(t) = a cos bt + c, find a, b and c.
(5)
(Total 14 marks)
(4)
(3)
(Total 7 marks)
## 20. (a) Expand (2 + x)4 and simplify your result.
(3)
IB Questionbank Maths SL 12
1
(b) Hence, find the term in x2 in (2 + x)4 1 2 .
x
(3)
(Total 6 marks)
3
21. The straight line with equation y = x makes an acute angle θ with the x-axis.
4
(1)
## (ii) cos 2θ.
(6)
(Total 7 marks)
22. Consider the events A and B, where P(A) = 0.5, P(B) = 0.7 and P(A ∩ B) = 0.3.
The Venn diagram below shows the events A and B, and the probabilities p, q and r.
IB Questionbank Maths SL 13
(a) Write down the value of
(i) p;
(ii) q;
(iii) r.
(3)
## (b) Find the value of P(A | B′).
(2)
(c) Hence, or otherwise, show that the events A and B are not independent.
(1)
(Total 6 marks)
IB Questionbank Maths SL 14
23. The graph of f(x) = 16 4 x 2 , for –2 ≤ x ≤ 2, is shown below.
The region enclosed by the curve of f and the x-axis is rotated 360° about the x-axis.
Find the volume of the solid formed.
(Total 6 marks)
(2)
## (b) Write down the range of f–1.
(1)
IB Questionbank Maths SL 15
Let g(x) = log3 x, for x > 0.
## (c) Find the value of (f –1 ° g)(2), giving your answer as an integer.
(4)
(Total 7 marks)
1 3
25. Let f(x) = x x 2 3x . Part of the graph of f is shown below.
3
## (a) Find the coordinates of A.
(8)
IB Questionbank Maths SL 16
(b) Write down the coordinates of
## (i) the image of B after reflection in the y-axis;
2
(ii) the image of B after translation by the vector ;
5
(iii) the image of B after reflection in the x-axis followed by a horizontal stretch with
1
scale factor .
2
(6)
(Total 14 marks)
cos x
26. Let f(x) = , for sin x ≠ 0.
sin x
1
(a) Use the quotient rule to show that f′(x) = .
sin 2 x
(5)
## (b) Find f′′(x).
(3)
π π
In the following table, f′ = p and f′′ = q. The table also gives approximate values of
2 2
π
f′(x) and f′′(x) near x = .
2
π π π
x 0.1 0.1
2 2 2
f′(x) –1.01 p –1.01
f″(x) 0.203 q –0.203
## (c) Find the value of p and of q.
(3)
IB Questionbank Maths SL 17
(d) Use information from the table to explain why there is a point of inflexion on the graph of
π
f where x = .
2
(2)
(Total 13 marks)
3 2
27. The line L1 is represented by the vector equation r = 1 p 1 .
25 8
A second line L2 is parallel to L1 and passes through the point B(–8, –5, 25).
## (a) Write down a vector equation for L2 in the form r = a + tb.
(2)
5 7
A third line L3 is perpendicular to L1 and is represented by r = 0 q 2 .
3 k
(5)
## (c) Find the coordinates of A.
(6)
IB Questionbank Maths SL 18
6
The lines L2 and L3 intersect at point C where BC 3 .
24
## (ii) Hence, find | AC |.
(5)
(Total 18 marks)
28. Let f(x) = p(x – q)(x – r). Part of the graph of f is shown below.
The graph passes through the points (–2, 0), (0, –4) and (4, 0).
(2)
(1)
## (c) Find the value of p.
(3)
(Total 6 marks)
IB Questionbank Maths SL 19
6 2
29. Let AB 2 and AC 3 .
3 2
(2)
(3)
(3)
(Total 8 marks)
## 30. Let f(x) = cos 2x and g(x) = 2x2 – 1.
π
(a) Find f .
2
(2)
π
(b) Find (g ° f) .
2
(2)
(c) Given that (g ° f)(x) can be written as cos (kx), find the value of k, k .
(3)
(Total 7 marks)
31. Let f(x) = kx4. The point P(1, k) lies on the curve of f. At P, the normal to the curve is parallel to
1
y = x . Find the value of k.
8
(Total 6 marks)
## 32. Solve log2x + log2(x – 2) = 3, for x > 2.
(Total 7 marks)
IB Questionbank Maths SL 20
33. A function f is defined for –4 ≤ x ≤ 3. The graph of f is given below.
The graph has a local maximum when x = 0, and local minima when x = –3, x = 2.
(a) Write down the x-intercepts of the graph of the derivative function, f′.
(2)
## (b) Write down all values of x for which f′(x) is positive.
(2)
(c) At point D on the graph of f, the x-coordinate is –0.5. Explain why f′′(x) < 0 at D.
(2)
(Total 6 marks)
34. Consider the function f with second derivative f′′(x) = 3x – 1. The graph of f has a minimum
4 358
point at A(2, 4) and a maximum point at B , .
3 27
## (a) Use the second derivative to justify that B is a maximum.
(3)
3 2
(b) Given that f′ = x – x + p, show that p = –4.
2
(4)
IB Questionbank Maths SL 21
(c) Find f(x).
(7)
(Total 14 marks)
1
35. José travels to school on a bus. On any day, the probability that José will miss the bus is .
3
7
If he misses his bus, the probability that he will be late for school is .
8
3
If he does not miss his bus, the probability that he will be late is .
8
Let E be the event “he misses his bus” and F the event “he is late for school”.
The information above is shown on the following tree diagram.
(a) Find
## (i) P(E ∩ F);
(ii) P(F).
(4)
IB Questionbank Maths SL 22
(b) Find the probability that
(i) José misses his bus and is not late for school;;
(ii) José missed his bus, given that he is late for school.
(5)
The cost for each day that José catches the bus is 3 euros. José goes to school on Monday and
Tuesday.
## (c) Copy and complete the probability distribution table.
X (cost in euros) 0 3 6
1
P (X)
9
(3)
(d) Find the expected cost for José for both days.
(2)
(Total 14 marks)
## 36. Let f(x) = 6 + 6sinx. Part of the graph of f is shown below.
The shaded region is enclosed by the curve of f, the x-axis, and the y-axis.
IB Questionbank Maths SL 23
(a) Solve for 0 ≤ x < 2π.
(i) 6 + 6sin x = 6;
(ii) 6 + 6 sin x = 0.
(5)
(b) Write down the exact value of the x-intercept of f, for 0 ≤ x < 2.
(1)
(c) The area of the shaded region is k. Find the value of k, giving your answer in terms of π.
(6)
π
Let g(x) = 6 + 6sin x . The graph of f is transformed to the graph of g.
2
## (d) Give a full geometric description of this transformation.
(2)
p
(e) Given that
p
2 g ( x)dx = k and 0 ≤ p < 2π, write down the two values of p.
(3)
(Total 17 marks)
37. The first three terms of an infinite geometric sequence are 32, 16 and 8.
(1)
(2)
## (c) Find the sum to infinity of this sequence.
(2)
(Total 5 marks)
IB Questionbank Maths SL 24
38. Let g(x) = 2x sin x.
(4)
(3)
(Total 7 marks)
## diagram not to scale
The radius of the smaller circle is 8 cm and the radius of the larger circle is 10 cm.
π
Points A, B and C are on the circumference of the larger circle such that AÔB is radians.
3
(2)
## (b) Find the area of the shaded region.
(4)
(Total 6 marks)
IB Questionbank Maths SL 25
40. The diagram below shows the probabilities for events A and B, with P(A′) = p.
(1)
(3)
(3)
(Total 7 marks)
(2)
## (b) Hence, solve the equation 4 – cos 2θ + 5 sin θ = 0 for 0 ≤ θ ≤ 2π.
(5)
(Total 7 marks)
IB Questionbank Maths SL 26
3
42. The graph of the function y = f(x) passes through the point , 4 . The gradient function of f is
2
given as f′(x) = sin (2x – 3). Find f(x).
(Total 6 marks)
43. The diagram shows quadrilateral ABCD with vertices A(1, 0), B(1, 5), C(5, 2) and D(4, –1).
2
(ii) Find BD .
(5)
## (ii) Find a vector equation for the line (BD).
(4)
The lines (AC) and (BD) intersect at the point P(3, k).
IB Questionbank Maths SL 27
(c) Show that k = 1.
(3)
(5)
(Total 17 marks)
## (a) Find (f ° g)(x).
(2)
3
The vector translates the graph of (f ° g) to the graph of h.
1
(3)
## (c) Show that h(x) = x2 – 8x + 19.
(2)
(d) The line y = 2x – 6 is a tangent to the graph of h at the point P. Find the x-coordinate of P.
(5)
(Total 12 marks)
IB Questionbank Maths SL 28
45. Let f(x) = x3. The following diagram shows part of the graph of f.
## diagram not to scale
The point P (a, f(a)), where a > 0, lies on the graph of f. The tangent at P crosses the x-axis at
2
the point Q , 0 . This tangent intersects the graph of f at the point R(–2, –8).
3
a3
(a) (i) Show that the gradient of [PQ] is .
2
a
3
## (iii) Hence show that a = 1.
(7)
IB Questionbank Maths SL 29
The equation of the tangent at P is y = 3x – 2. Let T be the region enclosed by the graph of f, the
tangent [PR] and the line x = k, between x = –2 and x = k where –2 < k < 1. This is shown in the
diagram below.
## diagram not to scale
(b) Given that the area of T is 2k + 4, show that k satisfies the equation k4 – 6k2 + 8 = 0.
(9)
(Total 16 marks)
46. The letters of the word PROBABILITY are written on 11 cards as shown below.
## Two cards are drawn at random without replacement.
Let A be the event the first card drawn is the letter A.
Let B be the event the second card drawn is the letter B.
## (a) Find P(A).
(1)
IB Questionbank Maths SL 30
(b) Find P(B│A).
(2)
## (c) Find P(A ∩ B).
(3)
(Total 6 marks)
47. Let f(x) = ex cos x. Find the gradient of the normal to the curve of f at x = π.
(Total 6 marks)
48. The following diagram shows the graphs of the displacement, velocity and acceleration of a
moving object as functions of time, t.
(a) Complete the following table by noting which graph A, B or C corresponds to each
function.
Function Graph
displacement
acceleration
(4)
## (b) Write down the value of t when the velocity is greatest.
(2)
IB Questionbank Maths SL 31
(Total 6 marks)
## 49. Let f(x) = x2 and g(x) = 2(x – 1)2.
(a) The graph of g can be obtained from the graph of f using two transformations.
Give a full geometric description of each of the two transformations.
(2)
3
(b) The graph of g is translated by the vector to give the graph of h.
2
The point (–1, 1) on the graph of f is translated to the point P on the graph of h.
Find the coordinates of P.
(4)
(Total 6 marks)
## (ii) Write down the domain of f–1.
(3)
1
(b) Solve the equation f–1(x) = ln .
x
(4)
(Total 7 marks)
51. The graph of y = x between x = 0 and x = a is rotated 360° about the x-axis.
The volume of the solid formed is 32π. Find the value of a.
(Total 7 marks)
IB Questionbank Maths SL 32
52. A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.
The point P(x, y) is a vertex of the rectangle and also lies on the circle. The angle between (OP)
π
and the x-axis is θ radians, where 0 ≤ θ ≤ .
2
(i) x;
(ii) y.
(2)
## (b) Show that A = 18 sin 2θ.
(3)
IB Questionbank Maths SL 33
dA
(c) (i) Find .
d
(ii) Hence, find the exact value of θ which maximizes the area of the rectangle.
(iii) Use the second derivative to justify that this value of θ does give a maximum.
(8)
(Total 13 marks)
53. The vertices of the triangle PQR are defined by the position vectors
4 3 6
OP 3 , OQ 1 and OR 1 .
1 2 5
(a) Find
(i) PQ ;
(ii) PR .
(3)
1
(b) Show that cos RP̂Q .
2
(7)
## (c) (i) Find sin RP̂Q .
(ii) Hence, find the area of triangle PQR, giving your answer in the form a 3 .
(6)
(Total 16 marks)
IB Questionbank Maths SL 34
ax
54. Let f(x) = 2
, –8 ≤ x ≤ 8, a . The graph of f is shown below.
x 1
## (a) Show that f(–x) = –f(x).
(2)
2ax( x 2 3)
(b) Given that f′′(x) = , find the coordinates of all points of inflexion.
( x 2 1) 3
(7)
a
f ( x)dx 2 ln(x
2
(c) It is given that 1) C .
(i) Find the area of the shaded region, giving your answer in the form p ln q.
8
(ii) Find the value of 2 f ( x 1)dx .
4
(7)
(Total 16 marks)
IB Questionbank Maths SL 35
55. Let f(x) = x2 and g(x) = 2x – 3.
(2)
## (b) Find (f ° g)(4).
(3)
(Total 5 marks)
56. Find the cosine of the angle between the two vectors 3i + 4j + 5k and 4i – 5j – 3k.
(Total 6 marks)
10
57. The fifth term in the expansion of the binomial (a + b)n is given by p 6 (2q) 4 .
4
(1)
## (b) Write down a and b, in terms of p and/or q.
(2)
(c) Write down an expression for the sixth term in the expansion.
(3)
(Total 6 marks)
## 58. (a) Find log2 32.
(1)
IB Questionbank Maths SL 36
32 x
(b) Given that log2 y can be written as px + qy, find the value of p and of q.
8
(4)
(Total 5 marks)
(2)
## (b) Find f′(3) and f′′(3).
(1)
(c) The point P on the graph of f has x-coordinate 3. Explain why P is not a point of
inflexion.
(2)
(Total 5 marks)
π 1
60. Let f(x) = 3e 2 x sin x + e2x cos x, for 0 ≤ x ≤ π. Given that tan , solve the equation
6 3
f(x) = 0.
(Total 6 marks)
π
61. Let f(x) = e–3x and g(x) = sin x .
3
## (a) Write down
(i) f′(x);
(ii) g′(x).
(2)
IB Questionbank Maths SL 37
π π
(b) Let h(x) = e–3x sin x . Find the exact value of h′ .
3 3
(4)
(Total 6 marks)
## Two cards are drawn at random, one from each box.
(a) Copy and complete the table below to show all nine equally likely outcomes.
3, 9
3, 10
3, 10
(2)
(2)
(2)
## (d) Find the expected value of S.
(3)
IB Questionbank Maths SL 38
(e) Anna plays a game where she wins \$50 if S is even and loses \$30 if S is odd.
Anna plays the game 36 times. Find the amount she expects to have at the end of the 36
games.
(3)
(Total 12 marks)
8
63. The line L1 is parallel to the z-axis. The point P has position vector 1 and lies on L1.
0
(2)
## 2 2 6
The line L2 has equation r = 4 s 1 . The point A has position vector 2 .
1 5 9
(4)
## Let B be the point of intersection of lines L1 and L2.
9
(c) (i) Show that OB 1 .
14
(ii) Find AB .
(7)
(d) The point C is at (2, 1, –4). Let D be the point such that ABCD is a parallelogram.
Find OD .
(3)
(Total 16 marks)
IB Questionbank Maths SL 39
64. In this question s represents displacement in metres and t represents time in seconds.
## (ii) If s = 0 when t = 0, write down an expression for s in terms of a and t.
(6)
Trains approaching a station start to slow down when they pass a point P. As a train slows
down, its velocity is given by v = 40 – at, where t = 0 at P. The station is 500 m from P.
## (b) A train M slows down so that it comes to a stop at the station.
(i) Find the time it takes train M to come to a stop, giving your answer in terms of a.
8
(ii) Hence show that a = .
5
(6)
## (c) For a different train N, the value of a is 4.
Show that this train will stop before it reaches the station.
(5)
(Total 17 marks)
(5)
## (b) Find f–1(x).
(3)
(Total 8 marks)
IB Questionbank Maths SL 40
2 3
66. (a) Let u = 3 and w = 1 . Given that u is perpendicular to w, find the value of p.
1 p
(3)
1
(b) Let v = q . Given that v 42 , find the possible values of q.
5
(3)
(Total 6 marks)
67. Let X be normally distributed with mean 100 cm and standard deviation 5 cm.
(a) On the diagram below, shade the region representing P(X > 105).
(2)
(b) Given that P(X < d) = P(X > 105), find the value of d.
(2)
(c) Given that P(X > 105) = 0.16 (correct to two significant figures), find P(d < X < 105).
(2)
(Total 6 marks)
IB Questionbank Maths SL 41
68. The diagram below shows the graph of a function f(x), for –2 ≤ x ≤ 4.
IB Questionbank Maths SL 42
(a) Let h(x) = f(–x). Sketch the graph of h on the grid below.
(2)
1
(b) Let g(x) = f(x – 1). The point A(3, 2) on the graph of f is transformed to the point P on
2
the graph of g. Find the coordinates of P.
(3)
(Total 5 marks)
p
69. Consider f(x) = x2 + , x ≠ 0, where p is a constant.
x
## (a) Find f′(x).
(2)
IB Questionbank Maths SL 43
(b) There is a minimum value of f(x) when x = –2. Find the value of p.
(4)
(Total 6 marks)
(Total 7 marks)
## (a) Given that f–1(1) = 8, find the value of k.
(3)
2
(b) Find f–1 .
3
(4)
(Total 7 marks)
72. In a class of 100 boys, 55 boys play football and 75 boys play rugby. Each boy must play at
least one sport from football and rugby.
(a) (i) Find the number of boys who play both sports.
(ii) Write down the number of boys who play only rugby.
(3)
## (i) Find the probability that he plays only one sport.
(ii) Given that the boy selected plays only one sport, find the probability that he plays
rugby.
(4)
IB Questionbank Maths SL 44
Let A be the event that a boy plays football and B be the event that a boy plays rugby.
(2)
## (d) Show that A and B are not independent.
(3)
(Total 12 marks)
20
73. Let f(x) = 3 + 2
, for x ≠ ±2. The graph of f is given below.
x 4
## (ii) Show that f′(x) = 0 at A.
(7)
IB Questionbank Maths SL 45
40(3x 2 4)
(b) The second derivative f′′(x) = . Use this to
( x 2 4) 3
## (i) justify that the graph of f has a local maximum at A;
(ii) explain why the graph of f does not have a point of inflexion.
(6)
(1)
## (d) Write down the range of f.
(2)
(Total 16 marks)
74. Let f(x) = x . Line L is the normal to the graph of f at the point (4, 2).
(4)
## (b) Point A is the x-intercept of L. Find the x-coordinate of A.
(2)
IB Questionbank Maths SL 46
In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the
line L.
## (c) Find an expression for the area of R.
(3)
(d) The region R is rotated 360° about the x-axis. Find the volume of the solid formed, giving
(8)
(Total 17 marks)
75. Let p = sin40 and q = cos110. Give your answers to the following in terms of p and/or q.
(i) sin140;
(ii) cos70.
(2)
(3)
## (c) Find an expression for tan140.
(1)
(Total 6 marks)
IB Questionbank Maths SL 47
76. Consider the arithmetic sequence 2, 5, 8, 11, ....
(3)
(3)
(Total 6 marks)
(1)
## (b) On the diagram below, sketch the curve of g, for 0 x 2.
y
4
3
2
1
0 x
π π 3π 2π
–1 2 2
–2
–3
–4
(3)
(c) Write down the number of solutions to the equation g (x) = 2, for 0 x 2.
(2)
(Total 6 marks)
IB Questionbank Maths SL 48
1
78. (a) Find 2 x 3 dx.
(2)
3 1
(b) Given that
0 2x 3
dx = ln P , find the value of P.
(4)
(Total 6 marks)
79. A particle moves along a straight line so that its velocity, v m s−1 at time t seconds is given by
v = 6e3t + 4. When t = 0, the displacement, s, of the particle is 7 metres. Find an expression for s
in terms of t.
(Total 7 marks)
(4)
## Let g (x) = ex.
(b) Find (g ◦ f) (x), giving your answer in the form ax + b, where a, b, .
(3)
(Total 7 marks)
IB Questionbank Maths SL 49
1 3
81. Consider f (x) = x + 2x2 – 5x. Part of the graph of f is shown below. There is a maximum
3
point at M, and a point of inflexion at N.
(3)
(4)
## (c) Find the x-coordinate of N.
(3)
(d) The line L is the tangent to the curve of f at (3, 12). Find the equation of L in the form
y = ax + b.
(4)
(Total 14 marks)
IB Questionbank Maths SL 50
82. Let f (x) = 3(x + 1)2 – 12.
(2)
(8)
## (c) Hence sketch the graph of f.
(2)
(d) Let g (x) = x2. The graph of f may be obtained from the graph of g by the two
transformations:
## a stretch of scale factor t in the y-direction
followed by
p
a translation of .
q
p
Find and the value of t.
q
(3)
(Total 15 marks)
83. A four-sided die has three blue faces and one red face. The die is rolled.
Let B be the event a blue face lands down, and R be the event a red face lands down.
## (a) Write down
(i) P (B);
(ii) P (R).
(2)
IB Questionbank Maths SL 51
(b) If the blue face lands down, the die is not rolled again. If the red face lands down, the die
is rolled once again. This is represented by the following tree diagram, where p, s, t are
probabilities.
## Find the value of p, of s and of t.
(2)
Guiseppi plays a game where he rolls the die. If a blue face lands down, he scores 2 and is
finished. If the red face lands down, he scores 1 and rolls one more time. Let X be the total score
obtained.
3
(c) (i) Show that P (X = 3) = .
16
(3)
## (ii) Calculate the expected value of X.
(5)
(e) If the total score is 3, Guiseppi wins \$10. If the total score is 2, Guiseppi gets nothing.
Guiseppi plays the game twice. Find the probability that he wins exactly \$10.
(4)
(Total 16 marks)
IB Questionbank Maths SL 52
84. A box contains 100 cards. Each card has a number between one and six written on it.
The following table shows the frequencies for each number.
Number 1 2 3 4 5 6
Frequency 26 10 20 k 29 11
(2)
(b) Find
## (ii) the interquartile range.
(5)
(Total 7 marks)
85. The following diagram shows part of the graph of f, where f (x) = x2 − x − 2.
(4)
## (b) Find the x-coordinate of the vertex.
(2)
(Total 6 marks)
IB Questionbank Maths SL 53
1 π
86. (a) Given that cos A = and 0 A , find cos 2A.
3 2
(3)
2 π
(b) Given that sin B = and B , find cos B.
3 2
(3)
(Total 6 marks)
## 87. Part of the graph of a function f is shown in the diagram below.
–2 –1 0 1 2 3 4 x
–1
–2
–3
–4
(2)
## (i) Find g (−3).
(ii) Describe fully the transformation that maps the graph of f to the graph of g.
(4)
(Total 6 marks)
IB Questionbank Maths SL 54
88. There are 20 students in a classroom. Each student plays only one sport. The table below gives
their sport and gender.
Female 5 3 3
Male 4 2 3
## (a) One student is selected at random.
(i) Calculate the probability that the student is a male or is a tennis player.
(ii) Given that the student selected is female, calculate the probability that the student
does not play football.
(4)
(b) Two students are selected at random. Calculate the probability that neither student plays
football.
(3)
(Total 7 marks)
5
89. Let
1
3 f ( x) dx 12.
1
(a) Show that 5
f ( x) dx 4.
(2)
2 5
(b) Find the value of x f ( x) dx x f ( x) dx.
1 2
(5)
(Total 7 marks)
IB Questionbank Maths SL 55
90. Consider the points A (1, 5, 4), B (3, 1, 2) and D (3, k, 2), with (AD) perpendicular to (AB).
(a) Find
(i) AB ;
(3)
## (b) Show that k = 7.
(3)
1
The point C is such that BC = AD.
2
(4)
(3)
(Total 13 marks)
## (a) (i) Write down the range of the function f.
(ii) Consider f (x) =1, 0 x 2. Write down the number of solutions to this equation.
(5)
(b) Find f ′ (x), giving your answer in the form a sinp x cosq x where a, p, q .
(2)
1
π
(c) Let g (x) = 3 sin x (cos x) 2 for 0 x
. Find the volume generated when the curve
2
of g is revolved through 2 about the x-axis.
(7)
(Total 14 marks)
IB Questionbank Maths SL 56
92. The following diagram shows a semicircle centre O, diameter [AB], with radius 2.
Let P be a point on the circumference, with PÔB = radians.
## (a) Find the area of the triangle OPB, in terms of .
(2)
(b) Explain why the area of triangle OPA is the same as the area triangle OPB.
(3)
Let S be the total area of the two segments shaded in the diagram below.
## (c) Show that S = 2( − 2 sin ).
(3)
(d) Find the value of when S is a local minimum, justifying that it is a minimum.
(8)
(2)
(Total 18 marks)
## 93. Consider the infinite geometric sequence 3, 3(0.9), 3(0.9)2, 3(0.9)3, … .
(a) Write down the 10th term of the sequence. Do not simplify your answer.
(1)
IB Questionbank Maths SL 57
(b) Find the sum of the infinite sequence.
(4)
(Total 5 marks)
94. A particle is moving with a constant velocity along line L. Its initial position is A(6, –2, 10).
After one second the particle has moved to B(9, –6, 15).
(4)
## (b) Write down an equation of the line L.
(2)
(Total 6 marks)
95. Let f be the function given by f(x) = e0.5x, 0 ≤ x ≤ 3.5. The diagram shows the graph of f.
## (a) On the same diagram, sketch the graph of f–1.
(3)
IB Questionbank Maths SL 58
(b) Write down the range of f–1.
(1)
## (c) Find f–1(x).
(3)
(Total 7 marks)
96. Let A and B be independent events, where P(A) = 0.6 and P(B) = x.
(1)
(i) find x;
(4)
## (c) Hence, explain why A and B are not mutually exclusive.
(1)
(Total 6 marks)
IB Questionbank Maths SL 59
97. The diagram shows part of the graph of y = f′(x). The x-intercepts are at points A and C.
There is a minimum at B, and a maximum at D.
## (a) (i) Write down the value of f′(x) at C.
(ii) Hence, show that C corresponds to a minimum on the graph of f, i.e. it has the
same x-coordinate.
(3)
(1)
## (c) Show that B corresponds to a point of inflexion on the graph of f.
(3)
(Total 7 marks)
π
98. Let f(x) = sin3 x + cos3 x tan x, < x < π.
2
## (a) Show that f(x) = sin x.
(2)
IB Questionbank Maths SL 60
2 4 5
(b) Let sin x = . Show that f(2x) = .
3 9
(5)
(Total 7 marks)
99. Two standard six-sided dice are tossed. A diagram representing the sample space is shown
below.
## Score on second die
1 2 3 4 5 6
1 • • • • • •
2 • • • • • •
Score on first die 3 • • • • • •
4 • • • • • •
5 • • • • • •
6 • • • • • •
(a) Find
(6)
## If the sum is 6, she wins 3 points.
If the sum is greater than 6, she wins 1 point.
If the sum is less than 6, she loses k points.
Find the value of k for which Elena’s expected number of points is zero.
(7)
(Total 13 marks)
IB Questionbank Maths SL 61
100. The acceleration, a m s–2, of a particle at time t seconds is given by a = 2t + cost.
## (a) Find the acceleration of the particle at t = 0.
(2)
(b) Find the velocity, v, at time t, given that the initial velocity of the particle is 2 m s–1.
(5)
3
(c) Find vdt , giving your answer in the form p – q cos 3.
0
(7)
(d) What information does the answer to part (c) give about the motion of the particle?
(2)
(Total 16 marks)
101. Let f(t) = a cos b (t – c) + d, t ≥ 0. Part of the graph of y = f(t) is given below.
## When t = 3, there is a maximum value of 29, at M.
When t = 9 , there is a minimum value of 15.
IB Questionbank Maths SL 62
(a) (i) Find the value of a.
π
(ii) Show that b = .
6
## (iv) Write down a value for c.
(7)
1
The transformation P is given by a horizontal stretch of a scale factor of , followed by a
2
3
translation of .
10
(2)
## (c) Find g(t) in the form g(t) = 7 cos B(t – C) + D.
(4)
(d) Give a full geometric description of the transformation that maps the graph of g to the
graph of f.
(3)
(Total 16 marks)
(a) Find
## (ii) the first term.
(4)
IB Questionbank Maths SL 63
(b) Find S10.
(3)
(Total 7 marks)
## (a) Write down the value of u1, u2, and u3.
(3)
20
(b) Find (3 2n) .
n 1
(3)
(Total 6 marks)
(a) Find
(i) f(11);
(ii) f(86);
(iii) f(5).
(3)
(2)
## (c) Let g(x) = x2. Find (g ° f)(x).
(2)
(Total 7 marks)
IB Questionbank Maths SL 64
105. The quadratic function f is defined by f(x) = 3x2 – 12x + 11.
## (a) Write f in the form f(x) = 3(x – h)2 – k.
(3)
(b) The graph of f is translated 3 units in the positive x-direction and 5 units in the positive
y-direction. Find the function g for the translated graph, giving your answer in the form
g(x) = 3(x – p)2 + q.
(3)
(Total 6 marks)
106. The graph of a function of the form y = p cos qx is given in the diagram below.
(2)
## (b) Calculate the value of q.
(4)
(Total 6 marks)
IB Questionbank Maths SL 65
π 12
107. Given that π and that cosθ = , find
2 13
(a) sin θ;
(3)
(3)
## (c) sin (θ + π).
(1)
(Total 7 marks)
108. (a) Given that 2 sin2 θ + sinθ – 1 = 0, find the two values for sin θ.
(4)
(b) Given that 0° ≤ θ ≤ 360° and that one solution for θ is 30°, find the other two possible
values for θ.
(2)
(Total 6 marks)
109. Consider the points A(5, 8), B(3, 5) and C(8, 6).
(a) Find
(i) AB ;
(ii) AC .
(3)
## (ii) Find the sine of the angle between AB and AC .
(3)
(Total 6 marks)
IB Questionbank Maths SL 66
110. A test marked out of 100 is written by 800 students. The cumulative frequency graph for the
marks is given below.
(a) Write down the number of students who scored 40 marks or less on the test.
(2)
(b) The middle 50 % of test results lie between marks a and b, where a < b.
Find a and b.
(4)
(Total 6 marks)
111. A random variable X is distributed normally with a mean of 100 and a variance of 100.
(a) Find the value of X that is 1.12 standard deviations above the mean.
(4)
IB Questionbank Maths SL 67
(b) Find the value of X that is 1.12 standard deviations below the mean.
(2)
(Total 6 marks)
112. In a game a player rolls a biased four-faced die. The probability of each possible score is shown
below.
Score 1 2 3 4
1 2 1
Probability x
5 5 10
(2)
## (b) Find E(X).
(3)
(c) The die is rolled twice. Find the probability of obtaining two scores of 3.
(2)
(Total 7 marks)
113. Find the equation of the tangent to the curve y = e2x at the point where x = 1.
(Total 6 marks)
2
114. (a) Find
1
(3x 2 2) dx.
(4)
IB Questionbank Maths SL 68
1
2e
2x
(b) Find dx .
0
(3)
(Total 7 marks)
115. The velocity v m s–1 of a moving body at time t seconds is given by v = 50 – 10t.
(2)
(4)
(Total 6 marks)
(a) logx 49 = 2
(3)
(b) log2 8 = x
(2)
1
(c) log25 x =
2
(3)
## (d) log2 x + log2(x – 7) = 3
(5)
(Total 13 marks)
IB Questionbank Maths SL 69
117. Let f (x) = 2x2 – 12x + 5.
(3)
## (b) Write down the vertex of the graph of f.
(2)
(c) Write down the equation of the axis of symmetry of the graph of f.
(1)
## (d) Find the y-intercept of the graph of f.
(2)
p q
(e) The x-intercepts of f can be written as , where p, q, r .
r
Find the value of p, of q, and of r.
(7)
(Total 15 marks)
1
118. Let f(x) = , x ≠ 0.
x
## (a) Sketch the graph of f.
(2)
2
The graph of f is transformed to the graph of g by a translation of .
3
## (b) Find an expression for g(x).
(2)
IB Questionbank Maths SL 70
(c) (i) Find the intercepts of g.
## (iii) Sketch the graph of g.
(10)
(Total 14 marks)
119. A spring is suspended from the ceiling. It is pulled down and released, and then oscillates up
and down. Its length, l centimetres, is modelled by the function l = 33 + 5cos((720t)°), where t is
time in seconds after release.
(2)
(3)
(3)
## (d) What is the period of the motion?
(2)
(Total 10 marks)
9 2 1 6
120. Two lines L1 and L2 are given by r1 = 4 s 6 and r2 = 20 t 10 .
6 10 2 2
52
(a) Let θ be the acute angle between L1 and L2. Show that cosθ = .
140
(5)
IB Questionbank Maths SL 71
(b) (i) P is the point on L1 when s = 1. Find the position vector of P.
## (ii) Show that P is also on L2.
(8)
6
(c) A third line L3 has direction vector x . If L1 and L3 are parallel, find the value of x.
30
(3)
(Total 16 marks)
121. The heights of trees in a forest are normally distributed with mean height 17 metres. One tree is
selected at random. The probability that a selected tree has a height greater than 24 metres is
0.06.
(a) Find the probability that the tree selected has a height less than 24 metres.
(2)
(b) The probability that the tree has a height less than D metres is 0.06.
Find the value of D.
(3)
(c) A woodcutter randomly selects 200 trees. Find the expected number of trees whose height
lies between 17 metres and 24 metres.
(4)
(Total 9 marks)
1
122. The probability of obtaining heads on a biased coin is .
3
(a) Sammy tosses the coin three times. Find the probability of getting
## (ii) two heads and one tail.
(5)
IB Questionbank Maths SL 72
(b) Amir plays a game in which he tosses the coin 12 times.
## (i) Find the expected number of heads.
(ii) Amir wins \$ 10 for each head obtained, and loses \$ 6 for each tail.
Find his expected winnings.
(5)
(Total 10 marks)
## 123. Let g(x) = x3 – 3x2 – 9x + 5.
(a) Find the two values of x at which the tangent to the graph of g is horizontal.
(8)
## (b) For each of these values, determine whether it is a maximum or a minimum.
(6)
(Total 14 marks)
124. The diagram below shows part of the graph of y = sin 2x. The shaded region is between x = 0
and x = m.
(2)
## (b) Hence or otherwise write down the value of m.
(2)
IB Questionbank Maths SL 73
(c) Find the area of the shaded region.
(6)
(Total 10 marks)
125. Consider the infinite geometric sequence 25, 5, 1, 0.2, … .
(b) Find
## (c) Find the sum of the infinite sequence.
(Total 6 marks)
2 1 7
126. Consider the events A and B, where P(A) = , P(B′) = and P(A B) = .
5 4 8
(Total 6 marks)
Calculate
## (b) the area of triangle PQR.
(Total 6 marks)
IB Questionbank Maths SL 74
128. The cumulative frequency graph below shows the heights of 120 girls in a school.
130
120
110
100
90
Cumulative frequency
80
70
60
50
40
30
20
10
0
150 155 160 165 170 175 180 185
Height in centimetres
## (ii) find the interquartile range.
(b) Given that 60 of the girls are taller than a cm, find the value of a.
(Total 6 marks)
129. Given that p = loga 5, q = loga 2, express the following in terms of p and/or q.
(a) loga 10
IB Questionbank Maths SL 75
(b) loga 8
## (c) loga 2.5
(Total 6 marks)
130. The following diagram shows a triangle ABC, where AĈB is 90, AB = 3, AC = 2 and BÂC
is .
5
(a) Show that sin = .
3
4 5
(b) Show that sin 2 = .
9
## (c) Find the exact value of cos 2.
(Total 6 marks)
131. The functions f (x) and g (x) are defined by f (x) = ex and g (x) = ln (1+ 2x).
## (ii) Find ( f ◦ g)−1 (x).
(Total 6 marks)
IB Questionbank Maths SL 76
132. The velocity v of a particle at time t is given by v = e−2t + 12t. The displacement of the particle
at time t is s. Given that s = 2 when t = 0, express s in terms of t.
(Total 6 marks)
133. The heights of boys at a particular school follow a normal distribution with a standard deviation
of 5 cm. The probability of a boy being shorter than 153 cm is 0.705.
## (a) Calculate the mean height of the boys.
(b) Find the probability of a boy being taller than 156 cm.
(Total 6 marks)
## (a) Write down the value of
(i) f ′ (1);
(ii) f ′ (3).
IB Questionbank Maths SL 77
(b) On the diagram below, draw the graph of y = 3 f (−x).
## (c) On the diagram below, draw the graph of y = f (2x).
(Total 6 marks)
IB Questionbank Maths SL 78
135. Consider the expansion of the expression (x3 − 3x)6.
(Total 6 marks)
## 136. Let f (x) = x3 − 3x2 − 24x +1.
The tangents to the curve of f at the points P and Q are parallel to the x-axis, where P is to the
left of Q.
## (ii) the tangent at Q intersects N1.
(Total 6 marks)
3
137. It is given that
1
f (x)dx = 5.
3
(a) Write down
1
2 f (x)dx.
3
(b) Find the value of 1
(3x2 + f (x))dx.
(Total 6 marks)
IB Questionbank Maths SL 79
138. (a) Given that (2x)2 + (2x) −12 can be written as (2x + a)(2x + b), where a, b , find the
value of a and of b.
(b) Hence find the exact solution of the equation (2x)2 + (2x) −12 = 0, and explain why there
is only one solution.
(Total 6 marks)
139. The population of a city at the end of 1972 was 250 000. The population increases by 1.3 per
year.
## (b) Find the population at the end of 2002.
(Total 6 marks)
140. One of the terms of the expansion of (x + 2y)10 is ax8 y2. Find the value of a.
(Total 6 marks)
## (c) Write down the domain of f −1.
(Total 6 marks)
IB Questionbank Maths SL 80
142. The eye colour of 97 students is recorded in the chart below.
Male 21 16 9
Female 19 19 13
## (a) Write down the probability that the student is a male.
(b) Write down the probability that the student has green eyes, given that the student is a
female.
(c) Find the probability that the student has green eyes or is male.
(Total 6 marks)
(Total 6 marks)
## (b) Given that q u =14, find the value of q.
(Total 6 marks)
145. The weights of a group of children are normally distributed with a mean of 22.5 kg and a
standard deviation of 2.2 kg.
(a) Write down the probability that a child selected at random has a weight more than 25.8
kg.
(b) Of the group 95 weigh less than k kilograms. Find the value of k.
IB Questionbank Maths SL 81
(c) The diagram below shows a normal curve.
On the diagram, shade the region that represents the following information:
## 87 of the children weigh less than 25 kg
(Total 6 marks)
146. The velocity, v, in m s−1 of a particle moving in a straight line is given by v = e3t−2, where t is
the time in seconds.
## (a) Find the acceleration of the particle at t =1.
(b) At what value of t does the particle have a velocity of 22.3 m s−1?
(Total 6 marks)
## 147. A set of data is
18, 18, 19, 19, 20, 22, 22, 23, 27, 28, 28, 31, 34, 34, 36.
The box and whisker plot for this data is shown below.
## (b) Find the interquartile range.
(Total 6 marks)
IB Questionbank Maths SL 82
148. The following diagram shows a sector of a circle of radius r cm, and angle at the centre. The
perimeter of the sector is 20 cm.
20 2r
(a) Show that = .
r
## (b) The area of the sector is 25 cm2. Find the value of r.
(Total 6 marks)
149. Consider two different quadratic functions of the form f (x) = 4x2 − qx + 25. The graph of each
function has its vertex on the x-axis.
## (b) For the greater value of q, solve f (x) = 0.
(c) Find the coordinates of the point of intersection of the two graphs.
(Total 6 marks)
## (ii) Find the range of f.
(c) Find the coordinates of the point of intersection of the graphs of f and g.
(Total 6 marks)
IB Questionbank Maths SL 83
151. The graph of a function f is shown in the diagram below. The point A (–1, 1) is on the graph,
and y = −1 is a horizontal asymptote.
## (b) Write down the equation of the horizontal asymptote of g.
(c) Let A′ be the point on the graph of g corresponding to point A. Write down the
coordinates of A′.
(Total 6 marks)
## (a) Show that f ′ (x) = −5 sin 2x.
π 3π
(b) In the interval x , one normal to the graph of f has equation x = k.
4 4
Find the value of k.
(Total 6 marks)
IB Questionbank Maths SL 84
153. The histogram below represents the ages of 270 people in a village.
## (a) Use the histogram to complete the table below.
Mid-interval
Age range Frequency
value
0 age 20 40 10
20 ≤ age 40
40 ≤ age 60
60 ≤ age 80
80 ≤ age ≤100
(2)
## (b) Hence, calculate an estimate of the mean age.
(4)
(Total 6 marks)
IB Questionbank Maths SL 85
154. The Venn diagram below shows information about 120 students in a school. Of these, 40 study
Chinese (C), 35 study Japanese (J), and 30 study Spanish (S).
A student is chosen at random from the group. Find the probability that the student
(1)
(2)
## (c) does not study any of these languages.
(3)
(Total 6 marks)
IB Questionbank Maths SL 86
155. A bag contains four apples (A) and six bananas (B). A fruit is taken from the bag and eaten.
Then a second fruit is taken and eaten.
(a) Complete the tree diagram below by writing probabilities in the spaces provided.
(3)
(b) Find the probability that one of each type of fruit was eaten.
(3)
(Total 6 marks)
156. The following diagram shows part of the graph of y = cos x for 0 x 2. Regions A and B are
(1)
## (b) Calculate the area of A.
(1)
IB Questionbank Maths SL 87
(c) Find the total area of the shaded regions.
(4)
(Total 6 marks)
157. The first four terms of a sequence are 18, 54, 162, 486.
(a) Use all four terms to show that this is a geometric sequence.
(2)
(b) (i) Find an expression for the nth term of this geometric sequence.
(ii) If the nth term of the sequence is 1062 882, find the value of n.
(4)
(Total 6 marks)
158. (a) Write down the first three terms of the sequence un = 3n, for n 1.
(1)
(b) Find
20
(i) 3n ;
n 1
100
(ii) 3n .
n 21
(5)
(Total 6 marks)
IB Questionbank Maths SL 88
159. Let f (x) = loga x, x 0.
(i) f (a);
(ii) f (1);
(iii) f (a4 ).
(3)
y
1
f
–2 –1 0 1 2 x
–1
–2
## On the same diagram, sketch the graph of f−1.
(3)
(Total 6 marks)
160. Consider the function f (x) = 4x3 + 2x. Find the equation of the normal to the curve of f at the
point where x =1.
(Total 6 marks)
IB Questionbank Maths SL 89
161. Differentiate each of the following with respect to x.
(a) y = sin 3x
(1)
(b) y = x tan x
(2)
ln x
(c) y=
x
(3)
(Total 6 marks)
x
162. The diagram below shows the graph of f (x) = 1 + tan for −360 x 360.
2
(2)
## (ii) the value of f (90).
(2)
IB Questionbank Maths SL 90
(c) Solve f (x) = 0 for −360 x 360.
(2)
(Total 6 marks)
163. The following diagram shows part of the graph of f (x) = 5 − x2 with vertex V (0, 5).
h
Its image y = g (x) after a translation with vector has vertex T (3, 6).
k
(i) h;
(ii) k.
(2)
(2)
## (c) On the same diagram, sketch the graph of y = g (−x).
(2)
(Total 6 marks)
IB Questionbank Maths SL 91
164. A discrete random variable X has a probability distribution as shown in the table below.
x 0 1 2 3
(2)
## (b) Given that E(X) =1.5, find the value of a and of b.
(4)
(Total 6 marks)
4
1
165. (a) Expand e in terms of e.
e
(4)
4 4
1 1
(b) Express e + e as the sum of three terms.
e e
(2)
(Total 6 marks)
166. The area A km2 affected by a forest fire at time t hours is given by A = A0 ekt.
When t = 5, the area affected is 1 km2 and the rate of change of the area is 0.2 km2 h−1.
## (a) Show that k = 0.2.
(4)
1
(b) Given that A0 = , find the value of t when 100 km2 are affected.
e
(2)
(Total 6 marks)
IB Questionbank Maths SL 92
167. On the axes below, sketch a curve y = f (x) which satisfies the following conditions.
## x f (x) f ′ (x) f ′′ (x)
−2 x 0 negative positive
0 –1 0 positive
0 x 1 positive positive
1 2 positive 0
## 1x2 positive negative
(Total 6 marks)
IB Questionbank Maths SL 93
168. Let A and B be independent events such that P(A) = 0.3 and P(B) = 0.8.
## (b) Find P(A B).
(Total 6 marks)
IB Questionbank Maths SL 94
169. The following diagram shows part of the graph of f (x).
## (b) Which diagram is the graph of – f (x) ?
IB Questionbank Maths SL 95
(c) Which diagram is the graph of f (–x)
(Total 6 marks)
170. The heights of a group of students are normally distributed with a mean of 160 cm and a
standard deviation of 20 cm.
(a) A student is chosen at random. Find the probability that the student’s height is greater
than 180 cm.
(b) In this group of students, 11.9 have heights less than d cm. Find the value of d.
(Total 6 marks)
(Total 6 marks)
(Total 6 marks)
## (b) Find f −1 (x).
(Total 6 marks)
IB Questionbank Maths SL 96
174. Consider the four numbers a, b, c, d with a b c ≤ d, where a, b, c, d .
The mean of the four numbers is 4.
The mode is 3.
The median is 3.
The range is 6.
## Find the value of a, of b, of c and of d.
(Total 6 marks)
175. (a) Let logc 3 = p and logc 5 = q. Find an expression in terms of p and q for
## (ii) log c 25.
1
(b) Find the value of d if log d 6 = .
2
(Total 6 marks)
IB Questionbank Maths SL 97
176. The following diagram shows part of the curve of a function ƒ. The points A, B, C, D and E lie
on the curve, where B is a minimum point and D is a maximum point.
(a) Complete the following table, noting whether ƒ′(x) is positive, negative or zero at the
given points.
A B E
f ′ (x)
(b) Complete the following table, noting whether ƒ′′(x) is positive, negative or zero at the
given points.
A C E
ƒ′′ (x)
(Total 6 marks)
177. The velocity, v m s−1, of a moving object at time t seconds is given by v = 4t3 − 2t.
When t = 2, the displacement, s, of the object is 8 metres.
## Find an expression for s in terms of t.
(Total 6 marks)
IB Questionbank Maths SL 98
178. The following diagram shows a circle with radius r and centre O. The points A, B and C are on
the circle and AÔC =.
4 2
The area of sector OABC is and the length of arc ABC is .
3 3
## Find the value of r and of .
(Total 6 marks)
179. Let ƒ (x) = a sin b (x − c). Part of the graph of ƒ is given below.
## Given that a, b and c are positive, find the value of a, of b and of c.
(Total 6 marks)
IB Questionbank Maths SL 99
180. Let ƒ (x) = 3 sin 2x, for 1 x 4 and g (x) = −5x2 + 27x − 35 for 1 x 4. The graph of ƒ is
shown below.
## (a) On the same diagram, sketch the graph of g.
(b) One solution of ƒ (x) = g (x) is 1.89. Write down the other solution.
(c) Let h (x) = g (x) − ƒ (x). Given that h (x) 0 for p x q, write down the value of p and
of q.
(Total 6 marks)
181. Consider the infinite geometric series 405 + 270 + 180 +....
(a) For this series, find the common ratio, giving your answer as a fraction in its simplest
form.
## (b) Find the fifteenth term of this series.
(c) Find the exact value of the sum of the infinite series.
(Total 6 marks)
## IB Questionbank Maths SL 100
182. The points P(−2, 4), Q (3, 1) and R (1, 6) are shown in the diagram below.
## (a) Find the vector PQ .
(b) Find a vector equation for the line through R parallel to the line (PQ).
(Total 6 marks)
(i) r;
(ii) s.
(Total 6 marks)
## 184. Solve the following equations.
(a) ln (x + 2) = 3.
(Total 6 marks)
## IB Questionbank Maths SL 101
185. The probability distribution of the discrete random variable X is given by the following table.
x 1 2 3 4 5
(Total 6 marks)
## 186. The graph of a function g is given in the diagram below.
The gradient of the curve has its maximum value at point B and its minimum value at point D.
The tangent is horizontal at points C and E.
(a) Complete the table below, by stating whether the first derivative g′ is positive or negative,
and whether the second derivative g′′ is positive or negative.
Interval g′ g′′
axb
exƒ
## IB Questionbank Maths SL 102
(b) Complete the table below by noting the points on the graph described by the following
conditions.
Conditions Point
(Total 6 marks)
## 187. (a) Express y = 2x2 – 12x + 23 in the form y = 2(x – c)2 + d.
The graph of y = x2 is transformed into the graph of y = 2x2 – 12x + 23 by the transformations
## a vertical stretch with scale factor k followed by
a horizontal translation of p units followed by
a vertical translation of q units.
(i) k;
(ii) p;
(iii) q.
(Total 6 marks)
## IB Questionbank Maths SL 103
188. The diagram below shows a circle of radius r and centre O. The angle AÔB = .
The length of the arc AB is 24 cm. The area of the sector OAB is 180 cm2.
(Total 6 marks)
(Total 6 marks)
## IB Questionbank Maths SL 104
190. Consider the function ƒ : x 3x2 – 5x + k.
## (a) Write down ƒ′ (x).
The equation of the tangent to the graph of ƒ at x = p is y = 7x – 9. Find the value of
(b) p;
(c) k.
(Total 6 marks)
191. In a class, 40 students take chemistry only, 30 take physics only, 20 take both chemistry and
physics, and 60 take neither.
(a) Find the probability that a student takes physics given that the student takes chemistry.
(b) Find the probability that a student takes physics given that the student does not take
chemistry.
(c) State whether the events “taking chemistry” and “taking physics” are mutually exclusive,
(Total 6 marks)
## IB Questionbank Maths SL 105
192. The diagram below shows the graph of ƒ (x) = x2 e–x for 0 x 6. There are points of inflexion
at A and C and there is a maximum at B.
(a) Using the product rule for differentiation, find ƒ′ (x).
## (b) Find the exact value of the y-coordinate of B.
(c) The second derivative of ƒ is ƒ′′ (x) = (x2 – 4x + 2) e–x. Use this result to find the exact
value of the x-coordinate of C.
(Total 6 marks)
## (b) Find the acceleration, a, at time t.
(c) Find the value of t when the maximum value of a first occurs.
(Total 6 marks)
## IB Questionbank Maths SL 106
194. The four populations A, B, C and D are the same size and have the same range.
Frequency histograms for the four populations are given below.
(a) Each of the three box and whisker plots below corresponds to one of the four populations.
Write the letter of the correct population under each plot.
## ...... ...... ......
(b) Each of the three cumulative frequency diagrams below corresponds to one of the four
populations. Write the letter of the correct population under each diagram.
(Total 6 marks)
## IB Questionbank Maths SL 107
195. Let ln a = p, ln b = q. Write the following expressions in terms of p and q.
(a) ln a3b
a
(b) ln
b
(Total 6 marks)
196. The box and whisker diagram shown below represents the marks received by 32 students.
## (b) Write down the value of the upper quartile.
(c) Estimate the number of students who received a mark greater than 6.
(Total 6 marks)
## IB Questionbank Maths SL 108
197. The following diagram shows the graph of a function f.
## Consider the following diagrams.
Complete the table below, noting which one of the diagrams above represents the graph of
(a) f ′(x);
(b) f ′′(x).
Graph Diagram
(a) f ′ (x)
(Total 6 marks)
## IB Questionbank Maths SL 109
2 1
198. Events E and F are independent, with P(E) = and P(E F) = . Calculate
3 3
(a) P(F);
## (b) P(E F).
(Total 6 marks)
199. Part of the graph of the function y = d (x −m)2 + p is given in the diagram below.
The x-intercepts are (1, 0) and (5, 0). The vertex is V(m, 2).
## (a) Write down the value of
(i) m;
(ii) p.
(b) Find d.
(Total 6 marks)
200. The line L passes through the points A (3, 2, 1) and B (1, 5, 3).
## (a) Find the vector AB .
(b) Write down a vector equation of the line L in the form r = a + tb.
(Total 6 marks)
## IB Questionbank Maths SL 110
201. Find the exact value of x in each of the following equations.
## (b) loga (3x + 5) = 2
(Total 6 marks)
5x
202. Let g (x) = 3x – 2, h (x) = , x 4.
x4
## (b) Solve the equation (h ◦ g) (x) = 0.
(Total 6 marks)
203. The velocity v in m s−1 of a moving body at time t seconds is given by v = e2t−1. When t = 0 5.
the displacement of the body is 10 m. Find the displacement when t =1.
(Total 6 marks)
204. The line L passes through A (0, 3) and B (1, 0). The origin is at O. The point R (x, 3 − 3x) is on
L, and (OR) is perpendicular to L.
(Total 6 marks)
## IB Questionbank Maths SL 111
205. A fair coin is tossed five times. Calculate the probability of obtaining
## (b) at least one head.
(Total 6 marks)
π
206. The function f is defined by f : x 30 sin 3x cos 3x, 0 x .
3
(a) Write down an expression for f (x) in the form a sin 6x, where a is an integer.
## (b) Solve f (x) = 0, giving your answers in terms of .
(Total 6 marks)
207. The heights of certain flowers follow a normal distribution. It is known that 20 of these
flowers have a height less than 3 cm and 10 have a height greater than 8 cm.
## Find the value of the mean and the standard deviation .
(Total 6 marks)
208. The shaded region in the diagram below is bounded by f (x) = x , x = a, and the x-axis. The
shaded region is revolved around the x-axis through 360. The volume of the solid formed is
0.845.
(Total 6 marks)
## IB Questionbank Maths SL 112
209. A test marked out of 100 is written by 800 students. The cumulative frequency graph for the
marks is given below.
800
700
600
Number
of 500
candidates
400
300
200
100
10 20 30 40 50 60 70 80 90 100
Mark
(a) Write down the number of students who scored 40 marks or less on the test.
(b) The middle 50% of test results lie between marks a and b, where a < b. Find a and b.
(Total 6 marks)
210. A theatre has 20 rows of seats. There are 15 seats in the first row, 17 seats in the second row,
and each successive row of seats has two more seats in it than the previous row.
(Total 6 marks)
## IB Questionbank Maths SL 113
211. The function f is given by f (x) = 2sin (5x – 3).
## (b) Write down f ( x)dx .
(Total 6 marks)
212. The diagram shows a cube, OABCDEFG where the length of each edge is 5cm. Express the
following vectors in terms of i, j and k.
E
G
y
F
C
k
j
O
i B
(a) OG ;
(b) BD ;
(c) EB .
(Total 6 marks)
## IB Questionbank Maths SL 114
213. The graph of a function of the form y = p cos qx is given in the diagram below.
y
40
30
20
10
x
/2
–10
–20
–30
–40
## (b) Calculate the value of q.
(Total 6 marks)
214. The velocity v m s–1 of a moving body at time t seconds is given by v = 50 – 10t.
(Total 6 marks)
(Total 6 marks)
## IB Questionbank Maths SL 115
216. A sum of \$5000 is invested at a compound interest rate of 6.3% per annum.
(a) Write down an expression for the value of the investment after n full years.
(b) What will be the value of the investment at the end of five years?
(c) The value of the investment will exceed \$10000 after n full years,
## (ii) Calculate the minimum value of n.
(Total 6 marks)
3
217. The function f is defined by f ( x) , for –3 < x < 3.
9 – x2
(Total 6 marks)
## IB Questionbank Maths SL 116
218. A triangle has its vertices at A(–1, 3), B(3, 6) and C(–4, 4).
## (b) Show that, to three significant figures, cos BÂC – 0.569.
(Total 6 marks)
219. A factory makes calculators. Over a long period, 2% of them are found to be faulty. A random
sample of 100 calculators is tested.
(a) Write down the expected number of faulty calculators in the sample.
## (b) Find the probability that three calculators are faulty.
(c) Find the probability that more than one calculator is faulty.
(Total 6 marks)
## (a) Write f in the form f (x) = 3(x – h)2 – k.
(b) The graph of f is translated 3 units in the positive x-direction and 5 units in the positive
y-direction. Find the function g for the translated graph, giving your answer in the form
g (x) = 3(x – p)2 + q.
(Total 6 marks)
221. The speeds of cars at a certain point on a straight road are normally distributed with mean and
standard deviation . 15% of the cars travelled at speeds greater than 90 km h–1 and 12% of
them at speeds less than 40 km h–1. Find and .
(Total 6 marks)
## IB Questionbank Maths SL 117
222. Let Sn be the sum of the first n terms of an arithmetic sequence, whose first three terms are u1,
u2 and u3. It is known that S1 = 7, and S2 = 18.
## (c) Calculate u4.
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 118
223. Consider the line L with equation y + 2x = 3. The line L1 is parallel to L and passes through the
point (6, –4).
## (b) Find the equation of L1 in the form y = mx + b.
(c) Find the x-coordinate of the point where line L1 crosses the x-axis.
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 119
224. Consider the expansion of (x2 – 2)5.
## (a) Write down the number of terms in this expansion.
(b) The first four terms of the expansion in descending powers of x are
## Find the value of A.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
225. The following diagram shows a circle of centre O, and radius r. The shaded sector OACB has
an area of 27 cm2. Angle AÔB = θ = 1.5 radians.
A
C
B
r
O
## (b) Calculate the length of the minor arc ACB.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 121
226. Two unbiased 6-sided dice are rolled, a red one and a black one. Let E and F be the events
Find
(a) P(E);
(b) P(F);
## (c) P(E F).
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 122
227. Let f (x) = (3x + 4)5. Find
(a) f (x);
(b) f (x)dx.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## 228. Find the exact solution of the equation 92x = 27(1–x).
Working:
....……………………………………..........
(Total 6 marks)
## IB Questionbank Maths SL 123
229. The curve y = f (x) passes through the point (2, 6).
dy
Given that = 3x2 – 5, find y in terms of x.
dx
Working:
....……………………………………..........
(Total 6 marks)
## (b) Hence or otherwise, solve the equation log3 x – log3 (x – 5) = 1.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 124
3 2
231. Find the cosine of the angle between the two vectors and .
4 1
Working:
....……………………………………..........
(Total 6 marks)
232. The 45 students in a class each recorded the number of whole minutes, x, spent doing
experiments on Monday. The results are x = 2230.
(a) Find the mean number of minutes the students spent doing experiments on Monday.
Two new students joined the class and reported that they spent 37 minutes and 30 minutes
respectively.
## (b) Calculate the new mean including these two students.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 125
(a) Find f –1(x).
## (b) Write down the domain of f –l(x).
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## 234. The graph of y = f (x) is shown in the diagram.
y
2
–2 –1 0 1 2 3 4 5 6 7 8 x
–1
–2
## IB Questionbank Maths SL 126
(a) On each of the following diagrams draw the required graph,
(i) y = 2 f (x);
y
2
–2 –1 0 1 2 3 4 5 6 7 8 x
–1
–2
(ii) y = f (x – 3).
y
2
–2 –1 0 1 2 3 4 5 6 7 8 x
–1
–2
## IB Questionbank Maths SL 127
(b) The point A (3, –1) is on the graph of f. The point A is the corresponding point on the
graph of y = –f (x) + 1. Find the coordinates of A.
Working:
(b) ..................................................................
(Total 6 marks)
## 235. Consider y = sin x .
9
(a) The graph of y intersects the x-axis at point A. Find the x-coordinate of A, where
0 x π.
## (b) Solve the equation sin x = – 1 , for 0 x 2.
9 2
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 128
236. The table below shows some values of two functions, f and g, and of their derivatives f and g .
x 1 2 3 4
f (x) 5 4 –1 3
g (x) 1 –2 2 –5
f (x) 5 6 0 7
g (x) –6 –4 –3 4
## (a) d (f (x) + g (x)), when x = 4;
dx
3
(b) g'( x) 6dx .
1
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 129
237. In triangle PQR, PQ is 10 cm, QR is 8 cm and angle PQR is acute. The area of the triangle is 20
cm2. Find the size of angle PQ̂R.
Working:
........................................................
(Total 6 marks)
238. A class contains 13 girls and 11 boys. The teacher randomly selects four students. Determine
the probability that all four students selected are girls.
Working:
........................................................
(Total 6 marks)
## IB Questionbank Maths SL 130
239. Given that 3 7
3
= p + q 7 where p and q are integers, find
(a) p;
(b) q.
Working:
(a) .................................................
(b) .................................................
(Total 6 marks)
## IB Questionbank Maths SL 131
240. The following table shows the mathematics marks scored by students.
Mark 1 2 3 4 5 6 7
Frequency 0 4 6 k 8 6 6
## (b) Write down the mode.
Working:
(a) .................................................
(b) .................................................
(Total 6 marks)
## IB Questionbank Maths SL 132
241. Part of the graph of f (x) = (x – p) (x – q) is shown below.
## (c) Write down the y-coordinate of B.
Working:
(a) .....................................................
(b) .....................................................
(c) .....................................................
(Total 6 marks)
## IB Questionbank Maths SL 133
1
242. Consider the functions f (x) = 2x and g (x) = , x 3.
x 3
## (c) Write down the domain of g−1.
Working:
(a) .....................................................
(b) .....................................................
(c) .....................................................
(Total 6 marks)
## IB Questionbank Maths SL 134
243. The first term of an infinite geometric sequence is 18, while the third term is 8. There are two
possible sequences. Find the sum of each sequence.
Working:
.........................................................
.........................................................
(Total 6 marks)
k 1
244. Given 3 x2
dx = ln 7, find the value of k.
Working:
........................................................
(Total 6 marks)
## IB Questionbank Maths SL 135
245. A machine was purchased for \$10000. Its value V after t years is given by V =100000e−0.3t. The
machine must be replaced at the end of the year in which its value drops below \$1500.
Determine in how many years the machine will need to be replaced.
Working:
........................................................
(Total 6 marks)
## IB Questionbank Maths SL 136
246. Let f (x) = 6 sin x , and g (x) = 6e–x – 3 , for 0 x 2. The graph of f is shown on the diagram
below. There is a maximum value at B (0.5, b).
y
B
x
0 1 2
## (c) Solve f (x) = g (x) , 0.5 x 1.5.
Working:
(a) .................................................
(b) .................................................
(Total 6 marks)
## IB Questionbank Maths SL 137
247. Let f (x) = (2x + 7)3 and g (x) cos2 (4x). Find
(a) f ′ (x);
(b) g′ (x).
Working:
(a) .................................................
(b) .................................................
(Total 6 marks)
248. The events A and B are independent such that P(B) = 3P(A) and P(AB) = 0.68. Find P(B)
Working:
........................................................
(Total 6 marks)
## IB Questionbank Maths SL 138
249. A boat B moves with constant velocity along a straight line. Its velocity vector is given by
4
v = .
3
## (b) Find the coordinates of B when t = 2.
(c) Write down a vector equation representing the position of B, giving your answer in the
form r = a + tb.
Working:
(a) .....................................................
(b) .....................................................
(c) .....................................................
(Total 6 marks)
## IB Questionbank Maths SL 139
250. Consider the equation 3 cos 2x + sin x = 1
(a) Write this equation in the form f (x) = 0 , where f (x) = p sin2 x + q sin x + r , and p , q , r
.
## (c) Write down the number of solutions of f (x) = 0, for 0 x 2.
Working:
(a) .....................................................
(b) .....................................................
(c) .....................................................
(Total 6 marks)
## IB Questionbank Maths SL 140
251. The following diagram shows a rectangular area ABCD enclosed on three sides by 60 m of
fencing, and on the fourth by a wall AB.
Find the width of the rectangle that gives its maximum area.
Working:
........................................................
(Total 6 marks)
## IB Questionbank Maths SL 141
252. Let f (x) = x3 – 2x2 – 1.
## (a) Find f (x).
(b) Find the gradient of the curve of f (x) at the point (2, –1).
Working:
(a) …………………………………………
(b) …………………………………………
(Total 6 marks)
253. The diagram below shows two circles which have the same centre O and radii 16 cm and 10 cm
respectively. The two arcs AB and CD have the same sector angle = 1.5 radians.
A B
C D
## IB Questionbank Maths SL 142
Find the area of the shaded region.
Working:
…………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 143
254. The cumulative frequency curve below shows the marks obtained in an examination by a group
of 200 students.
200
190
180
170
160
150
140
Number
of 130
students
120
110
100
90
80
70
60
50
40
30
20
10
0
10 20 30 40 50 60 70 80 90 100
Mark obtained
## IB Questionbank Maths SL 144
(a) Use the cumulative frequency curve to complete the frequency table below.
## Mark (x) 0 x < 20 20 x < 40 40 x < 60 60 x < 80 80 x < 100
Number of
22 20
students
(b) Forty percent of the students fail. Find the pass mark.
Working:
(b) …………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 145
255. Let f (x) = 2x + 1.
## (a) On the grid below draw the graph of f (x) for 0 x 2.
(b) Let g (x) = f (x +3) –2. On the grid below draw the graph of g (x) for –3 x –1.
y
6
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x
–1
–2
–3
–4
–5
–6
## IB Questionbank Maths SL 146
Working:
(Total 6 marks)
1 3 7
256. Let A and B be events such that P(A) = , P(B) = and P(A B) = .
2 4 8
## (b) Calculate P(AB).
(c) Are the events A and B independent? Give a reason for your answer.
Working:
(a) …………………………………………..
(b) …………………………………………..
(c) ……………………………………..........
(Total 6 marks)
## IB Questionbank Maths SL 147
257. Let f (x) = sin (2x + 1), 0 x π.
## (a) Sketch the curve of y = f (x) on the grid below.
y
2
1.5
0.5
0
0.5 1 1.5 2 2.5 3 3.5 x
–0.5
–1
–1.5
–2
(b) Find the x-coordinates of the maximum and minimum points of f (x), giving your answers
correct to one decimal place.
Working:
(b) …………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 148
258. Let p = log10 x, q = log10 y and r = log10 z.
x
Write the expression log10 2 in terms of p, q and r.
y z
Working:
…………………………………………..
(Total 6 marks)
259. In a triangle ABC, AB = 4 cm, AC = 3 cm and the area of the triangle is 4.5 cm2.
## Find the two possible values of the angle BÂC .
Working:
…………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 149
260. Solve the equation 2 cos2 x = sin 2x for 0 x π, giving your answers in terms of π.
Working:
…………………………………………..
(Total 6 marks)
261. A car starts by moving from a fixed point A. Its velocity, v m s–1 after t seconds is given by
v = 4t + 5 – 5e–t. Let d be the displacement from A when t = 4.
## (b) Calculate the value of d.
Working:
(a) …………………………………………..
(b) …………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 150
262. The following table shows four series of numbers. One of these series is geometric, one of the
series is arithmetic and the other two are neither geometric nor arithmetic.
(a) Complete the table by stating the type of series that is shown.
## Series Type of series
(i) 1 11 111 1111 11111 …
(ii) 3 9 27
1 …
4 16 64
(iii) 0.9 0.875 0.85 0.825 0.8 …
(iv) 1 2 3 4 5
2 3 4 5 6
(b) The geometric series can be summed to infinity. Find this sum.
Working:
(b) …………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 151
263. The table below shows the marks gained in a test by a group of students.
Mark 1 2 3 4 5
Number of students 5 10 p 6 2
The median is 3 and the mode is 2. Find the two possible values of p.
Working:
…………………………………………..
(Total 6 marks)
## 264. Two lines L1 and L2 have these vector equations.
L1 : r = 2i + 3j + t(i– 3j)
L2 : r = i + 2j + s(i – j)
## The angle between L1 and L2 is . Find the cosine of the angle .
Working:
…………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 152
265. The equation x2 – 2kx + 1 = 0 has two distinct real roots. Find the set of all possible values of k.
Working:
…………………………………………..
(Total 6 marks)
266. When the expression (2 + ax)10 is expanded, the coefficient of the term in x3 is 414 720. Find
the value of a.
Working:
…………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 153
267. The following diagram shows a triangle ABC, where BC = 5 cm, B̂ = 60°, Ĉ = 40°.
60° 40°
B C
5 cm
## (b) Find the area of the triangle.
Working:
(a) …………………………………………..
(b) …………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 154
268. Let f (x) = 6 3 x 2 . Find f (x).
Working:
…………………………………………........
(Total 6 marks)
## IB Questionbank Maths SL 155
269. The cumulative frequency curve below shows the heights of 120 basketball players in
centimetres.
120
110
100
90
80
70
60
Number of players
50
40
30
20
10
0
160 165 170 175 180 185 190 195 200
Height in centimetres
## IB Questionbank Maths SL 156
Use the curve to estimate
## (b) the interquartile range.
Working:
(a) …………………………………………..
(b) …………………………………………..
(Total 6 marks)
## 270. Find the term containing x3 in the expansion of (2 – 3x)8.
Working:
…………………………………………........
(Total 6 marks)
## IB Questionbank Maths SL 157
271. The following diagram shows a circle divided into three sectors A, B and C. The angles at the
centre of the circle are 90°, 120° and 150°. Sectors A and B are shaded as shown.
C 90°
150°
120°
The arrow is spun. It cannot land on the lines between the sectors. Let A, B, C and S be the
events defined by
## A: Arrow lands in sector A
B: Arrow lands in sector B
C: Arrow lands in sector C
S: Arrow lands in a shaded region.
## IB Questionbank Maths SL 158
Find
(a) P(B);
(b) P(S);
(c) P(AS).
Working:
(a) …………………………………………..
(b) …………………………………………..
(c) …………………………………………..
(Total 6 marks)
## 272. Let a = log x, b = log y, and c = log z.
x2 y
Write log 3 in terms of a, b and c.
z
Working:
…………………………………………........
(Total 6 marks)
## IB Questionbank Maths SL 159
273. Let a, b, c and d be integers such that a < b, b < c and c = d.
## The mode of these four numbers is 11.
The range of these four numbers is 8.
The mean of these four numbers is 8.
## Calculate the value of each of the integers a, b, c, d.
Working:
a = ............................., b = .............................
c = ............................., d = .............................
(Total 6 marks)
## IB Questionbank Maths SL 160
274. Let f (x) = 2x + 1 and g (x) = 3x2 – 4.
Find
(a) f –1(x);
(b) (g f ) (–2);
(c) ( f g) (x).
Working:
(a) …………………………………………..
(b) …………………………………………..
(c) …………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 161
275. The displacement s metres of a car, t seconds after leaving a fixed point A, is given by
s = 10t – 0.5t2.
## (b) Calculate the value of t when the velocity is zero.
(c) Calculate the displacement of the car from A when the velocity is zero.
Working:
(a) …………………………………………..
(b) …………………………………………..
(c) …………………………………………..
(Total 6 marks)
276. Let f (x) = 2 + cos (2x) – 2 sin (0.5x) for 0 x 3, where x is in radians.
(a) On the grid below, sketch the curve of y = f (x), indicating clearly the point P on the curve
where the derivative is zero.
y
4
3
2
1
0
0.5 1 1.5 2 2.5 3 x
–1
–2
–3
–4
## IB Questionbank Maths SL 162
(b) Write down the solutions of f (x) = 0.
Working:
(b) …………………………………………..
(Total 6 marks)
Calculate
## (b) the rate of increase of the population when t = 10.
Working:
(a) …………………………………………..
(b) …………………………………………..
(Total 6 marks)
## IB Questionbank Maths SL 163
1
278. The derivative of the function f is given by f (x) = e–2x + , x < 1.
1 x
The graph of y = f (x) passes through the point (0, 4). Find an expression for f (x).
Working:
…………………………………………........
(Total 6 marks)
## IB Questionbank Maths SL 164
3
279. Let f be a function such that
0
f ( x ) dx 8 .
## (a) Deduce the value of
3
(i) 0
2 f ( x) dx;
3
(ii) f ( x) 2dx.
0
d
(b) If
c
f ( x 2)dx 8 , write down the value of c and of d.
Working:
## (a) (i) ........................................................
(ii) .......................................................
(b) c = ......................., d = .......................
(Total 6 marks)
280. The diagram below shows a circle of radius 5 cm with centre O. Points A and B are on the
circle, and AÔB is 0.8 radians. The point N is on [OB] such that [AN] is perpendicular to
[OB].
5 cm
0.8
O B
N
## IB Questionbank Maths SL 165
Find the area of the shaded region.
Working:
…………………………………………........
(Total 6 marks)
## IB Questionbank Maths SL 166
281. Part of the graph of the periodic function f is shown below. The domain of f is 0 x 15 and
the period is 3.
f(x)
4
0 x
0 1 2 3 4 5 6 7 8 9 10
(a) Find
(i) f (2);
## (iii) f (14).
(b) How many solutions are there to the equation f (x) = 1 over the given domain?
Working:
## (a) (i) ………………………………………
(ii) ………………………………………
(iii) ………………………………………
(b) ……………………………………………
(Total 6 marks)
## IB Questionbank Maths SL 167
282. Gwendolyn added the multiples of 3, from 3 to 3750 and found that
3 + 6 + 9 + … + 3750 = s.
Calculate s.
Working:
..................................................................
(Total 6 marks)
## 283. Find the term containing x10 in the expansion of (5 + 2x2)7.
Working:
..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 168
284. The number of hours of sleep of 21 students are shown in the frequency table below.
4 2
5 5
6 4
7 3
8 4
10 2
12 1
Find
## (c) the interquartile range.
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 169
285. Part of the graph of y = p + q cos x is shown below. The graph passes through the points (0, 3)
and (, –1).
0
2 x
–1
## Find the value of
(a) p;
(b) q.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 170
x
286. Let f (x) = e 3 + 5 cos 2x. Find f (x).
Working:
..................................................................
(Total 6 marks)
287. Find all solutions of the equation cos 3x = cos (0.5x), for 0 x .
Working:
..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 171
288. The vector equations of two lines are given below.
5 3 – 2 4
r1 = + , r2 = + t
1 – 2 2 1
## The lines intersect at the point P. Find the position vector of P.
Working:
..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 172
289. Consider events A, B such that P (A) 0, P (A) 1, P (B) 0, and P (B) 1.
In each of the situations (a), (b), (c) below state whether A and B are
independent (I);
neither (N).
(b) P(A B) = 0
## (c) P(A B) = P(A)
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 173
3
290. Given that 1
g ( x)dx = 10, deduce the value of
3 1
(a) 1 2
g ( x)dx;
3
(b) 1
( g ( x) 4)dx.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 174
291. Given that log5 x = y, express each of the following in terms of y.
(a) log5 x2
(b) log5 1
x
(c) log25 x
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 175
x
292. Let f (x) = e–x, and g (x) = , x –1. Find
1 x
(a) f –1 (x);
(b) (g ° f ) (x).
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## Calculate the scalar product cd.
Working:
..................................................................
(Total 2 marks)
## IB Questionbank Maths SL 176
294. A family of functions is given by
## f (x) = x2 + 3x + k, where k {1, 2, 3, 4, 5, 6, 7}.
One of these functions is chosen at random. Calculate the probability that the curve of this
function crosses the x-axis.
Working:
..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 177
295. The diagram below shows a triangle and two arcs of circles.
The triangle ABC is a right-angled isosceles triangle, with AB = AC = 2. The point P is the
midpoint of [BC].
E
B
P
2
A 2 C
## (b) Calculate the area of the shaded region BECD.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 178
296. Consider the following relations between two variables x and y.
A. y = sin x
B. y is directly proportional to x
C. y = 1 + tan x
E. y = 2x
## F. distance y as a function of time x, velocity decreasing
Each sketch below could represent exactly two of the above relations on a certain interval.
## (i) (ii) (iii)
y y y
x x x
Complete the table below, by writing the letter for the two relations that each sketch could
represent.
(i)
(ii)
(iii)
## IB Questionbank Maths SL 179
297. A student measured the diameters of 80 snail shells. His results are shown in the following
cumulative frequency graph. The lower quartile (LQ) is 14 mm and is marked clearly on the
graph.
90
80
70
Cumulative frequency
60
50
40
30
20
10
0
0 5 10 15 20 25 30 35 40 45
LQ = 14
Diameter (mm)
(a) On the graph, mark clearly in the same way and write down the value of
## (b) Write down the interquartile range.
Working:
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 180
298. The graph of the function f (x) = 3x – 4 intersects the x-axis at A and the y-axis at B.
## (a) Find the coordinates of
(i) A;
(ii) B.
(b) Let O denote the origin. Find the area of triangle OAB.
Working:
## (a) (i) ...........................................................
(ii) ...........................................................
(b) ..................................................................
(Total 6 marks)
299. The equation kx2 + 3x + 1 = 0 has exactly one solution. Find the value of k.
Working:
..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 181
300. A painter has 12 tins of paint. Seven tins are red and five tins are yellow. Two tins are chosen at
random. Calculate the probability that both tins are the same colour.
Working:
..................................................................
(Total 6 marks)
## 301. Complete the following expansion.
(2 + ax)4 = 16 + 32ax + …
Working:
..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 182
302. Arturo goes swimming every week. He swims 200 metres in the first week. Each week he
swims 30 metres more than the previous week. He continues for one year (52 weeks).
## (b) How far does he swim altogether?
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 183
4 3
303. A vector equation for the line L is r = + t .
4 1
Which of the following are also vector equations for the same line L?
4 2
A. r = + t .
4 1
4 6
B. r = + t .
4 2
0 1
C. r = + t .
1 3
7 3
D. r = + t .
5 1
Working:
……..................................................................
……..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 184
q
304. (a) The diagram shows part of the graph of the function f (x) = . The curve passes
x– p
through the point A (3, 10). The line (CD) is an asymptote.
y
C
15
10 A
–15 –10 –5 0 5 10 15 x
-5
-10
-15
D
(i) p;
(ii) q.
## IB Questionbank Maths SL 185
(b) The graph of f (x) is transformed as shown in the following diagram. The point A is
transformed to A (3, –10).
y
C
15
10
–15 –10 –5 0 5 10 15 x
–5
–10
A
–15
D
## IB Questionbank Maths SL 186
Give a full geometric description of the transformation.
Working:
## (a) (i) ...........................................................
(ii) ...........................................................
(b) ..................................................................
..................................................................
(Total 6 marks)
60 – 30
305. (a) Find the scalar product of the vectors and .
25 40
(b) Two markers are at the points P (60, 25) and Q (–30, 40). A surveyor stands at O (0, 0)
and looks at marker P. Find the angle she turns through to look at marker Q.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 187
306. The mass m kg of a radio-active substance at time t hours is given by
m = 4e–0.2t.
## (a) Write down the initial mass.
(b) The mass is reduced to 1.5 kg. How long does this take?
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
dy
307. It is given that = x3+2x – 1 and that y = 13 when x = 2.
dx
Find y in terms of x.
Working:
..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 188
308. The function f is given by f (x) = x2 – 6x + 13, for x 3.
## (c) State the domain of f –1.
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 189
309. (a) Factorize the expression 3 sin2 x – 11 sin x + 6.
## (b) Consider the equation 3 sin2 x – 11 sin x + 6 = 0.
(i) Find the two values of sin x which satisfy this equation,
## (ii) Solve the equation, for 0° x 180°.
Working:
(a) ..................................................................
(b) (i) ...........................................................
(ii) ...........................................................
(Total 6 marks)
## IB Questionbank Maths SL 190
310. (a) Find (1 + 3 sin (x + 2))dx.
(b) The diagram shows part of the graph of the function f (x) = 1 + 3 sin (x + 2).
a
The area of the shaded region is given by 0
f ( x)dx .
–4 –2 0 2 4 x
–2
## Find the value of a.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 191
311. The diagram shows the graph of y = f (x).
y
0 x
## IB Questionbank Maths SL 192
On the grid below sketch the graph of y = f (x).
y
0 x
(Total 6 marks)
312. From January to September, the mean number of car accidents per month was 630. From
October to December, the mean was 810 accidents per month.
What was the mean number of car accidents per month for the whole year?
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 193
313. In an arithmetic sequence, the first term is –2, the fourth term is 16, and the nth term is 11 998.
## (b) Find the value of n.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 194
x
314. Let f (x) = 2x, and g (x) = , (x 2).
x–2
Find
(a) (g f ) (3);
## (b) g–1 (5).
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 195
315. The following diagram shows a circle of centre O, and radius 15 cm. The arc ACB subtends an
angle of 2 radians at the centre O.
C
A B
15
Diagram not to scale
cm
OA = 15 cm
Find
## (b) the area of the shaded region.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 196
x 1 – 2
316. A vector equation of a line is t , t .
y 2 3
## Find the equation of this line in the form ax + by = c, where a, b, and c .
Working:
......................................................................
(Total 6 marks)
317. Two boats A and B start moving from the same point P. Boat A moves in a straight line at
20 km h–1 and boat B moves in a straight line at 32 km h–1. The angle between their paths is
70°.
## Find the distance between the boats after 2.5 hours.
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 197
9
1
318. Consider the expansion of 3x 2 – .
x
## (b) Find the constant term in this expansion.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 198
319. Let f (x) = sin 2x and g (x) = sin (0.5x).
## (b) Consider the equation f (x) = g (x).
Find the number of solutions to this equation, for 0 x .
2
Working:
## (a) (i) ..........................................................
(ii) ..........................................................
(b) .................................................................
(Total 6 marks)
## IB Questionbank Maths SL 199
320. Solve the equation log27 x = 1 – log27 (x – 0.4).
Working:
......................................................................
(Total 6 marks)
1
321. The derivative of the function f is given by f (x) = – 0.5 sin x, for x –1.
1 x
The graph of f passes through the point (0, 2). Find an expression for f (x).
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 200
322. A box contains 22 red apples and 3 green apples. Three apples are selected at random, one after
the other, without replacement.
(a) The first two apples are green. What is the probability that the third apple is red?
(b) What is the probability that exactly two of the three apples are red?
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 201
323. The diagram shows part of the graph of y = a (x – h)2 + k. The graph has its vertex at P, and
passes through the point A with coordinates (1, 0).
y
P
2
A
–1 0 1 x
(i) h;
(ii) k.
Working:
## (a) (i) ..........................................................
(ii) ..........................................................
(b) .................................................................
(Total 6 marks)
## IB Questionbank Maths SL 202
324. Figure 1 shows the graphs of the functions f1, f2, f3, f4.
Figure 2 includes the graphs of the derivatives of the functions shown in Figure 1, eg the
derivative of f1 is shown in diagram (d).
Figure 1 Figure 2
y y
f1 (a)
O x
x
O
y y
f2 (b)
x O x
O
y y
f3 (c)
x O x
O
y y
f4 (d)
O
x O x
y
(e)
O x
## IB Questionbank Maths SL 203
Complete the table below by matching each function with its derivative.
f1 (d)
f2
f3
f4
Working:
(Total 6 marks)
## B: –0.5 cos (0.5x) 0.5.
π π
C: – arctan x .
2 2
(a) Determine which statements are true for all real numbers x. Write your answers (yes or
no) in the table below.
Statement (a) Is the statement true for all (b) If not true, example
real numbers x? (Yes/No)
A
B
C
## IB Questionbank Maths SL 204
(b) If a statement is not true for all x, complete the last column by giving an example of one
value of x for which the statement is false.
Working:
(Total 6 marks)
## 326. Let f (x) = x 3 . Find
(a) f (x);
(b) f ( x)dx.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 205
327. In triangle ABC, AC = 5, BC = 7, Â = 48°, as shown in the diagram.
## 5 7 diagram not to scale
48°
A B
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 206
328. Consider the function f (x) = 2x2 – 8x + 5.
## (b) Find the minimum value of f (x).
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## 329. Find the coefficient of x3 in the expansion of (2 – x)5.
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 207
330. Solve the equation ex = 5 – 2x, giving your answer correct to four significant figures.
Working:
......................................................................
(Total 6 marks)
1
331. Given that sin x = , where x is an acute angle, find the exact value of
3
(a) cos x;
## (b) cos 2x.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 208
3 4
332. For events A and B, the probabilities are P (A) = , P (B) = .
11 11
## Calculate the value of P (A B) if
6
(a) P (A B) = ;
11
## (b) events A and B are independent.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
333. The graph of y = x3 – 10x2 +12x + 23 has a maximum point between x = –1 and x = 3. Find the
coordinates of this maximum point.
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 209
334. Three positive integers a, b, and c, where a < b < c, are such that their median is 11, their mean
is 9 and their range is 10. Find the value of a.
Working:
......................................................................
(Total 6 marks)
6–x
335. Consider the functions f : x 4(x – 1) and g : x .
2
## (b) Solve the equation ( f ° g–1) (x) = 4.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 210
336. Calculate the acute angle between the lines with equations
4 4 2 1
r = + s and r = + t
– 1 3 4 – 1
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 211
337. The diagram shows part of the curve y = sin x. The shaded region is bounded by the curve and
the lines y = 0 and x = .
4
y
3 x
4
3π 2 3π 2
Given that sin = and cos =– , calculate the exact area of the shaded
4 2 4 2
region.
Working:
......................................................................
(Total 6 marks)
## IB Questionbank Maths SL 212
338. \$1000 is invested at 15% per annum interest, compounded monthly. Calculate the minimum
number of months required for the value of the investment to exceed \$3000.
Working:
......................................................................
(Total 6 marks)
## 339. Consider the trigonometric equation 2 sin2 x = 1 + cos x.
(a) Write this equation in the form f (x) = 0, where f (x) = a cos2 x + b cos x + c,
and a, b, c .
## (c) Solve f (x) = 0 for 0° x 360°.
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 6 marks)
## IB Questionbank Maths SL 213
340. The sketch shows part of the graph of y = f (x) which passes through the points A(–1, 3), B(0,
2), C(l, 0), D(2, 1) and E(3, 5).
6
E
5
4
A
3
B
2
D
1
C
–4 –3 –2 –1 0 1 2 3 4 5
–1
–2
## (a) Calculate g (0), g (1), g (2) and g (3).
(b) On the same axes, sketch the graph of the function g (x).
Working:
(a) ..................................................................
..................................................................
(Total 6 marks)
(a) the median;
## (b) the mean.
Number (x) 1 2 3 4 5 6
Frequency (f ) 5 9 16 18 20 7
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
342. The diagram shows part of the graph with equation y = x2 + px + q. The graph cuts the x-axis at
–2 and 3.
x
–3 –2 –1 0 1 2 3 4
–2
–4
–6
## IB Questionbank Maths SL 215
Find the value of
(a) p;
(b) q.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
343. Each year for the past five years the population of a certain country has increased at a steady
rate of 2.7% per annum. The present population is 15.2 million.
## (b) What was the population five years ago?
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 216
344. The following diagram shows a triangle with sides 5 cm, 7 cm, 8 cm.
5 7
Diagram not to scale
Find
## (b) the area of the triangle.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 217
345. The point P ( 1 , 0 ) lies on the graph of the curve of y = sin (2x –1).
2
## Find the gradient of the tangent to the curve at P.
Working:
.......................................................................
(Total 4 marks)
## (3x + 2y)4 = 81x4 + 216x3 y +...
Working:
.......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 218
347. A bag contains 10 red balls, 10 green balls and 6 white balls. Two balls are drawn at random
from the bag without replacement. What is the probability that they are of different colours?
Working:
.......................................................................
(Total 4 marks)
348. Find
## (a) sin (3x 7)dx;
e
–4 x
(b) dx .
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 219
349. Find the angle between the following vectors a and b, giving your answer to the nearest degree.
a = –4i – 2j
b = i – 7j
Working:
.......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 220
350. (a) On the following diagram, sketch the graphs of y = ex and y = cos x
for –2 x 1.
y
2
x
–2 –1 0 1
–1
–2
## Find this solution.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 221
351. The function f is defined by
3
f :xa 3 – 2x , x .
2
Evaluate f –1(5).
Working:
.......................................................................
(Total 4 marks)
352. (a) Write the expression 3 sin2 x + 4 cos x in the form a cos2 x + b cos x + c.
## 3 sin2 x + 4 cos x – 4 = 0, 0 x 90.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 222
353. The following diagram shows the graph of y = f (x). It has minimum and maximum points at
1
(0, 0) and ( 1, ).
2
3.5
2.5
1.5
0.5
–2 –1 0 1 2 3 x
–0.5
–1
–1.5
–2
–2.5
3
(a) On the same diagram, draw the graph of y f ( x – 1) .
2
(b) What are the coordinates of the minimum and maximum points of
3
y f ( x – 1) ?
2
Working:
(b) ................................................................
(Total 4 marks)
## IB Questionbank Maths SL 223
354. In the following diagram, O is the centre of the circle and (AT) is the tangent to the circle at T.
O A
## Diagram not to scale
If OA = 12 cm, and the circle has a radius of 6 cm, find the area of the shaded region.
Working:
.......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 224
355. The first three terms of an arithmetic sequence are 7, 9.5, 12.
## (a) What is the 41st term of the sequence?
(b) What is the sum of the first 101 terms of the sequence?
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 225
356. The diagram below shows a line passing through the points (1, 3) and (6, 5).
y
(6,5)
(1,3)
0 x
Find a vector equation for the line, giving your answer in the form
x a c
t , where t is any real number.
y b d
Working:
.......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 226
357. The diagram shows parts of the graphs of y = x2 and y = 5 – 3(x – 4)2.
y = x2
8
6 2
y = 5 – 3(x–4)
x
–2 0 2 4 6
## IB Questionbank Maths SL 227
The graph of y = x2 may be transformed into the graph of y = 5 – 3(x – 4)2 by these
transformations.
## A reflection in the line y = 0 followed by
a vertical stretch with scale factor k followed by
a horizontal translation of p units followed by
a vertical translation of q units.
## Write down the value of
(a) k;
(b) p;
(c) q.
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 228
358. The diagram below shows a sector AOB of a circle of radius 15 cm and centre O. The angle at
the centre of the circle is 2 radians.
A B
## (b) Calculate the area of the shaded region.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 229
1
359. Solve the equation log9 81 + log9 + log9 3 = log9 x.
9
Working:
.......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 230
360. The following Venn diagram shows the universal set U and the sets A and B.
U B
A
(a) Shade the area in the diagram which represents the set B A'.
## (b) Find n(B A′).
(c) An element is selected at random from U. What is the probability that this element is
in B A′ ?
Working:
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 231
361. Consider the function f (x) = k sin x + 3x, where k is a constant.
## (a) Find f (x).
(b) When x = , the gradient of the curve of f (x) is 8. Find the value of k.
3
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 232
x
362. The diagram below shows the graph of y = x sin , for 0 x < m, and 0 y < n, where x is in
3
radians and m and n are integers.
y
n
n–1
0 m–1 m x
## Find the value of
(a) m;
(b) n.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 233
363. Given that f (x) = 2e3x, find the inverse function f –1(x).
Working:
.......................................................................
(Total 4 marks)
4 4 4
364. Consider the binomial expansion (1 x) 4 1 x x 2 x 3 x 4 .
1 2 3
4 4 4
(a) By substituting x = 1 into both sides, or otherwise, evaluate .
1 2 3
## IB Questionbank Maths SL 234
9 9 9 9 9 9 9 9
(b) Evaluate .
1 2 3 4 5 6 7 8
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
2x x 1
365. The vectors and are perpendicular for two values of x.
x – 3 5
(a) Write down the quadratic equation which the two values of x must satisfy.
## (b) Find the two values of x.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 235
366. The diagrams below show two triangles both satisfying the conditions
## AB = 20 cm, AC = 17 cm, AB̂C = 50°.
Diagrams not
to scale
Triangle 1 Triangle 2
A A
B C B C
## (b) Calculate the area of Triangle 1.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 236
367. The events B and C are dependent, where C is the event “a student takes Chemistry”, and B is
the event “a student takes Biology”. It is known that
## (a) Complete the following tree diagram.
Chemistry Biology
0.4 C
B
C
B
## (b) Calculate the probability that a student takes Biology.
(c) Given that a student takes Biology, what is the probability that the student takes
Chemistry?
Working:
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 237
368. The depth, y metres, of sea water in a bay t hours after midnight may be represented by the
function
2
y a b cos t , where a, b and k are constants.
k
The water is at a maximum depth of 14.3 m at midnight and noon, and is at a minimum depth of
10.3 m at 06:00 and at 18:00.
## Write down the value of
(a) a;
(b) b;
(c) k.
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 238
369. (a) Express f (x) = x2 – 6x + 14 in the form f (x) = (x – h)2 + k, where h and k are to be
determined.
(b) Hence, or otherwise, write down the coordinates of the vertex of the parabola with
equation y – x2 – 6x + 14.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## 370. Let f (x) = 1 – x2. Given that f (3) = 0, find f (x).
Working:
....................................................................
(Total 4 marks)
## IB Questionbank Maths SL 239
371. Town A is 48 km from town B and 32 km from town C as shown in the diagram.
32km
A
48km B
Given that town B is 56 km from town C, find the size of angle CÂB to the nearest degree.
Working:
....................................................................
(Total 4 marks)
372. In a survey of 200 people, 90 of whom were female, it was found that 60 people were
unemployed, including 20 males.
Unemployed
Employed
Totals 200
## IB Questionbank Maths SL 240
(b) If a person is selected at random from this group of 200, find the probability that this
person is
Working:
## (b) (i) ..........................................................
(ii) ..........................................................
(Total 4 marks)
1 6
373. Find the size of the angle between the two vectors and . Give your answer to the
2 – 8
nearest degree.
Working:
....................................................................
(Total 4 marks)
## IB Questionbank Maths SL 241
374. A group of ten leopards is introduced into a game park. After t years the number of leopards, N,
is modelled by N = 10 e0.4t.
## (a) How many leopards are there after 2 years?
(b) How long will it take for the number of leopards to reach 100? Give your answers to an
appropriate degree of accuracy.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
375. Given the function f (x) = x2 – 3bx + (c + 2), determine the values of b and c such that f (1) = 0
and f (3) = 0.
Working:
....................................................................
(Total 4 marks)
## IB Questionbank Maths SL 242
2
376. A line passes through the point (4,–1) and its direction is perpendicular to the vector . Find
3
the equation of the line in the form ax + by = p, where a, b and p are integers to be determined.
Working:
....................................................................
(Total 4 marks)
## (b) What is the domain of f –1?
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
2
P
378. Let log10P = x , log10Q = y and log10R = z. Express log10
3
in terms of x , y and z.
QR
## IB Questionbank Maths SL 243
Working:
....................................................................
(Total 4 marks)
379. The diagram shows the graph of y = f (x), with the x-axis as an asymptote.
B(5, 4)
A(–5, –4)
(a) On the same axes, draw the graph of y =f (x + 2) – 3, indicating the coordinates of the
images of the points A and B.
## IB Questionbank Maths SL 244
(b) Write down the equation of the asymptote to the graph of y = f (x + 2) – 3.
Working:
(b)
...................................................................
.
(Total 4 marks)
## 380. (a) Express 2 cos2 x + sin x in terms of sin x only.
(b) Solve the equation 2 cos2 x + sin x = 2 for x in the interval 0 x , giving your answers
exactly.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 245
381. Each day a runner trains for a 10 km race. On the first day she runs 1000 m, and then increases
the distance by 250 m on each subsequent day.
## (a) On which day does she run a distance of 10 km in training?
(b) What is the total distance she will have run in training by the end of that day? Give your
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
9
2
382. Determine the constant term in the expansion of x – 2 .
x
Working:
....................................................................
(Total 4 marks)
## IB Questionbank Maths SL 246
383. (a) Sketch, on the given axes, the graphs of y = x2 and y – sin x for –1 x 2.
## IB Questionbank Maths SL 247
(b) Find the positive solution of the equation
x2 = sin x,
Working:
(b)
...................................................................
.
(Total 4 marks)
384. In an arithmetic sequence, the first term is 5 and the fourth term is 40. Find the second term.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 248
385. Two functions f, g are defined as follows:
f : x 3x + 5
g : x 2(1 – x)
Find
(a) f –1(2);
(b) (g f )(–4).
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 249
386. In a survey, 100 students were asked “do you prefer to watch television or play sport?” Of the
46 boys in the survey, 33 said they would choose sport, while 29 girls made this choice.
Television
Sport 33 29
Total 46 100
## (a) a student selected at random prefers to watch television;
(b) a student prefers to watch television, given that the student is a boy.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 250
387. The vectors u, v are given by u = 3i + 5j, v = i – 2j.
## Find scalars a, b such that a(u + v) = 8i + (b – 2)j.
Working:
......................................................................
(Total 4 marks)
(a) log2 5;
## (b) loga 20.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 251
389. Solve the equation 3 cos x = 5 sin x, for x in the interval 0° x 360°, giving your answers to
the nearest degree.
Working:
......................................................................
(Total 4 marks)
390. Find a vector equation of the line passing through (–1, 4) and (3, –1). Give your answer in the
form r = p + td, where t .
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 252
391. Find the coordinates of the point on the graph of y = x2 – x at which the tangent is parallel to the
line y = 5x.
Working:
......................................................................
(Total 4 marks)
## 392. If f (x) = cos x, and f = – 2, find f (x).
2
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 253
393. Find the sum of the infinite geometric series
2 4 8 16 ...
3 9 27 81
Working:
......................................................................
(Total 4 marks)
## 394. Find the coefficient of a5b7 in the expansion of (a + b)12.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 254
395. If A is an obtuse angle in a triangle and sin A = 5 , calculate the exact value of sin 2A.
13
Working:
......................................................................
(Total 4 marks)
396. The quadratic equation 4x2 + 4kx + 9 = 0, k > 0 has exactly one solution for x.
Find the value of k.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 255
397. The diagram shows three graphs.
B
y
A
Write down
## (b) the coordinates of the point where C cuts the x-axis.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 256
398. Let f (x) = x3.
f (5 h) f (5)
(a) Evaluate for h = 0.1.
h
f (5 h) f (5)
(b) What number does approach as h approaches zero?
h
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
399. Two ordinary, 6-sided dice are rolled and the total score is noted.
(a) Complete the tree diagram by entering probabilities and listing outcomes.
Outcomes
6 ...............
.......
6
.......
....... not 6 ...............
6 ...............
....... .......
not 6
## IB Questionbank Maths SL 257
(b) Find the probability of getting one or more sixes.
Working:
(b) ...............................................................
(Total 4 marks)
## 400. The table shows the scores of competitors in a competition.
Score 10 20 30 40 50
Number of competitors
1 2 5 k 3
with this score
## The mean score is 34. Find the value of k.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 258
401. A curve with equation y =f (x) passes through the point (1, 1). Its gradient function is
f (x) = –2x + 3.
## Find the equation of the curve.
Working:
......................................................................
(Total 4 marks)
3
402. Given that sin θ = 1 , cos θ = – and 0° ≤ θ ≤ 360°,
2 2
## (b) write down the exact value of tan θ.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 259
403. The line L passes through the origin and is parallel to the vector 2i + 3j.
Write down a vector equation for L.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 260
404. The following Venn diagram shows a sample space U and events A and B.
U A B
(b) Find
Working:
## (b) (i) ...........................................................
(ii) ...........................................................
(c) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 261
405. The diagram shows a vertical pole PQ, which is supported by two wires fixed to the horizontal
ground at A and B.
36 B
30
Q 70
A
BQ = 40 m
PB̂Q = 36°
BÂQ = 70°
AB̂Q = 30°
Find
## (b) the distance between A and B.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 262
406. Given that f (x) = (2x + 5)3 find
(a) f (x);
(b) f ( x)dx.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 263
407. The diagrams show how the graph of y = x2 is transformed to the graph of y = f (x) in three
steps.
## For each diagram give the equation of the curve.
y y
(a)
1
0 x 0 x
y=x2 1
(b) (c) 7
y
0 1 x 0 1 x
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 264
408. The diagram shows a circle of radius 5 cm.
## Find the perimeter of the shaded region.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 265
409. f (x) = 4 sin 3x .
2
## For what values of k will the equation f (x) = k have no solutions?
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 266
410. \$1000 is invested at the beginning of each year for 10 years.
The rate of interest is fixed at 7.5% per annum. Interest is compounded annually.
(a) how much the first \$1000 is worth at the end of the ten years;
(b) the total value of the investments at the end of the ten years.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 267
411. The triangle ABC is defined by the following information
2 3 0
OA = , AB = , AB BC = 0, AC is parallel to .
3 4 1
## (a) On the grid below, draw an accurate diagram of triangle ABC.
O x
–2 –1 1 2 3 4 5 6
–1
–2
–3
–4
## IB Questionbank Maths SL 268
(b) Write down the vector OC .
Working:
(b) ..................................................................
(Total 4 marks)
y
## IB Questionbank Maths SL 269
Complete the table below to show whether each expression is positive, negative or zero.
a
c
b2 – 4ac
b
Working:
(Total 4 marks)
## IB Questionbank Maths SL 270
413. The diagram shows the graph of the function y = 1 + 1 , 0 < x 4. Find the exact value of the
x
4
1
y = 1+ –
x
3
2
1 13
1
0 1 2 3 4
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 271
414. (a) Factorize x2 – 3x – 10.
## (b) Solve the equation x2 – 3x – 10 = 0.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## 415. The diagram represents the graph of the function
f : x (x – p)(x – q).
y
– 12 2 x
## IB Questionbank Maths SL 272
(b) The function has a minimum value at the point C. Find the x-coordinate of C.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
416. Find the equation of the normal to the curve with equation
y = x3 + 1
## at the point (1, 2).
Working:
.........................................................................
(Total 4 marks)
## IB Questionbank Maths SL 273
417. Find the sum of the arithmetic series
17 + 27 + 37 +...+ 417.
Working:
.........................................................................
(Total 4 marks)
418. A triangle has sides of length 4, 5, 7 units. Find, to the nearest tenth of a degree, the size of the
largest angle.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 274
419. The graph represents the function
f : x p cos x, p .
y
3
x
–3
Find
## (b) the area of the shaded region.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 275
2p 3
420. If A = and det A = 14, find the possible values of p.
4 p p
Working:
......................................................................
(Total 4 marks)
421. O is the centre of the circle which has a radius of 5.4 cm.
A B
## IB Questionbank Maths SL 276
The area of the shaded sector OAB is 21.6 cm2. Find the length of the minor arc AB.
Working:
......................................................................
(Total 4 marks)
D C
A O B
(a) CD
(b) OA
## IB Questionbank Maths SL 277
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
2x
423. Solve the equation 9x–1 = 1 .
3
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 278
424. Find the coefficient of x5 in the expansion of (3x – 2)8.
Working:
......................................................................
(Total 4 marks)
## 425. Differentiate with respect to x
(a) 3 4x
(b) esin x
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 279
426. The vectors i , j are unit vectors along the x-axis and y-axis respectively.
The vectors u = – i + 2 j and v = 3 i + 5 j are given.
(a) Find u + 2 v in terms of i and j .
A vector w has the same direction as u + 2 v , and has a magnitude of 26.
(b) Find w in terms of i and j .
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 280
427. Two functions f and g are defined as follows:
## f (x) = cos x, 0 x 2;
g (x) = 2x + 1, x .
## Solve the equation (g f)(x) = 0.
Working:
......................................................................
(Total 4 marks)
428. An arithmetic series has five terms. The first term is 2 and the last term is 32. Find the sum of
the series.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 281
429. The diagram shows the parabola y = (7 – x)(l + x). The points A and C are the x-intercepts and
the point B is the maximum point.
A 0 C x
## Find the coordinates of A, B and C.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 282
430. For the events A and B, p(A) = 0.6, p(B) = 0.8 and p(A B) = 1.
Find
(a) p(AB);
(b) p( A B).
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
431. At a conference of 100 mathematicians there are 72 men and 28 women. The men have a mean
height of 1.79 m and the women have a mean height of 1.62 m. Find the mean height of the 100
mathematicians.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 283
432. The diagram shows part of the graph of y = 12x2(1 – x).
0 x
(a) Write down an integral which represents the area of the shaded region.
## (b) Find the area of the shaded region.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 284
433. Differentiate with respect to x:
## (b) 1n(3x – 1).
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 285
25
434. The mean of the population x1, x2, ........ , x25 is m. Given that x
i 1
i = 300 and
25
( x – m)
i 1
i
2
= 625, find
## (b) the standard deviation of the population.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## 435. Solve the equation 3 sin2 x = cos2 x, for 0° x 180°.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 286
5 2 11 2
436. A and B are 2 × 2 matrices, where A = and BA = . Find B.
2 0 44 8
Working:
......................................................................
(Total 4 marks)
437. The quadrilateral OABC has vertices with coordinates O(0, 0), A(5, 1), B(10, 5) and C(2, 7).
## (a) Find the vectors OB and AC .
(b) Find the angle between the diagonals of the quadrilateral OABC.
Working:
(a) ..................................................................
(b) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 287
438. Find the coefficient of a3b4 in the expansion of (5a + b)7.
Working:
......................................................................
(Total 4 marks)
439. Three of the following diagrams I, II, III, IV represent the graphs of
(a) y = 3 + cos 2x
(b) y = 3 cos (x + 2)
(c) y = 2 cos x + 3.
## IB Questionbank Maths SL 288
Identify which diagram represents which graph.
I y II y
4
2
1
2
x
–
– –
–
–
–1
x
–
– –
–
– –2
y y
III IV
3
5
2
4
1
3
x
2
– – – – –
1
x
–
– –
–
–
–3
Working:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 4 marks)
## IB Questionbank Maths SL 289
440. Solve the equation 43x–1 = 1.5625 × 10–2.
Working:
......................................................................
(Total 4 marks)
441. The diagram shows part of the graph of y = 1 . The area of the shaded region is 2 units.
x
y
0 1 a x
## IB Questionbank Maths SL 290
Find the exact value of a.
Working:
......................................................................
(Total 4 marks)
442. The function f is given by f (x) = 1n ( x 2) . Find the domain of the function.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 291
443. A population of bacteria is growing at the rate of 2.3% per minute. How long will it take for the
size of the population to double? Give your answer to the nearest minute.
Working:
......................................................................
(Total 4 marks)
## 444. Let f (x) = x , and g (x) = 2x. Solve the equation
(f –1 g)(x) = 0.25.
Working:
......................................................................
(Total 4 marks)
## IB Questionbank Maths SL 292
445. The diagrams show a circular sector of radius 10 cm and angle θ radians which is formed into a
cone of slant height 10 cm. The vertical height h of the cone is equal to the radius r of its base.
10cm
10cm
h
Working: | 42,008 | 131,783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-16 | latest | en | 0.816815 |
https://www.gamedev.net/forums/topic/355929-hlsl-point-lights/ | 1,547,747,832,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659056.44/warc/CC-MAIN-20190117163938-20190117185938-00300.warc.gz | 804,418,680 | 26,105 | # HLSL Point Lights
This topic is 4823 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hello, I was wondering how do you calculate a point light in HLSL? Thanks in Advance
##### Share on other sites
2D or 3D?
3D -
In the vertex shader's output, you include the untransformed position of the vertex too. It will linearly interpolate it between vertices, which is just fine as the pixels are also linearly interpolated in the same place. So, in the vertex shader, output the transformed position (untransformed position multiplied by the view matrix), the texcoord and the untransformed position.
In the pixel shader, sample a texture using the input texcoord, and multiply it based on the distance between the untransformed position and the light.
Example hlsl shader, also includes bumpmapping
2D - If your using 2D, which chances are you aren't, post here and I'll tell you how.
Hope this helps :)
##### Share on other sites
You can do this all in the vertex shader, or partially in both the vertex and pixel shader.
For per pixel light, typically you pass vertex position as a UV from the vertex shader to the pixel shader, as the previous poster said. You'll also need to pass the normal. In the pixel shader you can compute the direction and distance:
Regardless of pixel or vertex shader, you'd do this:
lightvec = (vertexpos - lightpos);
lightdist = length(lightvec);
lightvec = normalize(lightvec);
You can then apply attenutation based on the distance, do a directional light calculation based on the lightvector and the normal. The pointlight color value is (lightvec dot normal) * attenuation * lightcolor * materialcolor.
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× | 456 | 1,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-04 | latest | en | 0.867665 |
https://brilliant.org/problems/top-pipe-falls/ | 1,508,521,495,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824293.62/warc/CC-MAIN-20171020173404-20171020193404-00853.warc.gz | 658,497,393 | 17,995 | # Top pipe falls
There is a system consisting of 3 pipes(hollow cylinders as shown in the figure). Now, the system is little disturbed. Let the initial downward acceleration of the top pipe given the conditions be
$$a_1$$ - When there is friction between the 2 cylinders and the ground but there is no friction between any of the pipes.
$$a_2$$ - When there is no friction between the 2 cylinders and the ground but there is friction between the pipes.
Find $$\displaystyle \frac{a_1}{a_2}$$.
If your answer is of the form $$\frac{p}{q}$$(for co-prime $$p,q$$), give your answer as $$p+q$$
× | 156 | 597 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2017-43 | longest | en | 0.929708 |
https://codereview.stackexchange.com/questions/210103/map-coloring-with-mrv-and-degree-heuristics-in-python | 1,563,604,296,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526446.61/warc/CC-MAIN-20190720045157-20190720071157-00261.warc.gz | 351,106,067 | 37,152 | # Map coloring with MRV and Degree heuristics in Python
I am relatively new to Python. I wrote this solution to the well known map coloring problem and also implemented the MRV and Degree heuristics. Here, I am considering the map of Australia - ['WA', 'NT', 'SA', 'Q', 'NSW', 'V', 'T'] and 3 given colors - ['R','G', 'B']
# choosing first node with degree heruistics
# applying MRV with backtracking
from enum import Enum
import pdb
class Color(Enum):
RED = 1
GREEN = 2
BLUE = 3
class Graph:
self.totalNodes = totalNodes
self.color = color
self.nodeSequence = [""]*totalNodes
def isSafe(self, node, c):
return False
return True
def graphColorUtil(self, node, colorLimit):
if node == '':
# check and color any uncolored node
for key, value in self.color.items():
if value == 0:
self.graphColorUtil(key, colorLimit)
return True
# pdb.set_trace()
for c in range(1, colorLimit+1):
if(self.isSafe(node, c) == True):
self.color[node] = c
nextNode = self.getNodeWithMRV(node, colorLimit)
if(self.graphColorUtil(nextNode, colorLimit) == True):
return True
else:
self.color[node] = 0
return False
def graphColoring(self, colorLimit):
# pdb.set_trace()
startNode = self.pickNode('')
if(self.graphColorUtil(startNode, colorLimit) == True):
return True
else:
print("Solution does not exists")
return False
# pick node using MRV
def pickNode(self, initialNode):
maxCount = 0
selectedNode = ''
# the very first node
if (initialNode == ''):
if (len(neighbourList) > maxCount and self.color[node] == 0):
maxCount = len(neighbourList)
selectedNode = node
# the other nodes
else:
if (self.color[childNode] == 0 and len(self.adjacencyList[childNode]) > maxCount):
selectedNode = childNode
return selectedNode
def getNodeWithMRV(self, parentNode, colorLimit):
selectedNode = ''
countColor = 0
for c in range(1, colorLimit+1):
if(self.isSafe(childNode, c) == True):
countColor += 1
if (countColor < minCount):
selectedNode = childNode
return selectedNode
# driver code
def main():
'WA': ['NT', 'SA'],
'NT': ['WA', 'SA', 'Q'],
'SA': ['WA', 'NT', 'Q', 'NSW', 'V'],
'Q': ['NT', 'SA', 'NSW'],
'NSW': ['SA', 'Q', 'V'],
'V': ['SA', 'T', 'NSW'],
'T': ['V']
};
color = {
'WA': 0,
'NT': 0,
'SA': 0,
'Q': 0,
'NSW': 0,
'V': 0,
'T': 0
};
colorLimit = 3
g.graphColoring(colorLimit)
for node, color in g.color.items():
print(node, Color(color).name)
main()
What could be the possible ways to refactor this code? I am also interested for feedback on Python code style in general.
• You seem to lose something while copying. How minCount ever changes? – vnp Dec 21 '18 at 18:24
• You mean maxCount in the pickNode() function? – Abrar Dec 22 '18 at 3:19
• No. I mean minCount in getNodeWithMRV – vnp Dec 22 '18 at 3:31
• sorry, minCount is unnecessary there. Updated code. – Abrar Dec 22 '18 at 3:40
• if (countColor < minCount): is still there. – vnp Dec 22 '18 at 4:14
Although not familiar with MRV and degree heuristics, i can make some remarks about the Python code style:
Loops can be made more Pythonic
for i in range(len(self.adjacencyList[initialNode])):
should be written as:
for childNode in self.adjacencyList[initialNode]:
Conditionals
if(self.isSafe(childNode, c) == True):
should be
if self.isSafe(childNode, c):
method isSafe
def isSafe(self, node, c):
return False
return True
could be:
def isSafe(self, node, c):
return False
return True
or even, more Pythonic, but a bit cryptic:
def isSafe(self, node, c):
# return True if all node neighbours colors differ from c
Data structure
The repetition of the keys in the adjacencyList and color suggest a data structure like the following, although this requires a lot changes in the existing code:
nodes = {
'WA' : {'color' : 0, 'neighbours' : ['NT', 'SA']},
'NT' : {'color' : 0, 'neighbours' : ['WA', 'SA', 'Q']},
'SA' : {'color' : 0, 'neighbours' : ['WA', 'NT', 'Q', 'NSW', 'V']},
'Q' : {'color' : 0, 'neighbours' : ['NT', 'SA', 'NSW']},
'NSW': {'color' : 0, 'neighbours' : ['SA', 'Q', 'V']},
'V' : {'color' : 0, 'neighbours' : ['SA', 'T', 'NSW']},
'T' : {'color' : 0, 'neighbours' : ['V']},
}
Others:
• self.nodeSequence is not used
• self.totalNodes is not used
• minCount = 0 was edited out in getNodeWithMRV but should be there, or if (countColor < minCount) should be if (countColor < 0)
• pickNode is called only once with a constant argument '', and can therefor be made simpler
• getNodeWithMRV will always return '' because countColor will never be smaller than 0.
• the ; at the end of adjacencyList = ... and color = ... an origin in another language :-) | 1,360 | 4,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2019-30 | longest | en | 0.555853 |
http://dtsheet.com/doc/93672/ad-ad2s100ap | 1,498,283,452,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320226.61/warc/CC-MAIN-20170624050312-20170624070312-00639.warc.gz | 130,715,379 | 19,135 | a
AC Vector Processor
FEATURES
Complete Vector Coordinate Transformation on Silicon
Mixed Signal Data Acquisition
Three-Phase 1208 and Orthogonal 908 Signal
Transformation
Three-Phase Balance Diagnostic–Homopolar Output
FUNCTIONAL BLOCK DIAGRAM
Cosθ Sinθ
Cosθ
APPLICATIONS
AC Induction and DC Permanent
Magnet Motor Control
HVAC, Pump, Fan Control
Material Handling
Robotics
Spindle Drives
Gyroscopes
Dryers
Washing Machines
Electric Cars
Actuator
Three-Phase Power Measurement
Digital-to-Resolver & Synchro Conversion
GENERAL DESCRIPTION
The AD2S100 performs the vector rotation of three-phase 120
degree or two-phase 90 degree sine and cosine signals by transferring these inputs into a new reference frame which is controlled
by the digital input angle φ. Two transforms are included in the
AD2S100. The first is the Clarke transform which computes
the sine and cosine orthogonal components of a three-phase
input. These signals represent real and imaginary components
which then form the input to the Park transform. The Park
transform relates the angle of the input signals to a reference
frame controlled by the digital input port. The digital input
port is a 12-bit parallel binary representation.
If the input signals are represented by Vds and Vqs, respectively,
where Vds and Vqs are the real and imaginary components, then
the transformation can be described as follows:
Vds' = Vds Cosφ – Vqs Sinφ
Vqs' = Vds Sinφ + Vqs Cosφ
Cos (θ + 120°)
Cos (θ + 240°)
Sinθ
CONV1
Ia
Vds
INPUT
DATA
STROBE
φ POSITION
PARALLEL
DATA
12 BITS
SINE AND
SECTOR
COSINE
MULTIPLIER MULTIPLIER
BUSY
Va
Vds'
Ib
2φ -3φ
30-20
Ic
Vqs
Vb
Cos θ + φ
Cos (θ + 120° + φ)
Vc Cos (θ + 240° + φ)
SINE AND
SECTOR
COSINE
MULTIPLIER MULTIPLIER
Vqs'
Sin θ + φ
Ia + Ib + Ic
3
DECODE
CONV2
HOMOPOLAR
OUTPUT
HOMOPOLAR
REFERENCE
+5V GND –5V
The digital input section will accept a resolution of up to 12 bits
(AD2S100). An input data strobe signal is required to synchronize the position data and load this information into the device
counters. A busy output is provided to identify the conversion
status of the AD2S100. The busy period represents the conversion time of the vector rotation.
Two analog output formats are available. A two-phase rotated
output facilitates multiple rotation blocks. Three phase format
signals are available for use with a PWM inverter.
PRODUCT HIGHLIGHTS
Hardware Peripheral for Standard Microcontrollers and
DSP Systems
The AD2S100 removes the time consuming cartesian transformations from digital processors and benchmarks a speed improvement of 30:1 on standard 20 MHz processors. AD2S100
transformation time = 2 µs (typ).
Field Oriented Control of AC and DC Brushless Motors
The AD2S100 accommodates all the necessary functions to
provide a hardware solution for ac vector control of induction
motors and dc brushless motors.
Where Vds' and Vqs' are the output of the Park transform
and Sinφ, and Cosφ are the values internally derived by the
AD2S100 from the binary digital data.
Three-Phase Imbalance Detection
The input section of the device can be configured to accept
either three-phase inputs, two-phase inputs of a three-phase
system, or two 90 degree input signals. The homopolar output
detects the imbalance of a three-phase input only. Under normal conditions, this output will be zero.
Resolver-to-Digital Converter Interface
The AD2S100 can be used to sense overcurrent situations or
imbalances in a three-phase system via the homopolar output.
The AD2S100 provides general purpose interface for position
sensors used in the application of dc brushless and ac induction
motor control.
REV. A
Information furnished by Analog Devices is believed to be accurate and
reliable. However, no responsibility is assumed by Analog Devices for its
use, nor for any infringements of patents or other rights of third parties
which may result from its use. No license is granted by implication or
otherwise under any patent or patent rights of Analog Devices.
One Technology Way, P.O. Box 9106, Norwood, MA 02062-9106, U.S.A.
Tel: 617/329-4700
Fax: 617/326-8703
(VDD = +5 V 6 5%; VSS = –5 V 6 5% AGND = DGND = O V; TA = –408C to
Parameter
Min
SIGNAL INPUTS
PH/IP1, 2, 3, 4 Voltage Level
PH/IPH1, 2, 3 Voltage Level
Input Impedance
PH/IP1, 2, 3
PH/IPH1, 2, 3
PH/IP1, 4
Gain
PH/IP1, 2, 3, 4
PH/IPH1, 2, 3
Conditions
± 2.8
63.3
± 4.25
V p-p
V p-p
DC to 50 kHz
DC to 50 kHz
kΩ
kΩ
MΩ
Mode 1 Only (2 Phase) Sin & Cos
0.98
1
0.56
1.02
0.35
9
0.7
18
24
50
200
± 2.8
2
2
1
3.0
2
± 3.3
5
Positive Pulse
µs
V dc
V dc
Conversion in Process
IOH = 0.5 mA
IOL = 0.5 mA
2.5
1.5
610
V dc
V dc
µA
pF
1.5
100
V dc
V dc
µA
pF
10
3.5
10
CONV2
DGND
VDD
V DD
1° Input to Settle to ± 1 LSB
(Input to Output)
ns
kHz
3.5
VDD
PH/IP, PH/IPH INPUTS
DC to 50 kHz
Inputs = 0 V
366
1
DGND
V p-p
mV
V/µs
µs
50
1.7
CONVERT MODE
(CONV1, CONV2)
VIH
VIL
Input Current
Input Capacitance
DC to 600 Hz
DC to 600 Hz
DC to 600 Hz
Guaranteed Monotonic
Ω
mA
kΩ
pF
15
4.0
4
DIGITAL INPUTS
DB1–DB12
VIH
VIL
Input Current, IIN
Input Capacitance, CIN
%
arc min
arc min
kHz
kHz
100
BUSY
Pulse Width
VOH
VOL
CONVERT LOGIC
CONV1
NO CONNECT
Units
10
18
ANALOG SIGNAL OUTPUTS
PH/OP1, 2, 3, 4
Output Voltage3
Offset Voltage
Slew Rate
Small Signal Step Response
STROBE
Write
Max Update Rate
Max
7.5
13.5
1
VECTOR PERFORMANCE
3θ Input-Output
Angular Error1, 2 (PH/IP)
(PH/IPH)
Monotonicity
Full Power Bandwidth
Small Signal Bandwidth
Output Resistance
Output Drive Current
Typ
Outputs to AGND
Internal 50 kΩ Pull-Up Resistor
2-Phase Orthogonal with 2 Inputs
Nominal Input Level
3-Phase (0°, 120°, 240°) with 3 Inputs
Nominal Input Level
3-Phase (0°, 120°, 240°) with 2 Inputs
Nominal Input Level
–2–
REV. A
Parameter
Min
HOMOPOLAR OUTPUT
HPOP–Output
VOH
VOL
HPREF–REFERENCE
4
HPFILT-FILTER
POWER SUPPLY
VDD
VSS
IDD
ISS
4.75
–5.25
Typ
Max
Units
Conditions
1
0.5
V dc
V dc
V dc
100
kΩ
IOH = 0.5 mA
IOL = 0.5 mA
Homopolar Output-Internal
ISOURCE = 25 µA and 20 kΩ
to AGND
Internal Resistor with External
Capacitor = 220 nF
5
–5
4
4
5.25
–4.75
10
10
V dc
V dc
mA
mA
Quiescent Current
Quiescent Current
NOTES
1
Angular accuracy includes offset and gain errors. Stationary digital input and maximum analog frequency inputs.
2
Included in the angular error is an allowance for the additional error caused by the phase delay as a function of input frequency. For example, if
fINPUT = 600 Hz, the contribution to the error due to phase delay is: 650 ns × fINPUT × 60 × 360 = 8.4 arc minutes.
3
Output subject to input voltage and gain.
Specifications in boldface are production tested.
Specifications subject to change without notice.
RECOMMENDED OPERATING CONDITIONS
ABSOLUTE MAXIMUM RATINGS (TA = +25°C)
Power Supply Voltage (+VDD, –VSS) . . . . . . . . . ± 5 V dc ± 5%
Analog Input Voltage (PH/IP1, 2, 3, 4) . . . . . . 2 V rms ± 10%
Analog Input Voltage (PH/IPH1, 2, 3) . . . . . . 3 V rms ± 10%
Ambient Operating Temperature Range
Industrial (AP) . . . . . . . . . . . . . . . . . . . . . . . –40°C to +85°C
VDD to AGND . . . . . . . . . . . . . . . . . . . . . . . –0.3 V to +7 V dc
VSS to AGND . . . . . . . . . . . . . . . . . . . . . . . +0.3 V to –7 V dc
AGND to DGND . . . . . . . . . . . . . . . . . . . . . . . . . . . ± 0.3 V dc
Analog Input Voltage to AGND . . . . . . . . . . . . . . . VSS to VDD
Digital Input Voltage to DGND . . . . –0.3 V to VDD + 0.3 V dc
Digital Output Voltage to DGND . . . –0.3 V to VDD + 0.3 V dc
Analog Output Voltage to AGND
. . . . . . . . . . . . . . . . . . . . . . VSS – 0.3 V to VDD + 0.3 V dc
Analog Output Load Condition (PH/OP1, 2, 3, 4
Sinθ, Cosθ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 kΩ
Power Dissipation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 mW
Operating Temperature
Industrial (AP) . . . . . . . . . . . . . . . . . . . . . . . –40°C to +85°C
Storage Temperature . . . . . . . . . . . . . . . . . –65°C to +150°C
Lead Temperature (Soldering, 10 sec) . . . . . . . . . . . . . +300°C
ORDERING GUIDE
Model
Temperature Range
Accuracy
Option*
–40°C to +85°C
18 arc min
P-44A
*P = Plastic Leaded Chip Carrier.
CAUTION
1. Absolute Maximum Ratings are those values beyond which
damage to the device may occur.
2. Correct polarity voltages must be maintained on the +VDD
and –VSS pins.
CAUTION
ESD (electrostatic discharge) sensitive device. Electrostatic charges as high as 4000 V readily
accumulate on the human body and test equipment and can discharge without detection.
Although the AD2S100 features proprietary ESD protection circuitry, permanent damage may
occur on devices subjected to high energy electrostatic discharges. Therefore, proper ESD
precautions are recommended to avoid performance degradation or loss of functionality.
REV. A
–3–
WARNING!
ESD SENSITIVE DEVICE
PIN DESIGNATIONS1, 2, 3
VDD
STROBE
NC
NC
BUSY
NC
DGND
VDD
NC
Begin Conversion
Positive Power Supply
Negative Power Supply
Sin (θ + φ)
Cos (θ + φ)
Cos (θ + 240° + φ)
Cos (θ + 120° + φ)
Analog Ground
Sin θ Input
High Level Cos (θ + 240°) Input
Cos (θ + 240°) Input
High Level Cos (θ + 120°) Input
Cos (θ + 120°) Input
High Level Cos θ Input
Cos (θ) Input
Negative Power Supply
Homopolar Reference
Homopolar Output
Homopolar Filter
Select Input Format (3 Phase/3 Wire, Sin θ
Cos θ/Input, 3 Phase/2 Wire)
Cos Output
Sin Output
(DB1 = MSB, DB12 = LSB
Parallel Input Data)
Positive Power Supply
Digital Ground
Conversion in Progress
VSS
6
5
4
3
2
1
44
43
42
41
40
PH/OP1 7
39 NC
PH/OP3
8
38 DB1
PH/OP2
9
37 DB2
AGND
10
PH/IP4
11
35 DB4
12
TOP VIEW
(NOT TO SCALE)
34 DB5
PH/IPH3
36 DB3
PH/IP3 13
PH/IPH2
33 DB6
14
32 DB7
PH/IP2 15
31 DB8
PH/IPH1 16
30 DB9
PH/IP1 17
22
23
24
25
26
27
28
COS
SIN
DB12
DB11
21
CONV2
20
CONV1
19
HPOP
18
HPFILT
29 DB10
HPREF
STROBE
VDD
VSS
PH/OP4
PH/OP1
PH/OP3
PH/OP2
AGND
PH/IP4
PH/IPH3
PH/IP3
PH/IPH2
PH/IP2
PH/IPH1
PH/IP1
VSS
HPREF
HPOP
HPFILT
CONV1
CONV2
COS
SIN
DB12
DB1
VDD
DGND
BUSY
PH/OP4
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
19
20
21
22
23
24
25
26
27
38
41
42
44
NC
Mnemonic Description
VSS
Pin
PIN CONFIGURATION
NC = NO CONNECT
NOTES
Signal Inputs Ph/IP and PH/IPH on Pin Nos 11 through 17.
1
90° orthogonal signals = Sin θ, Cos θ (Resolver) = PH/IP4 and PH/IP1.
2
Three phase, 120°, three-wire signals
= Cos θ, Cos (θ + 120°), Cos (θ + 240°).
= PH/IP1, PH/IP2, PH/IP3
High Level = PH/IPH1, PH/IPH2, PH/IPH3.
3
Three Phase, 120°, two-wire signals = Cos (θ + 120°), Cos (θ + 240°)
= PH/IP2, PH/IP3.
In all cases where any of the input Pins 11 through 17 are not used, they must
be left unconnected.
–4–
REV. A
To relate these stator current to the reference frame the rotor
currents assume the same rectangular coordinates, but are now
rotated by the operator ejf, where ejf = Cos f + jSin f.
THEORY OF OPERATION
A fundamental requirement for high quality induction motor
drives is that the magnitude and position of the rotating air-gap
rotor flux be known. This is normally carried out by measuring
the rotor position via a position sensor and establishing a rotor
reference frame that can be related to stator current coordinates.
Here the term vector rotator comes into play where the stator
current vector can be represented in rotor-based coordinates or
vice versa.
To generate a flux component in the rotor, stator current is applied. A build-up of rotor flux is concluded which must be
maintained by controlling the stator current, ids, parallel to the
rotor flux. The rotor flux current component is the magnetizing
current, imr.
The AD2S100 uses ejf as the core operator. Here f represents
the digital position angle which rotates as the rotor moves. In
terms of the mathematical function, it rotates the orthogonal ids
and iqs components as follows:
Torque is generated by applying a current component which is
perpendicular to the magnetizing current. This current is normally called the torque generating current, iqs.
where ids', iqs' = stator currents in the rotor reference frame. And
ids' + jiqs' = (Ids + jIqs) ejf
ejf = Cos f + jSin f
= (Ids + jIqs)(Cos f + jSin f)
To orient and control both the torque and flux stator current
vectors, a coordinate transformation is carried out to establish a
new reference frame related to the rotor. This complex calculation is carried out by the AD2S100 vector processor.
The output from the AD2S100 takes the form of:
ids' = Ids Cos f – Iqs Sin f
iqs' = Ids Sin f + Iqs Cos f
To expand upon the vector operator a description of a single
vector rotation is of assistance. If it is considered that the moduli of a vector is OP and that through the movement of rotor
position by f, we require the new position of this vector it can
be deduced as follows:
The matrix equation is:
[ ] [
ids'
iqs'
Let original vector OP = A (Cos u + jSIN u) where A is a
constant;
so if OQ = OP ejf
=
Cos f
– Sin f
Sin f
Cos f
] [ ]
Ids
Iqs
and it is shown in Figure 2.
φ
(1)
and: ejf = Cos f + jSin f
OQ = A (Cos (u + f) + jSin (u + f))
= A [Cos u Cos φ – Sin u Sin φ + jSin u Cos φ + jCos u Sin φ]
= A [(Cos u + jSin u) (Cos f + jSin f)]
ids
ids'
e jφ
(2)
iqs
iqs'
a
Q
Figure 2. AD2S100 Vector Rotation Operation
θ+φ
P
φ
INPUT CLARK
θ
COSθ
COSθ + 120°
COSθ + 240°
SINθ
d
O
3φ + 2φ
TRANSFORMATION
SINE AND
COSINE
MULTIPLIER
(DAC)
LATCH
Cos(θ + φ)
Figure 1. Vector Rotation in Polar Coordinate
DIGITAL
φ
The complex stator current vector can be represented as is = ias
j 2π
j 4π
and a2 = e
. This can be re3
3
placed by rectangular coordinates as
PARK
SINE AND
COSINE
MULTIPLIER
(DAC)
Cos(θ +(120° + φ))
Cos(θ +(240° + φ))
OUTPUT CLARK
(3)
Figure 3. Converter Operation Diagram
In this equation ids and iqs represent the equivalent of a twophase stator winding which establishes the same magnitude of
MMF in a three-phase system. These inputs can be seen after
the three-phase to two-phase transformation in the AD2S100
block diagram. Equation (3) therefore represents a three-phase
to two-phase conversion.
REV. A
2φ–3φ
LATCH
+ aibs + a2ics where a = e
is = ids + jiqs
LATCH
–5–
CONVERTER OPERATION
ANALOG SIGNAL INPUT AND OUTPUT CONNECTIONS
Input Analog Signals
The architecture of the AD2S100 is illustrated in Figure 3. The
AD2S100 is configured in the forward transformation which rotates the stator coordinates to the rotor reference frame.
All analog signal inputs to AD2S100 are voltages. There are two
different voltage levels of three-phase (0°, 120°, 240°) signal inputs. One is the nominal level, which is ± 2.8 V dc or 2 V rms
and the corresponding input pins are PH/IP1 (Pin 17), PH/IP2
(Pin 15), PH/IP3 (Pin 13) and PH/IP4 (Pin 11).
Forward Rotation
In this configuration the 3φ–2φ Clark is bypassed, and inputs are
fed directly into the quadrature (PH/IP4) and direct (PH/ IPI)
inputs to the Park transform, eiφ, where φ is defined by the
AD2S100’s digital input. Position data, φ, is loaded into the input latch on the positive edge of the strobe pulse. (For detail on
the timing, please refer to the “timing diagram.”) The negative
edge of the strobe signifies that conversion has commenced. A
busy pulse is subsequently produced as data is passed from the
of the multiplier, the busy pulse remains high to ensure simultaneous setting of φ in both the Sin and Cos registers.
The high level inputs can accommodate voltages from nominal
up to a maximum of ± VDD/VSS. The corresponding input pins
are PH/IPH1 (Pin 16), PH/IPH2 (Pin 14) and PH/IPH3 (Pin
12). The homopolar output can only be used in the three-phase
connection mode.
The converter can accept both two-phase format and threephase format input signals. For the two-phase format input, the
two inputs must be orthogonal to each other. For the threephase format input, there is the choice of using all three inputs
or using two of the three inputs. In the latter case, the third input signal will be generated internally by using the information
of other two inputs. The high level input mode, however, can
only be selected with three-phase/three-input format. All these
different conversion modes, including nominal/high input level
and two/three-phase input format can be selected using two select pins (Pin 23, Pin 24). The functions are summarized in
Table I.
The negative edge of the busy pulse signifies that the multipliers
time. The resultant two outputs are accessed via the PH/OPI
(Pin 7) and PH/OP4 (Pin 6), alternatively they can be directly
applied to the output Clark transform. The Clark output is the
vector sum of the analog input vector (Cosθ (PH/OPl), Cos (θ +
120°) (PH/OP2), Cos (θ + 240°) (PH/OP3) and the digital input vector φ.
For other configurations, please refer to “Forward and Reverse
Transformation.”
Table I. Conversion Mode Selection
CONV1
(Pin 23)
CONNECTING THE CONVERTER
Power Supply Connection
Mode
The power supply voltages connected to VDD and VSS pins
should be +5 V dc and –5 V dc and must not be reversed. Pin 4
(VDD) and Pin 41 (VDD) should both be connected to +5 V;
similarly, Pin 5 (VSS) and Pin 19 (VSS) should both be connected to –5 V dc.
MODE1 2-Phase Orthogonal with 2 Inputs
NC
Nominal Input Level
MODE2 3-Phase (0°, 120°, 240°) with 3 Inputs DGND
Nominal/High Input Level*
MODE3 3-Phase (0°, 120°, 240°) with 2 Inputs VDD
Nominal Input Level
It is recommended that decoupling capacitors, 100 nF (ceramic)
and 10 µF (tantalum) or other high quality capacitors, are connected in parallel between the power line VDD, VSS and AGND
adjacent to the converter. Separate decoupling capacitors should
be used for each converter. The connections are shown in Figure 4.
In this mode, either nominal or high level inputs can be used.
For nominal level input operation, PH/IP1, PH/IP2 and PH/IP3
are the inputs, and there should be no connections to PH/IPH1,
PH/IPH2 and PH/IPH3; similarly, for high level input operation, the PH/IPH1, PH/IPH2 and PH/IPH3 are the inputs, and
there should be no connections to PH/IP1, PH/IP2 and PH/IP3.
In both cases, the PH/IP4 should be left unconnected. For high
level signal input operation, select MODE2 only.
AGND
TOP VIEW
10µF
100nF
12
VDD
MODE2: 3-Phase/3 Inputs with Nominal/High Input Level
10µF
+
VDD
In this mode, PH/IP1 and PH/IP4 are the inputs and the Pins
12 through 16 must be left unconnected.
100nF
GND
DGND
MODE1: 2-Phase/2 Inputs with Nominal Input Level
VDD
VSS
VDD
1
CONV2
(Pin 24)
*The high level input mode can only be selected with MODE2.
+5V
+
Description
34
VSS
MODE3: 3-Phase/2 Inputs with Nominal Input Level
In this mode, PH/IP2 and PH/IP3 are the inputs and the third
signal will be generated internally by using the information of
other two inputs. It is recommended that PH/IP1, PH/IPH1,
PH/IPH2, PH/IP4 and PH/IPH3 should be left unconnected.
23
–5V
Figure 4. AD2S100 Power Supply Connection
–6–
REV. A
Output Analog Signals
Example: From the equivalent circuit, it can be seen that the inclusion of a 20 kΩ resistor will reduce Vts to ± 0.25 V dc. This
corresponds to an imbalance of ± 0.75 V dc in the inputs.
There are three forms of analog output from the AD2S100.
Sin/Cos orthogonal output signals are derived from the Clark/
three-to-two-phase conversion before the Park angle rotation.
These signals are available on Pin 25 (Cos u) and Pin 26 (Sin
u), and occur before Park angle rotation.
Homopolar Filtering
The equation VSUM = Cosu + Cos (u + 120°) + Cos (u + 240°)
= 0 denotes an imbalance when VSUM ≠ 0. There are conditions,
however, when an actual imbalance will occur and the conditions as defined by VSUM will be valid. For example, if the first
phase was open circuit when u = 90° or 270°, the first phase is
valid at 0 V dc. VSUM is valid, therefore, when Cosu is close to 0.
In order to detect an imbalance u has to move away from 90° or
270°, i.e., when on a balanced line Cos u ≠ 0.
Three-Phase Output Signals
(Cos (θ + φ), Cos (φ + θ + 120°), Cos (φ + θ + 240°)), where
φ represents digital input angle. These signals are available on
Pin 7 (PH/OP1), Pin 9 (PH/OP2) and Pin 8 (PH/OP3),
respectively.
Two-Phase (Sin (θ + φ), Cos (θ + φ)) Signals
Line imbalance is detected as a function of HPREF, either set
by the user or internally set at ± 0.5 V dc. This corresponds to a
dead zone when f = 90° or 270° ± 30°, i.e., VSUM = 0, and,
therefore, no indicated imbalance. If an external 20 kΩ resistor
is added, this halves Vts and reduces the zone to ± 15°. Note this
example only applies if the first phase is detached.
These represent the output of the coordinate transformation.
These signals are available on Pin 6 (PH/OP4, Sin (θ + φ)) and
Pin 7 (PH/OP1, Cos (θ + φ)).
HOMOPOLAR OUTPUT
HOMOPOLAR Reference
In a three-phase ac system, the sum of the three inputs to the
converter can be used to indicate whether or not the phases are
balanced.
In order to prevent this false triggering an external capacitor
needs to be placed from HPFILT to ground, as shown in Figure
5. This averages out the perceived imbalance over a complete
cycle and will prevent the HPOP from alternatively indicating
balance and imbalance over u = 0° to 360°.
If VSUM = PH/IP1 + PH/IP2 + PH/IP3 (or PH/IPH1 +
PH/IPH2 + PH/IPH3) this can be rewritten as VSUM = [Cosu, +
Cos (u + 120°) + Cos (u + 240°)] = 0. Any imbalances in the
line will cause the sum VSUM ≠ 0. The AD2S100 homopolar
output (HPOP) goes high when VSUM > 3 × Vts. The voltage
level at which the HPOP indicates an imbalance is determined
by the HPREF threshold, Vts. This is set internally at ± 0.5 V dc
(± 0.1 V dc). The HPOP goes high when
For
dθ
dt
dθ
(Cosθ + Cos(θ + 120° ) + Cos(θ + 240° ))
V ts <
×V
3
dt
= 100 rpm C EXT = 2.2 µF
Note: The slower the input rotational speed, the larger the time
constant required over which to average the HPOP output. Use
of the homopolar output at slow rotational speeds becomes
impractical with respect to the increased value for CEXT.
where V is the nominal input voltage.
With no external components VSUM must exceed ± 1.5 V dc in
order for HPOP to indicate an imbalance. The sensitivity of the
threshold can be reduced by connecting an external resistor between HPOP and ground in Figure 5 where,
V ts =
= 1000 rpm C EXT = 200 nF
34
0.5 REXT
DGND
REXT + 20000
TOP VIEW
REXT = Ω
Vts = V dc.
1
23
CEXT
220nF
HPFILT
HPOP
HPOP
AGND
HPREF
25µA
HOMOPOLAR
REFERENCE
TO TRIGGER
12
REXT
GND
Figure 6. AD2S100 Homopolar Output Connections
20kΩ
EXTERNAL
RESISTOR
Figure 5. The Equivalent Homopolar Reference Input
Circuitry
REV. A
HPREF
–7–
TIMING DIAGRAMS
Busy Output
TYPICAL CIRCUIT CONFIGURATION
The state of converter is indicated by the state of the BUSY output (Pin 44). The BUSY output will go HI at the negative edge
of the STROBE input. This is used to synchronize digital input
data and load the digital angular rotation information into the
device counter. The BUSY output will remain HI for 2 µs, and
go LO until the next strobe negative edge occurs.
TWO/THREE PHASE
OUTPUT
Figure 8 shows a typical circuit configuration for the AD2S100
in a three phase, nominal level input mode (MODE2).
38
TOP VIEW
MSB
34
16
PH/IP1
23
SIN
COS
PH/IP3
PH/IP2
30
27
STROBE
+5V
DIGITAL ANGLE INPUT
BUSY
41
10µF
PH/IP4
12
HPFILT
t4
1
STROBE
PH/OP2
AGND
THREE PHASE INPUT
The width of the positive STROBE pulse should be at least
100 ns, in order to successfully start the conversion. The maximum frequency of STROBE input is 366 kHz, i.e., there should
be at least 2.73 µs from the negative edge of one STROBE pulse
to the next rising edge. This is illustrated by the following timing diagram and table.
PH/OP1
PH/OP3
HPREF
HPOP
Strobe Input
100nF
LSB
t1
–5V
t2
tf
tr
BUSY
100nF
GND
10µF
t3
Figure 8. Typical Circuit Configuration
APPLICATIONS
Forward and ReverseTransformation
Note: Digital data should be stable 25 ns before and after positive strobe edge.
The AD2S100 can perform both forward and reverse transformations. The section “Theory of Operation” explains how the
chip operates with the core operator e+jφ, which performs a forward transformation. The reverse transformation, e–jφ, is not
mentioned in the above sections of the data sheet simply to
avoid the confusion in the functionality and pinout. However,
the reverse transformation is very useful in many different applications, and the AD2S100 can be easily configured in a reverse
transformation configuration. Figure 9 shows four different
phase input/output connections for AD2S100 reverse transformation operation.
Parameter
Min
t1
t2
t3
t4
tr
100 ns
Typ
Max
30 ns
1.7 µs
2.5 µs
100 ns
20 ns
150 ns
tf
10 ns
120 ns
Condition
STROBE Pulse Width
STROBE ↓ to BUSY ↑
BUSY Pulse Width
BUSY ↓ to STROBE ↑
BUSY Pulse Rise Time
BUSY Pulse Rise Time
BUSY Pulse Fall Time
BUSY Pulse Fall Time
REVERSE
TRANSFORMATION
FORWARD
TRANSFORMATION
3 PHASE – 3 PHASE
Cosθ
Cos(θ + 120°)
Cos(θ + 240°)
e+jφ
Cos(θ + φ)
Cosθ
2 PHASE – 2 PHASE
Sinθ
e+jφ
Cosθ
2 PHASE – 3 PHASE
3 PHASE – 2 PHASE
Sinθ
Cosθ
Cos(θ + 120°)
Cos(θ + 240°)
Cos(θ +φ)
Cos(θ + φ + 120°)
Cos(θ + φ + 240°)
e+jφ
Sin(θ + φ)
Cos(θ + φ)
Cos(θ + φ + 120°)
Cos(θ + φ + 240°)
Cos(θ + φ)
e+jφ
Sin(θ + φ)
Cosθ
Cos(θ + 120°)
Cos(θ + 240°)
e–jφ
e–jφ
Cosθ
Sinθ
e–jφ
–1
Cosθ
Cos(θ + 120°)
Cos(θ + 240°)
Cos(θ – φ + 240°)
Cos(θ – φ)
Cosθ
Sinθ
Cos(θ – φ)
Cos(θ – φ + 120°)
Sin(θ – φ)
Cos(θ – φ)
Cos(θ – φ + 120°)
Cos(θ – φ + 240°)
Cos(θ – φ)
e–jφ
Sin(θ – φ)
–1
Figure 9. Reverse Transformation Connections
–8–
REV. A
calculating power required in the rotor reference frame is significantly reduced because the currents and flux are rotating at the
slip frequency. This permits calculations to be carried out in
time frames of, 100 µs, or under by a fixed-point DSP. Benchmark timing in this type of architecture can attain floating-point
speed processing with a fixed-point processor. Perhaps the largest advantage is in the ease with which the rotor flux position
can be obtained. A large amount of computation time is, therefore, removed by the AD2S100 vector processors due to the
split architecture shown in Figure 11. Motor control systems
employing one DSP to carry out the cartesian to polar transformations required for vector control are, therefore, tasked with
additional duties due to the fact that they normally operate in
the flux reference frame.
√In Figure 9, “–1” operator performs a 180° phase shift operation. It can be illustrated by a 2-phase-to-3-phase reverse transformation. An example is shown in Figure 10.
Cosθ
PH/OP1
Cos(θ + φ)
PH/IP1 (Cosθ)
R
Cos(θ – φ)
PH/OP3
Cos(θ + 240° + φ)
Cos(θ + 120° – φ)
PH/IP4 (Sinθ)
PH/OP2
Cos(θ + 120° + φ)
Cos(θ + 240° – φ)
R
Sinθ
R
2
φ
Figure 10. Two-Phase to Three-Phase Reverse
Transformation
The robustness of the control system can also be increased by
carrying out the control in the rotor reference frame. This is
achieved through the ability to increase and improve both the
algorithm quality in nonlinear calculations attributed to magnetizing inductance and rotor time constant for example. An
increase in sampling time can also be concluded with this architecture by avoiding the additional computing associated with
number truncation and rounding errors which reduce the signalto-noise rejection ratio.
Field Oriented Control of AC Induction Machine in a Rotor
Flux Frame
The architecture shown in Figure 11 identifies a simplified
scheme where the AD2S100 permits the DSP computing core
to execute the motor control in what is normally termed the
rotor reference frame. This reference frame actually operates in
synchronism with the rotor of a motor. This has significant
benefits regarding motor control efficiency and economics. The
POSITION
FEEDBACK
VELOCITY
FEEDBACK v
v
POSITION
v SET POINT
SPEED
CONTROL
+
VECTOR
is1
CO-PROCESSOR
v
ε'
LIMIT
+ ω'
–
–
ε
(a + jb)e–jρ
Cm
TORQUE
CONTROL
LIMIT
Vqs'
+ iqs'
md'
–
ω
+ imr'
+ ids'
–
–
imr
+
VECTOR
CO-PROCESSOR
iqs
FIELD
WEAKENING
ω+
is2
imrmax
Vds'
Vs1
(a + jb)e'–jρ'
ids
ω1
Vs2
ρ'
ω2
Tr
θ2
+
REVERSE
ROTATION
+
ε
FORWARD
ROTATION
iqs
is3
ids
ρ
Figure 11. Rotor Reference Frame Architecture
REV. A
Vs3
iqs
–9–
SIMPLE SLIP CONTROL
In an adjustable-frequency drive, the control strategy must ensure that motor operation is restricted to low slip frequencies,
resulting in stable operation with a high power factor and a high
torque per stator ampere. Figure 12 shows the block diagram of
simple slip control using the AD2S100. Here, the slip frequency
command ω2 and the current amplitude command are sent to
the microprocessor to generate two orthogonal signals, |I| Sin θ
and |I| Cos θ here (θ = ω2.) With the actual shaft position angle,
φ, (resolver-to-digital converter) and the orthogonal signals from
Ia
I Sinθ
(I) SET
SLIP
FREQ
Ib
Ic
dθ
ω2 = dt
PWM
+
INVERTER
AC
INDUCTION
MTR
RESOLVER
φ
Figure 12. Slip Control of AC Induction Motor with
the µP, the AD2S100 generates the inverter frequency and amplitude command into a three-phase format. The three-phase
sine wave reference currents are reproduced in the stator phases.
bibliography at the end of the data sheet.
Electronically commutated permanent magnet synchronous
motors (PMSM) are used in high performance drives for
machine tools and robotics. When a field orientated control
scheme is deployed, the resulting brushless drive has all the
properties required for servo applications in machine tool fed
drives, industrial robots, and spindle drives. These properties
include large torque/inertia ratio, a high peak torque capability
for fast acceleration and deceleration with high torsional stiffness at standstill.
Figure 13 shows the AD2S100 configured for both forward and
reverse transformations. This architecture concludes both flux
and torque current components independently. The additional
control of Vd (flux component) allows for the implementation
of field weakening schemes and maintenance of power factor.
ω ref
+
PI
Iqref
+
PI
–
–
Idref
+
Vq
e–jφ
Vd
2/3
INV +
PWM
PI
PMSM
φ
–
Id
Iq
e+jφ
3/2
Va
Vb
Vc
For more detailed information, please refer to the application
note “Vector Control Using a Single Vector Rotation Semiconductor for Induction and Permanent Magnet Motors.”
MOTION CONTROL DSP COPROCESSOR
AC induction motors are superior to dc motors with respect to
size/power ratio, weight, rotor inertia, maximum rotating velocity, efficiency and cost for motor ratings greater than 5 HP.
However, because of nonlinear and the highly interactive multivariable control structure, ac induction motors have been considered difficult to control in applications demanding variable
speed and torque.
Field orientated control theory and practice, under development
since 1975, has offered the same level of control enjoyed by traditional dc machines. Practical implementation of these algorithms involves the use of DSP and microprocessor based
architectures. The AD2S100 removes the needs for software
implementation of the rotor-to-stator and stator-to-rotor transformations in the DSP or µP. The reduction in throughput
times from typically 100 µs (µP) and 40 µs (DSP) to 2 µs increases system bandwidths while also allowing additional features to be added to the CPU. The combination of the fixed
control engine” shown in Figure 14, enables bandwidths previously attainable only through the use of floating point devices.
For more detailed information on the AD2S100 vector control
refer to application notes “Vector Control Using a Single Vector
Rotation Semiconductor for Induction and Permanent Magnet
Motors.”
MEASUREMENT OF HARMONICS
Three-phase ac power systems are widely used in power generation, transmission and electric drive. The quality of the electricity supply is affected by harmonics injected into the power main.
In inverter fed ac machines, fluxes and currents of various frequencies are produced. Predominantly in ac machines the 5th
and 7th harmonics are the most damaging; their reaction with
the fundamental flux component produces 6th harmonic torque
pulsations. The subsequent pulsating torque output may result
in uneven motion of the motor, especially at low speeds.
The AD2S100 can be used to monitor and detect the presence
and magnitude of a particular harmonic on a three-phase line.
Figure 15 shows the implementation of such a scheme using the
AD2S100. Note, the actual line voltages will have to be scaled
before applying to the three-phase input of the AD2S100.
Selecting a harmonic is achieved by synchronizing the rotational
frequency of the park digital input, φ, with the frequency of the
fundamental flux component and the integer harmonic selected.
The update rate, r, of the counters is determined by:
ω
r = 4096
Figure 13. PMSM Servo Control Using AD2S100
n×ω
2π
Here, r = input clock pulse rate (pulses/second);
n = the order of harmonics to be measured;
ω = fundamental angular frequency of the ac signal.
–10–
REV. A
HOST COMPUTER
VECTOR
COPROCESSOR
VECTOR
COPROCESSOR
DAC
DAC-8412
INV
+
PWM
INDUCTION
MOTOR
θ
R/D
CONVERTER
ia, ib, ic
Figure 14. Advanced Motion Control Engine
The magnitude of the n-th harmonic as well as the fundamental
component in the power line is represented by the output of the
low-pass filter, ak. In concert with magnitude of the harmonic
the AD2S100 homopolar output will indicate whether the
application, refer to the related application note listed in the
bibliography.
MSB
MSB-1
.
.
.
MSB – (n–1)
.
.
.
LSB + (n–1)
n = POLES
.
.
.
.
Va
Vb
Vc
Vd1
Vd
TWO-TO-THREE
CLARK
TRANSFORMATION Vq
e–jφ PARK
TRANSFORMATION
LOW PASS
FILTER
MSB
MSB-1
MSB-2
.
.
.
.
.
.
.
LSB
ak
Vq1
12,14 OR 16-BIT RESOLUTION MODE
HOMOPOLAR
OUTPUT
12-BIT UP/DOWN
COUNTER
Figure 16. A General Consideration in Connecting R/D
Converter and AD2S100 for Multiple Pole Motors
PULSE INPUTS
DIRECTION
Figure 17 shows the AD2S80A configured for use with a four
pole motor, where n = 2. Using the formula described the MSB
is shifted left once
Figure 15. Harmonics Measurement Using AD2S100
For multi-pole motor applications where a single speed resolver
is used, the AD2S100 input has to be configured to match the
electrical cycle of the resolver with the phasing of the motor
windings. The input to the AD2S100 is the output of a resolverto-digital converter, e.g., AD2S80A series. The parallel output
of the converter needs to be multiplied by 2n–1, where
n = the number of pole parts of the motor. In practice this is
implemented by shifting the parallel output of the converter left
relative to the number of pole pairs.
Figure 16 shows the generic configuration of the AD2S80A with
the AD2S100 for a motor with n pole pairs. The MSB of the
AD2S100 is connected to MSB-(n-1) bit of the AD2S80A digital output, MSB-1 bit to MSB-(n-2) bit, . . ., LSB bit to LSB
REV. A
MULTIPLE POLE MOTORS
–11–
(MSB)
(LSB)
BIT1
BIT2
.
.
.
.
.
.
BIT13
BIT14
.
.
.
.
.
.
MSB
MSB-1
.
.
.
.
.
.
.
LSB
14-BIT RESOLUTION MODE
Figure 17. Connecting of R/D Converter AD2S80A and
The AD2S100 can be configured for use as a 12-bit digital-toresolver (DRC) or synchro converter (DSC). DRCs and DSCs
are used to simulate the outputs of a resolver or a synchro. The
simulated outputs are represented by the transforms outlined
below.
Resolver Outputs
Asinωt.cosφ
Asinωt.sinφ
DRC—Must Select Mode 1
Inputs
PH/IP4
PH/IP1
Outputs PH/OP1
PH/OP4
Pin 11
Pin 1
Pin 7
Pin 6
AGND
Reference Asinωt
Asinωt Cosφ
Asinωt Sinφ
DSC—Must Select Mode 1
Inputs
PH/IP4
PH/IP1
Outputs PH/OP1
PH/OP2
PH/OP3
Synchro Outputs
Asinωt.sinφ
Asinωt.sin (φ + 120°)
Asinωt.sin (φ + 240°)
where:
Configuring the AD2S100 for DRC and DSC operation is done
by the following.
C1938–18–7/94
DIGITAL-TO-RESOLVER AND SYNCHRO CONVERSION
Pin 11
Pin 17
Pin 7
Pin 9
Pin 8
Reference Asinωt
AGND
–Asinωt Sinφ
–Asinωt Sin (φ + 120°)
–Asinωt Sin (φ + 240°)
NOTES
1. Valid information is only available after the strobe pulse and BUSY go low.
2. To correct for inverse phasing of the DSC outputs the reference should be
inverted, or the MSB can be inverted.
Asinωt = fixed ac reference
φ = digital input angle, i.e., shaft position
The waveforms are shown in Figures 18 and 19.
APPLICATION NOTES LIST
1. “Vector Control Using a Single Vector Rotation Semiconductor for Induction and Permanent Magnet Motors,” by F. P.
Flett, Analog Devices.
S2 TO S4
(COS)
2. “Gamana – DSP Vector Coprocessor for Brushless Motor
Control,” by Analog Devices and Infosys Manufacturing
System.
S3 TO S1
(SIN)
3. “Silicon Control Algorithms for Brushless Permanent Magnet
Synchronous Machines,” by F. P. Flett.
4. “Single Chip Vector Rotation Blocks and Induction Motor
Field Oriented Control,” by A. P. M. Van den Bossche and
P. J. M. Coussens.
R2 TO R4
(REF)
0°
90 °
180 °
270°
360°
θ
Figure 18. Electrical Representation and Typical Resolver
Signals
5. “Three Phase Measurements with Vector Rotation Blocks in
Mains and Motion Control,” P. J. M. Coussens, et al.
6. “Digital to Synchro and Resolver Conversion with the AC
Vector Processor AD2S100,” by Dennis Fu.
7. “Experiment with the AD2S100 Evaluation Board,” by
Dennis Fu.
S1 TO S2
OUTLINE DIMENSIONS
Dimensions shown in inches and (mm).
S2 TO S3
0.056 (1.42)
0.042 (1.07)
6
0°
90°
180 °
270 °
0.025 (0.63)
0.015 (0.38)
40
PIN 1
IDENTIFIER
7
0.048 (1.21)
0.042 (1.07)
R1 TO R2
0.180 (4.57)
0.165 (4.19)
39
0.021 (0.53)
0.013 (0.33)
0.63 (16.00)
0.59 (14.99)
360 °
θ
0.032 (0.81)
0.026 (0.66)
TOP VIEW
Figure 19. Electrical Representation and Typical Synchro
Signals
0.050
(1.27)
BSC
29
17
18
0.020
(0.50)
R
28
0.695 (17.65)
SQ
0.685 (17.40)
–12–
0.040 (1.01)
0.025 (0.64)
0.656 (16.66)
SQ
0.650 (16.51)
0.110 (2.79)
0.085 (2.16)
REV. A
PRINTED IN U.S.A.
0.048 (1.21)
0.042 (1.07)
S3 TO S1 | 11,573 | 37,168 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-26 | longest | en | 0.725384 |
https://renewabletechy.com/what-is-the-difference-between-thermal-energy-and-heat/ | 1,716,256,650,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058342.37/warc/CC-MAIN-20240520234822-20240521024822-00645.warc.gz | 428,084,063 | 39,577 | # What Is The Difference Between Thermal Energy And Heat?
Thermal energy and heat are related concepts in physics and thermodynamics. While they are often used interchangeably in everyday language, they have distinct scientific definitions. Thermal energy refers to the total internal energy of an object due to the random motion of its atoms and molecules. Heat, on the other hand, is the transfer of thermal energy between objects at different temperatures.
This article will examine the precise definitions of thermal energy and heat, explain the relationship between the two, and provide examples to illustrate the differences. We will also discuss how thermal energy and heat are measured, as well as applications like heat transfer and thermodynamics where distinguishing between the two concepts is essential.
## Definitions
When discussing energy, it’s important to understand the differences between two related but distinct concepts – thermal energy and heat.
Thermal energy refers to the total internal kinetic energy of all the molecules within a substance. This internal energy consists of the random motions and vibrations of the atoms, molecules, and other particles that make up a substance. The faster these particles move, the more thermal energy the substance contains. Thermal energy is measured in joules and is an intrinsic property of matter that exists on a microscopic scale.
In contrast, heat is the transfer of thermal energy between substances due to temperature differences. It flows spontaneously from a substance at a higher temperature to a substance at a lower temperature. Heat itself is energy in transit and does not remain with the substance. Heat has the same units as thermal energy (joules) and can be measured by the temperature change it produces.
So in summary:
• Thermal energy is the total internal kinetic energy of the molecules in a substance.
• Heat is the transfer of thermal energy between substances.
## Relationship Between Thermal Energy and Heat
Although the terms thermal energy and heat are sometimes used interchangeably, they refer to distinct but related concepts. Thermal energy is the total internal kinetic and potential energy of all the molecules within a substance. It is related to the temperature of the substance and the movements and vibrations of its molecules and atoms.
Heat, on the other hand, refers specifically to the transfer of thermal energy between substances or objects that are at different temperatures. When two objects with different temperatures come into contact, thermal energy will spontaneously transfer from the higher temperature object to the lower temperature object. This flow of thermal energy is heat.
So in summary, thermal energy is contained within a substance, while heat is the movement of thermal energy between substances. Heat is the transfer of thermal energy driven by temperature differences, and results in energy being lost by the hotter substance and gained by the colder substance.
## Temperature vs Thermal Energy
Though related, temperature and thermal energy are distinct scientific concepts. Temperature measures how hot or cold an object is, while thermal energy refers to the total energy of microscopic motions of particles in a system.
Temperature is measured on a scale, such as Celsius or Fahrenheit. It describes the average kinetic energy of molecules within a substance. Higher temperatures mean the molecules have more kinetic energy and are moving faster on average.
Thermal energy, also called internal energy, is the total potential and kinetic energy associated with the random motions of atoms and molecules in an object. It depends on the number of atoms, their mass, and how vigorously they are moving. More thermal energy means more total microscopic motion within the molecules.
For example, a pot of boiling water and a swimming pool at the same temperature of 100°C have vastly different amounts of thermal energy. The swimming pool has far more water molecules, thus greater total kinetic energy, even though their average motions, as measured by temperature, are the same.
Understanding the distinction between temperature and thermal energy is key for fields like thermodynamics and engineering. While related, temperature measures molecular motion on average, while thermal energy quantifies the total kinetic energy from molecular motion within a system.
## Measuring Thermal Energy
Thermal energy is measured in joules or calories. A joule is the SI unit of energy, while a calorie is based on the energy needed to raise one gram of water by one degree Celsius. Specifically, one calorie is the amount of thermal energy required to raise one gram of water from 14.5°C to 15.5°C under standard atmospheric pressure.
In the metric system, the joule is the standard unit used to measure energy. One joule (J) is defined as the work required to produce one watt of power for one second. For thermal energy, this translates to the amount of heat required to raise one gram of water by 0.24 degrees Celsius. The calorie and joule are convertible units, with 1 calorie equaling 4.184 joules.
To measure the thermal energy of an object or system, the total heat energy can be quantified in joules or calories. This is done by measuring the temperature change as energy is added or removed. The specific heat capacity of the material must also be known, which indicates how much energy is required to change its temperature by one degree. By calculating the mass, temperature change, and specific heat, the total thermal energy can be determined through the equation:
Thermal Energy = Mass x Specific Heat Capacity x Temperature Change
This allows the thermal energy contained within or transferred to/from an object or system to be numerically measured in joules or calories. This quantification is essential for thermodynamic analysis and calculations in science and engineering.
## Measuring Heat
Heat is measured in units of energy. The two main units used to measure heat are joules and calories:
Joules – The joule is the SI unit of energy in the International System of Units. One joule is defined as the amount of work required to produce one watt of power for one second. Joules are commonly used by scientists to measure heat and energy in various contexts.
Calories – A calorie is defined as the amount of heat energy needed to raise the temperature of 1 gram of water by 1 degree Celsius. Calories are commonly used to measure the energy content in foods. Nutrition labels on food packages often list the calorie content per serving.
In science, joules are more commonly used than calories when measuring heat energy. However, in everyday contexts, calories are still frequently used, especially when discussing food energy content. For example, the amount of heat needed to boil a pot of water could be measured in joules, while the amount of energy contained in a candy bar would be measured in calories.
## Heat Transfer
Heat can be transferred between objects or systems through three main methods: conduction, convection, and radiation.
Conduction is the transfer of heat between objects or systems that are in direct contact with each other. It occurs when atoms and molecules with higher kinetic energy transfer energy to neighboring atoms and molecules with lower energy. Metals are good conductors of heat because their atoms can easily vibrate against each other and pass kinetic energy.
Convection is the transfer of heat by the movement of heated fluid particles between objects or systems. It often involves an intermediary substance like air or water. For example, as air is heated it becomes less dense and rises, allowing cooler denser air to take its place, creating convection currents. This allows heat to be dispersed through the movement and mixing of the air.
Radiation is the transfer of heat via electromagnetic waves or photons. It does not rely on direct contact or any intermediary substance. For example, the sun radiates heat to the Earth across the vacuum of space. The heat we feel from a fireplace is also radiant heat transferred by infrared waves. Radiant heat transfer decreases with increasing distance, as radiation spreads out and loses intensity further away from the heat source.
Understanding these three heat transfer mechanisms is important for designing effective heating and cooling systems, predicting weather patterns, and much more.
## Examples
Hot tubs illustrate the difference between thermal energy and heat well. The water in a hot tub contains thermal energy due to the motion of its molecules. The motors and heaters in the hot tub’s equipment provide the energy needed to increase the water’s thermal energy, raising its temperature. The thermal energy in the now hot water can transfer heat to people sitting in the hot tub. The people absorb the heat, feeling warmer as it transfers from the water to their bodies. However, the total thermal energy in the system (the hot tub water) does not change during this heat transfer.
Another example is a pot of water on the stove. Adding heat from the stove burner increases the water’s thermal energy, making the water molecules move faster. This adds heat to the water, raising its temperature. However, if the pot is not insulated, heat transfers from the hot water to the surrounding air. As heat leaves the water, the water’s temperature decreases, but its total thermal energy remains constant. This illustrates how heat can be gained or lost, while thermal energy remains fixed.
A final example is a glass of ice water with ice cubes floating in it. The glass contains thermal energy due to the motion of the water molecules. The thermal energy is lower than it would be without the ice, so the drink is cold. As heat transfers from the surrounding air into the drink, the ice absorbs it, keeping the drink cold. The ice acts as a heat sink, absorbing heat without changing its own thermal energy. It melts as it absorbs the heat entering the glass. This example shows how heat transfer impacts temperature, while thermal energy remains unchanged.
## Practical Applications
Thermal energy and heat are central to many practical applications that power modern society, especially energy generation and engines.
One key example is the internal combustion engine, used in most cars and trucks. This engine burns gasoline, converting the chemical energy in the fuel into thermal energy through combustion. This thermal energy rapidly heats and expands the air in the engine’s cylinders, pushing the pistons and generating mechanical energy that powers the vehicle.
The principle behind most electric power generation also relies on thermal energy and heat. In coal, gas, or nuclear power plants, thermal energy released from burning fuel or nuclear reactions produces steam. This high-pressure steam then spins a turbine connected to an electrical generator, converting the thermal energy into electricity.
Solar thermal systems harness heat from the sun’s rays to warm water or spaces. The sunlight striking the solar panels contains thermal energy that gets transferred to a liquid, which can then heat buildings, swimming pools, or water for domestic use.
Understanding thermal energy flows and heat transfer is also crucial in designing effective heating and cooling systems, insulating buildings, and increasing energy efficiency in homes and industry.
## Conclusion
In summary, thermal energy and heat are related but distinct concepts. Thermal energy refers to the total kinetic energy of molecules within a substance, which relates to the temperature of the substance. Heat refers specifically to the transfer of thermal energy between objects or systems due to a temperature difference. While thermal energy is contained within a system, heat is energy in transit. Measuring thermal energy requires complex calculations using thermodynamic properties. Heat is easier to measure based on temperature change and the heat capacity of a system. Understanding the nuanced difference between thermal and heat is key for applications like designing heat exchangers, solar panels, and other thermodynamic systems. | 2,252 | 12,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2024-22 | latest | en | 0.923081 |
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# Q38: In one state, all cities and most towns have
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Q38: In one state, all cities and most towns have [#permalink]
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Q38:
In one state, all cities and most towns have antismoking ordinances. A petition entitled “Petition for Statewide Smoking Restrictionâ€
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hey ya......
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07 Oct 2005, 15:05
B, I think.
It seems to most directly address the fact that the actual law being promoted doesn't actually have statewide restrictions.
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Re: CR # Petition for statewied Smoking restriction [#permalink]
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07 Oct 2005, 21:31
[quote="nakib77"]Q38:
In one state, all cities and most towns have antismoking ordinances. A petition entitled “Petition for Statewide Smoking Restrictionâ€
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07 Oct 2005, 22:03
A. Premiums are out of scope
B. My pick. If there is hardly any establishment in rural areas that is open to public and if smoking restrictions in all establishments in urban areas (cities and towns) are taken care of by the existing ordinances, it makes the argument misleading as it would not achieve anything.
C. This strengthens the argument. We are asked to weaken it.
D. Out of scope
E. Out of scope
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09 Oct 2005, 12:00
OA is E.
Can anybody explain that..... I am simply lost here.....
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10 Oct 2005, 19:23
Typo?
Seriously, I can't make sense of it either!
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10 Oct 2005, 21:39
i don't agree with E because it is out of scope. fire hazards are not the issue here at all. is the an OE? where is this question form? My answer is B.
Kudos [?]: 12 [0], given: 0
10 Oct 2005, 21:39
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