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https://tech-story.net/circuit-connections-in-resistors/ | 1,638,090,160,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358480.10/warc/CC-MAIN-20211128073830-20211128103830-00611.warc.gz | 613,966,789 | 35,608 | Circuit Connections in Resistors
Circuit Connections in Resistors
A Resistor when connected in a circuit, that connection can be either series or parallel. Let us now know what will happen to the total current, voltage and resistance values if they are connected in series as well, when connected in parallel.
Resistors in Series
Let us observe what happens, when few resistors are connected in Series. Let us consider three resistors with different values, as shown in the figure below.
Resistance
The total resistance of a circuit having series resistors is equal to the sum of the individual resistances. That means, in the above figure there are three resistors having the values 1KΩ, 5KΩ and 9KΩ respectively.
Total resistance value of the resistor network is −
R=R1+R2+R3R=R1+R2+R3
Which means 1 + 5 + 9 = 15KΩ is the total resistance.
Where R1 is the resistance of 1st resistor, R2 is the resistance of 2nd resistor and R3 is the resistance of 3rd resistor in the above resistor network.
Voltage
The total voltage that appears across a series resistors network is the addition of voltage drops at each individual resistances. In the above figure we have three different resistors which have three different values of voltage drops at each stage.
Total voltage that appears across the circuit −
V=V1+V2+V3V=V1+V2+V3
Which means 1v + 5v + 9v = 15v is the total voltage.
Where V1 is the voltage drop of 1st resistor, V2 is the voltage drop of 2nd resistor and V3 is the voltage drop of 3rd resistor in the above resistor network.
Current
The total amount of Current that flows through a set of resistors connected in series is the same at all the points throughout the resistor network. Hence the current is same 5A when measured at the input or at any point between the resistors or even at the output.
Current through the network −
I=I1=I2=I3I=I1=I2=I3
Which means that current at all points is 5A.
Where I1 is the current through the 1st resistor, I2 is the current through the 2nd resistor and I3 is the current through the 3rd resistor in the above resistor network.
Resistors in Parallel
Let us observe what happens, when few resistors are connected in Parallel. Let us consider three resistors with different values, as shown in the figure below.
Resistance
The total resistance of a circuit having Parallel resistors is calculated differently from the series resistor network method. Here, the reciprocal 1/R1/R value of individual resistances are added with the inverse of algebraic sum to get the total resistance value.
Total resistance value of the resistor network is −
1R=1R1+1R2+1R31R=1R1+1R2+1R3
Where R1 is the resistance of 1st resistor, R2 is the resistance of 2nd resistor and R3 is the resistance of 3rd resistor in the above resistor network.
For example, if the resistance values of previous example are considered, which means R1 = 1KΩ, R2 = 5KΩ and R3 = 9KΩ. The total resistance of parallel resistor network will be −
1R=11+15+191R=11+15+19
=45+9+545=5945=45+9+545=5945
From the method we have for calculating parallel resistance, we can derive a simple equation for two-resistor parallel network. It is −
R=R1×R2R1+R2R=R1×R2R1+R2
Voltage
The total voltage that appears across a Parallel resistors network is same as the voltage drops at each individual resistance.
The Voltage that appears across the circuit −
V=V1=V2=V3V=V1=V2=V3
Where V1 is the voltage drop of 1st resistor, V2 is the voltage drop of 2nd resistor and V3 is the voltage drop of 3rd resistor in the above resistor network. Hence the voltage is same at all the points of a parallel resistor network.
Current
The total amount of current entering a Parallel resistive network is the sum of all individual currents flowing in all the Parallel branches. The resistance value of each branch determines the value of current that flows through it. The total current through the network is
I=I1+I2+I3I=I1+I2+I3
Where I1 is the current through the 1st resistor, I2 is the current through the 2nd resistor and I3 is the current through the 3rd resistor in the above resistor network. Hence the sum of individual currents in different branches obtain the total current in a parallel resistive network.
A Resistor is particularly used as a load in the output of many circuits. If at all the resistive load is not used, a resistor is placed before a load. Resistor is usually a basic component in any circuit. | 1,230 | 4,501 | {"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} | 4.59375 | 5 | CC-MAIN-2021-49 | latest | en | 0.927965 |
http://www.millersville.edu/~bikenaga/calculus/parfrac/parfrac.html | 1,448,722,517,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398453553.36/warc/CC-MAIN-20151124205413-00260-ip-10-71-132-137.ec2.internal.warc.gz | 560,968,205 | 9,103 | # Partial Fractions
Partial fractions is the opposite of adding fractions over a common denominator. It applies to integrals of the form
The idea is to break into a sum of smaller terms which are easier to integrate.
(A function of the form , where and are polynomials, is called a rational function.)
I'll start by doing an example to give you a feel for the procedure. Then I'll go back and explain the steps in the method. The procedure is a bit long, and requires a substantial amount of algebra. Therefore, before using partial fractions, you should be sure that there isn't an easier way to do the integral.
First, I want to mention a formula that often comes up in these problems:
(Do you see how to work it out? Substitute , so .) For example,
Example. Compute .
. Write
Multiply both sides by to clear denominators:
Let . I get
Let . I get
Therefore,
So
Now I'll describe the steps in the method. Some of this will seem a little abstract until you see some examples.
Consider an integral of the form
where and are polynomials.
Recall that the degree of a polynomial is the highest power of the variable that occurs in it. Nonzero constants have degree 0; by convention, 0 has degree .
Step 1. If the degree of the top is greater than or equal to the degree of the bottom, divide the bottom into the top.
Step 2. You will now have an integral that looks like
where and are polynomials, and the top is smaller in degree than the bottom.
Factor the bottom of the fraction into a product of linear terms and irreducible quadratic terms.
• A linear term is a term where the variable occurs only to the first power. Here are some linear terms:
• An irreducible quadratic term is a quadratic term with only imaginary (complex) roots. That is, it is a quadratic which "doesn't factor". Here are some irreducible quadratic terms:
You can check that a quadratic is irreducible by using the general quadratic formula to find its roots. If the roots are complex numbers, the quadratic does not factor.
Warning: is not irreducible:
("Ugly" factors are allowed.) And is not considered an irreducible quadratic: It is , the square of a linear term. This distinction will become important in Step 3.
Step 3. Obtain the partial fractions decomposition for the fraction.
This is the heart of the partial fractions method. It is basically a lot of algebra, but it's sufficiently complicated that the best way to describe it is by doing some examples.
Example. Compute .
The top has degree 3 while the bottom has degree 2. Divide the bottom into the top:
So
Integrate by substitution:
Hence,
In this problem, division simplified the integral enough that it wasn't necessary to go any further in the partial fractions procedure.
Example. Consider the integral . The top and the bottom have the same degree. Divide the bottom into the top:
Therefore,
I can do easily; I need to integrate the second term. Factor the denominator: . Then
Let . I get , so .
Let . I get , so .
Thus,
Therefore,
Long division in Step 1 is a preliminary operation which puts the integral into the right form for the rest of the procedure. If you do a division, check before going on to see whether you can use a simple technique (like substitution) to do the integrals you've obtained. Sometimes you can complete the integration immediately; otherwise, you'll need to go on to Step 2.
Example. Compute .
This example will show how to handle repeated factors --- in this case, . Here's what you do:
For a repeated factor, you have one term for each power up to the power the factor is raised to. In this case, you have a term for and a term for .
Multiply to clear denominators:
Let . I get .
Let . I get .
Plug the a and c values back in:
With only b left, I can plug in any number and solve for b. I'll let :
Therefore,
Hence,
Alternatively, take equation (*). Multiply out the b-term:
Since this is an identity, I can differentiate both sides:
Now I can recycle . Plugging in gives , so as before.
At any point, you can plug in an arbitrary number for x, or you can differentiate both sides of the equation.
Example. Compute .
In this case, the repeated factor is the " ". I want a, b, and c so that
Clear denominators:
Let . I obtain , so .
Next, let . Then so .
Put and back into the equation:
Let . I get , so .
Substitute the a, b, and c values into the original decomposition:
Finally, do the integral:
Example. How would you try to decompose
using partial fractions? That is, what is the initial partial fractions equation?
The linear factor is repeated 4 times, and the linear factor is repeated 2 times. So you use
You could do the terms first instead. Notice that the numerator has no effect on the decomposition.
Example. How would you try to decompose
using partial fractions? That is, what is the initial partial fractions equation?
The linear factor x is repeated 3 times and the linear factor is repeated twice. Therefore, you should try to solve
Notice that the top of the fraction is irrelevant in deciding how to set up the decomposition. It only comes in during the solution process.
Notice also that " " is considered a linear term (x) raised to the third power. You get one term on the right for x, one for , and one for --- no "skipping"!
Example. Compute .
The quadratic factor is irreducible. Here's how I try to decompose the fraction:
Thus, a quadratic factor (or a quadratic factor to a power) will produce terms on the right with "two letters" on top.
The rationale is the same as the one I gave for repeated factors. I don't know what kind of fraction to expect, so I have to take the most general case.
Clear denominators:
Let . I get , so . Plug back in:
Let . I get , so , or . Plug back in:
At this point, I've run out of "nice" numbers to plug in for x. There are several ways to proceed.
First, since I only have one letter to find (b), I can plug in a random value for x and solve. For example, plugging in gives . Simplifying gives , so , or .
Second, since the equation is an identity, I can {\it differentiate both sides}. Doing so, I get
(I got two b-terms by applying the Product Rule to .) Now I can recycle an old x-value, say . Plugging this in gives , or , so again.
You can use any combination of differentiating and plugging in that you wish.
Thus,
The first and third terms come from basic formulas. I integrated the second term using the substitution .
You handle repeated quadratic factors just as you handle repeated linear factors.
Example. How would you try to decompose
using partial fractions? That is, what is the initial partial fractions equation?
The quadratic is irreducible. Try the decomposition
Example. Compute .
Try the decomposition
Let . I get , so . Plug back in to obtain
Let . I get , so . Plug back in to obtain
I can differentiate both sides, or I can plug in another value of x. I'll plug in . Doing so gives , or , so .
Thus, the integral becomes
The first and third terms are integrated using basic formulas; the second term is integrated using the substitution .
Example. Compute .
Try the decomposition
In this case, there is no value of x I can plug in which will allow me to solve for one of a, b, c, d immediately. Therefore, I'll have to plug in and get equations, which I'll solve simultaneously later.
Let . This gives .
Differentiate the equation to obtain
Let . This gives .
Before I differentiate again, I'll multiply the -terms in, so that I can avoid using the Product Rule:
Differentiate:
Let . This gives , or , so . Plug this into to obtain , or . Then gives .
Differentiate one more time:
Thus, , or . Solve simultaneously with :
Therefore, . Plugging into gives .
Therefore, the integral becomes
The first and third terms are integrated using basic formulas. The second term is integrated using the substitution .
Example. Compute .
The quadratic does not factor. Try the decomposition
Let . I get , so . Plug back in to obtain
Let . I get , so . Plug back in to obtain
Differentiate:
Let . I get . so .
Therefore, the integral becomes
I'll do the second term separately, since the first term is easy. The idea is to complete the square, then do a substitution:
Hence,
Example. Compute .
First,
In this example, there's an irreducible quadratic factor . I try the decomposition
Clear denominators:
Let . Then , so . Plug it back in:
Let . I get , so . Plug it back in:
Now I can either plug in a value for x at random, or differentiate. I'll differentiate:
Differentiate again:
This gives .
Plug the values back into the original fractional decomposition:
The integral is
I'll do the integrals separately. First,
Next,
I can do the first integral using a substitution:
The second requires completing the square:
Putting the two together,
Finally, answer to the original problem is
What a mess! This is why you should consider other methods before you turn to partial fractions! | 1,995 | 8,976 | {"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} | 4.71875 | 5 | CC-MAIN-2015-48 | latest | en | 0.944756 |
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# Handler’s Classification in Computer Architecture
In 1977, Wolfgang Handler presented a computer architectural classification scheme for determining the degree of parallelism and pipelining built into the computer system hardware. Parallel systems are complicated to the program as compared to the single processor system because parallel system architecture varies according to the multiple CPUs and they must be coordinated or synchronized.
In Handler’s classification pipeline processing systems are divided into three subsystems:
1. Processor Control Unit (PCU): Each PCU corresponds to one processor or one CPU.
2. Arithmetic Logic Unit (ALU): ALU is equivalent to the processing element (PE).
3. Bit Level Circuit (BLC): BLC corresponds to the combinational logic circuit required for 1-bit operations in ALU.
ALU is a small unit that has fewer features than the processor and it works under the instructions of the processor. ALU is designed for performing arithmetic and logical calculations as per its name. A system comprises multiple ALUs to run parallel and increase the performance of the system. BLC(Bit-level circuit) is required to perform single or bit operations in ALU.
Handler’s classification uses three pairs of integers containing 6 independent entities that describe the computer system:
Computer =
where K = number of processors (PCUs) within the computer
• K’ = number of PCUs that can be pipelined
• D = number of ALUs (PEs) under the control of PCU
• D‘ = number of PEs that can be pipelined
• W = word length of a PE
• W’ = number of pipeline stages in all PEs
Example 1: Let us consider an example of Texas Instruments’ Advanced Scientific Computer (TI ASC), which has one controller that controls 4 arithmetic pipelines, each having a 64-bit word length and 8 pipeline stages.
From this data, we get K = 1, K’ = 1, D = 4, D’ = 1, W = 64, W’ = 8 . So, we can represent TI ASC according to Handler’s classification as follows :
TI ASC =
Let us now look at one more exercise on Handler’s classification.
Example 2: CDC 6600 has only a single CPU with an ALU, that has 10 specialized hardware functions each of 60-bit word length and up to 10 of these functions can be linked into a longer pipeline. It also has 10 peripheral I/O processors which operate in parallel with the CPU and with each other also. Each of the I/O processors has 1 ALU with 12 bits of word length.
Here, we have two parts to consider, i.e. Central processor(CP) and I/O Processor(IP). So, the representation is given by,
CDC 6600 =
From the given information it can be observed that for the Central processor, K = 1, K’ = 1, D = 1, D’ = 10, W = 60, W’ = 1 and for the I/O processor, K = 10, K’ = 1, D = 1, D’ = 1, W = 12, W’ = 1
Hence, the expression becomes,
CDC 6600 =
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To clarify this is for an economics course.
$$x \in \mathbb R^n$$ For a firm with two outputs with cost function where
$$C(q)=q_1^2+q_2^2+10$$ given output levels $q=(q_1,q_2) \ge 0$.
If output prices given and fixed, find profit maximizing solution.
I tried to find the global max, but ended up with global min at $(0,0)$, and am not sure how to proceed given just this one equation. Any help would be greatly appreciated.
• Sorry for the formatting, was trying to figure out how to do subscript and super script. – Shaner Oct 9 '16 at 21:47
• What eactly is the profoit in this case? I only see a cost function. – 6005 Oct 9 '16 at 21:49
• @Shaner I changed the formatting to Latex – Jannik Pitt Oct 9 '16 at 21:49
• Thanks for the help with the formatting! – Shaner Oct 9 '16 at 21:50
• I have written the problem exactly as I have it. I can add more assumptions , if more information is necessary please let me know and I'll search my book. – Shaner Oct 9 '16 at 21:53
You can't solve this problem without knowing the output prices. The higher the prices, the higher the manufactured quantities should be. You are minimizing the cost function, which you have done correctly. The minimum cost is $10$ at $q_1=q_2=0$ To maximize profit, you need to be able to compute it. Alternately, you could assume the price of item $1$ is $p_1$ and the price of item $2$ is $p_2$, find that revenue is $p_1q_1+p_2q_2$ and the profit is $p_1q_1+p_2q_2-q_1^2-q_2^2-10$, and find the maximum the same way you did viewing $p_1,p_2$ as parameters. You should find $q_1=p_1/2, q_2=p_2/2$ | 497 | 1,621 | {"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} | 3.75 | 4 | CC-MAIN-2019-39 | latest | en | 0.903443 |
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# Regression Analysis to Stock Price
See attached data file.
Use simple regression to determine the extent to which your stock's closing price can be determined from the closing value of the DJIA. As before, you should include a copy of the data you've collected to date. In your SLP upload, include a screen shot of your Excel output.
SLP Assignment Expectations:
In general, SLPs are expected to have the attributes of precision, clarity, breadth, depth, and applicability. However, not all of these attributes are applicable to all SLPs, especially in math and science courses. For this SLP, the expectations are:
Accurately state the null and alternative hypotheses for this test
Submit data in Excel format, with proper labels and headings
Correctly stimate a simple regression model using these data
Discuss the meaning of the regression results for these data
#### Solution Preview
Simple regression to determine the extent to which your stock's closing price can be determined from the closing value of the DJIA.
Null and alternative hypotheses
H0: There is no linear relationship between closing price of NYSE and the closing value of the DJIA.
H1: There is linear relationship between closing price of NYSE and the closing value of the DJIA.
Scatter diagram ...
#### Solution Summary
Step by step method for regression analysis is discussed here. Regression coefficients, coefficient of determination, scatter diagram and significance of regression model are explained in the solution.
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http://www.retro11.de/ouxr/211bsd/usr/src/games/hack/hack.pri.c | 1,713,426,167,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-18/segments/1712296817200.22/warc/CC-MAIN-20240418061950-20240418091950-00720.warc.gz | 54,880,665 | 5,123 | /* Copyright (c) Stichting Mathematisch Centrum, Amsterdam, 1985. */ /* hack.pri.c - version 1.0.3 */ #include "hack.h" #include xchar scrlx, scrhx, scrly, scrhy; /* corners of new area on screen */ extern char *hu_stat[]; /* in eat.c */ extern char *CD; swallowed() { char *ulook = "|@|"; ulook[1] = u.usym; cls(); curs(u.ux-1, u.uy+1); fputs("/-\\", stdout); curx = u.ux+2; curs(u.ux-1, u.uy+2); fputs(ulook, stdout); curx = u.ux+2; curs(u.ux-1, u.uy+3); fputs("\\-/", stdout); curx = u.ux+2; u.udispl = 1; u.udisx = u.ux; u.udisy = u.uy; } /*VARARGS1*/ boolean panicking; panic(str,a1,a2,a3,a4,a5,a6) char *str; { if(panicking++) exit(1); /* avoid loops - this should never happen*/ home(); puts(" Suddenly, the dungeon collapses."); fputs(" ERROR: ", stdout); printf(str,a1,a2,a3,a4,a5,a6); #ifdef DEBUG #ifdef UNIX if(!fork()) abort(); /* generate core dump */ #endif UNIX #endif DEBUG more(); /* contains a fflush() */ done("panicked"); } atl(x,y,ch) register x,y; { register struct rm *crm = &levl[x][y]; if(x<0 || x>COLNO-1 || y<0 || y>ROWNO-1){ impossible("atl(%d,%d,%c)",x,y,ch); return; } if(crm->seen && crm->scrsym == ch) return; crm->scrsym = ch; crm->new = 1; on_scr(x,y); } on_scr(x,y) register x,y; { if(x < scrlx) scrlx = x; if(x > scrhx) scrhx = x; if(y < scrly) scrly = y; if(y > scrhy) scrhy = y; } /* call: (x,y) - display (-1,0) - close (leave last symbol) (-1,-1)- close (undo last symbol) (-1,let)-open: initialize symbol (-2,let)-change let */ tmp_at(x,y) schar x,y; { static schar prevx, prevy; static char let; if((int)x == -2){ /* change let call */ let = y; return; } if((int)x == -1 && (int)y >= 0){ /* open or close call */ let = y; prevx = -1; return; } if(prevx >= 0 && cansee(prevx,prevy)) { delay_output(); prl(prevx, prevy); /* in case there was a monster */ at(prevx, prevy, levl[prevx][prevy].scrsym); } if(x >= 0){ /* normal call */ if(cansee(x,y)) at(x,y,let); prevx = x; prevy = y; } else { /* close call */ let = 0; prevx = -1; } } /* like the previous, but the symbols are first erased on completion */ Tmp_at(x,y) schar x,y; { static char let; static xchar cnt; static coord tc[COLNO]; /* but watch reflecting beams! */ register xx,yy; if((int)x == -1) { if(y > 0) { /* open call */ let = y; cnt = 0; return; } /* close call (do not distinguish y==0 and y==-1) */ while(cnt--) { xx = tc[cnt].x; yy = tc[cnt].y; prl(xx, yy); at(xx, yy, levl[xx][yy].scrsym); } cnt = let = 0; /* superfluous */ return; } if((int)x == -2) { /* change let call */ let = y; return; } /* normal call */ if(cansee(x,y)) { if(cnt) delay_output(); at(x,y,let); tc[cnt].x = x; tc[cnt].y = y; if(++cnt >= COLNO) panic("Tmp_at overflow?"); levl[x][y].new = 0; /* prevent pline-nscr erasing --- */ } } setclipped(){ error("Hack needs a screen of size at least %d by %d.\n", ROWNO+2, COLNO); } at(x,y,ch) register xchar x,y; char ch; { #ifndef lint /* if xchar is unsigned, lint will complain about if(x < 0) */ if(x < 0 || x > COLNO-1 || y < 0 || y > ROWNO-1) { impossible("At gets 0%o at %d %d.", ch, x, y); return; } #endif lint if(!ch) { impossible("At gets null at %d %d.", x, y); return; } y += 2; curs(x,y); (void) putchar(ch); curx++; } prme(){ if(!Invisible) at(u.ux,u.uy,u.usym); } doredraw() { docrt(); return(0); } docrt() { register x,y; register struct rm *room; register struct monst *mtmp; if(u.uswallow) { swallowed(); return; } cls(); /* Some ridiculous code to get display of @ and monsters (almost) right */ if(!Invisible) { levl[(u.udisx = u.ux)][(u.udisy = u.uy)].scrsym = u.usym; levl[u.udisx][u.udisy].seen = 1; u.udispl = 1; } else u.udispl = 0; seemons(); /* reset old positions */ for(mtmp = fmon; mtmp; mtmp = mtmp->nmon) mtmp->mdispl = 0; seemons(); /* force new positions to be shown */ /* This nonsense should disappear soon --------------------------------- */ for(y = 0; y < ROWNO; y++) for(x = 0; x < COLNO; x++) if((room = &levl[x][y])->new) { room->new = 0; at(x,y,room->scrsym); } else if(room->seen) at(x,y,room->scrsym); scrlx = COLNO; scrly = ROWNO; scrhx = scrhy = 0; flags.botlx = 1; bot(); } docorner(xmin,ymax) register xmin,ymax; { register x,y; register struct rm *room; register struct monst *mtmp; if(u.uswallow) { /* Can be done more efficiently */ swallowed(); return; } seemons(); /* reset old positions */ for(mtmp = fmon; mtmp; mtmp = mtmp->nmon) if(mtmp->mx >= xmin && mtmp->my < ymax) mtmp->mdispl = 0; seemons(); /* force new positions to be shown */ for(y = 0; y < ymax; y++) { if(y > ROWNO && CD) break; curs(xmin,y+2); cl_end(); if(y < ROWNO) { for(x = xmin; x < COLNO; x++) { if((room = &levl[x][y])->new) { room->new = 0; at(x,y,room->scrsym); } else if(room->seen) at(x,y,room->scrsym); } } } if(ymax > ROWNO) { cornbot(xmin-1); if(ymax > ROWNO+1 && CD) { curs(1,ROWNO+3); cl_eos(); } } } curs_on_u(){ curs(u.ux, u.uy+2); } pru() { if(u.udispl && (Invisible || u.udisx != u.ux || u.udisy != u.uy)) /* if(! levl[u.udisx][u.udisy].new) */ if(!vism_at(u.udisx, u.udisy)) newsym(u.udisx, u.udisy); if(Invisible) { u.udispl = 0; prl(u.ux,u.uy); } else if(!u.udispl || u.udisx != u.ux || u.udisy != u.uy) { atl(u.ux, u.uy, u.usym); u.udispl = 1; u.udisx = u.ux; u.udisy = u.uy; } levl[u.ux][u.uy].seen = 1; } #ifndef NOWORM #include "def.wseg.h" extern struct wseg *m_atseg; #endif NOWORM /* print a position that is visible for @ */ prl(x,y) { register struct rm *room; register struct monst *mtmp; register struct obj *otmp; if(x == u.ux && y == u.uy && (!Invisible)) { pru(); return; } if(!isok(x,y)) return; room = &levl[x][y]; if((!room->typ) || (IS_ROCK(room->typ) && levl[u.ux][u.uy].typ == CORR)) return; if((mtmp = m_at(x,y)) && !mtmp->mhide && (!mtmp->minvis || See_invisible)) { #ifndef NOWORM if(m_atseg) pwseg(m_atseg); else #endif NOWORM pmon(mtmp); } else if((otmp = o_at(x,y)) && room->typ != POOL) atl(x,y,otmp->olet); else if(mtmp && (!mtmp->minvis || See_invisible)) { /* must be a hiding monster, but not hiding right now */ /* assume for the moment that long worms do not hide */ pmon(mtmp); } else if(g_at(x,y) && room->typ != POOL) atl(x,y,'\$'); else if(!room->seen || room->scrsym == ' ') { room->new = room->seen = 1; newsym(x,y); on_scr(x,y); } room->seen = 1; } char news0(x,y) register xchar x,y; { register struct obj *otmp; register struct trap *ttmp; struct rm *room; register char tmp; room = &levl[x][y]; if(!room->seen) tmp = ' '; else if(room->typ == POOL) tmp = POOL_SYM; else if(!Blind && (otmp = o_at(x,y))) tmp = otmp->olet; else if(!Blind && g_at(x,y)) tmp = '\$'; else if(x == xupstair && y == yupstair) tmp = '<'; else if(x == xdnstair && y == ydnstair) tmp = '>'; else if((ttmp = t_at(x,y)) && ttmp->tseen) tmp = '^'; else switch(room->typ) { case SCORR: case SDOOR: tmp = room->scrsym; /* %% wrong after killing mimic ! */ break; case HWALL: tmp = '-'; break; case VWALL: tmp = '|'; break; case LDOOR: case DOOR: tmp = '+'; break; case CORR: tmp = CORR_SYM; break; case ROOM: if(room->lit || cansee(x,y) || Blind) tmp = '.'; else tmp = ' '; break; /* case POOL: tmp = POOL_SYM; break; */ default: tmp = ERRCHAR; } return(tmp); } newsym(x,y) register x,y; { atl(x,y,news0(x,y)); } /* used with wand of digging (or pick-axe): fill scrsym and force display */ /* also when a POOL evaporates */ mnewsym(x,y) register x,y; { register struct rm *room; char newscrsym; if(!vism_at(x,y)) { room = &levl[x][y]; newscrsym = news0(x,y); if(room->scrsym != newscrsym) { room->scrsym = newscrsym; room->seen = 0; } } } nosee(x,y) register x,y; { register struct rm *room; if(!isok(x,y)) return; room = &levl[x][y]; if(room->scrsym == '.' && !room->lit && !Blind) { room->scrsym = ' '; room->new = 1; on_scr(x,y); } } #ifndef QUEST prl1(x,y) register x,y; { if(u.dx) { if(u.dy) { prl(x-(2*u.dx),y); prl(x-u.dx,y); prl(x,y); prl(x,y-u.dy); prl(x,y-(2*u.dy)); } else { prl(x,y-1); prl(x,y); prl(x,y+1); } } else { prl(x-1,y); prl(x,y); prl(x+1,y); } } nose1(x,y) register x,y; { if(u.dx) { if(u.dy) { nosee(x,u.uy); nosee(x,u.uy-u.dy); nosee(x,y); nosee(u.ux-u.dx,y); nosee(u.ux,y); } else { nosee(x,y-1); nosee(x,y); nosee(x,y+1); } } else { nosee(x-1,y); nosee(x,y); nosee(x+1,y); } } #endif QUEST vism_at(x,y) register x,y; { register struct monst *mtmp; return((x == u.ux && y == u.uy && !Invisible) ? 1 : (mtmp = m_at(x,y)) ? ((Blind && Telepat) || canseemon(mtmp)) : 0); } #ifdef NEWSCR pobj(obj) register struct obj *obj; { register int show = (!obj->oinvis || See_invisible) && cansee(obj->ox,obj->oy); if(obj->odispl){ if(obj->odx != obj->ox || obj->ody != obj->oy || !show) if(!vism_at(obj->odx,obj->ody)){ newsym(obj->odx, obj->ody); obj->odispl = 0; } } if(show && !vism_at(obj->ox,obj->oy)){ atl(obj->ox,obj->oy,obj->olet); obj->odispl = 1; obj->odx = obj->ox; obj->ody = obj->oy; } } #endif NEWSCR unpobj(obj) register struct obj *obj; { /* if(obj->odispl){ if(!vism_at(obj->odx, obj->ody)) newsym(obj->odx, obj->ody); obj->odispl = 0; } */ if(!vism_at(obj->ox,obj->oy)) newsym(obj->ox,obj->oy); } seeobjs(){ register struct obj *obj, *obj2; for(obj = fobj; obj; obj = obj2) { obj2 = obj->nobj; if(obj->olet == FOOD_SYM && obj->otyp >= CORPSE && obj->age + 250 < moves) delobj(obj); } for(obj = invent; obj; obj = obj2) { obj2 = obj->nobj; if(obj->olet == FOOD_SYM && obj->otyp >= CORPSE && obj->age + 250 < moves) useup(obj); } } seemons(){ register struct monst *mtmp; for(mtmp = fmon; mtmp; mtmp = mtmp->nmon){ if(mtmp->data->mlet == ';') mtmp->minvis = (u.ustuck != mtmp && levl[mtmp->mx][mtmp->my].typ == POOL); pmon(mtmp); #ifndef NOWORM if(mtmp->wormno) wormsee(mtmp->wormno); #endif NOWORM } } pmon(mon) register struct monst *mon; { register int show = (Blind && Telepat) || canseemon(mon); if(mon->mdispl){ if(mon->mdx != mon->mx || mon->mdy != mon->my || !show) unpmon(mon); } if(show && !mon->mdispl){ atl(mon->mx,mon->my, (!mon->mappearance || u.uprops[PROP(RIN_PROTECTION_FROM_SHAPE_CHANGERS)].p_flgs ) ? mon->data->mlet : mon->mappearance); mon->mdispl = 1; mon->mdx = mon->mx; mon->mdy = mon->my; } } unpmon(mon) register struct monst *mon; { if(mon->mdispl){ newsym(mon->mdx, mon->mdy); mon->mdispl = 0; } } nscr() { register x,y; register struct rm *room; if(u.uswallow || u.ux == FAR || flags.nscrinh) return; pru(); for(y = scrly; y <= scrhy; y++) for(x = scrlx; x <= scrhx; x++) if((room = &levl[x][y])->new) { room->new = 0; at(x,y,room->scrsym); } scrhx = scrhy = 0; scrlx = COLNO; scrly = ROWNO; } /* 100 suffices for bot(); no relation with COLNO */ char oldbot[100], newbot[100]; cornbot(lth) register int lth; { if(lth < sizeof(oldbot)) { oldbot[lth] = 0; flags.botl = 1; } } bot() { register char *ob = oldbot, *nb = newbot; register int i; extern char *eos(); if(flags.botlx) *ob = 0; flags.botl = flags.botlx = 0; #ifdef GOLD_ON_BOTL (void) sprintf(newbot, "Level %-2d Gold %-5lu Hp %3d(%d) Ac %-2d Str ", dlevel, u.ugold, u.uhp, u.uhpmax, u.uac); #else (void) sprintf(newbot, "Level %-2d Hp %3d(%d) Ac %-2d Str ", dlevel, u.uhp, u.uhpmax, u.uac); #endif GOLD_ON_BOTL if(u.ustr>18) { if(u.ustr>117) (void) strcat(newbot,"18/**"); else (void) sprintf(eos(newbot), "18/%02d",u.ustr-18); } else (void) sprintf(eos(newbot), "%-2d ",u.ustr); #ifdef EXP_ON_BOTL (void) sprintf(eos(newbot), " Exp %2d/%-5lu ", u.ulevel,u.uexp); #else (void) sprintf(eos(newbot), " Exp %2u ", u.ulevel); #endif EXP_ON_BOTL (void) strcat(newbot, hu_stat[u.uhs]); if(flags.time) (void) sprintf(eos(newbot), " %ld", moves); if(strlen(newbot) >= COLNO) { register char *bp0, *bp1; bp0 = bp1 = newbot; do { if(*bp0 != ' ' || bp0[1] != ' ' || bp0[2] != ' ') *bp1++ = *bp0; } while(*bp0++); } for(i = 1; idata->mlevel, mtmp->mgold, mtmp->mhp, mtmp->mhpmax, mtmp->data->ac, (mtmp->data->damn + 1) * (mtmp->data->damd + 1)); } #endif WAN_PROBING cls(){ if(flags.toplin == 1) more(); flags.toplin = 0; clear_screen(); flags.botlx = 1; } | 4,379 | 11,744 | {"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} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.30885 |
https://www.jobilize.com/online/course/quadratic-functions-by-openstax?qcr=www.quizover.com&page=5 | 1,607,160,538,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141747323.98/warc/CC-MAIN-20201205074417-20201205104417-00582.warc.gz | 713,933,327 | 21,673 | # Quadratic concepts -- multiplying binomials (Page 6/1)
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This module teaches about multiplying binomials. Specifically about common patterns that can be memorized and using the "FOIL" method.
The following three formulae should be memorized.
${\left(x+a\right)}^{2}={x}^{2}+2\text{ax}+{a}^{2}$
${\left(x-a\right)}^{2}={x}^{2}-2\text{ax}+{a}^{2}$
$\left(x+a\right)\left(x-a\right)={x}^{2}-{a}^{2}$
It is important to have these three formulae on the top of your head. It is also nice to be able to show why these formulae work, for instance by using FOIL. But the most important thing of all is knowing what these three formulae mean, and how to use them .
These three are all “algebraic generalizations,” as discussed in the first unit on functions. That is, they are equations that hold true for any values of $x$ and $a$ . It may help if you think of the second equation above as standing for:
${\left(\text{Anthing}-\text{Anything Else}\right)}^{2}={\text{Anything}}^{2}-2{\left(\text{Anything Else}\right)}^{2}$ For instance, suppose the Anything (or $x$ ) is 5, and the Anything Else (or $a$ ) is 3.
${\left(x-a\right)}^{2}={x}^{2}-2\text{ax}+{a}^{2}$ , when $x=5$ , $a=3$ .
• $5-{3}^{2}\stackrel{?}{=}{5}^{2}-2\left(3\right)\left(5\right)+{3}^{2}$
• ${2}^{2}\stackrel{?}{=}\text{25}-\text{30}+9$
• $4=4$
It worked! Now, let’s leave the Anything as $x$ , but play with different values of $a$ .
More examples of ${\left(x-a\right)}^{2}={x}^{2}-2\text{ax}+{a}^{2}$
$\begin{array}{cccc}a=1:& \left(x-1{\right)}^{2}& =& {x}^{2}-2x+1\\ a=2:& \left(x-2{\right)}^{2}& =& {x}^{2}-4x+4\\ a=3:& \left(x-3{\right)}^{2}& =& {x}^{2}-6x+9\\ a=5:& \left(x-5{\right)}^{2}& =& {x}^{2}-10x+25\\ a=10:& \left(x-10{\right)}^{2}& =& {x}^{2}-20x+100\end{array}$
Once you’ve seen a few of these, the pattern becomes evident: the number doubles to create the middle term (the coefficient of $x$ ), and squares to create the final term (the number).
${\left(2y-6\right)}^{2}$
There are three ways you can approach this.
${\left(2y-6\right)}^{2}$ , computed three different ways
Square each term FOIL Using the formula above
$\begin{array}{cc}& {\left(2y-6\right)}^{2}\\ & {\left(2y\right)}^{2}-2\left(6\right)\left(2y\right)+{6}^{2}\\ & {4y}^{2}-\text{24}y+\text{36}\end{array}$ $\begin{array}{cc}& \left(2y-6\right)\left(2y-6\right)\\ & \left(2y\right)\left(2y\right)-\left(2y\right)6-\left(2y\right)6+\text{36}\\ & {4y}^{2}-\text{12}y-\text{12}y+\text{36}\\ & {4y}^{2}-\text{24}y+\text{36}\end{array}$ $\begin{array}{cc}& {\left(2y-6\right)}^{2}\\ & {\left(2y\right)}^{2}-2\left(6\right)\left(2y\right)+{6}^{2}\\ & {4y}^{2}-\text{24}y+\text{36}\end{array}$
Did it work? If a formula is true, it should work for any $y$ -value; let’s test each one with $y=5$ . (Note that the second two methods got the same answer, so we only need to test that once.)
$\begin{array}{ccc}{\left(2y-6\right)}^{2}& \stackrel{?}{=}& {4y}^{2}-\text{36}\\ {\left(2\cdot 5-6\right)}^{2}& \stackrel{?}{=}& {4y}^{2}-\text{36}\\ {\left(\text{10}-6\right)}^{2}& \stackrel{?}{=}& \text{100}-\text{36}\\ {4}^{2}& \stackrel{?}{=}& \text{64}✗\end{array}$ $\begin{array}{ccc}{\left(2y-6\right)}^{2}& \stackrel{?}{=}& {4y}^{2}-\text{24}y+\text{36}\\ {\left(2\cdot 5-6\right)}^{2}& \stackrel{?}{=}& 4{\left(5\right)}^{2}-\text{24}\cdot 5+\text{36}\\ {\left(\text{10}-6\right)}^{2}& \stackrel{?}{=}& \text{100}-\text{120}+\text{36}\\ {4}^{2}& \stackrel{?}{=}& \text{16}✓\end{array}$
We conclude that squaring each term individually does not work. The other two methods both give the same answer, which works.
The first method is the easiest, of course. And it looks good. ${\left(2y\right)}^{2}$ is indeed ${4y}^{2}$ . And ${6}^{2}$ is indeed 36. But as you can see, it led us to a false answer —an algebraic generalization that did not hold up.
I just can’t stress this point enough. It sounds like a detail, but it causes errors all through Algebra II and beyond. When you’re adding or subtracting things, and then squaring them, you can’t just square them one at a time. Mathematically, ${\left(x+a\right)}^{2}\ne {x}^{2}+{a}^{2}$ . You can confirm this with numbers all day. ${\left(7+3\right)}^{2}=\text{100}$ , but ${7}^{2}+{3}^{2}=\text{58}$ . They’re not the same.
So that leaves the other two methods. FOIL will never lead you astray. But the third approach, the formula, has three distinct advantages.
1. The formula is faster than FOIL.
2. Using these formulae is a specific case of the vital mathematical skill of using any formula—learning how to plug numbers and variables into some equation that you’ve been given, and therefore understanding the abstraction that formulae represent.
3. Before this unit is done, we will be completing the square, which requires running that particular formula backward —which you cannot do with FOIL.
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
in a comparison of the stages of meiosis to the stage of mitosis, which stages are unique to meiosis and which stages have the same event in botg meiosis and mitosis
Got questions? Join the online conversation and get instant answers! | 2,499 | 8,235 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 34, "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} | 4.6875 | 5 | CC-MAIN-2020-50 | latest | en | 0.790079 |
https://searchessayhelp.com/game-of-cards/ | 1,656,440,209,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103573995.30/warc/CC-MAIN-20220628173131-20220628203131-00349.warc.gz | 575,589,148 | 30,545 | BETWEENIES CARD GAME
Introduction
The exact rules vary, the main concept is that the player is dealt two cards and bets on whether the value of a third card dealt about to be dealt will be between the values of the two previously dealt cards. In this work we calculate the probabilities associated with the game, namely, the probability that a given hand is dealt and the probability of winning given the hand dealt, and i suggest a betting strategy based on Kelly’s criterion (KC).
HOW TO PLAY
Each player is first dealt 2 cards face down. Taking it in turns, players turns their cards over. Each player must bet a certain number of fingers of drink on whether they think the next card turned over from the pack will have a value in-between the two cards they have in front of them (aces are low in this game).
If the card turned over is equal to either of the two cards then the player must drink double the stake bet. This is called “Hitting the posts”.
Cards are left turned up on the table until everyone has had one turn each. The whole deck is then re-shuffled.
Strategy
To maximize your bets, bet when there are at least 8 cards between you two. For example, 2 & J…3 & Q….4 & K…5 & A.
If your cards are closer together, pass or bet zero.
The Rules
The game is played in rounds and with a standard deck of 52 cards. The cards from 2 to 10 are associated with their face values, while Jack, Queen, and King with the values 11, 12, and 13, respectively. Aces can be associated with either 1 or 14, subject to further rules stated below. In the beginning of the round, every player contributes a fixed amount of money (henceforth assumed to be equal to 1), known as the “ante,” to the pot in order to play. Subsequently, each player is dealt two cards, one at a time, face up.
Assuming that the first card is an ace, the player must declare it “low” (of value 1) or “high” (of value 14).
Assuming that the second card is an ace, it is automatically declared “high.” After the two cards are dealt to a player, the player bets an amount of money, ranging from 0 to the pot amount, that the third card dealt will be strictly between the two already dealt cards.
If the player wins, the player receives the amount of money bet from the pot; otherwise, the player contributes the amount of money bet to the pot. If a player’s wealth becomes zero, the player quits the game
Kelly’s Criterion
Put, assuming independent trials in a game of chance, it suggests a betting strategy, based on which a player can expect an exponential increase of his wealth. The rate of this increase is, more precisely, equal to the information gain between the two underlying probability distributions of the game: true outcome probabilities and projected outcome probabilities, as suggested by the advertised odds. We demonstrate it here by repeating the example in Kelly’s original paper , which is worth working out in full detail, as the original exposition is rather terse, omitting, however, the extra information provided by the “private wire” considered therein, which is an additional complication not directly relevant to our discussion.
Description of a General Game and the Classical Approach
Consider a random variable X with n possible mutually exclusive outcomes xi with probabilities pi, i = 1,…..,n. Assume further that the odds placed on x(i) are t(i) ÷ 1, whereby it is meant that, if a player places a bet bi on xi, and xi is indeed the outcome, he will receive a wealth of t(i)b(i) back including the original bet hence the use of the new symbol “÷” instead of the previous “:”); otherwise the bet is lost. How should the various bi be determined? One possible approach is to maximize the expected wealth: assuming that the player’s initial total wealth is w, the total wealth after betting and assuming outcome xi)
Assuming p(i)t(i) ≤ 1 for some i, for any bet with bi = b > 0, the new bet with all bj , j/= i left
unchanged and bi = b −epsilon > 0, e> 0 is at least as profitable; hence the optimal bet can be taken to have bi = 0. Focusing on an i such that p t > 1, for any bet with bi = b < w, the bet with all b j , j/_ i left unchanged and bi _ b _ _ 0 is at least as profitable
This strategy is, however, highly risky, as, with probability 1−pi∗ the bet is lost and the player is ruined. Furthermore, the probability that the player is not ruined after m rounds of the game is _pi∗_m; assuming that pi∗ 0 may actually be advantageous.
Simulations of card Gameplay and Short-Term Considerations to explain reasonable.
The mean wealth after one round given a hand of spread s ≥ 7 and initial wealth w is
Fraction eventually decreases towards 29151/28561 ≈ 1.02066 as n increases.
Figure A sequence of rounds of Betweenies (party version) leading to ruin (a) and a large wealth (b), with a starting wealth of w = 50. Note the large lost bet in the final rounds of (b).
Figure: shows two actual games of the party version, one of which resulted in ruin and one in a very large wealth over 1000 rounds (which could have been even larger had it not been for a large lost bet in the final rounds.)
We assume that the player played solo with an infinite pot. The player’s initial wealth was taken to be 50. The simulation on the left is typical of games resulting in ruin and gives some insight into the mechanisms that cause ruin. More specifically, we see that the player started by doing well, and three successful bets in rounds 61, 65, and 71 boosted his wealth to 386, an almosteight-fold increase. In round 77, however, an unsuccessful large bet reduced his wealth to 152, followed by two more unsuccessful sizeable bets in rounds 78 and 79, finally reducing his wealth to a meager 74. In short, ruin was partly caused by large unsuccessful bets, namely, localized large losses. Furthermore, long periods where the wealth slope is equal to −1 are clearly visible in the figure, and they correspond to the periods where the player places no bets due to unfavorable hands, but pays the ante in the beginning of each round
These considerations bring us back to the criticism of KC. Is it possible to reduce uncertainty of wealth growth conceding a decrease of the expected final wealth? Let us consider the following two sets of simulations, each consisting of 10,000 games of (at most) 100 rounds each, and where the initial wealth is always w = 50. In the first set bets are placed according to KC: the probability of ruin is 52.13%, the mean wealth is 525.9, and the probability of no gain (end wealth is less than or equal to the initial wealth) is 60.55%, but the median wealth assuming no loss is 392 and the 5% and 95% quantiles lie at 67 and 4,699, respectively. In the second set the bets placed are double the bets suggested by KC, but the player has moderate expectations and withdraws at any point his wealth exceeds 80: the probability of ruin is 35.62%, the mean wealth is 64.34, the probability of no gain is 35.72%, but the median wealth assuming no loss is 94 and the 5% and 95% quantiles lie at 82 and 138, respectively.
Violating KC and hyper betting at low wealth allows the player to leave faster the low wealth zone where ruin is likely to occur due to gradual loss of wealth. Alas, it makes ruin due to sudden loss of wealth much more likely when wealth is high and bets are high… except that now the player pulls out of the game before such high values of wealth are reached! This strategy then outperforms KC in short term, leading to smaller probability of ruin and loss: gain is now small but almost certain, and its effective value is constrained in a much narrower range than before, so that the variation in the expected wealth is smaller.If the player gets greedy and over bets without withdrawing when wealth exceeds 80, the probability of ruin increases to 77.83%.
References
Article on In Between, http://www.onlinepoker.net/Card-Games/In-Between.php
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,831 | 7,978 | {"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} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.973149 |
https://www.physicsforums.com/threads/rolles-theorem-showing-two-distinct-points.581167/ | 1,545,159,493,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376829568.86/warc/CC-MAIN-20181218184418-20181218210418-00101.warc.gz | 989,125,472 | 13,019 | # Rolles Theorem, showing two distinct points.
1. Feb 25, 2012
### srhjnmrg
Have the following question and just wondering if my solution is correct
Let g(x)= x^5+3x-1. Show that there are no distinct points x_1, x_2 in R such that g(x_1)=g(x_2).
Proof by contradiction. Assume we have two solution x_1<x_2 in ℝ, i,e g(x_1)+g(x_2)=0, since g is differentiable on (x_1,x_2) and continuous on [x_1,x_2], then we can apply rolles theorem, there exits a C belonging to (x_1, x_2) such that df/dx=0, however df/dx=5x^4+3>0 Hence we have a contradiction and only one solution to f(x)=0.
Last edited by a moderator: Feb 25, 2012
2. Feb 25, 2012
### micromass
Sounds correct!!
3. Feb 25, 2012
### HallsofIvy
I presume you mean g(x_1)- g(x_2)= 0, not the sum.
4. Feb 25, 2012
### Staff: Mentor
Or you could just take the direct approach. Since g'(x) = 5x4 + 3, we see that g'(x) > 0 for all x, which means that the graph of g is increasing on the entire real line, hence g is one-to-one. This fact implies that if g(x1) = g(x2), then x1 = x2. | 364 | 1,047 | {"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} | 3.515625 | 4 | CC-MAIN-2018-51 | latest | en | 0.907597 |
https://byjus.com/question-answer/what-is-the-sum-of-additive-inverse-and-multiplicative-inverse-of-7/ | 1,685,249,007,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643462.13/warc/CC-MAIN-20230528015553-20230528045553-00309.warc.gz | 186,677,693 | 21,474 | Question
# What is the sum of additive inverse and multiplicative inverse of −7.
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Solution
## The additive inverse of a number a is the number that, when added to a, yields zero.Additive inverse of −7=7The multiplicative inverse of a number for any n is simply 1n.The reciprocal of a number obtained is such that when it is multiplied with the original number the value equals to identity 1.Multiplicative inverse of −7=−17Therefore, required sum is 7−17=487
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## Number 48244
Numbers > Numbers from 48001 to 49000
### mathematical properties and Trigonometric Properties of 48244
48244 is the natural number which follows 48243 and precedes 48245.
48244 It's a even number.
### 48244 on distinct languages
(en) - fourty-eight thousand two hundred and fourty-four
(fr) - quarante huit mille deux cents quarante quatre
(de) - achtundviertausendzweihundertvierundvierzig
(pt) - quarenta e oito mil duzentos quarenta e quatro
### mathematical properties of 48244
Square Root of 48244: 219.64516839667
Square of 48244 :2327483536
Square of 48244 :2327483536
prime factors of 48244 :
divisors of 48244: 1 2 4 7 14 28 1723 3446 6892 12061 24122 48244
### Number 48244 on other bases
48244 in Binary: 1011110001110100
48244 in octal :136164
48244 in hex: bc74
log(e) de 48244 : 10.784026746703
log(10) 48244 : 4.6834433087902
log(1 + 48244 ): 10.784047474454
### Trigonometric Properties of 48244
Sine of 48244 : 0.99124591647493
Cosine of 48244 : -0.13202853127935
Tangent of 48244 : -7.5078159763638
48244 NOT es multiple de 3.
48244 NOT multiple of 5.
48244 es multiple of 7.
48244 NOT multiple of 9.
48244 NOT multiple of 11.
Telefonos por 648244
Quien me llama 948244000
#### Numbers generated Randomly
91 48341 19116 36665 29741 24456 39833 25574 48440 39836 21269 2428 21206 31154 12571 27994 21413 38505 47372 15796 10044 1552 23832 7029 41848 23558 39860 17438 42825 42782 40392 47415 32720 16794 36985 9653 49347 6540 1312 30403 19368 44659 30623 6383 14314 7570 27572 20570 23517 22423 42755 35651 22457 3268 28399 1470 10753 31023 7553 29228 1093 13239 19484 5231 36036 124 5793 46976 3269 9130 29682 37052 17196 27155 21821 4465 35276 47930 25200 15890 11049 26425 15231 30816 31082 39392 19359 23801 39772 22711 7381 42075 4164 11939 28481 4270 31559 20496 23322 46374 19881 36397 22648 40229 28075 16363 39110 24857 48606 43383 41167 40005 44049 21204 34862 311 43970 27829 9472 29363 | 958 | 2,601 | {"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-2020-34 | latest | en | 0.07123 |
https://www.aqua-calc.com/convert/density/stone-per-us-pint-to-gram-per-metric-cup | 1,603,359,415,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879362.3/warc/CC-MAIN-20201022082653-20201022112653-00044.warc.gz | 621,012,781 | 10,855 | Convert stones per pint to grams per (metric cup)
[st/pt to g/metric cup (g/metric c) (st:stone, pt:pint, g:gram, c:cup)
stones per pint to grams per metric cup conversion cards
• 1
through
20
stones per pint
• 1 st/pt to g/metric c = 3 355.13997 g/metric c
• 2 st/pt to g/metric c = 6 710.27994 g/metric c
• 3 st/pt to g/metric c = 10 065.41991 g/metric c
• 4 st/pt to g/metric c = 13 420.55988 g/metric c
• 5 st/pt to g/metric c = 16 775.69985 g/metric c
• 6 st/pt to g/metric c = 20 130.83982 g/metric c
• 7 st/pt to g/metric c = 23 485.97979 g/metric c
• 8 st/pt to g/metric c = 26 841.11976 g/metric c
• 9 st/pt to g/metric c = 30 196.25973 g/metric c
• 10 st/pt to g/metric c = 33 551.3997 g/metric c
• 11 st/pt to g/metric c = 36 906.53967 g/metric c
• 12 st/pt to g/metric c = 40 261.67964 g/metric c
• 13 st/pt to g/metric c = 43 616.81961 g/metric c
• 14 st/pt to g/metric c = 46 971.95958 g/metric c
• 15 st/pt to g/metric c = 50 327.09955 g/metric c
• 16 st/pt to g/metric c = 53 682.23952 g/metric c
• 17 st/pt to g/metric c = 57 037.37949 g/metric c
• 18 st/pt to g/metric c = 60 392.51946 g/metric c
• 19 st/pt to g/metric c = 63 747.65943 g/metric c
• 20 st/pt to g/metric c = 67 102.7994 g/metric c
• 21
through
40
stones per pint
• 21 st/pt to g/metric c = 70 457.93937 g/metric c
• 22 st/pt to g/metric c = 73 813.07934 g/metric c
• 23 st/pt to g/metric c = 77 168.21931 g/metric c
• 24 st/pt to g/metric c = 80 523.35928 g/metric c
• 25 st/pt to g/metric c = 83 878.49925 g/metric c
• 26 st/pt to g/metric c = 87 233.63922 g/metric c
• 27 st/pt to g/metric c = 90 588.77919 g/metric c
• 28 st/pt to g/metric c = 93 943.91916 g/metric c
• 29 st/pt to g/metric c = 97 299.05913 g/metric c
• 30 st/pt to g/metric c = 100 654.1991 g/metric c
• 31 st/pt to g/metric c = 104 009.33907 g/metric c
• 32 st/pt to g/metric c = 107 364.47904 g/metric c
• 33 st/pt to g/metric c = 110 719.61901 g/metric c
• 34 st/pt to g/metric c = 114 074.75898 g/metric c
• 35 st/pt to g/metric c = 117 429.89895 g/metric c
• 36 st/pt to g/metric c = 120 785.03892 g/metric c
• 37 st/pt to g/metric c = 124 140.17889 g/metric c
• 38 st/pt to g/metric c = 127 495.31886 g/metric c
• 39 st/pt to g/metric c = 130 850.45883 g/metric c
• 40 st/pt to g/metric c = 134 205.5988 g/metric c
• 41
through
60
stones per pint
• 41 st/pt to g/metric c = 137 560.73877 g/metric c
• 42 st/pt to g/metric c = 140 915.87874 g/metric c
• 43 st/pt to g/metric c = 144 271.01871 g/metric c
• 44 st/pt to g/metric c = 147 626.15868 g/metric c
• 45 st/pt to g/metric c = 150 981.29865 g/metric c
• 46 st/pt to g/metric c = 154 336.43862 g/metric c
• 47 st/pt to g/metric c = 157 691.57859 g/metric c
• 48 st/pt to g/metric c = 161 046.71856 g/metric c
• 49 st/pt to g/metric c = 164 401.85853 g/metric c
• 50 st/pt to g/metric c = 167 756.9985 g/metric c
• 51 st/pt to g/metric c = 171 112.13847 g/metric c
• 52 st/pt to g/metric c = 174 467.27844 g/metric c
• 53 st/pt to g/metric c = 177 822.41841 g/metric c
• 54 st/pt to g/metric c = 181 177.55838 g/metric c
• 55 st/pt to g/metric c = 184 532.69835 g/metric c
• 56 st/pt to g/metric c = 187 887.83832 g/metric c
• 57 st/pt to g/metric c = 191 242.97829 g/metric c
• 58 st/pt to g/metric c = 194 598.11826 g/metric c
• 59 st/pt to g/metric c = 197 953.25823 g/metric c
• 60 st/pt to g/metric c = 201 308.3982 g/metric c
• 61
through
80
stones per pint
• 61 st/pt to g/metric c = 204 663.53817 g/metric c
• 62 st/pt to g/metric c = 208 018.67814 g/metric c
• 63 st/pt to g/metric c = 211 373.81811 g/metric c
• 64 st/pt to g/metric c = 214 728.95808 g/metric c
• 65 st/pt to g/metric c = 218 084.09805 g/metric c
• 66 st/pt to g/metric c = 221 439.23802 g/metric c
• 67 st/pt to g/metric c = 224 794.37799 g/metric c
• 68 st/pt to g/metric c = 228 149.51796 g/metric c
• 69 st/pt to g/metric c = 231 504.65793 g/metric c
• 70 st/pt to g/metric c = 234 859.7979 g/metric c
• 71 st/pt to g/metric c = 238 214.93787 g/metric c
• 72 st/pt to g/metric c = 241 570.07784 g/metric c
• 73 st/pt to g/metric c = 244 925.21781 g/metric c
• 74 st/pt to g/metric c = 248 280.35778 g/metric c
• 75 st/pt to g/metric c = 251 635.49775 g/metric c
• 76 st/pt to g/metric c = 254 990.63772 g/metric c
• 77 st/pt to g/metric c = 258 345.77769 g/metric c
• 78 st/pt to g/metric c = 261 700.91766 g/metric c
• 79 st/pt to g/metric c = 265 056.05763 g/metric c
• 80 st/pt to g/metric c = 268 411.1976 g/metric c
• 81
through
100
stones per pint
• 81 st/pt to g/metric c = 271 766.33757 g/metric c
• 82 st/pt to g/metric c = 275 121.47754 g/metric c
• 83 st/pt to g/metric c = 278 476.61751 g/metric c
• 84 st/pt to g/metric c = 281 831.75748 g/metric c
• 85 st/pt to g/metric c = 285 186.89745 g/metric c
• 86 st/pt to g/metric c = 288 542.03742 g/metric c
• 87 st/pt to g/metric c = 291 897.17739 g/metric c
• 88 st/pt to g/metric c = 295 252.31736 g/metric c
• 89 st/pt to g/metric c = 298 607.45733 g/metric c
• 90 st/pt to g/metric c = 301 962.5973 g/metric c
• 91 st/pt to g/metric c = 305 317.73727 g/metric c
• 92 st/pt to g/metric c = 308 672.87724 g/metric c
• 93 st/pt to g/metric c = 312 028.01721 g/metric c
• 94 st/pt to g/metric c = 315 383.15718 g/metric c
• 95 st/pt to g/metric c = 318 738.29715 g/metric c
• 96 st/pt to g/metric c = 322 093.43712 g/metric c
• 97 st/pt to g/metric c = 325 448.57709 g/metric c
• 98 st/pt to g/metric c = 328 803.71706 g/metric c
• 99 st/pt to g/metric c = 332 158.85703 g/metric c
• 100 st/pt to g/metric c = 335 513.997 g/metric c
• stone per pint stands for stone per US pint
• stones per pint stands for stones per US pint
Foods, Nutrients and Calories
MICHIGAN CHERRY PIE, UPC: 076185007738 contain(s) 232 calories per 100 grams or ≈3.527 ounces [ price ]
Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Indo-Pacific Black weighs 1 441.7 kg/m³ (90.00239 lb/ft³) with specific gravity of 1.4417 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Clay organic very soft, dry weighs 693 kg/m³ (43.26258 lb/ft³) [ weight to volume | volume to weight | price | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-502, liquid (R502) with temperature in the range of -51.12°C (-60.016°F) to 60°C (140°F)
Weights and Measurements
microjoule per hour (μJ/h) is the SI multiple of the derived power measurement unit joule per second (J/sec) or watt (W).
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kg/l to troy/fl.oz conversion table, kg/l to troy/fl.oz unit converter or convert between all units of density measurement.
Calculators
Calculate volume of a right circular cone and its surface area | 3,059 | 7,034 | {"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-2020-45 | latest | en | 0.18324 |
https://www.coursehero.com/file/6596750/a7/ | 1,496,038,839,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463612018.97/warc/CC-MAIN-20170529053338-20170529073338-00301.warc.gz | 1,050,438,493 | 41,770 | # a7 - xy z 2 on the surface x 2 y 2 z 2 = 1 7 Let f x,y = x...
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Math 237 Assignment 7 Due: Friday, Due Mar 11th * 1. Find the absolute maximum and minimum values of f ( x,y ) = 2 x + x 2 + y 2 - xy 2 in the region bounded by the parabola 2 x + 4 = y 2 and the line x = 2. * 2. Find the points on the surface z = x 2 + y 2 that is closest to the point (1 , 1 , 0). * 3. Use the method of Lagrange multipliers to find the maximum and minimum values of f ( x,y ) = x on the piriform curve defined by y 2 + x 4 - x 3 = 0. 4. Find the maximum and minimum of f ( x,y ) = x 2 - y 2 on the region x 2 + y 2 1. 5. Find the maximum and minimum of f ( x,y ) = x 3 - 3 x + y 2 + 2 y on the region bounded by the lines x = 0, y = 0, x + y = 1. 6. Use the method of Lagrange multipliers to find the maximum and minimum values of
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Unformatted text preview: xy + z 2 on the surface x 2 + y 2 + z 2 = 1. 7. Let f ( x,y ) = x 2 + y 2-1 2 y . a) Use the method of Lagrange multipliers to find the maximum and minimum points of f ( x,y ) on the curve y = √ 1-2 x 2 . b) Let R be the region bounded by the curve y = √ 1-2 x 2 and the x-axis. Find the maximum and minimum value of f ( x,y ) on the region R . NOTE: Only * questions will be graded...
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## This note was uploaded on 12/05/2011 for the course MATH 237 taught by Professor Wolczuk during the Winter '08 term at Waterloo.
Ask a homework question - tutors are online | 529 | 1,559 | {"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} | 3.65625 | 4 | CC-MAIN-2017-22 | longest | en | 0.787829 |
http://www.algebra.com/cgi-bin/show-question-source.mpl?solution=67620 | 1,371,623,372,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368708142617/warc/CC-MAIN-20130516124222-00073-ip-10-60-113-184.ec2.internal.warc.gz | 319,755,456 | 786 | ```Question 91873
Similarly with second-degree terms, you can just foil out (x + y)^3:
= (x + y)(x + y)(x + y)
= x^2 + 2xy + y^2(x + y)
= You try to solve it.
Alternatively, you could use the Binomial Theorem.``` | 74 | 216 | {"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-2013-20 | latest | en | 0.908845 |
https://forums.unrealengine.com/t/best-way-to-go-about-creating-realistic-ballistics/441717 | 1,716,613,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00478.warc.gz | 221,106,890 | 6,339 | # Best way to go about creating realistic ballistics?
Hi, I’d like to create a realistic ballistic system for my first person shooter, not 100% real life simulation, just some simple penetration and velocity sim.
So far I have my weapon base with a few weapons. I can’t decide between using “projectiles” or “Predict Projectile Path By TraceChannel” (virtual projectiles). Projectiles simulate bullet drop already, not sure if the second option does.
My questions are:
1. Projectiles or virtual projectiles (line trace)? Which one will give me the ability to do something like this:
1. I’ve tried using projectiles with Event Hit to draw a plane that shows the direction of the impact, but i can only get it to trigger one event hit. I can’t get the projectile to keep moving through the object as shown in the image above. Should I use overlap events instead?
If anyone has any suggestions/input that would help me out a lot. Thanks! I hope I explained what I’m trying to do clearly.
There are many ways to do this, but here are two possible methods:
Use overlap all on your projectile. Then on each actor (walls, props, other players) in your game have a value between 0 and 1 that tells how much it stops the projectile. Each time you overlap, add that value to the projectile. Each time you overlap also check to see that the value is less than 1. You can also modify the projectile velocity to slow down the bullet if you wish. When it is equal to or greater than 1, then destroy the projectile.
Another method is to use a multi-line trace. The line trace will return an array of hit objects. You can then loop through in order and pretty much do the same thing we did for the projectile. This will then tell you the last thing it hit and you can ignore any hits in your array that go beyond where the value on the projectile was equal to or greater than 1.
There’s a great tutorial for realistic projectile physics by MipMapGames. It’s pretty realistic and includes air resistance, penetration, etc
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I followed video steps in making the Compute Exit Location function, but it isn’t working. My bullets are just disappearing after they hit something!! لیزر موهای زائد | 471 | 2,190 | {"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-2024-22 | latest | en | 0.934141 |
http://www.slideshare.net/Tommy96/lesson-5-predictability-of-events-date-mining-m30p | 1,469,845,564,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257832475.43/warc/CC-MAIN-20160723071032-00090-ip-10-185-27-174.ec2.internal.warc.gz | 672,974,929 | 28,036 | Upcoming SlideShare
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# Lesson 5: Predictability of Events (date mining) (M30P)
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### Lesson 5: Predictability of Events (date mining) (M30P)
1. 1. Alberta Ingenuity & CMASTE Lesson 5: Predictability of Events Purpose: To see if predictability of events can be improved by having access to a larger data set. Problem: Consider a 64 team basketball tournament with single knockout, and the probability that you can predict the winning team in the final as well as the winner of all 63 games. Hypothesis: Forecasts made by prediction groups tend to be more accurate than those made by individuals. Prediction: The Machine Learning strategy of data mining will be able to improve the success rate of picking game winners by at least 20%. Design: This model is similar to but not identical to the NCAA championship; however we will use the NCAA tournament for the model. (The 2008 tournament format with first round teams is included). Since there are 64 teams in a single knockout tournament, the number of games played forms the geometric series 32 + 16 + 8 + 4 + 2 + 1 , with a sum 1 of 63. The probability of picking the correct final game winner is , if all teams are 64 considered equal, and the probability of picking the winner in every game is the very 63 1 1 small value of or . Do you think the predictability could be improved 2 9.22 × 1018 by adding some other information to you? Specifically, you will receive the predictions of all the other participants, as percentages. The number of different predictions possible for the entire tournament is 2 63 , but this is substantially more than there are people on the planet. Still, if this contest was web-based the number of prediction sheets to be analyzed would be very large. This is where the machine learning technique of data mining can be used to analyze the picks of all participants. Data mining is the process of sorting through large amounts of data using computer-designed queries to find relevant information and/or patterns. Since we are only using a classroom of students this process can be done by hand. Once students’ first prediction sheets are handed in, the data will be compiled by the teacher and these results will be given to each participant to look over. Then each student will fill out a second identical form with their new picks; they may change their selections or not, based on the group predictions. After the tournament is complete, we want to compare the success rate between making your picks alone and making the picks with the extra information from all the other participant selections. Materials: Students are given two tournament sheets with all competing teams paired to show the first round of 32 games one week before the tournament begins. (Note: the sheets do not show the slots for the 2 semifinal games and the final game, but these can easily be fit into the centre area of the sheet.) AICML5PredictabilityofEvents Centre for Machine Learning 1/9
2. 2. Alberta Ingenuity & CMASTE Procedure: 1) Students may do any research they want on team strengths, rankings, etc., but must hand in their first selection sheet completed 3 days before the tournament begins. 2) The teacher will compile all the prediction data from the first selection sheets and distribute this information to students.(The “data mining” compilation sheets are included). Perhaps giving the percentage of students choosing each team winning at each bracket all the way to the final game would be appropriate. 3) Students would then use this data to fill out the second prediction sheet and hand this in the day before the tournament begins. Evidence: After the tournament has concluded, the teacher will count the number of correct predictions for each student on both the first and second prediction sheet and record these at the bottom of the sheets. Analysis: 1) The teacher would then analyze the results of success rates on sheet 1 compared to success rates on sheet 2, student by student. 2) The teacher would find how many students had their rate of success in predictions from sheet 1 to sheet 2: a) increase b) decrease or c) stay the same. As well, the teacher would find the total change for the class, and the percent change for the class as a whole ( ± %) Evaluation: The teacher would then decide if aggregating opinions was more efficient than individual forecasting, for this group of students and if collaborative problem solving produced better results. Synthesis: This tournament is followed by millions in North America and in other parts of the world. If this activity was done on a scale where a very large number of participants took part, the amount of data analysis would be onerous, but this is the nature of true data mining. Using computer-designed queries, the predictions of the large group can readily be made available to each individual. This idea can also be extended beyond sports predictions to areas such as timely and effective use of medical records, better prediction of products that will be in demand, future fashion trends, forecasting stock market trends, etc. Sources: 1. http://research.yahoo.com/node/1898, “Predicting the Future with Basketball Bets”, Chen, Yiling et al, 2008 2. Data Mining and Decision Support, Mladenic, Dunja, et al, pg. 3 – 8 3. www.SampleWords.com, 64 Slot Regional Layout Tournament Brackets 4. http://en.wikipedia.org/wiki/Data_mining AICML5PredictabilityofEvents Centre for Machine Learning 2/9
3. 3. Alberta Ingenuity & CMASTE Here is the tournament format from 2008 with first round teams shown: The next pages are: pg 4) the first blank copy of the 64–team tournament format, to be completed by each individual student and handed in several days before the tournament, pg 5-7) the teacher “data mining” compilation sheets, pg 8) the second blank copy of the 64–team tournament format to be completed after the data from the first sheets has been compiled and distributed to all students, and pg 9) the final evaluation sheet for the teacher. AICML5PredictabilityofEvents Centre for Machine Learning 3/9
4. 4. Alberta Ingenuity & CMASTE Sheet #1 Due Date ________________ Name _____________________ Number Correct Results: = = ______ % 63 games 63 AICML5PredictabilityofEvents Centre for Machine Learning 4/9
5. 5. Alberta Ingenuity & CMASTE Teacher “Data Mining” Compilation Sheets Round 1 Round 2 Round 3 Semi Final Final ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% AICML5PredictabilityofEvents Centre for Machine Learning 5/9
6. 6. Alberta Ingenuity & CMASTE Round 1 Round 2 Round 3 Semi Final Final ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% AICML5PredictabilityofEvents Centre for Machine Learning 6/9
7. 7. Alberta Ingenuity & CMASTE Round 1 Round 2 Round 3 Semi Final Final ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% ___________ ___% __________ ___% __________ ___% __________ ___% ___________ ___% Teacher Note: These sheets allow for all 64 teams to be accounted for at each stage of the tournament, but many of the blanks should not be needed as many teams will not be selected by any students as the tournament progresses. AICML5PredictabilityofEvents Centre for Machine Learning 7/9
8. 8. Alberta Ingenuity & CMASTE Sheet #2 Due Date ________________ Name ____________________ Number Correct Results: = = ______ % 63 games 63 AICML5PredictabilityofEvents Centre for Machine Learning 8/9
9. 9. Alberta Ingenuity & CMASTE Final Evaluation Sheet Number of students in activity _________ Total correct predictions on sheet 1 _________ Total correct predictions on sheet 2 _________ Net change from sheet 1 to sheet 2 ( ± ) __________ Percent change from sheet 1 to sheet 2 ________% Number of students who increased their success rate from sheet 1 to sheet 2 __________ Number of students who decreased their success rate from sheet 1 to sheet 2 __________ Conclusion (individual predictions vs. group predictions) _______________________________________________________________________ _______________________________________________________________________ _______________________________________________________________________ _______________________________________________________________________ AICML5PredictabilityofEvents Centre for Machine Learning 9/9 | 3,254 | 13,589 | {"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 | 4 | CC-MAIN-2016-30 | latest | en | 0.932537 |
http://math.stackexchange.com/questions/282056/from-a-mathematicians-point-of-view-what-is-the-purpose-of-dx-in-int-fx | 1,469,371,875,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257824109.37/warc/CC-MAIN-20160723071024-00128-ip-10-185-27-174.ec2.internal.warc.gz | 159,524,260 | 21,523 | # From a mathematician's point of view, what is the purpose of '$dx$' in $\int f(x)\ dx$?
I've done a bit of searching and found a fairly well written explanation, but at the end, the author noted that this explanation seems to work fine for a physicist's purposes - but a mathematician would groan at it due to oversimplifications or inaccuracies.
Since I first posted this paper, two different people have emailed me to tell me that Real Mathematicians don't do this. Playing with dx in the ways described in this paper is apparently one of those smarmy tricks that physicists use to give headaches to mathematicians.
http://www4.ncsu.edu/unity/lockers/users/f/felder/public/kenny/papers/dx.html
I was also confused when reading it because my Calculus prof last semester said that the chain rule
$$\frac{dy}{dz}=\frac{dy}{dx}\cdot \frac{dx}{dz}$$
cannot be treated as fractions, despite the fact that they look like it, and the $dx$ would not cancel out between the two since you can't do that with differentials. But this article said just the opposite.
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You can think of it as just a symbol to say that your integrating function is a function of the variable $x$. So, you know what I'm saying if I write $\int x^2y+\cos(yx) dx$. – Sigur Jan 19 '13 at 16:18
duplicate of math.stackexchange.com/q/200393/264 – Zev Chonoles Jan 19 '13 at 16:23
Your calculus professor is right. A lot of these little "tricks", however, are allowed in a branch of mathematics known as non-standard analysis. – George V. Williams Jan 19 '13 at 16:25
@ZevChonoles: I disagree with closing this as a duplicate since this question seems to contrast the physicists and the mathematicians way of understanding $dx$. I think something useful can be said here. – Thomas Jan 19 '13 at 16:47
I have heard from several mathematicians that one thing that they didn't like about physics was the "abuse" of mathematics. Physicists are not necessarily bad mathematicians, they (IMO) just use math differently. For them, the precise nature isn't (always) at the core of what they are doing. Since the physicist works with the "real world", he/she doesn't often work with approximations and therefore is not passionately interested in the exact nature of equations and such (hopefully I didn't offend anyone).
I see on the site that you give reference to, that the person thinks about $dx$ as a finite quantity. He talks for example about $x + dx$ as being a real number. This is not necessarily a bad way to think about it, but again it isn't the exact definition.
About the integral: When we write for example $$\int_a^b f(x)\; dx$$ we have defined this symbol as is. We haven't defined the symbol as made up of different individual component that can we written on their own. So even though $f(x)$ in the expression indeed does have meaning on its own, we are not allowed to remove it and just write: $$\int_a^b dx.$$ Now, someone might actually use this notation from time to time, but in "standard basic calculus" this hasn't been defined, and so it doesn't have any meaning. (It doesn't mean that we can think about it in a certain way).
Now all that said, this doesn't mean that there isn't a rationale behind the choice of the different components of the symbol. So, of course, the $f(x)$ is the function. The $a$ and the $b$ are the limits of the integral and the $dx$ tells us to treat $x$ as the variable (we say that we integrate with respect to $x$). Why is this important? As mentioned in another answer, if you consider the integral: $$\int_{a}^{b} x^2y.$$ you wouldn't know if $x$ is the variable of $y$ is the variable (or if both are constants). So that is why we like to "add" the $dx$ or $dy$ to the end.
This isn't the only reason. The definite integral $$\int_a^b f(x)\; dx$$ is defined as a limit: $$\lim_{n\to \infty}\sum_{i=1}^{n} f(x_i)\Delta x.$$ Here $\Delta x$ is a finite number defined as $\Delta x = \frac{b - a}{n}$. And taking the limit $n\to \infty$ we have $\Delta x \to 0$. So we "think" of $dx$ as this infinitely small number. We can "think" of the $\Delta$ as becomming the $d$, even though the $dx$ can't be written on its own.
The same applies to the derivative $$\frac{dx}{dy}.$$ Here again we have just defined this one symbol where the parts of the symbol ($dx$ and $dy$) are not well defined on their own.
All this said, one can actually make a meaning (i.e. define) of $dx$, but when we are just taking about basic calculus, we don't usually do that.
Also note that some mathematicians will just write $\int f(x)$ when they mean $\int f(x) \; dx$ because it is "obvious" that we are integrating with respect to $x$.
You can find more things about the $dx$ in this question: What is $dx$ in integration?
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A possibility is to see $dx$ as a differential form. Then, the integration for differential forms is well defined.
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There are a number of ways to view the $dx$. As Seirios pointed out it might be a differential form, which is the usual way to have it "make sense.", but it's often simply defined as the notation of an integral. Integrating a function $f$ with some measure $\mu$ can be described by the notation $\int f d\mu$ and when the measure used is the Lebesgue measure it is customary to replace $d\mu$ with $dx$. In this way the notation "makes sense" because it is well defined, but the $dx$ itself has no meaning without the explicit context of the integral.
$\frac{dy}{dx}$ means something entirely different, that is the derivative of $y$ with respect to $x$. The reuse of the symbols, and the fact that $\frac{dy}{dx}$ looks like a fraction is only because the symbols were in use long before the appropriate formalism was invented for calculus to make sense rigorously, and at that point there was no turning back.
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Remember, that from a historical perspective, differential calculus was developed first from the "physics" point of view, that is, $\int f(x) dx$ was originally thought of as a sum (the $\int$ sign is actually an $s$): $$\sum f(x_i)\Delta x$$ Similarly, the chain rule makes a lot of sense if used as a fraction, and that was it's original meaning. The problem is that when trying to set mathematics on a rigorous foundation, one could no longer use the original "intuitive" definitions. These "physical" methods help us think about what the definitions mean, but we cannot use them in mathematical proofs.
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Trang chủ » Sequential Output Tracing Questions? Top 72 Best Answers
# Sequential Output Tracing Questions? Top 72 Best Answers
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Sequential Output Tracing Questions
• 1. White and purple bottles are not at either extremes.
• 2. Brown bottle is to the immediate right of violet bottle and to the immediate left of the white bottle.
• 3. White, violet and brown bottles are not at the center.
• 4. Black bottle is to the immediate left of purple bottle.
• 5. Blue bottle is neither second last nor at either extremes.
## Sequential Output Tracing | PART-I: VERBAL REASONING | Section-I: General Mental Ability | Chapter-7
Sequential Output Tracing | PART-I: VERBAL REASONING | Section-I: General Mental Ability | Chapter-7
Sequential Output Tracing | PART-I: VERBAL REASONING | Section-I: General Mental Ability | Chapter-7
## Which of the following steps will be the last but one?
Which of the following step will be the last but one? The numbers are arranged in descending order while the words are arranged in alphabetical order alternately. The position of only one term is altered at each step. Clearly, step V is the last step and step IV is the last but one.
## What is the basic element of a sequential circuit?
The basis of all sequential circuits is a circuit called a flip-flop, and the simplest flip-flop circuit is called the S-R flip-flop, (set and reset). A flip-flop is a circuit, whose output(s) change state for some sequence of inputs, and which remain unchanged until another sequence of inputs is used.
What are sequential elements?
A sequential circuit is a logical circuit, where the output depends on the present value of the input signal as well as the sequence of past inputs. While a combinational circuit is a function of present input only. A sequential circuit is a combination of a combinational circuit and a storage element.
What makes a circuit sequential?
The sequential circuit is a special type of circuit that has a series of inputs and outputs. The outputs of the sequential circuits depend on both the combination of present inputs and previous outputs. The previous output is treated as the present state.
What is the sequential circuit?
Sequential circuits refer to the combinational logic circuits that consist of input variables (X) and logic gates (or Computational circuits) along with the output variable (Z). For example, flip-flops, counter, register, clocks, etc.
What is storage element in sequential circuit?
The storage elements (memory) used in clocked sequential circuits are called flip-flops. A flip-flop is a binary storage device capable of storing one bit of information. In a stable state, the output of a flip-flop is either 0 or 1.
## What is sequential output?
Sequential output is basically output of a machine which is working on a fixed sequence. We start with an input and following a fixed pattern or sequence machine gives us the final output.
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Using the Bureau of Metrology (BOM) raw temperature data given for the 8 cities, as a group we have chosen
3 cities being Adelaide, Sydney and Melbourne to compare and analyse the aims of the project. The aim of the project in a wrap is to analyse the variability and error in weather forecasts for the cities we have chosen cities. Along with the set of data we are also given 5 different multiple linear regression models that might help us improve the temperature of the forecast to get it as close as possible to the observed temperatures given by BOM. We have used the 5 models for every one of the 3 cities and used excel and Minitab to help us get closest results of the forecast temperatures to the observed temperatures. Along with the regression model Minitab produced, it also has given us a P-value; the P-value helps us to find out the significance of that particular constant/coefficient to that specific model, if the P-value is greater than the chosen alpha being alpha=0.05 in our case, that particular constant/coefficient is not significant to that corresponding model. The following is summary and brief descriptions of the outcomes produced by excel:
Y = a+b*ecsp, is the first model we have used to try and improve the temperatures of the forecast, where Y is the improved temperatures, a is a constant and b is the coefficient to the variable ecsp, where ecsp is the European Central Model (given to us part of the BOM data), we went ahead and took the difference between the observed temperatures and the improved temperatures, squared that difference, then took the Sum Of Squares (SOS) and finally calculated the Root Mean Square (RMS) by taking the square root of the SOS and dividing it by the count of the SOS, we then used solver in excel to give us the minimum constant/coefficient for that model, the results are as follows: Y = 2.4603+1.0148*ecsp SOS = 1608.6942 RMS = 2.1139 Y = a+b*laps, is the second model we have used to help us improve the forecast temperatures, where Y is the improved temperatures, a is a constant and b is the coefficient to the variable laps, where laps is the Limited Area Prediction Scheme (also given to us also part of the BOM data), as for the first model we have followed the same steps and got the following results: Y = -0.0362+1.1026*laps SOS = 1865.6611 RMS = 2.2765 Y = a+b*ecsp+c*laps, is the third model we have used, where Y is the improved temperatures, a is a constant, b is the coefficient to the variable ecsp and c is the coefficient to the variable laps, we have went ahead and followed the usual steps, the only difference is that with this model we have combined both the ecsp and laps temperatures to see if it will help us improve the forecast temperatures better than the first 2 models, we got the subsequent results: Y = 1.0492+0.6227*ecsp+0.4476*laps SOS = 1414.5925 RMS = 1.9823 Y = a+b*ecsp+c*laps+d*[T(obs)-3], is the fourth model we have used to help us improve the forecast temperatures, where Y is the improved temperatures, a is a constant, b is the
coefficient to the variable ecsp, c is the coefficient to the variable laps and d is the coefficient to the variable [T(obs)-3], where [T(obs)-3] is the observed temperature 3 days beforehand, as with the other models we followed the usual steps, but in addition to model 3 we added the d coefficient to the variable [T(obs)-3] to further analyse if it were to help us improve the forecast temperatures, the following results are as follows: Y = 1.2504+0.6331*ecsp+0.4553*laps-0.0249*[T(obs)-3] SOS = 1408.1738 RMS = 1.9861 Y = a+b*ecsp+c*laps+d*[T(obs)-3}+e*[T(obs)-4], is the fifth and final model we used to help us improve the forecast temperatures, this model is exactly like model 4, the only difference is that we added an extra coefficient/variable being e and [T(obs)-4] respectively, where [T(obs)-4] is the observed temperature 4 days beforehand, this model is used to also improve on the other models, the following are the results: Y = 1.3021+0.6382*ecsp+0.4522*laps-0.0156*[T(obs)-3]-0.0130*[T(obs)-4] SOS = 1405.4852 RMS = 1.9870 Model 1 2 3 4 5 a/P-value 2.4601/0.000 -0.0363/0.935 1.0491/0.009 1.2503/0.004 1.3020/0.003 b/P-value 1.01480/0.000 1.10263/0.000 0.62272/0.000 0.63309/0.000 0.63823/0.000 c/P-value N/A N/A 0.44764/0.000 0.45532/0.000 0.45216/0.000 d/P-value N/A N/A N/A -0.02490/0.224 -0.01563/0.584 e/P-value N/A N/A N/A N/A -0.01303/0.649
Analysing the results from both excel and Minitab and comparing the 5 models we can see that the fourth model was the most effective model as it has given us the lowest RMS out of all the 5 models, however we can also see that model 5 was also an effective model to the given data as it was only had a difference of 0.009 in the RMS. Having a look at the coefficients/constants in the given Minitab results we can see that the corresponding results for the P-values for model 2 constant a, model 4 coefficient d to the variable [T(obs)-3] and model 5 coefficients d and e to their variables [T(obs)-3] and [T(obs)-4] respectively are all not significant to the respective models, as it has given a P-value greater than the chosen alpha=0.05. Like we have done for Adelaide we have also done for Melbourne and Sydney, the results of the 5 models in order 1-5 are:
Melbourne
Y = 1.5196+1.0520*ecsp SOS = 1104.8829
RMS = 1.7519 Y = 0.4826+1.0678*laps SOS = 1881.2783 RMS = 2.2860 Y = 1.1162+0.8798*ecsp+0.1870*laps SOS = 1071.2407 RMS = 1.7250 Y = 1.3447+0.8871*ecsp+0.1998*laps-0.0287*[T(obs)-3] SOS = 1060.6494 RMS = 1.7237 Y = 1.4750+0.8946*ecsp+0.2037*laps-0.0014*[T(obs)-3}-0.0439*[T(obs)-4] SOS = 1049.0557 RMS = 1.7166 Model 1 2 3 4 5 a/P-value 1.5194/0.000 0.4824/ 0.243 1.1162/0.000 1.3448/0.000 1.4750/0.000 b/P-value 1.05196/0.000 1.06784/0.000 0.87983/0.000 0.88717/0.000 0.89471/0.000 c/P-value N/A N/A 0.18699/0.001 0.19970/0.000 0.20361/0.000 d/P-value N/A N/A N/A -0.02871/0.101 -0.00135/ 0.952 e/P-value N/A N/A N/A N/A -0.01303/ 0.050
Having a look at the results for the 5 models of Melbourne from excel we can see that the most effective model was model 5 as it has given us the lowest RMS. Looking at the results produced by Minitab i.e. the constants/coefficient to their corresponding P-value, we can say that the following constants/coefficients are not significant to their respective models: model 2 constant a, model 4 coefficient d to the variable [T(obs)-3] and model 5 coefficients d and e to their variables [T(obs)-3] and [T(obs)-4] respectively.
Sydney
Y = 0.9217+1.0401*ecsp SOS = 1046.8208 RMS = 1.7052 Y = 1.2618+1.0340*laps SOS = 1788.9680 RMS = 2.2292 Y = 0.7047+0.9383*ecsp+0.1128*laps SOS = 1036.8434 RMS = 1.6971
Y = 0.9822+0.9404*ecsp+0.1271*laps-0.0269*[T(obs)-3] SOS = 1026.9851 RMS = 1.6961 Y = 1.1900+0.9363*ecsp+0.14103*laps-0.0100*[T(obs)-3}-0.0346*[T(obs)-4] SOS = 1020.4495 RMS = 1.6931 Model 1 2 3 4 5 a/P-value 0.9212/ 0.037 1.2618/ 0.034 0.7044/ 0.121 0.9819/ 0.052 1.1897/0.023 b/P-value 1.04015/0.000 1.03398/0.000 0.93842/0.000 0.94058/0.000 0.93648/0.000 c/P-value N/A N/A 0.11274/ 0.065 0.12694/0.040 0.14087/0.025 d/P-value N/A N/A N/A -0.02690/0.202 -0.01005/0.674 e/P-value N/A N/A N/A N/A -0.03460/0.144
Like Melbourne, we can see from the results produced by excel that model 5 was the most effective model for Sydney too as it has given us the lowest RMS figure. Now having a look at the results produced by Minitab and seeing the significance of the constant/variable to the particular model we look at the corresponding P-value and concluded that it was not significant if the P-value was higher than the chosen alpha being 0.05, unlike Adelaide and Melbourne, Sydney had more constant/coefficient that were not significant: model 3 constant a and coefficient c to the variable laps, model 4 constant a and coefficient d to the variable [T(obs)-3] and model 5 coefficients d and e to their variables [T(obs)-3] and [T(obs)-4] respectively. In summary of the results, having a look at all 3 cities and comparing all 5 models and the significance of the constant/coefficients they have given us, we can conclude that model 5 was the most effective in giving us the closest temperatures of the forecast to the observed temperatures given to us by BOM (except for Adelaide where model 4 was the most effective), we can also conclude that the significance of coefficients d and d and e to their corresponding models 4 and 5 were not significant to all 3 cities we have analysed. The city that had the most effect to the regression model 5 was Sydney as it has given the lowest RMS, and for that particular reason it resulted to giving us the closest forecast temperatures to the observed temperatures. | 2,671 | 8,707 | {"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-2019-51 | latest | en | 0.941647 |
https://community.boredofstudies.org/threads/general-thoughts-general-mathematics.269078/page-14 | 1,685,854,639,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00571.warc.gz | 215,884,588 | 19,824 | # General Thoughts: General Mathematics (1 Viewer)
#### luke de celis
##### New Member
Question 19 from multiple choice was hard to comprehend considering the possible answers provided. The question asked us to calculate the total pay a mechanic receives when he works 40 hours @ \$22.35/hour, plus a special allowance of 150/week and double time for 'emergency hours'. However, it states that he only receives the double time when he works for a minimum of 4 hours. It states that he works for 5 hours on one emergency occasion and 1.5 on another thus the double time on 1.5 would not be counted,Thus, many people were getting the answer '1267.5' however it was not provided in the options. I would like to know how people found the real answer
yeh i didnt like this one, but i thought it said, gets a minimum payment of 4hours every call and i think it worked out :s
#### littlemonster24
##### Member
This is exactly how I felt! For someone who didn't study that much, I found it surprisingly easy, which I suppose will bring down my mark because I'm expecting a mark of between 75-80, and if everyone else does well, I'll be scaled down badly. Funfunfun. Oh well. At the end of the day, it's just a bunch of exams, and really, compared to everything else you have in your life, and what you're gonna have, the HSC is really nothing. There are so many more exciting and fascinating things in life, in one's future, and a whole year of our life is wasted away to what'll be a few numbers on a piece of paper when we get our results back.
+1 I agree too!
+ 1
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#### FallopianTubez
##### New Member
1.c
2.c
3.c
4.c
5.c
6.c
7.c
8.c
9.c
10.c
11.c
12.c
13.c
14.c
15.c
16.c
17.c
18.c
19.c
20.c
21.c
22.c
Cheers dude, real funny. Suck me left
#### chauchihuahua
##### New Member
Not saying that I am right, but this is what I did:
22.35 * 40 gives a total of \$894 per week.
He gets \$150 for a call out.
The total number of hours of the call out are 5+1.5 = 6.5
6.5 * 44.70 (which is the double time rate)= 290.55
So, 894+150+290.55 = \$1334.55 which was (c)
Hope that helps! (Hope it's right, haha!) God Bless!
I remember this question and its from the Preliminary Financial Maths syllabus.
With this question, i believe this person is a mechanic who can only work 40 hours a week. The emergency call out is when he needs to perform the work as a priority (like get the work done quicker) so he gets paid more for the "urgent" work he needs to carry out.
I believe he cannot possibly stay another 6.5 hours over 40 hours as he will probably be exhausted idk.
I calculated:
40 hours - 6.5 emergency work hours = 33.5 hours normal pay times by \$22.35 p/h = \$748.725 (normal pay)
6.5 emergency hours double pay times by \$44.70 p/h = \$290.55 (double pay)
33.5 hours + 6.5 hours = 40 hours in total.
Add \$150 allowance for emergency call out
Therefore, \$748.725 + \$290.55 + \$150.00 = \$1189.28 (a) - i think it was.
I could be wrong but i don't know.
#### chauchihuahua
##### New Member
Question 19 from multiple choice was hard to comprehend considering the possible answers provided. The question asked us to calculate the total pay a mechanic receives when he works 40 hours @ \$22.35/hour, plus a special allowance of 150/week and double time for 'emergency hours'. However, it states that he only receives the double time when he works for a minimum of 4 hours. It states that he works for 5 hours on one emergency occasion and 1.5 on another thus the double time on 1.5 would not be counted,Thus, many people were getting the answer '1267.5' however it was not provided in the options. I would like to know how people found the real answer
I remember this question and its from the Preliminary Financial Maths syllabus.
With this question, i believe this person is a mechanic who can only work 40 hours a week. The emergency call out is when he needs to perform the work as a priority (like get the work done quicker) so he gets paid more for the "urgent" work he needs to carry out.
I believe he cannot possibly stay another 6.5 hours over 40 hours as he will probably be exhausted idk.
I calculated:
40 hours - 6.5 emergency work hours = 33.5 hours normal pay times by \$22.35 p/h = \$748.725 (normal pay)
6.5 emergency hours double pay times by \$44.70 p/h = \$290.55 (double pay)
33.5 hours + 6.5 hours = 40 hours in total.
Add \$150 allowance for emergency call out
Therefore, \$748.725 + \$290.55 + \$150.00 = \$1189.28 (a) - i think it was.
I could be wrong but i don't know.
#### Jack Cody
##### New Member
For question 20, and correct me if I'm wrong, but is the answer not a?
I got A as the income at the breakeven point is \$0. And the income at the 500 people attending point is the income minus the cost which would be \$15000. Thus the difference in income is \$15000. Or did I interpret the question incorrectly?
I got the same
#### dydx
##### Member
People over here had trouble remembering how many minutes were in a degree
Serious shit's going down LOL.
By the looks of it mainly problem solving based questions are involved. So unless they have decent English skills and cutting away the fat to get @ the meat of a Q they've got no chance.
But for 2U+ it would be pretty easy; no Calculus
STFU RAAQIM YOU SHIT CUNT 24/120 for trials and 6/40 for half yearly.
#### electricckid
##### New Member
could someone tell me how many i got correct??
1.D
2.A
3.B
4.C
5.B
6.D
7.B
8.D
9.C
10.A
11.B
12.B (had no idea how to do this)
13.A
14.B
15.B
16.D
17.C
18.A
19.D
20.C
21.B (Didnt know how to do this one either)
22.A
Would be heaps appreciated if someone could let me know
#### joezia
##### Member
i think i fucked up badly, what would a raw mark of 65 align to?
#### miss_forgettful
##### New Member
could someone tell me how many i got correct??
1.D
2.A
3.B
4.C
5.B
6.D
7.B
8.D
9.C
10.A
11.B
12.B (had no idea how to do this)
13.A
14.B
15.B
16.D
17.C
18.A
19.D
20.C
21.B (Didnt know how to do this one either)
22.A
Would be heaps appreciated if someone could let me know
Question 10 is actually C. \$2100 * (19.74/100/365) * 39 days = \$2144.29
Question 12 is actually A. You flip the 2y/5 around so it's now 5/2y. Then you change the division sign to a multiplication sign. Now just multiply the fractions together like normal and then simplify.
Question 21 is actually A. You have to find the distance between town A and town B. Remember that Distance/Speed/Time triangle? Find the distance using the stats from Train 1. D = S * T = 90 * 2 = 180km. Now that you have the distance, you can work out the speed of train 2. S = D / T = 180 / 80 minutes. But you have to convert 80 minutes to hours because it's km/h. So 80/60 = 1.3333.... 180 / 1.33.. = 135 km/h
So you got 19/22 congrats ^^
#### miss_forgettful
##### New Member
Does anyone know Question 26 c) ??
"Furniture priced at 20 000 is purchased. Deposit of 15% is paid. The balance is borrowed using a FLAT-RATE LOAN at 19% per annum interest, to be repaid in equal monthly instalments over five years. What will be the amount of each monthly instalment?
So is this annuities and you had to find M? But the formulas include compound interest but this is flat rate... help!
#### 2011_
##### Member
Question 10 is actually C. \$2100 * (19.74/100/365) * 39 days = \$2144.29
Question 12 is actually A. You flip the 2y/5 around so it's now 5/2y. Then you change the division sign to a multiplication sign. Now just multiply the fractions together like normal and then simplify.
Question 21 is actually A. You have to find the distance between town A and town B. Remember that Distance/Speed/Time triangle? Find the distance using the stats from Train 1. D = S * T = 90 * 2 = 180km. Now that you have the distance, you can work out the speed of train 2. S = D / T = 180 / 80 minutes. But you have to convert 80 minutes to hours because it's km/h. So 80/60 = 1.3333.... 180 / 1.33.. = 135 km/h
So you got 19/22 congrats ^^
Agreed. Well done on 19 / 22
#### 2011_
##### Member
Does anyone know Question 26 c) ??
"Furniture priced at 20 000 is purchased. Deposit of 15% is paid. The balance is borrowed using a FLAT-RATE LOAN at 19% per annum interest, to be repaid in equal monthly instalments over five years. What will be the amount of each monthly instalment?
So is this annuities and you had to find M? But the formulas include compound interest but this is flat rate... help!
I can't remember exactly. I think I may have used I = Prn (simple interest formula), then calculated total amount to be repaid (Principal + Interest) then divided that by n number of periods. That may be wrong, I'm not even 100% sure it is what I did on the day - can't remember
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#### miss_forgettful
##### New Member
I can't remember exactly. I think I may have used I = Prn (simple interest formula), then calculated total amount to be repaid (Principal + Interest) then divided that by n number of periods. That may be wrong, I'm not even 100% sure it is what I did on the day - can't remember
Damn! Yeah I'm pretty sure that's right... >.<
#### 2011_
##### Member
I can't remember exactly. I think I may have used I = Prn (simple interest formula), then calculated total amount to be repaid (Principal + Interest) then divided that by n number of periods. That may be wrong, I'm not even 100% sure it is what I did on the day - can't remember
I'm hoping that is right.
#### A-Nonentity
##### Member
I stuffed up financial and bearings because of a mind blank, other then thattttt I think It was pretty fair. | 2,629 | 9,545 | {"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} | 2.890625 | 3 | CC-MAIN-2023-23 | longest | en | 0.9563 |
https://www.fmaths.com/tips/readers-ask-what-is-geometry-in-mathematics.html | 1,623,706,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00386.warc.gz | 715,251,040 | 11,433 | ## What is geometry in simple words?
Geometry is a branch of mathematics that studies the sizes, shapes, positions angles and dimensions of things. Flat shapes like squares, circles, and triangles are a part of flat geometry and are called 2D shapes. These shapes have only 2 dimensions, the length and the width.
## What are the 3 types of geometry?
There are three basic types of geometry: Euclidean, hyperbolic and elliptical.
## What geometry means?
1a: a branch of mathematics that deals with the measurement, properties, and relationships of points, lines, angles, surfaces, and solids broadly: the study of properties of given elements that remain invariant under specified transformations.
## What is geometry and why is it important?
Geometry allows students to connect mapping objects in the classroom to real-world contexts regarding direction and place. Understanding of spatial relationships is also considered important in the role of problem solving and higher-order thinking skills.
## What are 10 geometric concepts?
Mathplanet hopes that you will enjoy studying Geometry online with us!
• Points, Lines, Planes and Angles.
• Proof.
• Perpendicular and parallel.
• Triangles.
• Similarity.
• Right triangles and trigonometry.
• Transformations.
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## How is geometry important?
Geometry helps us in deciding what materials to use, what design to make and also plays a vital role in the construction process itself. Geometrical tools like the protractor, ruler, measuring tape, and much more are used in construction work, astronomy, for measurements, drawing etc.
## What kind of jobs use geometry?
Career Information for Jobs Involving Geometry
• Architect.
• Cartographer and Photogrammetrist.
• Drafter.
• Mechanical Engineer.
• Surveyor.
• Urban and Regional Planner.
## How do you understand geometry?
Geometry is the math of shapes and angles. To understand geometry, it is easier to visualize the problem and then draw a diagram. If you’re asked about some angles, draw them. Relationships like vertical angles are much easier to see in a diagram; if one isn’t provided, draw it yourself.
## How is geometry used today?
Applications of geometry in the real world include computer-aided design for construction blueprints, the design of assembly systems in manufacturing, nanotechnology, computer graphics, visual graphs, video game programming and virtual reality creation.
## Why is it called geometry?
It is one of the oldest branches of mathematics, having arisen in response to such practical problems as those found in surveying, and its name is derived from Greek words meaning “Earth measurement.” Eventually it was realized that geometry need not be limited to the study of flat surfaces (plane geometry ) and rigid
## Who uses geometry?
Aerospace engineers use geometric principles to design military aircraft and spacecraft that will operate well in hazardous conditions. Mechanical engineers design, construct and install mechanical devices. One way they use geometry is to calculate the volume of tanks used in water pumping stations.
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## Why is geometry so hard?
They are required to use their spatial and logical skills instead of the analytical skills they were accustomed to using in Algebra. There are 3 major reasons students struggle with Geometry: 1. They don’t understand and can’t apply the vocabulary to decode the problem.
## Is geometry useful in real life?
How is geometry used in real life? Geometry has many practical uses in everyday life, such as measuring circumference, area and volume, when you need to build or create something. Geometric shapes also play an important role in common recreational activities, such as video games, sports, quilting and food design.
## How do jewelers use geometry?
How they use Geometry. Jewelers use geometry when they cut diamonds. They use angles to make the diamond as valuable as possible. | 796 | 4,070 | {"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} | 4.09375 | 4 | CC-MAIN-2021-25 | longest | en | 0.911789 |
https://www.pcdandf.com/pcdesign/index.php/magazine/10604-thermal-modeling-1602 | 1,718,236,731,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861319.37/warc/CC-MAIN-20240612234213-20240613024213-00858.warc.gz | 854,381,501 | 16,774 | ### MAGAZINE
What determines the temperature of a via? Hint: It’s not the current.
In work published in October 2015, Johannes Adam and I estimated the internal temperature of a via using simulation techniques.1 What we concluded defied all conventional wisdom. It turns out the temperature of a via does not depend on the level of current through it. The via temperature depends on the temperature of the trace(s) it is connected to. This is because the thermal conductivity between via and trace is so good that any excess heat is conducted away from the via to the trace(s). Thus, a large trace carrying a large current needs only a single small via for connection to another similar trace.
If there ever was a conclusion that cried out “show me!” this is it! This month, we report on an empirical investigation that confirms the prior simulation.
The Test Board
The relevant portion of the test board is shown in FIGURE 1. The board is approximately 60-mils thick FR-4. The board contains 0.5 oz. copper nominally plated with 1.0 oz. additional copper. Two traces were compared, one nominally 27 mils wide, the other 200 mils wide. Each trace is 6.0" in length, one-half on the top layer and one-half on the bottom layer. Each trace has a single 10 mil diameter plated via connecting top to bottom. It is important to note the via structure is identical in each trace. The board was supported 2.5" above a plywood surface in still air by four screws at the corners. The board was microsectioned after testing to measure the actual dimensions for the simulation.2
Figure 1. Relevant portion of via test board.
The 10 mil diameter via plated to 1.0 oz. has roughly the same conducting cross-sectional area as the 27 mil trace. That is why the 27 mil wide trace was chosen for this study.
The Simulation
It is difficult to simulate a via test on a board this size. The reason is the small via wall thickness. Think of the board as a three-dimensional object. We will model it in 3D using a wire grid made of many small cubes. Each individual cube is considered homogeneous. Therefore, the set of equations solves for the boundary conditions associated with each cube. The dimension of the cube must be smaller than the smallest dimension modeled (the via wall). The smaller the cube, the more cubes in a given volume. The sizes of the calculational matrices are determined by the number of cubes. The load on the computer CPU is determined by the size of the matrices. To model the portion of the board shown in Figure 1 places a very large burden on even a moderately powerful modern desktop computer.
Therefore, for practical purposes we need to “trim” the model. We do this primarily by trimming two dimensions in the simulation (note: trace width and board thickness are not changed in the model):
1. Shorten the overall trace to 3.5" from 6.0".
2. Reduce the overall width of the board material.
The practical impact of these modifications is to reduce the cooling of the trace being simulated. Therefore, we expect a result that is slightly higher (hotter) than actual, particularly at higher currents.
An important parameter in any thermal model is the thermal conductivity coefficient for the dielectric. Rather than trust estimates on the Web and the sometimes incomplete measures found on datasheets, a sample of board material was sent to C-Therm in Canada. It measured the thermal conductivities as (W/mK):
• In-plane 0.679
• Through-plane 0.512
Simulation Results
TABLE 1 shows the results of the simulation. They are consistent with the results presented in Brooks and Adam in October 2015.1 For a small trace, approximately the same size as the via, the via is cooler than the trace because the via cools more effectively into the dielectric layer of the board than does the trace. For the larger trace carrying more current, heat generated in the via is conducted away from the via onto the trace, resulting in the via being a little hotter than the trace (but not a lot).
Table 1. Via Temperature Simulation Results
FIGURES 2 and 3 are the simulated thermal profiles of the 200 mil trace carrying 8.55A and the 27 mil trace carrying 6.65A, respectively. Note the thermal profile cools a little near the via for the 27 mil trace. The via is the hottest point on the 200 mil trace, but only 7.4% hotter than the trace itself.
Figure 2. Thermal profile of 200 mil trace carrying 8.55A.
Figure 3. Thermal profile of 27 mil trace carrying 6.65A.
Test Procedure
The test equipment list included the following: constant current generator, Dr. Meter DC Power Supply HY3010E; temperature measurements, Omega Instruments data logger model OM-EL-USB-TC-LCD with type K probe; Pico Technologies portable oscilloscope model 2204A; TekPower model TP9605BT digital meter (to check current levels).
The test procedure is fairly straightforward. A constant current is applied to the trace. The current level is checked and confirmed by a second digital meter. Temperature is measured with a thermocouple probe whose tip diameter is specified at 30 Ga. (approximately 10 mil diameter). The probe response time is approximately 1 sec. The probe was calibrated to an ice cube and to boiling water at an elevation of 360' and found to be perfectly calibrated. Various tests were run to verify the probe does not affect the temperature of a trace being measured.
For each test, a constant current is applied to the trace until the temperature stabilizes (approximately 6 min.). Then the temperature is measured and recorded. After the measurements were taken, the board was returned to Prototron Circuits for microsectioning to confirm all dimensions.
Measured Test Results
The measured test results are provided in TABLE 2. Two especially important observations should be made:
1. First, a 6.6A current through the 27 mil wide trace results in a via temperature of 109°C, while a higher current of 8.6A results in a larger (200 mil wide) board results in a much lower via temperature of only 44.5°C. This confirms the trace is controlling the via temperature.
2. The simulation data are very close to the measured data. This gives us confidence the simulation approach is a viable approach for predicting temperatures in complex situations.
Table 2. Measured Results of Via Tests
Conclusion
The results of this evaluation are consistent with, and seem to confirm, the results reported in our earlier paper. It is not the current that determines the temperature of a via; it is the temperature of the associated traces. If the traces are sized correctly for the current level, much smaller and fewer vias are required to transition between trace layers than has been previously believed. This means designers can have much greater flexibility in freeing routing channels underneath the traces on their boards.
References
1. Douglas G. Brooks and Johannes Adam, “How Hot Is My Via?” PCD&F, October 2015, vol. 32, no. 10.
2. The trace widths were essentially as designed. The thicknesses were (top layer) 2.1 mil and (bottom layer) 2.9 mil.
Acknowledgments
This type of study would not have been possible without the cooperation of several people and organizations. In particular, I want to thank my longtime partner Dave Graves (now with Monsoon Solutions in Bellevue, WA) for helping prepare the final artwork for the test board. C-Therm Technologies (Fredericton, New Brunswick) graciously measured the thermal conductivity of the board material to facilitate the simulation. And a special thanks to Prototron Circuits, Inc. (Redmond, WA), which provided the test boards and also the microsectioning work and measurements. And my collaborator on trace thermal issues, Johannes Adam (Leimen, Germany), continues to be a great help in evaluating results.
Douglas Brooks, Ph.D., is owner of UltraCAD Design, a PCB design service bureau and author of PCB Currents: How They Flow, How They React; This email address is being protected from spambots. You need JavaScript enabled to view it.. | 1,736 | 8,018 | {"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} | 2.875 | 3 | CC-MAIN-2024-26 | latest | en | 0.94713 |
https://www.coursehero.com/file/p5l0nj9/The-least-squares-estimates-are-The-estimated-response-of-sales-to-advertising/ | 1,619,063,608,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039560245.87/warc/CC-MAIN-20210422013104-20210422043104-00261.warc.gz | 789,640,132 | 572,120 | The least squares estimates are The estimated response of sales to advertising
The least squares estimates are the estimated
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The least squares estimates are:The estimated response of sales to advertising is:What is the marginal effect of advertising on sales if the advertising expenditure level is \$500?What is the marginal effect of advertising on sales if the advertising expenditure level is \$2000?±2109.727.64012.1512.768se 6.80 1.046 3.556 0.941SALESPRICEADVERTADVERTEq. 5.24Extending the Model for Burger Barn Sales±12.1515.536SALESADVERTADVERT 5.6 Polynomial Equations 5.6.2
Principles of Econometrics, 4t h Edition Page 19 Chapter 5: The Multiple Regression Model 5.6 Polynomial Equations
Principles of Econometrics, 4t h Edition Page 22 Chapter 5: The Multiple Regression Model 5.7 Interaction Variables Suppose that we wish to study the effect of income and age on an individual’s expenditure on pizza An initial model would be: Implications of this model are: 1. : For a given level of income, the expected expenditure on pizza changes by the amount β 2 with an additional year of age 2. : For individuals of a given age, an increase in income of \$1,000 increases expected expenditures on pizza by β 3 1 2 3 β β β PIZZA AGE INCOME e Eq. 5.27 2 β E PIZZA AGE 3 β E PIZZA INCOME Interaction Term
Principles of Econometrics, 4t h Edition Page 23 Chapter 5: The Multiple Regression Model Table 5.4 Pizza Expenditure Data 5.7 Interaction Variables
Principles of Econometrics, 4t h Edition Page 24 Chapter 5: The Multiple Regression Model The estimated model is:The signs of the estimated parameters are as we anticipated: both AGEand INCOMEhave significant coefficients, based on their tstatisticsIs it reasonable to expect that, regardless of the age of the individual, an increase in income by \$1,000 should lead to an increase in pizza expenditure by \$1.83?342.887.5761.832(t) -3.27 3.95PIZZAAGEINCOME - 5.7 Interaction Variables
Principles of Econometrics, 4t h Edition Page 25 Chapter 5: The Multiple Regression Model It is more reasonable to assume that as a person grows older, his marginal propensity to spend on pizza declines That is, the effect of income depends on the age of the individual. One way of accounting for such interactions is to include an interaction variable that is the product of the two variables involved. 5.7 Interaction Variables
Principles of Econometrics, 4t h Edition Page 26 Chapter 5: The Multiple Regression Model We will add the interaction variable ( AGE × INCOME ) to the regression model The new model is: Implications of this revised model are: 1. | 674 | 2,669 | {"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} | 3.546875 | 4 | CC-MAIN-2021-17 | latest | en | 0.842065 |
https://wirings-diagram.com/honda-motorcycle-wiring-diagram/dans-motorcycle-wiring-diagrams-honda-motorcycle-wiring-diagram/ | 1,623,606,003,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487610196.46/warc/CC-MAIN-20210613161945-20210613191945-00176.warc.gz | 576,433,947 | 26,514 | # Dan's Motorcycle "wiring Diagrams" – Honda Motorcycle Wiring Diagram
Dan's Motorcycle "wiring Diagrams" – Honda Motorcycle Wiring Diagram
Honda Motorcycle Wiring Diagram – hero honda motorcycle wiring diagram, honda 125 motorcycle wiring diagram, honda motorcycle alarm wiring diagram, Every electrical arrangement is composed of various different parts. Each part should be set and linked to different parts in specific way. If not, the arrangement won’t function as it should be. In order to be certain that the electric circuit is constructed correctly, Honda Motorcycle Wiring Diagram is demanded. How can this diagram help with circuit construction?
The diagram provides visual representation of an electrical structure. However, the diagram is a simplified version of this structure. It makes the procedure for building circuit simpler. This diagram gives advice of circuit components as well as their own placements.
## Components of Honda Motorcycle Wiring Diagram and Some Tips
There are just two things which are going to be present in any Honda Motorcycle Wiring Diagram. The first element is emblem that indicate electric component in the circuit. A circuit is usually composed by various components. The other thing that you will come across a circuit diagram would be lines. Lines in the diagram show exactly how each element connects to a another.
The rankings of circuit’s elements are relative, not accurate. The order is also not plausible, unlike wiring schematics. Diagram only reveals where to place component at a spot relative to other elements within the circuit. Though it is exemplary, diagram is a fantastic basis for anyone to construct their own circuit.
One thing you have to learn before reading a circuit diagram would be your symbols. Every symbol that’s exhibited on the diagram shows specific circuit element. The most common elements are capacitor, resistorbattery. There are also other elements such as ground, switch, motor, and inductor. Everything depends on circuit that is being built.
As stated previous, the lines at a Honda Motorcycle Wiring Diagram signifies wires. At times, the wires will cross. However, it doesn’t mean link between the cables. Injunction of 2 wires is usually indicated by black dot on the junction of 2 lines. There’ll be main lines that are represented by L1, L2, L3, and so on. Colours can also be utilized to differentiate cables.
Ordinarily, there are two main types of circuit links. The primary one is known as series connection. It is the easier type of relationship because circuit’s elements are placed inside a singular line. Because of the electrical current in each component is comparable while voltage of this circuit is total of voltage in each component.
Parallel link is more complex compared to string one. Unlike in string connection, the voltage of every element is similar. It is because the component is directly linked to electricity resource. This circuit includes branches which are passed by different electrical current levels. The present joins together when the branches meet.
There are several things that an engineer should pay attention to when drawing wirings diagram. To start with, the symbols used in the diagram should be accurate. It should represent the specific element required to build a planned circuit. After the symbol is incorrect or unclear, the circuit won’t function since it’s supposed to.
It’s also highly suggested that engineer brings favorable supply and damaging source symbols for better interpretation. Ordinarily positive supply emblem (+) is located above the line. Meanwhile the negative supply symbol is put below it. The current flows in the left to right.
In addition to this, diagram drawer is advised to limit the amount of line crossing. The line and part placement should be designed to lessen it. But if it is unavoidable, use universal symbol to indicate whether there is a junction or when the lines are not really connected.
Because you can see drawing and interpreting Honda Motorcycle Wiring Diagram can be a complicated job on itself. The information and tips that were elaborated above should be a fantastic kick start, however. Honda Motorcycle Wiring Diagram
Dan's Motorcycle "wiring Diagrams" – Honda Motorcycle Wiring Diagram Uploaded by Hadir on Thursday, February 14th, 2019 in category Wiring Diagram.
Here we have another image Honda Motorcycle Wiring Diagrams | Bikes | Honda, Motorcycle, Honda – Honda Motorcycle Wiring Diagram featured under Dan's Motorcycle "wiring Diagrams" – Honda Motorcycle Wiring Diagram. We hope you enjoyed it and if you want to download the pictures in high quality, simply right click the image and choose "Save As". Thanks for reading Dan's Motorcycle "wiring Diagrams" – Honda Motorcycle Wiring Diagram. | 915 | 4,819 | {"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-2021-25 | latest | en | 0.92541 |
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Sun May 20, 2012 By:
# Prove the given :-
Mon May 21, 2012
Apply C1 -> C1 - C3 and C2 -> C2 - C3
On simplifying the so obtained determinant, you will get,
|(b+c-a) 0 a2|
(a+b+c)2 |0 (c+a-b) b2|
|(c-a-b) (c-a-b) (a+b)2|
Apply R3 -> R3 - (R1 + R2)
And then apply C1 -> aC1+ C3, C2 -> bC2 + C3
On simplification, you will get the required result.
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# If the matrix
Sat July 29, 2017
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# What is perspective?
Wiki User
2011-09-07 11:39:41
Mathematics
Perspective is a mathematical system for projecting the three-dimensional world onto a two-dimensional surface. It allows a flat image to convey a three-dimensional state.
The linear perpective relies on lines of sight converging on either one or two vanishing points. Our position in relation to the base (on which an object sits) affects how much of the objects we see, and what view we have of the space around them.
As a Subjective View
Perspective is an individual's "point of view". It can subjectively color his or her interpretation of events and policies, because he sees how they affect his particular area or group.
Putting art aside for a moment, according to one dictionary, the word perspective originally comes from the Latin words per meaning "through" and specere which means "to look." These are combined to mean "to look through" or "to look at."
However, the meaning of a word can change and usually even splits into several meanings over time. The conventional "art definition" of perspective specifically describes creating the appearance of distance into our art. This emphasis on distance stems from it being a difficult and impressive effect to achieve, especially on paper that is completely flat. Here we are attempting to convey a sense of reality with space and depth on something which has none. As such, the most typical "art definition" of perspective has evolved into:
"The technique of representing a three-dimensional image on a two-dimensional surface."
But being three-dimensional means that an object has height and width, not just depth alone. Despite this, perspective became less about three-dimensional form than obsessing almost exclusively on that third dimension of depth. This is so much the case that it is commonly referred to as depth perspective.
Furthermore, perspective already exists while seeing in reality where no kind of flat surface is involved. There are also perspective art forms that make no use of flat surfaces in their final states such as interior design, landscape design, stage set design, sculpture, architecture or in any kind of display or exhibit.
With that said, "the technique of representing a three-dimensional image on a two-dimensional surface" does not actually explain what perspective is at all, despite any true importance that depth may have.
First understand that our viewpoint is simply that position we are seeing things from. With that in mind, perspective basically means the same as "viewpoint" and "position." For example, "It looks good from my viewpoint," "It looks good from my position" and "It looks good from my perspective."
Oddly, this meaning of perspective is primarily used outside of art.
So the most general definition of perspective is "a position in relation to different positions." For example, this includes the position of our eye in relation to the positions of objects in a scene.
Applying this to art, we do not necessarily mean the viewpoint of the artist in relation to the subject. More specific, what matters is the best perspective for the audience. A more universal "art definition" of perspective, therefore, is "Creating viewpoints that best communicate a subject to an audience."
Perspective is really about establishing "an eye" in your art through which your audience sees. So although it has been considered the most difficult subject in all of art, its concept is quite simple.
Wiki User
2011-09-07 11:39:41
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## Biological basis of culture
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## Which statement about culture is true?
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http://scirundocwiki.sci.utah.edu/SCIRunDocs/index.php/CIBC:Documentation:SCIRun:Reference:SCIRun:BuildTDCSMatrix | 1,620,422,000,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988828.76/warc/CC-MAIN-20210507211141-20210508001141-00061.warc.gz | 41,286,301 | 7,010 | # CIBC:Documentation:SCIRun:Reference:SCIRun:BuildTDCSMatrix
## BuildTDCSMatrix
### Information
Package: SCIRun
Catagory: FiniteElements
Author(s): Moritz Dannhauer
Version: 4.7
### Description
The module BuildTDCSMatrix generates a matrix which formulates the forward problem of transcranial direct current stimulation (tDCS).
#### Detailed Description
The module has six inputs and one output. In order to formulate the tDCS forward problem the description of the FEM mesh, the associated FEM stiffness matrix, the definition of the injecting electrodes and surface impedance information are needed to perform this task.
All inputs can be generated by SCIRun, or created else where or loaded from Matlab files or the Matlab interface.
##### Inputs
1. FEM Stiffness Matrix: $S \in N \times N$, where N = number of nodes, which can be generated with the BuildFEMatrix module, the mesh (second input port) and conductivity information.
2. Tetrahedral Mesh Field:
3. Electrode Element definition Matrix: this column matrix $(E \times 1)$ maps electrode elements to the distinct electrodes that they belong to. Using a simple, artificial two electrode problem as an example, if electrode 1 has 5 elements and electrode 2 has 3 elements, the matrix (E = 8) would be:
$\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ \end{bmatrix} $
4. Matrix Electrode Element Type: a column matrix $(E \times 1)$ which describes the mapping of each Electrode Element to a specific Element Type. Elements can be points (1), triangles (2) or tetrahedral elements (3). Continuing the sample problem, if electrode 1 contains tetrahedral elements and electrode 2 contains a set of points, the matrix (E = 8) would be:
$\begin{bmatrix} 3 \\ 3 \\ 3 \\ 3 \\ 3 \\ 1 \\ 1 \\ 1 \\ \end{bmatrix} $
5. Electrode Element Definition Matrix: this matrix $(E \times 4)$ has a fixed size which is independent of the Electrode Elements Types, and which describes the points that define (based on the mesh supplied to the module from the second input port) the Electrode element. Continuing the artificial sample problem, the matrix (E = 8) would be:
$\begin{bmatrix} 0 & 25 & 0 & 152 \\ 5 & 0 & 0 & 151 \\ 147 & 237 & 0 & 148 \\ 148 & 237 & 0 & 152 \\ 151 & 0 & 0 & 152 \\ 24 & 0 & 0 & 0 \\ 143 & 0 & 0 & 0 \\ 232 & 0 & 0 & 0 \\ \end{bmatrix} $
For last 3 lines only the first element is used, since it is defined as a point electrode. Therefore an arbitrary combination of electrode definitions can be used which should be part of of the input mesh.
6. Contact impedance Matrix: this column matrix $(E \times 1)$ describes the impedance of the electrodes to the surrounding material. For each electrode element a specific impedance can be used. Continuing the artificial sample problem, the matrix (E = 8) would be:
$\begin{bmatrix} 5 \\ 5 \\ 5 \\ 5 \\ 5 \\ 10 \\ 10 \\ 10 \\ \end{bmatrix} $
Hint: Use standard SI units for mesh and electrode definitions (i.e., meters (m)), stiffness matrix conductivity (S/m), contact impedance (Ohm).
##### Output
1. tDCS Forward Matrix
##### Computation
For the 2 electrode sample problem introduced in the previous section, the module generates the following matrices:
$tDCS = \begin{bmatrix} A* & B \\ B' & C \\ \end{bmatrix} $
$A* \in N \times N = S + A$
$B \in N \times 2$
$B' \in 2 \times N$
$C \in 2 \times 2$
$tDCS \in (N + 2) \times (N + 2)$
See this paper for more details.
The module checks the plausibility of the inputs, and looks to see if an reference node is set to 0 after output generation. In that case, an error message will be generated, and the tDCS matrix will not referenced. However, it also contains information how to set up a reference node.
After the tDCS matrix is generated, a linear equation solver (e.g. module: SolveLinearSystem) can be used to solve the unknown potentials u1 (at mesh nodes) for a given injected current pattern r2.
$tDCS * u = rhs$
$u = \begin{bmatrix} u1 & u2 \end{bmatrix}'$
The right hand side vector for the example could be:
$\begin{bmatrix} r1 & r2 \end{bmatrix}' \in (N + 2) \times 1$
$r1 = \begin{bmatrix} 0 & \cdots & 0 \end{bmatrix}' \in N \times 1$
$r2 = \begin{bmatrix} 1 & -1 \end{bmatrix}' \in 2 \times 1$
#### Summary
The very flexible module BuildTDCSMatrix computes the tDCS forward matrix using a serial algorithm.
### Recent Changes
Go back to Documentation:SCIRun:Reference:SCIRun | 1,229 | 4,403 | {"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} | 2.578125 | 3 | CC-MAIN-2021-21 | latest | en | 0.727821 |
http://stackoverflow.com/questions/19475729/fastest-way-to-find-euclidean-distance-in-python | 1,394,920,788,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678699721/warc/CC-MAIN-20140313024459-00096-ip-10-183-142-35.ec2.internal.warc.gz | 127,728,296 | 16,178 | # fastest way to find euclidean distance in python
I have 2 sets of 2D points (A and B), each set have about 540 points. I need to find the points in set B that are farther than a defined distance alpha from all the points in A.
I have a solution, but is not fast enough
``````# find the closest point of each of the new point to the target set
def find_closest_point( self, A, B):
outliers = []
for i in range(len(B)):
# find all the euclidean distances
temp = distance.cdist([B[i]],A)
minimum = numpy.min(temp)
# if point is too far away from the rest is consider outlier
if minimum > self.alpha :
outliers.append([i, B[i]])
else:
continue
return outliers
``````
I am using python 2.7 with numpy and scipy. Is there another way to do this that I may gain a considerable increase in speed?
-
– Inbar Rose Oct 20 '13 at 9:21
Since you seem to look for outliers only, which is a nearest neighbour search. I think you can (and should) use `scipy.spatial.KDTree` (the old cKDTree probably does not support that, so use a newer scipy where cKDTree does). – seberg Oct 20 '13 at 10:09
– ali_m Oct 20 '13 at 11:07
``````>>> from scipy.spatial.distance import cdist
>>> A = np.random.randn(540, 2)
>>> B = np.random.randn(540, 2)
>>> alpha = 1.
>>> ind = np.all(cdist(A, B) > alpha, axis=0)
>>> outliers = B[ind]
``````
gives you the points you want.
-
Actually this gives me a set of True and False, though I think that I got an idea of what you wanted to do, and actually is a huge speed improvement. – api55 Oct 20 '13 at 9:58
@api55: updated the answer. – larsmans Oct 20 '13 at 10:05
now it works fine, and i got a big increase in speed aprox. from 9.5 to 0.6 for 10 tries of the whole algorithm, thank you – api55 Oct 20 '13 at 10:20 | 511 | 1,746 | {"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} | 3.15625 | 3 | CC-MAIN-2014-10 | latest | en | 0.87289 |
https://control.com/textbook/chemistry/periodic-table-of-the-elements/ | 1,701,591,639,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100489.16/warc/CC-MAIN-20231203062445-20231203092445-00170.warc.gz | 208,121,025 | 20,496 | Chapter Chemistry in Industrial Instrumentation
# Periodic Table of the Elements
All substances are comprised of various elements in various combinations and proportions. Elements may thus be thought of as the building-blocks of matter. A Periodic Table of the Elements is a table listing the known elements in order of their atomic numbers.
Multiple attributes appear for each element in the table. Two of these attributes – atomic number and atomic mass – are directly related to the number of particles in the nucleus of each atom. We will examine the table’s entry for the element potassium (K) to explore these concepts.
Potassium has an atomic number (number of protons in the nucleus of each potassium atom) of 19. This number defines the element. If we were somehow to add or subtract protons from the nucleus of a potassium atom, it would cease being potassium and transmutate into a different element. Note how every element in the table has its own unique atomic number, and how each of these numbers is whole (no fractions or decimals).
The atomic mass or atomic weight shown for potassium is 39.0983. This quantity is the sum of protons and neutrons found in the nucleus of each potassium atom. Like the atomic number (19), we would logically expect the atomic mass to be a whole number as well, since protons and neutrons only come in whole quantities. The primary reason we see a non-whole number for potassium’s atomic mass is that this table reflects the average atomic mass of potassium atoms as found in nature. Some potassium atoms have atomic masses greater than 39, and some have atomic masses less than 39. We know that the number of protons in every potassium atom is fixed (which is what gives potassium its elemental identity), which means the only quantity that may cause the atomic mass to vary is the number of neutrons in the nucleus. The most common form of potassium ($$^{39}$$K) atom possesses 19 protons and 20 neutrons in its nucleus, giving it an atomic mass of 39 (19 + 20). The next most common form of potassium found on Earth is ($$^{41}$$K), possessing 19 protons and 22 neutrons.
We refer to atoms of the same element with differing atomic masses as isotopes. From a chemical perspective, isotopes are identical. That is to say, they engage in the exact same chemical reactions in the exact same manner. To use potassium as an example, an atom of $$^{39}$$K will join with a chlorine atom (Cl) to form the compound potassium chloride (KCl) just as readily as an atom of $$^{41}$$K will join with a chlorine atom to form the same compound. The three isotopes of hydrogen ($$^{1}$$H, $$^{2}$$H, and $$^{3}$$H: hydrogen, deuterium, and tritium, respectively) are all chemically identical: all are highly flammable, combining with oxygen to create water (H$$_{2}$$O). However, deuterium ($$^{2}$$H) has twice the density of normal hydrogen ($$^{1}$$H), while tritium ($$^{3}$$H) has three times the density of normal hydrogen and is highly radioactive! Isotopes only differ in their mass and in their nuclear properties (such as radioactivity: the tendency for a nucleus to spontaneously decay, usually resulting in a loss or gain of protons that subsequently alters the identity of the decayed atom).
The Periodic Table is called “periodic” because its configuration reveals a repeating pattern of chemical behaviors approximately following atomic number. Horizontal rows in the table are called periods, while vertical columns are called groups. Elements in the same group (vertical column) share similar chemical reactivities – that is, they tend to engage in the same types of chemical reactions – despite having different masses and physical properties such as melting point, boiling point, etc. This periodicity is a function of how electrons are arranged around the nucleus of each atom, a subject we will explore in more detail later in this chapter. As mentioned previously, chemistry is the study of how atoms bond together to form molecules, and this bonding takes place through the interaction of the electrons surrounding the atoms’ nuclei. It makes perfect sense, then, that the configuration of those electrons determine the chemical (bonding) properties of atoms.
Some periodic tables show the first ionization energy value for each element – the amount of energy required to force the first electron of an electrically balanced atom to separate from that atom – in addition to other attributes such as atomic number and atomic mass. If we note the ionization energies of the elements, reading each element in turn from left-to-right, starting with period 1 (hydrogen and helium) and progressing to subsequent periods, we see an interesting pattern:
& & (measured in electron-volts'') \cr
Element Period First ionization energy Hydrogen (H) 1 13.5984 Helium (He) 1 24.5874 Lithium (Li) 2 5.3917 Beryllium (Be) 2 9.3227 Boron (B) 2 8.298 Carbon (C) 2 11.2603 Nitrogen (N) 2 14.5341 Oxygen (O) 2 13.6181 Fluorine (F) 2 17.4228 Neon (Ne) 2 21.5645 Sodium (Na) 3 5.1391 Magnesium (Mg) 3 7.6462 Aluminum (Al) 3 5.9858 Silicon (Si) 3 8.1517 Phosphorus (P) 3 10.4867 Sulfur (S) 3 10.36 Chlorine (Cl) 3 12.9676 Argon (Ar) 3 15.7596 Potassium (K) 4 4.3407
First ionization energy represents the relative stability of the last electron balancing the electrical charge of an atom. We see from this table that 24.5874 electron-volts of energy is needed to remove one electron from an electrically-balanced atom of helium (changing He into He$$^{1+}$$), while only 13.5984 electron-volts of energy is required to do the same to an atom of hydrogen. This tells us the electron configuration of helium is at a lower energy (and therefore more stable) than that of hydrogen.
The ionization energies increase with increasing atomic number (with an occasional down-step) until the last column of the period is reached, and then there is a comparatively enormous down-step in energy at the first column of a new period. This pattern is clearly evident when the first ionization energies are plotted against atomic number:
This periodicity suggests that as atoms grow in atomic number, the additional electrons do not simply pile on in random fashion or in a plain and simple progression from inner orbits to outer orbits. Rather, they “fill in” a structured energy pattern, with major changes in structure at the start of each new period. More details of this structured pattern will be explored later in this chapter.
The low ionization energy values for all the “Group 1” elements (far left-hand column) suggest they are relatively easy to positively ionize, and indeed we find this to be the case through experimentation. Hydrogen, lithium, sodium, potassium, and the rest all readily become positively-charged ions upon interaction with other atoms, since their low ionization energy values means they may easily lose an electron.
The high ionization energy values for all the “Group 18” elements (far right-hand column) suggest they possess a very stable electron structure, which is also verified by experiment. These are the noble elements, possessing very little reactive potential.
Looking at the “Group 17” column, just to the left of the noble elements, we notice that they are all just one electron shy of the stable electron structure enjoyed by the noble atoms when in their electrically-balanced states. This suggests it might be easy to add one more electron to atoms of these elements, which (once again!) is a principle validated by experiment. Fluorine, chlorine, bromine, iodine, and even astatine all readily ionize negatively, readily accepting an extra electron from surrounding atoms. As one might expect from this tendency, these elements readily bond through electrostatic attraction with the “Group 1” elements (hydrogen, lithium, sodium, potassium, etc.), each “Group 17” atom accepting an extra electron from each “Group 1” atom which readily provides it. Ordinary table salt (sodium chloride, or NaCl) is an example of a compound formed by this sort of bond.
Thus, Group 1 and Group 17 elements are both highly reactive in a chemical sense, but in different ways. Group 1 elements easily form bonds with Group 17 elements, but Group 1 elements do not generally bond (solely) with other Group 1 elements, and likewise Group 17 elements do not generally bond (solely) with other Group 17 elements. It is the structure of the electrons surrounding each atom’s nucleus that determines how those atoms will bond with other atoms.
#### Lessons in Industrial Automation
Published under the terms and conditions of the Creative Commons Attribution 4.0 International Public License | 1,950 | 8,656 | {"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} | 2.734375 | 3 | CC-MAIN-2023-50 | latest | en | 0.89633 |
https://www.jiskha.com/display.cgi?id=1326487102 | 1,516,745,729,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892699.72/warc/CC-MAIN-20180123211127-20180123231127-00157.warc.gz | 911,140,171 | 4,092 | # Finance
posted by .
Find the future value one year from now of a \$7,000 investment at
a 3 percent annual compound interest rate. Also calculate the future
value if the investment is made for two years.
• Finance -
Pt = Po(1+r)^n.
r = 3%/100% = 0.03 = APR expressed as a decimal.
n = 1comp/yr * 1yr = 1 = # of compound-
ing period.
Pt = 7000(1.03)^1 =
Pt = 7000(1.03)^2 =
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find the formula for calculating compound interest. If Mr. John Chrystal invests \$6,000 today (Present Value) at a compound interest of 9 percent, calculate the Future Value of the investment after 30 years using the compound interest … | 572 | 2,272 | {"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} | 3.09375 | 3 | CC-MAIN-2018-05 | latest | en | 0.89542 |
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This question was previously asked in
JSSC JE Civil Re-Exam 31 Oct 2022 Official Paper-II
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2. Laminar flow
3. Non uniform flow
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JSSC JE Full Test 1 (Paper 1)
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120 Questions 360 Marks 120 Mins
## Detailed Solution
Explanation:
Steady flow: A flow is considered to be steady when conditions at any point in the fluid do not change with time i.e.
$$\frac{{\partial V}}{{\partial t}} = 0$$
But it may vary from one point to another and also the properties do not change with time, i.e.
$$\frac{{\partial p}}{{\partial t}} = 0,\frac{{\partial ρ }}{{\partial t}} = 0,\frac{{\partial T}}{{\partial t}} = 0$$
Unsteady flow: A flow is said to be unsteady when the flow parameters at any point change with time.
For unsteady flow, $$\frac{{\partial V}}{{\partial t}} \ne 0,\frac{{\partial ρ }}{{\partial t}} \ne 0,\frac{{\partial T}}{{\partial t}} \ne 0$$
Laminar flow:
Laminar Flow: Laminar flow is the flow in which fluid particles flow in layers. Each layer moves smoothly past the adjacent layer with little or no intermixing. There is no intermingling of fluid particles across the cross-section. So laminar flow resembles streamline flow.
Turbulent flow – Turbulent flow occurs at relatively larger velocities and is characterized by chaotic behaviour, irregular motion, large mixing and eddies. For such flows, inertial effects are more pronounced than the viscous effects.
• Mathematically the velocity field of turbulent flow is represented as$$V = \bar V + V'$$ or the velocity fluctuates at small time scales around a large time-averaged velocity.
• Similarly $$P = \bar P + P',T = \bar T + T'$$etc.
• Parameter Reynolds number $$\left( {\frac{{\rho vd}}{\mu }} \right)$$ is used to characterize Laminar and Turbulent flow.
• If Re < 2100 for pipe flow the flow is laminar and if Re > 104 the flow is turbulent. | 549 | 1,971 | {"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} | 3.90625 | 4 | CC-MAIN-2024-18 | latest | en | 0.805816 |
https://socratic.org/questions/how-do-you-factor-completely-10r-3s-2-25r-2s-2-15r-2s-3 | 1,579,568,668,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250601040.47/warc/CC-MAIN-20200120224950-20200121013950-00426.warc.gz | 673,538,702 | 6,059 | # How do you factor completely 10r^3s^2 + 25r^2s^2 – 15r^2s^3?
May 15, 2017
$10 {r}^{3} {s}^{2} + 25 {r}^{2} {s}^{2} - 15 {r}^{2} {s}^{3} = 5 {r}^{2} {s}^{2} \left(2 r + 5 - 3 s\right)$
#### Explanation:
Given:
$10 {r}^{3} {s}^{2} + 25 {r}^{2} {s}^{2} - 15 {r}^{2} {s}^{3}$
Note that all of the terms are divisible by $5$, ${r}^{2}$ and ${s}^{2}$, so by $5 {r}^{2} {s}^{2}$.
So we can separate that out as a factor:
$10 {r}^{3} {s}^{2} + 25 {r}^{2} {s}^{2} - 15 {r}^{2} {s}^{3} = 5 {r}^{2} {s}^{2} \left(2 r + 5 - 3 s\right)$
The remaining factor is already linear, so will not factor any further. | 309 | 606 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "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} | 4.5 | 4 | CC-MAIN-2020-05 | latest | en | 0.723159 |
https://www.gradesaver.com/textbooks/math/algebra/college-algebra-6th-edition/chapter-p-prerequisites-fundamental-concepts-of-algebra-exercise-set-p-2-page-33/61 | 1,555,798,372,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578530060.34/warc/CC-MAIN-20190420220657-20190421002657-00546.warc.gz | 711,983,321 | 13,427 | ## College Algebra (6th Edition)
$\frac{-27b^{15}}{a^{18}}$
$$(\frac{-15a^{4}b^{2}}{5a^{10}b^{-3}})^{3}$$ Group factors with like bases: $$=[(\frac{-15}{5})(\frac{a^{4}}{a^{10}})(\frac{b^{2}}{b^{-3}})]^{3}$$ $$=[-3a^{(4-10)}b^{(2-(-3))}]^{3}$$ $$=[-3a^{-6}b^{(2+3)}]^{3}$$ $$=[-3a^{-6}b^{5}]^{3}$$ When raising a product to a power, raise each factor to that power: $$=(-3)^{3}a^{(-6\times3)}b^{(5\times3)}$$ $$=-27a^{-18}b^{15}$$ Write as a fraction, move the base with the negative exponent to the other side of the fraction, and make the exponent positive: $$=\frac{-27b^{15}}{a^{18}}$$ | 249 | 590 | {"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} | 4.375 | 4 | CC-MAIN-2019-18 | latest | en | 0.613841 |
https://www.uibk.ac.at/mechatronik/mikroelektronik/lehre/webapps/ieee754.html.en | 1,627,656,526,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00245.warc.gz | 1,113,906,478 | 11,581 | # Digital technology and semiconductor circuit design
## IEEE-754WebApp
Fixed-point arithmentics reach their limits when it comes to an exact represenation of numbers and values to many decimal places. Floating-point arithmetics, which was introduced by IEEE Standard 754 in 1989, significantly enlarge the representable numerical range. The -format consits of 32 Bits. The first Bit $$S$$ represents the sign. The next 8 Bits describe the exponent $$E$$. The remaining 21 places represent the Mantisse's value $$M_\text{f}$$. The number $$x$$ is calculated from the given Bit pattern $$x = (-1)^S \cdot 2^{E-OS} \cdot (1 + M_\text{f})$$ Here, the Offset of $$OS = 127$$ is applied so that also negative numbers are allowed.
For the conversion of the number $$x$$ = into the Single Precision-format, the binary represenatation of the integer part and of the fractional part is done seperately. (see calculation Integer part and Fractional part)
The result of the conversion is:
Integer part:
12310 = 11110112
Fractional part:
0.45610 = 0.011101001011111⋯2
The resulting number $$x_2=$$(1111011.0111⋯)2 has to be normalized. This means $$x_2$$ has to be multiplied with the power of two 2$$-e$$ = 26 whereby the comma is offset by six places to the left. Thus, in the normalized form, there is always a 1 before the comma. The first 23 digits of the fractional part are stored as Mantisse $$M_f$$. The integer part is the same with 1. for every number in the normalzed form and does not have to be stored. The exponent's sign is taken into account with the Offset $$OS = 127$$. The biased exponent $$E = OS + e$$ (hier 13310 = 100001012) is stored.
Mantisse:
(123.456)10 = (1111011.1110110111)2
(1.1110110111101)2 · 26
biased exponent:
127 + (6) = 13310 = 100001012
Binary represenation:
0 | 10000101 | 111011011101001011111001
### Second variant
For the conversion of a number $$x$$ = into the Single Precision-format first the Exponent $$e$$ has to be normalized. The number $$x$$ can be represented as $$x=M\cdot 2^e$$. M is with $$e = \lfloor \log_2 x \rfloor$$ a number between 1 and 2.
log2(123.456) = 6.9478rounded down: $$e =$$ 6
$$x =$$ 123.456 = 1.929 · 26
Here it is important to consider that negative exponents are rounded up according to their amount. The fractional part $$M_f$$ of the Mantisse $$M$$ can now be converted into the binary system. (see here) Further, the Offset $$OS=127$$ is added to the exponent. This so-called biased exponent can also be transferred into the binary system.
Mantisse:
$$1 + M_f$$ = (1.929)10
=
1 + (0.111011⋯)2
biased exponent:
$$E = OS + e$$ = 127 + 6 = 13310
=
100001012
Binary represenation:
0 | 10000101 | 111011011101001011111001
Nach oben scrollen | 792 | 2,716 | {"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": 2, "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} | 3.125 | 3 | CC-MAIN-2021-31 | latest | en | 0.854218 |
http://es.wikidoc.org/index.php/Quartile | 1,585,846,801,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00086.warc.gz | 58,358,302 | 9,573 | # Quartile
In descriptive statistics, a quartile is any of the three values which divide the sorted data set into four equal parts, so that each part represents 1/4th of the sampled population.
Thus:
• first quartile (designated Q1) = lower quartile = cuts off lowest 25% of data = 25th percentile
• second quartile (designated Q2) = median = cuts data set in half = 50th percentile
• third quartile (designated Q3) = upper quartile = cuts off highest 25% of data, or lowest 75% = 75th percentile
The difference between the upper and lower quartiles is called the interquartile range.
There is no universal agreement on choosing the quartile values.[1] One possible rule (employed by the TI-83 calculator boxplot and 1-Var Stats functions) is as follows:
1. Use the median to divide the ordered data set into two halves. Do not include the median into the halves.
2. The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data.
The examples below assume this rule.
Example 1
Data Set: 6, 47, 49, 15, 42, 41, 7, 39, 43, 40, 36
Ordered Data Set: 6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49
${\displaystyle {\begin{cases}Q_{1}=25.5\\Q_{2}=40\\Q_{3}=42.5\end{cases}}}$
Example 2
Ordered Data Set: 7, 15, 36, 39, 40, 41
${\displaystyle {\begin{cases}Q_{1}=20.25\\Q_{2}=37.5\\Q_{3}=39.75\end{cases}}}$
Example 3
Ordered Data Set: 1 2 3 4
${\displaystyle {\begin{cases}Q_{1}=1.75\\Q_{2}=2.5\\Q_{3}=3.25\end{cases}}}$
range]] | 493 | 1,507 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 3, "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} | 4.625 | 5 | CC-MAIN-2020-16 | latest | en | 0.812284 |
https://www.physicsforums.com/threads/letters-to-numbers.84504/ | 1,555,958,758,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578577686.60/warc/CC-MAIN-20190422175312-20190422201312-00552.warc.gz | 792,248,458 | 20,929 | # Letters to Numbers?
#### aricho
Letters to Numbers???
Hey, I have no idea how to do this....
Replace all the letters with the respective digits in such a way that the calculation is correct
SEND+MORE=MONEY
the answer is 9567+1085=10652, but i dont know how to get there.
Related Introductory Physics Homework News on Phys.org
#### arildno
Homework Helper
Gold Member
Dearly Missed
All right, let us say S, E, N, D, M,R,O, Y stands for different digits s,e,n,d, m,r,o,y between 0 and 9.
Hence, SEND=s*1000+e*100+n*10+d*1
Do you agree to this?
sorry, lost
#### arildno
Homework Helper
Gold Member
Dearly Missed
aricho said:
sorry, lost
Why are you lost now?
Try to formulate what EXACTLY you are struggling with understanding; that's difficult, I know, but the only way to actually start achieving understanding.
#### aricho
SEND=s*1000+e*100+n*10+d*1
i would have thougt it would have been s*1000+e*100+n*10+d
#### arildno
Homework Helper
Gold Member
Dearly Missed
And your thought is perfectly correct and valid!
But, since any number multiplied with 1 equals itself, d=d*1, right?
So that means we agree on our expression after all..
The reason why I put in the *1 notation, is that it is conventional (and systematic) to do so, not because your idea is wrong (which it isn't)!
OK?
#### VietDao29
Homework Helper
This problem have so many solutions to it... I'll give you another one:
9342 + 1093 = 10435. Is that also correct?
Looking thoroughly at that, you will notice: m = 1, o = 0, r = 8 (or 9). The other number can be wisely chosen to fit the SEND + MORE = MONEY.
Viet Dao,
#### aricho
arildno, yes, got it now.
Whats next?
#### arildno
Homework Helper
Gold Member
Dearly Missed
VietDao29 said:
This problem have so many solutions to it... I'll give you another one:
9342 + 1093 = 10435. Is that also correct?
Looking thoroughly at that, you will notice: m = 1, o = 0, r = 8 (or 9). The other number can be wisely chosen to fit the SEND + MORE = MONEY.
Viet Dao,
WOW, YOU ARE VERY SMART!!!
IT IS SO GREAT THAT YOU MAKE O.P. A LOT MORE CONFUSED THAN HE ALREADY WAS!!!!
#### arildno
Homework Helper
Gold Member
Dearly Missed
aricho said:
arildno, yes, got it now.
Whats next?
Okay, now you can make a similar decomposition into sum expressions of MORE and MONEY as well, right? (Do that!)
Verify therefore that SEND MORE=MONEY can be written as:
(s+m)*1000+(e+o)*100+(n+r)*10+(d+e)*1=m*10000+o*1000+n*100+e*10+y*1
Ok?
#### VietDao29
Homework Helper
arildno said:
IT IS SO GREAT THAT YOU MAKE O.P. A LOT MORE CONFUSED THAN HE ALREADY WAS!!!!
Whoops, I thought I will have a chance to explain more when he continues asking questions. I hate reading long, long posts... therefore I just shorten everything.
Viet Dao,
#### arildno
Homework Helper
Gold Member
Dearly Missed
VietDao29 said:
Whoops, I thought I will have a chance to explain more when he continues asking questions. I hate reading long, long posts... therefore I just shorten everything.
Viet Dao,
He's all yours now, if you like.
I'm logging off..
#### VietDao29
Homework Helper
Just stay there,... I don't like, anyway. Maybe I can learn a different way from you.
Viet Dao,
yer, kinda....
#### arildno
Homework Helper
Gold Member
Dearly Missed
What do you mean by "kinda"?
Did you get what I was doing , but don't understand why I've done it like that?
#### aricho
sorry, i just wrote it down, agree-got it
#### arildno
Homework Helper
Gold Member
Dearly Missed
All right:
1. Now, you agree that whatever digits the letters represent, each of the numbers SEND and MORE must be less than or equal to 9999, right?
2. But that must mean that their sum must be less than or equal to 9999+9999=19998, agreed?
3. Now, look at the MONEY-side of your equation:
The first term there is m*10000
Thus, if you combine this with the insight you've gained in the above argument, what digit must "m" be if we assume that "m" is different from zero?
If you have problems with this post, please pinpoint what you didn't understand too well.
Last edited:
#### aricho
.....................1?
#### arildno
Homework Helper
Gold Member
Dearly Missed
Okay, 9 is the biggest digit we've got, right?
So, the number 9999 must be bigger than any other number with 4 digits, whatever digit a given letter might represent.
Get it?
(I made a writing error in 1., I've fixed it now)
#### aricho
yer.............got that
"Letters to Numbers?"
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• Solo and co-op problem solving | 1,286 | 4,761 | {"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} | 3.515625 | 4 | CC-MAIN-2019-18 | longest | en | 0.913145 |
http://slideplayer.com/slide/3172178/ | 1,524,226,113,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937440.13/warc/CC-MAIN-20180420100911-20180420120911-00527.warc.gz | 291,838,773 | 26,793 | # Calculus.
## Presentation on theme: "Calculus."— Presentation transcript:
Calculus
Calculus Calculus (differentiation and integration) was developed to improve this understanding. Differentiation and integration can help us solve many types of real-world problems. We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.). Derivatives are met in many engineering and science problems, especially when modelling the behaviour of moving objects.
differentiation dy/dx = 1 y = 3 x = 3
y = 2x y = 4 x = 2
differentiation
Y = x2 dy/dx =2x Video The normal is at right angles to the tangent
The derivative of ex is ex (itself)
dy/dx = ex
The derivative of ex is ex (itself)
The table of derivatives
y = f(x) dy/dx f′(x) k, any constant x x x x x 2 xn, any constant n nxn−1
The table of derivatives
y = f(x) dy/dx f′(x) ex ex ekx kekx ln x = logex 1/x
The table of derivatives
y = f(x) dy/dx f′(x) sin x cos x sin kx k cos kx cos x −sin x cos kx −k sin kx tan x sec2 x tan kx k sec2 kx Watch the minus sign
differentiation y = axn dy/dx = axnxn-1
differentiation y = 4 x3 + 2x2 +3 dy/dx = 12x2 +4x
differentiation y = (x3 + x) / x2 x3/x2 + x/x2 = x + 1/x = x + x-1 dy/dx = 1 - x-2 = 1 - 1/x2
differentiation y = √x + 1/√x = x1/2 + x-1/2 dy/dx = 1/2 x-1/2 - 1/2x-3/2) =1/(2√x) - 1/(2√x3)
Finding a tangent the gradient of the curve at the points quoted
Find the tangent of y = 3x2 – 2x + 4 when x = 0 and 3 1) When x = 0 y = +4 (substituting in the y equation) Gradient dy/dx = 6x -2 = -2 (when x = 0) Also (y – 4)/(x – 0) = m (gradient)
Finding a tangent So (y – 4)/x = -2 (y – 4) = - 2x y = -2x +4 The equation for the gradient at x = 0, y=4 is y = -2x + 4
Finding a tangent Find the tangent of y = 3x2 – 2x + 4 when x = 0 and 3 When x = 3 y = = 25 (substituting in the y equation) Gradient dy/dx = 6x -2 = 16 (when x = 3) Also (y – 25)/(x – 3) = m (gradient)
Finding a tangent So (y – 25)/(x -3) =16 (y – 25) = 16(x – 3) = 16x – 48 y = 16x – The equation for the gradient at x = 0, y=4 y = 16x - 23
Chain rule Differentiate y=sin(x2 +3) let: u = x 2 +3 So du/dx = 2x let y = sin u dy/du =cos u dy/dx = 2xcos(x2 +3)
Differentiate with respect to x:
Chain rule Differentiate with respect to x: y= 3√e1-x = e(1-x)/3 = e1/3-x/3 let: u = 1/3-x/3 du/dx -1/3 dy/du = e(1-x)/3 dy/dx = -1/3 e(1-x)/3 Chain rule
Derivatives of sine cosine and tangent
Example y = sin 3x (dy)/(dx) = 3 cos 3x Example y = 5 sin 3x . ` (dy)/(dx)` = 15 cos 3x
Derivatives of sine cosine and tangent
Example y =cos3x. dy/dx = - 3sin 3x Example y= tan 3x dy/dx = 3sec23x
Derivatives of sine cosine and tangent
Example: Differentiate y=sin(x 2 +3) let: u = x 2 +3 So du/dx = 2x let y = sin u dy/du =cos u dy/dx = 2xcos(x2 +3)
The product rule: if y = uv then dy/dx = u.dv/dx + v.du/dx
The product rule y = x3sin2x let u = x3; and v = sin2x dy/dx = v.du/dx + u.dv/x du/dx = 3x2 and dv/dx = 2Cos2x Therefore dy/dx = sin2x.3x2 + x3 2cos2x
The quotient rule: if y = u/v then dy/dx = (v.du/dx - u.dv/dx)/v2
dy/dx = (v.du/dx - u.dv/x) / v2
The quotient rule Minus when divided y = (e4x)/(x2 +2) let u = e4x; and v = x2 + 2 dy/dx = (v.du/dx - u.dv/x) / v2 du/dx = 4e4x and dv/dx = 2x Therefore dy/dx = [(x2+2).4e4x] - [e4x.2x]/(x2 +2)2 = e4x (2x2 +8 – 2x) /(x2 +2)2 Don’t forget
Velocity and acceleration
An object travels a distance s = 2t3 - 2t2 +8t + 6 metres in t seconds. Find both velocity and acceleration of the body at time 5 seconds. velocity = ds/dt = 6t2 – 4t +8 When t =5 velocity = m/s continue
Velocity and acceleration
Acceleration = dv/dt = 12t – 4 From v = 6t2 -4t +8 (last slide) Acceleration at 5t sec = 60-4 = 56m/s2
Turning points The equation of a curve is y = x3 – 4.5x2 - 12x +15. Find by calculation the value of the maximum and minimum turning points on the curve. Turning points occur at dy/dx = 0 Therefore:- dy/dx = 3x2 - 9x – 12 = 0 = x2 - 3x – 4 (dividing by common factor 3) = 0 Factorising x2 - 3x - 4; then (x - 4)(x+1) = 0. Turning points are at x = + 4 and x = -1 continue
Turning points Check for max and min. d2(3x2-9x-12)y/dx2 = 6x - 9 Substituting for x = +4: 6x – 9 = = 15. Since this is positive it must be a minimum y point Substituting for x = -1: 6x + (-9) = (-9) = -15. Since this is negative it must be a maximum y point continue
Turning points Therefore ymin = x3 – 4.5x2 - 12x +15 = = - 41 (as x = +4) Therefore ymax = x3 – 4.5x2 - 12x +15 = -1 – = (x = -1)
Integration (opposite of differentiation)
xn dx = [(xn+1)/n+1] + c x2 dx = [(x2+1)/2+1] + c = x3/3 +c
Integration (opposite of differentiation)
∫1/x dx = (ln x) + c ∫ 1/ax + b dx = 1/a ln |ax + b| + c ∫ ex dx = ex + c
Integration (opposite of differentiation)
∫ emx dx =1/memx + c ∫cos x dx = sin x + c ∫ cos nx dx =1/n(sin nx) + c
Integration (opposite of differentiation)
∫ sin x dx = −cos nx + c ∫ sin nx dx = −1/n cos nx + c
Integration (opposite of differentiation)
∫ sec2 x dx = tan x + c ∫ sec2 nx dx =1/n tan nx + c
IntEgraTIon BETWEEN LIMITS
Example Evaluate ∫ ( 6x2 - 2x + 2) dx, between the limits +4 and +1 ∫ ( 6x2 - 2x + 2) dx = 6x3/3 – 2x2/2 +2x +c = 2x3 – x2 +2x +c continue
IntEgraTING BETWEEN LIMITS
When x = +4 2x3 – x2 +2x +c = 128 – c = 120 +c (equation 1) When x =+1 2x3 – x2 +2x +c = 2 – c =3 + c (equation 2) continue
IntEgraTIon BETWEEN LIMITS
To find ∫ ( 6x2 - 2x + 2) dx, between the limits +4 and +1 subtract equation 2 from equation c c =117 | 2,188 | 5,496 | {"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} | 4.6875 | 5 | CC-MAIN-2018-17 | latest | en | 0.802948 |
https://brainly.in/question/71119 | 1,484,936,989,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280850.30/warc/CC-MAIN-20170116095120-00191-ip-10-171-10-70.ec2.internal.warc.gz | 801,444,456 | 10,515 | # What is the figure of merit of a glavanometer
2
by AShish872
2015-01-12T21:09:22+05:30
K=i/theta; where k is the figure of merit.. i refers to the current supplied and theta stands for the number of divisions of deflection
2015-01-13T03:58:15+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
The figure of merit for a galvanometer is the amount of current it takes to deflect the needle by one unit angle (radian) on the scale. The deflection is proportional to the input current through the galvanometer.
The deflection by a unit angle (radian) is given as
deflection θ = I / K or, K = I / θ
So it is the current per unit value o f θ.
K = figure of merit.
deflection θ can be taken in terms of the number of units in to which the scale on the galvanometer is divided. The scale on it may be divided in to say 10 units. Each of it is divided further into 10 sub units or fractional parts. Then θ here is number of units moved. | 314 | 1,215 | {"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-2017-04 | latest | en | 0.912821 |
http://forums.wolfram.com/mathgroup/archive/1992/Dec/msg00033.html | 1,529,859,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00559.warc.gz | 121,019,555 | 7,121 | Vector multiplication??
• To: mathgroup at yoda.physics.unc.edu
• Subject: Vector multiplication??
• From: William M. MacDonald <mcdonald at fafner.umd.edu>
• Date: Tue, 8 Dec 92 15:45:56 -0500
```The following was not sent to this group but may be of general
interest so I am sending it.
To: "Katherine (Williams) Derbyshire" <kewms at kew.com>
Subject: Re: Vector multiplication??
Regarding A^2 where A is a vector.
There is a very powerful element-by-element processing built into
Mathematica that makes it unnecessary to write explicit DO loops for
many things. If A and B are vectors of the same length then
1. A^2 gives the list of the elements squared, A[i]^2;
2. A*B gives the list with elements A[i]*B[i]
3. A + B gives the list with elements A[i]+B[i];
4. A^B is the list with elements A[i]^B[i];
5. A/B is the list with elements A[i]/B[i].
Also if one defines
n=Range[0,10]
then
x^n
is the list
{1,x,x^2,x^3,...,x^10}
All this is very obvious to anyone who knows APL, so its just a
matter of learning the language.
```
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• Next by thread: Vector multiplication?? | 346 | 1,216 | {"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-2018-26 | latest | en | 0.867526 |
https://www.java-forums.org/new-java/54966-two-java-equations-print.html | 1,493,567,882,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125654.80/warc/CC-MAIN-20170423031205-00481-ip-10-145-167-34.ec2.internal.warc.gz | 933,237,219 | 2,180 | # Two Java Equations
Printable View
• 02-02-2012, 06:14 PM
Army
Two Java Equations
Someone please help me with this ASAP,, I don't understand this at all
1. Solve the expression and find the output of Z = ((X > Y) && (Y < X))? X:Y when int X=5 and Y=10.
5
10
15
20
2. Solve the expression and find the output of Z = ((X > Y) || (Y < X))? X:Y when int X=15 and Y=10.
5
10
15
25
• 02-02-2012, 06:23 PM
Army
Re: Two Java Equations
Another question I don't understand
If J=true and K=false, what is J&&K?
True
no
False
yes
• 02-02-2012, 06:31 PM
JosAH
Re: Two Java Equations
If X > Y then Y < X so X > Y && Y < X are identical. (both expressions are equivalent). If X > Y or Y < X is a tautology (both expressions are equivalent again).
kind regards,
Jos | 263 | 757 | {"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-2017-17 | longest | en | 0.782334 |
https://la.mathworks.com/matlabcentral/cody/problems/44836-iccanobif-numbers-1/solutions/1710132 | 1,586,133,171,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00141.warc.gz | 546,070,860 | 15,833 | Cody
# Problem 44836. Iccanobif numbers 1
Solution 1710132
Submitted on 19 Jan 2019 by William
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 = 1;y_correct = 1; assert(isequal(iccanobif(x),y_correct))
2 Pass
x = 9;y_correct = 124; assert(isequal(iccanobif(x),y_correct))
3 Pass
x = 43;y_correct=36181429034; assert(isequal(iccanobif(x),y_correct))
4 Pass
for flag=1:50 y(flag)=iccanobif(flag); end dy=diff(y); assert(isequal(max(dy(1:25))+min(dy(1:25)),250750)) assert(isequal(max(dy)+min(dy),19910139546138)) sdy=sign(dy); assert(isequal(sum(sdy==-1),8)) [m1,w1]=min(y(1:10)); [m2,w2]=min(y(11:20)); [m3,w3]=min(y(21:30)); [m4,w4]=min(y(31:40)); [m5,w5]=min(y(41:50)); assert(isequal([m1 m2 m3 m4 m5],[1 836 113815 106496242 21807674140])) assert(isequal(w1*w2*w3*w4*w5,15)) | 347 | 910 | {"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-2020-16 | latest | en | 0.415122 |
http://jwilson.coe.uga.edu/EMT668/EMAT6680.2002/Greenwood/Assignment5/htmls/asgn5.html | 1,542,552,144,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744381.73/warc/CC-MAIN-20181118135147-20181118161147-00247.warc.gz | 178,305,229 | 5,351 | By Donna Greenwood
Here is my library of GSP 4.0 tools.Feel free to bring them up and play with them if your browser has GSP 4.0 set up as a helper application.If you have the earlier version of GSP, these tools won't work.These are grouped as follows.Just pick a type of construction that interests you and take a look!
Triangle Centers
If you want to run any of these tools, the input is always three points.
`Centroid (intersection of medians) `
Circumcenter (intersection of perpendicular bisectors of the sides)
Orthocenter (intersection of the altitudes)
Incenter (intersection of the angle bisectors)
Four Common Triangle Centers
Circles and Triangles Based Upon An Input Triangle
Again, the input for these constructions is the three points necessary to define a triangle.
Incircle
Circumcircle
Medial Triangle (formed by connecting the midpoints of the sides of a given triangle)
Orthocenter, Mid-segment triangle
Nine Point Circle (constructed by finding the center of a circle that passes through the 3 mid-points of the sides, the 3 feet of the altitudes and the 3 midpoints of the segments from the triangle vertices to the orthocenter)
Orthic Triangle (constructed by connecting the feet of the altitudes)
Pedal Triangle (for more information on pedal triangles, see write up #9)
Polygon Constructions
Equilateral Triangle Given a Side AB
Square Given a Side AB
Regular Hexagon Given a Side AB
Regular Octagon Given a Side AB | 328 | 1,460 | {"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} | 3.125 | 3 | CC-MAIN-2018-47 | latest | en | 0.874249 |
http://stackoverflow.com/questions/4566835/computing-average-of-two-colors | 1,394,875,577,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394678696864/warc/CC-MAIN-20140313024456-00067-ip-10-183-142-35.ec2.internal.warc.gz | 136,436,617 | 17,074 | # Computing “average” of two colors
This is only marginally programming related - has much more to do w/ colors and their representation.
I am working on a very low level app. I have an array of bytes in memory. Those are characters. They were rendered with anti-aliasing: they have values from 0 to 255, 0 being fully transparent and 255 totally opaque (alpha, if you wish).
I am having trouble conceiving an algorithm for the rendering of this font. I'm doing the following for each pixel:
`````` // intensity is the weight I talked about: 0 to 255
intensity = glyphs[text[i]][x + GLYPH_WIDTH*y];
if (intensity == 255)
continue; // Don't draw it, fully transparent
else if (intensity == 0)
setPixel(x + xi, y + yi, color, base); // Fully opaque, can draw original color
else { // Here's the tricky part
// Get the pixel in the destination for averaging purposes
pixel = getPixel(x + xi, y + yi, base);
// transfer is an int for calculations
transfer = (int) ((float)((float) (255.0 - (float) intensity/255.0) * (float) color.red + (float) pixel.red)/2); // This is my attempt at averaging
newPixel.red = (Byte) transfer;
transfer = (int) ((float)((float) (255.0 - (float) intensity/255.0) * (float) color.green + (float) pixel.green)/2);
newPixel.green = (Byte) transfer;
// transfer = (int) ((float) ((float) 255.0 - (float) intensity)/255.0 * (((float) color.blue) + (float) pixel.blue)/2);
transfer = (int) ((float)((float) (255.0 - (float) intensity/255.0) * (float) color.blue + (float) pixel.blue)/2);
newPixel.blue = (Byte) transfer;
// Set the newpixel in the desired mem. position
setPixel(x+xi, y+yi, newPixel, base);
}
``````
The results, as you can see, are less than desirable. That is a very zoomed in image, at 1:1 scale it looks like the text has a green "aura".
Any idea for how to properly compute this would be greatly appreciated.
-
You need to blend the background and foreground colours. A-la:
``````pixelColour = newColour * intensity + backgroundColour * (1 - intensity)
``````
By the way, this is a really slow way of rendering and blending fonts. You should instead render all the characters of the font to an off-screen surface with all the properties you need, and then use that as a texture to render to other surfaces when you need text.
Edit:
This doesn't look right:
``````(255.0 - (float) intensity/255.0)
``````
``````(255.0 - (float) intensity)/255.0
``````
-
+1 linear interpolation is probably better. – EnabrenTane Dec 30 '10 at 22:54
Or, to put it in the terms of the original code: `transfer = (int) ((1.0 - intensity/255.0) * color.red + intensity/255.0 * pixel.red);` – Mark Ransom Dec 30 '10 at 23:02
P.S. "render all the characters of the font to an off-screen surface" - I think that's what `glyphs` is. – Mark Ransom Dec 30 '10 at 23:04
I don't care much about performance, since this is a proof-of-concept academic app. But thanks for your input! I'll try that. – Francisco P. Dec 31 '10 at 0:10
I believe that "aura" is caused by anti aliasing. The technique averages pixels with their neighbors.
I realize you don't seem to be using OpenGL but this chapter might help explain some of the theory. Wish I had a better answer, but hopefully this points you in the right direction. My first attempt would be to disable Antialiasing since it seems to do more harm than good. There is probably a better solution than that though.
-
It may be too much complicated to doing alpha blending pixel by pixel because current pixel value modifies next pixel value.
I would redesign the algorithm with the thinking of box wise blending.
With many getPixel calling for a single glyph, you can't produce proper target image.
- | 983 | 3,699 | {"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-2014-10 | latest | en | 0.798273 |
https://calculus.subwiki.org/wiki/First_derivative_test_is_conclusive_for_function_with_algebraic_derivative | 1,591,454,387,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348513321.91/warc/CC-MAIN-20200606124655-20200606154655-00573.warc.gz | 272,302,676 | 9,289 | # First derivative test is conclusive for function with algebraic derivative
This article describes a situation, or broad range of situations, where a particular test or criterion is conclusive, i.e., it works as intended to help us determine what we would like to determine.
The test is first derivative test. See more conclusive cases for first derivative test | inconclusive cases for first derivative test
## Statement
### Single definition case
For any function of the following type, the first derivative test is always conclusive, i.e., it can be used to definitively determine whether the function has a local extreme value at a given critical point, and if so, what the nature of the local extreme value is:
• Nonconstant polynomial function on the real line, or on an interval or union of finitely many intervals in the real line
• Nonconstant rational function on the real line (whatever subset it's defined on), or on an interval or union of finitely many intervals in its maximum possible domain
• Nonconstant function defined on the real line, or on an interval or union of finitely many intervals in the real line, such that the derivative of the function is a rational function on its domain
Note that we omit constant functions from consideration because the derivative is identically zero and the analysis of local extreme values does not require the use of derivatives.
## Facts used
1. Nonzero polynomial function has finitely many zeros
2. Any finite set is a discrete set, and all points in that set are isolated
3. First derivative test is conclusive for differentiable function at isolated critical point
## Proof
The proofs in all cases are essentially the same, but we present them separately to make them accessible to people who are not familiar with the more advanced cases.
### Case of polynomial function on the real line
Given: A nonconstant polynomial function $f$, a critical point $c$ for $f$
To prove: The first derivative test is conclusive for $f$ at $c$. In other words, $f'$ has constant sign on the immediate left of $c$, and it has constant sign on the immediate right of $c$.
Proof:
Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The derivative $f'$ is a nonzero polynomial function. (rules for polynomial differentiation) $f$ is a nonconstant polynomial function Direct. Note that the fact that $f$ is nonconstant forces $f'$ to not be identically zero.
2 $f'$ is defined everywhere and has finitely many zeros, so $f$ has finitely many critical points. Fact (1) Step (1) Step-fact combination direct
3 All the critical points of $f$ are isolated critical points. In particular, $c$ is an isolated critical point for $f$. Fact (2) Step (2) Step-fact combination direct
4 The first derivative test is conclusive for $f$ at $c$. Fact (3) Step (3) Step-fact combination direct
### Case of rational function on its maximum possible domain
Given: A (reduced form, i.e., no common factors between numerator and denominator) nonconstant rational function $f$, a critical point $c$ for $f$
To prove: The first derivative test is conclusive for $f$ at $c$. In other words, $f'$ has constant sign on the immediate left of $c$, and it has constant sign on the immediate right of $c$.
Proof:
Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The derivative $f'$ is a nonzero rational function. (rules for polynomial differentiation, plus quotient rule for differentiation) $f$ is a rational function Direct. Note that the derivative is nonzero because the function is not constant.
2 $f'$ is defined everywhere on the domain of $f$ and has finitely many zeros, so $f$ has finitely many critical points given by the zeros of $f'$. Fact (1) Step (1) [SHOW MORE]
3 All the critical points of $f$ are isolated critical points. In particular, $c$ is an isolated critical point for $f$. Fact (2) Step (2) Step-fact combination direct
4 The first derivative test is conclusive for $f$ at $c$. Fact (3) Step (3) Step-fact combination direct
### Case of function whose derivative is a rational function on its domain
The proof is similar to the rational function case, except that we don't have to do Step (1) of the proof above and can proceed directly from Step (2) onward. | 958 | 4,309 | {"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": 42, "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} | 4.34375 | 4 | CC-MAIN-2020-24 | longest | en | 0.888295 |
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Topic: Matlab
Replies: 11 Last Post: Dec 13, 2012 12:12 AM
Messages: [ Previous | Next ]
Tanner Handa Posts: 7 Registered: 12/12/12
Re: Matlab
Posted: Dec 12, 2012 10:41 PM
bartekltg <bartekltg@gmail.com> wrote in message <kabhvn\$qq8\$1@node1.news.atman.pl>...
> W dniu 2012-12-13 03:52, Tanner Handa pisze:
> > bartekltg <bartekltg@gmail.com> wrote in message
>
> > so how would I be able to make my happy birthday song sound better to
> > make the sounds shorter??
> > This is what I have and now when i do t = linspace(0, 0.5, 0.5 *
> > freqsamp); it make the frequencies really high pitched.
>
> I don't see difference in tone between
>
> t = linspace(0,1,freqsamp);
> a = (sin(1.5*pi*220*t) + sin(1.5*pi*440*t));
>
> and
>
> t2= linspace(0,0.5,freqsamp/2);
> a2 = (sin(1.5*pi*220*t2) + sin(1.5*pi*440*t2));
>
>
> sound(a, freqsamp),sound(a2, freqsamp)
>
>
>
> > clear
> > freqsamp = 10000;
> > t = linspace(0,1,freqsamp);
>
> Add 'note length'
>
> nl = 0.4;
>
> t = linspace(0,nl,nl*freqsamp);
>
>
> > a = sin(1.5*pi*220*t) + sin(1.5*pi*440*t);
> > b = sin(2*pi*220*t) + sin(2*pi*440*t);
> > c = sin(2.5*pi*220*t) + sin(2.5*pi*440*t); d = sin(3*pi*220*t) +
> > sin(3*pi*440*t); e = sin(4*pi*220*t) + sin(4*pi*440*t);
> > f = sin(4.5*pi*220*t) + sin(4.5*pi*440*t);
> > g = (sin(5*pi*220*t) + sin(5*pi*440*t))/1.5; h = sin(5.5*pi*220*t) +
> > sin(5.5*pi*440*t);
> > i = sin(6*pi*220*t) + sin(6*pi*440*t);
> > j = sin(6.5*pi*220*t) + sin(6.5*pi*440*t);
>
> OK, I think.
> But you can create function to do this:
>
> a= foo (t, 1.5);
> b= foo (t, 2) ;
> ...
> g= foo(t,5)/1.5;
> ...
>
>
> > sound(a, freqsamp) %HA %3
> > sound(a, freqsamp) %ppy %3
>
> And there is very long pause between sounds.
> You want to play whole song.
>
> sound ( [a,a,b,a,f,e,a,a,b,a,g,f], freqsamp );
>
> But now you haven't any pause. Create short 'pause'
> (vector of silent;) I mean zeros) and add between every note.
> Better: add this to function foo.
>
>
> bartekltg
Im not sure what foo is doing. This isn't working when I try to program this in MATLAB. I believe I understand everything else though.
Date Subject Author
12/12/12 Tanner Handa
12/12/12 bartekltg
12/12/12 Tanner Handa
12/12/12 bartekltg
12/12/12 Tanner Handa
12/12/12 bartekltg
12/12/12 Tanner Handa
12/12/12 Tanner Handa
12/12/12 bartekltg
12/12/12 Tanner Handa
12/13/12 bartekltg
12/13/12 Tanner Handa | 963 | 2,486 | {"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} | 3.28125 | 3 | CC-MAIN-2017-30 | latest | en | 0.660865 |
https://math.stackexchange.com/questions/2066454/arranging-people-in-a-line-without-specific-people-at-either-end/2066484 | 1,579,813,587,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250613416.54/warc/CC-MAIN-20200123191130-20200123220130-00124.warc.gz | 549,967,102 | 31,514 | # Arranging people in a line without specific people at either end
In how many ways can ten people be arranged in a line if neither of two particular people can sit on either end of the row?
What i thought was find how many ways one particular person must sit at either end then multiply that value by 2 then subtract it from how many ways ten people can be arranged without restriction
• I've noticed that for the past hour or so, you've been posting questions all on elementary combinatorics, with very similar names, showing only your questions, without demonstrating any attempt at finding an answer. Maybe you should put in a little more effort into finding an answer yourself, or at least showing where you have trouble, before posting a question. Also, you should give your questions more specific names to distinguish them. – Kevin Long Dec 20 '16 at 21:23
• Can you clarify what you mean by "if neither of two particular people can sit on either end of the row"? Can I assume that then you have 8 choices in the front end then 7 choices in the back end? – h94 Dec 20 '16 at 21:25
• Does 'two particular people sitting on either end' mean configurations like AxxxxB and BxxxxA only, or also ABxxxx, BAxxxx, xxxxAB and xxxxBA? – CiaPan Dec 20 '16 at 21:25
• Hint: choose a person for the left, choose a person for the right, then permute the remainder. – lulu Dec 20 '16 at 21:29
• As a suggestion for working on problems like these: if you think you have a good method, try it with a smaller collection. In this case, say, suppose you had four people, $A,B,C,D$ and that neither $A$ nor $B$ can be on an end. Now, it is easy to simply list all the cases so you can test your method out. Just asking other people to do it for you is a terrible way to learn anything. – lulu Dec 20 '16 at 21:34
In how many ways can ten people be arranged in a line if neither of two particular people can sit on either end of the row?
Without any restrictions, $10!$ ways you can arrange them
$9!\times 2$ ways where first person is at any end $9!\times 2$ ways where second person is at any end
$8!\times 2$ ways where both persons are at end
$10! - (9!\times 2 + 9!\times 2) + 8!\times 2$
You have $8$ choices for the person sitting on one end (because there are only $8$ people that can be on either end), you have $7$ choices for the person sitting on the other end (because you already put a person on the end and you still can't put two people on the end), and then you have $8!$ for the rest of the people. Your total is therefore $8*7*8!$. | 649 | 2,545 | {"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} | 3.546875 | 4 | CC-MAIN-2020-05 | latest | en | 0.938053 |
http://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-7-section-7-4-applications-of-percent-exercise-set-page-501/1 | 1,524,292,711,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945037.60/warc/CC-MAIN-20180421051736-20180421071736-00169.warc.gz | 421,285,004 | 13,130 | ## Prealgebra (7th Edition)
percent * base = amount $1 .5\%\times b=24$ $0.015b=24$ $0.015b\div0.015=24\div0.015$ $b=1600$ | 58 | 123 | {"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} | 3.8125 | 4 | CC-MAIN-2018-17 | latest | en | 0.600806 |
http://cycling74.com/forums/topic/nth-root/ | 1,397,855,901,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00274-ip-10-147-4-33.ec2.internal.warc.gz | 55,955,214 | 13,895 | nth root?
Sep 8, 2010 at 5:07am
nth root?
Hi all,
For equal divisions of the octave into n divisions, the equation is to find the nth root of 2.
Given that I can’t find a symbol/funtion for [expr] to take the ‘nth root’ of something, how do I calculate it?
Any suggestions greatly appreciated.
Thanks,
Steven
#52224
Sep 8, 2010 at 6:17am
[expr pow(\$f1,1.0/n)] where n is the nth root (because the nth root of n is the same as finding the power of 1/n). That example would ind the nth root of the input. If you want the input to [expr] to be n in finding the nth root of 2, you would use [expr pow(2,1.0/\$f1)]
#187699
Sep 8, 2010 at 8:53am
Excellent…I figured it was something so straightforward but I just wasn’t getting it…thank you!
#187700
You must be logged in to reply to this topic. | 242 | 804 | {"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} | 2.921875 | 3 | CC-MAIN-2014-15 | longest | en | 0.908018 |
https://cracku.in/blog/xat-probability-questions-pdf/ | 1,717,025,505,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059412.27/warc/CC-MAIN-20240529230852-20240530020852-00121.warc.gz | 152,837,133 | 36,314 | 0
538
# XAT Probability Questions [Important PDF]
Download Probability Questions for XAT PDF – XAT Probability questions pdf by Cracku. Practice XAT solved Probability Questions paper tests, and these are the practice question to have a firm grasp on the Probability topic in the XAT exam. Top 20 very Important Probability Questions for XAT based on asked questions in previous exam papers. The XAT question papers contain actual questions asked with answers and solutions.
Question 1: In a box carrying one dozen of oranges one third have become bad.If 3 oranges taken out from the box random ,what is the probability that at least one orange out of the 3 oranges picked up is good ?
a) 1/55
b) 54/55
c) 45/55
d) 3/55
e) None of these
Solution:
Total number of oranges in the box = 12
Number of ways of selecting 3 oranges out of 12 oranges, n(S) = $C^{12}_3$
= $\frac{12 \times 11 \times 10}{1 \times 2 \times 3} = 220$
Number of oranges which became bad = $\frac{12}{3}=4$
Number of ways of selecting 3 oranges out of 4 bad oranges = $C^4_3 = C^4_1 = 4$
Number of desired selection of oranges, n(E) = 220 – 4 = 216
$\therefore$ $P(E) = \frac{n(E)}{n(S)}$
= $\frac{216}{220}= \frac{54}{55}$
=> Ans – (B)
Instructions
Study the information carefully to answer the following questions:
A bucket contains 8 red, 3 blue and 5 green marbles.
Question 2: If 3 marbles are drawn at random, what is the probability that none is red ?
a) ${3 \over 8}$
b) ${1 \over {16}}$
c) ${1 \over {10}}$
d) ${3 \over {16}}$
e) None of these
Solution:
Number of ways of drawing 3 marbles out of 16
$n(S) = C^{16}_3 = \frac{16 \times 15 \times 14}{1 \times 2 \times 3}$
= $560$
Out of the three drawn marbles, none is red, i.e., they will be either blue or green.
=> $n(E) = C^8_3 = \frac{8 \times 7 \times 6}{1 \times 2 \times 3}$
= $56$
$\therefore$ Required probability = $\frac{n(E)}{n(S)}$
= $\frac{56}{560} = \frac{1}{10}$
Question 3: If 2 marbles are drawn at random, what is the probability that both are green?
a) ${1 \over 8}$
b) ${5 \over {16}}$
c) ${2 \over 7}$
d) ${3 \over 8}$
e) None of these
Solution:
Number of ways of drawing 2 marbles out of 16
$n(S) = C^{16}_2 = \frac{16 \times 15}{1 \times 2}$
= $120$
Out of the two drawn marbles, both are green
=> $n(E) = C^5_2 = \frac{5 \times 4}{1 \times 2}$
= $10$
$\therefore$ Required probability = $\frac{n(E)}{n(S)}$
= $\frac{10}{120} = \frac{1}{12}$
Question 4: If 4 marbles are drawn at random, what is the probability that 2 are red and 2 are blue ?
a) ${{11} \over {16}}$
b) ${3 \over {16}}$
c) ${11 \over {72}}$
d) ${3 \over {65}}$
e) None of these
Solution:
Number of ways of drawing 4 marbles out of 16
=> $n(S) = C^{16}_4 = \frac{16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4}$
= $1820$
Out of the four drawn marbles, 2 are red and 2 are blue.
=> $n(E) = C^8_2 \times C^3_2 = \frac{8 \times 7}{1 \times 2} \times \frac{3 \times 2}{1 \times 2}$
= $28 \times 3 = 84$
$\therefore$ Required probability = $\frac{n(E)}{n(S)}$
= $\frac{84}{1820} = \frac{3}{65}$
Question 5: A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one
ball is drawn from each bag, and the probability that both are green.
a) 13/70
b) 1/4
c) 6/35
d) 8/35
e) None of these
Solution:
Total balls in bag A = 4 + 6 = 10
Probability that ball is green = $\frac{4}{10}$
Total balls in bag B = 3 + 4 = 7
Probability that ball is green = $\frac{3}{7}$
=> Required probability = $\frac{4}{10} \times \frac{3}{7}$
= $\frac{6}{35}$
Question 6: A bag A contains 4 green and 6 red balls. Another bag B contains 3 green and 4 red balls. If one
ball is drawn from each bag, and the probability that both are green.
a) 13/70
b) 1/4
c) 6/35
d) 8/35
e) None of these
Solution:
Total balls in bag A = 4 + 6 = 10
Probability that ball is green = $\frac{4}{10}$
Total balls in bag B = 3 + 4 = 7
Probability that ball is green = $\frac{3}{7}$
=> Required probability = $\frac{4}{10} \times \frac{3}{7}$
= $\frac{6}{35}$
Question 7: A bag contains 2 red, 3 green and 2 blue balls. 2 balls are to be drawn randomly. What is the probability that the balls drawn contain no blue ball ?
a) 5/7
b) 10/21
c) 2/7
d) 11/21
e) None of these
Solution:
Total number of balls = 2 + 3 + 2 = 7
Total number of outcomes = Drawing 2 balls out of 7
= $C^7_2 = \frac{7 \times 6}{1 \times 2} = 21$
Favourable outcomes = Drawing 2 balls out of 5 (so that none is blue)
= $C^5_2 = \frac{5 \times 4}{1 \times 2} = 10$
=> Required probability = $\frac{10}{21}$
Question 8: There are 8 brown balls, 4 orange balls and 5 black balls in a bag. Five balls are chosen at random. What is the probability of their being 2 brown balls, 1 orange ball and 2 black balls ?
a) $\frac{191}{1547}$
b) $\frac{180}{1547}$
c) $\frac{280}{1547}$
d) $\frac{189}{1547}$
e) None of these
Solution:
Total number of balls in the bag = 8 + 4 + 5 = 17
P(S) = Total possible outcomes
= Selecting 5 balls at random out of 17
=> $P(S) = C^{17}_5 = \frac{17 \times 16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4 \times 5}$
= $6188$
P(E) = Favorable outcomes
= Selecting 2 brown, 1 orange and 2 black balls.
=> $P(E) = C^8_2 \times C^4_1 \times C^5_2$
= $\frac{8 \times 7}{1 \times 2} \times 4 \times \frac{5 \times 4}{1 \times 2}$
= $28 \times 4 \times 10 = 1120$
$\therefore$ Required probability = $\frac{P(E)}{P(S)}$
= $\frac{1120}{6188} = \frac{280}{1547}$
Question 9: In a bag there are 4 white, 4 red and 2 green balls. Two balls are drawn at random. What is the probability that at least one ball is of green colour ?
a) $\frac{4}{5}$
b) $\frac{3}{5}$
c) $\frac{1}{5}$
d) $\frac{2}{5}$
e) None of these
Solution:
There are 4 white, 4 red and 2 green balls and two balls are drawn at random.
Total possible outcomes = Selection of 2 balls out of 10 balls
= $C^{10}_2 = \frac{10 * 9}{1 * 2} = 45$
Favourable outcomes = 1 green ball and 1 ball of other colour + 2 green balls
= $C^2_1 \times C^8_1 + C^2_2$
= 2*8 + 2 = 18
$\therefore$ Required probability = $\frac{18}{45} = \frac{2}{5}$
Question 10: A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked at random. What is the probability that the ball drawn is white ?
a) $\frac{7}{11}$
b) $\frac{7}{30}$
c) $\frac{5}{11}$
d) $\frac{7}{15}$
e) $\frac{8}{15}$
Solution:
Probability of choosing bag 1 = (1/2)
Probability of choosing bag 2 = (1/2)
Probability of choosing white ball from bag 1 = 3/5
Probability of choosing white ball from bag 2 = 2/6
Probability of choosing bag 1 and white ball from it = (1/2)(3/5) = 3/10
Probability of choosing bag 2 and white ball from it = (1/2)(2/6) = 2/12
Probability of choosing a bag and drawing a white ball = (3/10) + (2/12) = (28/60) = (7/15)
Option D is the correct answer.
Question 11: A bag contains 3 white balls and 2 black balls. Another bag contains 2 white and 4 black balls. A bag and a ball are picked at random. What is the probability that the ball drawn is white ?
a) $\frac{7}{11}$
b) $\frac{7}{30}$
c) $\frac{5}{11}$
d) $\frac{7}{15}$
e) $\frac{8}{15}$
Solution:
Probability of choosing bag 1 = (1/2)
Probability of choosing bag 2 = (1/2)
Probability of choosing white ball from bag 1 = 3/5
Probability of choosing white ball from bag 2 = 2/6
Probability of choosing bag 1 and white ball from it = (1/2)(3/5) = 3/10
Probability of choosing bag 2 and white ball from it = (1/2)(2/6) = 2/12
Probability of choosing a bag and drawing a white ball = (3/10) + (2/12) = (28/60) = (7/15)
Option D is the correct answer.
Question 12: In a bag, there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red ?
a) 29/35
b) 7/15
c) 23/35
d) 2/5
e) 19/35
Solution:
Probability that at least 1 ball is red = 1 – probability that none of them is red.
Probability that none if the two balls is red = (9/15)(8/14)
Probability that at least 1 ball is red = 1 – probability that none of them is red. = 1- [(9/15)(8/14)] = (210-72)/210
= 138/210
=23/35
Option C is the correct answer.
Question 13: In a bag, there are 6 red balls and 9 green balls. Two balls are drawn at random, what is the probability that at least one of the balls drawn is red ?
a) 29/35
b) 7/15
c) 23/35
d) 2/5
e) 19/35 | 2,972 | 8,426 | {"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} | 4.40625 | 4 | CC-MAIN-2024-22 | latest | en | 0.822485 |
https://amazoniamovie.com/qa/what-is-the-value-of-2-power-12.html | 1,620,469,195,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988858.72/warc/CC-MAIN-20210508091446-20210508121446-00408.warc.gz | 123,122,999 | 9,248 | What Is The Value Of 2 Power 12?
What is the value of 2 power?
Computer science Two to the power of n, written as 2n, is the number of ways the bits in a binary word of length n can be arranged.
A word, interpreted as an unsigned integer, can represent values from 0 (000…0002) to 2n − 1 (111…1112) inclusively..
How do you find the value of exponents?
An exponent is a number that tells how many times the base number is used as a factor. For example, 34 indicates that the base number 3 is used as a factor 4 times. To determine the value of 34, multiply 3*3*3*3 which would give the result 81.
What is 2 by the power of 5?
5 to p…” Exponents are just repeated multiplication. 5 to power 2 is 5 time itself twice, or 5×5. 5 to power 3 is 5 times itself thrice, or 5x5x5.
How do you solve powers?
When raising a power to a power in an exponential expression, you find the new power by multiplying the two powers together. For example, in the following expression, x to the power of 3 is being raised to the power of 6, and so you would multiply 3 and 6 to find the new power.
What is the fastest way to find the power of 2?
When multiplying 2x × 2y, remember that you simply add the exponents together. For example, 23 (8) × 27 (128) = 27 + 3 = 210 (1024). Similarly, you can break up a single power of 2 into two powers which add up to the original power, such as 29 (512) = 26 + 3 = 26 (64) × 23 (8).
What power of 2 is 32?
Powers of 2 Table. Powers of 2 TableBit Line #Power of 2 Expo- nentBinary Bit Weight in Decimal30229536,870,912312301,073,741,824322312,147,483,64832 more rows•May 31, 2020
Is 1 raised to any power always 1?
The answer is yes. The reason it is always one because “to the power of” just means how many times you would multiply the number by itself. This is the reason 1 to the power of any number is 1.
What is the power of 12?
trillionPositive powersNamePowerNumberbillion (milliard)91,000,000,000trillion (billion)121,000,000,000,000quadrillion (billiard)151,000,000,000,000,000quintillion (trillion)181,000,000,000,000,000,00025 more rows
What is the value of 2 Power 50?
1,125,899,906,842,624Answer and Explanation: 2 to the 50th power equals 1,125,899,906,842,624.
What is the value of 2 Power 64?
PowerValue624,611,686,018,427,387,904639,223,372,036,854,775,8086418,446,744,073,709,551,6166536,893,488,147,419,103,23297 more rows
What is to the power of 1?
Any number raised to the power of one equals the number itself. Any number raised to the power of zero, except zero, equals one. This multiplication rule tells us that we can simply add the exponents when multiplying two powers with the same base.
What is the rule for exponents?
Only move the negative exponents. Product Rule: am ∙ an = am + n, this says that to multiply two exponents with the same base, you keep the base and add the powers. , this says that to divide two exponents with the same base, you keep the base and subtract the powers.
How do you find the nearest power of 2?
next = pow(2, ceil(log(x)/log(2))); This works by finding the number you’d have raise 2 by to get x (take the log of the number, and divide by the log of the desired base, see wikipedia for more). Then round that up with ceil to get the nearest whole number power.
Why is any base to the power of 0 1?
In short, the multiplicative identity is the number 1, because for any other number x, 1*x = x. So, the reason that any number to the zero power is one is because any number to the zero power is just the product of no numbers at all, which is the multiplicative identity, 1.
What is the value of 2 Power 100?
If you are asking for 100^2, or 100 squared, the answer is 10,000. 2 raised to the power of 100 is equal to 1,267,650,600,228,229,401,496,703,205,376.
What does 2 to the power of 8 mean?
The exponent of a number says how many times to use the number in a multiplication. In 82 the “2” says to use 8 twice in a multiplication, so 82 = 8 × 8 = 64. In words: 82 could be called “8 to the power 2” or “8 to the second power”, or simply “8 squared” Exponents are also called Powers or Indices.
Why is 12 the perfect number?
Twelve is a sublime number, a number that has a perfect number of divisors, and the sum of its divisors is also a perfect number. Since there is a subset of 12’s proper divisors that add up to 12 (all of them but with 4 excluded), 12 is a semiperfect number. … Twelve is also the kissing number in three dimensions.
Is number 12 a lucky number?
In numerology, twelve is the number of completeness. … It’s widely used in the Bible (the number of apostles, the Tribes of Israel, etc.). In general, even numbers are considered lucky, since it is believed that good luck comes in pairs. So with 12/12/12 is a like a triple dose of good luck!
What is 10 with the power of 0?
When n is less than 0, the power of 10 is the number 1 n places after the decimal point; for example, 10−2 is written 0.01. When n is equal to 0, the power of 10 is 1; that is, 100 = 1. | 1,441 | 5,002 | {"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} | 4.4375 | 4 | CC-MAIN-2021-21 | latest | en | 0.916026 |
http://math.stackexchange.com/questions/439934/quadratic-expression-that-generate-primes | 1,466,981,403,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783395560.69/warc/CC-MAIN-20160624154955-00113-ip-10-164-35-72.ec2.internal.warc.gz | 200,630,501 | 19,330 | # Quadratic expression that generate primes
I recently learned that there exist quadratic expression that generate some primes and some of these equations generate more primes than others. In the following video, the person shows the following expression
$$4x^2 -2x+1$$
that generates some primes. So I was wondering, are there infinitely many expression that exist that can produce primes and what makes one quadratic produce more primes than another? In other words, do some quadratic expressions have a special property that causes them to generate more primes than others?
Thanks!
-
@anon Sorry, which is more correct? I'm not too sure. – Jeel Shah Jul 9 '13 at 19:46
$4x^2-2x+1$ is an expression, while $4x^2-2x+1=13$ is an equation. – Tomas Jul 9 '13 at 19:47
Take any three primes. There is a quadratic which takes those three values at $x=0, x=1, x=2$ respectively - or with a little adjustment $x=1, x=2, x=3$. I am presuming you mean that primes are attained at integer values. For any prime $p$, $y=x^2$ attains the value $y=p$ for some real value of $x$. – Mark Bennet Jul 9 '13 at 19:47
I think there is more to it. $x^2 + x + 41$ is surprisingly good at generating primes. $4x^2 - 2x + 1$ doesn't seem exceptional, though – Cocopuffs Jul 9 '13 at 19:51
Can you specify, whether "are there infinitely many expression that exist that can produce primes " means, that there are infintely many functions that attain 1) one prime 2) a certain number of primes 3) a certain number of primes at consecutive values for $x$ 4) any given number of primes 5) any given number of primes at consecutive values for $x$ 6) infinitely many primes 7) infinitely many primes at consecutive values for $x$ 8) all primes ? – Tomas Jul 9 '13 at 19:55
There are infinitely many linear polynomials $f(n)=an+b$ such that $f(n)$ yields prime values at infinitely many integer arguments - I assume this is what "producing primes" means. There are also infinitely many quadratic polynomials that obtain at least one prime value, e.g. $x^2+ax+p$ for various integers $a$ and primes $p$ at $x=0$. Whether or not we know of any quadratics that provably obtain an infinite number of prime values at integer arguments, I'm not sure.
If increase the degree and number of variables, though, we do know of such polynomials that take infinitely many prime values, in fact whose positive values are only prime numbers! See this section of Wikipedia for more information.
Beyond linear polynomials (the subject of the quantitative version of Dirichlet's theorems on primes in arithmetic progressions, which generalizes in a very different direction in algebraic number theory on Cheboratev density), capturing the asymptotic frequency with which systems of polynomials simultaneously yield primes is a very hard and at the moment speculative business.
Bateman-Horn conjecture. Let $f_i(x)$ be a finite family of irreducible polynomials (say with positive leading coefficients). Let $P(n)$ count $k\le n$ for which $f_i(k)$ are prime for all $i$. Then $$P(n)\sim \frac{\displaystyle \prod_p \frac{1-N(p)/p}{(1-1/p)^m}}{(\deg f_1)\cdots(\deg f_m)}\int_2^n\frac{dt}{(\log t)^m}$$ where $N(p)$ counts the solutions to $\{ f_i(x)=0$ in ${\bf F}_p$ (the finite field of integers mod $p$).
"Heuristic" here - ubiquitous in analytic number theory - effectively means techniques based on experience, empirical data, intuition, educated guessing, and informal statistical reasoning.
Note that a very particular case of Bateman-Horn is the quantitative version of the twin prime conjecture, the $1$st Littlewood-Hardy conjecture. It also implies the Bunyakovsky_conjecture, the one-polynomial case of BH, and is a quantitative refinement of Shinzel's hypothesis, itself an extension of Bunyakovsky to multiple polynomials. Thus BH is a massively general statement.
Even so, I do not think any particular case of BH has been proven. That includes any quadratic polynomials, which are the specific subject of your question. As for intuition as to why some expressions may generate them more frequently then others (detectable in the difference in the multiplicative constants in the giant formula above), the form of the conjecture suggests that the answer may lie in local-global thinking (another thing that turns up a lot in number theory).
That is to say that there may be good reason to expect that long-term "global" nature of $P(n)$ can be reliably statistically forecasted by the "local" nature of it (the counting function $N(p)$). Given this expectation, "polynomials yield primes at different rates" reduces to "polynomials don't always have the same number of zeros mod every prime $p$," which seems (to me at least) to be not quite as surprising (since it is more finitistic, I guess).
- | 1,174 | 4,777 | {"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} | 4.09375 | 4 | CC-MAIN-2016-26 | latest | en | 0.909058 |
https://math.stackexchange.com/questions/415754/prove-that-the-limit-exist-ii | 1,632,018,917,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056656.6/warc/CC-MAIN-20210919005057-20210919035057-00625.warc.gz | 437,337,382 | 38,479 | # Prove that the limit exist II
First question was here. I add one new condition.
Let $T : H \rightarrow H$ is a linear continuous unitary ($T^*=T^{-1}$) operator, $H$ is a Hilbert space (not necessary). Suppose that
1. $\forall h \in H \Rightarrow Th=h$.
2. $\forall n \in \mathbb{N} \ || T_n ||\leq 1$. NEW CONDITION
3. $T_n$ - a sequence of linear operators $T_n \underset{n\rightarrow \infty}{\longrightarrow} T$, $|| T_n - T||\rightarrow 0$. So the sequence $T_n$ tends to $T$ in the norm and we have $\forall h \in H \ \ || T_n h - T h|| \leq || T_n - T|| \ ||h||\rightarrow 0$.
Can we prove that there exists a limit of the sum $S_n= \frac{1}{n}\left( T_1 h + T_1 T_2 h + \dots + T_1 \dots T_n h \right)$ where $n \rightarrow \infty$?
• So 1 is saying that $T=Id$, right...? Then why $T$? Jun 9 '13 at 23:53
• Because in my homework $Th=h$ only in some subspace not in entire space H. I'm asking only about this particular case.
– user59928
Jun 10 '13 at 12:57
The limit does not necessarily exist.
Assume that $H$ is a complex Hilbert space and define $T_k=e^{i\alpha_k} Id$, where $(\alpha_k)$ is a sequence of real numbers tending to $0$. Then all $T_k$'s are unitary and $\Vert T_k-Id\Vert\to 0$. On the other hand, we have $$S_n=\frac1n\left(e^{i A_1}+\dots +e^{i A_n} \right) Id ,$$ where $A_k=\alpha_1+\dots +\alpha_k$.
So one just needs to choose $(\alpha_k)$ in such a way that the sequence $(e^{iA_k})$ is not convergent in the Cesaro sense.
For example, one can proceed as follows (this is probably not the simplest way to do so). Let $(I_p)$ be a sequence of consecutive intervals of $\mathbb N$ and denote by $N_p$ the cardinality of the interval $I_p$. Put $\alpha_k=0$ if $k\in I_{p}$ for some odd $p$, and $\alpha_k=\frac{2\pi}{N_p}$ if $k\in I_p$ for some even $p$. Then $e^{iA_k}=1$ if $k\in I_p$ for some odd $p$, and the $e^{iA_k}$, $k\in I_p$ enumerate the $N_p$ roots of $1$ is $p$ is even (starting with $e^{i{2\pi}/{N_p}}$ and ending with $1$). Hence $$\sum_{k\in I_p} e^{iA_k}=\left\{ \begin{matrix} N_p&p\;{\rm odd}\cr 0&p\; {\rm even} \end{matrix}\right.$$ It follows that if the sequence $(N_p)$ is very fast increasing (say $N_1+\dots N_p=o(N_{p+1})$) and if we put $n_p=N_1+\dots +N_p$, then $\frac{1}{n_{2l}}\sum_{k=1}^{n_{2l}} e^{iA_k}\to 0$ as $l\to\infty$, whereas $\frac{1}{n_{2l+1}}\sum_{k=1}^{n_{2l+1}} e^{iA_k}\to 1$.
• I was trying to figure out: $a_k$ sequence of real numbers tending to 0. Then in your example it seems to me, that $a_k$ can't be tending to 0. Can you explain to me this moment? Sorry if this is clear.
– user59928
Jun 11 '13 at 8:06
• The sequence $(\alpha_k)$ does tend to $0$: if $k$ is large, the index $p=p(k)$ such that $\alpha_k\in I_p$ is large too, so $N_{p(k)}\to\infty$ as $k\to\infty$. Jun 11 '13 at 10:39 | 1,021 | 2,792 | {"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 | 0.7025 |
https://fongphysics.com/physics-contests/page/2/ | 1,656,179,526,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103036077.8/warc/CC-MAIN-20220625160220-20220625190220-00730.warc.gz | 301,097,511 | 23,512 | # Velocity-Time Graphs
Before Class Preparation:
• Finish Non-Uniform motion handout
• If possible, print this worksheet and
• Complete Part A (Remember these are Velocity-Time Graphs)
In Class Practice:
• PollEV Discussions for motion
• conduct an inquiry into the uniform motion of an object
• solve problems involving uniform linear motion in one dimension using graphical analysis
• Using Motion Sensors to investigate v-t graphs
• PollEV Discussions
After Class Work:
• Read & take notes on this Powerpoint if you like taking notes
# Quiz – Standing/Sound Waves
### Before Class Preparation:
• Finish Assigned Braingenies & Self Quizzes
### In Class Work:
• 15 minute group discussion
• 30 minute quiz: Application of Standing/Sound Waves
Can you:
• Determine a wave’s period & frequency when given a graph of the wave.
• Calculate the frequency of a standing wave in a medium that is:
• Open at both ends
• Closed at both ends
• Open at one end
• Calculate the speed of sound from a standing wave patterns in an air column closed at one end.
• Draw a standing wave pattern for an air column or string instrument and
• Calculate frequencies/lengths of the instrument
• Describe and calculate how sound changes (frequency, wavelength, velocity) through different mediums
### After quiz:
Take up the quiz immediately using Sage & Scribe
# Non-uniform motion (Multiple Days)
Before Class Preparation:
• Print & Complete Page 1: “Part A) – Changing Velocity” from this document
• Read & Print the rest of you can
In Class Practice:
• Day 1:
• Displacement vs Position vs Distance
• Non-Uniform Motion Group Work, Parts B) to D)
• Learning to use Labquest & Motion sensors
• Predicting shapes of d-t graphs
• Sketching d-t Graphs
• Day 2:
• Complete Non-uniform motion
After Class Work:
# Introduction to Motion
Before Class Preparation:
Videos:
In Class Activity:
• Buggy/Pull Back Car Activity
• Describe the difference between uniform & non-uniform motions
• Explain how you would predict the position of the object, 1.0 s after your last measurement
• Assessment Criteria (Thinking/Inquiry):
After Class Work:
# Waves wrap up
Before Class Preparation:
• Read and print (if you want to)
In Class Activity:
• PollEV
• Paper Lab submission check on Google Classroom
After Class Work:
# Sound Demonstrations
Before Class Preparation:
• Complete air column data tables
In Class Activity:
• Sound Demo Activities:
• investigate, in qualitative and quantitative terms, the properties of mechanical waves and sound, and solve related problems
• Simulations:
After Class Work:
# Intro to Physics Experiments: Waves
Before Class Preparation:
In Class Activity:
• Lesson: Averaging raw data and its uncertainties
• Complete Air Column Resonance Raw Data Table in Google Classroom
After Class Work:
# Resonance in an air column
Before Class Preparation:
• Read this overview of Resonance in an Air Column
• What is the distance between different parts of a standing wave (nodes, anti-node) in terms of wavelength?
• Read this handout: Air Column Resonance
• Print if possible
• Identify the variables on the handout
• We’ll be working with water, so computers are not a good idea
In Class Activity:
• What is the goal in a science experiment?
• Resonance of standing waves
• Resonance in an air column
• Verifying the speed of sound in air
• Record your results here or scan the QR code:
After Class Work:
# Quiz – Waves
### Before Class Preparation:
• Finish Assigned Braingenies & Self Quizzes on Google Classroom
### In Class Work:
• 15 minute quiz
Can you:
• State the wavelength, amplitude, velocity, frequency of a wave
• Solve problems involving the wave equation v=fλ
• Use the principle of superposition and draw a graph to represent interference of waves
### After quiz:
Take up the quiz immediately using Sage & Scribe
# Making Standing Waves
Before Class Preparation:
In Class Activity:
After Class Work: | 900 | 3,971 | {"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-2022-27 | latest | en | 0.749638 |
https://classroomsecrets.co.uk/decimals-year-5-multiply-by-10100-and-1000-discussion-problems/ | 1,713,394,923,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817181.55/warc/CC-MAIN-20240417204934-20240417234934-00020.warc.gz | 149,568,522 | 79,482 | Decimals Year 5 Multiply by 10,100 and 1,000 Discussion Problems – Classroom Secrets | Classroom Secrets
Maths Resources & WorksheetsYear 5 Maths Resources & WorksheetsSummer Block 1 (Decimals)11 Multiply by 10, 100 and 1,000 › Decimals Year 5 Multiply by 10,100 and 1,000 Discussion Problems
# Decimals Year 5 Multiply by 10,100 and 1,000 Discussion Problems
## Step 11: Year 5 Multiply by 10,100 and 1,000 Discussion Problems
Year 5 Multiply by 10,100 and 1,000 Discussion Problems includes two discussion problems which can be used in pairs or small groups to further pupils' understanding of the concepts taught in the Year 5 Multiply by 10,100 and 1,000 Resource Pack.
More resources for Summer Block 1 Step 11.
### What's included in the pack?
This pack includes:
• Year 5 Multiply by 10,100 and 1,000 Discussion Problems with answers. | 222 | 850 | {"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} | 2.515625 | 3 | CC-MAIN-2024-18 | latest | en | 0.855481 |
https://converter.ninja/volume/liters-to-metric-teaspoons/239-l-to-brteaspoon/ | 1,723,009,433,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00426.warc.gz | 140,835,577 | 5,413 | # 239 liters in metric teaspoons
## Conversion
239 liters is equivalent to 47800 metric teaspoons.[1]
## Conversion formula How to convert 239 liters to metric teaspoons?
We know (by definition) that: $1\mathrm{liter}\approx 200\mathrm{brteaspoon}$
We can set up a proportion to solve for the number of metric teaspoons.
$1 liter 239 liter ≈ 200 brteaspoon x brteaspoon$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{brteaspoon}\approx \frac{239\mathrm{liter}}{1\mathrm{liter}}*200\mathrm{brteaspoon}\to x\mathrm{brteaspoon}\approx 47800\mathrm{brteaspoon}$
Conclusion: $239 liter ≈ 47800 brteaspoon$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 metric teaspoon is equal to 2.09205020920502e-05 times 239 liters.
It can also be expressed as: 239 liters is equal to $\frac{1}{\mathrm{2.09205020920502e-05}}$ metric teaspoons.
## Approximation
An approximate numerical result would be: two hundred and thirty-nine liters is about forty-seven thousand, eight hundred metric teaspoons, or alternatively, a metric teaspoon is about zero times two hundred and thirty-nine liters.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 357 | 1,348 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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} | 3.953125 | 4 | CC-MAIN-2024-33 | latest | en | 0.697122 |
http://www.brainteaserbay.com/page/49/ | 1,701,390,124,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100258.29/warc/CC-MAIN-20231130225634-20231201015634-00398.warc.gz | 63,678,018 | 11,099 | ### The Gambler Three Coins
A gambler is holding three coins.
One coin is an ordinary quarter, the second has two heads, the third has two tails.
The gambler randomly chooses one of the coins and flips it, the coin shows head.
What is the chance that the other side shows tail?
The chance that the other is tail is one over three (1/3) as only one coin (the ordinary coin) in the three coins has head on one side and tail on the other side.
The gambler’s first toss shows heads. The question was to find the probability that the other side of the coin will show tail and not to find the probability of tails coming up. 😛
### The Barber
The barber of Seville shaves all of the men living in Seville.
No man living in Seville is allowed by law to shave himself.
The barber of Seville lives in Seville and is the only barber in Seville.
Who shaves the barber of Seville? | 206 | 878 | {"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} | 3.515625 | 4 | CC-MAIN-2023-50 | latest | en | 0.954003 |
http://stackoverflow.com/questions/2079948/normal-of-an-equilateral-triangle-that-has-been-arbitrarily-transformed-in-3d-sp | 1,444,489,798,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443737956371.97/warc/CC-MAIN-20151001221916-00095-ip-10-137-6-227.ec2.internal.warc.gz | 296,915,936 | 21,146 | # Normal of an equilateral triangle that has been arbitrarily transformed in 3D space
I have an issue that I can't quite seem to find a starting point on; I'm not even sure I can describe it well enough to get an answer.
I need to find the normal of an equilateral triangle in 3D space without knowing the points of the triangle beforehand. Think about taking a photo of a triangular "yield" street sign from any angle, and determining the out-facing normal of the sign from that photo. (I'm not doing that, exactly, but similar - so I'll use the sign/photo metaphor from here on).
** UPDATE **: This same question came up the day before I asked my version, which you can view here. Thanks to BlueRaja for pointing me there. I think the discussion there will answer the question. However, a computational approach is presented below which is also very interesting.
I know how to find the normal of a triangle when I create the triangle in code, but am unsure of how to map points to the triangle in the photo in 3D. I know the length of each side, so I know how far apart the points should be in any orientation. I can build an interactive tool that I can overlay a triangle and rotate it into location and get the points from that, but I need to do this without interaction. Doing that also doesn't help me figure out the math involved.
I'm not even sure I need to determine the points as much as just finding the correct rotation matrix.
I'm just not able to figure out where to start... Searches for the concept come up empty or just not what I'm looking to do (e.g.: they are 2D transforms not 3D)
It's also possible I'm overly complicating things and there is a simple transform equation that would do this in its sleep.
-
Be prepared to get 2 results from your calculations; one for the triangle "in front" of the projection plane, and one for "behind". – Ignacio Vazquez-Abrams Jan 17 '10 at 4:47
Indeed, I'm used to dealing with this for 2D line normals too. But thanks for the reminder. – user252466 Jan 17 '10 at 4:51
Have you tried searching for computer vision resources? This seems like the sort of problem they'd tackle often. There's also a computer-vision tag on SO that might be appropriate. – celion Jan 17 '10 at 14:54
@celion: Good point - added CV tag. Thanks. I had searched many many resources before asking here; hard to search for what i'm not exactly sure i'm looking for :-) – user252466 Jan 17 '10 at 15:28
This can be solved (aside from the built-in quadruple ambiguities) either mathematically or computationally. Since this is SO, I'll describe a computational approach.
In overview, the approach is to look at the projected angles, and since you know the true angles you can calculate the orientation. To get specific, start with the following visualization: imagine the triangle flat in the x-y plane and it's normal along the z-axis, and put a sphere here that touches all corners of the triangle, with everything centered at the origin. Now rotate the normal to all points of the sphere and note the projected angles. The key point here is that now for each possible projected angle you can draw an iso-angle path on the sphere (i.e. the path of the normal that indicates all the positions for which you observe the same projected angle -- which is probably a circle, but I'm not certain without working out the math). So to solve the original problem, take two of the observed angles, draw the iso-angle paths, and the possible solutions will be the intersection of these paths.
Computationally, construct your iso-angle paths by moving the normal in, say, 1 degree increments over the sphere, and note the three angles for each position, and then rearrange this data into iso-angle paths, by sorting it by angle. Then for two of the angles in an observed projection, find where the two iso-angle paths intersect. Note that the paths will have two intersections which corresponds to the built-in ambiguity of whether a particular corner is near or far from the observer, and also, depending on how you choose deal with reflections of the normal, the paths my be disconnected (though other than reflections I think the iso-angle paths will not be disconnected).
-
Neat! This is how I think, and this I can relate to. This brings up a new, different path (no pun intended) in my head that I'll try out on paper and see how it works. Thanks! Will update in a bit. – user252466 Jan 18 '10 at 22:06
I feel like I just answered this yesterday.
-
No, that answer is for three known points. Here the points are completely unknown. All you have to go on is the foreshortened projection onto a plane. – Ignacio Vazquez-Abrams Jan 17 '10 at 4:43
Read the comments - I tell you how to find the points in 3d, knowing what they are in 2d (ie. knowing the distance between them) – BlueRaja - Danny Pflughoeft Jan 17 '10 at 4:54
Thanks! That post didn't come up in searches (i may have used different terms). That definitely sounds similar to my problem, and there are several paths to explore there-in. In regards to your comments in that post, can you expand on what you say there so i can better grasp the full concept? (Specifically, how does that work out to 9 equations/unknowns to solve, just based on the equation you present) – user252466 Jan 17 '10 at 5:17
@gw: Ian Boyd has edited his response (in the post I linked to); I believe it is correct and very well written, I refer you to his answer now. – BlueRaja - Danny Pflughoeft Jan 20 '10 at 19:27
Wow - thorough. Thanks for pointing that out. I've been in client meetings the past week so I've not had time to look at everything. – user252466 Jan 25 '10 at 17:48 | 1,333 | 5,669 | {"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-2015-40 | latest | en | 0.956088 |
https://www.nist.gov/el/materials-and-structural-systems-division-73100/inorganic-materials-group-73103/part-iii-2 | 1,718,352,912,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861545.42/warc/CC-MAIN-20240614075213-20240614105213-00737.warc.gz | 837,087,252 | 23,306 | An official website of the United States government
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# Part III Chapter 4. Microstructural Development and Probes
## Share
This chapter covers models, in 2-D and 3-D, for simulating the development of microstructure and for probing the developing microstructure. These models include computing the curvature of digital surfaces, simulating sintering, and simulating mercury porosimetry.
This section discusses how to compute curvatures in 3D, and then applies the curvature algorithm to simulating sintering in 2D. The model is surface-attachment-limited in its kinetics, and used a square template method of measuring local curvature.
This section discusses application of the digital-image based sintering model to three dimensions. It is surface-attachment-limited in its kinetics, and used a spherical template method of measuring local curvature.
This section examines the theoretical underpinnings of the template method of curvature computation, in 2D and 3D, for an arbitrary surface.
This section discusses modelling of mercury injection in two dimensions. There is a description of the basic algorithm, and several applications.
This section discusses the validity of the Katz-Thompson approach for modelling the permeability of porous materials using parameters measured using mercury injection.
This section discusses obtaining three-dimensional brick microstructures from x-ray tomography, and then computing various transport properties to compare with experimental measurements, to see how well the tomographic image compares with real microstructure.
This section discusses reconstruction techniques, wherein a 2-D slice of a material is used to generate a 3-D approximate image of the material. The limitations of this technique is explored used a 3-D model, whose microstructure is exactly known. The 3-D reconstructured microstructure is then compared to the known 3-D microstructure both visually, and using percolation and transport properties.
Go back to Part III Chapter 3. Percolation theory
References
(1) P.J.P. Pimienta, W.C. Carter, and E.J. Garboczi, Computational Materials Science 1, 63-77 (1992).
(2) D.P. Bentz, P.J.P. Pimienta, E.J. Garboczi, and W.C. Carter, in Synthesis and Processing of Ceramics: Scientific Issues, edited by W.E. Rhine, T.M. Shaw, R.J. Gottschall, and Y. Chen (Materials Research Society Vol. 249, Pittsburgh, 1992), pp. 413-418.
(3) J.W. Bullard, E.J. Garboczi, W.C. Carter, and E.R. Fuller, Computational Materials Science 4, 103-116 (1995).
(4) E.J. Garboczi and D.P. Bentz, in Advances in Cementitious Materials, edited by S. Mindess, Ceramics Transactions 16, 365-380 (1991).
(5) E.J. Garboczi, Powder Technology 67, 121 (1991).
(6) D.P. Bentz, D.A. Quenard, H.M. Kunzel, J. Baruchel, F. Peyrin, N.S. Martys, and E.J. Garboczi, Materials and Structures, 33 , 147-153 (2000).
(7) D.P. Bentz and N.S. Martys, Transport in Porous Media 17, 221-238 (1995). | 788 | 3,206 | {"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} | 2.6875 | 3 | CC-MAIN-2024-26 | latest | en | 0.855857 |
http://electricalobjectivequestion.blogspot.com/2013/02/basic-electrical-engineering-part-11.html?showComment=1392394428099 | 1,556,293,287,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578841544.98/warc/CC-MAIN-20190426153423-20190426175423-00384.warc.gz | 51,443,749 | 21,147 | ## Basic Electrical Engineering Objective Questions
[1] The unit which indicates the rate at which energy is expended?
A. Watt
B. Ampere-hour
C. Coulomb
D. Volt
[2] The peak voltage in an ac wave is always
A. Greater than the average voltage
B. Less than the average voltage
C. Greater than or equal to average voltage
D. Less than or equal to the average voltage
[3] When an electrical charge exists but there is no flow of current, the charge is said to be
A. Ionizing
B. Electronic
C. Static
D. Atomic
[4] As the number of turns in a coil that carries ac increases without limit, the current in the coil will
A. Eventually become very large
B. Stay the same
C. Decrease, approaching zero
D. Be stored in the core material
[5] As the number of turns in a coil increases, the reactance at a constant frequency
A. Increases
B. Decreases
C. Stays the same
D. Is stored in the core material
[6] In an RL circuit, as the ratio of inductive reactance to resistance (XL/R) decreases, the phase angle
A. Increases
B. Decreases
C. Stays the same
D. Becomes alternately positive and negative
[7] The best filter for a power supply is
A. A capacitor in series
B. A choke in series
C. A capacitor in series and a choke in parallel
D. A capacitor in parallel and a choke in series
[8] Voltage regulation can be achieved by a zener diode connected in
A. Parallel with the filter output, forward biased
B. Parallel with the filter output, reverse biased
C. Series with the filter output, forward biased
D. Series with the filter output, reverse biased
[9] A current surge takes place when a power supply is first turned on because
A. The transformer core is suddenly magnetized
B. The diodes suddenly start to conduct
C. The filter capacitor must be initially charged
D. Arching takes place in the power switch
[10] A dc electromagnet
A. Has constant polarity
B. Requires an air core
C. Cannot be used to permanently magnetize anything
D. Does not attract or repel permanent magnet
[11] In a multi-stage RC coupled amplifier the coupling capacitor______________
A) Limits the low frequency response
B) Limits the high frequency response
C) Does not affect the frequency response
D) Block the DC component without affecting the frequency response
[12] It is required to measure temperature in the range of 13000 deg C to 15000 deg c. The most suitable thermocouple to be used as a transducer would be?
a) Chromel - constantan
b) Iron - constantan
c) Chromel - alumel
d) Platinum- rhodium
[13] Telemetering is a method of?
a) Counting pulses sent over long distances
b) Transmitting pictures from one place to another
c) Transmitting information concerning a process over a distance
d) None
[14] A dc to dc converter having an efficiency of 80% is delivering 16W to a load. If the converter is generating an output of 200V from an input source of 20V, then the current drawn from the source will be?
a) 0.1A
b) 0.5A
c) 1.0A
d) 10.0A
[15] The location of lighting arrestor is?
a) Near the transformer
b) Near the circuit breaker
c) Away from the transformer
d) None
12:45 PM
1. 7) The best filter for a power supply is capacitor in parallel and a choke in series
1. At high frequencies, inductive impedance is high, but capacitive rectance is low. so unwanted high frequency noise is bypassed through capacitor.
2. XL==2*pie*F*L
while
XC==1/2*pie*F*C
2. In a multi-stage RC coupled amplifier the coupling capacitor______________
is Limits the low frequency response
can you explain this sir..?
3. [15] The location of lighting arrestor is?
1. near the trafo
2. Lightning arrestors are used on transmission lines not on transformers...
4. Sir send me basic electrical ....machine....measurment......trasmission.....switchgear and protection....iluumination....genration...ac fundamental mcq question sir plz your all question are good for comptatiove exam point of view sir send question on my gmail pawandeep172@gmail.com plz sir
5. Sir plz send all mcq question pawandeep172@gmail.com
6. location of lighting arrestor near trasformer is rt plzzz correct that 1 | 1,061 | 4,087 | {"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-2019-18 | longest | en | 0.896919 |
http://math.stackexchange.com/questions/204181/definite-integral-of-expaxbxc/204220 | 1,394,551,845,000,000,000 | text/html | crawl-data/CC-MAIN-2014-10/segments/1394011217448/warc/CC-MAIN-20140305092017-00033-ip-10-183-142-35.ec2.internal.warc.gz | 116,599,022 | 15,959 | # Definite Integral of $\exp(ax+bx^c)$
Can you please provide some thoughts / ideas / help in computing this definite integral? Any help will be great...I am so stuck with this one.
$$\int_0^{\infty}\exp\left(ax+bx^c\right)\,dx,$$
where $a < 0$, $b < 0$ and $c > 0$.
It looks like this one might not have a clean analytical solution but is there any standard form that this reduces to?
Thanks a lot for your help
Trambak
-
I don't think there will be a general standard form since this highly depends on the value of $c$. – Patrick Li Sep 28 '12 at 21:18
In general, there will be no non-recursive expressions. For example, even the simplest case:
$$\int_0^{\infty} \!\!e^{-x^k} \, dx = \frac{1}{k}\Gamma\left(\frac{1}{k}\right) .$$
Here $\Gamma : \mathbb{C} \to \overline{\mathbb{C}}$ denotes Euler's Gamma function, defined by
$$\Gamma(z) := \int_0^{\infty} e^{-t} \, t^{z-1} \, dt \, .$$
Of course, there are some special values of $k$ which give closed form expressions, e.g. $k = 2$ gives $\sqrt{\pi}/2$, but in general you have no hope of finding a nice expression.
(If there were then it'd be in the calculus books by now!)
-
Last sentence is a perfect example of "proof by lack of discovery by really smart people." :P /teasing – anorton Aug 12 '13 at 14:53
@anorton :o) The problem is because of the limited array of functions that we call elementary. Outside of trigonometric, exponential, logarithmic, and a jumble of all of these, there isn't too much left. Why should $\sin(\operatorname{e}^x)$ be allowable as a closed-form, when other functions aren't? Indeed, why not include $\Gamma$ in the set of allowable elementary functions, and then all of the problems go away, $\Gamma(z)$ would trivially be a closed form in itself. – Fly by Night Aug 12 '13 at 19:27
I tried the integration by parts bit. Here is how it looks:
say $$I = \int_0^{\infty}e^{ax+bx^c}dx$$ $$= \left(\dfrac{e^{ax+bx^c}}{a}\right)_{0}^{\infty} - \int_0^{\infty}\dfrac{bc}{a}x^{c-1}e^{ax+bx^c}dx$$ $$= -\dfrac{1}{a} -\dfrac{1}{a}\int_0^{\infty}(a+bcx^{c-1})e^{ax+bx^c}dx + I$$
For $a < 0$, $b < 0$ and $c > 1$, this thing results in the trivial identity $0=0$. For $c=1$, it is easily computable. Am I completely off here?
Thanks Trambak
-
$\int_0^\infty e^{ax+bx^c}~dx$
$=\int_0^\infty e^{ax}e^{bx^c}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{b^nx^{cn}e^{ax}}{n!}dx$
$=\sum\limits_{n=0}^\infty\dfrac{b^n\Gamma(cn+1)}{(-a)^{cn+1}n!}dx$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions#definite_integrals) | 868 | 2,554 | {"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": 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} | 3.328125 | 3 | CC-MAIN-2014-10 | latest | en | 0.860479 |
https://www.keele.ac.uk/catalogue/current/mat-30001.htm | 1,627,056,253,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046149929.88/warc/CC-MAIN-20210723143921-20210723173921-00192.warc.gz | 885,260,638 | 3,133 | MAT-30001 - Graph Theory
Coordinator: Raymond N Turner Room: MAC2.27 Tel: +44 1782 7 33739
Lecture Time: See Timetable...
Level: Level 6
Credits: 15
Study Hours: 150
School Office: 01782 733075
Programme/Approved Electives for 2021/22
None
Available as a Free Standing Elective
No
Co-requisites
None
Prerequisites
None
Barred Combinations
None
Description for 2021/22
This module introduces the concept of a graph as a pictorial representation of a symmetric relation. A variety of topics are investigated and, for each one, at least one of the major theorems is proved. The emphasis is on pure graph theory although some applications are explored via worked examples and coursework.
Aims
The aim of this module is to study various topics in graph theory, together with a number of applications.
Intended Learning Outcomes
recognise, and establish properties, of different types of graphs such as trees, bipartite and complete graphs: 1,2
prove, and apply, conditions for a graph to be Eulerian or Hamiltonian: 1,2
prove, and apply, results related to the colouring of the vertices or edges of a graph: 1,2
prove, and apply, results related to properties of extremal graphs: 2
prove, and apply, results concerning planar graphs: 2
Study hours
Learning/teaching comprises 30 hours video lectures, 5 hours flipped examples classes, and 2 hours final exam.
Independent study comprises 30 hours examples class preparation, 10 hours for completion of assignment, 20 hours preparation for examination, and 53 hours consolidation of lecture material.
School Rules
None
Description of Module Assessment
1: Assignment weighted 20%
On-line, take-home assignment
One take-home, written assignments to be completed on-line. The assignment consists of a set of questions with pre-allocated space for written solutions which will be uploaded to the KLE. Students should expect to spend 10 hours on the assessment.
2: Open Book Examination weighted 80%
On-line, open book examination
The examination paper will consist of no less than five and not more than eight questions all of which are compulsory. In response to Covid, the examination will be online and open book. A well-prepared student should expect to complete the assessment in two hours. | 517 | 2,256 | {"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} | 2.59375 | 3 | CC-MAIN-2021-31 | longest | en | 0.907198 |
https://mathlesstraveled.com/category/recursion/ | 1,680,440,124,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296950528.96/warc/CC-MAIN-20230402105054-20230402135054-00524.warc.gz | 426,243,188 | 23,153 | Category Archives: recursion
Computing optimal play for the greedy coins game, part 4
Last time I explained a method for computing best play for instances of the greedy coins game, which is feasible even for large games. This general approach is known as dynamic programming and is applicable whenever we have some recursively … Continue reading
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Computing optimal play for the greedy coins game, part 3
In a previous post we saw how we can organize play sequences in the greedy coins game into a tree. Then in the last post, we saw how to work our way from the bottom of the tree upward and … Continue reading
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Computing optimal play for the greedy coins game, part 2
I want to explain in more detail how we can think about computing the best possible score for Alice in the greedy coins game, assuming best play on the part of both players. I glossed over this too quickly in … Continue reading
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Computing optimal play for the greedy coins game
Recall the greedy coins game, in which two players alternate removing one of the coins from either end of a row, and the player with the highest total at the end is the winner. What if we wanted to play … Continue reading
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The Recamán sequence
I recently learned about a really interesting sequence of integers, called the Recamán sequence (it’s sequence A005132 in the Online Encyclopedia of Integer Sequences). It is very simple to define, but the resulting complexity shows how powerful self-reference is (for … Continue reading
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Apollonian gaskets and Descartes’ Theorem II
In a few previous posts I wrote about “kissing sets” of four mutually tangent circles, and the fact that their signed bends satisfy Descartes’ Theorem, (Remember that the signed bend of a circle is like the curvature , except that … Continue reading
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In my previous post, I explained a recursive procedure for drawing Apollonian gaskets. Given any three mutually tangent circles, there are exactly two other circles which are mutually tangent to all three (forming what we called a “kissing set”). This … Continue reading
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In my last post I showed off this tantalizing picture: This pattern of infinitely nested circles is called an Apollonian gasket. Over the next post or two I’ll explain some cool math behind actually constructing them. Mostly I will state … Continue reading
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More factorization diagrams
My post on factorization diagrams from a month ago turned out to be (unexpectedly) quite popular! I got ten times as many hits as usual the day it was published, and since then quite a few other people have created … Continue reading
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Factorization diagrams
In an idle moment a while ago I wrote a program to generate "factorization diagrams". Here’s 700: It’s easy to see (I hope), just by looking at the arrangement of dots, that there are in total. Here’s how I did … Continue reading
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https://www.cours-et-exercices.com/2018/01/magnetic-effects-of-electric-currents.html | 1,723,105,814,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00590.warc.gz | 565,359,639 | 36,264 | ### Magnetic Effects Of Electric Currents, And Electrical Measurements
Chapter XII. Magnetic Effects Of Electric Currents,And Electrical Measurements.
### (1) The Magnetic Effect of Electric Currents
255. The Magnetic Effect.?Of all the effects of electric currents, it is generally conceded that the magnetic effect is the one of greatest practical importance, and it is also the one most extensively used. An experiment illustrating this effect has been described in Art. 239. This experiment shows that an electric current, if parallel to a magnetic needle, and near it will deflect the north-seeking pole of the needle to the right or left depending upon the direction of the current flow. This deflection of the magnetic needle is due to the fact that surrounding every electric current are magnetic lines of force. It is this magnetic field of the current that causes the needle to turn. The position taken by the needle is the resultant of the forces of two magnetic fields; one, the earth's field, the other, that of the current.
256. Right-hand Rule for a Conductor.?To show the presence of the magnetic field about a current, pass a thick copper wire vertically through a sheet of paper, and connect the ends of the wire to a source of current. While the current (this should be as much as 10 amperes if possible) is flowing, sprinkle iron filings upon the paper and tap gently. The filings will arrange themselves in circles about the wire showing the magnetic field. (See Fig. 229.) The needle of a magnetoscope tends to place itself parallel to the lines of force of this field and from this action or[Pg 280] tendency the direction of the magnetic lines about a current may be determined. The following rule is helpful and should be memorized: Grasp the conductor with the right hand with the outstretched thumb in the direction that the current is flowing. The fingers will then encircle the wire in the direction of the lines of force. This rule may be reversed, for, if the fingers of the right hand grasp the wire so as to point with the magnetic field, then the current flows in the direction in which the thumb points. (See Fig. 230.)
Fig. 229.?Magnetic field about a wire carrying an electric current.
Fig. 230.?Right-hand rule for the magnetic field of a current.
257. Magnetic Field of a Helix.?If a wire be wound about a cylinder to form a cylindrical coil with parallel turns, it forms a helix or solenoid. The shape of the magnetic field about a current depends upon the form of the conductor. If the latter is in the form of a helix its magnetic field resembles that of a straight bar magnet. (See Fig. 231). In fact the helix has the properties of a magnet with north- and south-seeking poles while a current is flowing through it. If such a coil is suspended so as to turn freely, it tends to turn until the field within it is parallel to the earth's magnetic field. Such a suspended helix may therefore be used as a compass. In order to strengthen the magnetic field of a helix or solenoid, the space within its turns is filled with iron, often in the form[Pg 281] of small soft-iron wires. This bundle of iron wire is called the core of the helix. The core becomes strongly magnetized by the field of the helix while the current is flowing and quickly loses its magnetic force when the current is stopped. The direction of the current in a helix (Fig. 232) or the polarity of its core may be determined by another right-hand rule. If the helix is grasped with the right hand so that the fingers point in the direction in which the current is flowing, the extended thumb will point in the direction of the north pole of the helix. On the other hand, if the poles of the helix are known, then, when the helix is grasped with the right hand so that the thumb points to the north-seeking pole, the current is flowing in the wires in the direction that the fingers point.
Fig. 231.?The magnetic field of a helix.
Fig. 232.?Right-hand rule for a helix.
258. The Electromagnet.?These "right-hand" rules are applied in many different devices. Among these, perhaps the most important is the electromagnet, which is used in the electric bell, the telegraph, the telephone, the dynamo, the motor, and many other electric contrivances.
The electromagnet is defined as a mass of iron around which is placed a helix for conducting an electric current. On account of its large permeability, the iron core of the helix adds greatly to the effectiveness of the electromagnet, since the magnetism of the iron is added to that of the current in the helix. The magnetism remaining[Pg 282] in the iron after the current stops is called the residual magnetism. The residual magnetism is small when the core is made of small wires or thin plates, but is larger when the iron core is solid. Like artificial steel magnets, electromagnets are usually of two forms, bar and horseshoe. (See Figs. 233 and 234.) For most purposes the horseshoe form is the more effective since it permits a complete iron circuit for the magnetic lines of force. (See Fig. 235.) This is the form used in the electric bell, in the telegraph sounder, and in lifting magnets. (See Fig. 236.)
Fig. 233.?A bar electromagnet.
Fig. 234.?A horseshoe electromagnet.
Fig. 235.?A horseshoe electromagnet may have a complete iron circuit for its lines of force.
Fig. 236.?A lifting magnet.
259. Effective Electromagnets.?The magnetic effect of a current in a helix is small, hence the force usually is increased by inserting a core of iron. When at first man[Pg 283] tried to signal with electromagnets at a distance it was found that the current would not work the electromagnet. An American by the name of Joseph Henry discovered the remedy for this condition. He found that if the copper wire was insulated by wrapping silk thread about it, and then many layers of the silk insulated wire were wound upon a spool with an iron core, that the magnet would work at a great distance from the source of current. If the current is increased, the magnet is stronger than at first. Thus an electromagnet may be made stronger by (a) increasing the number of turns of wire in its coils and by (b) sending a stronger current through it.
Fig. 237.?A simple telegraph circuit.
260. The Telegraph.?The invention of an effective electromagnet by Henry made possible the electric telegraph. In its simplest form it consists of a battery, C, a key, K, and a sounder, S, with connecting wires. (See Fig. 237.) The sounder (Fig. 238) contains a horseshoe electromagnet and a bar of soft iron across its poles called an armature, A, attached to a lever L. When the key is closed, the electromagnet draws down the armature and lever until the latter hits a stop O, making a click. When the key is raised, the magnet releases the armature which is raised by the action of a spring at S until the lever hits a stop at T making another click. Closing and opening the circuit at K will start and stop the current which operates S which may be 100 miles or more from K. One voltaic cell will work a sounder in the same room. But if many miles of wire are in the circuit, the E.M.F. of a single cell will not force sufficient current through the long wire to operate the sounder.
[Pg 284]
Fig. 238.?A telegraph sounder.
Fig. 239.?A telegraph relay.
Fig. 240.?How the relay is used.
Samuel F. B. Morse (1791-1872). Inventor of the electromagnetic recording telegraph and of the dot and dash alphabet.
Samuel F. B. Morse
"From Appleton's Cyclopedia of American Biography, Copyright 1888 by D. Appleton & Co."
Thomas A. Edison, Orange, New Jersey. Invented the incandescent lamp; phonograph; moving picture; most noted inventor of electrical appliances of the present day.
Thomas A. Edison
"Copyright, Photographische Gessellschaft," and "By Permission of the Berlin Photographic Co., New York."
A battery of several cells is then required. Even a large battery is insufficient to operate a long line containing many sounders in circuit. Recourse is therefore usually made to a more sensitive device called a relay. (See Fig. 239.) In the relay a very small current will magnetize its electromagnet enough to draw toward it the delicately hung armature thereby closing a second circuit which contains a sounder and a battery. (See Fig. 240.) when the current in the main circuit is stopped, the armature of the relay is drawn back by a light spring. This opens the local circuit. Thus the local circuit is closed and opened by the relay just in time with the starting and stopping of the current in the main line. It is thus possible for a small current in the main line by the use of a relay, to close and open a second local circuit containing a local battery and sounder. Modern telegraph lines are operated in this manner.
Fig. 241.?An electric bell and its circuit.
261. The electric bell (see Fig. 241), consists of an electromagnet, M, a soft iron armature, A, attached to the tapper, T, and a post, R. When no current is flowing a spring at S holds the armature against the post R. When current flows through the helix, its core becomes magnetized and attracts the armature, drawing it away from the post, R, and causing the tapper to hit the bell. Drawing A away from the post, however, breaks the circuit at R and the current stops. The magnetism in the core disappears releasing the armature, which is then pulled back by the spring S against the post R. This completes the circuit and the process repeats itself several times a second as long as the current flows.
Fig. 242.?Magnetizing by the discharge of a Leyden jar.
262. Static and Current Electricity Compared.?The likeness between a discharge of static electricity and an electric current may be shown by winding a coil of insulated wire about a glass tube which contains a steel needle. If a Leyden jar (see Fig. 242) is discharged through the coil the steel needle is usually found to be magnetized, showing that the discharge of the static electricity has a magnetic effect similar to that of an electric current. Sometimes a given end of the needle has a north pole and at other times a south pole. This is believed to indicate[Pg 288] that the charge of the Leyden jar is oscillatory, and that in different discharges sometimes a surge in one direction and at other times a surge in the reverse direction has been most effective in magnetizing the needle. Compare this action with that described in Art. 233.
#### Important Topics
1. Right-hand rules, for conductor, for helix.
2. The electromagnet, two forms, where used?
3. Likeness between static and current electricity.
4. The electric bell, parts, action.
5. The telegraph, key, sounder, relay.
#### Exercises
1. What is the difference between an electric charge and a current?
2. How can a magnetic effect be produced from an electric charge?
3. What is a magnetic field? Give two evidences of a magnetic field about a current in a wire?
4. A current is flowing north in trolley wire, what is the direction of the magnetic field under the wire? Explain.
5. What would be the result if a hard steel core were placed in the electromagnet? Explain.
6. If the north-seeking pole of a helix is facing you, does the current in the coils before you move in a clockwise or in a counter-clockwise direction? Explain.
7. A helix is placed horizontally with its north-seeking pole toward the north. Does the current in the wire at the top of the helix move east or west? Explain.
8. State at least six conditions any one of which will put an electric bell circuit out of commission.
9. If one desires to insert a battery into a telegraph circuit already in operation, how will he determine the direction of the current in the wire?
10. If a boy who had magnetized his knife blade in a physics laboratory, pointed end south-seeking, should lose his way in the woods on a cloudy day, how could he determine his way out?
11. At a certain point the earth's field acts north, that of an electric current, east. The magnetoscope needle points exactly northeast when placed at that point. How do the two magnetic fields compare?
[Pg 289]
### (2) Electrical Measurements
263. Galvanometers.?In using electric currents it is often necessary or desirable to be able to know not only that a given current is weak or strong, but precisely what its strength is. We can determine the relative strengths of two currents by the use of a galvanometer.
Fig. 243.?The magnet is at the center of the coil.
Fig. 244.?A moving-magnet (tangent) galvanometer.
The older or moving-magnet type of galvanometer is similar to the galvanoscope mentioned in Art. 239. It consists of a magnetic needle mounted at the center of a coil of wire. The coil is placed facing east and west, so that the needle will be held by the earth's magnetic field parallel to the plane of the coil. When a current is sent through the coil a magnetic field is produced within it. This deflects the needle, its north end turning east or west depending upon the direction of the current. (See Fig. 243.) The coils of a moving-magnet or tangent galvanometer (see Fig. 244) are large and firmly fastened to the base, while the magnet is small.
The moving-coil type of galvanometer (see Fig. 245) consists of a large magnet fastened to the frame of the device. The magnet usually has a horseshoe form to[Pg 290] produce as strong a field as possible. The coil is wound on a light rectangular frame and is suspended between the two poles of the magnet. To concentrate the magnetic field, a cylinder of soft iron is usually placed within the coil. Fig. 246 represents a common form of moving-coil galvanometer.
Fig. 245.?To illustrate the principle of the moving-coil galvanometer.
Fig. 246.?A moving-coil (D'Arsonval) galvanometer.
264. Measurement of Electric Currents.?A galvanometer enables one to compare electric currents. To measure electric currents it is necessary to employ a unit of electrical quantity, just as in measuring the quantity of water delivered by a pipe, a unit of liquid measure is employed; thus, e.g., the current delivered by a given pipe may be 2 gallons of water per second, so in measuring the flow of an electric current one may speak of two coulombs per second. The coulomb is the unit quantity of electricity just as the unit of quantity of water is the gallon.
For most practical purposes, however, we are more interested in the rate or intensity of flow of current than in[Pg 291] the actual quantity delivered. The unit of rate of flow or current is called the ampere.
In determining the exact quantity of an electric current, physicists make use of a device called a coulomb meter. (See Fig. 247.) This contains a solution of silver nitrate in which are placed two silver plates. The current to be measured is sent through the solution, in at one plate and out at the other. The plate where the current goes in, the anode, A (Fig. 247), loses in weight since some of the silver is dissolved. The plate where the current goes out, the cathode, C, increases in weight since some of the silver is deposited. By an international agreement, the intensity of the current which deposits silver at the rate of 0.001118 g. per second is 1 ampere. This is equal to 4.025 g. per hour.
Fig. 247.?A coulomb meter, the anode A is separated from the cathode C by a porous cup.
The coulomb is defined as the quantity of electricity delivered by a current of one ampere during one second.
A 40-watt-incandescent lamp takes about 0.4 ampere of current. An arc lamp takes from 6 to 15 amperes. A new dry cell may send 20 amperes through a testing meter. A street car may take from 50 to 100 amperes.
265. The Ammeter.?The method described above is not used ordinarily for measuring current strengths on account of its inconvenience. The usual device employed is an ammeter. This instrument is a moving-coil galvanometer.[Pg 292] It contains, wound on a light form, a coil of fine copper wire. The form is mounted on jewel bearings between the poles of a strong permanent horseshoe magnet. (See Fig. 248.) As in other moving-coil galvanometers, a soft iron cylinder within the form concentrates the field of the magnet. The form and its coil is held in balance by two spiral springs which also conduct current into and out of the coil.
Only a small part of the whole current measured, in some cases only 0.0001 passes through the coil, the larger part of the current passing through a metal wire or strip called a shunt[L] (see Fig. 248) connecting the binding posts of the instrument. A fixed fraction of the whole current flows through the coil. Its field crossing the field of the horseshoe magnet, tends to turn until its turning force is balanced by the spiral springs. As the coil turns it moves a pointer attached to it across a scale graduated to indicate the number of amperes in the whole current.
Fig. 248.?Diagram of a commercial ammeter. S is the shunt.
It should be noted that while all of the current measured passed through the ammeter, but a small part goes through the coil.
[Pg 293]
266. Resistance of Conductors.?With an ammeter one may study the change produced in the amount of current flowing in a wire when a change is made in the wire conducting the current. For example, if one measures with an ammeter the current flowing from a dry cell through a long and then through a short piece of fine copper wire, it will be seen that less current flows when the long piece is used. That is, the long wire seems to hinder or to resist the passing of the current more than the short piece. In other words, the long wire is said to have more resistance.
The resistance of a conducting body is affected by several conditions.
(a) It is directly proportional to the length of the conductor, one hundred feet of wire having twice the resistance of fifty feet.
(b) It is inversely proportional to the square of the diameter; a wire 0.1 inch in diameter has four times the resistance of a wire 0.2 inch in diameter.
(c) It differs with different substances, iron having about six times as much as copper.
(d) It varies with the temperature, metals having greater resistance at a higher temperature.
Since silver is the best conductor known, the resistances of other substances are compared with it as a standard.
The ratio of the resistance of a wire of any substance as compared to the resistance of a silver wire of exactly the same diameter and length is called its relative resistance.
Purified substances arranged in order of increasing resistance for the same length and sectional area (Ayrton-Mather) are given on p. 294.
[Pg 294]
Silver annealed 1 Copper annealed from 1.04 (Copper annealed) to 1.09 Aluminum annealed 1.64 Nickel annealed 4.69 Platinum annealed 6.09 Iron annealed 6.56 German Silver from 12.8 (German Silver) to 20.2 Mercury 63.3 Nichrome 67.5 Carbon from 2700 (Carbon) to 6700
267. The ohm, the unit of resistance, is defined by international agreement as follows: An ohm is the resistance of a column of pure mercury, 106.3 cm. long with a cross-section of a square millimeter and at a temperature of 0?C.
It should be noted that each of the four conditions affecting resistance is mentioned in the definition, viz., length, cross-section, material, and temperature. Since it is inconvenient to handle mercury, standard resistance coils, made of an alloy of high resistance are used in comparing and measuring resistances.
A piece of copper wire No. 22 (diameter 0.644 mm.) 60. 5 ft. long has a resistance of 1 ohm. See table p. 296.
The resistance of some telephone receivers is 75 ohms, of a telegraph sounder, 4 ohms, of a relay 200 ohms.
268. Resistance of Circuits.?Every part of an electrical circuit possesses resistance. In an electric-bell circuit, for instance, the wires, the bell, the push-button, and the cell itself, each offers a definite resistance to the passage of the current. The resistance within the cell is termed internal resistance, while the resistance of the parts outside of the electric generator is called external resistance.
[Pg 295]
269. Electromotive Force.?In order to set in motion anything, some force must be applied. This is as true of electricity as of solids, liquids, or gases. By analogy that which is exerted by a battery or by a dynamo in causing current to flow is called an electromotive force. The unit of electromotive force, the volt, may be defined as the electromotive force that will drive a current of 1 ampere through the resistance of 1 ohm. The electromotive force of a dry cell is about 1.5 volts, of a Daniell cell 1.08 volts. Most electric light circuits in buildings carry current at 110 or 220 volts pressure. Currents for street cars have an electromotive force of from 550 to 660 volts.
Fig. 249.?Diagram of a commercial voltmeter.
270. The Voltmeter.?An instrument for measuring the electromotive force of electric currents is called a voltmeter (Fig. 249). It is usually a moving-coil galvanometer, and is always of high resistance. It is like an ammeter in construction and appearance. In fact, a voltmeter is an ammeter which has had its shunt removed or disconnected. In place of a shunt, the voltmeter uses a coil of wire of high resistance (see R, Fig. 249) in series with the galvanometer coil. The high resistance of the voltmeter permits but a very small current to flow through it. Hence a voltmeter must be placed across a circuit and[Pg 296] not in it. In other words a voltmeter is connected in shunt, while an ammeter is in series with the circuit as is shown in Fig. 250.
Dimensions and Functions of Copper Wires
B. & S. gauge number Diameter Circular mils Sectional area in square millimeters Weight and length, Density = 8.9, feet per pound Resistance at 24?C., feet per ohm Capacity in amperes Mils Millimeters 0000 460.000 11.684 211,600.00 107.219 1.56 19,929.700 312.0 000 409.640 10.405 167,805.00 85.028 1.97 15,804.900 262.0 00 364.800 9.266 133,079.40 67.431 2.49 12,534.200 220.0 0 324.950 8.254 105,592.50 53.470 3.13 9,945.300 185.0 2 257.630 6.544 66,373.00 33.631 4.99 6,251.400 131.0 4 204.310 5.189 41,742.00 21.151 7.93 3,931.600 92.3 6 162.020 4.115 26,250.50 13.301 12.61 2,472.400 65.2 8 128.490 3.264 16,509.00 8.366 20.05 1,555.000 46.1 10 101.890 2.588 10,381.00 5.260 31.38 977.800 32.5 12 80.808 2.053 6,529.90 3.309 50.69 615.020 23.0 14 64.084 1.628 4,106.80 2.081 80.59 386.800 16.2 16 50.820 1.291 2,582.90 1.309 128.14 243.250 11.5 18 40.303 1.024 1,624.30 0.823 203.76 152.990 8.1 20 31.961 0.812 1,021.50 0.5176 324.00 96.210 5.7 22 25.347 0.644 642.70 0.3255 515.15 60.510 4.0 24 20.100 0.511 504.01 0.2047 819.21 38.050 2.8 26 15.940 0.405 254.01 0.1288 1,302.61 23.930 2.0 28 12.641 0.321 159.79 0.08097 2,071.22 15.050 1.4 30 10.025 0.255 100.50 0.05092 3,293.97 9.466 1.0 32 7.950 0.202 63.20 0.03203 5,236.66 5.952 0.70 34 6.304 0.160 39.74 0.02014 8,328.30 3.743 0.50 36 5.000 0.127 25.00 0.01267 13,238.83 2.355 0.35 38 3.965 0.101 15.72 0.00797 20,854.65 1.481 0.25 40 3.144 0.080 9.89 0.00501 33,175.94 0.931 0.17
#### Important Topics
(1) Galvanometers: (1) moving magnet, fixed coil; (2) moving coil, fixed magnet, ammeter, voltmeter.
(2) Unit of quantity, coulomb.
(3) Unit of current, ampere.
(4) Unit of resistance, ohm.
(5) Unit of electromotive force, volt.
[Pg 297]
#### Exercises
1. How will the resistance of 20 ft. of No. 22 German silver wire compare with that of 10 ft. of No. 22 copper wire? Explain.
2. Where in a circuit is copper wire desirable? Where should German silver wire be used?
3. Explain the action of the ammeter. Why does not the needle or coil swing the full distance with a small current?
4. Why is a telegraph sounder more apt to work on a short line than upon a long one?
Fig. 250.?The ammeter is connected in series and the voltmeter in shunt.
5. Find the resistance of 15 miles of copper telephone wire No. 12. (See table p. 296.)
6. What will be the weight and resistance of 1,000 feet of No. 20 copper wire?
7. A storage battery sends 4 amperes of current through a plating solution. How much silver will it deposit in 2 hours?
8. (a) Compare the diameters of No. 22 and No. 16 copper wire.
(b) Compare the lengths of the same wires giving 1 ohm resistance.
(c) What relation exists between (a) and (b)?
9. Why is an electric bell circuit usually open while a telegraph line circuit is usually closed?
10. A copper wire and an iron wire of the same length are found to have the same resistance. Which is thicker? Why?
11. Why are electric bells usually arranged in parallel instead of in series?
12. What would happen if a voltmeter were put in series in a line?
[Pg 298]
### (3) Ohm's Law and Electrical Circuits
271. Conditions Affecting Current Flow.?Sometimes over a long circuit one cell will not work a telegraph sounder. In such a case, two, three, or more cells are connected so that the zinc of one is joined to the copper plate of the other. When connected in this way the cells are said to be in series (Fig. 251). In the figure A represents a voltmeter. It is found that when cells are in series the E.M.F. of the battery is the sum of the electromotive forces of the cells. An ammeter in the circuit shows increased current as the cells are added. Hence if the resistance of the circuit remains unchanged, the greater the E.M.F. the greater is the current strength. In this respect, the movement of electricity in a circuit is similar to the flow of water in a small pipe under pressure, as in the latter the flow of water increases as the pressure becomes greater. The current in a circuit may also be increased by lessening the resistance, since the current through a long wire is less than that through a short one, just as the flow of water will be greater through a short pipe than through a long one. To increase the current flowing in an electric circuit, one may therefore either increase the E.M.F. or decrease the resistance.
Fig. 251.?Diagram of cells connected in series.
272. Ohm's Law.?The relation between the electromotive force applied to a circuit, its resistance, and the current produced was discovered in 1827 by George Ohm. Ohm's law, one of the most important laws of electricity, states that, in any circuit, the current in amperes equals the electromotive force in volts divided by the resistance in ohms.
[Pg 299]
This principle is usually expressed thus:
Current intensity = electromotive force/resistance or
Amperes = volts/ohms or I = E/R
Fig. 252.?The street cars are connected in parallel with each other.
273. Resistance of Conductors in Series.?A study of the resistance of conductors when alone and when grouped in various ways is of importance since, the current flow through any circuit is dependent upon its resistance. The two most common methods of combining several conductors in a circuit are in series and in parallel. Conductors are in series when all of the current passes through each of the conductors in turn (Fig. 218), thus the cell, push-button, wires, and electric bell in an electric-bell circuit are in series. Conductors are in parallel when they are so connected that they are side by side and a part of the whole current goes through each. None of the current that passes through one conductor can go through the conductors in parallel with it. Thus the electric street cars are in parallel with each other. (See Fig. 252.) It is easily seen that none of the current passing through one car can go through any of the others. When the conductors are in series the combined resistance is the sum of the several resistances. Thus in an electric-bell circuit if the battery has a resistance of 1 ohm, the bell of 2 ohms, and the wire 1 ohm, the total resistance in the circuit is 4 ohms. When conductors are in parallel the combined resistance is always less than the separate resistances. Just as a crowd of people meets less resistance in leaving a building through several exits, so electricity[Pg 300] finds less resistance in moving from one point to another along several parallel lines, than along one of the lines.
274. Resistance of Conductors in Parallel.?If three conductors of equal resistance are in parallel, the combined resistance is just one-third the resistance of each separately (Fig. 253). The rule that states the relation between the combined resistance of conductors in parallel and the separate resistances is as follows:The combined resistance of conductors in parallel is the reciprocal of the sum of the reciprocals of the several resistances. For example, find the combined resistance of three unequal resistances in parallel; the first being 4 ohms, the second, 6 ohms, and third 3 ohms. The reciprocals of the three resistances are 1/4, 1/6, and 1/3. Their sum equals 6/24 + 4/24 + 8/24 = 18/24. The reciprocal of this is 24/18 which equals 1-1/3 ohms, the combined resistance.
Fig. 253.?The three conductors are connected in parallel.
This rule may be understood better if we consider the conductance of the conductors in parallel. Since the conductance of a two ohm wire is just one-half that of a one-ohm wire, we say that the conductance of a body is inversely as the resistance, or that it is the reciprocal of the resistance. The conductance of the 4-, 6-, and 3-ohm coils will therefore be respectively 1/4, 1/6, and 1/3, and since the combined conductance is the sum of the several conductances, the total conductance is 18/24. Also since this is the reciprocal of the total resistance, the latter is 24/18 or 1-1/3 ohms.
When two or more conductors are connected in parallel each one is said to be a shunt of the others. Many circuits are connected in shunt or in parallel. Fig. 254 represents four lamps in parallel. Incandescent lamps in buildings are usually connected in parallel, while arc lamps are[Pg 301] usually connected in series. Fig. 255 represents four lamps in series.
#### Important Topics
1. Conditions affecting current flow, (a) E.M.F., (b) resistance.
2. Ohm's law, three forms for formula.
3. Resistance of conductors: (a) in series, (b) in parallel; how computed, illustrations.
Fig. 255.?The four lamps are connected in series.
Fig. 254.?The four lamps are connected in parallel.
#### Exercises
1. What current flows through a circuit if its E.M.F. is 110 volts and the resistance is 220 ohms?
2. A circuit contains four conductors in series with resistances of 10, 15, 6, and 9 ohms respectively. What current will flow through this circuit at 110 volts pressure? What will be the resistance of these four conductors in parallel?
3. What is the combined resistance of 8 conductors in parallel if each is 220 ohms? What current will flow through these 8 conductors at 110 volts pressure?
4. What is the resistance of a circuit carrying 22 amperes, if the E.M.F. is 20 volts?
5. What E.M.F. will send 8 amperes of current through a circuit of 75 ohms resistance?
6. How does the voltmeter differ from the ammeter?
7. How can one determine the resistance of a conductor?
8. The resistance of a hot incandescent lamp is 100 ohms. The current used is 1.1 amperes. Find the E.M.F. applied.
9. What is the resistance of the wires in an electric heater if the current used is 10 amperes, the voltage being 110?
10. The resistance of 1000 ft. of No. 36 copper wire is 424 ohms. How many feet should be used in winding a 200 ohms relay?
11. The resistance of No. 00 trolley wire is 0.80 ohm per 1000 ft. What is the resistance of a line 1 mile long?
12. A wire has a resistance of 20 ohms. It is joined in parallel with another wire of 6 ohms, find their combined resistance.
[Pg 302]
13. The separate resistances of two incandescent lamps are 200 ohms and 70 ohms. What is their combined resistance when joined in parallel? When joined in series?
### (4) Methods of Grouping Cells and Measuring Resistance
275. Internal Resistance of a Voltaic Cell.?The current produced by a voltaic cell is affected by the resistance that the current meets in passing from one plate to another through the liquid of the cell. This is called the internal resistance of the cell. A Daniell cell has several (1-5) ohms internal resistance. The resistance of dry cells varies from less than 0.1 of an ohm when new to several ohms when old. If cells are joined together their combined internal resistance depends upon the method of grouping the cells.
Fig. 256.?The four cans exert four times the water pressure that one can will exert.
276. Cells Grouped in Series and in Parallel.?When in series the copper or carbon plate of one cell is joined to the zinc of another and so on. (See Fig. 251.) The effect of connecting, say four cells, in series may be illustrated by taking four cans of water, placed one above another. (See Fig. 256.) The combined water pressure of the series is the sum of the several pressures of the cans of water, while the opposition offered to the movement of a quantity of water through the group of cans is the sum of the several resistances of the cans. In applying this illustration to the voltaic cell, we make[Pg 303] use of Ohm's law. Let E represent the e.m.f. of a single cell, r the internal resistance of the cell, and R the external resistance or the resistance of the rest of the circuit. Consider a group of cells in series. If n represents the number of cells in series, then Ohm's law becomes
I = nE/(nr + R).
Cells are grouped in series when large E.M.F. is required to force a current through a large external resistance such as through a long telegraph line. Cells are connected in parallel when it is desired to send a large current through a small external resistance. To connect cells in parallel all the copper plates are joined and also all the zinc plates. (See Fig. 257.) To illustrate the effect of this mode of grouping cells, suppose several cans of water are placed side by side (Fig. 258). It is easily seen that the pressure of the group is the same as that of a single cell, while the resistance to the flow is less than that of a single cell. Applying this reasoning to the electric circuit we have by Ohm's law the formula for the current flow of a group of
n cells arranged in parallel I = E/((r/n) + R).
Fig. 257.?Four cells connected in parallel.
Fig. 258.?The water pressure of the group in parallel is the same as that of one.
[Pg 304]
277. Illustrative Problems.?Suppose that four cells are grouped in parallel, each with an E.M.F. of 1.5 volts and an internal resistance of 2 ohms. What current will flow in the circuit if the external resistance is 2.5 ohms? Substitute in the formula for cells in parallel the values given above, and we have I = 1.5/(0.5 + 2.5) = 1.5/3 = 0.5 ampere. Suppose again that these four cells were grouped in series with the same external resistance, substituting the values in the formula for cells in series we have I = 4(1.5)/(4 ? 2 + 2.5) = 6/10.5 = 0.57 ampere.
278. Volt-ammeter Method for Finding Resistance.?Measurements of the resistance of conductors are often made. One of these methods depends upon an application of Ohm's law. It is called the volt-ammeter method since it employs both a voltmeter and an ammeter. If the conductor whose resistance is to be measured is made a part of an electric circuit, being connected in series with the ammeter and in shunt with the voltmeter, the resistance may easily be determined, since R = E/I. (See Fig. 250.) If, for example, the difference in E.M.F., or as it is often called, the fall of potential between the ends of the wire as read on the voltmeter is 2 volts, and the current is 0.5 ampere, then the resistance of the wire is 4 ohms. This method may be readily applied to find the resistance of any wire that is a part of an electric circuit.
279. The Wheatstone Bridge.?To find the resistance of a separate wire or of an electrical device another method devised by an Englishman named Wheatstone is commonly employed. This method requires that three known resistances, a, b, c, in addition to the unknown resistance[Pg 305] x be taken. These four resistances are arranged in the form of a parallelogram. (See Fig. 259.) A voltaic cell is joined to the parallelogram at the extremities of one diagonal while a moving-coil galvanometer is connected across the extremities of the other diagonal. The known resistances are changed until when on pressing the keys at E and K no current flows through the galvanometer. when this condition is reached, the four resistances form a true proportion, thus a: b = c: x.
Since the values of a, b, and c are known, x is readily computed. Thus if a = 10, b = 100, and c = 1.8 ohms, then x, the unknown resistance, equals 18 ohms, since 10: 100 = 1.8: 18. This method devised by Wheatstone may be employed to find the resistance of a great variety of objects. It is the one most commonly employed by scientists and practical electricians.
Fig. 259.?Diagram of a Wheatstone bridge.
#### Important Topics
1. The internal resistance of voltaic cells.
2. Ohm's law applied to groups of cells. (a) Cells in series, (b) cells in parallel.
3. Measurement of resistance: (a) volt-ammeter method, (b) Wheatstone bridge method.
#### Exercises
1. What is the resistance of an electric bell circuit where the E.M.F. is 3 volts and the current is 0.6 ampere?
2. A telegraph wire is broken somewhere, the ends lying upon damp ground. If an E.M.F. of 30 volts is applied from the ground[Pg 306] to the wire and a current of 0.1 of an ampere flows, what is the resistance of the part connected to the ammeter. (The earth which completes the circuit from the end of the wire has very small resistance.) Why?
3. How far away is the break in the wire if the latter has a resistance of 80 ohms to the mile? Diagram.
4. What current will flow through a bell circuit of 8 ohms resistance if it contains three cells in series each with an E.M.F. of 1.5 volts and an internal resistance of 1/3 ohm?
5. If the same three cells are connected in parallel on the same circuit what current flows? Is the current in problem 4 or 5 the larger? Why?
6. If four cells each with 1.5 volts E.M.F. and an internal resistance of 0.4 ohm are connected with a circuit having an external resistance of 0.8 ohm, what current will the parallel connection give? The series connection? Which gives the larger current? Why?
7. Four Daniell cells each having 1 volt E.M.F. and 3 ohms internal resistance are connected in series with 2 telegraph sounders of 4 ohms each. The connecting wires have 6 ohms resistance. Find the current intensity.
8. A battery of 2 cells arranged in series is used to ring a door bell. The E.M.F. of each cell is 1.5 volts, internal resistance 0.3 ohm, and the resistance of the bell is 4 ohms. What is the current in amperes?
9. In the above problem find the current if the cells are connected in parallel. | 9,634 | 38,744 | {"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} | 3.03125 | 3 | CC-MAIN-2024-33 | latest | en | 0.916756 |
http://software.intel.com/de-de/forums/topic/283791 | 1,386,443,899,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163055701/warc/CC-MAIN-20131204131735-00041-ip-10-33-133-15.ec2.internal.warc.gz | 233,931,036 | 10,831 | # Clarification request: Encoding
## Clarification request: Encoding
The solution for
1 2 1 1 1 0
is
1 2 1 1 1
How would this be encoded?
1. (1 2 1) (1 1)
2. (1 2 1) (1) (1)
From the problem description I would expect it to be #1. But I came across a problem here that uses #2.
(235,139,151)(127,12)(163)(125,79,31)(110,48)(46,14,19)(211)(9,5)(4,20)(13)(10,3)(23)(165,16)(149)
In this case (46, 14, 19) and (211) are on the same row. Can someone verify please?
Just want to be sure.
Thanks.
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Nähere Informationen zur Compiler-Optimierung finden Sie in unserem Optimierungshinweis.
Hi,
Yes, for the problem you indicated "1 2 1 1 1", it would be encloded as indicated in # 1: (1 2 1) (1 1).
In your 2nd problem description, can you indicate the tiling picture and the dimensions of the rectangle tile?
Thanks
-Rama
My output is #1
dimensions: 525 x 525
(235 139 151)(127 12)(163)(125 79 31)(110 48)(46 14 19 211)(9 5)(4 20)(13)(10 3)(23)(165 16)(149)
Quoting
My output is #1
dimensions: 525 x 525
(235 139 151)(127 12)(163)(125 79 31)(110 48)(46 14 19 211)(9 5)(4 20)(13)(10 3)(23)(165 16)(149)
Thanks Rama, ,
Here's the visual representation (click image to enlarge):
Cool...Thanks
-Rama | 452 | 1,232 | {"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-2013-48 | latest | en | 0.737954 |
https://www.goodtecher.com/leetcode-1672-richest-customer-wealth/ | 1,723,432,408,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00012.warc.gz | 602,708,204 | 11,311 | # LeetCode 1672. Richest Customer Wealth
## Description
https://leetcode.com/problems/richest-customer-wealth/
You are given an `m x n` integer grid `accounts` where `accounts[i][j]` is the amount of money the `ith` customer has in the `jth` bank. Return the wealth that the richest customer has.
A customer’s wealth is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum wealth.
Example 1:
```Input: accounts = [[1,2,3],[3,2,1]]
Output: 6
Explanation:
```1st customer has wealth = 1 + 2 + 3 = 6
``````2nd customer has wealth = 3 + 2 + 1 = 6
```Both customers are considered the richest with a wealth of 6 each, so return 6.
```
Example 2:
```Input: accounts = [[1,5],[7,3],[3,5]]
Output: 10
Explanation:
1st customer has wealth = 6
2nd customer has wealth = 10
3rd customer has wealth = 8
The 2nd customer is the richest with a wealth of 10.```
Example 3:
```Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
Output: 17
```
Constraints:
• `m == accounts.length`
• `n == accounts[i].length`
• `1 <= m, n <= 50`
• `1 <= accounts[i][j] <= 100`
## Explanation
Compare account sum with global maximum sum.
## Python Solution
``````class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
max_sum = -1
for account in accounts:
account_sum = sum(account)
if account_sum > max_sum:
max_sum = account_sum
return max_sum``````
• Time complexity: O(N).
• Space complexity: O(N). | 471 | 1,492 | {"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-2024-33 | latest | en | 0.806168 |
https://www.patheos.com/blogs/jesuscreed/2018/09/11/math-theology-rjs/ | 1,542,131,087,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741340.11/warc/CC-MAIN-20181113173927-20181113195927-00089.warc.gz | 956,081,954 | 22,013 | # Math + Theology = ? (RJS)
Math + Theology = ? (RJS) September 11, 2018
What does math have to do with theology? Andy Walsh, Faith Across the Multiverse, makes some interesting connections. The logic of mathematics can help us think more clearly about theology. In Chapter 1 (the second chapter since he starts with 0) Walsh explores paradox, logic, math, and the way we view the world (this is where theology comes in). This is a fascinating chapter that considers the Bible, Math, Geometry, Paul Revere, the Matrix, the X-men, Raiders of the Lost Ark, Contact … and gets us to consider some very important points about God and the nature of our faith in God. If this chapter is an accurate indicator of what is to come, the book should be an excellent conversation starter. I will highlight only a few points that particularly struck me as I read the chapter.
Human language is vague and ambiguous. Yes it can be used to communicate clearly – but this is not what always happens. Math uses axioms, rules and proofs to avoid ambiguity. “Math helps us communicate precisely about abstract ideas, which is why I think it can help us wrap our minds around abstract qualities of God and our relationship with him.” (p. 26) There is a reason why many current Philosophy professors were math majors as undergraduates (a fact that surprised me when I first realized it.) Math is great preparation for Philosophy.
Axioms. Math, with its careful logic and precise rules, teaches us that we need to start with unprovable axioms. Every system starts with some assumptions taken to be true. From these initial postulates a wide range of other propositions can be proven. Math also teaches us that in any construct some conclusions, even if true, will be unprovable. There is a proof of incompleteness …
Remarkably, not only did the proposed basis for math fail to be proven complete, it was actually explicitly proven incomplete; some questions were demonstrably unanswerable. A simplified explanation of the proof of incompleteness is that you can create an expression whose meaning is essentially “This expression cannot be proved:’ And yet, since it is built up from proven expressions according to the rules of the language, it should be considered proved. In other words, what was proved was the unprovability of the statement. This essentially rendered that expression undecidable, meaning there is no way to know if it is true or false, given the axioms we chose. (p. 31)
One example Walsh uses is the distinction between Euclidean and non-Euclidean geometries (although he doesn’t use the word Euclidean). Two lines are defined as parallel if they both intersect a third at right angles. We commonly think that parallel lines remain equidistant and never intersect. But this is an axiom, not a provable fact. The axiom defines a flat space. It is perfectly logical to construct geometries where parallel lines may intersect or diverge – these define a curved space. Both kinds of geometries are useful. In fact, the geometry of curved spaces turns out to be far more useful in physics than the geometry of flat spaces (image credit NASA). But … the truth of some statements about parallel lines and other objects depend on the nature of the geometry chosen, i.e. the axioms on which the logical framework is based. These axioms are chosen, they are not “self-evident” or provable.
What does this have to do with faith and theology? Quite simply, it is necessary to choose the assumptions on which we base our lives. Belief in God is an axiom. It is consistent with observation, but is not provable a priori i.e. “from first principles.” It is one of the first principles. Walsh’s discussion says it well:
This perspective provides us with an operational definition of faith. Instead of defining it in terms of dogma or rejection of evidence, let’s say that faith is choosing a set of assumptions, or axioms, for understanding the world. And if you prefer, we can further refine this definition to state that faith is specifically choosing assumptions that either explicitly include a God or gods, or at least do not explicitly disallow the existence of such a being or beings. Many atheists and other areligious folks bristle at the idea of calling their choice of assumptions faith, and that’s understandable given the general usage of the words. I don’t see any need to insist on that broader definition of “faith;’ so long as we all understand that at some point we are all making a choice of assumptions, and that no particular set of assumptions is privileged a priori nor the only option for a consistent view of the world.
Assuming God rather than proving him might seem like dodging any requirement to provide evidence. Axioms can certainly be informed by evidence, and my belief in God is definitely informed by historical corroboration of the Bible. But axioms cannot themselves be deductively proven; as with pudding, the proof is in the tasting. I am primarily interested with what conclusions follow from my belief in God and how useful they are in my real life. This is comparable to the situation in geometry, where multiple geometries are logically and mathematically valid but the ones where parallel lines intersect are useful for describing a wider range of real world experiences. (p. 35)
Information. Walsh goes on to dig into information theory and the ways in which information can be conveyed.
01100111 01101111 01100100 00100000 01101001 01110011 00100000 01101100 01101111 01110110 01100101 00101110
47 6F 64 20 69 73 20 6C 6F 76 65 2E
Information does not exist in a vacuum and is not intrinsic to the symbol that carries information … “its meaning is extrinsic and relational.” He uses Paul Revere (one if by land, two if by sea) and David and Jonathan (1 Samuel 20: If I say to him, ‘Look, the arrows are on this side of you; bring them here,’ then come, because, as surely as the Lord lives, you are safe; there is no danger. But if I say to the boy, ‘Look, the arrows are beyond you,’ then you must go, because the Lord has sent you away.) as examples. Information is carried because both parties understand the potential messages. The information is not intrinsic to the lamps in the Old North Church or to Jonathon’s specific words to the boy he sends to fetch the arrows. The meaning is extrinsic to these bits – and is known in the relationship between the conveyor and recipient. After working through several biblical stories Walsh concludes “I think the idea that meaning is extrinsic and relational is on solid ground biblically, rather than being antithetical to the teaching of the Bible. And so we should not be surprised when others look at the world around them and claim that it is meaningless. They are looking at the bits of the message, but the meaning isn’t in the bits.” (p. 39-40)
Completeness and God. Walsh speculates on the relationship between completeness and God – The axiom system defined by God is complete and infinite. We are finite and will never know all truth – but that doesn’t mean that God doesn’t or can’t. It is also clear that knowing the axioms is not enough. Just as we will not be good chess players simply because we know the rules – we will not be good Christians simply by choosing God. In both cases we must actually play the game, in fact we must spend many hours playing the game and practicing.
Translating that to our definition of faith as choosing God for an axiom, we see that the book of James says something very similar. “What good is it, my brothers and sisters, if someone claims to have faith but does not have works?” (James 2:14). In our terms, choosing the axiom of faith is not enough to know God; it is simply a prerequisite. Knowing God is the process of taking that axiom and figuring out what truths it contains. That means playing the game-living your life according to the theorems, the true statements, that follow from belief in God. This is what James means by showing faith through works, and what separates knowing God from knowing of God, which, as James notes in verse 19, even the demons know of God.
And so we have a picture of faith that is not inherently uninquisitive or unempirical. Instead, faith represents a particular set of assumptions about the world, including that God is the ultimate author of the universe and the Bible. These assumptions are not privileged a priori or deductively superior; they are a matter of volition. The work of faith is to discern the meaning in what we observe in the world around us in a way that is consistent with those assumptions and then act accordingly.
This process of living our faith should then be highly exploratory, requiring active engagement and not passive acceptance. For, as we noted, even if God is Truth, that doesn’t guarantee that we will know God or Truth perfectly. We need to regularly check to see if the theorems we are choosing, the actions we take, the way we live our lives, are consistent with the axioms we have chosen. (p. 44)
Faith in God is not illogical, irrational, or “unscientific.” Nor should we expect to be able to prove the existence of God from first principles.
We are called as Christians to play the game – to live out our faith.
What do you think of Walsh’s discussion of the necessity of axioms in any logical framework?
How does this relate to faith in God?
Does it challenge some of your assumptions – or those of skeptical friends, family, and colleagues?
If you wish to contact me directly, you may do so at rjs4mail[at]att.net.
If interested you can subscribe to a full text feed of my posts at Musings on Science and Theology.
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Browse Our Archives | 2,205 | 10,113 | {"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} | 2.984375 | 3 | CC-MAIN-2018-47 | latest | en | 0.953764 |
https://carreersupport.com/project-roi/ | 1,685,504,474,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00648.warc.gz | 171,408,251 | 20,517 | # How To Calculate Project ROI and Why It Matters (With Example)
Actual ROI is the true return on investment generated from a project. This number is typically calculated after a project has concluded, and uses final costs and revenues to determine how much profit a project produced compared to what was estimated.
1. ROI = (Net profit / cost of investment) x 100.
2. Net profit = expected revenue – total expenses.
It is prudent to conduct an ROI analysis on specific projects to ascertain which project types yield the highest returns. The figures don’t give a breakdown for upcoming strategic planning if you just look at ROI across all finished projects. However, going through the numbers individually gives you the chance to really delve into which projects were a success while also examining failures and ways to move forward.
Make an assessment of all project costs before beginning the calculation process. Obtain the receipts and create a thorough cost analysis that includes a breakdown for each category. When you want to evaluate various costs to develop strategies for cost-reduction for a higher ROI on future projects, the category breakdown will be helpful.
Costs typically include materials, supplies, labor, overhead costs for employees, fuel, equipment, and contracted services. This varies based on the business and type of project. Simple ROI analyses that focus on a single project’s costs do not account for recurring costs like building leases and capital expenditures. The formula only considers the costs and returns related to that specific project and excludes any other projects.
There are two different ways to calculate return on investment. Both return the same result with a different interpretation. The ROI is frequently presented to investors as a percentage, but business owners and managers need actual profit figures to incorporate into the overall business plan.
For instance, if you invest \$5,000 in a project and earn \$10,000 once it is finished, you will receive a \$5,000 return. To calculate the return, multiply the \$5,000 initial investment by \$5,000. The result is 1 or 100 percent as a return. A 100 percent return is fantastic in many businesses.
To subtract the total of expenses from the total of profits, use the last row. This will quickly show the project’s return on investment. Tracking expenses and profits during the course of the project benefits from maintaining a running tally. It can reveal areas where you are overspending in real-time. Taking immediate corrective measures can increase the return on investment.
Zach Lazzari is a freelance writer with a wealth of startup and digital marketing experience. He has a varied background and is well-known in the world of digital marketing. Zach manages marketing for numerous clients in the outdoor industry as well as developing and selling numerous successful web properties. He has written business articles for Angling Trade Magazine and numerous corporate partners’ white papers and case studies.
## Types of ROI to use for projects
Your choice of ROI depends on the timing of your calculations. For instance, your ROI can be used to forecast performance if a project hasn’t yet been initiated. On the other hand, if the project is already complete, you can use the ROI you calculate to assess the project’s success, learn from errors, and make plans for subsequent endeavors. Applying the following ROI types can help you estimate project costs and assess a project once it is finished.
### Anticipated ROI
This kind of ROI, also known as expected ROI, is typically calculated by a financial planner before a team starts working on a project. A project’s potential for profit and other potential outcomes are predicted using anticipated ROI, which combines anticipated costs and revenues. To assess risk and decide whether pursuing a project is worthwhile, managers and executives frequently consult the anticipated ROI.
### Actual ROI
Actual ROI, as the name implies, is the true profit a project makes after it is finished. Financial planners can use this ROI to compare a project’s actual profit to their prediction by combining recorded costs and revenue. Businesses can increase profitability over time by making more informed investments by routinely comparing actual ROI to anticipated ROI.
### Positive ROI
Positive ROI is a type of actual ROI that describes a successful project. When revenue generated by a project outpaces production costs, financial planners determine that the project is profitable. Project managers can find trends in efficient budgeting by recognizing positive ROI, which they can then apply to future projects.
### Negative ROI
When a project’s anticipated costs exceed the amount of revenue it generated, this is known as a negative ROI, in contrast to positive ROI. Financial planners can better manage an organization’s finances by preventing unnecessary costs in future projects by keeping track of negative ROI.
## What is return on investment for a project?
Return on investment (ROI) is a technique used in financial planning to assess the worth of a project and forecast its potential performance. Making the right investments with the aid of accurate ROI calculations can help businesses increase their profit margins. Regardless of your position within a company, knowing the ROI for a specific project can help you make wise choices, set priorities, and increase productivity. Here are a few instances of ROI in the workplace:
Executives and management may learn more about the kinds of projects that are successful by demonstrating a project’s return on investment, which could enhance the business’ long-term investment strategy.
## Why is a project’s ROI important?
Project ROI is valuable for a variety of reasons. Regardless of the sector you work in, understanding a project’s potential ROI can help you adjust upcoming costs and clearly communicate the financial benefits to stakeholders and upper management. Additional justifications for why calculating your project’s ROI is crucial for success include the following:
## How to calculate project ROI
To calculate a projects ROI, consider the formula below:
ROI = (Net profit / cost of investment) x 100
Subtract the anticipated project costs from the anticipated revenue to arrive at your net profit:
Net profit = expected revenue – total expenses
Financial planners frequently break down projects into manageable tasks in order to calculate the total costs, making sure to account for each stage of the process. The price of the materials, the estimated number of hours needed to complete the project, the number of employees required, and their hourly wages are then taken into account. They also take into account the price of purchasing or leasing machinery, software, and buildings.
Total costs are calculated as follows: material costs plus (project hours multiplied by the number of workers and their hourly wages) plus equipment, software, building, and other costs.
Even though it can be difficult to estimate the cost and value of a project before you start working on it, you can make your calculations simpler by making a list of the elements you already know and consulting records of actual ROI of related projects.
## Project ROI example
You can use the following example to better comprehend project ROI:
An area used bookstore’s inventory sourcing is handled by Erica. She has the chance to buy 1,000 books from a rival bookstore that is closing down. The books are currently priced at \$1 each, but Erica intends to raise their price to \$4 each. Since she lacks a car, she intends to spend \$50 on a book delivery service. She anticipates spending around four hours, or \$50 in wages, choosing the inventory, coordinating the delivery service, cataloguing the new books, and making sure they are stored properly.
Erica performs the calculations below to calculate the project’s anticipated return on investment:
Expected revenue is equal to \$1,000 books multiplied by \$4 per book, or \$4,000.
Total costs equal \$1,100 (1,000 books at \$1 each plus \$50 for delivery and \$50 for labor).
She then calculates her potential net profit by deducting the anticipated revenue from the total costs, or cost of investment:
Potential net profit = \$4,000 – \$1,100 = \$3,900
In order to calculate the ROI, she divides the net profit by the total costs, or cost of investment, and multiplies that result by 100:
ROI = (\$3,900 / \$1,100) x 100 = 354%
Using this calculation, Erica predicts significantly positive ROI. She offers the chance to her manager, who sees the financial advantages and accepts the project.
## FAQ
How do you calculate ROI for projects?
How to calculate ROI
1. T = time required for the process.
2. V = Volume or quantity of units, transactions, people, etc. required.
3. D = Dollars or cost required.
4. Current = current value.
5. Project is the value a project will have if it is successful.
What ROI means?
Return on Investment (ROI)
What is a good ROI example?
According to conventional wisdom, a good return on investment (ROI) for an investment in stocks is an annual ROI of about 7% or higher.
What does an ROI of 30% mean?
For instance, a 30% ROI from one store appears better than a 20% ROI from another. The one-year investment is clearly preferable because the 30% may be spread out over three years as opposed to the 20% from just one. | 1,842 | 9,463 | {"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} | 2.890625 | 3 | CC-MAIN-2023-23 | longest | en | 0.942363 |
https://origin.geeksforgeeks.org/count-pairs-from-1-to-n-such-that-their-sum-is-divisible-by-their-xor/ | 1,660,577,340,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572192.79/warc/CC-MAIN-20220815145459-20220815175459-00047.warc.gz | 419,090,907 | 30,612 | # Count pairs from 1 to N such that their Sum is divisible by their XOR
• Last Updated : 28 May, 2022
Given a number , the task is to count pairs (x, y) such that their sum (x+y) is divisible by their xor value (x^y) and the condition 1 ≤ x < y < N holds true.
Examples
```Input: N = 3
Output: 3
Explanation:
(1, 2), (1, 3), (2, 3) are the valid pairs
Input: N = 6
Output: 11```
Approach:
• After taking the array as input, first we need to find out all the possible pairs in that array.
• So, find out the pairs from the array
• Then for each pair, check whether the sum of the pair is divisible by the xor value of the pair. If it is, then increase the required count by one.
• When all the pairs have been checked, return or print the count of such pair.
Below is the implementation of the above approach:
## C++
`// C++ program to count pairs from 1 to N` `// such that their Sum is divisible by their XOR` `#include ` `using` `namespace` `std;` `// Function to count pairs` `int` `countPairs(``int` `n)` `{` ` ``// variable to store count` ` ``int` `count = 0;` ` ``// Generate all possible pairs such that` ` ``// 1 <= x < y < n` ` ``for` `(``int` `x = 1; x < n; x++) {` ` ``for` `(``int` `y = x + 1; y <= n; y++) {` ` ``if` `((y + x) % (y ^ x) == 0)` ` ``count++;` ` ``}` ` ``}` ` ``return` `count;` `}` `// Driver code` `int` `main()` `{` ` ``int` `n = 6;` ` ``cout << countPairs(n);` ` ``return` `0;` `}`
## Java
`// Java program to count pairs from 1 to N ` `// such that their Sum is divisible by their XOR ` `class` `GFG ` `{` ` ` ` ``// Function to count pairs ` ` ``static` `int` `countPairs(``int` `n) ` ` ``{ ` ` ``// variable to store count ` ` ``int` `count = ``0``; ` ` ` ` ``// Generate all possible pairs such that ` ` ``// 1 <= x < y < n ` ` ``for` `(``int` `x = ``1``; x < n; x++) ` ` ``{ ` ` ``for` `(``int` `y = x + ``1``; y <= n; y++) ` ` ``{ ` ` ``if` `((y + x) % (y ^ x) == ``0``) ` ` ``count++; ` ` ``} ` ` ``} ` ` ``return` `count; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `main (String[] args)` ` ``{ ` ` ``int` `n = ``6``; ` ` ``System.out.println(countPairs(n)); ` ` ``} ` `}` `// This code is contributed by AnkitRai01`
## Python3
`# Python3 program to count pairs from 1 to N ` `# such that their Sum is divisible by their XOR ` `# Function to count pairs ` `def` `countPairs(n) : ` ` ``# variable to store count ` ` ``count ``=` `0``; ` ` ``# Generate all possible pairs such that ` ` ``# 1 <= x < y < n ` ` ``for` `x ``in` `range``(``1``, n) :` ` ``for` `y ``in` `range``(x ``+` `1``, n ``+` `1``) : ` ` ``if` `((y ``+` `x) ``%` `(y ^ x) ``=``=` `0``) :` ` ``count ``+``=` `1``; ` ` ``return` `count; ` `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: ` ` ``n ``=` `6``; ` ` ``print``(countPairs(n)); ` `# This code is contributed by AnkitRai01`
## C#
`// C# program to count pairs from 1 to N ` `// such that their Sum is divisible by their XOR ` `using` `System;` `public` `class` `GFG ` `{` ` ` ` ``// Function to count pairs ` ` ``static` `int` `countPairs(``int` `n) ` ` ``{ ` ` ``// variable to store count ` ` ``int` `count = 0; ` ` ` ` ``// Generate all possible pairs such that ` ` ``// 1 <= x < y < n ` ` ``for` `(``int` `x = 1; x < n; x++) ` ` ``{ ` ` ``for` `(``int` `y = x + 1; y <= n; y++) ` ` ``{ ` ` ``if` `((y + x) % (y ^ x) == 0) ` ` ``count++; ` ` ``} ` ` ``} ` ` ``return` `count; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main()` ` ``{ ` ` ``int` `n = 6; ` ` ``Console.WriteLine(countPairs(n)); ` ` ``} ` `}` `// This code is contributed by AnkitRai01`
## Javascript
``
Output:
`11`
Time Complexity: O(N2), as we are using nested loops to traverse N*N times.
Auxiliary Space: O(1), as we are not using any extra space.
My Personal Notes arrow_drop_up
Recommended Articles
Page : | 1,532 | 4,323 | {"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} | 3.21875 | 3 | CC-MAIN-2022-33 | latest | en | 0.661936 |
https://quant.stackexchange.com/questions/24519/modelling-callable-bonds-in-a-risk-model-historical-simulation | 1,722,979,811,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640508059.30/warc/CC-MAIN-20240806192936-20240806222936-00198.warc.gz | 388,411,987 | 41,855 | # Modelling callable bonds in a risk model (historical simulation)
What is a best-practice example on how to model callable bonds in a risk model - I focus on historical simulation (HS).
For plain-vanilla bonds the input factors for historical simulation could be
• the zero curve of the market (the currency)
Then HS would model changes in interst rates of the currency (as systematic risk) and spreads either in issuer level (idiosyncratic) or rating level (rather systematic risk). Then we could reprice the bond in these scenarios.
Looking at callable bonds on the other hand we have to simulate/estimate the chance that the bond is called and when. To do this we could use an interest rate model which we would have to calibrate on future interest rate uncertainty. Then we can simulate the future and price the bond in these scenarios.
Market data that reflect this that I know are swaptions and captions. But these are instruments for the money market/capital market of a currency. However, the decisin of the issuer to call the bonds will depend on the interest rate level of the currency and the issuers spread.
How can we find a risk model that can be calibrated to readily available market data and that models the systematic as well as the idiosyncratic part of the call risk? How do industry solutions look like?
Callable bonds are exposed to interest rates, spreads, and your interest rate model. You could link your spreads to interest rates, but then you will need a systematic spread model. In most pricing models that I have seen, spreads are not evolving through time (which is incorrect). The problem is one doesn't have any market instrument to calibrate the evolution of spreads in a risk neutral manner, and pricing is done in risk neutral world. For rates, rate evolution is calibrated on other instruments like swaptions or futures, which are market instruments, thus rate evolution models can be made risk neutral.
• What would you say dominates for the decision to call: rates or spreads? Commented Feb 25, 2016 at 10:26
It says I have to have 50 reputations points to comment, and I don't, and actually don't even know what that means so here is my "Answer." I think you're getting into apples and oranges. There is no risk neutral anything here. For historical simulation, you already have all of the rates to apply to your prospective cash flows, and in your simulation you have the amounts and dates of those cash flows from the bond description. Fine. Now the only addition is to add a decision function that terminates the cash flows to principal plus call spread if the bond qualifies under certain conditions: price justifies call and call date is far enough in the future to justify being triggered, etc. You are making this way more difficult than it should be.
• There are two things here. HS requires risk factor simulation, then you need to price the callable bond at each scenario of HS. Pricing is risk neutral | 613 | 2,963 | {"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} | 2.515625 | 3 | CC-MAIN-2024-33 | latest | en | 0.941327 |
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Although trains may use energy more (i)_____ than do automobiles, the latter move only when they contain at least one occupant, whereas railway carriages spend a considerable amount of time running up and down the tracks (ii)_____, or nearly so.
Observers of modern presidential campaigns who (i)_____ the highly (ii)_____ productions that pass for campaigns these days do sometimes find reason for hope in the occasional mix-ups that (iii)_____ candidates on the trail despite the presence of political strategists plotting every event with the tactical precision of military commanders.
When delivering 104 total pieces of newspaper for a community, the number of households that subscribe to 3 pieces of newspaper is twice as many as the number of household that subscribe to 1 piece of newspaper, while the number of households that subscribe to 2 pieces of newspaper is three times as many as the number of household that subscribe to 1 piece of newspaper. What is the number of households that subscribe to 2 pieces of newspaper?
y > 1,001
#### Quantity A
$\sqrt[3]{y}$
#### Quantity B
$\frac{y}{100}$
In a right isosceles triangle, the area of the shaded region is 44,BD=CE=4,and AC=AB,what is the length of AC?
QA=$\frac{3}{2}$QE
n=75, m=50
#### Quantity A
The length of arc ACB
#### Quantity B
The length of arc EDF
The tick marks shown on the number line are evenly spaced. Points D and E have coordinates of $4^{10}$ and $4^{11}$, respectively. The point that has a coordinate of $4^{9}$ is?
x and y are both integers If 4 ≤ x < 7 < y ≤ 12, then whats the range of $(x-y)^{2}$?
A normal distribution of variable X has a mean of 56 and a standard deviation of 4.
#### Quantity A
The percentage of variable X ranging from 60 to 62
#### Quantity B
The percentage of variable X ranging from 62 to 64
There are 5!, or 120, ways of arranging 5 different solid-colored flags side by side. If the colors of the flags are red, blue, yellow, green, and orange, how many of those arrangements have either the red flag or the blue flag in the middle position?
What is the probability that a number comprised of at least one 6 on all digits is selected when selecting a number from 1 to 1,000 (inclusive)?
If -1< y < 0, what is the relationship between y, $y^{2}$, $y^{3}$, and $y^{4}$?
What is the units digit of $23^{25}$-23?
#### Quantity A
The number of positive factors of 87
#### Quantity B
The number of positive factors of 97
What's the number of integers between 1 and 2,000, inclusive, that can be transformed into not only a perfect square number and also a perfect cubic number?
A box contains 6 cards, numbered 1, 2, 3, 4, 5, and 6, respectively. If one of the 6 cards is to be selected at random, what is the probability that the number on the card selected will be greater than 3, or even, or both?
Which of the following statements are true for all integers a and b?
Indicate all such statements.
#### Quantity A
The tens digit of
($4^{100}$)($5^{99}$)
#### Quantity B
The tens digit of
($4^{100}$)($5^{101}$)
In the figure shown, what is the length of line segment BD?
25000 +道题目
132本备考书籍 | 810 | 3,145 | {"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 | 3 | CC-MAIN-2021-39 | latest | en | 0.9116 |
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In Depth Information
The preceding formula has four sets of parentheses. Three sets are nested inside the fourth set. Excel evaluates
each nested set of parentheses and then sums the three results. This sum is then multiplied by the value in B6.
Make liberal use of parentheses in your formulas even when they aren't necessary. Using parentheses clarifies
the order of operations and makes the formula easier to read. For example, if you want to add 1 to the product
of two cells, the following formula does the job:
=A1*A2+1
Because of Excel's operator precedence rules, the multiplication will be performed before the addition. There-
fore, parentheses are not necessary. You may find it much clearer, however, to use the following formula even
though it contains superfluous parentheses:
=(A1*A2)+1
Every left parenthesis, of course, must have a matching right parenthesis. If you have
many levels of nested parentheses, you may find it difficult to keep them straight. For-
tunately, Excel lends a hand in helping you match parentheses. When editing a formula,
matching parentheses are colored the same, although the colors can be difficult to dis-
tinguish if you have a lot of parentheses. Also, when the cursor moves over a paren-
thesis, Excel momentarily displays the parenthesis and its matching parenthesis in
bold. This lasts for less than a second, so watch carefully.
In some cases, if your formula contains mismatched parentheses, Excel may propose a correction to your for-
mula. Figure 2-3 shows an example of Excel's AutoCorrect feature in action.
Simply accepting the correction proposed in the dialog box is tempting, but be careful.
In many cases, the proposed formula, although syntactically correct, isn't the formula
that you want. In the following example, I omitted the closing parenthesis after January.
In Figure 2-3, Excel proposed this correction:
=SUM(January/SUM(Total)
In fact, the correct formula is
=SUM(January)/SUM(Total)
Search JabSto ::
Custom Search | 437 | 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} | 2.578125 | 3 | CC-MAIN-2019-09 | latest | en | 0.904602 |
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### S. Silver
#29) "Your friend is using Descartes's Rule of Signs to find the number of negative real roots of x^3+x^2+x+1=0. Describe and correct the error."
"P(-x) = (-x)^3 +(-x)^2 + (-x) +1
= -x^3 - x^2 - x +1"
"Because there is only on sign change in P(-x), there must be one negative real root."
## Here is my work:
The error that my friend made was that they kept a negative x^2 instead of changing it to a positive x^2. My friend also made an error with how many sign changes there are, they said that it was only one sign change when there are actually three sign changes. There is only one negative real root, which my friend got right.
## Then this is the graphs of x^3+x^2+x+1=0 and its one negative real root, which was -1.
#31) A gardener is designing a new garden in the shape of a trapezoid. She wants the shorter base to be twice the height and the longer base to be 4 feet longer than the shorter base. If she has enough topsoil to create a 60ft^2 garden, what dimensions should she use for her garden? | 298 | 1,064 | {"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-2017-26 | longest | en | 0.940248 |
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### Author Topic: bee speed (Read 2876 times)
#### randydrivesabus
• Queen Bee
• Posts: 1072
##### bee speed
« on: May 23, 2006, 07:06:21 AM »
so how fast do they fly?
#### Michael Bush
• Universal Bee
• Posts: 16166
• Gender:
##### bee speed
« Reply #1 on: May 23, 2006, 08:38:46 AM »
http://www.beesource.com/pov/wenner/bsjun1992.htm
"during field work we use a simple formula: x = 150y - 500 (straight line in Figure 1). That is, to estimate distance (x = meters or yards) to each colony, we multiply complete round trip time (y = time between arrivals) by 150 and subtract 500 from the result. (The constant value of 500 represents the time spent filling at the station and unloading in the colony - see Wenner 1963). "
It'd too early to do the math right now. See if you can figure it out.
My website: bushfarms.com/bees.htm en espanol: bushfarms.com/es_bees.htm auf deutsche: bushfarms.com/de_bees.htm
My book: ThePracticalBeekeeper.com
-------------------
"Everything works if you let it."--James "Big Boy" Medlin
#### Jerrymac
• Galactic Bee
• Posts: 6047
• Gender:
##### bee speed
« Reply #2 on: May 23, 2006, 10:09:03 AM »
:rainbowflower: Light travels faster than sound. This is why some people appear bright until you hear them speak. :rainbowflower:
:jerry:
http://photobucket.com/albums/v225/Jerry-mac/
#### Finsky
• Super Bee
• Posts: 2791
• Gender:
##### bee speed
« Reply #3 on: May 23, 2006, 11:19:54 AM »
I remember that bee fly without load 20 km/h and with load 25 km/h.
It's wings beets 400 per second.
http://www.physorg.com/news8616.html
.
#### Scott Derrick
• Expert Bee Handler
• New Bee
• Posts: 35
• Gender:
• Go Gamecocks!!
##### bee speed
« Reply #4 on: May 23, 2006, 06:24:57 PM »
12 MPH.
"You're born. You suffer. You die. Fortunately, there's a loophole."
Billy Graham
#### randydrivesabus
• Queen Bee
• Posts: 1072
##### bee speed
« Reply #5 on: May 23, 2006, 06:40:09 PM »
they seem to be going so much faster than that.....
#### Scott Derrick
• Expert Bee Handler
• New Bee
• Posts: 35
• Gender:
• Go Gamecocks!!
##### bee speed
« Reply #6 on: May 23, 2006, 07:02:59 PM »
Well...really I've seen their speed published all over the net differently. Some say 12, 15, 22 mph. It's hard to say I guess. I know when they come out of the hive at me when I don't have my suit on they fly about 150 mph... :D
Scott
"You're born. You suffer. You die. Fortunately, there's a loophole."
Billy Graham
#### yoderski
• House Bee
• Posts: 59
##### bee speed
« Reply #7 on: July 06, 2006, 10:39:20 AM »
I know this has been addressed before, but I was able to observe my bees yesterday flying out over an open field. The sky was overcast, and it was about 5 o'clock in the evening, and I could follow their flight for 100-200 meters before they disappeared. I noticed 2 things. Most of them would make a circle or two up in front of the hive before they headed out on their way--I don't know what the function of that is, probably to get their bearings. The other thing is that the speed of bees published is much too slow. It is no exaggeration to say that they were covering 100 meters in 4-6 seconds which is over 30 mph, closer to 40 mph. Which explains the observation of them flying 100mph when they are flying out at you...
Jon Y.
Atmore, AL
#### yoderski
• House Bee
• Posts: 59
##### bee speed
« Reply #8 on: July 06, 2006, 10:40:08 AM »
Except, I don't know about the wind--they may have had a little tailwind!
Jon Y.
Atmore, AL
#### thegolfpsycho
• Field Bee
• Posts: 583
##### bee speed
« Reply #9 on: July 06, 2006, 01:03:04 PM »
Holy smokes!! You can see a bee in flight from 100 to 200 meters??? I can barely make out a 747 at that distance. :lol: I find the bees can fly faster than I can run, and futher than I can run too! I have to agree with rsderrick. When they're mad at me, they are even faster!
#### yoderski
• House Bee
• Posts: 59
##### bee speed
« Reply #10 on: July 06, 2006, 07:22:55 PM »
Yes, the sky was just right, with a dull gray background, just the right amount of light--I had never been out there when the background was perfect like that. Normally, I can't see more than 20 feet or so, but I am certain of what I saw....
Jon Y.
Atmore, AL
#### fcderosa
• House Bee
• Posts: 132
##### bee speed
« Reply #11 on: July 06, 2006, 10:38:31 PM »
I’ve noticed that a mad Italian out of the hive will hit me in the forehead with about as much force as a junebug in the forehead at about 50 MPH while traveling on a motorcycle. I’d have to say 50 MPH.
The good life is honey on a Ritz.
#### Jerrymac
• Galactic Bee
• Posts: 6047
• Gender:
##### Re: bee speed
« Reply #12 on: July 06, 2006, 10:40:34 PM »
Quote from: randydrivesabus
so how fast do they fly?
They'll get there when they get there.
:rainbowflower: Light travels faster than sound. This is why some people appear bright until you hear them speak. :rainbowflower:
:jerry:
http://photobucket.com/albums/v225/Jerry-mac/
#### TwT
• Senior Forum
• Global Moderator
• Galactic Bee
• Posts: 3396
• Ted
##### bee speed
« Reply #13 on: July 06, 2006, 10:44:44 PM »
some how when they are coming after my fingers they move faster than i do so I will vote real fast ;)
THAT's ME TO THE LEFT JUST 5 MONTHS FROM NOW!!!!!!!!
Never be afraid to try something new.
Amateurs built the ark,
Professionals built the Titanic
#### Dick Allen
• House Bee
• Posts: 163
##### bee speed
« Reply #14 on: July 07, 2006, 02:50:06 AM »
The books give a figure of 24 km/hr, or about 15 mph for an average flight speed, but it's never that simple. Leslie Goodman writing in 'Form and Function in the Honey Bee':
"It is a long-held observation that honey bees fly lower on slow upwind flights than when moving fast downwind, and it is assumed that they try to maintain a constant preferred optical flow rate of images moving from front to back beneath them during flight. It has been calculated that the preferred optical flow rate is about 3.5 rad/s for the honey bee." | 1,806 | 6,042 | {"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} | 2.90625 | 3 | CC-MAIN-2017-26 | latest | en | 0.649543 |
https://www.physicsforums.com/threads/open-ended-pipe-harmonics-mastering-physics-question.219420/ | 1,555,757,446,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529606.64/warc/CC-MAIN-20190420100901-20190420122901-00418.warc.gz | 814,724,776 | 18,408 | # Open ended pipe Harmonics Mastering Physics Question
#### TFM
[SOLVED] Open ended pipe Harmonics Mastering Physics Question
1. The problem statement, all variables and given/known data
Consider a pipe 45.0cm long if the pipe is open at both ends. Use v = 344m/s.
Now pipe is closed at one end.
What is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20 Hz to 20000 Hz?
2. Relevant equations
$$f_n = (2n-1)\frac{v}{4L}$$
3. The attempt at a solution
I have an answer that works, but masteringphysics doesn't accept. I first rearranged the equation to give me:
$$(2n-1) = \frac{f_n * 4L}{v}$$
then:
$$2n = (\frac{f_n * 4L}{v})+1$$
and finally:
$$n = ((\frac{f_n * 4L}{v})+1)/2$$
inserting the values gives 52.5 so I inserted 52 as the answer. wrong, I have tried 51-54, all wrong. so I thought tpo go backwards, using:
$$(2n-1) = \frac{f_n * 4L}{v}$$
and inserting values, to find the value which is the closest to 20000, buit under it - guess what, the value that came out:
52!
Any ideas
TFM
Related Introductory Physics Homework News on Phys.org
#### Kurdt
Staff Emeritus
Gold Member
The harmonics of a pipe closed at one end are all odd. For n = 2 you have the 3rd harmonic. For n=52 what harmonic do you have?
#### TFM
It will be the 53rd Harmonic. The trouble is, I have put 53 in, and it says its the wrong answer!
#### Kurdt
Staff Emeritus
Gold Member
It will be the 53rd Harmonic. The trouble is, I have put 53 in, and it says its the wrong answer!
Sorry that third harmonic was a bad example. The harmonics are given by 2n-1. So if n is 52 what is the harmonic. An easier way to have thought about it would to have solved for:
$$f_n = \frac{nv}{4L}$$
for n = 1, 3, 5,.....
#### TFM
Using:
$$f_n = \frac{nv}{4L}$$
and using n = 103,
I get a frequency of 19684, which is the first odd number below 20000. would this be the harmonic number?
TFM
#### Kurdt
Staff Emeritus
Gold Member
I get a frequency of 19684, which is the first odd number below 20000. would this be the harmonic number?
TFM
Yes n is the harmonic number.
#### TFM
Success!! n = 103.
IOne thing does bother me slightly - where does my orginal answer of 52 fit in?
TFM
#### Kurdt
Staff Emeritus
Gold Member
2n - 1 is just another way of saying n = 1, 3, 5, .... . So if you stick n = 52 into 2n - 1 you get 103.
Last edited:
#### TFM
That makes sense.
Thanks,
TFM
#### Kurdt
Staff Emeritus
Gold Member
That makes sense.
Thanks,
TFM
What I was originally aiming at was for you to put the n = 52 into that equation and get 103 but I used a stupid example which probably mislead you slightly.
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• Solo and co-op problem solving | 890 | 3,075 | {"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} | 3.703125 | 4 | CC-MAIN-2019-18 | longest | en | 0.928309 |
http://www.math.grin.edu/~rebelsky/Courses/MAT115/2008S/R/activity-19-3.html | 1,512,971,116,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948512208.1/warc/CC-MAIN-20171211052406-20171211072406-00144.warc.gz | 416,697,280 | 3,179 | # R notes for Activity 19-3
## 19-3 c: Visualizing Sample Data
You can read and preview the data with
```BodyTemps = read.csv("/home/rebelsky/Stats115/Data/BodyTemps.csv")
summary(BodyTemps)
```
You'll note that the data have two columns: `BodyTemp` and `Sex`. We just want the first column, which we will select with `BodyTemps\$BodyTemp`.
We can build a quick histogram of those data with the following command. (Since R and Minitab make different decisions as to how to make intervals, this may look a bit different than the sample answer.)
```hist(BodyTemps\$BodyTemp)
```
But we should certainly label the x axis
```hist(BodyTemps\$BodyTemp,
main="Sample Body Temperatures",
xlab="Body Temperature in Degrees F"
)
```
If we'd rather do a dot plot, we can use
```library(BHH2, lib="/home/rebelsky/Stats115/Packages")
dotPlot(BodyTemps\$BodyTemp,
main="Sample Body Temperatures",
xlab="Body Temperature in Degrees F"
)
```
We can create the normal probability plot with
```qqnorm(BodyTemps\$BodyTemp, datax=T, ylab="Body Temperature in Degrees F")
```
## 19-3 f: Computing Confidence Intervals
Since you used some form of technology to compute these confidence intervals in activity 19-1, I'm not sure why they're asking you to do so again. But, hey, let's cooperate. One technique is to tell R the formula. We'll start by recording the values we know.
```x_bar = 98.249
s = .733
n = 130
```
We can use `qt` to compute t*. Unlike the table on p. 625, `qt` computes the appropriate t value given the area to the left of that t. Hence, for a 95% confidence interval, we use .975. (Why .975? Because there's 0.025 to the right, and therefore 0.975 to the left.) As you should recall from the reading, the degrees of freedom should be `n`-1.
```t_star = qt(0.975, n-1)
```
Now, we're ready to compute the lower bound and upper bounds of the confidence interval using the standard formula.
```ci_lower = x_bar - t_star*s/sqrt(n)
ci_upper = x_bar + t_star*s/sqrt(n)
c(ci_lower, ci_upper)
```
Of course, that's a lot of work. Hence, we might want to use the built-in `t.test` function, which provides not just the confidence interval, but also a lot of other data. However, we need to work from the original data set, rather than from the mean and standard deviation already computed from that data set. (If you only know mean, standard deviation, and sample size, you'll need to use the technique above.) To use the `t.test` function, you also need to provide a hypothesized population parameter (`mu`) and a desired confidence interval (`conf.level`). While you don't need mu to compute the confidence interval, the t-test computes more than just the confidence interval, and therefore requires a bit more.
```t.test(BodyTemps\$BodyTemp, mu=98.6, conf.level=0.95)
```
For the other two confidence intervals, we would use
```t.test(BodyTemps\$BodyTemp, mu=98.6, conf.level=0.90)
t.test(BodyTemps\$BodyTemp, mu=98.6, conf.level=0.99)
```
## 19-3 j: Computing Another CI
Since you don't have the original data set, you cannot use the `t.test` function. Hence, you must provide R with the formulae.
```x_bar = 98.249
s = .733
n = 13
t_star = qt(0.975, n-1)
ci_lower = x_bar - t_star*s/sqrt(n)
ci_upper = x_bar + t_star*s/sqrt(n)
c(ci_lower, ci_upper)
```
Samuel A. Rebelsky, rebelsky@grinnell.edu
Copyright (c) 2007-8 Samuel A. Rebelsky.
This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. To view a copy of this license, visit `http://creativecommons.org/licenses/by-nc/2.5/` or send a letter to Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA. | 1,003 | 3,640 | {"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} | 3.21875 | 3 | CC-MAIN-2017-51 | latest | en | 0.804997 |
https://web2.0calc.com/questions/sally-has-four-feet-of-ribbon-that-she-needs-to-cut | 1,618,138,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038061820.19/warc/CC-MAIN-20210411085610-20210411115610-00156.warc.gz | 700,975,893 | 5,561 | +0
# Sally has four feet of ribbon that she needs to cut into 1 inch sections. How many inches are in 4 feet?
0
45
2
Sally has four feet of ribbon that she needs to cut into 1 inch sections. How many inches are in 4 feet?
Mar 4, 2021
#1
+30523
+1
4 ft * 12 in/ ft = 48 inches
Mar 4, 2021
#2
+30523
0
Maybe this is where YOUR confusion lies:
The amount of cuts she will need to make (which your question did not ask)
will be 47 cuts to make 48 pieces
47 inches or 3 ft 11 inches
ElectricPavlov Mar 4, 2021 | 178 | 537 | {"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-2021-17 | longest | en | 0.944014 |
http://www.ck12.org/book/CK-12-Trigonometry-Concepts/r1/section/1.8/ | 1,490,426,709,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218188824.36/warc/CC-MAIN-20170322212948-00098-ip-10-233-31-227.ec2.internal.warc.gz | 452,555,690 | 42,046 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
1.8: Sine, Cosine, and Tangent Functions
Difficulty Level: At Grade Created by: CK-12
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Practice Sine, Cosine, and Tangent Functions
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You are helping your grandfather with some repairs around the house when he mentions that he could use some help painting some boards on the staircase of his front porch. When you go out to see what he means, you notice that the stairs are supported by a set of boards that are glued together so that they are shaped like a triangle, with one extra board placed over top of them for decoration.
As you are looking around for paint, you think about the situation and realize that this reminds you a lot of your math class. In your study of triangles, you are about to start a unit working with the relationships between the sides of a triangle. You begin to wonder: "How many possible relationships are there between sides of triangles, anyway?"
By the end of this Concept, you'll have studied three of these important relationships, as well as know how many relationships there are total.
Guidance
The first three trigonometric functions we will work with are the sine, cosine, and tangent functions. The elements of the domains of these functions are angles. We can define these functions in terms of a right triangle: The elements of the range of the functions are particular ratios of sides of triangles.
We define the sine function as follows: For an acute angle \begin{align*}x\end{align*} in a right triangle, the \begin{align*}sin x\end{align*} is equal to the ratio of the side opposite of the angle over the hypotenuse of the triangle. For example, using this triangle, we have: \begin{align*}\sin A = \frac{a}{c}\end{align*} and \begin{align*}\sin B = \frac{b}{c}\end{align*}.
Since all right triangles with the same acute angles are similar, this function will produce the same ratio, no matter which triangle is used. Thus, it is a well-defined function.
Similarly, the cosine of an angle is defined as the ratio of the side adjacent (next to) the angle over the hypotenuse of the triangle. Using this triangle, we have: \begin{align*}\cos A = \frac{b}{c}\end{align*} and \begin{align*}\cos B = \frac{a}{c}\end{align*}.
Finally, the tangent of an angle is defined as the ratio of the side opposite the angle to the side adjacent to the angle. In the triangle above, we have: \begin{align*}\tan A = \frac{a}{b}\end{align*} and \begin{align*}\tan B = \frac{b}{a}\end{align*}.
There are a few important things to note about the way we write these functions. First, keep in mind that the abbreviations \begin{align*}sin x, cos x\end{align*}, and \begin{align*}tan x\end{align*} are just like \begin{align*}f(x)\end{align*}. They simply stand for specific kinds of functions. Second, be careful when using the abbreviations that you still pronounce the full name of each function. When we write \begin{align*}sin x\end{align*} it is still pronounced sine, with a long \begin{align*}“i.”\end{align*} When we write \begin{align*}cos x\end{align*}, we still say co-sine. And when we write \begin{align*}tan x\end{align*}, we still say tangent.
We can use these definitions to find the sine, cosine, and tangent values for angles in a right triangle.
Example A
Find the sine, cosine, and tangent of \begin{align*}\angle{A}\end{align*}:
Solution:
\begin{align*}\sin A & = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{4}{5}\\ \cos A & = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{3}{5}\\ \tan A & = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{4}{3}\end{align*}
Example B
Find \begin{align*}\sin B\end{align*} using \begin{align*}\triangle ABC\end{align*} and \begin{align*}\triangle NAP.\end{align*}
Solution:
Using \begin{align*}\triangle ABC: \sin B = \frac{3}{5}\end{align*}
Using \begin{align*}\triangle NAP: \sin B = \frac{6}{10} = \frac{3}{5}\end{align*}
Example C
Find \begin{align*}\sin B\end{align*} and \begin{align*}\tan A\end{align*} in the triangle below:
Solution:
\begin{align*}\sin B & = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{12}{13}\\ \tan A & = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{5}{12}\end{align*}
Vocabulary
Sine: The sine of an angle in a right triangle is a relationship found by dividing the length of the side opposite the given angle by the length of the hypotenuse.
Cosine: The cosine of an angle in a right triangle is a relationship found by dividing the length of the side adjacent the given angle by the length of the hypotenuse.
Tangent: The tangent of an angle in a right triangle is a relationship found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.
Guided Practice
Using the triangle shown here:
Find
1. The sine of angle \begin{align*}\angle{A}\end{align*}
2. The cosine of angle \begin{align*}\angle{A}\end{align*}
3. The tangent of angle \begin{align*}\angle{A}\end{align*}
Solution: 1. The sine is equal to the opposite divided by the hypotenuse.
\begin{align*}\sin{A} = \frac{opposite}{hypotenuse} = \frac{11}{61} \approx .18\end{align*}
2. The cosine is equal to the adjacent divided by the hypotenuse.
\begin{align*}\cos{A} = \frac{adjacent}{hypotenuse} = \frac{60}{61} \approx .98\end{align*}
3. The tangent is equal to the opposite divided by the adjacent.
\begin{align*}\tan{A} = \frac{opposite}{adjacent} = \frac{61}{60} \approx 1.01\end{align*}
Concept Problem Solution
Looking at a triangle like the shape of your grandfather's staircase:
We can see that there are several ways to make relationships between the sides. In this case, we are only interested in ratios between the sides, which means one side will be divided by another. If we assume that dividing a side by itself doesn't count (since the answer would always be equal to one), let's look at the number of possible combinations:
If we use the angle labelled above, there is: 1) The side opposite the angle divided by the hypotenuse (the sine function) 2) The side adjacent the angle divided by the hypotenuse (the cosine function) 3) The side opposite the angle divided by adjacent side (the tangent function)
You can also imagine taking the same sides, except reversing the numerator and denominator: 4) The hypotenuse divided by the side opposite the angle 5) The hypotenuse divided by the side adjacent to the angle 6) The side adjacent to the angle divided by the side opposite to the angle
The first three functions are what we introduced in this Concept. The last three are other functions you'll learn about in a different Concept.
Practice
Use the diagram below for questions 1-3.
1. Find \begin{align*}\sin A\end{align*} and \begin{align*}\sin C\end{align*}.
2. Find \begin{align*}\cos A\end{align*} and \begin{align*}\cos C\end{align*}.
3. Find \begin{align*}\tan A\end{align*} and \begin{align*}\tan C\end{align*}.
Use the diagram to fill in the blanks below.
1. \begin{align*}\tan A = \frac{?}{?}\end{align*}
2. \begin{align*}\sin C = \frac{?}{?}\end{align*}
3. \begin{align*}\tan C = \frac{?}{?}\end{align*}
4. \begin{align*}\cos C = \frac{?}{?}\end{align*}
5. \begin{align*}\sin A = \frac{?}{?}\end{align*}
6. \begin{align*}\cos A = \frac{?}{?}\end{align*}
From questions 4-9, we can conclude the following. Fill in the blanks.
1. \begin{align*}\cos \underline{\;\;\;\;\;\;\;} = \sin A\end{align*} and \begin{align*}\sin \underline{\;\;\;\;\;\;\;} = \cos A\end{align*}.
2. \begin{align*}\tan A\end{align*} and \begin{align*}\tan C\end{align*} are _________ of each other.
3. Explain why the cosine of an angle will never be greater than 1.
4. Use your knowledge of 45-45-90 triangles to find the sine, cosine, and tangent of a 45 degree angle.
5. Use your knowledge of 30-60-90 triangles to find the sine, cosine, and tangent of a 30 degree angle.
6. Use your knowledge of 30-60-90 triangles to find the sine, cosine, and tangent of a 60 degree angle.
7. As the degree of an angle increases, will the tangent of the angle increase or decrease? Explain.
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English
TermDefinition
cosine The cosine of an angle in a right triangle is a value found by dividing the length of the side adjacent the given angle by the length of the hypotenuse.
sine The sine of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the hypotenuse.
Tangent The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.
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Subjects: | 2,503 | 9,081 | {"found_math": true, "script_math_tex": 47, "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} | 4.875 | 5 | CC-MAIN-2017-13 | latest | en | 0.909968 |
https://discuss.pytorch.org/t/how-to-apply-3d-convolution-to-stacked-video-frames/84342 | 1,652,728,757,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512229.26/warc/CC-MAIN-20220516172745-20220516202745-00506.warc.gz | 289,421,367 | 4,544 | # How to apply 3D convolution to stacked video frames
I am working on video classification for motion recognition.
I selected 10 frame from video and applied optical flow these sequantial frames.
After that, I found x and y direction of each flows and stack them together.
Finally, I have 3D matrix with shape H x W x 20.
H = height
W = weight
20 = flows from 10 frame (2 times because of x and y dimension of optical flow).
So, I want to apply 3D convolution layer this matrix above.
When I look at pytorch documentation for 3D convolution, I saw 5 dimensional input like that (N,C,D,H,W).
But my input is 4D dimensional like that (N,C,H,W) with N samples.
So, How can I apply 3D convolutional to my matrix?
N is the batch size, the number of sample. In your case, sample != single frame. Your sample is HxWx1x20., (c=1, you have only one channel) .
To map that to (N,C,D,H,W). C=1, D=20, H=H, and W=W. N depends on your batch size, and if it fits in memory or not.
I want to ask one more question.
As @ebarsoum mentioned, I have HxWx1x20 numpy uint8 array.
I want to implement a dataset class that inherit torch.utils.data.Dataset.
In getitem part of the function, I return this array by applying transforms like:
transforms.RandomCrop,
transforms.RandomHorizontalFlip(),
transforms.ColorJitter(),
transforms.ToTensor(),
transforms.Normalize([0.485, 0.456, 0.406], [0.229, 0.224, 0.225]).
To be able apply the first three of them requires PiL image.
How can handle this ? | 398 | 1,480 | {"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-2022-21 | latest | en | 0.854777 |
https://oeis.org/A322600 | 1,627,646,516,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.52/warc/CC-MAIN-20210730091645-20210730121645-00235.warc.gz | 440,512,367 | 4,189 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A322600 a(n) is the number of unlabeled rank-3 graded lattices with 5 coatoms and n atoms. 4
1, 5, 20, 68, 190, 441, 907, 1690, 2916, 4734, 7310, 10836, 15528, 21619, 29365, 39045, 50961, 65434, 82809, 103453, 127751, 156117, 188980, 226794, 270037, 319204, 374813, 437409, 507553, 585831, 672847, 769233, 875637, 992735, 1121218, 1261802 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Jukka Kohonen, Table of n, a(n) for n = 1..1000 J. Kohonen, Counting graded lattices of rank three that have few coatoms, arXiv:1804.03679 [math.CO] preprint (2018). FORMULA For n>=3: a(n) = (175/192)n^4 - (3079/480)n^3 + (11771/480)n^2 - [7268/160, 7273/160]n + [33600, 34019, 34072, 33627, 33152, 34915, 33624, 33947, 33472, 33507, 34520, 34459, 32832, 33827, 34072, 34395, 33344, 34147, 33432, 33947, 34240, 33699, 33752, 34267, 32832, 34595, 34264, 33627, 33152, 34147, 34200, 34139, 33472, 33507, 33752, 35035, 33024, 33827, 34072, 33627, 33920, 34339, 33432, 33947, 33472, 34275, 33944, 34267, 32832, 33827, 34840, 33819, 33152, 34147, 33432, 34715, 33664, 33507, 33752, 34267] / 960. The value of the first bracket depends on whether n is even or odd. The value of the second bracket depends on whether (n mod 60) is 0, 1, 2, ..., 59. Conjectures from Colin Barker, Dec 20 2018: (Start) G.f.: x*(1 + 4*x + 14*x^2 + 43*x^3 + 102*x^4 + 184*x^5 + 282*x^6 + 368*x^7 + 411*x^8 + 400*x^9 + 333*x^10 + 237*x^11 + 142*x^12 + 70*x^13 + 26*x^14 + 7*x^15 + x^16) / ((1 - x)^5*(1 + x)^2*(1 + x^2)*(1 + x + x^2)*(1 + x + x^2 + x^3 + x^4)). a(n) = a(n-1) + a(n-2) - a(n-5) - a(n-6) - a(n-7) + a(n-8) + a(n-9) + a(n-10) - a(n-13) - a(n-14) + a(n-15) for n>15. (End) CROSSREFS Fifth row of A300260. Previous rows are A322598, A322599. Sequence in context: A271599 A032286 A097552 * A084850 A270169 A007327 Adjacent sequences: A322597 A322598 A322599 * A322601 A322602 A322603 KEYWORD nonn,easy AUTHOR Jukka Kohonen, Dec 19 2018 STATUS approved
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Last modified July 30 07:58 EDT 2021. Contains 346348 sequences. (Running on oeis4.) | 1,057 | 2,484 | {"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-2021-31 | latest | en | 0.469807 |
https://www.teacherspayteachers.com/Product/math-Art-Activity-on-Area-and-Perimeter-Digital-Printable-Worksheets-1406854 | 1,527,497,681,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794872114.89/warc/CC-MAIN-20180528072218-20180528092218-00515.warc.gz | 844,840,623 | 21,614 | # math & Art Activity on Area and Perimeter (Digital & Printable Worksheets)
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Your students will enjoy building Heron, (aka Hero), The Egyptian Crocodile while learning about area, perimeter, and triangles at the same time. Each carefully designed template is drawn over a centimeter square grid so that the area and perimeter can be calculated easily. The templates here are considered level 4 because they include, not only squares but also triangles. This activity is geared to grades 5-10 but is adaptable to many grade levels as described below.
What you get…
Level 4: (Triangles)
Heron the Crocodile (with grid)
Heron the Crocodile (no grid)
Area & Perimeter Summary Sheet
Heron ‘Discovery Notes’
Materials:
Scissors
Card stock paper-white preferably (Note: the templates work best if you print them directly onto thicker paper. However, you can print to regular paper and glue it onto thicker paper. You can also just make these using regular paper. These will not be quite as durable.
Optional: colored pencils, crayons, markers for coloring creations.
Rationale for the Complete Booklet:
Here’s an ingenious and ‘seriously → fun’ way to learn about area & perimeter, (and much more). This activity booklet will provide your students with a clever progression of beautifully designed paper geometric creations that will provide hours of student inquiry, and deep understanding of a number of important math concepts. And yes, you can use this activity book equally well in a kindergarten class or in a calculus class.
Here’s how this works, starting at Level 1 students cut out and fold figures into geometric creatures like ‘Flat Cat’ & ‘Square Hare’. At Level 1, the figures are divided into squares so students can find the area and perimeter by adding up the number of squares total, (area), and the number of square sides along the edge, (perimeter).
Note: If you’re a teacher K-2 you’re already thinking this is too hard – and for most kids this age you are right! However, I believe there is tremendous value in having students working with math concepts in their hands, so to speak, well before they work with them in their heads. (Personally I know the hours I spent building stuff as a young boy helped me earn my physics degree years later. Students in early elementary can recognize the shapes used and count them. They also might be inspired to create their own creations using geometric shapes.
Note: If you’re an advanced math teacher you’re thinking, this is too easy . But, please read on…
Right from the get-go, students will see that a ‘cat’ cannot be accurately represented using only squares just as our early ancestors did many, many years ago. However, you can get a decent estimate using just squares. More interesting shapes can be constructed using triangles, Level 4, but this does require some ‘hidden knowledge’ about triangles and the use of a ruler to find the perimeter. (Or you might apply the Pythagorean theorem depending on the age and level of your students.)
Next we move on to circles, Level 5, which of course requires a ‘secret’ formula and again the use of a ruler to measure the diameter, (again another important historical development – right? See where I’m going?)
Finally, students will be presented with shapes whose area and perimeter can only be calculated using advanced math – basically Calculus. It is NOT important that the elementary school teacher know how to calculate the area of these shapes using calculus! It is important, however, for teachers and students to understand the ‘end result’ of math in its continuum. (Calculus was one of the most exciting discoveries in math because it allows you to find the area and perimeter of nearly any shape. That allows us to build airplanes and send people to the moon! However, the basis of calculus is really exactly what students do to find the area of ‘Flat Cat’ – add up squares! But really, really small ones.)
So you see, this booklet provides a visual presentation of the math continuum that parallels, pretty closely, historical discoveries. Students, and teachers can see where the study of math might take them; also why more challenging classes in mathematics are needed.
What you get:
35 pages plus some (total in complete booklet)
Level 1 (Contains Whole or Half Squares)
Flat Cat & Square Hare (with grid)
Flat Cat & Square Hare (no grid)
Area & Perimeter Summary Sheets
Level 1/2 (Contains Squares; Optional: can be used for higher level discussions such as symmetry)
Several Student cover sheets
Photo of final creation, (to help with construction)
Ulam – The Dizzy Snake (Clockwise Spiral, with grid)
Malu – The Dizzy Snake (Counter Clockwise Spiral, with grid)
Ulam & Malu – The Dizzy Snake (Both Spirals, no grid)
Area & Perimeter Summary Sheet
Dizzy ‘Discovery Notes’ (for advanced students – Differentiated learning)
Discovery Journal (Template for teacher personalization)
Level 2: (Contains Squares & Fractions of Squares)
Archie the Dog (with grid)
Archie the dog (no grid)
Area & Perimeter Summary Sheet
Archie ‘Discovery Notes’
Level 1-3 (Contains squares only but potential for experimentation – optional)
Whirly Bird Helicopter (with grid)
Whirly Bird Helicopter (no grid)
Whirly Bird ‘Discovery Notes’
Level 3: (Contains Squares, Fractions of Squares, & Cut-a-ways)
Graph Giraffe & Linear Llama (with grid)
Graph Giraffe & Linear Llama (no grid)
Area & Perimeter Summary Sheet
Level 1-4: (Contains only squares but may evoke deep questions about definitions)
Mobius Loop Template
Area & Perimeter Summary Sheet
Mobius loop ‘Discovery Notes’
Level 4: (Triangles)
Heron the Crocodile (with grid)
Heron the Crocodile (no grid)
Area & Perimeter Summary Sheet
Heron ‘Discovery Notes’
Level 5: (Circles)
Biblio-Birds [4 templates]
Area & Perimeter Summary Sheet
Area & Perimeter Summary Sheet
Level 6: (Parabolas)
3- Mini Boomerangs
(Turning Tern, Newton, & Leibniz)
Graphing workings
Area Calculations Summary Sheet
Launch Directions for Mini Boomerang
This unit activity meets or exceeds 21st Century and STEM learning expectations, and Common Core learning outcomes. (I think STEM, which stands for science, technology, engineering and math, should be changed to STEAM, where 'A' stands for Art. What do you think?)
Total Pages
7+
N/A
Teaching Duration
N/A
Report this Resource
\$2.75 | 1,433 | 6,452 | {"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} | 3.4375 | 3 | CC-MAIN-2018-22 | latest | en | 0.895344 |
https://converter.ninja/mass/kilograms-to-ounces/678-kg-to-oz/ | 1,601,248,976,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401582033.88/warc/CC-MAIN-20200927215009-20200928005009-00719.warc.gz | 349,692,395 | 5,660 | # 678 kilograms in ounces
## Conversion
678 kilograms is equivalent to 23915.7462018155 ounces.[1]
## Conversion formula How to convert 678 kilograms to ounces?
We know (by definition) that: $1\mathrm{kg}\approx 35.273962\mathrm{oz}$
We can set up a proportion to solve for the number of ounces.
$1 kg 678 kg ≈ 35.273962 oz x oz$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{oz}\approx \frac{678\mathrm{kg}}{1\mathrm{kg}}*35.273962\mathrm{oz}\to x\mathrm{oz}\approx 23915.746236\mathrm{oz}$
Conclusion: $678 kg ≈ 23915.746236 oz$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 ounce is equal to 4.18134559365782e-05 times 678 kilograms.
It can also be expressed as: 678 kilograms is equal to $\frac{1}{\mathrm{4.18134559365782e-05}}$ ounces.
## Approximation
An approximate numerical result would be: six hundred and seventy-eight kilograms is about twenty-three thousand, nine hundred and fifteen point seven five ounces, or alternatively, a ounce is about zero times six hundred and seventy-eight kilograms.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic. | 350 | 1,283 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "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} | 3.8125 | 4 | CC-MAIN-2020-40 | latest | en | 0.693745 |
https://atcoder.jp/contests/awtf2024-open/tasks/awtf2024_b | 1,726,565,140,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00591.warc.gz | 92,607,274 | 6,763 | Contest Duration: - (local time) (300 minutes) Back to Home
B - 01 Inversion Expected /
Time Limit: 2 sec / Memory Limit: 2048 MB
### 問題文
0, 1 からなる長さ N の文字列 S が与えられます． 整数の組 (i,j) (1 \leq i < j \leq N) であって，Si 文字目が 1, j 文字目が 0 であるようなものを転倒ペアと呼ぶことにします．
S 内に転倒ペアが存在する限り，以下の操作を行います．
• 転倒ペア (i,j) をランダムに一つ選ぶ．選択はこれ以前の選択と独立で，かつ一様ランダムであるとする． そして，Si 文字目と j 文字目を入れ替える．
### 制約
• 1 \leq N \leq 250000
• S0, 1 からなる長さ N の文字列
### 入力
N
S
### 入力例 1
2
10
### 出力例 1
1
### 入力例 2
3
110
### 出力例 2
499122178
### 入力例 3
1
0
### 出力例 3
0
### 入力例 4
10
1011000010
### 出力例 4
133099253
### 入力例 5
100
0101110010001000111000111001001101001100000111110001010010001010101100011001011011101101100001100111
### 出力例 5
407907276
Score : 1000 points
### Problem Statement
You are given a string S of length N consisting of 0 and 1. We define an inversion pair as a pair of integers (i, j) (1 \leq i < j \leq N) such that the i-th character of S is 1 and the j-th character of S is 0.
As long as there is an inversion pair in S, perform the following operation:
• Randomly choose an inversion pair (i, j). The choice is independent of previous choices and is uniformly random. Then, swap the i-th and j-th characters of S.
Find the expected number of operations, modulo 998244353.
What is the expected value modulo 998244353?
It can be proved that the sought expected value is always rational. Moreover, under the constraints of this problem, it can be proved that if the expected value is expressed as an irreducible fraction \frac{P}{Q}, then Q \neq 0 \pmod{998244353}. Thus, there exists a unique integer R such that R \times Q \equiv P \pmod{998244353}, 0 \leq R < 998244353. Report this R.
### Constraints
• 1 \leq N \leq 250000
• S is a string of length N consisting of 0 and 1.
### Input
The input is given from Standard Input in the following format:
N
S
### Sample Input 1
2
10
### Sample Output 1
1
The expected number of operations is 1.
### Sample Input 2
3
110
### Sample Output 2
499122178
The expected number of operations is 3/2.
### Sample Input 3
1
0
### Sample Output 3
0
### Sample Input 4
10
1011000010
### Sample Output 4
133099253
### Sample Input 5
100
0101110010001000111000111001001101001100000111110001010010001010101100011001011011101101100001100111
### Sample Output 5
407907276 | 844 | 2,350 | {"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} | 3.359375 | 3 | CC-MAIN-2024-38 | latest | en | 0.415421 |
https://library.fiveable.me/incompleteness-and-undecidability/unit-2/formal-proofs-inference-rules/study-guide/MCWcbjY3erAQqAiR | 1,726,440,566,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651668.26/warc/CC-MAIN-20240915220324-20240916010324-00052.warc.gz | 327,265,042 | 52,820 | # 2.3 Formal proofs and inference rules
are the backbone of logical reasoning in mathematics and philosophy. They provide a structured way to demonstrate the of arguments, using a series of and to connect premises to conclusions.
From basic inference rules like to more complex strategies like , formal proofs offer a powerful toolkit for logical analysis. By mastering these techniques, we can rigorously examine arguments and uncover the underlying structure of logical reasoning.
## Formal Proofs and Inference Rules
### Purpose and structure of formal proofs
• Demonstrate the of arguments by showing the conclusion logically follows from the premises
• Validity means if the premises are true, the conclusion must also be true ()
• Proofs reveal the logical structure and reasoning behind arguments
• Structure formal proofs as a sequence of well-formed formulas (wffs) or statements
• Each wff is either a premise, an assumption, or derived from previous wffs using inference rules
• The final wff in the sequence is the conclusion of the argument
• proofs use propositional variables ($p$, $q$, $r$) and logical connectives ($\neg$, $\land$, $\lor$, $\to$, $\leftrightarrow$)
• Prove the validity of arguments involving compound propositions (statements)
• proofs extend with predicates, ($\forall$, $\exists$), and individual variables ($x$, $y$, $z$)
• Prove the validity of arguments involving properties and relations of objects
### Application of basic inference rules
• Modus ponens (MP): $p \to q$, $p$ $\vdash$ $q$
• If $p$ implies $q$ and $p$ is true, then $q$ must be true ("affirming the antecedent")
• (MT): $p \to q$, $\neg q$ $\vdash$ $\neg p$
• If $p$ implies $q$ and $q$ is false, then $p$ must be false ("denying the consequent")
• (HS): $p \to q$, $q \to r$ $\vdash$ $p \to r$
• If $p$ implies $q$ and $q$ implies $r$, then $p$ implies $r$ (transitive property of implication)
• (CI): $p$, $q$ $\vdash$ $p \land q$
• If $p$ and $q$ are both true, then their conjunction $p \land q$ is true
• (CE): $p \land q$ $\vdash$ $p$; $p \land q$ $\vdash$ $q$
• If the conjunction $p \land q$ is true, then both $p$ and $q$ must be true individually
• (DI): $p$ $\vdash$ $p \lor q$; $q$ $\vdash$ $p \lor q$
• If either $p$ or $q$ is true, then their disjunction $p \lor q$ is true
• (DE): $p \lor q$, $p \to r$, $q \to r$ $\vdash$ $r$
• If $p \lor q$ is true, and both $p$ and $q$ individually imply $r$, then $r$ must be true
### Validity proofs using deduction systems
• uses a vertical layout with subproofs indented under assumptions
• Discharges assumptions when reaching the conclusion of a subproof
• Applies inference rules to wffs in the main proof or subproofs
• uses a tree-like structure with ($\Gamma \vdash \varphi$)
• : $\Gamma$ (set of assumptions) $\vdash$ $\varphi$ (conclusion)
• Applies inference rules to manipulate sequents and prove the validity of the argument
• Steps in proofs:
2. Apply inference rules to derive new wffs
3. Reach the conclusion by discharging assumptions or applying the final inference rule
### Common proof strategies and techniques
• Assume the negation of the desired conclusion
• Derive a contradiction ($p \land \neg p$) using the premises and the assumption
• Conclude that the negation of the assumption must be true
• (disjunction elimination)
• Divide the problem into exhaustive and mutually exclusive cases
• Prove the desired conclusion for each case
• Conclude that the desired conclusion holds in general
• (for properties of natural numbers)
• Base case: Prove the statement holds for the initial value ($n = 0$ or $n = 1$)
• Inductive step: Assume the statement holds for $n = k$ and prove it holds for $n = k + 1$
• Conclude that the statement holds for all natural numbers $n$
## Key Terms to Review (36)
Conjunction elimination: Conjunction elimination is a rule of inference in formal logic that allows one to infer a single conjunct from a conjunction. This means if you have a statement that is the combination of two or more propositions, you can derive each individual proposition as a valid conclusion. This rule is fundamental in formal proofs, as it simplifies complex statements into more manageable parts, making the proof process more efficient.
Conjunction Introduction: Conjunction Introduction is a logical inference rule that allows the derivation of a conjunction from two or more propositions. When you have established the truth of individual statements, this rule permits you to combine them into a single compound statement, thus forming a conjunction. This is fundamental in formal proofs as it helps to construct more complex arguments based on simpler, validated statements.
Conjunction introduction: Conjunction introduction is a fundamental rule in formal logic that allows one to derive a conjunction from two statements that have already been established as true. This rule is essential in building logical proofs, as it enables the combination of individual truths into a single, more complex assertion, maintaining logical validity throughout the proof process.
Conjunction symbol (∧): The conjunction symbol (∧) is a logical operator used to represent the logical conjunction, meaning 'and'. When two statements are combined using this symbol, the resulting compound statement is true only if both of the individual statements are true. This operator is fundamental in formal proofs and inference rules, as it helps establish relationships between propositions and allows for the derivation of new truths based on existing ones.
Disjunction Elimination: Disjunction elimination is a rule of inference in formal logic that allows one to conclude a statement based on the disjunction of two or more statements. This rule is particularly useful in formal proofs as it enables the derivation of a conclusion when at least one of the premises is true, thus simplifying the reasoning process by eliminating unnecessary branches of reasoning.
Disjunction Introduction: Disjunction Introduction is a rule of inference in propositional logic that allows one to derive a disjunction from a single proposition. If a statement 'P' is true, one can validly conclude that 'P or Q' is also true, regardless of the truth value of 'Q'. This rule is important for constructing formal proofs and manipulating logical expressions effectively.
Disjunction symbol (∨): The disjunction symbol (∨) is a logical operator used to represent the 'or' operation in propositional logic. This symbol connects two statements, indicating that at least one of the statements must be true for the entire expression to be considered true. Understanding this operator is essential for formal proofs and inference rules, as it lays the groundwork for evaluating logical expressions and determining their validity.
Existential quantifier (∃): The existential quantifier (∃) is a symbol used in formal logic to indicate that there exists at least one element in a given domain that satisfies a specified property or condition. It is often used in conjunction with predicates to express statements like 'There exists an x such that P(x) is true'. This quantifier plays a crucial role in formal proofs and inference rules, allowing the formulation of statements about the existence of elements within a logical framework.
First-order logic: First-order logic is a formal system used in mathematics, philosophy, linguistics, and computer science that extends propositional logic by allowing the use of quantifiers and predicates to express statements about objects and their relationships. It provides a structured way to represent facts and reason about them, connecting deeply with the limitations of formal systems, independence results in set theory, and the foundational aspects of mathematical logic.
First-order logic: First-order logic is a formal system used in mathematics, philosophy, linguistics, and computer science that enables reasoning about objects and their properties through quantifiers and predicates. It connects the concepts of syntax and semantics, allowing for the construction of complex statements and the evaluation of their truth within specific interpretations.
Fitch-style natural deduction: Fitch-style natural deduction is a formal proof system used in logic to derive conclusions from premises through the application of inference rules. This method emphasizes the structured presentation of logical reasoning, allowing each step of the proof to be clearly laid out and justified. It uses a tree-like format, where each line corresponds to a logical statement, and premises can be assumed temporarily within sub-proofs, enhancing clarity in the derivation process.
Formal proof: A formal proof is a logical argument that establishes the truth of a statement through a sequence of deductive reasoning steps, using axioms and inference rules within a formal system. This structure ensures that each step follows necessarily from previous statements or premises, allowing for a clear and unambiguous verification of the conclusion. Formal proofs are essential for understanding provability and the foundational aspects of mathematical logic.
Formal proofs: Formal proofs are logical derivations that utilize a specific set of inference rules and axioms to establish the truth of a statement within a formal system. They provide a structured framework for reasoning, ensuring that each step in the proof follows logically from previous steps and adheres to established rules. This rigor is crucial for validating mathematical claims and is foundational in fields like logic and computer science.
Gentzen-style natural deduction: Gentzen-style natural deduction is a formal system of logic that emphasizes the use of introduction and elimination rules for each logical connective, allowing for direct derivation of conclusions from premises. This approach to proof construction mirrors intuitive reasoning processes, making it easier to understand the flow of arguments. It plays a crucial role in developing formal proofs and inference rules by providing a structured yet flexible framework for reasoning about propositions.
Hypothetical syllogism: Hypothetical syllogism is a rule of inference that allows one to derive a conclusion from two conditional statements. If one statement asserts that 'if A then B' and another asserts 'if B then C', then it logically follows that 'if A then C'. This concept is essential for constructing formal proofs and understanding the structure of arguments within formal systems.
Individual Variables (x, y, z): Individual variables are symbols used in formal logic and mathematical expressions to represent unspecified elements or values within a given context. These variables are crucial for forming logical statements and performing deductions, as they allow for generalization and abstraction in proofs and inference rules.
Inference rules: Inference rules are formal logical constructs that dictate how new statements or conclusions can be derived from existing ones within a formal system. These rules play a crucial role in determining the validity of arguments and proofs, providing a structured method to progress from premises to conclusions. By establishing how propositions can interact, inference rules help define the framework for reasoning within axiomatic systems and support the construction of formal proofs.
Modus ponens: Modus ponens is a fundamental rule of inference in formal logic that allows one to derive a conclusion from a conditional statement and its antecedent. This rule asserts that if 'P implies Q' (if P, then Q) is true and P is also true, then Q must necessarily be true. It serves as a foundational mechanism in reasoning within formal systems, linking the semantics of statements to their logical structure and implications.
Modus tollens: Modus tollens is a fundamental rule of inference in formal logic stating that if a conditional statement is true, and its consequent is false, then the antecedent must also be false. This logical principle is crucial for constructing valid arguments and proofs, as it allows for the derivation of conclusions from given premises. Understanding modus tollens helps clarify how formal systems operate by providing a structured method for reasoning about implications.
Natural Deduction: Natural deduction is a formal proof system that enables the derivation of conclusions from premises using a set of inference rules. This system emphasizes the intuitive aspects of logical reasoning, allowing for direct manipulation of logical statements in a structured way. It serves as a foundational method in proof theory, providing a clear framework for establishing the validity of arguments through a sequence of justified steps.
Negation symbol (¬): The negation symbol (¬) is a logical operator used to represent the negation of a proposition, indicating that the statement is false when the original proposition is true and vice versa. This symbol plays a crucial role in formal proofs and inference rules by allowing one to derive new truths from existing statements through the process of negation. It helps to create a clearer understanding of logical relationships and is foundational in constructing valid arguments.
Proof by cases: Proof by cases is a mathematical technique used to establish the truth of a statement by dividing the problem into multiple scenarios or cases and proving the statement for each individual case. This method is especially useful when a proposition can be true under different conditions, allowing each case to cover a different aspect of the overall argument.
Proof by Cases: Proof by cases is a logical method used in formal proofs where a statement is proven by dividing it into multiple scenarios, or 'cases', and demonstrating that the statement holds true for each case individually. This technique is particularly useful when a proposition can be validated through distinct conditions that cover all possible situations, ensuring completeness in the argument.
Proof by contradiction: Proof by contradiction is a mathematical proof technique where one assumes the opposite of what is to be proven, shows that this assumption leads to a contradiction, and thus concludes that the original statement must be true. This method relies on the principle that if an assumption leads to an impossibility, then the assumption itself must be false. It's a powerful tool in various areas of logic and mathematics, especially in establishing results that involve self-reference or undecidability.
Proof by Induction: Proof by induction is a mathematical technique used to prove statements or formulas that are asserted to be true for all natural numbers. The process involves two main steps: the base case, where the statement is verified for the initial value (often 0 or 1), and the inductive step, where one assumes the statement is true for an arbitrary natural number 'k' and then proves it for 'k+1'. This method connects closely with formal proofs and inference rules as it provides a systematic way to establish the validity of infinite cases through finite reasoning.
Proof by induction: Proof by induction is a mathematical technique used to prove a statement or proposition that is asserted for all natural numbers. It consists of two main steps: the base case, where the statement is verified for the initial value (usually 0 or 1), and the inductive step, where it is shown that if the statement holds for an arbitrary natural number n, then it also holds for n+1. This method connects the structure of formal proofs and inference rules by demonstrating that if a property holds for one case, it can be extended to all cases.
Propositional logic: Propositional logic is a branch of logic that deals with propositions, which are statements that can either be true or false. It forms the foundation for more complex logical systems and allows for the formulation of logical arguments through the use of connectives such as 'and', 'or', and 'not'. Understanding propositional logic is essential as it connects to formal proofs, soundness and completeness, semantics, and the structure of axiomatic systems.
Propositional Logic: Propositional logic is a branch of logic that deals with propositions, which are declarative statements that can be either true or false. It focuses on the relationships between these statements and uses logical connectives to form more complex expressions. Understanding propositional logic lays the foundation for analyzing formal systems, establishing soundness and completeness, and constructing formal proofs using inference rules.
Quantifiers: Quantifiers are symbols or expressions used in logic and mathematics to indicate the quantity of specimens in a certain context, specifically relating to the truth of a statement involving variables. They help in expressing propositions about a whole set of objects, making them essential for formal proofs, syntax, and the foundations of mathematical logic. There are primarily two types: the universal quantifier, denoted as $$orall$$ (for all), and the existential quantifier, denoted as $$hereexists$$ (there exists).
Sequent: A sequent is a formal expression in logic that represents a relationship between a set of premises and a conclusion. It is typically written in the form 'Γ ⊢ φ', where Γ denotes a set of premises and φ denotes the conclusion that follows from those premises. Sequents are fundamental in the study of formal proofs and inference rules, as they help structure logical arguments and facilitate the application of various proof techniques.
Sequents: A sequent is a formal expression in logic that represents an implication between a set of premises and a conclusion, typically written in the form 'A_1, A_2, ..., A_n ⊢ B', where the A's are the premises and B is the conclusion. This notation is crucial in formal proofs and inference rules, as it provides a clear framework for reasoning about logical relationships and deriving conclusions from given premises.
Soundness: Soundness is a property of a formal system where if a statement can be proven within that system, then it is true in all interpretations of that system. This concept connects to the reliability of inference rules and formal proofs, ensuring that if something is derivable, it accurately reflects the intended meaning of the language and structure employed.
Universal Quantifier (∀): The universal quantifier, denoted as $$\forall$$, is a symbol used in mathematical logic to express that a statement applies to all elements within a specific set or domain. This powerful tool allows for concise formulations of statements that declare a property or condition holds universally, enabling formal proofs and inference rules to be structured systematically.
Validity: Validity refers to the property of an argument where, if the premises are true, the conclusion must also be true. It is a crucial concept in formal proofs and inference rules, as it ensures that logical reasoning leads to accurate outcomes. Validity is not concerned with the actual truth of the premises but rather with the logical connection between them and the conclusion drawn from them.
Validity: Validity refers to the property of an argument or inference that ensures the conclusion necessarily follows from the premises, meaning that if the premises are true, the conclusion must also be true. This concept is crucial in assessing the soundness of arguments and the strength of logical reasoning. It serves as a foundation for building formal proofs and employing inference rules effectively.
Well-formed formulas: Well-formed formulas (WFFs) are expressions in a formal language that are constructed according to specific syntactical rules. They play a crucial role in logic by ensuring that statements are meaningful and can be evaluated for truth values. WFFs are essential for building axiomatic systems, developing formal proofs, and understanding propositional logic, as they provide a clear structure for valid expressions that represent logical relationships. | 3,960 | 20,081 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 84, "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} | 4.4375 | 4 | CC-MAIN-2024-38 | latest | en | 0.73538 |
http://forums.wolfram.com/mathgroup/archive/1998/Jan/msg00267.html | 1,529,598,352,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864191.74/warc/CC-MAIN-20180621153153-20180621173153-00135.warc.gz | 125,888,984 | 7,701 | # Re: Plotting vector-valued functions
```Malcolm Boshier wrote:
>
> I have a problem which is related to the recent thread about
> plotting lists of functions. In the case when a vector-valued function
> is expensive or impossible to Evaluate before plotting, Plot apparently
> forces you to evaluate the function repeatedly at each value of the
> independent parameter. This can be very inefficient.
> As an example, suppose that f[z] returns the eigenvalues of a 5 x 5
> matrix which is a function of z. In general this function cannot be
> evaluated without a value for z, so
> Plot[ Evaluate[f[z]], {z, zmin, zmax}] doesn't work.
> The only way around this that I have found is something like:
>
> Plot[{f[z][[1]], f[z][[2]], f[z][[3]], f[z][[4]], f[z][[5]]}, {z, zmin,
> zmax}]
>
> which of course requires 5 evaluations of f[z] for each value of z.
> It seems that unless the head of the first argument to Plot is List,
> Plot assumes that it will evaluate to a real number and returns with an
> error if it later finds that it doesn't. Why can't Plot trust the user
> long enough to discover that the function will evaluate to a list?
> Thanks for any solutions or explanations, Malcolm
I tend to use the "function that remembers its values" construct. For
example:
f[x_]:=f[x]=Module[{},Return[{f1,f2,f3}]]
g[x_]:=Module[{},Return[{g1,g2,g3}]]
This function returns a vector. Suppose I want to graph all three
components on the same graph
Plot[{f[x][[1]],f[x][[2]],f[x][[3]]},{x,low,high}]
The f function executes once for every point. The second and third
values are just look-ups. On the other hand
Plot[{g[x][[1]],g[x][[2]],g[x][[3]]},{x,low,high}]
The g function executes 3 times for every point.
If memory is a problem, then you can Clear it after the graph.
--
Remove the _nospam_ in the return address to respond.
```
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• Next by thread: Re: Plotting vector-valued functions | 572 | 2,087 | {"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-2018-26 | latest | en | 0.838181 |
https://community.learngxp.com/t/toc-limit-calculation/4758 | 1,675,251,433,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499934.48/warc/CC-MAIN-20230201112816-20230201142816-00826.warc.gz | 192,296,480 | 4,685 | # TOC limit calculation
Can anyone tell there is a problem in TOC limit calculation as below?
background:
19 L spinner flask for cell culture, working volume 15 L, flask area 3516 cm2; Cleaning acceptance criteria 10 ppm, detection method TOC, LOD: 0.34ppb, LOQ: 1.16ppb; swab method: 25 cm2, extract in 40ml water
Calculation for swab limit:
Allowable residue (TOC) for next batch: 10ppm x 15L = 150 mg
Average residue on the surface: 150mg/3516cm2 = 0.043mg/cm2
One swab contents in 25 cm2: 0.043 mg/cm2 x 25 cm2 = 1.07mg
Conc. of TOC in 40 mL water: 1.07mg/40mL = 0.026 mg/mL = 26ppm
I tried fermentors with different size, the limit will increase while the fermentor is larger!
Is it right I’m thinking in this way?
I think you are forgetting to account for the percentage of carbon in your residue (for purified proteins this is typically about 53% - not sure about cell culture broth). This should be calculated to reduce your final limit.
I’m assuming that the working volume of 15L is always used? If the process allows for smaller volumes to be used in the flask then the smallest possible volume should be used instead, as this presents the true worst-case risk and will yield a lower amount of allowable residue into the next batch.
Also, the reason a larger vessel could have a higher limit is because the ratio of working volume to surface area is probably greater. For example, trying to make a very small batch in a very large tank will yield low results (and vice versa).
[quote=Hanks]Can anyone tell there is a problem in TOC limit calculation as below?
background:
19 L spinner flask for cell culture, working volume 15 L, flask area 3516 cm2; Cleaning acceptance criteria 10 ppm, detection method TOC, LOD: 0.34ppb, LOQ: 1.16ppb; swab method: 25 cm2, extract in 40ml water
Calculation for swab limit:
Allowable residue (TOC) for next batch: 10ppm x 15L = 150 mg
Average residue on the surface: 150mg/3516cm2 = 0.043mg/cm2
One swab contents in 25 cm2: 0.043 mg/cm2 x 25 cm2 = 1.07mg
Conc. of TOC in 40 mL water: 1.07mg/40mL = 0.026 mg/mL = 26ppm
I tried fermentors with different size, the limit will increase while the fermentor is larger!
Is it right I’m thinking in this way?[/quote]
Hello Hanks,
Your calculations are perfect. I had replied to another question of yours with calculations. The limit depends on surface area of the equipment as well as the working volume (or batch size). In your example, you have considered 15 L as the working volume for 19 L flask (79%). When you go for higher size fermentors, working volume might be increased, which might have caused the increase in your MACO value. If you considered 79% as the working volume for any size of the flask / fermentor, the results will not change. Try yourself!
VEERRAJU
nvr.veeru@gmail.com | 764 | 2,795 | {"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} | 3.15625 | 3 | CC-MAIN-2023-06 | latest | en | 0.872457 |
https://www.neetprep.com/questions/55-Physics/690-Waves?courseId=8&testId=2670046-Recommended-MCQs--NEW-NCERT-PATTERN&questionId=57261-displacement-particle-given-ysintxwhere-x-inmetres-t-seconds-velocity-wave--msec--msec--msec--msec | 1,721,427,298,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514928.31/warc/CC-MAIN-20240719200730-20240719230730-00590.warc.gz | 772,184,035 | 70,217 | # The displacement of a particle is given by $y=5×{10}^{-4}\mathrm{sin}\left(100t-50x\right)$, <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> where x is in metres and t is in seconds. The velocity of the wave is: 1. 5000 m/sec 2. 2 m/sec 3. 0.5 m/sec 4. 300 m/sec
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The wave equations of two particles are given by ${y}_{1}=a\mathrm{sin}\left(\omega \text{\hspace{0.17em}}t-kx\right)$, ${y}_{2}=a\mathrm{sin}\left(kx+\omega \text{\hspace{0.17em}}t\right)$, then:
1 they are moving in the opposite direction. 2 the phase between them is 90°. 3 the phase between them is 45°. 4 the phase between them is 0°.
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The equation of a progressive wave is given by $y=4\mathrm{sin}\left\{\pi \left(\frac{t}{5}-\frac{x}{9}\right)+\frac{\pi }{6}\right\}$.
Which of the following is correct?
1. v = 5 m / sec
2. λ = 18 m
3. a = 0.04 m
4. n = 50 Hz
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Two waves are propagating to the point P along a straight line produced by two sources, A and B, of simple harmonic and equal frequency. The amplitude of every wave at P is ‘a’ and the phase of A is ahead by π/3 than that of B, and the distance AP is greater than BP by 50 cm. If the wavelength is 1 meter, then the resultant amplitude at point P will be:
1. 2a
2. $a\sqrt{3}$
3. $a\sqrt{2}$
4. a
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Tuning fork $$F_1$$ has a frequency of 256 Hz and it is observed to produce 6 beats/second with another tuning fork $$F_2$$. When $$F_2$$ is loaded with wax, it still produces 6 beats/second with $$F_1$$. The frequency of $$F_2$$ before loading was:
1. 253 Hz
2. 262 Hz
3. 250 Hz
4. 259 Hz
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Two tuning forks, A and B, vibrating simultaneously produce 5 beats. The frequency of B is 512 Hz. It is seen that if one arm of A is filed a little, then the number of beats increases. The frequency of A in Hz will be:
1 502 2 507 3 517 4 522
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The equation of a stationary wave is $$y = 0.8 ~cos\left(\frac{\pi x}{20}\right)sin200(\pi t)$$, where $$x$$ is in cm and $$t$$ is in sec. The separation between consecutive nodes will be:
1. $$20~\text{cm}$$
2. $$10~\text{cm}$$
3. $$40~\text{cm}$$
4. $$30~\text{cm}$$
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A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio:
1 1 : 2 2 2 : 3 3 3 : 4 4 4 : 5
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Three waves of equal frequency having amplitudes of 10 μm, 4 μm and 7 μm arrive at a given point with a successive phase difference of $\frac{\pi }{2}$. The amplitude of the resulting wave in μm is given by:
1. 7
2. 6
3. 5
4. 4
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An organ pipe that is closed at one end has a fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe that a normal person can hear is:
1 14 2 13 3 6 4 9
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Hints | 1,470 | 4,678 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "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} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.828631 |
https://www.mooclab.club/resources/single-variable-calculus.1852/ | 1,575,612,066,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540484815.34/warc/CC-MAIN-20191206050236-20191206074236-00168.warc.gz | 818,552,670 | 15,411 | # Coursera Single Variable Calculus
University of Pennsylvania via Coursera
Tags:
• Overview
1. Platform:
Coursera
Provider:
University of Pennsylvania
Length:
5 weeks
Language:
English
Credentials:
Paid Certificate Available
Overview
Calculus is one of the grandest achievements of human thought, explaining everything from planetary orbits to the optimal size of a city to the periodicity of a heartbeat. This brisk course covers the core ideas of single-variable Calculus with emphases on conceptual understanding and applications. The course is ideal for students beginning in the engineering, physical, and social sciences. Distinguishing features of the course include: 1) the introduction and use of Taylor series and approximations from the beginning; 2) a novel synthesis of discrete and continuous forms of Calculus; 3) an emphasis on the conceptual over the computational; and 4) a clear, dynamic, unified approach.
In this fifth part--part five of five--we cover a calculus for sequences, numerical methods, series and convergence tests, power and Taylor series, and conclude the course with a final exam. Learners in this course can earn a certificate in the series by signing up for Coursera's verified certificate program and passing the series' final exam.
Syllabus
A Calculus for Sequences
It's time to redo calculus! Previously, all the calculus we have done is meant for functions with a continuous input and a continuous output. This time, we are going to retool calculus for functions with a discrete input. These are sequences, and they will occupy our attention for this last segment of the course. This first module will introduce the tools and terminologies for discrete calculus.
Introduction to Numerical Methods
That first module might have seemed a little...strange. It was! In this module, however, we will put that strangeness to good use, by giving a very brief introduction to the vast subjects of numerical analysis, answering such questions as "how do we approximate solutions to differential equations?" and "how do we approximate definite integals?" Perhaps unsurprisingly, Taylor expansion plays a pivotal role in these approximations.
Series and Convergence Tests
In "ordinary" calculus, we have seen the importance (and challenge!) of improper integrals over unbounded domains. Within discrete calculus, this converts to the problem of infinite sums, or series. The determination of convergence for such will occupy our attention for this module. I hope you haven't forgotten your big-O notation --- you are going to need it!
Power and Taylor Series
This course began with an exploration of Taylor series -- an exploration that was, sadly, not as rigorous as one would like. Now that we have at our disposal all the tests and tools of discrete and continuous calculus, we can finally close the loop and make sense of what we've been doing when we Talyor-expand. This module will cover power series in general, from we which specify to our beloved Taylor series.
Concluding Single Variable Calculus
Are we at the end? Yes, yes, we are. Standing on top of a high peak, looking back down on all that we have climbed together. Let's take one last look down and prepare for what lies above.
Taught by
Robert Ghrist | 665 | 3,258 | {"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-2019-51 | longest | en | 0.900906 |
http://mathhelpforum.com/differential-geometry/114956-how-does-series-behave.html | 1,527,076,129,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00232.warc.gz | 181,868,675 | 10,837 | # Thread: how does this series behave..
1. ## how does this series behave..
i was tought this trick to make series converge to 0
for example
f_n (x) =1 ,when x belongs to [n,(n+1)]
f_n (x) =0 when it doesnt
so no matter what x value we have
when n goes to infinity the value 1 section will "run away
and it always converge to 0
now i cant undestand if the following function is a such a function.
f_n(x)=(n)^0.5 ,when x belongs to [0,1/n]
f_n(x)=0 ,when it doesnt
herewhen n goes to infinty
i dont have a moving section like before
here the section shrinks
and static am i correct?
2. Originally Posted by transgalactic
i was tought this trick to make series converge to 0
for example
f_n (x) =1 ,when x belongs to [n,(n+1)]
f_n (x) =0 when it doesnt
so no matter what x value we have
when n goes to infinity the value 1 section will "run away
and it always converge to 0
now i cant undestand if the following function is a such a function.
f_n(x)=(n)^0.5 ,when x belongs to [0,1/n]
f_n(x)=0 ,when it doesnt
herewhen n goes to infinty
i dont have a moving section like before
here the section shrinks
and static am i correct?
But you do have a moving section. Given any positive number $\displaystyle x$ the Archimedean principle furnishes us with a $\displaystyle N\in \mathbb{N}$ such that $\displaystyle \frac{1}{N}<x$ thus for $\displaystyle N \le n\implies f_n(x)=0$. Thus for any positive $\displaystyle x$ we can see that $\displaystyle \lim_{n\to\infty}f_n(x)=0$.
3. i cant understand how its moving as n goes to infinty
what you said is just that it has a value of 0
but i think that its because the section moving
but because the section is static but shrinks till zero
for n=2
0<x<1/2
for n=100
0<x<1/100
am i correct?
4. Originally Posted by transgalactic
i cant understand how its moving as n goes to infinty
what you said is just that it has a value of 0
but i think that its because the section moving
but because the section is static but shrinks till zero
for n=2
0<x<1/2
for n=100
0<x<1/100
am i correct?
I haven't the slightest idea of what you mean. Given any positive $\displaystyle x$ by the Archimedean principle I know that eventually $\displaystyle \frac{1}{n}<x\implies x\notin \left[0,\frac{1}{n}\right]\implies f_n(x)=0$.
5. imagy a graph of that function
6. Perhaps the validation of another member will convince you. Maybe I am, in fact, wrong. In which case another member will chide me and we can move on. | 726 | 2,456 | {"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} | 3.96875 | 4 | CC-MAIN-2018-22 | latest | en | 0.898192 |
https://www.teacherspayteachers.com/Store/Jane-Feener?utm_source=blog&utm_medium=text&utm_campaign=TpTMilestones | 1,638,673,174,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363134.25/warc/CC-MAIN-20211205005314-20211205035314-00584.warc.gz | 1,135,637,746 | 36,222 | OVERVIEW
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# Jane Feener
(1,912)
5.0
My Products
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Simple machines: Looking for a great activity to supplement your unit on simple machines? Lapbooks are a great way for students to learn about a topic in a fun and hands on way.UPDATED - JAN. 2019 New fonts and clip art added.This unit contains
Subjects:
Physical Science
3rd, 4th, 5th
Types:
Graphic Organizers, Activities, Printables
\$4.00
483
Digestive System Note takerThis booklet is meant to be used with your science program on the digestive system. Students canuse information from their science resources, websites, and videos to complete the note taker. The pages act as graphic
Subjects:
Science, Anatomy, Health
4th, 5th
Types:
Graphic Organizers, Activities, Assessment
\$2.50
218
Place Value Interactive Math Journal/Notebook - Updated with new fonts and cover pages.This Interactive Math Journal for place value contains notes and foldables to teach your students about place value. These pages can be used to introduce a
Subjects:
Math, Decimals, Place Value
Homeschool, 4th, 5th
Types:
Activities, Assessment, Printables
CCSS:
4.NBT.A.1, 4.NBT.A.2, 4.NBT.A.3, 5.NBT.A.1
\$6.75
500
Looking for an engaging way to help your students learn the important concept of fractions? This lapbook containing many different fold-its is designed to do just that. Topics covered in this lapbook include:-What is a fraction definition? What
Subjects:
Math, Fractions
Homeschool, 3rd, 4th, 5th
Types:
Activities, Flash Cards, Printables
CCSS:
3.NF.A.1, 3.NF.A.2, 3.NF.A.2a, 3.NF.A.3, 3.NF.A.3b, 3.NF.A.3d, 4.NF.A.1, 4.NF.A.2
\$4.00
674
This Interactive Math Journal for operations contains notes and foldables to teach your students about addition, subtraction, multiplication and division. These pages can be used to introduce a concept or as an activity to review key points of a
Subjects:
Math, Basic Operations, Decimals
Homeschool, 4th, 5th
Types:
Activities, Fun Stuff, Assessment
CCSS:
4.NBT.B.4, 4.NBT.B.5, 4.NBT.B.6
\$8.75
288
We're All Wonders by R. J. Palacio is a great book to help teach your students about being kind and that we are all wonders in our own special way. This book companion helps to give your students a fun way to respond to the book by making a
Subjects:
English Language Arts, Classroom Community, Specialty
1st, 2nd
Types:
Graphic Organizers, Activities, Bulletin Board Ideas
\$3.50
245
Do Unto Otters is a writing activity and craftivity you can use after reading the book Do Unto Otters by Laurie Keller.This product includes:a graphic organizer to organize information given in the storytwo different writing prompts*I "otter" show
Subjects:
English Language Arts, Character Education, Classroom Community
2nd, 3rd
Types:
Activities, Printables, Bulletin Board Ideas
\$3.00
369
Weather Looking for a great activity to supplement your unit on weather? Lapbooks are a great way for students to learn about a topic in a fun and hands on way. This unit contains the following foldables that can be used to create a lapbook or
Subjects:
General Science, Science, Earth Sciences
Homeschool, 3rd, 4th, 5th
Types:
Activities, Fun Stuff, Printables
\$4.00
516
Class rules - visual reminder postersDo you find yourself constantly reminding students about class rules?These posters are designed to act as a visual reminder of some important class rules. Post them around your class or you could even introduce
Subjects:
Classroom Management, Character Education, Back to School
2nd, 3rd, 4th, 5th
Types:
Posters, Bulletin Board Ideas
\$3.00
214
Looking for some fun and engaging ways to practice comparing fractions? Try these activities, which use number lines and benchmarks to help your students’ master comparing fractions with like and unlike denominators. These centers can be used for
Subjects:
Math, Math Test Prep, Fractions
3rd, 4th, 5th
Types:
CCSS:
4.NF.A.2
\$5.75
166
Fraction Posters - These fraction posters are designed to use in your classroom to help students see various models used to represent the following fractions:one wholeone halfone thirdone fourth/ one quarterone sixthone eighthAlso included:
Subjects:
Math, Fractions
Homeschool, 3rd, 4th, 5th
Types:
Posters, Printables, Bulletin Board Ideas
CCSS:
3.NF.A.1
\$2.00
173
PatternsThese patterns posters were designed to help my students remember the vocabulary associated with patterns.Pattern posters included are:-patterns are all around us-pattern-core-term-element-repeating pattern-attribute-increasing/growing
Subjects:
Math, Vocabulary
1st, 2nd
Types:
Posters, Printables, Bulletin Board Ideas
\$3.00
182
What is a scientist and what do they do? This activity is a fun way to teach students about the inquiry method of science. Students will create their own scientist! There are a number of options for you to chose from to help differentiate this
Subjects:
Science, Other (Science)
1st, 2nd, 3rd
Types:
Activities, Printables, Bulletin Board Ideas
\$3.50
77
Are you teaching your students about Canadian money? These money posters and activities are a great place to start teaching younger students about Canadian coins. Display the posters, make a money anchor chart and complete a booklet about the
Subjects:
Math, Measurement, Other (Math)
1st, 2nd, 3rd
Types:
Worksheets, Posters, Activities
\$7.00
135
BOOM Cards are a super fun way to practice adding two digit numbers! Boom Cards are digital task cards that are self-checking so your students get immediate feedback. They work on an iPad, Chrome Book, Kindle Fire, other devices, or even your
Subjects:
Math
Homeschool, 2nd, 3rd
Types:
Task Cards, Flash Cards, Internet Activities
CCSS:
2.NBT.B.5
\$3.00
30
Have your students learn about or review how to represent numbers using this cute Elf craftivity. Using the colored elf cards included in this pack show students the different ways they can represent numbers and then let them practice by making
Subjects:
Math, Numbers, Christmas/ Chanukah/ Kwanzaa
1st, 2nd
Types:
Math Centers, Activities, Bulletin Board Ideas
\$3.00
92
Decimals- Looking for a way to help your students learn the important concept of decimals? This lapbook containing many different fold-its is designed to do just that. Topics covered in this lapbook include: -What is a decimal definition? -A
Subjects:
Math, Decimals
Homeschool, 4th, 5th, 6th
Types:
Activities, Flash Cards, Printables
\$4.00
316
End of Year Memory book/lapbook **Updated April 2016 - includes new fonts and Melonheadz clipart. Updated to include a "All About My Teachers" for classes with more than one teacher ** Are you looking for a fun activity for the end of the year
Subjects:
English Language Arts, Holidays/Seasonal, End of Year
Homeschool, 1st, 2nd, 3rd, 4th, 5th, 6th
Types:
Activities, Fun Stuff, Printables
\$4.00
277
Are you reading the book Memoirs of an Elf this year? This Christmas - Take An Elfieis a great companion activity to go along with the book. Students will be highly engaged as they make their elf and complete their written response.The finished
Subjects:
Writing, Christmas/ Chanukah/ Kwanzaa
Homeschool, 2nd, 3rd, 4th
Types:
Activities, Fun Stuff, Bulletin Board Ideas
\$2.50
133
This bundle includes the following:2-Digit Subtraction Strategy Posters and Booklet2-Digit Addition Strategy Posters and BookletDo not buy this bundle if you have purchased either of these products separately.Four posters and explanations for the
Subjects:
Math, Basic Operations, Mental Math
2nd, 3rd
Types:
Math Centers, Posters, Activities
\$7.00
\$5.60
75
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showing 1-20 of 207
TEACHING EXPERIENCE
I taught grade five for 14 years and now I am now teaching grade two. I am enjoying the switch to the primary grades and working with the little ones. In the past I have also worked as an IRT (Instructional resource teacher) providing support to students who just need a little extra help and time to become successful.
MY TEACHING STYLE
I think learning should be fun! I like my students to be active learners. I like to create lessons which incorporate a variety of learning styles and can easily be differentiated. I love to decorate my class and make it an inviting place to come each day to learn.
HONORS/AWARDS/SHINING TEACHER MOMENT
MY OWN EDUCATIONAL HISTORY
B.A. Ed (Elementary), B.A. Special Education, M. Ed (Curriculum, Teaching & Learning) | 2,164 | 8,434 | {"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-2021-49 | latest | en | 0.861002 |
https://www.nagwa.com/en/plans/467185703080/ | 1,685,924,083,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650409.64/warc/CC-MAIN-20230604225057-20230605015057-00642.warc.gz | 987,112,430 | 7,564 | # Lesson Plan: Adding and Subtracting Mixed Numbers Mathematics
This lesson plan includes the objectives, prerequisites, and exclusions of the lesson teaching students how to add and subtract mixed numbers with like and unlike denominators.
#### Objectives
Students will be able to
• add a proper fraction and a mixed number,
• subtract a proper fraction from a mixed number,
• add two mixed numbers with like denominators,
• subtract mixed numbers with like denominators,
• add two mixed numbers with unlike denominators,
• subtract mixed numbers with unlike denominators,
• solve problems involving adding and subtracting mixed numbers.
#### Prerequisites
Students should already be familiar with
• simplifying fractions,
• converting between mixed numbers and improper fractions,
• adding and subtracting fractions with like and unlike denominators.
#### Exclusions
Students will not cover
• adding or subtracting more than two mixed numbers,
• calculations involving a combination of addition and subtraction. | 191 | 1,023 | {"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-2023-23 | latest | en | 0.843099 |
http://www.chegg.com/homework-help/questions-and-answers/105-kg-monkey-starts-climb-rope-reach-banana-located-height-650-m-ropewill-snap-tension-ex-q793592 | 1,386,236,162,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163043224/warc/CC-MAIN-20131204131723-00022-ip-10-33-133-15.ec2.internal.warc.gz | 274,259,883 | 7,474 | # Force
0 pts ended
A 10.5 kg monkey starts to climb along rope to reach a banana located at a height of 6.50 m. The ropewill snap if the tension exceeds 115.5 N. Calculate the leastamount of time the monkey could take to reach the banana withoutbreaking the rope. | 69 | 265 | {"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-2013-48 | latest | en | 0.920794 |
http://www.gurufocus.com/term/deb2equity/SRI/Debt%252Bto%252BEquity/Stoneridge%252C%2BInc | 1,490,816,811,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218191353.5/warc/CC-MAIN-20170322212951-00255-ip-10-233-31-227.ec2.internal.warc.gz | 526,495,351 | 27,713 | Switch to:
GuruFocus has detected 3 Warning Signs with Stoneridge Inc \$SRI.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Stoneridge Inc (NYSE:SRI)
Debt-to-Equity
0.47 (As of Dec. 2016)
Stoneridge Inc's current portion of long-term debt for the quarter that ended in Dec. 2016 was \$8.6 Mil. Stoneridge Inc's long-term debt for the quarter that ended in Dec. 2016 was \$75.1 Mil. Stoneridge Inc's total equity for the quarter that ended in Dec. 2016 was \$178.3 Mil. Stoneridge Inc's debt to equity for the quarter that ended in Dec. 2016 was 0.47.
A high debt to equity ratio generally means that a company has been aggressive in financing its growth with debt. This can result in volatile earnings as a result of the additional interest expense.
Definition
Debt to Equity measures the financial leverage a company has.
Stoneridge Inc's Debt to Equity Ratio for the fiscal year that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (8.626 + 75.06) / 178.315 = 0.47
Stoneridge Inc's Debt to Equity Ratio for the quarter that ended in Dec. 2016 is calculated as
Debt to Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt) / Total Equity = (8.626 + 75.06) / 178.315 = 0.47
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
In the calculation of Debt to Equity, we use the total of Current Portion of Long-Term Debt and Long-Term Debt divided by Total Equity. In some calculations, Total Liabilities is used to for calculation.
Be Aware
Because a company can increase its Return on Equity by having more financial leverage, it is important to watch the leverage ratio when investing in high Return on Equity companies.
Related Terms
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Stoneridge Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 deb2equity 0.97 1.99 2.64 2.00 2.04 1.35 1.32 1.43 1.27 0.47
Stoneridge Inc Quarterly Data
Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 deb2equity 1.46 1.43 1.51 1.42 1.44 1.27 1.16 1.01 0.80 0.47
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More) | 699 | 2,593 | {"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-2017-13 | latest | en | 0.952764 |
https://www.physicsforums.com/threads/limit-error-mathematica.525728/ | 1,544,817,876,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826306.47/warc/CC-MAIN-20181214184754-20181214210754-00505.warc.gz | 1,003,429,463 | 12,646 | # Limit Error - Mathematica
1. Aug 30, 2011
### you878
I was using Mathematica to find the limit of the equation:
(x^3*Floor[x - 3])/(x - 3)
As x approaches 3.
Mathematica gave the answer as 0, but when I checked by hand, I did not get that.
As the function approaches 3 from the left side, it goes to positive infinity. As the function approaches 3 from the right side, it goes to 0. Since the two one-sided limits do not equal each other, shouldn't the limit at 3 not exist?
(The Floor[x-3] function I used was to represent the Step-function [[x-3]])
2. Aug 30, 2011
### HallsofIvy
Yes, for x close to but less than 3, "floor[x- 3]" is -1 so the limit, as x goes to 3, is the same as [itex]\lim_{x\to 3}x^3/(x- 3)[/tex] which does not exit. The limit itself does not exist.
I don't use Mathematica so I can't speak for how it tried to find that limit.
3. Aug 30, 2011
### jackmell
Tell you what, place your cursor over the Limit word (in Mathematica) and hit F1 to get help on the matter. Read that help carefully, then answer your own question.
Last edited: Aug 30, 2011 | 329 | 1,087 | {"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} | 2.921875 | 3 | CC-MAIN-2018-51 | latest | en | 0.943501 |
https://math.stackexchange.com/questions/269853/how-to-show-that-topological-groups-are-automatically-hausdorff | 1,656,907,183,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104293758.72/warc/CC-MAIN-20220704015700-20220704045700-00376.warc.gz | 423,150,970 | 63,068 | # How to show that topological groups are automatically Hausdorff?
On page 146, James Munkres' textbook Topology(2ed),
Show that $$G$$ (a topological group) is Hausdorff. In fact, show that if $$x \neq y$$, there is a neighborhood $$V$$ of $$e$$ such that $$V \cdot x$$ and $$V \cdot y$$ are disjoint.
Noticeably, the definition of topological group in Munkres's textbook differs from that in wikipedia.
A topological group $$G$$ is a group that is also a topological space satisfying the $$T_1$$ axiom, such that the map of $$G \times G$$ into $$G$$ sending $$x \times y$$ into $$x \cdot y$$ and the map of $$G$$ into $$G$$ sending $$x$$ into $$x^{-1}$$, are continuous maps.
• Just to clarify, some people use a definition of a topological group which does not include $T_1$, so those won't be Hausdorff. Jan 3, 2013 at 18:44
• @Sigur That points are closed. Jan 3, 2013 at 18:46
• @Sigur: All definitions of $T_1$ that I’ve seen are equivalent, so it doesn’t matter. Jan 3, 2013 at 18:52
• @BrianM.Scott, I know that. I've just asked to suggest him to read about $T_1$ spaces. Jan 3, 2013 at 18:53
• For Further Reading; If $G$ is a topologial group the following conditions are equivalent. i) $G$ is a $T_{0}$ space. ii) $G$ is a $T_{1}$ space. iii) $G$ is a $T_{2}$ space. iv) If $\beta_{e}$ is a fundamental system of neighborhoods of $e$ then $\cap \beta_{e} =\{e\}$. v) \{e\} is a closed subgroup of $G$. vi) For all $f:H\rightarrow G$ in $\tau g$, $Kerf$ is a closed subgroup of $H$. Jan 4, 2013 at 19:29
A space $X$ is Hausdorff if and only if the diagonal $\Delta_X\subseteq X\times X$ is closed. Consider the map $G\times G\rightarrow G$ given by $(x,y)\mapsto xy^{-1}$. It is continuous by the axioms for a topological group, and the diagonal is the inverse image of the the identity $\{e\}$, which is closed by assumption. So $G$ is Hausdorff if $\{e\}$ is closed, i.e., if $G$ is $T_1$ (by homogeneity, $T_1$ for $G$ is equivalent to $\{e\}$ being closed).
Given $x\neq y$ in a Hausdorff $G$, let $U_x$ and $U_y$ be disjoint opens around $x$ and $y$, respectively. Both $U_xx^{-1}$ and $U_yy^{-1}$ are opens around $e$, so we can find open $V$ with $e\in V\subseteq U_xx^{-1}\cap U_yy^{-1}$. Then $Vx$ and $Vy$ are disjoint neighborhoods of $x$ and $y$, as desired.
• Why are Vx and Vy closed? Sep 1, 2020 at 19:54
• My argument does not assert that they are closed. They are open. They may or may not be closed, but this is irrelevant. Sep 2, 2020 at 1:49
A topological space $G$ is hausdorff iff the diagonal in $G\times G$ is closed. Can you see how the diagonal is the inverse image of a closed set under a continuous map?
By the above argument, there are two cases:
1. If $$\{1\}$$ is closed then the diagonal is closed (as it is the inverse image of $$\{1\}$$ under the continuous map $$g_1 \times g_2\mapsto g_1g_2^{-1}$$). So $$G$$ is Hausdorff.
2. If $$\{1\}$$ is open then $$G$$ is discrete, so it's Hausdorff. | 970 | 2,946 | {"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": 22, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-27 | longest | en | 0.864323 |
https://help.alteryx.com/aac/en/trifacta-classic/wrangle-language/other-functions/iptoint-function.html | 1,716,840,069,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00156.warc.gz | 245,232,215 | 6,791 | # IPTOINT Function
Computes an integer value for a four-octet internet protocol (IP) address. Source value must be a valid IP address or a column reference to IP addresses.
IP addresses must be in the following format:
aaa.bbb.ccc.ddd
where aaa, bbb, ccc, and ddd, are integers 0 - 255, inclusive.
Note
The formula used to compute the integer equivalent of the above IP address is the following:
(aaa * 2563) + (bbb * 2562) + (ccc * 256) + (ddd)
As a result, each valid IP address has a unique integer equivalent.
Wrangle vs. SQL: This function is part of Wrangle, a proprietary data transformation language. Wrangle is not SQL. For more information, see Wrangle Language.
## Basic Usage
Numeric literal example:
iptoint('1.2.3.4' )
Output: Returns the integer value 16909060.
Column reference example:
Output: Returns the value of the IpAddr column converted to an integer value.
## Syntax and Arguments
Argument
Required?
Data Type
Description
Y
string
Column name or string literal identifying the IP address to convert to an integer value
Name of the column or IP address literal whose values are used to compute the equivalent integer value.
• Missing input values generate missing results.
• Multiple columns and wildcards are not supported.
Usage Notes:
Required?
Data Type
Example Value
Yes
String literal or column reference (IP address)
4.3.2.1
## Examples
Tip
### Example - Convert IP addresses to integers
This examples illustrates how you can convert IP addresses to numeric values for purposes of comparison and sorting.
Functions:
Item
Description
IPTOINT Function
Computes an integer value for a four-octet internet protocol (IP) address. Source value must be a valid IP address or a column reference to IP addresses.
IPFROMINT Function
Computes a four-octet internet protocol (IP) address from a 32-bit integer input.
Source:
192.0.0.1
10.10.10.10
1.2.3.4
1.2.3
http://12.13.14.15
https://16.17.18.19
Transformation:
When the above data is imported, the application initially types the column as URL values, due to the presence of the http:// and https:// protocol identifiers. Select the IP Address data type for the column. The last three values are listed as mismatched values. You can fix the issues with the last two entries by applying the following transform, which matches on both http:// and https:// strings:
Transformation Name Replace text or pattern IpAddr http%?:// ''
Note
The %?Wrangle matches zero or one time on any character, which enables the matching on both variants of the protocol identifier.
Now, only the 1.2.3 value is mismatched. Perhaps you know that there is a missing zero at the end of it. To add it back, you can do the following:
Transformation Name Replace text or pattern IpAddr 1.2.3[end] '1.2.3.0' true
All values in the column should be valid for the IP Address data type. To convert these values to their integer equivalents:
Transformation Name New formula Single row formula IPTOINT(IpAddr) 'ip_as_int'
You can now manipulate the data based on this numeric key. To convert the integer values back to IP addresses for checking purposes, use the following:
Transformation Name New formula Single row formula IPFROMINT(ip_as_int) 'ip_check'
Results:
X
ip_as_int
ip_check
192.0.0.1
3221225473
192.0.0.1
10.10.10.10
168430090
10.10.10.10
1.2.3.4
16909060
1.2.3.4
1.2.3.0
16909056
1.2.3.0
12.13.14.15
202182159
12.13.14.15
16.17.18.19
269554195
16.17.18.19 | 898 | 3,503 | {"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": 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} | 2.6875 | 3 | CC-MAIN-2024-22 | latest | en | 0.637213 |
https://physics.stackexchange.com/questions/256634/is-it-contradictory-with-any-theory-or-experimental-result-to-have-a-negative-gr | 1,652,953,974,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662526009.35/warc/CC-MAIN-20220519074217-20220519104217-00381.warc.gz | 530,546,971 | 66,065 | # Is it contradictory with any theory or experimental result to have a negative gravitational force mass?
I am aware that there are many similar questions here about this in this site, but most answers concentrate on negative inertial and gravitational energy. My question is more specific.
QM together with special relativity tells us that we can generate interactions in which same charges repel if using spin one force carriers (photons). And same charges attract if using spin 2 carriers (gravitons?). I do not see anything wrong if, by symmetry you could have two mass charges, negative and positive. Of course, I am assume they work only as charges, so the inertial mass will still be positive, same too with the active mass of general relativity responsible for curvature. Here the principle of equivalence would work only for positive masses, or best, for the absolute values of the masses regardless of sign.
So is there any reason in which general relativity forbids a gravitational force of opposite sign? (do not call it gravity is that helps). In such a scenario I suspect that same charges will clump separately from positive charges and repel each other. So perhaps we have in nature galaxies made of negative matter whose dynamics would be otherwise indistinguishable from galaxies with positive matter.
Is this scenario plausible or does GR forbids it?
• Dumb question by somebody not familiar with renormalization: would the existence of such hypothesized particles help in reducing the difficulties of renormalizing quantum gravity ?
– user83548
May 18, 2016 at 18:07
• More on negative mass and gravity. May 18, 2016 at 18:58
• that answer deals with negative inertial, as opposed to gravitational, mass
– user83548
May 18, 2016 at 19:04
Negative energy or mass is not forbidden in Relativity, but gravity is not a force but geometry, so if you have a negative mass it would repell positive mass as well as negative mass, just like positive mass would attract negative and positive mass all together.
If you place a positive and a negative mass near each other the positive mass would attract the negative one while the negative mass repells the positive one, and they would both accelerate in direction of the positive mass until they get close to the speed of light.
The technical term for this is runaway pair, see video, presentation and plot.
• This do not provide a answer, the OP explicitly says that inertial mass is positive, something violated in the video presentation
– user83548
May 18, 2016 at 18:26
• If inertial and gravitational mass were not equal it would be forbidden by Relativity. May 18, 2016 at 18:30
• I know it would be no longer standard GR, but would modifying the principle of equivalence just slightly, "would work for the absolute values of the masses regardless of sign", have any major consequences?
– user83548
May 18, 2016 at 18:36
• @brucesmitherson: You can't modify anything in science except for the purpose of hypothesis building, but then you have to immediately progress to destroying the hypothesis by testing it against all known data. If you modify the equivalence principle even "slightly", then you either run into problems with the data or you have created the case of a currently indistinguishable theory. May 18, 2016 at 21:19
• In Newtonian mechanics both are numerically the same, but one could in principle distinguish between the gravitational and the inertial mass when hypothesizing about exotic matter. In general relativity this is as far as I know not possible any more, they both have to have the same sign no matter what, may it be positive or negative. May 18, 2016 at 22:21 | 797 | 3,665 | {"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-2022-21 | longest | en | 0.929667 |
http://mymathforum.com/algebra/35697-face-down-paper-strategy.html | 1,544,840,571,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00040.warc.gz | 208,409,327 | 8,947 | My Math Forum Face down paper strategy
Algebra Pre-Algebra and Basic Algebra Math Forum
May 1st, 2013, 06:21 PM #1 Newbie Joined: Oct 2012 Posts: 8 Thanks: 0 Face down paper strategy Three pieces of paper each have one random number written on them. They are placed upside down on a table. The objective is to choose the slip with the highest number on it. The rules are: You may turn over any one of the three papers and look at its number. If you think it is the highest, then keep that paper and stop. Otherwise, discard that paper and choose a second one. You may then either keep the second one or discard it and take the last paper, in which case you must keep that one. 1. Is there any strategy you may use to increase your chances of winning, or will it make no difference how you play the game? 2. If there is a superior strategy, describe it. If there is none, then explain why.
May 1st, 2013, 06:54 PM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,885 Thanks: 1088 Math Focus: Elementary mathematics and beyond Re: Face down paper strategy If you turn over a sheet and discard it in favor of another, what do you know about the two numbers? From that, what is the probability that the third sheet, or the second sheet (given it is higher than the first) is the highest number? If you keep the first sheet, what is the probability it is the highest?
May 1st, 2013, 08:09 PM #3 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,622 Thanks: 954 Re: Face down paper strategy All numbers > 0 ? All numbers different? Is there a limit? Or is 100000000000000000000000000.........00000000000000 000000000000000 possible?
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Contact - Home - Forums - Cryptocurrency Forum - Top | 537 | 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} | 3.359375 | 3 | CC-MAIN-2018-51 | latest | en | 0.921541 |
https://universitywritings.com/2021/06/23/phy-151-lab-report-m0/ | 1,627,867,713,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154302.46/warc/CC-MAIN-20210802012641-20210802042641-00607.warc.gz | 582,774,368 | 9,647 | # Phy 151 lab report m0
Here is all Lab Report :
1. “You measured the diameter, d, of the pendulum bob using a ruler and a digital caliper. List all of the diameters of the bob as measured by both instruments. Determine the range of measured values for both instruments by calculating the difference between the maximum and minimum recorded values for the group. Compare the relative sizes of these two ranges and draw a conclusion about the precision of these two methods of measuring d. Repeat this analysis for the measurements of the thickness of the 100 gram mass.Compare the size of the range of ruler measurements of the pendulum bob diameter to the size of the range of ruler measurements of the thickness of the 100 g mass. If one is much larger than the other, which is it and why is it so much larger?
2. For the group of measurements of T in V-2(B), calculate T, vTand vT . vTis the standard deviation of the calculated values for T and vTis the standard deviation of the mean. Report T = T ± vT.
3. For each value of m used in V-4, you recorded three values of xm. Using x0 from posi-tion 2 in Figure M0-2, determine the three values of x for each value of m. Tabulate m, xm and x. Plot m vs. x. Whenever plotting a linear relation, be sure to include a trendline and the equation of the trendline in the plot. You should plot all three calculated values of x as a single set of data on one plot. Using LINEST in Excel, calculate the slope s, uncertainty in the slope vs, the intercept b and the uncertainty in the intercept vb. Report the slope and the intercept as s ±vs and b ±vb. From these, determine and report k ±vk.
4. Using one of the measurements of xo with a deliberate parallax error (e.g., posi-tion 1 or position 3 in Figure M0-2), recalculate all of your values of x. With these new values of x and using LINEST, recalculate the slope and intercept of the data. Tabulate m, xm and the recalculated values of x. Compare the value of the slope calculated with these data to the slope calculated in VI-3. Recalculate and report k ±vk. Compare the value for k determined in VI-3 with the value for k determined in VI-4″
M0 Powerpoint Video（Should be really help for lab report ,Let me know if you can’t open it）：
You are only responsible for completing the sections listed on the title page.（important）
I have already bought It All of the experimental steps are in it, and you need to read it. Also Lab report question are inside.
We have Data already, So use it for this lab report.
Pages (550 words)
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No matter what kind of academic paper you need and how urgent you need it, you are welcome to choose your academic level and the type of your paper at an affordable price. We take care of all your paper needs and give a 24/7 customer care support system. | 1,024 | 4,620 | {"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-2021-31 | latest | en | 0.888358 |
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