url stringlengths 6 1.61k | fetch_time int64 1,368,856,904B 1,726,893,854B | content_mime_type stringclasses 3 values | warc_filename stringlengths 108 138 | warc_record_offset int32 9.6k 1.74B | warc_record_length int32 664 793k | text stringlengths 45 1.04M | token_count int32 22 711k | char_count int32 45 1.04M | metadata stringlengths 439 443 | score float64 2.52 5.09 | int_score int64 3 5 | crawl stringclasses 93 values | snapshot_type stringclasses 2 values | language stringclasses 1 value | language_score float64 0.06 1 |
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https://republicofsouthossetia.org/question/6-2-4-a-2-9-7-7-3-0-4-5-16205984-91/ | 1,638,033,654,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00163.warc.gz | 586,549,584 | 12,870 | 6 = 2 4 a + 2) 9x + 7 = -7 3) 0 = 4 + 5
Question
6 = 2 4
a +
2) 9x + 7 = -7
3) 0 = 4 + 5
in progress 0
1 month 2021-10-20T22:18:17+00:00 1 Answer 0 views 0 | 99 | 158 | {"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-2021-49 | latest | en | 0.590496 |
https://math.stackexchange.com/questions/3009367/another-proof-of-euclids-theorem-infinite-number-of-primes | 1,569,246,923,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514576965.71/warc/CC-MAIN-20190923125729-20190923151729-00423.warc.gz | 568,778,152 | 33,139 | # Another Proof of Euclid's Theorem (infinite number of primes)?
Here $$\mathbb N = \{2,3,4,\dots\}$$ with the binary operation of addition.
If $$m \in \mathbb N$$ we denote by $$G_{\mathbb N} (m)$$ the semigroup generated by $$m$$.
Definition: A number $$p$$ is said to be prime if for all $$m \lt p$$, $$\;p \notin G_{\mathbb N} (m)$$.
We denote the set of non-empty finite subsets of $$\mathbb N$$ by $$\mathcal F (\mathbb N)$$.
Let $$\mathtt E$$ be a function
$$\quad \mathtt E: \mathbb N \to \mathcal F (\mathbb N)$$
satisfying the following:
$$\quad \quad\quad\forall n \in \mathbb N$$
$$\tag 0 \mathtt E (2) = \{2\}$$
$$\tag 1 \text{ If } (\forall \text{ prime } p \lt n) \; n \notin G_{\mathbb N} (p) \text{ then } \mathtt E (n) = \{n\}$$
$$\tag 2 \text{ If } \, (\exists \text{ prime } p \lt n) \; n \in G_{\mathbb N} (p) \text{ then } \mathtt E (n) \text{ is the union of all such primes}$$
$$\tag 3 \mathtt E (n+1) \cap \mathtt E (n) = \emptyset$$
We have the following result:
Theorem 1: There exist one and only one function $$\mathtt E$$ satisfying $$\text{(0)}$$ thru $$\text{(2)}$$; it will also satisfy $$\text{(3)}$$. Moreover, for every $$n$$, all the numbers in the set $$\mathtt E (n)$$ are prime (the prime 'factors').
Question: Can the theorem be proved in this $$(\mathbb N,+)$$ setting?
If yes, we can continue.
Theorem 2: The set of all prime numbers is an infinite set.
Proof
If $$a_1$$ is any number, consider the 'next further out' number
$$\tag 4 a_2 = \sum_{i=1}^{a_1+1}\, a_1 = \sum_{i=1}^{a_1}\,( a_1 + 1)$$.
A simple argument using $$\text{(3)}$$ shows that $$\mathtt E (a_1) \subsetneq \mathtt E (a_2)\;$$ (c.f. Bill Dubuque's remark).
Employing recursion we get a sequence $$a_1, a_2, a_3,\dots$$ with a corresponding chain of strictly increasing sets
$$\quad \mathtt E (a_1) \subsetneq \mathtt E (a_2) \subsetneq E (a_3) \dots$$
So there are sets of prime numbers with more elements than any finite set. $$\blacksquare$$
My Work
Using the recursion theorem to implement the Sieve of Eratosthenes.
The proof of theorem 2 is along the lines found in the proof given by Filip Saidak. Also, if we set $$a_1$$ to $$1$$ in theorem 2 we get the researched OEIS sequence A007018.
Note that the proof supplied by Filip Saidak has most likely been known for many years; see Bill Dubuque's answer to the math.stackexchange.com question
Is there an intuitionist (i.e., constructive) proof of the infinitude of primes?
• Note, multiplication (of positive integers) is recursive addition. That is, we define $a\cdot1$ to be $a$ and $a\cdot(n+1)$ to be $a\cdot n+a$. – Barry Cipra Nov 22 '18 at 17:39
• @BarryCipra Yes I know. But I get excited when something that appears 'bound to multiplication' - the prime numbers - can get released into a more elementary framework. – CopyPasteIt Nov 22 '18 at 17:44
• The proof via $\,n(n\!+\!1)\,$ has more prime factors than $n$ for $\,n\ge 1$ is surely much older than Saidak's 2005 paper. The generated sequence is OEIS A007018. Note: adding $1$ yields Sylvestoer sequence $a_{n+1} = a_n^2 - a_n + 1 =$ OEIS A000058. See the OEIS notes for other connections. – Bill Dubuque Nov 22 '18 at 18:05
• @CopyPasteIt The least factor $> 1$ of $n$ is prime (i.e. irreducible) is already quite minimal. – Bill Dubuque Nov 22 '18 at 19:08
• @CopyPasteIt Well one can unwind everything down to Peano arithmetic, but that won't be very arithmetically enlightening. Why prefer assembly language over the beautiful higher-level language carefully crafted by number theorists over many centuries? – Bill Dubuque Nov 22 '18 at 19:37
It is valid, but seems to be fundamentally the same as the classical proof; both hinge upon the fact that $$p_1\times p_2\times \cdots\times p_n+1$$ is divisible by some prime not in $$\{p_1,p_2,\dots,p_n\}$$. | 1,254 | 3,820 | {"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": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-39 | latest | en | 0.690462 |
https://brainmass.com/statistics/analysis-of-variance/anova-at-lld-records-162754 | 1,495,635,281,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607846.35/warc/CC-MAIN-20170524131951-20170524151951-00063.warc.gz | 743,693,343 | 18,501 | Share
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# ANOVA - At LLD Records
1- At LLD Records, some of the market research of college students is done during promotions on college campuses, while other market research of college students is done through anonymous mail, phone, internet, and record store questionnaires. In all cases, for each new CD the company solicits an "intent-to-purchase" score from the student, with being the lowest score ("no intent to purchase") and being the highest score ("full intent to purchase").
The manager finds the following information for intent-to-purchase scores for a soon-to-be-released CD
Group Sample size Sample Mean Sample variance
On campus 25 64.3 116.1
By mail 25 66.1 82.7
By phone 25 60.2 94.4
By internet 25 62.5 137.7
In a store 25 68.2 78.5
The manager's next step is to conduct a one-way, independent-samples ANOVA test to decide if there is a difference in the mean intent-to-purchase score for this CD depending on the method of collecting the scores.
Answer the following, carrying your intermediate computations to at least three decimal places and rounding your responses to at least one decimal place.
a- What's the value of the mean square for error (the "between groups" mean square) that would be reported in the ANOVA test?
b- What's the value of the mean square for treatment (the "within groups" mean square) that would be reported in the ANOVA test?
\$2.19 | 343 | 1,402 | {"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-2017-22 | longest | en | 0.924374 |
https://forum.pyro.ai/t/why-does-scan-give-the-wrong-result-here/4361 | 1,657,079,062,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104660626.98/warc/CC-MAIN-20220706030209-20220706060209-00073.warc.gz | 318,719,328 | 9,191 | # Why does scan give the wrong result here?
## Short version of the problem
I have two implementations of a simple model. The one using numpyro’s scan gives wrong results and I would like to know what I’m doing wrong.
## Detailed explanation with MWE
I’m working on a Capture-Recapture-type model where the probability of being captured on each of the T capture occasions depends on whether the animal was captured in the previous occasion.
P(y_{it} = 1) = \begin{cases} p \quad \text{if } y_{it-1} = 0,\\ c \quad \text{if } y_{it-1} = 1. \end{cases}
The observed data looks like this for T=5:
Y = \begin{bmatrix} 1, 0, 0, 1, 0\\ 0, 0, 1, 0, 0\\ 1, 1, 1, 0, 0\\ \vdots \end{bmatrix}_{C \times T}
where each row is the capture history of each animal that was ever captured during the T capture occasions.
The model uses data augmentation – effectively appending rows of zeros to Y – and then including a latent binary indicator z_i which is 1 if that animal is included in the original sample and 0 otherwise (the purpose of this is to do inference on N, the true size of the population).
I’ve implemented a version of this model using numpyro’s scan which gives wrong results and another version using vectorized operation which gives the correct results. I would like to understand what I’m doing wrong with scan.
Here’s the scan version:
# Model using loop (scan) over capture occasions:
def model_6_2_3(nz=150, predict=False, yobs=None):
C, T = yobs.shape if yobs is not None else (100, 5)
M = C + nz
# Augment data by adding rows of zeros:
yaug = jnp.vstack([yobs, jnp.zeros(shape=(nz, T))]) if yobs is not None else None
# Priors:
omega = npr.sample("omega", dist.Beta(3, 3)) # Probability of inclusion, P(z=1).
p = npr.sample("p", dist.Beta(3, 3)) # Prob. of capture if not captured last period.
c = npr.sample("c", dist.Beta(3, 3)) # Prob. of capture if captured last period.
plate_animals = npr.plate("Animals", M, dim=-1)
# Determine which animal is part of the population:
with plate_animals:
z = npr.sample("z", dist.Bernoulli(omega), infer={"enumerate": "parallel"})
# The transition function will be applied to each column of yaug.
def transition_fn(yprev, y):
with plate_animals:
p_eff = z * ((1 - yprev) * p + yprev * c)
npr.sample("yaug", dist.Bernoulli(p_eff), obs=y)
return y, None
yinit = jnp.zeros((M,))
scan(
transition_fn,
yinit,
jnp.swapaxes(yaug, 0, 1)
)
And here’s the version without scan:
# Alternative version without looping:
def model_6_2_3_v2(nz=150, predict=False, yobs=None):
# Augment dataset:
C, T = yobs.shape if yobs is not None else (100, 5)
M = C + nz
# Augment data by adding rows of zeros:
yaug = jnp.vstack([yobs, jnp.zeros(shape=(nz, T))]) if yobs is not None else None
# Array with previous period catches:
yprev = jnp.hstack([jnp.zeros((M,1)), yaug[:, :-1]])
# Priors:
omega = npr.sample("omega", dist.Beta(3, 3))
p = npr.sample("p", dist.Beta(3, 3))
c = npr.sample("c", dist.Beta(3, 3))
plate_animals = npr.plate("Animals", M, dim=-2)
plate_time = npr.plate("Time", T, dim=-1)
# Determine which animal is part of the population:
with plate_animals:
z = npr.sample("z", dist.Bernoulli(omega), infer={"enumerate": "parallel"})
with plate_animals, plate_time:
p_eff = z * ((1 - yprev) * p + yprev * c)
npr.sample("yaug", dist.Bernoulli(p_eff), obs=yaug)
I simulate some data and run NUTS on each model like this:
import numpyro as npr
import numpyro.distributions as dist
from numpyro.infer import Predictive, NUTS, HMC, MCMC, SVI
from numpyro.contrib.control_flow import scan, cond
from numpyro import handlers
import jax
import jax.numpy as jnp
import jax.random as random
import jax.lax as lax
import numpy as np
from collections import namedtuple
# Define function to simulate data under Mb
def data_fn(N=200, T=5, p=0.3, c=0.4):
# First capture occasion:
yfull = np.zeros((N, T))
yfull[:,0] = np.random.binomial(n=1, p=p, size=N)
# Later capture occasions:
for t in range(1, T):
p_eff = (1 - yfull[:,t-1]) * p + yfull[:, t-1] * c
yfull[:,t] = np.random.binomial(n=1, p=p_eff, size=N)
ever_detected = yfull.max(1)
C = np.sum(ever_detected)
yobs = yfull[ever_detected==1]
return dict(N=N, p=p, c=c, C=C, T=T, yfull=yfull, yobs=yobs)
data = data_fn(N=200, T=5, p=0.3, c=0.4)
# Utility function for inference:
def run_inference(model, rng_key, args, **kwargs):
if args.algo == "NUTS":
kernel = NUTS(model)
elif args.algo == "HMC":
kernel = HMC(model)
mcmc = MCMC(
kernel,
num_warmup=args.num_warmup,
num_samples=args.num_samples,
num_chains=args.num_chains,
progress_bar=False if "NUMPYRO_SPHINXBUILD" in os.environ else True,
)
mcmc.run(rng_key, **kwargs)
mcmc.print_summary()
return mcmc.get_samples()
Args = namedtuple("Args", "algo num_warmup num_samples num_chains")
args = Args("NUTS", 500, 2000, 4)
# Create random key:
key = random.PRNGKey(0)
Inference on the scan model:
key, subkey = random.split(key)
posterior = run_inference(model_6_2_3, subkey, args, **{"nz": 150, "yobs": data["yobs"]})
# Returns wrong values (note in particular mean of c, which should be close to 0.40):
# mean std median 5.0% 95.0% n_eff r_hat
# c 0.79 0.05 0.79 0.70 0.88 4781.03 1.00
# omega 0.54 0.07 0.53 0.42 0.65 5128.32 1.00
# p 0.35 0.02 0.35 0.30 0.38 5357.59 1.00
# Number of divergences: 0
Inference on the vectorized model:
key, subkey = random.split(key)
posterior = run_inference(model_6_2_3_v2, subkey, args, **{"nz": 150, "yobs": data["yobs"]})
# Returns correct values:
# mean std median 5.0% 95.0% n_eff r_hat
# c 0.40 0.03 0.40 0.35 0.46 5190.41 1.00
# omega 0.62 0.04 0.62 0.55 0.69 5409.63 1.00
# p 0.29 0.03 0.29 0.25 0.33 4779.27 1.00
# Number of divergences: 0
Could you use this log_density utility to see if you can get the same log density given some c, omega, p values? Then you might look at the trace output of log_density(...) to see what’s the difference.
Thanks for the suggestion. I’m getting an error, probably because I’m not using the function log_density properly.
This is what I’m doing:
from numpyro.contrib.funsor.infer_util import log_density
log_density(
model=model_6_2_3, # Get the same error with the vectorized model.
model_args=(),
model_kwargs={"yobs": data["yobs"]},
params={"c": 0.4, "p": 0.3, "omega": 0.5})
This is the stack trace:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
/var/folders/2y/fh6x5yb519xfpyzyx6yrc9g00000gn/T/ipykernel_41048/2430114414.py in <module>
3 model_args=(),
4 model_kwargs={"yobs": data["yobs"]},
----> 5 params={"c": 0.4, "p": 0.3, "omega": 0.5})
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/infer_util.py in log_density(model, model_args, model_kwargs, params)
269 """
270 result, model_trace, _ = _enum_log_density(
272 )
273 return result.data, model_trace
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/infer_util.py in _enum_log_density(model, model_args, model_kwargs, params, sum_op, prod_op)
157 model = substitute(model, data=params)
158 with plate_to_enum_plate():
--> 159 model_trace = packed_trace(model).get_trace(*model_args, **model_kwargs)
160 log_factors = []
161 time_to_factors = defaultdict(list) # log prob factors
~/anaconda3/lib/python3.7/site-packages/numpyro/handlers.py in get_trace(self, *args, **kwargs)
169 :return: OrderedDict containing the execution trace.
170 """
--> 171 self(*args, **kwargs)
172 return self.trace
173
~/anaconda3/lib/python3.7/site-packages/numpyro/primitives.py in __call__(self, *args, **kwargs)
103 return self
104 with self:
--> 105 return self.fn(*args, **kwargs)
106
107
~/anaconda3/lib/python3.7/site-packages/numpyro/primitives.py in __call__(self, *args, **kwargs)
103 return self
104 with self:
--> 105 return self.fn(*args, **kwargs)
106
107
/var/folders/2y/fh6x5yb519xfpyzyx6yrc9g00000gn/T/ipykernel_41048/2983584856.py in model_6_2_3(nz, predict, yobs)
14
15 # Determine which animal is part of the population:
---> 16 with plate_animals:
17 z = npr.sample("z", dist.Bernoulli(omega), infer={"enumerate": "parallel"})
18
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in __enter__(self)
505 )
506 indices = to_data(
--> 507 self._indices, name_to_dim=name_to_dim, dim_type=DimType.VISIBLE
508 )
509 # extract the dimension allocated by to_data to match plate's current behavior
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in to_data(x, name_to_dim, dim_type)
707 }
708
--> 709 msg = apply_stack(initial_msg)
710 return msg["value"]
~/anaconda3/lib/python3.7/site-packages/numpyro/primitives.py in apply_stack(msg)
45 pointer = 0
46 for pointer, handler in enumerate(reversed(_PYRO_STACK)):
---> 47 handler.process_message(msg)
48 # When a Messenger sets the "stop" field of a message,
49 # it prevents any Messengers above it on the stack from being applied.
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in process_message(self, msg)
521 def process_message(self, msg):
522 if msg["type"] in ["to_funsor", "to_data"]:
--> 523 return super().process_message(msg)
524 return OrigPlateMessenger.process_message(self, msg)
525
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in process_message(self, msg)
234 self._pyro_to_funsor(msg)
235 elif msg["type"] == "to_data":
--> 236 self._pyro_to_data(msg)
237
238 @staticmethod
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in _pyro_to_data(cls, msg)
264 name_to_dim.update(
265 cls._get_name_to_dim(
--> 266 batch_names, name_to_dim=name_to_dim, dim_type=dim_type
267 )
268 )
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in _get_name_to_dim(batch_names, name_to_dim, dim_type)
249 # read dimensions and allocate fresh dimensions as necessary
250 for name, dim_request in name_to_dim.items():
--> 251 name_to_dim[name] = _DIM_STACK.request(name, dim_request)[1]
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in request(self, name, dim)
162 name, dim = self._gendim(
--> 163 NameRequest(name, dim_type), DimRequest(dim, dim_type)
164 )
165
~/anaconda3/lib/python3.7/site-packages/numpyro/contrib/funsor/enum_messenger.py in _gendim(self, name_request, dim_request)
132 or (dim_type == DimType.VISIBLE and fresh_dim <= self._first_available_dim)
133 ):
--> 134 raise ValueError(f"Ran out of free dims during allocation for {fresh_name}")
135
136 return fresh_name, fresh_dim
ValueError: Ran out of free dims during allocation for Animals
Maybe you need enum(model), as in the log_density docstring?
I get the same error with enum.
log_density(
model=enum(model_6_2_3), # Same error as without enum
# model=enum(config_enumerate(model_6_2_3)), # <-- Get the same error, too.
model_args=(),
model_kwargs={"yobs": data["yobs"]},
params={"c": 0.4, "p": 0.3, "omega": 0.5})
Sorry about this - currently, we need to specify first_available_dim
enum(model_6_2_3, first_available_dim=-2)
Thank you @fehiepsi, I can get log_density to run now. The returned log densities for the scan and the vectorized models are definitely different, as we expected. Looking at the trace values, nothing jumps out to me – but I also don’t quite understand some of the output there. I’m pasting the traces below in the hopes something is obviously wrong to you.
Could the fact that in the scan model (model_6_2_3) includes a plate outside the the scan’s function be the problem?
Here are the traces using log_density:
ld_scan, trace_scan = log_density(
model=enum(model_6_2_3, first_available_dim=-2),
# model=enum(config_enumerate(model_6_2_3), first_available_dim=-2),
model_args=(),
model_kwargs={"yobs": data["yobs"]},
params={"c": 0.4, "p": 0.3, "omega": 0.5})
which produces trace_scan
OrderedDict(
[('omega',
{'type': 'sample',
'name': 'omega',
'fn': <numpyro.distributions.continuous.Beta at 0x7fbb299457d0>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': 0.5,
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [],
'infer': {'dim_to_name': OrderedDict(), 'name_to_dim': {}}}),
('p',
{'type': 'sample',
'name': 'p',
'fn': <numpyro.distributions.continuous.Beta at 0x7fbb01ac1c50>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': 0.3,
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [],
'infer': {'dim_to_name': OrderedDict(), 'name_to_dim': {}}}),
('c',
{'type': 'sample',
'name': 'c',
'fn': <numpyro.distributions.continuous.Beta at 0x7fbaea1b4bd0>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': 0.4,
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [],
'infer': {'dim_to_name': OrderedDict(), 'name_to_dim': {}}}),
('z',
{'type': 'sample',
'name': 'z',
'fn': <numpyro.distributions.distribution.ExpandedDistribution at 0x7fbb2993bc90>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': DeviceArray([[0],
[1]], dtype=int32),
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [CondIndepStackFrame(name='Animals', dim=-1, size=312)],
'infer': {'enumerate': 'parallel',
'dim_to_name': OrderedDict([(-2, 'z'), (-1, 'Animals')]),
'name_to_dim': {'z': -2, 'Animals': -1}},
'done': True}),
('_PREV_yaug',
{'type': 'sample',
'name': '_PREV_yaug',
'fn': <numpyro.distributions.discrete.BernoulliProbs at 0x7fbaea147c10>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': DeviceArray([0., 0., 0., 1., 0., 0., 1., 0., 0., 1., 1., 0., 0., 0., 0.,
0., 0., 0., 0., 1., 1., 0., 1., 0., 0., 1., 0., 0., 1., 0.,
1., 1., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1., 1.,
0., 0., 1., 1., 0., 1., 0., 1., 1., 1., 1., 1., 0., 0., 0.,
0., 0., 1., 0., 0., 1., 1., 0., 0., 0., 1., 0., 0., 1., 1.,
1., 0., 0., 0., 1., 0., 0., 0., 0., 0., 1., 0., 1., 0., 0.,
1., 0., 0., 1., 0., 0., 1., 0., 0., 0., 1., 1., 0., 0., 0.,
0., 0., 1., 0., 0., 0., 1., 0., 1., 1., 0., 0., 1., 0., 0.,
0., 0., 0., 1., 1., 0., 0., 0., 1., 0., 1., 0., 0., 0., 0.,
1., 0., 0., 0., 0., 0., 0., 0., 0., 1., 1., 1., 0., 1., 1.,
0., 0., 0., 1., 0., 0., 1., 1., 1., 0., 1., 1., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.], dtype=float32),
'scale': None,
'is_observed': True,
'intermediates': [],
'cond_indep_stack': [CondIndepStackFrame(name='Animals', dim=-1, size=312)],
'infer': {'_scan_current_index': 0,
'dim_to_name': OrderedDict([(-2, 'z'), (-1, 'Animals')]),
'name_to_dim': {'z': -2, 'Animals': -1}}}),
('yaug',
{'args': (),
'fn': <numpyro.distributions.discrete.BernoulliProbs at 0x7fbb1a690e90>,
'intermediates': [],
'value': DeviceArray([[[0., 0., 0., ..., 0., 0., 0.]],
[[1., 0., 0., ..., 0., 0., 0.]],
[[1., 1., 1., ..., 0., 0., 0.]],
[[1., 0., 1., ..., 0., 0., 0.]]], dtype=float32),
'_control_flow_done': True,
'type': 'sample',
'name': 'yaug',
'kwargs': {'rng_key': None, 'sample_shape': ()},
'scale': None,
'is_observed': True,
'cond_indep_stack': [CondIndepStackFrame(name='Animals', dim=-1, size=312)],
'infer': {'_scan_current_index': Traced<ShapedArray(int32[], weak_type=True)>with<DynamicJaxprTrace(level=1/0)>,
'dim_to_name': OrderedDict([(-3, '_time_yaug'),
(-2, 'z'),
(-1, 'Animals')]),
'name_to_dim': {'_time_yaug': -3, 'z': -2, 'Animals': -1}}})])
And for the vectorized model,
ld_vec, trace_vec = log_density(
model=enum(model_6_2_3_v2, first_available_dim=-3),
# model=enum(config_enumerate(model_6_2_3), first_available_dim=-2),
model_args=(),
model_kwargs={"yobs": data["yobs"]},
params={"c": 0.4, "p": 0.3, "omega": 0.5})
which produces trace_vec
OrderedDict(
[('omega',
{'type': 'sample',
'name': 'omega',
'fn': <numpyro.distributions.continuous.Beta at 0x7fbb298b5890>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': 0.5,
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [],
'infer': {'dim_to_name': OrderedDict(), 'name_to_dim': {}}}),
('p',
{'type': 'sample',
'name': 'p',
'fn': <numpyro.distributions.continuous.Beta at 0x7fbb1a59b3d0>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': 0.3,
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [],
'infer': {'dim_to_name': OrderedDict(), 'name_to_dim': {}}}),
('c',
{'type': 'sample',
'name': 'c',
'fn': <numpyro.distributions.continuous.Beta at 0x7fbb2993b290>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': 0.4,
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [],
'infer': {'dim_to_name': OrderedDict(), 'name_to_dim': {}}}),
('z',
{'type': 'sample',
'name': 'z',
'fn': <numpyro.distributions.distribution.ExpandedDistribution at 0x7fbb39409510>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': DeviceArray([[[0]],
[[1]]], dtype=int32),
'scale': None,
'is_observed': False,
'intermediates': [],
'cond_indep_stack': [CondIndepStackFrame(name='Animals', dim=-2, size=312)],
'infer': {'enumerate': 'parallel',
'dim_to_name': OrderedDict([(-3, 'z'), (-2, 'Animals')]),
'name_to_dim': {'z': -3, 'Animals': -2}},
'done': True}),
('yaug',
{'type': 'sample',
'name': 'yaug',
'fn': <numpyro.distributions.discrete.BernoulliProbs at 0x7fbb298ec810>,
'args': (),
'kwargs': {'rng_key': None, 'sample_shape': ()},
'value': DeviceArray([[0., 0., 1., 1., 1.],
[0., 0., 0., 1., 0.],
[0., 0., 0., 1., 1.],
...,
[0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0.]], dtype=float32),
'scale': None,
'is_observed': True,
'intermediates': [],
'cond_indep_stack': [CondIndepStackFrame(name='Time', dim=-1, size=5),
CondIndepStackFrame(name='Animals', dim=-2, size=312)],
'infer': {'dim_to_name': OrderedDict([(-3, 'z'),
(-2, 'Animals'),
(-1, 'Time')]),
'name_to_dim': {'z': -3, 'Animals': -2, 'Time': -1}}})])
Thanks @Maturin! Turns out that this is a bug, could you report this on Github? enumeration requires all batch dimensions declared using plate. scan creates a batch time dimension but we don’t have plate for it.
Great! Just opened the following issue on Github #1442.
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Double Dates (Posted on 2005-09-14)
Steve is in charge of designing a wall-hanging calendar. Each month is allocated a grid of 5 X 7 squares, labeled Sunday thru Saturday across the top. The problem is, Steve hates to put two dates in the same square on the calendar, necessary when the month spans parts of six weeks. Is it possible for Steve to find a year when he never has to put two dates in the same square? What is the most double-date squares he would ever need for a single year?
See The Solution Submitted by Josh70679 Rating: 3.3333 (3 votes)
Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Pesky months -- alternative count Comment 4 of 4 |
(In reply to re: Pesky months by Charlie)
If we count each month that requires a double date as requiring only one double date (doubling up at the beginning of the month if necessary to avoid two double date boxes in a month), the table becomes:
`Su31 We28 We31 Sa30* Mo31 Th30 Sa31* Tu31 Fr30 Su31 We30 Fr31* 3Mo31 Th28 Th31 Su30 Tu31 Fr30 Su31 We31 Sa30* Mo31 Th30 Sa31* 2Tu31 Fr28 Fr31* Mo30 We31 Sa30* Mo31 Th31 Su30 Tu31 Fr30 Su31 2We31 Sa28 Sa31* Tu30 Th31 Su30 Tu31 Fr31* Mo30 We31 Sa30* Mo31 3Th31 Su28 Su31 We30 Fr31* Mo30 We31 Sa31* Tu30 Th31 Su30 Tu31 2Fr31* Mo28 Mo31 Th30 Sa31* Tu30 Th31 Su31 We30 Fr31* Mo30 We31 3Sa31* Tu28 Tu31 Fr30 Su31 We30 Fr31* Mo31 Th30 Sa31* Tu30 Th31 3Su31 We29 Th31 Su30 Tu31 Fr30 Su31 We31 Sa30* Mo31 Th30 Sa31* 2Mo31 Th29 Fr31* Mo30 We31 Sa30* Mo31 Th31 Su30 Tu31 Fr30 Su31 2Tu31 Fr29 Sa31* Tu30 Th31 Su30 Tu31 Fr31* Mo30 We31 Sa30* Mo31 3We31 Sa29 Su31 We30 Fr31* Mo30 We31 Sa31* Tu30 Th31 Su30 Tu31 2Th31 Su29 Mo31 Th30 Sa31* Tu30 Th31 Su31 We30 Fr31* Mo30 We31 2Fr31* Mo29 Tu31 Fr30 Su31 We30 Fr31* Mo31 Th30 Sa31* Tu30 Th31 3Sa31* Tu29 We31 Sa30* Mo31 Th30 Sa31* Tu31 Fr30 Su31 We30 Fr31* 4`
with a minimum of 2 and a maximum of 4 double dates in the year.
Posted by Charlie on 2005-09-14 14:07:39
Search: Search body:
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OpenStudy (anonymous):
-1/2x =-3/4 solve using the multiplication principle
OpenStudy (anonymous):
OpenStudy (hoblos):
-1/2x =-3/4 multiply both sides by 2 =>x=3/2
OpenStudy (anonymous):
so do multiply ny 2 or -2 ?
OpenStudy (hoblos):
sorry -2
OpenStudy (anonymous):
thanks hoblos and moorsy! | 102 | 324 | {"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-2023-40 | latest | en | 0.827978 |
https://www.diyaudio.com/community/threads/quad-63-and-later-delay-line-inductors.338927/ | 1,723,354,044,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640975657.64/warc/CC-MAIN-20240811044203-20240811074203-00065.warc.gz | 554,267,541 | 32,548 | # QUAD 63 (and later) Delay Line Inductors
#### stokessd
There was some interest in another thread about the modern Quad inductors that form a portion of the delay line in the 63's and newer Quad electrostatic loudspeakers.
Strangely, the British electronics in these speakers is quite robust. But over the 24+ years I've been fixing quads, i've seen just about every failure you can have. I've only seen a couple damaged inductors and they were the results of massive overdriving of the speaker. Below is my knowledge of these inductors, although I haven't really dug into them very deeply (both metaphorically and physically) which you'd want to do if you were going to reproduce them.
The inductors are air core units with a second shorting winding that increases the inductance dramatically beyond what a single coil of this size and wire gauge is capable of. The inductance is about 2.7H each if my measurements are correct. The attachment below shows my method and results. (NOTE: The DCR is in Kohms not ohms as noted on the sheet. And I went back and re-measured the inductor resistance and got the same numbers as the first two inductors; I beleive all inductors are identical). I used a function generator, a resistor and a voltmeter to measure the inductor by forming an AC voltage divider. Using the fact that and inductor has an impedance of: Z = 2 * PI * f * L, I was able to calculate L because I know, Z, and f.
As seen in the pictures, the copper coil is sitting in a plastic "bobbin" with three leads which are soldered to the delay line circuit board. The entire bobbin, and actually the entire delay line circuit board is encapsulated in bee's wax which makes these things easy to work on.
Two of the leads are connected via a buss bar inside the bobbin, and the third (middle) lead runs from the center of the coil. The shorting coil is attached to both the left and right attachment points on the buss bar (forming a short through that bar. And the main inductor coil is attached to one side of the buss bar, at the outside of the coil and the other end is attached to the center lead coming from the middle of the coil.
I don't have a feel for how the shorting coil is arranged relative to the main coil. I'd think it would be bifilar wound, but I have never completely unpotted one of these to check it out.
The way I've seen these fail is that the very fine wire breaks near the solder connection at the buss bar. It appears that the wire overheads and melts, but on come specimens the wire looks thinned which doesn't make sense with a pure melting failure mechanism.
To fix them, I melt the wax around the buss bar and in the center of the coil and using a short piece of jumper wire, I re-solder the fine magnet wire to the jumper and then the jumper to the buss bar. I use the jumper wire because without un-potting the entire coil, the wire where it breaks is too short to reach the buss bar.
Here's a couple pictures of the inductor including one through my stereo-microscope eyepiece. Taking a picture with my cell phone through the eyepiece is a real kludge but works reasonably well to document things.
Sheldon
1 user
#### MarcelvdG
2.7 H air cored? Wow!
You have probably known this for decades, but according to Peter Baxandall (in chapter 3 of Loudspeaker and Headphone Handbook, J. Borwick, ed., Butterworths, London, 1988), the shorted windings are meant to damp the transmission line for high audio frequencies, so treble is only produced by the middle of the loudspeaker. That has something to do with preventing edge effects due to the finite size of the ESL-63.
1 user
#### esl 63
Reading about the induktor, it is basically 18H, but to work ideally it was calculated that they needed a "lossy" induktor so a shorted induktor was added. And the net result became 18H*(22/132)=3H
#### esl 63
The placement of the delay induktors is not a coincidence either... well nothing that Peter Walker did was. Each Induktor is affecting the nearby ones, all is taking care off in the design, measured and calculated and solved! Peter Walker, soo ahead of it´s time.
#### bolserst
…The way I've seen these fail is that the very fine wire breaks near the solder connection at the buss bar. It appears that the wire overheads and melts, but on come specimens the wire looks thinned which doesn't make sense with a pure melting failure mechanism.
Thanks muchly for posting these pics!
Your mentioning the wire looking thinned, makes me wonder if the wire was unintentionally stretched beyond yield tension before or after wires were soldered terminated, and “necked” down. The increased resistance at that point would certainly lead to local heating. Alternatively, it is not uncommon for thin wires to fail near terminals when the solder joint is poor(ie of high resistance).
#### Hans Polak
@Sheldon,
Thanks for the information.
May be I can integrate this in my replacement circuit diagram.
Hans
#### stokessd
Why not?
Easy to apply, safe, effective, cheap. The only downside is that the melting point is low.
Sheldon
#### Hans Polak
Hi Sheldon,
I've tried to capture the ESL63 in a LTSpice simulation.
Maybe you're interested in seeing the results, resp. freq. range and step response for the 8 sections and the impedance as seen from the Power Amp.
Hans
#### Attachments
• Graph1.jpg
559.3 KB · Views: 1,062
• Graph2.jpg
267.7 KB · Views: 1,031
#### MarcelvdG
How do you model the inductor losses and parasitic capacitances?
#### stokessd
That is very interesting, it seems that in your simulation, there isn't significant rolloff in the first few sections. That doesn't seem to follow what my ears are telling me, but I've never measured each section individually. To do it right, I'd have to do more than disconnect the sections, you'd need to replace the removed sections with an equivalent capacitance.
I could do that experiment with my test jig and a second uncharged mid panel. I have a panel test fixture (as seen here: SDS Audio Labs - Quad Electrostatic Loudspeaker Measurements) that would work well for such a test.
Sheldon
#### Hans Polak
That is very interesting, it seems that in your simulation, there isn't significant rolloff in the first few sections. That doesn't seem to follow what my ears are telling me, but I've never measured each section individually. To do it right, I'd have to do more than disconnect the sections, you'd need to replace the removed sections with an equivalent capacitance.
I could do that experiment with my test jig and a second uncharged mid panel. I have a panel test fixture (as seen here: SDS Audio Labs - Quad Electrostatic Loudspeaker Measurements) that would work well for such a test.
Sheldon
However there are still a few points to clear.
1) The most obvious is that the LF impedance curve being strongly dependent on the level.
The only component to my feeling that can cause this effect, are the 12 delay line inductors.
If I had one, I could measure their behaviour at various levels with my vector network analyzer.
2) The other component is the used audio transformer, but without knowing the various parameters.
When you have the possibility, it would be great if you could measure:
1) the input inductance (L1=Ls+Lp) with the secondary open
2) input inductance (Ls) with secondary short circuited and
3) output inductance (L2) with the input open.
That would give me some important parameters of this trafo.
Hans
#### Hans Polak
How do you model the inductor losses and parasitic capacitances?
The capacitances may be neglected with two identical trafo's correctly connected, see image below.
For the inductances see my question to Sheldon.
Hans
#### Attachments
• trafo1.jpg
56.9 KB · Views: 605
#### bolserst
However there are still a few points to clear.
1) The most obvious is that the LF impedance curve being strongly dependent on the level.
The LF impedance changes with signal level because the primary inductance of the transformer is a function of core permeability which increases with signal level until core saturation is reached. I have measured the inductance as a function of Vrms for a few frequencies and will post once I track down my measurements.
2) The other component is the used audio transformer, but without knowing the various parameters.
I also measured step-up ratio and parasitics of the transformer(leakage inductance, winding capacitance, and resistance) and will post those as well.
For proper modeling of the HF roll-off behavior that stokessd mentioned, you would need to incorporate the losses intentionally introduced into the lattice inductors with shorted turns. You would also need to include the inherent winding capacitance in parallel with each inductor.
Peter Baxandall had a nice technical write-up of the ESL63 design, pages 179 – 194 of “Loudspeaker and Headphone Handbook”, J. Borwick, Ed., which is currently available for free preview on Google Books. Loudspeaker and Headphone Handbook - Google Books
Attached is a screen grab highlighting a couple sections related to your modeling effort.
#### Attachments
• ESL63-Model_notes.png
529.5 KB · Views: 620
#### Hans Polak
The LF impedance changes with signal level because the primary inductance of the transformer is a function of core permeability which increases with signal level until core saturation is reached. I have measured the inductance as a function of Vrms for a few frequencies and will post once I track down my measurements.
I also measured step-up ratio and parasitics of the transformer(leakage inductance, winding capacitance, and resistance) and will post those as well.
For proper modeling of the HF roll-off behavior that stokessd mentioned, you would need to incorporate the losses intentionally introduced into the lattice inductors with shorted turns. You would also need to include the inherent winding capacitance in parallel with each inductor.
Peter Baxandall had a nice technical write-up of the ESL63 design, pages 179 – 194 of “Loudspeaker and Headphone Handbook”, J. Borwick, Ed., which is currently available for free preview on Google Books. Loudspeaker and Headphone Handbook - Google Books
Attached is a screen grab highlighting a couple sections related to your modeling effort.
I’m looking forward to see the results of your measurements.
Hans
#### Hans Polak
Sheldon,
In the calculation in posting #1, Z = wL was used instead of Z = jwL.
This gives a wrong outcome for L.
When simulating the values for V1 and V2 of resp 2.08V and 117mV that you measured, I get a value for L = 3.7 Henry.
Hans
#### bolserst
Your transient analysis wasn't long enough to let the DC offset die down, so the 117mV RMS valuse is erroneous. You would need to continue the transient analysis for a second or so and then only analyze the last few mSec. Alternatively, you could use AC analysis instead. Also, be sure to include the DC resistance of the inductors which is 5680 ohm, so Z = R + jwL = 5680 + jwL. Using the data posted, I calculate value of L = 2.651 H
Last edited:
#### Hans Polak
Your transient analysis wasn't long enough to let the DC offset die down, so the 117mV RMS valuse is erroneous. You would need to continue the transient analysis for a second or so and then only analyze the last few mSec. Alternatively, you could use AC analysis instead. Also, be sure to include the DC resistance of the inductors which is 5680 ohm, so Z = R + jwL = 5680 + jwL. Using the data posted, I calculate value of L = 2.651 H
Thanks a lot for giving Rser, this changes a lot.
Again using Sheldons values of 2.08V and 117mV, 1010 Ohm and 1Khz, I get 2.67 Henry.
I will integrate this Rser in my LTSpice diagram.
Hans
#### bolserst
… will post once I track down my measurements.
Finally located the measurements I took back in 2016. Thank again to stokessd for letting me borrow one of the interface units he was purchasing. My main interest at the time was the transformer design, but took some additional measurements of the LC transmission line with the intention of working up a correlated LTspice model at some point, but it has been low my the priority list. Hopefully you will find the measurements sufficient for the purpose.
Attachment #1: The transformer primary inductance was measured for increasing input voltages applied directly to the primary(ie bypassing the 1.65 ohm resistors) showing the expected trend of increasing inductance with a sudden drop off as core saturation is approached.
Attachment #2: Parasitics were also measured. Step up ratio was 1:122 and nominal values of primary and secondary inductance are used. Note the unusually large leakage inductance and unusually small winding capacitance. Golfnut had posted some comments regarding why the transformer was designed this way.
About to take the ESL plunge
esl transformer.
Attachment #3: Transfer function of the unloaded transformer in red shows the expected peak where the leakage inductance resonates with the winding capacitance. The green line shows the response with the LC transmission line circuit is attached, but no ESL panels are connected. The orange line shows the response when 22pF capacitors are used to load the line.
Attachment #4, #5: Impulse response trends for the LC transmission line unloaded and then loaded with 22pF capacitors. Notice the increase in delay when capacitors are added.
Attachment #6, #7: Frequency response trends for the LC transmission line unloaded and then loaded with 22pF capacitors.
I can post a set of *.wav files containing the impulse responses if that would be helpful, just let me know.
***NOTE***
Meant to mention that the transformer output and LC line measurements were taken with a balanced HV probe having an input impedance of 2.7Mohm.
#### Attachments
• ESL63_TLine_04.png
207.9 KB · Views: 249
• ESL63_TLine_03.png
207.1 KB · Views: 268
• ESL63_TLine_02.png
131.5 KB · Views: 273
• ESL63_TLine_01.png
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• ESL63_Transformer_03.png
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• ESL63_Transformer_02.png
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• ESL63_Transformer_01.png
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Last edited:
#### Hans Polak
In the version below of my LTSpice sim, I have made the following changes:
1: for the 12 Inductances: L=2.67H Rser=5680 Ohm and Cpar = 22p.
2: the first 6 HF sections are set to 12 x 44pF , giving a series value of 22pF.
3: I have changed the coupling factor of the audio transformer from 1 to 0.9998. Seemingly a small step, but this seems a correct value to take the serial inductance into account.
Although FR and step response are much better as before, I fail to get the impedance peak ar ca 23Khz of the real thing.
Hans
#### Attachments
• Graph3.jpg
612.6 KB · Views: 225 | 3,478 | 14,733 | {"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-2024-33 | latest | en | 0.967553 |
https://www.shortform.com/blog/the-runs-created-formula-does-it-really-work/ | 1,723,294,793,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640808362.59/warc/CC-MAIN-20240810124327-20240810154327-00717.warc.gz | 762,502,944 | 27,433 | What is the runs created formula? How does the runs created formula work?
The runs created formula is a mathematical equation developed to analyze the ability of a team to produce runs and wins. The formula’s creator, Bill James, argues that the most important measurement of predicting wins is not traditional statistics like batting average and RBI, but a new metric called the runs created formula, with a runs created calculator that helped them determine player value.
## The Runs Created Formula
James wants to see exactly which plays most contribute to a team’s offensive production. He sets out to construct a model to predict how many runs a team might expect to score, based on their accumulated walks, singles, doubles, stolen bases, and other offense-generating plays. He pores over statistics from previous baseball seasons to see which components of offense are most correlated with runs. This allows him to assign a statistical “weight” to each, in effect determining how many runs (or portions of a run) a walk, hit, or stolen base is worth.
He distills it down to a Runs Created formula:
Runs Created = (Hits + Walks) x Total Bases/(At Bats + Walks)
After statistical analysis, James finds that this formula is remarkably effective in predicting a team’s run production. It shows that walks are a valuable (yet overlooked) part of offense, while batting average matters hardly at all. You can create your own runs created calculator based on this formula.
## Removing Luck from Baseball Analysis
James begins publishing annual treatises, in which he highlights the uselessness of commonly-cited baseball statistics like errors, runs batted in (RBIs). Another of James’s key insights is that much of the data that managers, scouts, and GMs use to evaluate players’ skill is based on luck and subjective opinion. In publishing his annual baseball abstracts, James attempts to strip as much luck out of baseball analysis as possible, and develops the runs created formula.
James also points out the absurdity of that statistic revered by baseball’s old guard: the RBI. A batter earns an RBI when they score a hit that results in another player scoring a run. By definition, it is a product of luck: a batter has to have the good fortune to be at bat when other players are already on base. The batter will, of course, have had nothing to do with their teammates getting on base before them. It is a matter of luck to even have the opportunity to score an RBI. A batter who hits a single with a runner in scoring position and a batter who hits a single with no one on base have performed the same athletic feat—but one will be credited with an RBI, while the other won’t. James believes his runs created calculator can provide a better measurement.
## Rejection by Baseball Insiders
Although James starts out with a small audience in 1977, his annual treatises on baseball, The Bill James Baseball Abstracts, reach a growing and influential audience throughout the 1980s. People are compelled by his sharp and pointed critiques of the ludicrous measures by which baseball teams measure success, and begin urging major league clubs to take James’s ideas more seriously. People with backgrounds in economics, mathematics, and statistical analysis are especially drawn to his rationalistic approach to analyzing something that everyone thinks they already understand.
Despite this, there is a marked hostility on the part of baseball insiders toward those they perceive as baseball outsiders, particularly this new army of data-focused, statistics-obsessed observers who dare to tell the big clubs that they are doing their jobs all wrong.
The small cadre of people who actually run professional baseball teams is determined to exclude outside voices and opinions, even if that means losing games. They prefer to do things their way than take advice from a bunch of pencil-pushers droning on about on-base percentage and runs created formulas. They are profoundly uninterested in using baseball statistics in an analytical way. In fact, in the early 1980s, the earliest participants in what later becomes the phenomenon of fantasy sports are more interested in getting their hands on good baseball data than the teams themselves—even though these fans stand to gain far less from such knowledge.
On the rare occasion when major league teams of this era do try to implement sabermetric principles in the front office, these attempts are met with hostility by baseball’s old guard, including sports journalists and ordinary fans. When the Boston Red Sox hire a James-inspired number-cruncher to advise the general manager on personnel decisions, the sports media in Boston treats the hiring as a scandal. They point to this person’s unathletic appearance, his non-baseball background, and his residence in New York City (home of the hated Yankees) as disqualifying factors. They excoriate the Boston GM for being under the sway of a computer geek who (to them) doesn’t know the first thing about baseball and claims to avoid even watching the games. After public outcry in Boston, the team is forced to fire this sabermetrician.
Exasperated James acolytes find their ideas rejected across the league. The ideological conservatism of baseball’s ruling elite seems to have a stranglehold on the game. Despite his growing popularity with the general public, James finds that he has made no headway with the most influential part of what he thinks would be his natural audience—the people who actually run professional baseball teams.
But he does have one devoted reader, who will eventually put his ideas into practice at baseball’s highest level—Billy Beane.
The runs created formula helped advocates of Sabermetrics weed out valuable players who offered win opportunities for a lower price than well-known all-stars.
The Runs Created Formula: Does It Really Work?
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#### Carrie Cabral
Carrie has been reading and writing for as long as she can remember, and has always been open to reading anything put in front of her. She wrote her first short story at the age of six, about a lost dog who meets animal friends on his journey home. Surprisingly, it was never picked up by any major publishers, but did spark her passion for books. Carrie worked in book publishing for several years before getting an MFA in Creative Writing. She especially loves literary fiction, historical fiction, and social, cultural, and historical nonfiction that gets into the weeds of daily life. | 1,364 | 6,924 | {"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} | 3.625 | 4 | CC-MAIN-2024-33 | latest | en | 0.958672 |
https://www.jiskha.com/display.cgi?id=1345656606 | 1,498,710,832,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128323864.76/warc/CC-MAIN-20170629033356-20170629053356-00575.warc.gz | 901,815,546 | 3,654 | # math
posted by .
A 60–ft building casts a shadow that is 12 ft long.
How tall is a building that casts a shadow of 17 ft at the same time of day?
• math -
Cross multiply and solve for x.
60/12 = x/17 | 60 | 207 | {"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} | 3.015625 | 3 | CC-MAIN-2017-26 | longest | en | 0.950867 |
http://electron6.phys.utk.edu/PhysicsProblems/Concepts%20and%20formulas/identical%20particles.html | 1,725,949,391,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651224.59/warc/CC-MAIN-20240910061537-20240910091537-00516.warc.gz | 10,847,211 | 1,952 | ## Identical particles
Postulate: Every elementary particle is either a fermion or a boson. A state of many identical particles is totally antisymmetric with respect to the interchange of any two particles if they are fermions, and it is totally symmetric if they are bosons. No two identical fermions can have exactly the same set of quantum numbers. This is called the Pauli exclusion principle.
### Atomic Spectra
LS coupling
We assume that the non-central part of the electrostatic interaction is much bigger than the spin-orbit interaction. (This is usually true for light multi-electron atoms.) The electrostatic interaction leads to a splitting of the level corresponding to a given electron configuration into a number of sublevels characterized by different values of the total orbital angular momentum of the electrons, L, and their total spin, S. The operator for the electrostatic interaction commutes with L = l1+l2+l2+... and S = s1+s2+s3+... . Some of the values of L and S obtained from the general rules for addition of angular momenta can correspond to states forbidden by the Pauli principle. Filled shells do not contribute to the total orbital angular momentum L and the total spin S. To each allowed term LS belong (2L+1)(2S+1) states, differing by the values of ML and MS. The spin-orbit interaction leads to a splitting of the term LS into a number of components corresponding to different values of the total angular momentum J. But it does not completely remove the degeneracy. Each J component is degenerate with a multiplicity of 2J+1.
J(2J+1) = (2L+1)(2S+1).
In the LS coupling scheme, a term is designated by 2S+1LJ. 2S+1 is called the multiplicity of the term.
Which terms corresponding to a given configuration have the lowest energy?
Hund's Rule
(established empirically to find the ground state term)
• The level with the largest multiplicity has the lowest energy.
• For a given multiplicity, the level with the largest value of L has the lowest energy.
• For less than half-filled shells:
• The component with the smallest value of J has the lowest energy (normal order).
Examples of less than half-filled shells: np2, nd2
• For more than half-filled shells:
• The component with the largest value of J has the lowest energy (inverted order).
Examples of more than half-filled shells: np4, nd8
• When the number of electrons is 2l + 1, i.e. when the shell is half filled, there is no multiplet splitting. | 574 | 2,459 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2024-38 | latest | en | 0.84333 |
https://www.accountingformanagement.org/internal-rate-of-return-method/ | 1,606,282,558,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141181179.12/warc/CC-MAIN-20201125041943-20201125071943-00513.warc.gz | 576,750,228 | 15,382 | # Internal rate of return method
Like net present value method, internal rate of return (IRR) method also takes into account the time value of money. It analyzes an investment project by comparing the internal rate of return to the minimum required rate of return of the company.
The internal rate of return sometime known as yield on project is the rate at which an investment project promises to generate a return during its useful life. It is the discount rate at which the present value of a project’s net cash inflows becomes equal to the present value of its net cash outflows. In other words, internal rate of return is the discount rate at which a project’s net present value becomes equal to zero.
The minimum required rate of return is set by management. Most of the time, it is the cost of capital of the company.
Under this method, If the internal rate of return promised by the investment project is greater than or equal to the minimum required rate of return, the project is considered acceptable otherwise the project is rejected. Internal rate of return method is also known as time-adjusted rate of return method.
To understand how computations are made and how a proposed investment is accepted or rejected under this method, consider the following example:
## Example:
The management of VGA Textile Company is considering to replace an old machine with a new one. The new machine will be capable of performing some tasks much faster than the old one. The installation of machine will cost \$8,475 and will reduce the annual labor cost by \$1,500. The useful life of the machine will be 10 years with no salvage value. The minimum required rate of return is 15%.
Required: Should VGA Textile Company purchase the machine? Use internal rate of return (IRR) method for your conclusion.
### Solution:
To conclude whether the proposal should be accepted or not, the internal rate of return promised by machine would be found out first and then compared to the company’s minimum required rate of return.
The first step in finding out the internal rate of return is to compute a discount factor called internal rate of return factor. It is computed by dividing the investment required for the project by net annual cash inflow to be generated by the project. The formula is given below:
Formula of internal rate of return factor:
In our example, the required investment is \$8,475 and the net annual cost saving is \$1,500. The cost saving is equivalent to revenue and would, therefore, be treated as net cash inflow. Using this information, the internal rate of return factor can be computed as follows:
Internal rate of return factor = \$8,475 /\$1,500
= 5.650
After computing the internal rate of return factor, the next step is to locate this discount factor in “present value of an annuity of \$1 in arrears table“. Since the useful life of the machine is 10 years, the factor would be found in 10-period line or row. After finding this factor, see the rate of return written at the top of the column in which factor 5.650 is written. It is 12%. It means the internal rate of return promised by the project is 12%. The final step is to compare it with the minimum required rate of return of the VGA Textile Company. That is 15%.
Conclusion:
According to internal rate of return method, the proposal is not acceptable because the internal rate of return promised by the proposal (12%) is less than the minimum required rate of return (15%).
Notice that the internal rate of return promised by the proposal is a discount rate that equates the present value of cash inflows with the present value of cash out flows as proved by the following computation:
*Value from “present value of an annuity of \$1 in arrears table“.
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### 24 Comments on Internal rate of return method
1. tejaswini
Thanking u but I want some more examples.
2. Ameli
What about how to compare to exclusive projects?
3. Mercy
how do i calculate the IRR when given the salvage value.
4. abel afore markson
what is the best formula for calculating internal rate of return?
5. Buster
It seems as though all the examples are on investment in fixed asset. What about investment in services e.g. investing in the stock market. Does the same approach holds?
6. Rose Anne
what if i’m not going to use the table in locating the discount factor?
7. Wasem solomon
De formula u gave n eg. Is rather compounding issues
8. kate jerry
How do I calculate for 19% discount rate using four figure table, so as to ascertain irr?
9. Elida
It’s a plsreuae to find someone who can think so clearly
10. sumit
You can solve like our class does?
11. James
how do we compute for IRR and NPV if no rate is given, only the current amount of a planned purchase of a machine, its life in years, savings it can give the company and its depreciation amount?
12. nehecristy
a project cost N556,000 to initiate and it will generate annual cashflows(receipt less payment) of N200,000 for a period of 5 yrs. the expected scrap value is 56000. depreciation is provided on straight line method.cost of capital is 22%
required: arr, npv, irr, pbp, discounted pvb.
pls help me. dis is an assignment for 15 marks in my college tanks.
13. Siva Kumar
How can i get Rate of Return for indefinite period?
14. George
It’s strange to see internal rate of return computed analytically, are you sure this is the correct approach? Seems to be different from what is claimed in a lot of other sources: that it can’t be computed analytically and you need approximations to make it work, e.g. https://www.gigacalculator.com/calculators/irr-calculator.php . Is your “internal rate of return factor” the same or different than the classic “internal rate of return” concept???
15. Kulwa
where can I get examples that show estimates flow of the net benefits for for two projects with the the same discount rate? example 5% and 10%
16. NAGGITA VICTORIA
More examples are needed
17. banky lawrence
suppose you have a 4yr project that costs \$500. the cash flow over the 4 yr life will be \$100,\$200,\$300 and \$400 from yr 1-4 respectively. whats the IRR of the project?
18. julius moses
thanks alot for that precise example
19. Etimu Simon.S.E
What is the cost of capital of the company? How does it relate to the minimum rate of return?
What is the realistic way for the company to determine the minimum rate of return?
20. Manoharan
IRR will work out for our loan availment – example if I get a loan of Rs.1 crore repayment in 10 equal instalment on 18% PA flat interest, what will be by nett IRR
21. Dillikumar
When the interest on term loan increases what will happen to IRR | 1,532 | 6,741 | {"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} | 3.78125 | 4 | CC-MAIN-2020-50 | longest | en | 0.933193 |
https://finanalys.com/blog/wacc-example-oil-gas-usa | 1,579,977,296,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251678287.60/warc/CC-MAIN-20200125161753-20200125190753-00220.warc.gz | 433,821,869 | 3,060 | Example of WACC calculation for any project in oil and gas production and exploration in USA
11/08/2019
Example of WACC calculation for any project in oil and gas production and exploration in USA
Weighted average cost of capital is considered as measure used to evaluate:
• financial and business performance (WACC is compared with economic return on assets).
• effectiveness of investment projects (WACC indicator is used as the discount rate as determining Net present value;
• effectiveness of Mergers and Acquisitions (WACC of combined company is compared with the sum of WACC of all companies before their merger);
• business valuation (WACC is used as a discount rate when determining company business value and economic profit)
Example of WACC calculation
Sector: oil and gas production and exploration
Location of project: United States
Currency of investment: USD
Indicator Source of information 2019 Tax rate of United States https://home.kpmg/xx/en/home/services/tax/tax-tools-and-resources/tax-rates-online/corporate-tax-rates-table.html 27% Cost of Equity Risk-free rate in USD http://pages.stern.nyu.edu/~adamodar/ 2.62% Equity risk premium http://pages.stern.nyu.edu/~adamodar/ 4.19% Unlevered beta for oil and gas http://pages.stern.nyu.edu/~adamodar/ 1.03 Debt/Equity for oil and gas http://pages.stern.nyu.edu/~adamodar/ 0.55 Relevered beta = Unlevered beta for oil and gas * (1+(1-tax rate) * Debt/Equity for oil and gas) 1,44 Cost of Equity in USD 8.7% Cost of Debt in USD Cost of Debt Assumption 12.0% After-tax cost of debt = Cost of Debt * (1 - Tax rate) 8.8% Capital structure Weight of equity as % of capital = 1 - Weight of debt as % of capital 64.4% Weight of debt as % of capital = Debt/Equity for agricultural sector / (1 + Debt/Equity for agricultural sector) 39.9% WACC in USD = Weight of equity as % of capital * Cost of Equity in USD + Weight of debt as % of capital * After-tax cost of debt 8.7% | 492 | 1,931 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-05 | longest | en | 0.888891 |
http://peterhalfpennytuition.com/algebra/factorising/simple-common-factor | 1,563,853,215,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195528687.63/warc/CC-MAIN-20190723022935-20190723044935-00553.warc.gz | 116,721,549 | 4,823 | ### Question 1
Factorise completely:
(a) $$5x + 15$$ (b) $$3x - 12$$ (c) $$7x + 21$$ (d) $$12x + 18$$ (e) $$4x - 6$$ (f) $$8x + 20$$ (g) $$6x - 9$$ (h) $$25x - 15$$ (i) $$5x + 5$$ (j) $$14x - 21$$ (k) $$8a + 3$$ (l) $$9x + 3$$
### Question 2
Factorise completely:
(a) $$x^2 + 4x$$ (b) $$y^2 - y$$ (c) $$7y^2 - 3y$$ (d) $$x^2 - 5x$$ (e) $$3x^2 + 2xy$$ (f) $$8x^2y + 3x$$ (g) $$a^3 - 4a$$ (h) $$bx^2 - ax$$ (i) $$a^2x + 4ab$$ (j) $$3x^2 - 5xy + 2x$$ (k) $$4x^3 + 7x^2$$ (l) $$9x^3y - 5bx^2 + 2bx$$ | 301 | 502 | {"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} | 2.828125 | 3 | CC-MAIN-2019-30 | latest | en | 0.231089 |
https://www.teacherspayteachers.com/Product/Grade-5-NBT2-Interactive-Math-Notebook-2551753 | 1,511,526,361,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934807650.44/warc/CC-MAIN-20171124104142-20171124124142-00428.warc.gz | 872,464,960 | 24,342 | Total:
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# {Grade 5} NBT.2 Interactive Math Notebook
Common Core Standards
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***SAVE 20% by purchasing my NBT INB Bundle!***
Looking for a way to make your math lessons/centers more engaging for your students? Looking to increase the amount of writing in math and application of skills? Use interactive notebooks!
This Common Core Aligned bundle includes activities for the following standard:
5.NBT.2: Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10.
What's Included?
* 7 Activities:
- Understanding Powers of 10
- Powers of 10 Interactive Slider
- Multiplying by Powers of 10 Flip
- Dividing by Powers of 10 Flip
- Power of 10 Sort (includes whole number version and decimal version)
- Power of 10 Word Problems
- Power of 10 Roll and Spin
* A Teacher Guide for EACH activity that includes directions, colored photograph samples, and answer keys
This set can also be found in my NBT Bundle
These activities can be used as guided practice, independent practice, morning work, homework, or in centers. Each activity varies in rigor and helps students dig deeper into the standard in an engaging and meaningful way.
Be sure to download the preview to see how I use them in my classroom and for a sample of each activity!
Thank you! :)
NOTE: While I always strive for accuracy, I am human and can make editing mistakes. If you find anything that needs correcting, please send me a message immediately. I would be happy to correct and upload a new file before you rate it! :) Thank you in advance.
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\$3.00 | 467 | 1,975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-47 | longest | en | 0.897266 |
https://www.assignmentexpert.com/homework-answers/engineering/civil-and-environmental-engineering/question-231200 | 1,721,710,797,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518014.29/warc/CC-MAIN-20240723041947-20240723071947-00537.warc.gz | 564,956,672 | 66,289 | 107 775
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# Answer to Question #231200 in Civil and Environmental Engineering for Alan Enrico V Tuib
Question #231200
Compute for the distance between the theodolite and the subtense bar if the subtended angle observed is 1degree 24 minutes and 32 seconds.
1
2021-08-31T01:42:33-0400
Distance = "H= \\frac{D}{2} \\cot \\frac{\\alpha}{2}"
"\\frac{\\alpha}{2}= \\frac{1^0+(\\frac{24}{60})^0+(\\frac{32}{60*60})^0}{2}=\\frac{\\frac{317}{225}}{2}=0.70444"
If D = 2 m
"H= \\frac{2}{2} \\cot 0.70444= 1.17659 m"
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for any assignment or question with DETAILED EXPLANATIONS! | 240 | 710 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2024-30 | latest | en | 0.754943 |
https://academy.mithilanchalgroup.com/rules-to-check-the-divisibility-of-numbers/ | 1,620,852,507,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989705.28/warc/CC-MAIN-20210512193253-20210512223253-00611.warc.gz | 95,672,559 | 19,286 | # Rules to Check the Divisibility of Numbers
• A number divided by zero is undefined.
• Every number is divisible by 1.
• A number is divisible by 2 if the unit digit has either 0 or 2 or 4 or 6 or 8.
• Any number is divisible by 3 if the sum of digits will be divisible by 3.
• You can say a number is divisible by 4 if the number created by last two digits will be divisible by 4.
• A number is divisible by 5 if at the unit place, it has 5 or 0.
• Any number is divisible by 8 if a number created by the digit of unit’s, ten’s and hundreds is divisible by 8.
• A number is divisible by 9 if the sum of all the digits will be divisible by 9.
• You can say a number is divisible by 10 if it has 0 at its unit’s place.
• A number is divisible by 11 if
The sum of digits at odd places – The sum of digits at even place (Vice-versa, whichever number is larger) = 0 or multiple of 11/divisible by 11.
You can check the following video for more clarification:
Try these questions once you are sure that you can check the divisibility of any number:
https://forms.gle/mTmWxUHwjY2MD3ne6 | 297 | 1,094 | {"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-2021-21 | latest | en | 0.91709 |
http://math.stackexchange.com/questions/364283/how-to-show-that-c-bigcup-i-in-i-a-i-is-a-supremum-of-a-subset-ca-i | 1,469,681,330,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00307-ip-10-185-27-174.ec2.internal.warc.gz | 171,195,802 | 20,241 | # How to show that $C(\bigcup _{i \in I} A_i)$ is a supremum of a subset $\{C(A_i): i \in I \}$ of the lattice $L_C$ of closed subsets?
According to Brris & Sankappanavar's "A course in universal algebra," the set $L_C$ of closed subsets of a set $A$ forms a complete lattice under $\subseteq$.
Here, a subset $X$ of $A$ is said to be closed if $C(X) = X$, where $C$ is a closure operator on $A$ in the sense that it satisfies C1 - C3 below:
(For any $X, Y \subseteq A$)
C1: $X \subseteq C(X)$
C2: $C^2(X) = C(X)$
C3: $X \subseteq Y \Rightarrow C(X) \subseteq C(Y)$.
They say that the supremum of a subset $\{C(A_i): i \in I\}$ of the lattice $\langle L_C, \subseteq \rangle$ is $C(\bigcup _{i \in I} A_i)$. If so, it must be that $$C(\bigcup _{i \in I} A_i) \subseteq \bigcup _{i \in I} C (A_i)$$ (since $\bigcup _{i \in I} C (A_i)$ is also an upper bound). But, I cannot so far show how this is so.
Postscript
It was an error to think that the above inclusion had to hold if $C(\bigcup _{i \in I} A_i)$ is $sup \{C(A_i): i \in I\}$. This inclusion does not follow, and its converse follows, actually, as pointed out by Brian and Abel. Still, $C(\bigcup _{i \in I} A_i)$ is the supremum of the set since, among the closed subsets of $A$, it is the set's smallest upper bound, as explained by Brian and Alexei.
This question was very poorly and misleadingly stated. I will delete it if it's requested.
-
I thank all three people who responded to this poorly & confusingly stated question. Specifically, I thank all of them for pointing out my error of having the inclusion backwards; thank Brian and Abel for diagnosing exactly where my reasoning went wrong in holding the wrong inclusion; thank Alexei for explaining why then C(U Ai) is the smallest upper bound in L_C. All three replies (with the comments) helped me to figure out the whole thing, and my "accepting" Alexei's reply simply means it was the last piece for me to end this (self-created) puzzle. My "acceptance" goes to all of you. – Seiichiro Apr 17 '13 at 23:12
I should have thanked Brian as well for helping me to see why C(U Ai) is the smallest upper bound in L_C. I will be more careful to study each reply more carefully before responding to them... – Seiichiro Apr 18 '13 at 1:42
HINT: The result given by B&S follows easily from the following useful fact:
Proposition. For any $X\subseteq A$, $C(X)=\bigcap\{Y\subseteq A:X\subseteq Y\text{ and }C(Y)=Y\}$.
Proof. Let $\mathscr{C}=\{Y\subseteq A:X\subseteq Y\text{ and }C(Y)=Y\}$, and let $Z=\bigcap\mathscr{C}$. If $Y\in\mathscr{C}$, then $Z\subseteq Y$ and hence $Z\subseteq C(Z)\subseteq C(Y)=Y$, so $Z\subseteq C(Z)\subseteq\bigcap\mathscr{C}=Z$, and $Z$ is therefore closed. And since $X\subseteq Y$ for each $Y\in\mathscr{C}$, it’s also true that $X\subseteq Z$, so $C(X)\subseteq C(Z)=Z$. On the other hand, $X\subseteq C(X)$ and $C\big(C(X)\big)=C(X)$, so $C(Z)\in\mathscr{C}$, and therefore $Z\subseteq C(X)$. Thus, $C(X)=Z$. $\dashv$
Added: I should probably note that your displayed inclusion is actually backwards: C3 ensures that $C(A_k)\subseteq C\left(\bigcup_{i\in I}A_i\right)$ for each $k\in I$, so $\bigcup_{i\in I}C(A_i)\subseteq C\left(\bigcup_{i\in I}A_i\right)$, but the reverse inclusion fails whenever $\bigcup_{i\in I}C(A_i)$ is not closed.
-
I think you misread the inclusion. – Abel Apr 17 '13 at 10:10
@Abel: No, I did not: I ignored it in favor of providing a tool to get the actual result. However, I probably should say something about the error. – Brian M. Scott Apr 17 '13 at 10:11
Indeed, I'm sorry. It was me who misread what you are proposing to prove. +1 for helping to aquire the actual result. – Abel Apr 17 '13 at 10:19
@Abel: That’s okay; your comment led me to improve my answer, so I reckon that it was a good thing. – Brian M. Scott Apr 17 '13 at 10:21
Thank you very much Brian, for realizing my error and giving this hint for the actual result. I really appreciated this. – Seiichiro Apr 17 '13 at 23:40
$\bigcup C(A_i)$ is not necessarily closed, and the smallest closed set containing it is $C[\bigcup C(A_i)]$. Now, $\bigcup A_i \subset \bigcup C(A_i)$, thus $C(\bigcup A_i) \subset C[\bigcup C(A_i)]$.
Conversely, $A_i \subset \bigcup A_i$, so $C(A_i) \subset C(\bigcup A_i)$. Therefore, $\bigcup C(A_i) \subset C(\bigcup A_i)$, so $C[\bigcup C(A_i)] \subset C(\bigcup A_i)$.
Thus, $C[\bigcup C(A_i)] = C(\bigcup A_i)$, Q.E.D.
-
Thank you very much, Alexei. I really appreciated your help, as well as Brian's and Abel's! – Seiichiro Apr 17 '13 at 23:38
Alexi: If you realize that I once "accepted" your reply and changed it to Brian's, I'm sorry for this awkward behavior. Due to my slowness, I could not see Brian's direct relevance to my problem at first. Still, I needed both your help and Brian's, to finally see how to get B&S's correct result. Thank you very much for your help. – Seiichiro Apr 18 '13 at 1:36
We can prove the following:
$A_i\subseteq \cup_{i\in I} A_i$, thus by the third property $C(A_i)\subseteq C\left(\cup_{i\in I} A_i\right)$ for all $i\in I$. Thus $$\cup_{i\in I}C(A_i)\subseteq C\left(\cup_{i\in I}A_i\right).$$
The formula in you question is about the converse inclusion and is false in general. Consider $I=\mathbb{N}$ and $A_i = [i^{-1},1-i^{-1}]\subseteq\mathbb{R}$. Clearly $C(A_i) = A_i$ and hence $\cup_{i\in I} C(A_i) = \cup_{i\in \mathbb{N}}[i^{-1},1-i^{-1}] = (0,1)$.
On the other hand $C(\cup_{i\in I} A_i) = C((0,1)) = [0,1]$.
The problem in your reasoning is that while $\cup_{i\in I} C(A_i)$ is an upper bound it is not in general an element of your lattice.
-
Thank you very much Abel, for pointing out exactly where I went wrong in making up the wrong question! Slow as usual, I finally realized my error thanks to this reply! – Seiichiro Apr 17 '13 at 23:42
@Seiichiro Glad I could help. – Abel Apr 18 '13 at 0:17 | 1,880 | 5,857 | {"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} | 3.625 | 4 | CC-MAIN-2016-30 | latest | en | 0.916517 |
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Category | 517 | 2,210 | {"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} | 3.015625 | 3 | CC-MAIN-2024-10 | latest | en | 0.906757 |
https://chrischona2015.org/how-many-miles-is-400-meters/ | 1,660,504,904,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00328.warc.gz | 170,011,856 | 4,733 | First and also foremost, this short article answers the question how plenty of miles is 400 meters or how much is 400 meters.
You are watching: How many miles is 400 meters
If you have been searching for 400 metres to miles, climate you are best here, too, since that is the global spelling for four hundred meter to miles.
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And make certain to inspect out our converter.
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## How plenty of Miles is 400 Meters
To price 400 meters is how many miles one merely has to look in ~ the 400 meters to mile formula = / 1609.344 which we have explained on our house page.
It complies with that:
400m to miles = 0.24855 mi 400 m come mi = 0.24855 miles400 meter to mile = 0.24855 miHow much is 400 meters? 4 hundred meters in miles is same to 0.24855 miles.Make sure to know that these results have actually been rounded come 10 decimals places.In case you require the result of the 400m to mile counter with much more decimal places, then usage our calculator in ~ the optimal of this page inserting 400 m together your length.Here’s all about 400 miles to meters.
## Convert 400 meters to Miles
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## 400 meter in Miles
Here room 400 meters in nautical miles, 400 meter in inspection miles and 400 meter in kilometers:
400 m = 0.21598 nautical miles 400 m = 0.24854797979798 inspection miles 400 meter = 0.4 kilometers
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Another possibility would be sending us an email with convert 400m come miles in the subject line. | 669 | 2,781 | {"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-2022-33 | latest | en | 0.921327 |
https://www.technicalbookspdf.com/absolute-value-equations/ | 1,563,691,724,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526931.25/warc/CC-MAIN-20190721061720-20190721083720-00524.warc.gz | 845,864,404 | 7,409 | # Absolute Value Equations
Follow these steps to solve an absolute value equality which contains one absolute value:
• Isolate the absolute value on one side of the equation.
• Is the number on the other side of the equation negative? If you answered yes, then the equation has no solution. If you answered no, then go on to step 3.
• Write two equations without absolute values. The first equation will set the quantity inside the bars equal to the number on the other side of the equal sign; the second equation will set the quantity inside the bars equal to the opposite of the number on the other side.
• Solve the two equations.
Follow these steps to solve an absolute value equality which contains two absolute values (one on each side of the equation):
• Write two equations without absolute values. The first equation will set the quantity inside the bars on the left side equal to the quantity inside the bars on the right side. The second equation will set the quantity inside the bars on the left side equal to the opposite of the quantity inside the bars on the right side.
• Solve the two equations.
Let’s look at some examples.
Example 1: Solve |2x – 1| + 3 = 6
Step 1: Isolate the absolute value
|2x – 1| + 3 = 6
|2x – 1| = 3
Step 2: Is the number on the other side of the equation negative?
No, it’s a positive number, 3, so continue on to step 3
Step 3: Write two equations without absolute value bars
2x – 1 = 3
2x – 1 = -3
Step 4: Solve both equations
2x – 1 = 3 : 2x – 1 = -3
2x = 4 : 2x = -2
x = 2 : x = -1
Example 2: Solve |3x – 6| – 9 = -3
Step 1: Isolate the absolute value
|3x – 6| – 9 = -3
|3x – 6| = 6
Step 2: Is the number on the other side of the equation negative?
No, it’s a positive number, 6, so continue on to step 3
Step 3: Write two equations without absolute value bars
3x – 6 = 6
3x – 6 = -6
Step 4: Solve both equations
3x – 6 = 6 : 3x – 6 = -6
3x = 12 : 3x = 0
x = 4 : x = 0
Example 3: Solve |5x + 4| + 10 = 2
Step 1: Isolate the absolute value
|5x + 4| + 10 = 2
|5x + 4| = -8
Step 2: Is the number on the other side of the equation negative?
Yes, it’s a negative number, -8. There is no solution to this problem.
Example 4: Solve |x – 7| = |2x – 2|
Step 1: Write two equations without absolute value bars
x – 7 = 2x – 2
x – 7 = -(2x – 2)
Step 4: Solve both equations
x – 7 = 2x – 2 : x – 7 = -2x + 2
-x – 7 = -2 : 3x – 7= 2
-x = 5 : 3x = 9
x = -5 : x = 3
Example 5: Solve |x – 3| = |x + 2|
Step 1: Write two equations without absolute value bars
x – 3 = x + 2
x – 3 = -(x + 2)
Step 4: Solve both equations
x – 3 = x + 2 : x – 3 = -x – 2
– 3 = -2 : 2x – 3= -2
false statement : 2x = 1
No solution from this equation : x = 1/2
So the only solution to this problem is x = 1/2
Example 6: Solve |x – 3| = |3 – x|
Step 1: Write two equations without absolute value bars
x – 3 = 3 – x
x – 3 = -(3 – x)
Step 4: Solve both equations
x – 3 = 3 – x : x – 3 = -(3 – x)
2x – 3 = 3 : x – 3= -3 + x
2x = 6 : -3 = -3
x = 3 : All real numbers are solutions to this equation
Since 3 is included in the set of real numbers, we will just say that the solution to this equation is All Real Numbers | 1,179 | 3,719 | {"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} | 4.84375 | 5 | CC-MAIN-2019-30 | latest | en | 0.742474 |
https://oeis.org/A139555 | 1,568,920,545,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573570.6/warc/CC-MAIN-20190919183843-20190919205843-00550.warc.gz | 597,166,572 | 4,449 | This site is supported by donations to The OEIS Foundation.
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A139555 a(n) = number of prime-powers (including 1) that each are <= n and are coprime to n. 8
1, 1, 2, 2, 4, 2, 5, 4, 6, 4, 8, 4, 9, 6, 7, 7, 11, 6, 12, 8, 10, 8, 13, 8, 13, 10, 13, 11, 16, 8, 17, 14, 15, 13, 16, 11, 19, 14, 16, 13, 20, 12, 21, 16, 17, 16, 22, 15, 22, 17, 20, 18, 24, 17, 22, 18, 21, 19, 25, 16, 26, 21, 22, 22, 25, 18, 28, 22, 25, 19, 29, 21, 30, 24, 26, 24 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,3 COMMENTS Indices of first occurrence of each natural number: 1, 3, 5, 7, 9, 15, 11, 13, 21, 17, 19, 23, 32, 33, ..., . - Robert G. Wilson v From Reinhard Zumkeller, Oct 27 2010: (Start) a(n) <= A000010(n); a(A051250(n)) = A000010(A051250(n)), 1 <= n <= 17; conjecture: a(n) < A000010(n) for n > 60, cf. A051250. (End) LINKS R. Zumkeller, Table of n, a(n) for n = 1..1000 FORMULA a(n) = Sum_{k=1..A000010(n)} A010055(A038566(n,k)). - Reinhard Zumkeller, Feb 23 2012 EXAMPLE All the positive integers <= 21 that are coprime to 21 are 1,2,4,5,8,10,11,13,16,17,19,20. Of these integers, only 1,2,4,5,8,11,13,16,17,19 are prime-powers. There are 10 of these prime-powers; so a(21) = 10. MAPLE isA000961 := proc(n) if n = 1 or isprime(n) then true; else RETURN(nops(ifactors(n)[2]) =1) ; fi ; end: A139555 := proc(n) local a, i; a := 0 ; for i from 1 to n do if isA000961(i) and gcd(i, n) = 1 then a := a+1 ; fi ; od: a ; end: seq(A139555(n), n=1..100) ; # R. J. Mathar, May 12 2008 MATHEMATICA f[n_] := Length@ Select[Range@ n, Length@ FactorInteger@ # == 1 == GCD[n, # ] &]; Array[f, 76] (* Robert G. Wilson v *) PROG (Haskell) a139555 = sum . map a010055 . a038566_row -- Reinhard Zumkeller, Feb 23 2012, Oct 27 2010 CROSSREFS Cf. A139556. Cf. A065515. - Reinhard Zumkeller, Oct 27 2010 Sequence in context: A243271 A232245 A121895 * A241814 A088371 A133181 Adjacent sequences: A139552 A139553 A139554 * A139556 A139557 A139558 KEYWORD nonn AUTHOR Leroy Quet, Apr 27 2008 EXTENSIONS More terms from R. J. Mathar and Robert G. Wilson v, May 12 2008 STATUS approved
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Last modified September 19 15:08 EDT 2019. Contains 327198 sequences. (Running on oeis4.) | 1,017 | 2,479 | {"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} | 3.796875 | 4 | CC-MAIN-2019-39 | latest | en | 0.682085 |
https://www.ipclaives.it/gold2/ti-84-plus-c-silver-edition-cross-product-inbuilt-function.html | 1,638,580,961,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362923.11/warc/CC-MAIN-20211204003045-20211204033045-00223.warc.gz | 892,977,318 | 5,114 | # ti 84 plus c silver edition cross product inbuilt function
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### Ti-84 Cross Product Program and Dot Product for Vectors Triple ...
6 Feb 2016 ... Vector Cross Products are a big thing in Calculus 3, but they can be tedious to calculate due ... Ti-84 Cross Product Program and Dot Product for Vectors Triple Scalar Product ... Using Your TI-84 Plus CE Part 2: Basic Graphing.
### TI-84 Plus C Silver Edition Guidebook - Texas Instruments Calculators
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### Vector Math Program for TI-84 Plus Calc King
19 Mar 2018 ... Calculate vector addition and subtract, and dot and cross products ... or take the dot or cross product, and receive the resulting vector along ... You can name it whatever you like, but it's wise to name it something related to its function. ... to use than the TI-Nspire CX; 2x faster than TI-84 Plus C Silver Edition ...
### TI 83/84 Cross Product Program - Engineering Fundamentals Program
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### How do I find a vector cross product on a TI-84? Socratic
9 Jan 2017 ... ⟨3,2,0⟩×⟨1,4,0⟩=⟨0,0,10⟩. On the TI- n spire cx we calculate the cross product using the crossP function: enter image source here.
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1. Jan 9, 2009
### Aeana
Hey guys, I'm really struggling with an equation that I have to use for a piece of coursework. I think I'm missing something really basic but I can't seem to get past it and wondered if somebody else could help.
I want to know if it's possible to find the Laplace transform of the following:
du/dt=-0.0291*u+0.0629*w-32*theta+(0.2/m)*deltaT
I've tried everything I can think of including taking the Laplace transform of each variable and then adding them together at the end which I don't think works. I've searched everywhere I can think of for an answer to my problem but I've found nothing.
Can somebody please help me?
2. Jan 9, 2009
### HallsofIvy
Staff Emeritus
Are you saying that u and w are independent variables? Then one equation is simply not enough to determine two variables. If you had two equations, then you could take the Laplace transform of both sides of both equations to get two algebraic equations in the Laplace transforms of u and w.
Assuming that the theta and deltaT are constants, taking the Laplace transform of both sides of this equation, and writing U(s) and W(s) for the transforms of u(t) and w(t), respectively,
sU(x)- u(0)= -0.0291U(s)+ 0.0629W(s)+ ((0.2/m)deltaT- 32theta)/s
Last edited: Jan 9, 2009
3. Jan 9, 2009
### Aeana
Thanks for replying HallsofIvy. u and w are independent variables yes, they are derived from other equations which are computed in simulink. However to be able to compute the values for u,w,etc. I must know the transfer function for the equation mentioned previously. It's very messy.
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https://ca.style.yahoo.com/blogs/shine-on/shrove-tuesday-mathematical-formula-perfect-pancake-161829964.html | 1,713,935,954,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819067.85/warc/CC-MAIN-20240424045636-20240424075636-00576.warc.gz | 134,118,210 | 159,048 | # Shrove Tuesday: A mathematical formula for the perfect pancake
It’s Shrove Tuesday, better known as Pancake Tuesday in these parts. For those of you planning on flipping up a nice stack of celebratory flapjacks, the Daily Mail reports that British mathematicians have cooked up a handy formula to help you cook the perfect batch.
First, here’s the rather intimidating equation:
100 - [10L - 7F + C(k - C) + T(m - T)]/(S - E)
Yes, these numbers do relate to pancakes. A team of math whizzes from the University of Wolverhampton in England decided that cooking the traditional breakfast food could and should be reduced to numbers.
The math breaks down like this:
L = the number of lumps in the batter (ideally 0)
C = the actual consistency of the batter (scored out of 10)
F = flipping score (scored out of 100)
K = the ideal consistency of the batter (ideally 5)
T = the actual temperature of the pan (ideally 377)
M = the perfect pan temperature (377)
S = the time the batter stands before cooking (ideally 30 minutes)
E = the time the pancake stands before eating (ideally 0 minutes)
Now recall those high school math skills, plug in your figures, and see how your own pancakes score. The closer you get to 100, the better the pancake.
Will this formula actually aid in the creation of the perfect pancake? Perhaps not, but there’s no denying that math is indeed a big part of cooking — there are numbers galore in your average recipe, and conversions, fractions and sums are all par for the course.
This formula, however, takes the math in cooking to a whole new level.
Also see:
#### Man spends more than \$58,000 tasting cornflakes around the world
How to go gluten-free and have your pancakes too
Wait, so skipping breakfast isn't so bad after all? | 415 | 1,776 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2024-18 | latest | en | 0.917897 |
https://www.sololearn.com/pt/Discuss/2654285/python-regular-expression-practice-quiz-authentication | 1,719,048,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862310.25/warc/CC-MAIN-20240622081408-20240622111408-00852.warc.gz | 860,303,808 | 196,753 | Python Regular Expression Practice Quiz "Authentication!" | Sololearn: Learn to code for FREE!
Novo curso! Todo programador deveria aprender IA generativa!
+ 2
# Python Regular Expression Practice Quiz "Authentication!"
4th Jan 2021, 6:49 PM
Jochen MĆ¼ller
21 Respostas
4th Jan 2021, 9:16 PM
JaScript
+ 2
Thanks for your replies I thought this is going to be a a easier Quiz because it is more at the beginning of the Regular Expression Chapter. The solutuion from Lisa maybe not that Elegant but at least it worked as it should. This helped me a lot. thanks this is now my code which is working in the right way: import re passwords = ["Aa1", "A1a", "1Aa", "1aA", "a1A", "aA1", "aa1", "aab"] for p in passwords: number = re.search("[0-9]",p) LETTER = re.search("[A-Z]",p) if bool(number) is True and bool(LETTER) is True: print("Password created") else: print("Wrong format") Still curious how the single regex would look like ;)
5th Jan 2021, 10:42 AM
Jochen MĆ¼ller
+ 2
Calvin Sheen Thomas Yeah that was the regex the OP used and was having issues. I ended up using two expressing numPat = rā[0-9]ā upperPat = rā[A-Z]ā if re. search(numPat, password) and re.search(upperPat, password) I havenāt sorted a way to get it all in one expression
16th Jan 2021, 2:35 AM
Cody Knight
+ 1
It tried around and now my regular expression is: pattern = r"[A-Z0-9]" and it works exactly as my previous!? ( aA1 and a1A ->Wrong format) So now I have a shorter Expression but still the same Problem :D
4th Jan 2021, 8:50 PM
Jochen MĆ¼ller
+ 1
Seems to be a pro quiz š¤ Is it necessary to use a single regex? If not we could work around by using one pattern to check for number and one for capital letters (though not very elegant) passwords = ["aA1", "1Aa", "A11", "a1a", "Aaq"] for p in passwords: number = re.search('[0-9]', p) LETTER = re.search('[A-Z]', p) print(p, bool(number), bool(LETTER))
4th Jan 2021, 9:28 PM
Lisa
+ 1
5th Jan 2021, 12:29 PM
Lisa
+ 1
Hey Jochen, Not sure if you sorted this out yet. Have you tried using re.search() instead of re.match()? re.match() starts at the beginning of the string and will return false if their isn't an immediate match. re.search() looks at entire string for a match. I think pattern = r"[A-Z0-9]" checks for the characters in that order. HAven't figured this out with single regex yet
16th Jan 2021, 12:42 AM
Cody Knight
+ 1
Cody Knight Your regex expression doesn't give the required results; it just searches for a string which has an upper-case letter OR an integer.
16th Jan 2021, 2:13 AM
Calvin Thomas
+ 1
Yup, doing this with two regex is easy. Here's the single regex version: r"([A-Z]+\w*\d)|(\d+\w*[A-Z])" This works but using two regex is the preferable method for me.
16th Jan 2021, 4:36 AM
Calvin Thomas
+ 1
21st May 2021, 3:24 AM
Wedja Souza
+ 1
25th Mar 2022, 7:33 PM
Rusel Bruff
0
Maybe? pattern = r"\d.*[A-Z]|[a-z].\*d"
4th Jan 2021, 6:57 PM
Lisa
0
with this Code Aa1 and A1a also shown as Wrong Format
4th Jan 2021, 8:46 PM
Jochen MĆ¼ller
0
Here you go: from re import search as a b,c=r"([A-Z]+\w*\d)|(\d+\w*[A-Z])",input() d="Verified" if a(b,c) else "Not Verified" print(d) #Might as well copy and paste it in the #playground.
6th Jan 2021, 10:03 AM
Calvin Thomas
0
23rd Jun 2021, 8:46 AM
Maikel
0
Maikel Your code doesn't work for inputs like "8A" and "234LPP". This pattern seems to work with re.search(): r"[A-Z].*\d|\d.*[A-Z]"
24th Jun 2021, 6:44 AM
Calvin Thomas
0
1st Aug 2021, 4:20 PM
Masoud Ramezani
0 | 1,157 | 3,480 | {"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} | 2.59375 | 3 | CC-MAIN-2024-26 | latest | en | 0.857068 |
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Hello everyone i have a enemy which i want to make explode when he dies, i have split him into two parts, and all ready have the explosion working, now i just need to figure out the math to make one of the pieces fly away on a smooth curve. At the moment i have it so the piece rotates about its Z axis and moves along its Y axis, this does look ok, and i get the right general idea with the piece moving, however it looks a bit pap, there must be a cheap and chearful way of modeling this behaviour?
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Do you have a physics system? One way would be to associate the detached part with a rigid body, and just start that with some initial velocity and let the physics system take care of it.
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Cheap and cheerful? If your appliation isn't large enough to justify using a full blown physics engine, you can whip it up using a bit of vector maths:
Try this:
step 1) Calculate the vector D between the middle of each piece and the centre of the explosion.
step 2) Create velocity V where V=(force_of_explosion / D.length).
step 2a) Your formula may vary, the point is that there is SOME relation between distance from explosion and force applied to object. Try (force / D^2) as an alternative.
step 5) Normalise D to create the direction Dn that each piece should travel in.
step 6) On each frame, add vector (Dn * V) to the position of the object. the object will move at velocity V along vector Dn away from the epicentre of the explosion. This assumes the explosion is spherical and the pressure is evenly distributed.
For dramatic effect why not:
7) Create a few particles and apply the same principle?
8) Make the object rotate by a random amount in two axis rather than 1? (for 3d simulations this looks as if they are rotating in 3 axis, without the hassle)
9) Apply wind resistance: multiply V by 0.9 every second. When V is sufficiently small, stop the object from moving.
For a 2d system this kind of thing is all you need. If you want to make a 3d simulation, try a physics engine. I'm currently using ODE.
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Quote:
Original post by speciesUnknownCheap and cheerful? If your appliation isn't large enough to justify using a full blown physics engine, you can whip it up using a bit of vector maths:Try this:step 1) Calculate the vector D between the middle of each piece and the centre of the explosion.step 2) Create velocity V where V=(force_of_explosion / D.length). step 2a) Your formula may vary, the point is that there is SOME relation between distance from explosion and force applied to object. Try (force / D^2) as an alternative.step 5) Normalise D to create the direction Dn that each piece should travel in.step 6) On each frame, add vector (Dn * V) to the position of the object. the object will move at velocity V along vector Dn away from the epicentre of the explosion. This assumes the explosion is spherical and the pressure is evenly distributed. For dramatic effect why not:7) Create a few particles and apply the same principle?8) Make the object rotate by a random amount in two axis rather than 1? (for 3d simulations this looks as if they are rotating in 3 axis, without the hassle)9) Apply wind resistance: multiply V by 0.9 every second. When V is sufficiently small, stop the object from moving.For a 2d system this kind of thing is all you need. If you want to make a 3d simulation, try a physics engine. I'm currently using ODE.
This works very well for what i need, i just need to figure out a way to incorperate gravity into the formula i need the tureet to move towards the ground, i've got the turret flying up using this formula but it never returns :) lol i'll see what i can do suggestions are greatly appreciated.
Im not using a physics engine since its a small project i am just learning :)
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Gravity is an acceleration of two objects towards each other. We can ignore the second object (the earth) because its mass is so huge, and just apply a force to one object (the one in question). This is an adequate video game approximation of gravity.
The method I use for this is simple.
step 0) So you have your object, with a normalised movement direction Dn. You also have the velocity V that the object moves at.
step 1) Take the direction vector Dn and multiply it by V to get the vector D back.
step 2) Take the vector G, representing gravity, divide it by the number of frames per second, and add this to the result. A good example is G=(0.0, -9.81, 0.0)/fps (earth) or G=(0.0, -1.635, 0.0)/fps (the moon, IIRC, has 1/6 earths gravity)
---Note that the number of frames per second is important to include in the gravity vector, adding it later is far harder since the end product of the entire calculation is also divided by the framerate.
--Note also that you can add together a series of environmental forces, such as wind, to this vector before continuing.
step 3) Calculate the length of the result and store it in V, and renormalise the vector and store it in Dn. Then continue the algorithm as per my last post.
n.b. I quickly get into a mess if I try to add extra variables to these calculations. So, if for example you want to divide V by the number of frames per second, therefore ensuring that V represents units per second, do this shortly before moving the object, rather than as part of the calculations up to that point. e.g.,
position = position + (Dn * V);
If you want a fixed framerate, use this in the final calculation:
position = position + (Dn * V * timestep);
If you want to have a speed control on the game, use this:
position = position + (Dn * V * timestep * speed_factor);
and so on. Try to integrate those into the previous calculations - pretty difficult.
[Edited by - speciesUnknown on February 7, 2008 6:20:26 PM]
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I welcome the correction of any mistakes I have made here, since I didn't pay attention in physics class and I flunked maths. This is a system I have used, and it works.
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You'r method works wonders for the gravity although i did have too tweak some values :)
The only problem i'm having now is i need to move a character between three or more waypoints, which i can do, although there is an instant turn involved, i'd like the characters to turn gradually whilst moving until they face the desired direction, i have done this by simply rotating the Y axis to make the character face the point whilst moving, this has one draw back though, the character could end up spiralling around a waypoint forever, is there any other alternatives to my method?
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Quote:
Original post by LordFalloutYou'r method works wonders for the gravity although i did have too tweak some values :)
Would you mind telling me what tweaks you made and posting a bit of code, since I did go to all that effort typing it out? It may help me touch up on my method. (I frequently recommend it to my friends at Uni when they ask similar questions)
Quote:
The only problem i'm having now is i need to move a character between three or more waypoints, which i can do, although there is an instant turn involved, i'd like the characters to turn gradually whilst moving until they face the desired direction, i have done this by simply rotating the Y axis to make the character face the point whilst moving, this has one draw back though, the character could end up spiralling around a waypoint forever, is there any other alternatives to my method?
That might belong in a separate thread, in the AI forum? Anyway, do you mean the AI navigation nodes? If you distinguish between primary and secondary nodes, you can have the character search via primary nodes, and travel along secondary nodes, and have the secondary nodes use a higher number of lines around corners, enough so it looks smooth. You can also provide more information on characters behaviour as they come to a corner (such as a node where they consider peering round the corner before going to the next node, rather than running blindly round and getting shot)
I've not done any AI navigation yet, but there are some pretty experienced people in the Artificial Intelligence forum.
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No problem, im away from my computer for a little while but as soon as i get back, i'll post my tweaks, although there mainly tweaks to fit your method around my code. It should be later tomorrow peel your eyes ? :)
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### Picturing Triangular Numbers
##### Stage: 3 Challenge Level:
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##### Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
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##### Stage: 2 Challenge Level:
In this problem we are looking at sets of parallel sticks that cross each other. What is the least number of crossings you can make? And the greatest?
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##### Stage: 3 Challenge Level:
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##### Stage: 2 Challenge Level:
Here are two kinds of spirals for you to explore. What do you notice?
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
### Walking the Squares
##### Stage: 2 Challenge Level:
Find a route from the outside to the inside of this square, stepping on as many tiles as possible.
### Cunning Card Trick
##### Stage: 3 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
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##### Stage: 3 Challenge Level:
Can you explain how this card trick works?
### Painted Cube
##### Stage: 3 Challenge Level:
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##### Stage: 2 Challenge Level:
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### Nim-7
##### Stage: 1, 2 and 3 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Three Dice
##### Stage: 2 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Hidden Rectangles
##### Stage: 3 Challenge Level:
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### Handshakes
##### Stage: 3 Challenge Level:
Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Route to Infinity
##### Stage: 3 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### Threesomes
##### Stage: 3 Challenge Level:
Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw?
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Intersecting Circles
##### Stage: 3 Challenge Level:
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### Squares in Rectangles
##### Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Nim-7 for Two
##### Stage: 1 and 2 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
### One O Five
##### Stage: 3 Challenge Level:
You can work out the number someone else is thinking of as follows. Ask a friend to think of any natural number less than 100. Then ask them to tell you the remainders when this number is divided by. . . .
### Is There a Theorem?
##### Stage: 3 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### What Numbers Can We Make Now?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Go Forth and Generalise
##### Stage: 3
Spotting patterns can be an important first step - explaining why it is appropriate to generalise is the next step, and often the most interesting and important.
### ...on the Wall
##### Stage: 3 Challenge Level:
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### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
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### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Shear Magic
##### Stage: 3 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Konigsberg Plus
##### Stage: 3 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Got it for Two
##### Stage: 2 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
### Odd Squares
##### Stage: 2 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### More Magic Potting Sheds
##### Stage: 3 Challenge Level:
The number of plants in Mr McGregor's magic potting shed increases overnight. He'd like to put the same number of plants in each of his gardens, planting one garden each day. How can he do it?
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### Build it Up
##### Stage: 2 Challenge Level:
Can you find all the ways to get 15 at the top of this triangle of numbers?
### More Number Pyramids
##### Stage: 3 Challenge Level:
When number pyramids have a sequence on the bottom layer, some interesting patterns emerge...
### Partitioning Revisited
##### Stage: 3 Challenge Level:
We can show that (x + 1)² = x² + 2x + 1 by considering the area of an (x + 1) by (x + 1) square. Show in a similar way that (x + 2)² = x² + 4x + 4 | 2,195 | 9,290 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-43 | latest | en | 0.85689 |
https://de.mathworks.com/matlabcentral/cody/problems/43594-find-adjacency-matrix/solutions/1028943 | 1,596,921,916,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738351.71/warc/CC-MAIN-20200808194923-20200808224923-00234.warc.gz | 276,068,118 | 15,715 | Cody
# Problem 43594. Find Adjacency Matrix
Solution 1028943
Submitted on 24 Oct 2016 by uu tsi
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
s = [1 1 2 2 3 3 4 5 5 6 7 8]; t = [2 4 4 8 4 6 8 6 8 7 8 8]; y_correct = [0 1 0 1 0 0 0 0 ; 1 0 0 1 0 0 0 1 ; 0 0 0 1 0 1 0 0 ; 1 1 1 0 0 0 0 1 ; 0 0 0 0 0 1 0 1 ; 0 0 1 0 1 0 1 0 ; 0 0 0 0 0 1 0 1 ; 0 1 0 1 1 0 1 1] assert(isequal(getMatrix(s,t),y_correct))
y_correct = 0 1 0 1 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 1 1 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 1 0 1 0 1 0 0 0 0 0 0 1 0 1 0 1 0 1 1 0 1 1
2 Pass
s = [1 ]; t = [8 ]; y_correct = [ 0 0 0 0 0 0 0 1; 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0; 0 0 0 0 0 0 0 0; 1 0 0 0 0 0 0 0 ] assert(isequal(getMatrix(s,t),y_correct))
y_correct = 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 | 706 | 1,031 | {"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-2020-34 | latest | en | 0.61959 |
https://socratic.org/questions/how-do-you-calculate-sum-n-1-4-3-2-n | 1,723,738,918,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641299002.97/warc/CC-MAIN-20240815141847-20240815171847-00424.warc.gz | 408,441,402 | 6,247 | # How do you calculate sum_(n=1)^(∞)(4/3)^(2-n)?
## How do you calculate sum_(n=1)^(∞)(4/3)^(2-n)?
Jan 24, 2018
$S = \frac{16}{3}$
#### Explanation:
.
First term $\left(n = 1\right)$, is:
${\left(\frac{4}{3}\right)}^{2 - 1} = {\left(\frac{4}{3}\right)}^{1} = \frac{4}{3}$
Second Term ($n = 2$), is:
${\left(\frac{4}{3}\right)}^{2 - 2} = {\left(\frac{4}{3}\right)}^{0} = 1$
Third term $\left(n = 3\right)$, is:
${\left(\frac{4}{3}\right)}^{2 - 3} = {\left(\frac{4}{3}\right)}^{- 1} = \frac{1}{\frac{4}{3}} ^ 1 = \frac{1}{\frac{4}{3}} = \frac{3}{4}$
Fourth Term $\left(n = 4\right)$, is:
${\left(\frac{4}{3}\right)}^{2 - 4} = {\left(\frac{4}{3}\right)}^{- 2} = \frac{1}{\frac{4}{3}} ^ 2 = \frac{1}{\frac{16}{9}} = \frac{9}{16}$
Fifth term $\left(n = 5\right)$, is:
${\left(\frac{4}{3}\right)}^{2 - 5} = {\left(\frac{4}{3}\right)}^{- 3} = \frac{1}{\frac{4}{3}} ^ 3 = \frac{1}{\frac{64}{27}} = \frac{27}{64}$
As can be seen, the first five terms of this series are:
$\frac{4}{3} , 1 , \frac{3}{4} , \frac{9}{16} , \frac{27}{64}$
This is a geometric series with a common ratio of $\left({a}_{n + 1} / {a}_{n}\right) = {a}_{2} / {a}_{1} = \frac{1}{\frac{4}{3}} = \frac{3}{4}$:
$r = \frac{3}{4}$
If $\left\mid r \right\mid < 1$ there is a formula for the sum of infinite geometric series:
$S = {a}_{1} / \left(1 - r\right)$
$S = \frac{\frac{4}{3}}{1 - \left(\frac{3}{4}\right)} = \frac{\frac{4}{3}}{\frac{1}{4}} = \frac{4}{3} \cdot \frac{4}{1} = \frac{16}{3}$
Jan 24, 2018
${\sum}_{n = 1}^{\infty} {\left(\frac{4}{3}\right)}^{2 - n} = \frac{16}{3}$
#### Explanation:
${\sum}_{n = 1}^{\infty} {\left(\frac{4}{3}\right)}^{2 - n}$
=${\sum}_{n = 1}^{\infty} {\left(\frac{3}{4}\right)}^{n - 2}$
=${\sum}_{n = 0}^{\infty} {\left(\frac{3}{4}\right)}^{n - 1}$
=${\left(\frac{3}{4}\right)}^{- 1} \cdot \frac{1}{1 - \frac{3}{4}}$
=$\frac{\frac{4}{3}}{\frac{1}{4}}$
=$\frac{16}{3}$ | 872 | 1,894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 25, "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} | 4.5625 | 5 | CC-MAIN-2024-33 | latest | en | 0.466141 |
https://schools.bchydro.com/activities/safety/safety-shuffle | 1,716,128,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057788.73/warc/CC-MAIN-20240519132049-20240519162049-00817.warc.gz | 436,846,190 | 17,871 | Safety shuffle | Grades 4-7 | BCHydro Power Smart for Schools
Safety
# Safety shuffle
Play a game to learn how to stay safe around power lines and shuffle away from danger.
4-7
Duration
30 mins
Type
Game
## Overview
Learn about the various ways electricity can travel that could potentially be hazardous or unsafe. Watch a video and play a game to learn how to stay safe around power lines.
### Instructions
#### What you'll need
• "Stay safe" video
• Skipping rope or similar rope about two metres in length. One for each pair of students
• "Safety shuffle" game teacher handout. Print one copy or view on tablet.
• Outdoor or indoor space large enough for students to space out in pairs with skipping ropes.
#### Introduction
1. Do the "Safety code" activity and/or play the video to review information about how electricity travels and how to stay safe around power lines.
#### Electric shuffle
1. Power down the classroom, take the skipping ropes and "Safety shuffle" game questions and go outside or to the gym to play a game.
2. Divide students into pairs, provide each pair with a skipping rope and have students space apart, holding the rope taut between them with arms outstretched horizontally. Explain that students represent the power poles, and the rope represents the power line between the poles. The object of the game is to prevent the power line (rope) from drooping and touching the ground.
3. Designate four areas as the 'safe zone', that are at least 10 metres away from the students.
4. Taking turns, ask students to choose from one of these categories: touch potential, step potential, and arcing. Read a question from that category. Have the class call out their answer ‘yes’ or ‘no’ at the same time. If any pair answers incorrectly they need to shuffle their feet, two shoe lengths each towards each other. The skipping rope will begin to sag. Continue with the questions and when a rope touches the ground, all the students shuffle to the safe zone and call out 911. The game can then begin again. Students whose rope never touches the ground are the winners.
### Curriculum Fit
#### Curricular competencies
##### Applying and innovating:
• Co-operatively design projects
• Transfer and apply learning to new situations
##### Communicating:
• Represent and communicate ideas and findings in a variety of ways, such as diagrams and simple reports, using digital technologies as appropriate
#### Content
• Strategies and skills to use in potentially hazardous or unsafe situations (Grade 4)
• Basic principles for responding to emergency situations: following safety guidelines (Grade 6, 7)
#### Curricular competencies
##### Social and community health
• Identify and describe strategies for avoiding and/or responding to potentially unsafe, abusive, or exploitive situations
#### Content
• Personal development
• Goal-setting strategies - emergent leadership skills, safety hazards and rules at school, at home, and in the community (Grade 4, 5)
#### Curricular competencies
• Demonstrate safe behaviours in a variety of environments (Grade 4, 5, & 6)
• Demonstrate safety skills in an experiential learning environment (Grade 7)
### Teaching Notes
#### Electrical safety tips
• Never put fingers or other objects in an outlet
• Keep metal objects out of toasters
• Never use anything with a cord or plug around water
• Never pull a plug out by its cord
• Stay away from substations and power lines
• Don't climb on power poles
• Never fly kites near power lines
• Stay away from broken or fallen power lines
• Never touch or climb trees that are near power lines
• Never touch big, metal transformer boxes with warning signs
• Obey warning signs
#### Electrical safety, electricity, and power lines
Power lines are conductive meaning the electrical current runs through them with the least resistance. However if something makes contact with a live power line like a tree, kite, or ladder, the electrical current may flow to the ground. The place where the current touches the ground is the highest voltage and from that point the electrical current spreads out in irregular concentric circles. The voltage or electrical intensity decreases as it moves further from the source. 10 metres or more from the source of contact, like a downed power line, is a safe distance.
Touch potential is the ability electricity has to find its way through touch to get to the ground. If a kite gets tangled in a power line, the electrical current could travel through the kite and you to reach the ground. Birds do not get zapped when standing on a power line, however they would if they straddled two power lines, or touch their beak to the ground while standing on the power line.
Step potential is the ability electricity has to move through your body as you step away from the source of electricity. As electrical current flows through the ground the voltage decreases in concentric rings or ripples as you move further from the source. However, if you move away by lifting one foot, the change in voltage between the concentric rings can travel up one leg and down the other. Instead by keeping your legs together and shuffling your feet, the electrical current will stay in the ground. 10 metres is the safe distance calculated based on the voltage in the power lines here in B.C.
Arcing is the ability electricity has to jump through the air to find a path to ground. The term arcing describes the shape the electrical current uses to get to the ground. Some jobs around your home may take you close to power lines, such as trimming trees, working on your roof, or doing exterior renovations. It is important to keep yourself and any tools you’re using at least three metres (the length of a four-door car) away from power lines near your home. This is because electricity can "arc" or jump from power lines across a gap, to tools or ladders that you're using.
### Assessment
• Assess students’ ability to respond safely to an electrical safety hazard like a downed power line.
• Assess students’ understanding and knowledge of touch potential, step potential and arcing.
• Assess students’ participation and cooperation in the game. | 1,290 | 6,186 | {"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} | 2.953125 | 3 | CC-MAIN-2024-22 | latest | en | 0.921908 |
https://www.experts-exchange.com/questions/10248359/Need-a-fast-way-to-get-a-lot-of-primzahlen-german-numbers-that-can-be-divided-only-by-1-and-itself.html | 1,534,406,953,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210559.6/warc/CC-MAIN-20180816074040-20180816094040-00155.warc.gz | 887,259,229 | 24,403 | # Need a fast way to get a lot of "primzahlen"(german) numbers that can be divided only by 1 and itself.
This is my first try:
const AnzPrimz = 10000;
bool Primzahl;
int Primzahlen[AnzPrimz];
int ArrayIndex = 0;
for(int i=3; (ArrayIndex < AnzPrimz) && (i < 64000); i = i+2)
{
Primzahl = true;
for(int j=2; (j <= i/2) && Primzahl; j++)
{
if((i % j) == 0)
{
Primzahl = false;
}
}
if(Primzahl)
{
Primzahlen[ArrayIndex] = i;
ArrayIndex++;
}
}
How can I do this faster?
###### Who is Participating?
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Author Commented:
Edited text of question.
0
Commented:
Primzahlen = Prime Numbers
What you've posted is called the Sieve of Erastothenes and is a classic way of testing numbers for "primeness". It's problem is that it is slow.
This is an area of much research as finding large prime numbers is a key component in many crypto systems.
You might try searching the web for the terms "prime number", "factoring", "cryptography", "encryption".
0
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Commented:
this is not the sieve of erastothenes... it's the test every odd number method...
if u wanna find ALL the primes from 1 to n i think the easiest & ok-ly fast way IS SoE... it's the 'cross-out' the multiples of prime method...
0
Commented:
lychee,
Yes, you are correct, I didn't look closely enough at the code...
0
Commented:
noprob... i'm really interested in how to find large prime numbers fast tho...
0
Commented:
> How can I do this faster?
There's several things you can do to make your code faster. First, you're not considering even numbers. So, why are you dividing by 2?
Also, since you're not considering even numbers, there's no reason to consider even divisors. That's because any number, even or odd, when multiplied by an even number, results in an even number.
Next, to see if i is prime, you're searching from 2 to i/2. That's too many tests; you should stop searching at sqrt(i). A number has no integer factors larger than its square root.
Here's your code with those optimizations:
for(int i=3; (ArrayIndex < AnzPrimz) && (i < 64000); i = i+2)
{
Primzahl = true;
int nStop = sqrt(i);
for(int j=3; (j <= nStop) && Primzahl; j += 2)
{
if((i % j) == 0)
{
Primzahl = false;
}
}
if(Primzahl)
{
Primzahlen[ArrayIndex] = i;
ArrayIndex++;
}
}
The optimizations I've made seem subtle, but they're quite expensive. You've gone from an O(n*n/2) algorithm to an O(n/2 * sqrt(n)/2) algorithm. For n = 64000, that's 505 times more performant!
If you're only looking for numbers from 1 thru 64000, then the Sieve of Erastothones is fast enough. For much larger numbers, it ends up using way too much memory and brute-force testing (what you're doing) is actually faster because the Seive inherently has very poor data locality.
..B ekiM
0
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C++
From novice to tech pro — start learning today. | 976 | 3,754 | {"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-2018-34 | latest | en | 0.89329 |
http://www.exceltip.com/formulas-and-functions/count-blanknon-blank-cells.html | 1,544,443,105,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823322.49/warc/CC-MAIN-20181210101954-20181210123454-00125.warc.gz | 365,199,301 | 15,441 | How to Count Blank and Non Blank Cells in Microsoft Excel
In this article, you will learn how to count the blank and how to count not blank cells in Microsoft Excel. In Excel we have functions to count the blank cells and non-blank cells.
COUNTBLANK: – When we want to count blank cells only within a range then we use COUNTBLANK functions.
Syntax:-
COUNTA: - This function we use to count the cells within a range which is not empty.
Syntax:-
Count Blank Cells
Let’s take an example:-
First we want to calculate the blank cells in Column A, follow below given steps:-
• Enter the formula in cell C2
• =COUNTBLANK(A2:A13)
• Press Enter
Now we count the blank cells in 2 columns together, follow below given steps:-
• Enter the formula in cell C3
• =COUNTBLANK(A2:B13)
• Press Enter
Count Non-Blank Cells
Let’s take an example:-
First we want to calculate the blank cells in Column A, follow below given steps:-
• Enter the formula in cell C2
• =COUNTA(A2:A13)
• Press Enter
Now we count the blank cells in 2 columns together, follow below given steps:-
• Enter the formula in cell C3
• =COUNTA(A2:B13)
• Press Enter
Key take away:-
• Count number of non blank cells in Excel by using COUNTA function.
• Count if cells are blank within a range. | 320 | 1,267 | {"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-2018-51 | latest | en | 0.780848 |
https://math.stackexchange.com/questions/307294/convex-sets-proof | 1,627,664,485,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153971.20/warc/CC-MAIN-20210730154005-20210730184005-00363.warc.gz | 405,682,015 | 39,209 | # Convex sets proof
Prove the following theorem:
Let $$V$$ be a linear space and $$D$$ a convex set. Let $$x_1,\ldots,x_k$$ be $$k$$ points in $$D$$. Let $$a_1,\ldots,a_k$$ be non-negative scalars such that $$\sum\limits_{i=1}^n a_i=1$$. Then the so called convex combination $$\sum\limits_{i=1}^k a_ix_i$$ is an element of $$D$$.
I tried looking up the definition of convex sets which is that if you draw a line between two points in the set that the entire line should line within the set and that this should hold for all points in the set. For the rest, since I am entirely new to proofs like these, I dont have a clue how to proceed. Can someone please help me? It would be highly appreciated.
Well, first note that if we only have two points $x_1$ and $x_2$, then all that's being said is whenever $a + b = 1$ the point $a*x_1 + b*x_2$ is in $D$. This is very clear though, because $b = 1-a$ and so the point in question is $a*x_1 + (1-a)*x_2$, which is a point on the line between $x_1$ and $x_2$.
Generally speaking, if we have points $x_1, ..., x_k$, and $\sum_{i=1}^k a_i = 1$, then you can write $a_1 + ... + a_{k-1} = 1 - a_k$ to get that
$\sum_{i=1}^k a_i x_i = a_k x_k + (1-a_k)\sum_{i=1}^{k-1} \frac{a_i}{1 - a_k} x_k$
The points $x_k$ and $\sum_{i=1}^{k-1} \frac{a_i}{1 - a_k} x_k$ may by induction be assumed to be points in $D$, so this forms the induction step of the proof.
• Thanks a lot. Your explanation is very clear and understandable. Could you show me how you can complete the induction proof? – dreamer Feb 18 '13 at 18:32
• The first two sentences form the $k=2$ case, the induction base case (I guess I ignored the k=1 case as trivial). The remainder of what I wrote forms the proof that if the statement is true for $k-1$ then it is also true for $k$. I believe that these two pieces together form a complete induction proof. – Eric Haengel Feb 18 '13 at 18:39
• Ok. Thank you very much :) – dreamer Feb 18 '13 at 18:43
You can proceed by induction on $k$, the case $k=1$ being trivial. If $k>1$, let $u=\sum_{i=1}^{k-1}a_i=1-a_{k}$. If $u=0$, then $\sum_{i=1}^k a_ix_i = x_k\in D$. Otherwise let $b_i=\frac{a_i}u$ and observe that $y:=\sum_{i=1}^{k-1}b_ix_i\in D$ by induction assumption because $\sum_{i=1}^{k-1}b_i=1$ and all $b_i\ge 0$. Then $$\sum_{i=1}^k a_ix_i = x_k+u(y-x_k)$$ is a point on the line segment from $x_k$ to $y$, hence in $D$. | 813 | 2,388 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2021-31 | latest | en | 0.885832 |
https://wiki.haskell.org/index.php?title=Haskell_Quiz/Grid_Folding/Solution_Dolio&diff=10219&oldid=7489 | 1,495,726,052,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608084.63/warc/CC-MAIN-20170525140724-20170525160724-00073.warc.gz | 812,015,375 | 6,922 | # Haskell Quiz/Grid Folding/Solution Dolio
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The basis for my solution is simple. Consider each square to be a single-element list initially. Then, a row of such squares is a list of such lists. An entire grid is then a list of those lists.
When folding horizontally, one splits each row, reverses the half to go on top (as well as its elements, since they'll be flipped), and zips the two halves together by appending corresponding elements. This results in a top-first list of the stacked squares.
When folding vertically, one splits the grid in half, reverses the half to go on top, and zips the two grid halves together. The function that combines corresponding rows is yet another zip that appends corresponding elements (again, reversing the ones on top).
```module Main where
import Control.Monad.Reader
import Data.Char
import System
grid n = break . map return \$ [1..(n*n)]
where break [] = []
break l = let (h,t) = splitAt n l in h : break t
fold 'T' = folder vzipper
fold 'B' = folder (flip vzipper)
fold 'L' = map (folder hzipper)
fold 'R' = map (folder \$ flip hzipper)
fold _ = error "Unrecognized letter."
vzipper = zipWith (zipWith \$ (++) . reverse) . reverse
hzipper = zipWith (++) . (map reverse . reverse)
folder z = uncurry z . ap (flip splitAt) (flip div 2 . length)
pretty = unlines . map unwords . map (map show)
output s = if filter (not . isSpace) s == []
then error "Invalid folding scheme"
else putStr s
main = do [n, s] <- getArgs
output . pretty . foldl (flip fold) (grid \$ read n) \$ s``` | 415 | 1,593 | {"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} | 2.953125 | 3 | CC-MAIN-2017-22 | latest | en | 0.835161 |
https://wulixb.iphy.ac.cn/article/doi/10.7498/aps.69.20191585 | 1,722,815,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640417235.15/warc/CC-MAIN-20240804230158-20240805020158-00789.warc.gz | 515,658,493 | 26,092 | x
## 留言板
引用本文:
Citation:
## A new method for selecting arbitrary Poincare section
Zhang Shi, Wang Pan, Zhang Rui-Hao, Chen Hong
PDF
HTML
• #### 摘要
针对用庞加莱截面分析非线性系统时较难选择合适截面的问题, 提出了一种投影时域法. 该方法可以在时域上直观地判断所选平面是否合适并实时准确地调整平面的方向与位置. 文中给出了投影时域法的完整定义并从理论上详细分析了该方法的原理; 同时研究了在时域上判断所选平面是否是合适的庞加莱截面的条件以及如何调整所选平面的方向与位置; 最后, 利用投影时域法对三种典型的三维或四维非线性系统进行了仿真实验, 实验结果证明了该方法的有效性和实用性.
#### Abstract
Poincare section is an important method for analyzing nonlinear systems. Choosing a suitable plane as the Poincare section is the key to using the Poincare section to analyze a nonlinear system. At present, it is still a difficult problem to select a suitable Poincare section when analyzing a nonlinear system. This is caused by two reasons. On the one hand, the classical method for selecting a partial Poincare section only applies to analyze a part of the nonlinear system orbit, whether the selected plane is a suitable Poincare section is affected by the different initial points. On the other hand, according to the actual situation, different researchers have different needs for Poincare section. In order to solve this problem, a new method named Projection Time Domain method is put forward in this paper. This method can help us not only directly reflect the intersection between the nonlinear system orbit and the selected plane, but also accurately adjust the direction and position of the selected plane in real time. It can be used to quickly find a plane which fully intersects the nonlinear system orbit or an arbitrary plane as a Poincare section. In this paper, the complete definition of Projection Time Domain method is given firstly. Then, the principle of Projection Time Domain method is theoretically analyzed in detail. At the same time, the rules for determining whether the selected plane is a suitable Poincare section in the time domain are also studied. Finally, it is introduced how to quantify the direction and position of the selected plane in the phase space. The simulation experiments are conducted with three typical three-dimensional and four-dimensional nonlinear systems by using this new method. The experimental results consistent with the theoretical analysis, which demonstrate the effectiveness and practicability of this method.
#### 作者及机构信息
###### 通信作者: 陈红, chenhongdeepred@163.com
• 基金项目: 国家级-国家自然科技基金(61471158)
#### 施引文献
• 图 1 轨道在平面上的投影
Fig. 1. The projection of a trajectory onto a plane.
图 2 $F({l_1}(t), {l_2}(t))$${l_3}(t)$的波形
Fig. 2. Waveform of $F({l_1}(t), {l_2}(t))$ and ${l_3}(t)$.
图 3 涡卷的相轨与波形 (a)相轨; (b)波形
Fig. 3. Phase trajectoy and waveform of vortex: (a) Phase trajectoy; (b) waveform.
图 4 平面一的实验结果 (a)时域波形; (b)相图
Fig. 4. Experimental results of the first plane: (a) Waveform; (b) phase diagram.
图 5 平面二的实验结果 (a)时域波形; (b)相图
Fig. 5. Experimental results of the second plane: (a) Waveform;(b) phase diagram.
图 6 平面三的实验结果 (a)时域波形; (b)相图
Fig. 6. Experimental results of the third plane: (a) Waveform; (b) phase diagram.
图 7 平面一的实验结果 (a) C = 0时的波形; (b) C = 0时的相图; (c) C = 1.5时的波形; (d) C = 1.5时的相图; (e) C = –1.5时的波形; (f) C = –1.5时的相图
Fig. 7. Experimental results of the first plane: (a) Waveform while C = 0; (b) phase diagram while C = 0; (c) waveform while C = 1.5; (d) phase diagram while C = 1.5; (e) waveform while C = –1.5; (f) phase diagram while C = –1.5.
图 8 平面二的实验结果 (a)波形; (b)相图
Fig. 8. Experimental results of the second plane: (a) Waveform; (b) phase diagram.
图 9 $w = 0$时的实验结果 (a)波形; (b)相图
Fig. 9. Experimental results while $w = 0$: (a) Waveform; (b) phase diagram.
图 10 $z = 0$时的实验结果 (a)波形; (b)相图
Fig. 10. Experimental results while $z = 0$: (a) Waveform; (b) phase diagram.
图 11 $y = 0$时的实验结果 (a)波形; (b)相图
Fig. 11. Experimental results while $y = 0$: (a) Waveform; (b) phase diagram.
图 12 $x = 0$时的实验结果 (a)波形; (b)相图
Fig. 12. Experimental results while $x = 0$: (a) Waveform; (b) phase diagram. | 1,356 | 3,908 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-33 | latest | en | 0.820908 |
http://wiki.portal.chalmers.se/cse/pmwiki.php/CTFP11/Presentations | 1,542,497,621,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743913.6/warc/CC-MAIN-20181117230600-20181118012600-00470.warc.gz | 388,140,511 | 5,351 | Search:
This site contains wiki-pages for the department of computer science and engineering at Chalmers and GU.
# Abstracts and schedule for the final presentations of the CTFP course 2011
## Thursday 2011-05-12: EDIT-room
### 10.00. Guilhem (slides)
For any inductive datatype (ie, that is the carrier of the initial F-algebra), Lambek's Lemma provides an iteration operator fold that can eg, be used to derive an induction rule for the type if the functor F is polynomial. I'll show how to bypass this restriction and find an inductive schema to prove that some notion of property holds for the inhabitants of any inductive type (without any condition on the functor F).
### 10.35. Jean-Philippe:
I will review the initial algebra interpretation of data, then explain the notion of catamorphism, as well as the "fusion/fisson" rule. Then I will show how this rule can be used to optimise certain functional programs on a couple of examples.
### 11.10. Anders:
I will present a framework of adjunctions and functors that can be used to rewrite many programs as folds and unfolds to make them suitable for optimization, e.g fusion. I will give examples showing how to apply the framework to currying and mutual value recursion.
### 13.00. Tobias
I will showcase an approach, relying on the Yoneda Lemma and adjunctions, to establish isomorphisms between types that focuses on showing existence rather then deriving witnesses. At the beginning of the talk I will show how to establish some very simple isomorphism (inspired by "high-school algebra" laws) using this approach. Then I will in briefly review the concept of fixpoints and present the "rolling rules" and the "diagonal rules". At the end I will show how we can use all of our new knowledge to establish the isomorphisms between recursive types. For example: BinaryTree ≅ List ∘ Rose and Rose ≅ Fork (rose trees and fork trees).
### 13.35. Ramona
Categorial and Categorical Grammars
## Tuesday 2011-05-17: EDIT room
### 9.00. Willard
Categorical Semantics for Arrows
### 9.35. Ann
Categorical De Morgan laws
### 10.10. Dan:
I will talk about some building primitives in dependently typed languages (Pi-types, Sigma-types, etc), show how typing rules for these types are established and how this connects with category theory: what context morphisms are, and how one can get a model of a given type theory using categories with families.
### 11.00. Simon
Each topos gives rise to an internal language which is sound for topos logic, an intuitionistic higher-order logic. This allows reasoning about objects and morphisms of a topos in a more set-theoretical language. After a short recap of the notion of a topos, I will introduce the internal language and topos logic; then I will give some examples of theorems which hold in topos logic.
### 11.35. Arnar
A Core Calculus of Dependency
## Registrations
Register here for a presentation of a paper/book chapter by writing your name and a reference to the document.
Jean-Philippe
Willard
Categorical Semantics for Arrows (direct link; access from within Chalmers network)
Guilhem
Ramona
Dan
Anders
Tobias
Arnar
Ann
Simon
Page last modified on May 12, 2011, at 05:14 PM | 760 | 3,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-47 | latest | en | 0.89552 |
https://graphstream-project.org/gs-algo/org/graphstream/algorithm/Dijkstra.html | 1,611,113,648,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00077.warc.gz | 367,201,205 | 8,249 | org.graphstream.algorithm
## Class Dijkstra
• All Implemented Interfaces:
Algorithm, SpanningTree
```public class Dijkstra
extends AbstractSpanningTree```
Dijkstra's algorithm computes the shortest paths from a given node called source to all the other nodes in a graph. It produces a shortest path tree rooted in the source. This algorithm works only for nonnegative lengths.
This implementation uses internally Fibonacci Heap, a data structure that makes it run faster for big graphs.
### Length of a path
Traditionally the length of a path is defined as the sum of the lengths of its edges. This implementation allows to take into account also the "lengths" of the nodes. This is done by a parameter of type `Dijkstra.Element` passed in the constructors.
The lengths of individual elements (edges or/and nodes) are defined using another constructor parameter called `lengthAttribute`. If this parameter is `null`, the elements are considered to have unit lengths. In other words, the length of a path is the number of its edges or/and nodes. If the parameter is not null, the elements are supposed to have a numeric attribute named `lengthAttribute` used to store their lengths.
### Solutions
Internal solution data is stored in attributes of the nodes of the underlying graph. The name of this attribute is another constructor parameter called `resultAttribute`. This name must be specified in order to avoid conflicts with existing attributes, but also to distinguish between solutions produced by different instances of this class working on the same graph (for example when computing shortest paths from two different sources). If not specified, a unique name is chosen automatically based on the hash code of the Dijkstra instance. The attributes store opaque internal objects and must not be accessed, modified or deleted. The only way to retrieve the solution is using different solution access methods.
### Usage
A typical usage of this class involves the following steps:
Note that if the graph changes after the call of `compute()` the computed solution is no longer valid. In this case the behavior of the different solution access methods is undefined.
### Example
``` Graph graph = ...;
// Edge lengths are stored in an attribute called "length"
// The length of a path is the sum of the lengths of its edges
// The algorithm will store its results in attribute called "result"
Dijkstra dijkstra = new Dijkstra(Dijkstra.Element.edge, "result", "length");
// Compute the shortest paths in g from A to all nodes
dijkstra.init(graph);
dijkstra.setSource(graph.getNode("A"));
dijkstra.compute();
// Print the lengths of all the shortest paths
for (Node node : graph)
System.out.printf("%s->%s:%6.2f%n", dijkstra.getSource(), node, dijkstra.getPathLength(node));
// Color in blue all the nodes on the shortest path form A to B
for (Node node : dijkstra.getPathNodes(graph.getNode("B")))
// Color in red all the edges in the shortest path tree
for (Edge edge : dijkstra.getTreeEdges())
// Print the shortest path from A to B
System.out.println(dijkstra.getPath(graph.getNode("B"));
// Build a list containing the nodes in the shortest path from A to B
// Note that nodes are added at the beginning of the list
// because the iterator traverses them in reverse order, from B to A
List <Node> list1 = new ArrayList<Node>();
for (Node node : dijkstra.getPathNodes(graph.getNode("B")))
// A shorter but less efficient way to do the same thing
List<Node> list2 = dijkstra.getPath(graph.getNode("B")).getNodePath();
```
Author:
Stefan Balev
• ### Nested Class Summary
Nested Classes
Modifier and Type Class and Description
`static class ` `Dijkstra.Element`
This enumeration is used to specify how the length of a path is computed
• ### Constructor Summary
Constructors
Constructor and Description
`Dijkstra()`
Constructs an instance in which the length of the path is considered to be the number of edges.
```Dijkstra(Dijkstra.Element element, java.lang.String resultAttribute, java.lang.String lengthAttribute)```
Constructs an instance with the specified parameters.
```Dijkstra(Dijkstra.Element element, java.lang.String resultAttribute, java.lang.String lengthAttribute, java.lang.String flagAttribute, java.lang.Object flagOn, java.lang.Object flagOff)```
Constructs an instance with the specified parameters.
• ### Method Summary
All Methods
Modifier and Type Method and Description
`void` `clear()`
Removes the attributes used to store internal solution data in the nodes of the graph.
`void` `compute()`
Computes the shortest paths from the source node to all nodes in the graph.
`java.lang.String` `defaultResult()`
`java.lang.Iterable<org.graphstream.graph.Path>` `getAllPaths(org.graphstream.graph.Node target)`
An iterable view of of all the shortest paths from the source node to a given target node.
`java.util.stream.Stream<org.graphstream.graph.Path>` `getAllPathsStream(org.graphstream.graph.Node target)`
This iterator traverses all the shortest paths from the source node to a given target node.
`org.graphstream.graph.Edge` `getEdgeFromParent(org.graphstream.graph.Node target)`
Returns the edge between the target node and the previous node in the shortest path from the source to the target.
`org.graphstream.graph.Node` `getParent(org.graphstream.graph.Node target)`
Returns the node preceding the target in the shortest path from the source to the target.
`org.graphstream.graph.Path` `getPath(org.graphstream.graph.Node target)`
Returns the shortest path from the source node to a given target node.
`java.lang.Iterable<org.graphstream.graph.Edge>` `getPathEdges(org.graphstream.graph.Node target)`
An iterable view of the edges on the shortest path from the source node to a given target node.
`java.util.stream.Stream<org.graphstream.graph.Edge>` `getPathEdgesStream(org.graphstream.graph.Node target)`
This iterator traverses the edges on the shortest path from the source node to a given target node.
`double` `getPathLength(org.graphstream.graph.Node target)`
Returns the length of the shortest path from the source node to a given target node.
`java.lang.Iterable<org.graphstream.graph.Node>` `getPathNodes(org.graphstream.graph.Node target)`
An iterable view of the nodes on the shortest path from the source node to a given target node.
`java.util.stream.Stream<org.graphstream.graph.Node>` `getPathNodesStream(org.graphstream.graph.Node target)`
This iterator traverses the nodes on the shortest path from the source node to a given target node.
`<T extends org.graphstream.graph.Node>T` `getSource()`
Dijkstra's algorithm computes shortest paths from a given source node to all nodes in a graph.
`java.util.stream.Stream<org.graphstream.graph.Edge>` `getTreeEdgesStream()`
Dijkstra's algorithm produces a shortest path tree rooted in the source node.
`double` `getTreeLength()`
Dijkstra's algorithm produces a shortest path tree rooted in the source node.
`void` `setSource(org.graphstream.graph.Node source)`
Dijkstra's algorithm computes shortest paths from a given source node to all nodes in a graph.
`void` `setSource(java.lang.String source)`
`void` `setTarget(java.lang.String target)`
• ### Methods inherited from class org.graphstream.algorithm.AbstractSpanningTree
`getFlagAttribute, getFlagOff, getFlagOn, getTreeEdges, init, setFlagAttribute, setFlagOff, setFlagOn`
• ### Methods inherited from class java.lang.Object
`equals, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`
• ### Constructor Detail
• #### Dijkstra
```public Dijkstra(Dijkstra.Element element,
java.lang.String resultAttribute,
java.lang.String lengthAttribute)```
Constructs an instance with the specified parameters. The edges of the shortest path tree are not tagged.
Parameters:
`element` - Graph elements (edges or/and nodes) used to compute the path lengths. If `null`, the length of the path is computed using edges.
`resultAttribute` - Attribute name used to store internal solution data in the nodes of the graph. If `null`, a unique name is chosen automatically.
`lengthAttribute` - Attribute name used to define individual element lengths. If `null` the length of the elements is considered to be one.
• #### Dijkstra
`public Dijkstra()`
Constructs an instance in which the length of the path is considered to be the number of edges. Unique result attribute is chosen automatically. The edges of the shortest path tree are not tagged.
• #### Dijkstra
```public Dijkstra(Dijkstra.Element element,
java.lang.String resultAttribute,
java.lang.String lengthAttribute,
java.lang.String flagAttribute,
java.lang.Object flagOn,
java.lang.Object flagOff)```
Constructs an instance with the specified parameters.
Parameters:
`element` - Graph elements (edges or/and nodes) used to compute the path lengths. If `null`, the length of the path is computed using edges.
`resultAttribute` - Attribute name used to store internal solution data in the nodes of the graph. If `null`, a unique name is chosen automatically.
`lengthAttribute` - Attribute name used to define individual element lengths. If `null` the length of the elements is considered to be one.
`flagAttribute` - attribute used to set if an edge is in the spanning tree
`flagOn` - value of the flagAttribute if edge is in the spanning tree
`flagOff` - value of the flagAttribute if edge is not in the spanning tree
• ### Method Detail
• #### getSource
`public <T extends org.graphstream.graph.Node> T getSource()`
Dijkstra's algorithm computes shortest paths from a given source node to all nodes in a graph. This method returns the source node.
Returns:
the source node
`setSource(Node)`
• #### setSource
`public void setSource(org.graphstream.graph.Node source)`
Dijkstra's algorithm computes shortest paths from a given source node to all nodes in a graph. This method sets the source node.
Parameters:
`source` - The new source node.
`getSource()`
• #### setSource
`public void setSource(java.lang.String source)`
• #### setTarget
`public void setTarget(java.lang.String target)`
• #### clear
`public void clear()`
Removes the attributes used to store internal solution data in the nodes of the graph. Use this method to free memory. Solution access methods must not be used after calling this method.
Specified by:
`clear` in interface `SpanningTree`
Overrides:
`clear` in class `AbstractSpanningTree`
• #### compute
`public void compute()`
Computes the shortest paths from the source node to all nodes in the graph.
Specified by:
`compute` in interface `Algorithm`
Overrides:
`compute` in class `AbstractSpanningTree`
Throws:
`java.lang.IllegalStateException` - if `AbstractSpanningTree.init(Graph)` or `setSource(Node)` have not been called before or if elements with negative lengths are discovered.
`Algorithm.compute()`
Computational Complexity:
O(m + nlogn) where m is the number of edges and n is the number of nodes in the graph.
• #### getPathLength
`public double getPathLength(org.graphstream.graph.Node target)`
Returns the length of the shortest path from the source node to a given target node.
Parameters:
`target` - A node
Returns:
the length of the shortest path or `Double.POSITIVE_INFINITY` if there is no path from the source to the target
Computational Complexity:
O(1)
• #### getTreeLength
`public double getTreeLength()`
Dijkstra's algorithm produces a shortest path tree rooted in the source node. This method returns the total length of the tree.
Returns:
the length of the shortest path tree
Computational Complexity:
O(n) where n is the number of nodes is the graph.
• #### getEdgeFromParent
`public org.graphstream.graph.Edge getEdgeFromParent(org.graphstream.graph.Node target)`
Returns the edge between the target node and the previous node in the shortest path from the source to the target. This is also the edge connecting the target to its parent in the shortest path tree.
Parameters:
`target` - a node
Returns:
the edge between the target and its predecessor in the shortest path, `null` if there is no path from the source to the target or if the target and the source are the same node.
`getParent(Node)`
Computational Complexity:
O(1)
• #### getParent
`public org.graphstream.graph.Node getParent(org.graphstream.graph.Node target)`
Returns the node preceding the target in the shortest path from the source to the target. This node is the parent of the target in the shortest path tree.
Parameters:
`target` - a node
Returns:
the predecessor of the target in the shortest path, `null` if there is no path from the source to the target or if the target and the source are the same node.
`getEdgeFromParent(Node)`
Computational Complexity:
O(1)
• #### getPathNodesStream
`public java.util.stream.Stream<org.graphstream.graph.Node> getPathNodesStream(org.graphstream.graph.Node target)`
This iterator traverses the nodes on the shortest path from the source node to a given target node. The nodes are traversed in reverse order: the target node first, then its predecessor, ... and finally the source node. If there is no path from the source to the target, no nodes are traversed. This iterator does not support `Iterator.remove()`.
Parameters:
`target` - a node
Returns:
an iterator on the nodes of the shortest path from the source to the target
`getPathNodes(Node)`
Computational Complexity:
Each call of `Iterator.next()` of this iterator takes O(1) time
• #### getPathNodes
`public java.lang.Iterable<org.graphstream.graph.Node> getPathNodes(org.graphstream.graph.Node target)`
An iterable view of the nodes on the shortest path from the source node to a given target node. Uses `getPathNodesStream(Node)`.
Parameters:
`target` - a node
Returns:
an iterable view of the nodes on the shortest path from the source to the target
`getPathNodesStream(Node)`
• #### getPathEdgesStream
`public java.util.stream.Stream<org.graphstream.graph.Edge> getPathEdgesStream(org.graphstream.graph.Node target)`
This iterator traverses the edges on the shortest path from the source node to a given target node. The edges are traversed in reverse order: first the edge between the target and its predecessor, ... and finally the edge between the source end its successor. If there is no path from the source to the target or if he source and the target are the same node, no edges are traversed. This iterator does not support `Iterator.remove()`.
Parameters:
`target` - a node
Returns:
an iterator on the edges of the shortest path from the source to the target
`getPathEdges(Node)`
Computational Complexity:
Each call of `Iterator.next()` of this iterator takes O(1) time
• #### getPathEdges
`public java.lang.Iterable<org.graphstream.graph.Edge> getPathEdges(org.graphstream.graph.Node target)`
An iterable view of the edges on the shortest path from the source node to a given target node. Uses `getPathEdgesStream(Node)`.
Parameters:
`target` - a node
Returns:
an iterable view of the edges on the shortest path from the source to the target
`getPathEdgesStream(Node)`
• #### getAllPathsStream
`public java.util.stream.Stream<org.graphstream.graph.Path> getAllPathsStream(org.graphstream.graph.Node target)`
This iterator traverses all the shortest paths from the source node to a given target node. If there is more than one shortest paths between the source and the target, other solution access methods choose one of them (the one from the shortest path tree). This iterator can be used if one needs to know all the paths. Each call to `Iterator.next()` method of this iterator returns a shortest path in the form of `Path` object. This iterator does not support `Iterator.remove()`.
Parameters:
`target` - a node
Returns:
an iterator on all the shortest paths from the source to the target
`getAllPaths(Node)`
Computational Complexity:
Each call of `Iterator.next()` of this iterator takes O(m) time in the worst case, where m is the number of edges in the graph
• #### getAllPaths
`public java.lang.Iterable<org.graphstream.graph.Path> getAllPaths(org.graphstream.graph.Node target)`
An iterable view of of all the shortest paths from the source node to a given target node. Uses `getAllPathsStream(Node)`
Parameters:
`target` - a node
Returns:
an iterable view of all the shortest paths from the source to the target
`getAllPathsStream(Node)`
• #### getTreeEdgesStream
`public java.util.stream.Stream<org.graphstream.graph.Edge> getTreeEdgesStream()`
Dijkstra's algorithm produces a shortest path tree rooted in the source node. This iterator traverses the edges of this tree. The edges are traversed in no particular order.
Specified by:
`getTreeEdgesStream` in interface `SpanningTree`
Specified by:
`getTreeEdgesStream` in class `AbstractSpanningTree`
Returns:
an iterator on the edges of the shortest path tree
`AbstractSpanningTree.getTreeEdges()`
Computational Complexity:
Each call of `Iterator.next()` of this iterator takes O(1) time
• #### getPath
`public org.graphstream.graph.Path getPath(org.graphstream.graph.Node target)`
Returns the shortest path from the source node to a given target node. If there is no path from the source to the target returns an empty path. This method constructs a `Path` object which consumes heap memory proportional to the number of edges and nodes in the path. When possible, prefer using `getPathNodes(Node)` and `getPathEdges(Node)` which are more memory- and time-efficient.
Parameters:
`target` - a node
Returns:
the shortest path from the source to the target
Computational Complexity:
O(p) where p is the number of the nodes in the path
• #### defaultResult
`public java.lang.String defaultResult()` | 3,898 | 17,544 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2021-04 | latest | en | 0.855345 |
https://www.denexa.com/blog/tag/counting-poker-chips/ | 1,726,671,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651899.75/warc/CC-MAIN-20240918133146-20240918163146-00656.warc.gz | 676,726,897 | 27,412 | # More chip counting exercises
If our first batch of chip counting exercises was no problem, here’s five more to test your chip-counting skills! Are you good enough at counting chips to quickly calculate the correct totals?
You can click the images to zoom in. Each stack shown contains the same number of chips as the stack splashed next to it. The chips use the standard denominations:
• Yellow—50¢
• White—\$1
• Red—\$5
• Green—\$25
• Black—\$100
• Purple—\$500
This time, we’re including some exercises using a chip rack. Each row of chips shown in the rack represents a full tube of 20 chips. Don’t forget to include these in your total!
## Exercises
1. 6 × 500 = 3000, 11 × 100 = 1100, 16 × 25 = 400, 60 × 5 = 300, 25 × 1 = 25, 3 × 50¢ = 1.50. 3000 + 1100 + 400 + 300 + 25 + 1.50 = 4826.50.
2. 2 × 500 = 1000, 28 × 100 = 2800, 26 × 25 = 650, 84 × 5 = 420, 32 × 1 = 32, 9 × 50¢ = 4.50. 1000 + 2800 + 650 + 420 + 32 + 4.50 = 4906.50.
3. 3 × 500 = 1500, 11 × 100 = 1100, 20 × 25 = 500, 19 × 5 = 95, 21 × 1 = 21. 1500 + 1100 + 500 + 95 + 21 = 3216.
4. 4 × 100 = 400, 3 × 25 = 75, 16 × 5 = 80, 2 × 50¢ = 1. 400 + 75 + 80 + 1 = 556.
5. 5 × 500 = 2500, 12 × 100 = 1200, 15 × 25 = 375, 46 × 5 = 230, 17 × 1 = 17, 7 × 50¢ = 3.50. 2500 + 1200 + 375 + 230 + 17 + 3.50 = 4325.50.
# Chip counting exercises
So you know the basics of how to count chips...but now you just need practice! Here are five chip counting problems to solve. How quickly can you get the correct totals?
You can click the images to zoom in. Each stack shown contains the same number of chips as the stack splashed next to it. The chips use the standard denominations:
• Yellow—50¢
• White—\$1
• Red—\$5
• Green—\$25
• Black—\$100
• Purple—\$500
## Exercises
1. 8 × 100 = 800, 8 × 25 = 200, 6 × 5 = 30, 3 × 1 = 3. 800 + 200 + 30 + 3 = 1033.
2. 2 × 500 = 1000, 6 × 100 = 600, 1 × 25 = 25, 4 × 5 = 20. 1000 + 600 + 25 + 20 = 1645.
3. 3 × 100 = 300, 2 × 5 = 10, 2 × 1 = 2, 3 × 50¢ = 1.50. 300 + 10 + 2 + 1.50 = 313.50.
4. 5 × 25 = 125, 3 × 5 = 15, 9 × 1 = 9. 25 + 15 + 9 = 149.
5. 13 × 100 = 1300, 5 × 25 = 125, 4 × 5 = 20, 1 × 1 = 1. 1300 + 125 + 20 + 1 = 1446.
# Using a chip rack to count poker chips
Last week, we showed you the basics of how to count poker chips. This week, we’ll show you how to use a chip rack to help count large amounts of chips.
Chip racks (sometimes called chip trays) are designed for standard casino chips. Most 11.5g composite and clay chips will fit in these racks. They are made of clear acrylic and can be found at many online retailers—Amazon has several offers of a ten-pack for \$12 to \$14.
If your chips are the same thickness as a standard chip, each tube of the rack will accommodate exactly 20 chips. Since a standard rack has five tubes, that means a full rack will contain 100 chips. So, if you have a large number of chips, just create a 20-chip rack stack (break it down as described last week to verify that it contains 20 chips) and use it to size into your remaining chips. When you create five big 20-chip stacks, you know you have 100 chips, a full rack.
What about when you’re getting the chips out of the rack? How do you verify that the rack contains 100 chips? One way is to actually get the chips out of the rack, verify that the five stacks are even, and break one down to show that each stack contains 20 chips, but there’s an easier method. All you have to do it remove one chip from one of the end tubes, and then run it along the length of the rack, pushing it against the chips, as shown in the picture. If any of the tubes contain only 19 chips, the chip will naturally fall into the tube.
One word of caution: the first time you put your chips in a rack, make sure that it does, in fact, hold exactly 100 chips. Particularly thin or thick (nonstandard) chips may fit 95 or 105 chips to a rack.
# How to count poker chips
Huge piles of chips are one of the first things that come to mind when someone mentions poker, but most people probably don’t stop to think why chips exist in the first place. Isn’t it a hassle converting all that cash into these weird play money discs, only to exchange them for money again later? But there’s an excellent reason for that—chips are easier to count than cash! Casino cash offices have millions of dollars of specialized equipment for counting cash, but the only equipment for counting chips are plastic racks, a smooth surface, and a clerk’s bare hands. Anyone can learn how to count poker chips like a pro!
## Selecting chips
The first step to counting chips is to get some chips that are easily countable. Not all chips are made alike! Casinos use chips that are flat, smooth, and made of clay that gives them some friction and “stickiness” to make them easily stackable. They have labels with custom-printed artwork to distinguish the chip from those from other casinos, and images which appear under a blacklight to deter counterfeiters. Such chips are expensive, costing more than \$1 for a single chip! The durability of these chips makes them cost-effective for the casinos, but hobbyists simply can’t afford to spend that much on chips.
Instead of clay chips, home poker enthusiasts must rely on cheaper plastic-based chips. The very cheapest of these are thin and lightweight plastic chips with interlocking ridges to keep stacks of chips from toppling over. Such chips are to be avoided; the interlocking feature of the chips makes them very difficult to count! Instead, you want something more like a casino chip, with smooth surfaces. Some texture is good, to help add clay-like friction that that is missing in a plastic chip. Many retailers offer a composite chip, which is composed of a metal slug (to add weight) with plastic molded around it. These chips often include artwork of dice engraved on their faces. These chips are reasonably-priced and readily available, and will do just fine for most players. For players wanting a more casino-like feel, generic clay chips are available on the Internet, such as Da Vinci chips (pictured), which are sold in batches of fifty for \$20.
## Assigning values
Now that you have your chips, you need to assign values to them. It’s important to use values which are conveniently spaced apart, so that chips can be colored up or colored down (changed between denominations) easily. You don’t want one chip to be worth twenty of the next color down! You also want your players to understand the easily understand value of the chips, and if your players have played in a casino (or other games) before, they will expect your chip colors to match what they’ve seen.
Here is one standard chip color scheme, used in many casinos:
• Yellow—50¢
• White—\$1
• Red—\$5
• Green—\$25
• Black—\$100
• Purple—\$500
Beyond the \$500 level, chip colors are not standard from casino to casino. Of course, if you play penny-ante poker, it hardly makes sense to have \$100 chips; instead, you can divide this chart by 100, and have your white chips valued as 1¢.
Note that each chip is worth either four or five of the next chip below it. This makes counting the chips easier!
Now you have your chips, and you know how much each is worth. You’re in a game, and you want to know how much money you have. Here’s how to count your chips:
1. First, separate your chips by color, and arrange each color into a stack on a smooth, flat surface. A felt table or chip count board works best, but any flat surface should do (avoid uneven surfaces like a bed or carpet).
2. Select a color to count (e.g. red chips).
3. Carefully count chips from the bottom of the stack, forming a smaller stack. Stop when you get to the number of chips which would equal the next higher chip (e.g. five red \$5 chips equals one green \$25 chip).
4. Place the main stack next to the small stack. Now, bracing the big stack with your thumb, slide your index finger across the short stack, then use it to tilt the big stack away from the small stack, as shown in the photo. This process, known as sizing into the big stack, should produce another small stack equal to the height of the first one.
5. Keep sizing into the stack repeatedly until you don’t have enough chips to make a full stack. Place these chips on the table individually next to the stacks.
6. Run the back of your index finger across the top of the chip stacks to verify that they are all the same height. If any stack has too many chips, you’ll knock it off, or if it’s missing one, you’ll feel your finger dip.
7. Splash the last chip stack out on the table. This is toppling the stack so that it’s fanned out on the table, as shown in the photo at the top of this post. This allows you to visually verify the number of chips each stack contains.
8. Perform this procedure for each color of chip, starting a new row for each color. It’s typical to have the highest-value chips closest toward you, with the value of each row further away from you diminishing (a procedure which is done in casinos to keep unscrupulous patrons from snagging the high-dollar chips after they have been counted, but is useful at home to keep things orderly).
9. Now, counting the chips is simple multiplication. If you have five stacks of five red \$5 chips each, then each stack is \$25, so you have \$125 worth of red chips.
10. If you have multiple denominations of chips to count, start with the largest denomination and work your way down to the smallest. It may help to use a calculator to count very large amounts of chips of separate denominations (add each denomination’s count to a running total in your calculator).
Before passing any quantity of chips to a player, it’s a good idea to break down the stacks of chips, as shown above, to allow them to visually verify that the correct amount of chips is present. You should do this when presenting a player with a buy-in, making change, splitting pots, etc.
It’s typical to make mistakes handling the chips at first, but repetition will help you become more familiar with the feel of your chips and the mechanics of sizing into stacks and counting. Keep practicing! | 2,689 | 10,080 | {"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} | 4.75 | 5 | CC-MAIN-2024-38 | latest | en | 0.796838 |
https://runyoncanyon-losangeles.com/other/what-is-a-glm-repeated-measures/ | 1,716,536,499,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058684.66/warc/CC-MAIN-20240524060414-20240524090414-00834.warc.gz | 434,677,017 | 9,742 | ## What is a GLM repeated measures?
The GLM Repeated Measures procedure provides analysis of variance when the same measurement is made several times on each subject or case. If between-subjects factors are specified, they divide the population into groups.
## How do you do a repeated measures Ancova in SPSS?
The repeated measures ANCOVA can be found in SPSS in the menu Analyze/General Linear Model/Repeated Measures… The dialog box that opens is different than the GLM module you might know from the MANCOVA. Before specifying the model we need to group the repeated measures. This is done by creating a within-subject factor.
What is a repeated measures factor?
Repeated-measures means that the same subject received more than one treatment and or more than one condition. When one of the factors is repeated-measures and the other is not, the analysis is sometimes called a mixed-model ANOVA (but watch out for that word mixed, which can have a variety of meanings in statistics).
What is a repeated measures Manova?
A one-way repeated measures multivariate analysis of variance (i.e., the one-way repeated measures MANOVA), also referred to as a doubly multivariate MANOVA, is used to determine whether there are any differences in multiple dependent variables over time or between treatments, where participants have been measured at …
### How to do repeated measures analysis with SPSS-Idre?
MIXED pulse BY exertype time /FIXED = exertype time exertype*time /REPEATED = time | SUBJECT (id) COVTYPE (un). From previous studies we suspect that our data might actually have an auto-regressive variance-covariance structure so this is the model we will look at next.
How to use mixed models in repeated measures?
Repeated measures analyse an introduction to the Mixed models (random effects) option in SPSS. Demonstrates different Covariance matrix types & how to use the Likelihood ratio test to evaluate different models.
Are there large sample differences in repeated measures ANOVA?
Large sample differences, however, are unlikely; these suggest that the population means weren’t equal after all. The simplest repeated measures ANOVA involves 3 outcome variables, all measured on 1 group of cases (often people). Whatever distinguishes these variables (sometimes just the time of measurement) is the within-subjects factor.
#### When to use a multivariate normal distribution in SPSS?
This is often -not always- satisfied by each case in SPSS representing a different person or other statistical unit. The test variables follow a multivariate normal distributionin the population. However, this assumption is not needed if the sample size >= 25. Sphericity. | 552 | 2,683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-22 | latest | en | 0.940769 |
http://slideplayer.com/slide/5185457/ | 1,638,032,050,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00363.warc.gz | 67,983,802 | 21,115 | The decimal point helps us to keep track of where the "ones" place is.
Presentation on theme: "The decimal point helps us to keep track of where the "ones" place is."— Presentation transcript:
The decimal point helps us to keep track of where the "ones" place is.
4/17/2017 In our number system, digits can be placed to the left and right of a decimal point, to indicate numbers greater than one or less than one. 1.0 or 0.1 The decimal point helps us to keep track of where the "ones" place is. It's placed just to the right of the ones place. As we move right from the decimal point, each number place is divided by 10. 1
Parts of the number 4/17/2017 The whole number portion of a decimal number are those digits to the left of the decimal place. Example: In the number 23.65, the whole number portion is 23. In the number 0.024, the whole number portion is 0. What are the decimal portions of these numbers? 2
4/17/2017 3
4/17/2017 4
Five hundred forty-eight thousandths 0.548
4/17/2017 Hint #1: Remember to read the decimal point as "and" -- notice in the last two problems what a difference that makes! Five hundred and forty-eight thousandths Hint #2: When writing a decimal number that is less than 1, a zero is normally used in the ones place: Five hundred forty-eight thousandths 0.548 0.548 not .548 5
Expanded Form of a Decimal Number
4/17/2017 Expanded Form of a Decimal Number Expanded Form of a Decimal Number The expanded form of a decimal number is the number written as the sum of its whole number and decimal place values. Example: 3 + 0.7 + 0.06 + = 100 + 3 + 0.06 = 6
Just a Note… BUT The number 14.504 is equal to 14.504000
4/17/2017 Just a Note… The number is equal to Adding extra zeros to the right of a decimal number does not change its value BUT The number 14,505 ≠14,504,000 7
4/17/2017 Adding Decimals = 8.4 + 1.22 9.62 12.924 + 3.6 = 12.924 +3.600 16.524 ALWAYS be sure to stack the numbers so the decimals stay above one another!!! 8
Subtracting Decimals 10.66 – 5.22 = 10.66 -5.22 5.44
4/17/2017 Subtracting Decimals 10.66 – 5.22 = 10.66 -5.22 5.44 = Add extra 0's, using the fact that 18.2 = 9
4/17/2017 Multiplying Decimals If not using a calculator, count the number of decimal places in all terms to be multiplied The answer should contain that many decimal places. 5.222 m x m m2 (5 decimal places) 10
Dividing with Decimals
4/17/2017 Dividing with Decimals 11
Dividing Decimals Dividend ÷ divisor = quotient divisor )dividend
4/17/2017 Dividing Decimals Dividend ÷ divisor = quotient divisor )dividend ÷ 4.76 = 23.4 12
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Similar presentations | 787 | 2,681 | {"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-2021-49 | longest | en | 0.831303 |
http://clay6.com/qa/43638/two-small-spheres-of-masses-m-1-and-m-2-are-suspended-by-weightless-insulat | 1,513,017,923,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513866.9/warc/CC-MAIN-20171211183649-20171211203649-00669.warc.gz | 66,807,950 | 25,863 | # Two small spheres of masses $M_1$ and $M_2$ are suspended by weightless insulating threads of lengths $L_1$ and $L_2$ The spheres carry charges $Q_1$ and $Q_2$ respectively. The spheres are suspended such that they are in level with one another and the threads are inclined to the vertical at angles of $\theta_ 1$ and $\theta_2$ as shown. Which one of the following conditions is essential, if $\theta_1 = \theta_2$?
## 1 Answer
$(B) \; M_1 =M_2$
Hence B is the correct answer.
answered Jun 6, 2014 by | 151 | 506 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-51 | longest | en | 0.942938 |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A100004 Theta series of Barnes-Wall lattice in 128 dimensions. 1
%I #18 Apr 26 2021 13:04:17
%S 1,0,0,0,1260230400,0,211691822284800,167823813692620800,
%T 478913396715566841600,800774658647826785894400,
%U 611331534352048834410577920,247750040775246869910100377600,59524143410351572596837132825600,9219467698757233237143045380505600,982518868151763901368740969501491200
%N Theta series of Barnes-Wall lattice in 128 dimensions.
%D J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag.
%H Andy Huchala, <a href="/A100004/b100004.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Ba#BW">Index entries for sequences related to Barnes-Wall lattices</a>
%e Theta series = 1+1260230400*q^8+211691822284800*q^12+167823813692620800*q^14+
%e 478913396715566841600*q^16+800774658647826785894400*q^18+
%e 611331534352048834410577920*q^20+247750040775246869910100377600*q^22+
%e 59524143410351572596837132825600*q^24+9219467698757233237143045380505600*q^26+
%e 982518868151763901368740969501491200*q^28+
%e 75861944796267943358105372399071395840*q^30+
%e 4424187851601897335090905879137082310400*q^32+
%e 201630718820389821054181008698644837171200*q^34+
%e 7386564136116987592917746020874575301836800*q^36+
%e 222716047981438421360112524163657218025062400*q^38+O(q^40)
%e or in terms of e8 and the cusp form:
%e e8^16-3840*e8^13*delta+4942080*e8^10*delta^2-2416128000*e8^7*delta^3
%e +364791398400*e8^4*delta^4-8259045949440*e8*delta^5.
%o (Sage) e8 = eisenstein_series_qexp(4,25,normalization = "integral");
%o delta = CuspForms(1,12).0.q_expansion(25);
%o f = e8^16-3840*e8^13*delta+4942080*e8^10*delta^2-2416128000*e8^7*delta^3+364791398400*e8^4*delta^4-8259045949440*e8*delta^5;
%o f[:25]
%K nonn
%O 0,5
%A Eric Rains and _N. J. A. Sloane_, Sep 27 2005
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Last modified March 3 08:35 EST 2024. Contains 370499 sequences. (Running on oeis4.) | 867 | 2,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-10 | latest | en | 0.355744 |
https://www.curriki.org/oer/Preview-Lesson-One-of-Six--54285 | 1,586,100,321,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371606067.71/warc/CC-MAIN-20200405150416-20200405180916-00523.warc.gz | 880,880,182 | 18,911 | ### This is Lesson One of the Rapid Still Imagination Lesson Plan for elementary grades K-5.The next five lessons, and other information, may be found in the Imagination Lesson Plan PDF ABOVE.
LESSON ONE: Identifying and Exploring Different Types of Movements and Their Possible Variations
Activity #1. Identifying Different Types of Movements
Ask students the following questions and write a list of their responses on chart paper on the wall. Title the list, “Different Types of Movements.”
• What are some different types of movements people can do?
Ask several students, one at a time, to physically try out some of the different movements
from the list.
• How would you describe what happens in the movement?
Choose one student’s type of movement to focus on (for example, a “jump”).
Ask the following questions and create a short list of responses:
• What do all jumps have in common, regardless of how you jump?
• What are some different ways of doing this movement?
• How can the movement be varied in some way without changing the type of movement it is?
Let students know that, as a class, you will be calling these different ways of doing the same type of movement, “Variations.” “Variations on a Movement.”
Lead a brief discussion on the words “Vary” or “Variation”—when students have heard this word before and in relation to what; and whether they relate it to any other subject area they study.
Activity #2. Coming up with Our Own Variation on a Type of Movement
In groups of 4, ask groups to choose one type of movement from the “Different Types of Movements” list and decide, as a group, how they would like to do that movement, and to remember specifically how they are doing it.
Once the groups have their movement, ask them to consider the following questions:
• Can you imagine another, very different way of doing this same type of movement?
• What would a variation on your movement be?
After students try out their variations, ask:
• How would you describe your variation?
• How is this variation different from your original movement?
Ask one group at a time to show their original movement and its variation.
Document responses to the following questions on a list titled, “How Movement Can Vary.”
• What do you notice about how the movement varied?
• In what ways is it different from the original version of the movement?
• What changed?
NOTE: Examples of responses might be things like: faster, slower, higher, lower, whole body moving vs. just the legs, etc.
Finish the lesson with the following questions, and document the responses:
• What are the different things we studied today?
• Are you relating in any way these ideas to other subject areas?
Non-profit Tax ID # 203478467 | 576 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2020-16 | latest | en | 0.952225 |
https://app.pandai.org/note/read/kssm-am-11-02-03/KSSM-F5-AM-02-03 | 1,723,216,034,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640767846.53/warc/CC-MAIN-20240809142005-20240809172005-00691.warc.gz | 78,766,690 | 14,445 | ## The Second Derivatives
2.3 The Second Derivatives
• The second derivative of a function $$y=f(x)$$ with respect to $$x$$ :
$$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \begin{pmatrix} \dfrac{dy}{dx}\end{pmatrix} \text{ or } f''(x) = \dfrac{d}{dx}[f'(x)]$$
Function
$$y=f(x)=x^3 - 2x^2+3x-5$$
First Derivative
$$\dfrac{dy}{dx}=f'(x)=3x^2 - 4x+3$$
Second Derivative
$$\dfrac{d^2y}{dx^2}=f''(x)=6x - 4$$
## The Second Derivatives
2.3 The Second Derivatives
• The second derivative of a function $$y=f(x)$$ with respect to $$x$$ :
$$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \begin{pmatrix} \dfrac{dy}{dx}\end{pmatrix} \text{ or } f''(x) = \dfrac{d}{dx}[f'(x)]$$
Function
$$y=f(x)=x^3 - 2x^2+3x-5$$
First Derivative
$$\dfrac{dy}{dx}=f'(x)=3x^2 - 4x+3$$
Second Derivative
$$\dfrac{d^2y}{dx^2}=f''(x)=6x - 4$$ | 380 | 808 | {"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} | 4.46875 | 4 | CC-MAIN-2024-33 | latest | en | 0.484636 |
https://docs.couchbase.com/server/current/n1ql/n1ql-language-reference/numericfun.html | 1,721,413,903,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514917.3/warc/CC-MAIN-20240719170235-20240719200235-00458.warc.gz | 182,951,798 | 6,365 | # Number Functions
• reference
+
Number functions are functions that are performed on a numeric field.
ABS(expression)
Returns absolute value of the number.
ACOS(expression)
ASIN(expression)
ATAN(expression)
ATAN2(expression1, expression2)
Returns arctangent of expression2/expression1.
CEIL(expression)
Returns smallest integer not less than the number.
COS(expression)
Returns cosine.
DEGREES(expression)
E()
Base of natural logarithms.
EXP(expression)
Returns eexpression.
LN(expression)
Returns log base e.
LOG(expression)
Returns log base 10.
FLOOR(expression)
Largest integer not greater than the number.
PI()
Returns PI.
POWER(expression1, expression2):
Returns expression1expression2.
RANDOM([ expression ])
Returns pseudo-random number with optional seed.
ROUND(expression [, digits ])
Rounds the value to the given number of integer digits to the right of the decimal point (left if digits is negative). Digits is 0 if not given.
SIGN(expression)
Valid values: -1, 0, or 1 for negative, zero, or positive numbers respectively.
SIN(expression)
Returns sine.
SQRT(expression)
Returns square root.
TAN(expression)
Returns tangent.
TRUNC(expression [, digits ])
Truncates the number to the given number of integer digits to the right of the decimal point (left if digits is negative). Digits is 0 if not given.
Example
```SELECT
AVG(reviews.rating) / 5 AS normalizedRating,
ROUND((avg(reviews.rating) / 5), 2) AS roundedRating,
TRUNC((avg(reviews.rating) / 5), 3) AS truncRating
FROM reviews AS reviews
WHERE reviews.customerId = "customer62"```
Returns
```{
"results": [
{
"normalizedRating": 0.42000000000000004,
"roundedRating": 0.42,
"truncRating": 0.42
}
]
}```
Output | 408 | 1,725 | {"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} | 2.75 | 3 | CC-MAIN-2024-30 | latest | en | 0.537244 |
https://algebra-calculators.com/rational-numbers/ | 1,679,362,443,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943589.10/warc/CC-MAIN-20230321002050-20230321032050-00164.warc.gz | 123,486,297 | 25,566 | ## Rational Numbers
Introduction
We work with basic math in our daily life like counting your fingers and toes, calculating your age, finding temperatures of a cold winter day in Michigan which would be below 0 degree, dividing a birthday cake among twelve of your friends, and many more. In each of these circumstances, we use different types of numbers. The types of numbers that we already know, are natural numbers, whole numbers and integers.
When you were studying fractions, you were just learning the way of writing a numerical quantity, with a numerator and a denominator. You also learned about using arithmetic operations with the fractions and about representing them as decimals. Now you are growing older, and are ready to dig mathematics a little deeper. You are going to learn about the numbers themselves and not just the way you write them on a piece of paper. Let us walk through a new type of number called rational numbers.
### Standard Form of Rational Numbers
A rational number is not just a fraction. It means that any number, be it natural number or integers, can be written in the form of
$\frac{p}{q}$
such that q is never equal to 0. But we cannot write every number in that way.
A rational number is said to be in a standard form if and only if
a) The denominator is an integer greater than 0
b) The only common divisor between the numerator and the denominator is 1.
Steps to express a rational number in the standard form:
a) Check if the denominator is positive or negative. If the denominator is negative, change it to positive by multiplying both numerator and denominator by -1.
b) Take the absolute values of the numerator and denominator and find their greatest common divisor.
c) Perform the division of each of the numerator and denominator by the obtained greatest common divisor. The resulting rational number is the standard form.
Positive and Negative Rational Numbers
A positive rational number is one which has the same signs (either positive or negative) in both its numerator and denominator.
Eg :
$\frac{-5}{-7},\frac{10}{13}$
A negative rational number is one which has opposite signs in both its numerator and denominator.
Eg :
$\frac{–5}{7},–\frac{10}{–13}$
Properties of Rational Numbers
Useful Properties-
1. If
$\frac{p}{q}$
is a rational number and r is any non-zero integer, then
$\frac{p}{q}=\frac{p×r}{q×r}$
This implies that the rational number remains the same if we multiply both the numerator and denominator by the same integer provided the integer is not equal to 0.
1. If
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers and
$\frac{p}{q}=\frac{r}{s}$
then,p×s=q×r and vice-versa.
Closure Property-
The closure property of rational numbers under addition states that if
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers,
$\frac{p}{q}+\frac{r}{s}$
is also a rational number.
• Subtraction
The closure property of rational numbers under subtraction states that if
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers,
$\frac{p}{q}–\frac{r}{s}$
is also a rational number.
• Multiplication
The closure property of rational numbers under multiplication states that if
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers,
$\frac{p}{q}×\frac{r}{s}$
is also a rational number
• Division
The closure property of rational numbers under division states that if
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers,
$\frac{p}{q}÷\frac{r}{s}=\frac{p}{q}×\frac{s}{r}$
is also a rational number
Commutative Property-
The commutative property of rational numbers under addition states that if
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers,
$\frac{p}{q}+\frac{r}{s}=\frac{r}{s}+\frac{p}{q}$
• Subtraction
The commutative property of rational numbers under subtraction states that if
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers,
$\frac{p}{q}–\frac{r}{s}\ne \frac{r}{s}–\frac{p}{q}$
. Hence, the commutative property does not hold for the subtraction of rational numbers.
• Multiplication
The commutative property of rational numbers under multiplication states that if
$\frac{p}{q}$
and
$\frac{r}{s}$
are any two rational numbers,
$\frac{p}{q}×\frac{r}{s}=\frac{r}{s}×\frac{p}{q}$
• Division
The commutative property of rational numbers under division states that if
pq
$\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
and
rs
$\frac{r}{s}$
are any two rational numbers,
pq÷rsrs÷pq
$\frac{p}{q}÷\frac{r}{s}\ne \frac{r}{s}÷\frac{p}{q}$
. Hence, the commutative property does not hold for division of any two rational numbers
Associative Property-
The associative property of rational numbers under addition states that if
pq,rs
$\frac{p}{q},\frac{r}{s}$
and
ut
$\frac{u}{t}$
are any three rational numbers,
pq+(rs+ut)=(pq+rs)+ut
$\frac{p}{q}+\left(\frac{r}{s}+\frac{u}{t}\right)=\left(\frac{p}{q}+\frac{r}{s}\right)+\frac{u}{t}$
.
• Subtraction
The associative property of rational numbers under subtraction states that if
pq,rs
$\frac{p}{q},\frac{r}{s}$
and
ut
$\frac{u}{t}$
are any three rational numbers,
pq(rsut)=(pqrs)ut
$\frac{p}{q}-\left(\frac{r}{s}-\frac{u}{t}\right)=\left(\frac{p}{q}-\frac{r}{s}\right)-\frac{u}{t}$
. Hence, the associative property does not hold for the subtraction of rational numbers.
• Multiplication
The associative property of rational numbers under multiplication states that if
pq,rs
$\frac{p}{q},\frac{r}{s}$
and
ut
$\frac{u}{t}$
are any three rational numbers,
pq×(rs×ut)=(pq×rs)×ut
$\frac{p}{q}×\left(\frac{r}{s}×\frac{u}{t}\right)=\left(\frac{p}{q}×\frac{r}{s}\right)×\frac{u}{t}$
.
• Division
The associative property of rational numbers under division states that if
pq,rs
$\frac{p}{q},\frac{r}{s}$
and
ut
$\frac{u}{t}$
are any three rational numbers,
pq÷(rs÷ut)=(pq÷rs)÷ut
$\frac{p}{q}÷\left(\frac{r}{s}÷\frac{u}{t}\right)=\left(\frac{p}{q}÷\frac{r}{s}\right)÷\frac{u}{t}$
. Hence, the associative property does not hold for division of any three or more rational numbers.
Identity Property-
The identity property of rational numbers under addition states that if
pq
$\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
is a rational number, then
pq+0=0+pq=pq
$\frac{p}{q}+0=0+\frac{p}{q}=\frac{p}{q}$
. Hence 0 is called the additive identity for rational numbers.
• Multiplication
The identity property of rational numbers under multiplication states that if
pq
$\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
is a rational number, then
pq×1=1×pq=pq
$\frac{p}{q}×1=1×\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}=\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
. Hence 1 is called the multiplicative identity for rational numbers.
Inverse Property-
The inverse property of rational numbers under addition states that if
pq
$\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
is a rational number, then there exists a rational number
pq
$-\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
such that
pq+(pq)=(pq)+pq=0
$\frac{p}{q}+\left(-\frac{p}{q}\right)=\left(-\frac{p}{q}\right)+\frac{p}{q}=0$
. Hence
pq
$-\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
is called the additive inverse of
pq
$\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
.
• Multiplication
The inverse property of rational numbers under multiplication states that if
pq
$\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
is a rational number, then there exists a rational number
qp
$\phantom{\rule{thickmathspace}{0ex}}\frac{q}{p}$
such that
pq×qp=qp×pq=1
$\frac{p}{q}×\frac{q}{p}=\frac{q}{p}×\frac{p}{q}=1$
. Hence
qp
$\frac{q}{p}$
is called the multiplicative inverse or reciprocal of
pq
$\phantom{\rule{thickmathspace}{0ex}}\frac{p}{q}$
.
Distributive Property of Multiplication over Addition
If
pq,rs
$\frac{p}{q},\frac{r}{s}$
and
ut
$\frac{u}{t}$
are any three rational numbers,
pq×(rs+ut)=pq×rs+pq×ut
$\frac{p}{q}×\left(\frac{r}{s}+\frac{u}{t}\right)=\frac{p}{q}×\frac{r}{s}+\frac{p}{q}×\frac{u}{t}$
. The distributive property of multiplication over addition holds good for all rational numbers.
Distributive Property of Multiplication over Subtraction
If
pq,rs
$\frac{p}{q},\frac{r}{s}$
and
ut
$\frac{u}{t}$
are any three rational numbers,
pq×(rsut)=pq×rspq×ut
$\frac{p}{q}×\left(\frac{r}{s}-\frac{u}{t}\right)=\frac{p}{q}×\frac{r}{s}-\frac{p}{q}×\frac{u}{t}$
. The distributive property of multiplication over subtraction holds good for all rational numbers.
Difference Between Rational and Irrational Numbers
S.No Rational Numbers Irrational Numbers 1 These numbers can be expressed in the form of a fraction pq $\frac{p}{q}$ where q=0 $q=0$ These numbers cannot be expressed in the form of a fraction pq $\frac{p}{q}$ 2 These numbers include numbers that are finite or recurring in nature. These numbers include numbers that are non-terminating and non-repeating in nature. 3 Perfect squares are always rational numbers Surds such as 2–√,3–√,5–√ $\sqrt{2},\sqrt{3},\sqrt{5}$ are always irrational numbers
Rational Numbers Examples
• 27
$\frac{2}{7}$
– Both the numerator 2 and the denominator 7 are integers.
• 5 – Can be expressed as
51
$\frac{5}{1}$
, wherein 5 is the quotient when we divide the integer 5 by the integer 1.
• 49
$\sqrt{49}$
– As the square root of 49 can be simplified to 7, which is also the quotient obtained when the integer 7 is divided by 1
• 0.5 – Can be written as
510
$\frac{5}{10}$
or
12
$\frac{1}{2}$
. Thus, all terminating decimals are rational numbers.
• 0.3333333333 – All recurring decimals are rational numbers. | 2,912 | 9,680 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 73, "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} | 4.96875 | 5 | CC-MAIN-2023-14 | latest | en | 0.935346 |
https://encyclopediaofmath.org/wiki/Kronecker_theorem | 1,716,835,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059045.25/warc/CC-MAIN-20240527175049-20240527205049-00528.warc.gz | 200,831,368 | 6,610 | # Kronecker theorem
Given $a _ {i} = ( a _ {i1} \dots a _ {in} ) \in \mathbf R ^ {n}$, $i = 1 \dots m$, and $b = ( b _ {1} \dots b _ {n} ) \in \mathbf R ^ {n}$; then for any $\epsilon > 0$ there exist integers $q _ {i}$, $i = 1 \dots m$, and $p _ {j}$, $j = 1 \dots n$, such that
$$\left | \sum _ {i = 1 } ^ { m } q _ {i} a _ {ij} - p _ {j} - b _ {j} \ \right | < \epsilon ,\ \ 1 \leq j \leq n,$$
if and only if for any $r _ {1} \dots r _ {n} \in \mathbf Z$ such that
$$\sum _ {j = 1 } ^ { n } a _ {ij} r _ {j} \in \mathbf Z ,\ \ i = 1 \dots m,$$
the number
$$\sum _ {j = 1 } ^ { n } b _ {j } r _ {j }$$
is also an integer. This theorem was first proved in 1884 by L. Kronecker (see [1]).
Kronecker's theorem is a special case of the following theorem [2], which describes the closure of the subgroup of the torus $T ^ {n} = \mathbf R ^ {n} / \mathbf Z ^ {n}$ generated by the elements $a _ {i} + \mathbf Z ^ {n}$, $i = 1 \dots m$: The closure is precisely the set of all classes $b + \mathbf Z ^ {n}$ such that, for any numbers $r _ {1} \dots r _ {n} \in \mathbf Z$ with
$$\sum _ {j = 1 } ^ { n } a _ {ij} r _ {j} \in \mathbf Z ,\ \ i = 1 \dots n,$$
one has also
$$\sum _ {j = 1 } ^ { n } b _ {j} r _ {j} \in \mathbf Z .$$
(Cf. [2].) Under the assumptions of Kronecker's theorem, this closure is simply $T ^ {n}$. This means that the subgroup of all elements of the form
$$\sum _ {i = 1 } ^ { m } q _ {i} ( a _ {i} + \mathbf Z ) ^ {n} ,$$
where $q _ {i} \in \mathbf Z$, is dense in $T ^ {n}$, while the subgroup of vectors
$$\sum _ {i = 1 } ^ { m } q _ {i} a _ {i} + p,$$
where $p \in \mathbf Z ^ {n}$, is dense in $\mathbf R ^ {n}$. Kronecker's theorem can be derived from the duality theory for commutative topological groups (cf. Topological group), [3].
In the case $m = 1$, Kronecker's theorem becomes the following proposition: A class $\omega + \mathbf Z ^ {n}$, where $\omega = ( \omega _ {1} \dots \omega _ {n} ) \in \mathbf R ^ {n}$, generates $T ^ {n}$ as a topological group if and only if the numbers $1, \omega _ {1} \dots \omega _ {n}$ are linearly independent over the field $\mathbf Q$ of rational numbers. In particular, the torus $T ^ {n}$ as a topological group is monothetic, i.e. is generated by a single element.
#### References
[1] L. Kronecker, "Näherungsweise ganzzahlige Auflösung linearer Gleichungen" , Werke , 3 , Chelsea, reprint (1968) pp. 47–109 [2] N. Bourbaki, "Elements of mathematics. General topology" , Addison-Wesley (1966) (Translated from French) [3] L.S. Pontryagin, "Topological groups" , Princeton Univ. Press (1958) (Translated from Russian)
The last statement above can be rephrased as: If $\omega _ {1} \dots \omega _ {n}$ are linearly independent over $\mathbf Q$, then the set $B= \{ ( \{ k \omega _ {1} \} \dots \{ k \omega _ {n} \} ) : k \in \mathbf Z \}$ is dense in $( 0, 1)$. Here $\{ x \} = x - [ x]$ denotes the fractional part of $x$( cf. Fractional part of a number). In fact, the set $B$ is even uniformly distributed, cf. Uniform distribution. | 1,112 | 3,026 | {"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} | 3.25 | 3 | CC-MAIN-2024-22 | latest | en | 0.694789 |
https://www.physicsforums.com/threads/equations-of-sets.845732/ | 1,508,440,926,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823360.1/warc/CC-MAIN-20171019175016-20171019195016-00449.warc.gz | 971,900,401 | 18,341 | Equations of sets
1. Nov 30, 2015
geoffrey159
1. The problem statement, all variables and given/known data
Let $A,B \in {\cal P}(E)$. Solve in ${\cal P}(E)$ the following equations:
1. $X\cup A = B$
2. $X\cap A = B$
3. $X - A = B$
2. Relevant equations
3. The attempt at a solution
1. We have $A\cup B = (A\cup X)\cup A = A\cup X = B$. So $A\subset B$ and the solution cannot be less than $C_B(A)$. So $X = C_B(A) \cup G,\ G\in{\cal P}(A)$
2. We have $A\cap B = A\cap X = B$. So $B\subset A$ and the solution cannot be less than $B$. So $X = B \cup G,\ G\in{\cal P}(C_E(A))$
3. We have $A\cap B = \emptyset$ and $A\cup B = A\cup X$. So the solution can't be less than $B$. Finally, $X = B\cup G,\ G\in{\cal P}(A)$
Is it correct ?
2. Nov 30, 2015
Ray Vickson
These equations are impossible if there are no restrictions on $A$ and $B$. In other words, it is easy to give examples of $A, B$ in which the equations cannot possibly hold, no matter how you try to choose $X$.
Draw some Venn diagrams and see this for yourself.
3. Nov 30, 2015
geoffrey159
Would you mind sharing your counterexamples for the 3 different cases? Assuming $A\subset B$ for q1, $B\subset A$ for q2, $A\cap B =\emptyset$ for q3.
4. Nov 30, 2015
Ray Vickson
No, I said that without some restrictions on $A$ and $B$ you can find cases where they are impossible.
Naturally, if you make some additional assumptions about $A$ and $B$ you CAN find solutions. It is just that as originally stated (with no restrictions on $A,B$) it may not always be possible to have any $X$. In other words, you either copied down the question incorrectly, or you were given a trick question.
Last edited: Nov 30, 2015
5. Nov 30, 2015
geoffrey159
Oh I see what you mean, I have been sloppy in my proof. In each cases, I assumed $X$ was a solution. With that assumption, I found restrictions for each cases, which are:
1. $A\subset B$
2. $B\subset A$
3. $A\cap B=\emptyset$.
So that outside these restrictions, there aren't any solution. Then I drew a diagram to find a minimal solution $X_{\text{min}}$ which are:
1. $C_B(A)$
2. $B$
3. $B$
Finally, I tried to find a 'maximal' solution $X = X_{\text{min}} \cup G$. In each case, it has to be true that $G$ belongs to
1. ${\cal P}(A)$
2. ${\cal P}(C_E(A))$
3. ${\cal P}(A)$
Is it ok then ?
6. Nov 30, 2015
Ray Vickson
I have no idea what $C_E (A)$ means.
7. Nov 30, 2015
geoffrey159
$C_E(A) = \{ x \in E : x\notin A\}$
8. Nov 30, 2015
Ray Vickson
9. Nov 30, 2015
Staff: Mentor
Yes, I am more familiar with $B - A$ or $B \backslash A$. I'm not sure why the author of the text in use in this thread felt the need to come up with new notation when there was existing notation that was clearer.
10. Dec 1, 2015 | 918 | 2,735 | {"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} | 4.15625 | 4 | CC-MAIN-2017-43 | longest | en | 0.907951 |
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Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A015438 Gaussian binomial coefficient [ n,12 ] for q=-13. 12
%I
%S 1,21633936185161,507029461102251552321630151,
%T 11807441196984503845077844573952807835871,
%U 275100402115798836253928241395289617394098490488956444,6409295323626866454933457428954320223001885025904687118646704057084
%N Gaussian binomial coefficient [ n,12 ] for q=-13.
%D J. Goldman and G.-C. Rota, The number of subspaces of a vector space, pp. 75-83 of W. T. Tutte, editor, Recent Progress in Combinatorics. Academic Press, NY, 1969.
%D I. P. Goulden and D. M. Jackson, Combinatorial Enumeration. Wiley, NY, 1983, p. 99.
%D M. Sved, Gaussians and binomials, Ars. Combinatoria, 17A (1984), 325-351.
%H Vincenzo Librandi, <a href="/A015438/b015438.txt">Table of n, a(n) for n = 12..80</a>
%H <a href="/index/Ga#Gaussian_binomial_coefficients">Index entries related to Gaussian binomial coefficients</a>.
%F a(n)=product_{i=1..12} ((-13)^(n-i+1)-1)/((-13)^i-1). - M. F. Hasler, Nov 03 2012
%t Table[QBinomial[n, 12, -13], {n, 12, 20}] (* _Vincenzo Librandi_, Nov 06 2012 *)
%o (Sage) [gaussian_binomial(n,12,-13) for n in range(12,17)] # _Zerinvary Lajos_, May 28 2009
%o (PARI) A015438(n,r=12,q=-13)=prod(i=1,r,(q^(n-i+1)-1)/(q^i-1)) \\ _M. F. Hasler_, Nov 03 2012
%o (MAGMA) r:=12; q:=-13; [&*[(1-q^(n-i+1))/(1-q^i): i in [1..r]]: n in [r..20]]; // _Vincenzo Librandi_, Nov 06 2012
%Y Cf. Gaussian binomial coefficients [n,r] for q=-13: A015265 (r=2), A015286 (r=3), A015303 (r=4), A015321 (r=5), A015337 (r=6), A015355 (r=7), A015370 (r=8), A015385 (r=9), A015402 (r=10), A015422 (r=11). - _M. F. Hasler_, Nov 03 2012
%K nonn,easy
%O 12,2 | 719 | 1,812 | {"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-2020-10 | latest | en | 0.468707 |
www.solarpowernotes.com | 1,386,851,024,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164583115/warc/CC-MAIN-20131204134303-00058-ip-10-33-133-15.ec2.internal.warc.gz | 536,603,812 | 6,673 | # How to determine wind direciton
We need no introduction of the wind; we all know that wind is the moving air with some force and pressure. So wind direction is defined as the area which is the origin of that air or from where wind originates is referred to as the wind direction. Wind direction plays an important role in determining the output and the generation of the different forms of energies from the wind resources. Basically wind direction is that quantity that represents in the form of degrees such as 90 or 180 degree.
How to Determine Wind Direction?
Wind direction is measured in the number of ways with the help of different types of technologies designed for them. Some important devices that are used for the measurement of the wind direction are as follows
Wind Sock:
A type of the measuring device that was very famous among the people in the past but nowadays many other modern technologies came into being. Wind sock is the types of the cloth piece that is of conical shape with both ends open. Usually it is seen on the airports and some times in the technological plants. It placed on the rod like structure stand and this sock moves towards that side where the power of the wind is stronger.
Anemometer:
A modern device that is used to detect the direction of the wind is called as the anemometer. The common type of the anemometer that is used to determine the direction of the wind power is the cup shaped anemometer. It has four cup arranged on the rod one behind the other. When the wind blows they rotate. By counting the number of turns we can easily determine the direction
Wind Direction Sensors:
A type of the sensing technology that also do the same job of detection of the wind direction, basically wind direction sensors are the device that can calculate the direction of the wind accurately and valuably. Son these are some methods that are used to measure the direction of the wind energy. Some other multi purpose devices are also used that perform two functions i.e. detect the speed and the direction of the wind such as wind sensors, ultrasonic sensors etc. Some times wind vanes a4re also used to calculate the direction of the wind
Without Technology:
Wind direction is also measured in the absence of the technology. We can use the index finger for that purpose of detection, wet the finger and take finger in the upward direction. The side of the finger that feels cooling is the area from where the wind originates.
Wind direction Map:
The map that is required for the calculation and study the direction and speed of the wind energy is called as the wind direction map. Basically the main purpose of constructing the map is to save the country from different types of storms and predict the weather conditions related to the wind. Wind direction map also helpful for the wind power generating turbines because we can increase the output or efficiency of the electrical energy by studying the map.
Importance of Wind Direction:
Wind direction is an important component because with the help of wind direction we are able to construct the wind turbines facing that area of the wind origin. This really helps to increase the production of energy. There are lots of benefits of the wind direction or the measuring the wind direction, such as it is very helpful for the airports in determination of portion of the stronger wind.
Factors Affecting Wind Direction:
Some important factors are also affects the direction of the wind and are as follows
1. Rotation Earth surface
2. Pressure of the Wind
3. Friction
Useful related articles on Wind energy | 705 | 3,603 | {"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-2013-48 | latest | en | 0.964419 |
https://devforum.roblox.com/t/hinge-constraint-servo-facing-mouse/1626594 | 1,723,768,467,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641319057.20/warc/CC-MAIN-20240815235528-20240816025528-00791.warc.gz | 145,764,942 | 7,205 | # Hinge constraint servo facing mouse
Hello.
I’ve been working on this T72B3 model recently, and have been having a lot of issues working out the turret.
Here’s my code for reference:
``````--MODULE:
local m={}
m.TurnTargetAngle = function(target: Vector3,origin: Vector3)
local delta = target - origin
local new_delta = Vector3.new(delta.x, 0, delta.z)
local unit_delta = new_delta.unit
local angle = math.atan2(unit_delta.z, unit_delta.x)
return (-math.deg(angle)+math.pi)
end
return m;
``````
``````--SCRIPT:
coroutine.resume(coroutine.create(function()
RunService.Heartbeat:Connect(function()
if currentdriver ~= nil then
local MP = MousePosition:InvokeClient(game.Players[tostring(currentdriver)])
local t = TargetModule.TurnTargetAngle(MP,Parts.TurretRing1.Position)
Parts.TurretRing2.Rotator.TargetAngle = t
end
end)
end))
``````
And here’s what’s happening:
Does anybody know what’s wrong?
I would love to help you but cannot see what the problem is from the images shown.
I believe the turret simply turns round eg has no tilt so images showing the view from above would show more about the rotation issue.
Can you provide one for me?
Let me sum it up as well as I can.
When I point the mouse towards a specific direction, the turret ring decides to turn to the side.
Here’s the best visualization I can make:
In terms of rotation, the bottom face points towards the “target”, and the front face points towards the sky.
(To clarify, this is using a hinge constraints target angle)
From what I can find it seems to be due to the fact that math.atan2 is relative to the positive world X axis. I’m no vector math genius so my solution is to mess around with negatives and subtracting rotation and stuff
Edit: I solved it by subtracting the WorldOrientation.Y of the Attachment0 of the turret base servo
Can you maybe explain it in a little more detail?
When I try to subtract the angle by `(A0.Y-TargetAngle)` it comes out to some absurd number.
When I try subtracting `(A0.Y-TargetAngle)` by the angle I get a weird assortment of random numbers.
I’m trying a few other methods and editing this as I go, but so far nothing has worked.
AS OF NOW THIS IS WHAT IT LOOKS LIKE:
SCRIPT:
`````` RunService.Heartbeat:Connect(function()
if currentdriver ~= nil then
local MP = MousePosition:InvokeClient(game.Players[tostring(currentdriver)])
local t = TargetModule.TurnTargetAngle(MP,Parts.TurretRing2.Armor.WorldPosition,(Parts.TurretRing2.Rotator.Attachment0.WorldOrientation.Y))
Parts.TurretRing2.Rotator.TargetAngle = t
end
end)
``````
MODULE:
``````local m={}
m.TurnTargetAngle = function(target: Vector3,origin: Vector3, sa)
local delta = target - origin
local new_delta = Vector3.new(delta.x, 0, delta.z)
local unit_delta = new_delta.unit
local angle = math.atan2(unit_delta.z, unit_delta.x)
print(sa-(-math.deg(angle)+math.pi))
return sa-(-math.deg(angle)+math.pi)
end
return m;
``````
You have to do it the other way around, such that sa is subtracting from the final angle.
Here is the code that has worked for me:
``````(-math.deg(AngleYaw)) - TurretAttach.WorldOrientation.Y
``````
where AngleYaw is angle and TurretAttach.WorldOrientation.Y is sa
1 Like | 791 | 3,195 | {"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-2024-33 | latest | en | 0.818059 |
https://www.javatpoint.com/karnaugh-map-in-digital-electronics | 1,726,635,232,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651836.76/warc/CC-MAIN-20240918032902-20240918062902-00078.warc.gz | 773,306,755 | 9,976 | # Karnaugh Map(K-Map) method
The K-map is a systematic way of simplifying Boolean expressions. With the help of the K-map method, we can find the simplest POS and SOP expression, which is known as the minimum expression. The K-map provides a cookbook for simplification.
Just like the truth table, a K-map contains all the possible values of input variables and their corresponding output values. However, in K-map, the values are stored in cells of the array. In each cell, a binary value of each input variable is stored.
The K-map method is used for expressions containing 2, 3, 4, and 5 variables. For a higher number of variables, there is another method used for simplification called the Quine-McClusky method. In K-map, the number of cells is similar to the total number of variable input combinations. For example, if the number of variables is three, the number of cells is 23=8, and if the number of variables is four, the number of cells is 24. The K-map takes the SOP and POS forms. The K-map grid is filled using 0's and 1's. The K-map is solved by making groups. There are the following steps used to solve the expressions using K-map:
1. First, we find the K-map as per the number of variables.
2. Find the maxterm and minterm in the given expression.
3. Fill cells of K-map for SOP with 1 respective to the minterms.
4. Fill cells of the block for POS with 0 respective to the maxterm.
5. Next, we create rectangular groups that contain total terms in the power of two like 2, 4, 8, … and try to cover as many elements as we can in one group.
6. With the help of these groups, we find the product terms and sum them up for the SOP form.
## 2 Variable K-map
There is a total of 4 variables in a 2-variable K-map. There are two variables in the 2-variable K-map. The following figure shows the structure of the 2-variable K-map:
• In the above figure, there is only one possibility of grouping four adjacent minterms.
• The possible combinations of grouping 2 adjacent minterms are {(m0, m1), (m2, m3), (m0, m2) and (m1, m3)}.
## 3-variable K-map
The 3-variable K-map is represented as an array of eight cells. In this case, we used A, B, and C for the variable. We can use any letter for the names of the variables. The binary values of variables A and B are along the left side, and the values of C are across the top. The value of the given cell is the binary values of A and B at left side in the same row combined with the value of C at the top in the same column. For example, the cell in the upper left corner has a binary value of 000, and the cell in the lower right corner has a binary value of 101.
## The 4-Variable Karnaugh Map
The 4-variable K-map is represented as an array of 16 cells. Binary values of A and B are along the left side, and the values of C and D are across the top. The value of the given cell is the binary values of A and B at left side in the same row combined with the binary values of C and D at the top in the same column. For example, the cell in the upper right corner has a binary value of 0010, and the cell in the lower right corner has a binary value of 1010
## 5-variable K-map
With the help of the 32- cell K-map, the boolean expression with 5 variables can be simplified. For constructing a 5-variable K-map, we use two 4-variable K-maps. The cell adjacencies within each of the 4- variable maps for the 5-variable map are similar to the 4- variable map.
A K-map for five variables (PQRST) can be constructed using two 4-variable maps. Each map contains 16 cells with all combinations of variables Q, R, S, and T. One map is for P = 0, and the other is for P = 1). | 919 | 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} | 3.609375 | 4 | CC-MAIN-2024-38 | latest | en | 0.898358 |
https://www.revision.co.ke/question-papers/kcse-cluster-tests-19/physics-paper-1 | 1,701,368,125,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100229.44/warc/CC-MAIN-20231130161920-20231130191920-00849.warc.gz | 1,112,241,798 | 38,611 | ### SECTION A (25 Marks)
1.
The density of sulphuric acid is 0.9g/cm3. Calculate the volume of 1.8kg of the acid. ....................................................................................................................................
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3 marks
2.
Draw a micrometer screw gauge showing a reading of 5.85mm if the pitch of the screw gauge is 0.5mm.
1 marks
3.
The figure 1 (a) below shows a soap bubble blown towards the wide end of a funnel. The narrow end is then closed.
i) On the figure 1 (b) indicate, the position of the bubble after the narrow end is opened.
ii) Explain this observation.
..........................................................................................................................
1 marks
4.
The figure 2 below shows air trapped in a J- shaped tube. What is the pressure exerted on the trapped air?( density of mercury 1600kg/m3 atmospheric pressure=1x105pa) (2 marks)
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2 marks
5.
A mass, m is suspended from a tight copper wire as shown in the figure below. The wire is then heated.
State and explain what was observed on the position of M as the wire was heated for sometime.
..........................................................................................................................................
2 marks
6.
Fig. 4 shows two possible arrangements of fins in a heat sink, a device that reduces accumulation of heat in electrical devices.
State with a reason which arrangement of fins makes the heat sink work better.
..........................................................................................................................................
2 marks
7.
Figure 5 shows a wheel spanner used to loosen a tight nut. The mechanic found it very difficult to use.
State with a reason the adjustment which he would make for the same force F to be sufficient to loosen the nut.
..........................................................................................................................................
2 marks
8.
Distinguish between stable and unstable equilibrium.
..........................................................................................................................................
1 marks
9.
The Figure 6(a) below shows a tennis ball hanging close to a water tap. When the tap is opened so that water flows, the ball is attracted to the stream as in figure (b)
Explain this observation
..........................................................................................................................................
2 marks
10.
Figure 7 shows a velocity-time graph for a moving object.
On the axes provided, sketch the displacement-time graph for the same motion.
1 marks
11.
Figure 8 shows a stack of coins. The one at the base is hit hard and ejected out of the stack. The remaining stack continues to slant in the same position,
Explain this observation
..........................................................................................................................................
1 marks
12.
Name the property of a spring that enables it to regain its original length when the load is removed.
..........................................................................................................................................
1 marks
13.
A column of air 24cm long is trapped by a column of mercury 10cm long in a horizontal tube with uniform cross-sectional area. If the atmospheric pressure is 76cm, determine the new height of air trapped in the tube when the tube is placed vertically.
..........................................................................................................................................
..........................................................................................................................................
3 marks
14.
The figure 9 below shows an inverted test-tube which floats in water enclosed in a plastic bottle that is well corked.
When the sides of the plastic bottle are squeezed, the test-tube sinks. Explain this observation.
..........................................................................................................................................
1 marks
### SECTION B (55 Marks)
15.
The figure 10 shows microscope M focused on smoke particles inside a smoke cell P.
i) What causes the Brownian motion observed?
..............................................................................................................................
..............................................................................................................................
ii) Identify the lenses marked A and B.
..............................................................................................................................
iii) What should be incorporated in the set up to make the observation clearer?
..............................................................................................................................
iv) Explain what would be observed if the particle of the air have the same size with smoke particles.
..............................................................................................................................
b) The figure 11 below shows an oil patch made by a drop of oil placed on a dusty surface of water. 100 similar drops have a volume of 2mm3.
Calculate the approximate diameter of one molecule of oil (the diagram is drawn to scale).
D is the diameter of the oil film.
9 marks
16.
a) The figure 12 shows a graph obtained when 0.5kg hot liquid Naphthalene in an Aluminium can of mass 0.4kg was allowed to cool to room temperature.
i) State and explain the shape of the graph and the happenings in the Aluminium can at each stage of the cooling process.
..............................................................................................................................
..............................................................................................................................
b) State three factors that affect the rate at which Naphthalene cools.
..............................................................................................................................
c) Using kinetic theory of gases, explain how the rise in temperature of a gas causes a rise in it’s the pressure when the volume is kept constant.
..............................................................................................................................
..............................................................................................................................
12 marks
17.
The diagram in the figure 13 below shows a hydraulic jack system.
a) Name the parts labelled W and X
..............................................................................................................................
..............................................................................................................................
b) Explain briefly how the device is used to raise the load at the position shown.
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
c) Part W is left open to the atmosphere as indicated. Explain
..............................................................................................................................
..............................................................................................................................
..............................................................................................................................
d) State two ways by which the mechanical advantage of the jack may be increased.
..............................................................................................................................
..............................................................................................................................
e) One such hydraulic jack was used to lift air of mass 1200kg The cross section area of Q was 500cm2 and that of P was 5cm2. Determine the force exerted on the pump piston.
..............................................................................................................................
12 marks
18.
i) On the figure, indicate all the forces acting on the sphere
ii) Find the tension in the cable.
b) A piece of metal weighing 3N in air,2N when totally immersed in water and 1.5N when totally immersed in liquid Q.
Calculate
i) the up thrust in water
ii) the upthrust in liquid Q.
iii) the relative density of the metal.
iv) the relative density of liquid Q
12 marks
19.
a) Define the term angular displacement.
..............................................................................................................................
b) Figure 15 shows a device that may be used to control the speed of a vehicle.
Explain how the device may be used to control the speed.
..............................................................................................................................
..............................................................................................................................
Determine
i) The maximum angular velocity.
ii) Angular acceleration during the last 1.5 minutes.
iii) How many revolutions the car made.
10 marks
Back Top | 1,354 | 10,125 | {"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-2023-50 | longest | en | 0.645971 |
https://metanumbers.com/56977 | 1,597,257,069,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738913.60/warc/CC-MAIN-20200812171125-20200812201125-00361.warc.gz | 396,082,061 | 7,481 | 56977
56,977 (fifty-six thousand nine hundred seventy-seven) is an odd five-digits composite number following 56976 and preceding 56978. In scientific notation, it is written as 5.6977 × 104. The sum of its digits is 34. It has a total of 2 prime factors and 4 positive divisors. There are 56,500 positive integers (up to 56977) that are relatively prime to 56977.
Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 34
• Digital Root 7
Name
Short name 56 thousand 977 fifty-six thousand nine hundred seventy-seven
Notation
Scientific notation 5.6977 × 104 56.977 × 103
Prime Factorization of 56977
Prime Factorization 227 × 251
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 56977 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 56,977 is 227 × 251. Since it has a total of 2 prime factors, 56,977 is a composite number.
Divisors of 56977
1, 227, 251, 56977
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 57456 Sum of all the positive divisors of n s(n) 479 Sum of the proper positive divisors of n A(n) 14364 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 238.699 Returns the nth root of the product of n divisors H(n) 3.96665 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 56,977 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 56,977) is 57,456, the average is 14,364.
Other Arithmetic Functions (n = 56977)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 56500 Total number of positive integers not greater than n that are coprime to n λ(n) 28250 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5769 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 56,500 positive integers (less than 56,977) that are coprime with 56,977. And there are approximately 5,769 prime numbers less than or equal to 56,977.
Divisibility of 56977
m n mod m 2 3 4 5 6 7 8 9 1 1 1 2 1 4 1 7
56,977 is not divisible by any number less than or equal to 9.
Classification of 56977
• Arithmetic
• Semiprime
• Deficient
Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
Other numbers
• LucasCarmichael
Base conversion (56977)
Base System Value
2 Binary 1101111010010001
3 Ternary 2220011021
4 Quaternary 31322101
5 Quinary 3310402
6 Senary 1115441
8 Octal 157221
10 Decimal 56977
12 Duodecimal 28b81
20 Vigesimal 728h
36 Base36 17yp
Basic calculations (n = 56977)
Multiplication
n×i
n×2 113954 170931 227908 284885
Division
ni
n⁄2 28488.5 18992.3 14244.2 11395.4
Exponentiation
ni
n2 3246378529 184968909446833 10538973553552203841 600479096160743918248657
Nth Root
i√n
2√n 238.699 38.4798 15.4499 8.93593
56977 as geometric shapes
Circle
Diameter 113954 357997 1.01988e+10
Sphere
Volume 7.74796e+14 4.07952e+10 357997
Square
Length = n
Perimeter 227908 3.24638e+09 80577.6
Cube
Length = n
Surface area 1.94783e+10 1.84969e+14 98687.1
Equilateral Triangle
Length = n
Perimeter 170931 1.40572e+09 49343.5
Triangular Pyramid
Length = n
Surface area 5.62289e+09 2.17988e+13 46521.5
Cryptographic Hash Functions
md5 2c8dfb88b2177d60e9a531815b5cbc49 971bf40a449388393e8165e297964bfde91a5861 ee5eac000103583f7b14ee23e77b21dcae96e676843748eb289fb2b2c41f23d8 bf061fac472e7928c6b8a77c1f7562971d11e3ce95e9eb03d6f33b70f862067215c5c5e9327234d584d793ef3d5be154747b13d7442295e95e04db1d72142a72 6d87b04b8058f601c122039092aff9cf205c94a1 | 1,440 | 4,138 | {"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} | 3.5625 | 4 | CC-MAIN-2020-34 | latest | en | 0.818711 |
https://social.msdn.microsoft.com/Forums/en-US/da1943dc-7820-429e-8467-bbddcc59b683/power-pivot?forum=sqlkjpowerpivotforexcel | 1,670,337,239,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711108.34/warc/CC-MAIN-20221206124909-20221206154909-00603.warc.gz | 547,856,875 | 15,680 | Power Pivot
• Question
• When using power pivot do you need to export all the transactions into a table first then use the pivots to create the reports?
Or can the pivot link directly to the power pivot query code?
It's possible that table could be >1Million records which I believe would be a problem in excel.
Thanks
D
• Moved by Thursday, September 29, 2016 1:36 AM PowerPivot
Wednesday, September 28, 2016 3:47 PM
• That limitation of 1,0458,576 rows applies to excel worksheets. Power Pivot is not bound by that limitation as the storage mechanism is completely different. Although they look similar, an Excel Worksheet is not the same thing as a Power Pivot or Tabular table. I'm not sure of the upper bound but the limitation is really dictated by how much on-board memory you have. You should be able to import 10's of millions of rows into a power pivot table (via direct sql query) quite easily if you have enough memory.
1,048,576 rows
Thursday, September 29, 2016 11:54 AM
• Hello,
Your question is not clear for me, may can you rephrase it, please?
Power Pivot has a highly compression algorithm and kann easily handle a few million fact data, as Long as your Client machine have enough memory and you are using 64 bit MS Excel; see About the Relativity of Large Data Volumes
Olaf Helper
[ Blog] [ Xing] [ MVP]
Thursday, September 29, 2016 6:33 AM
• Hi David,
When creating a Pivot table from a PowerPivot data model, the data was load to the worksheet first, and then you can create table or chart from it.
For the row limitation, the max row numbers in the table is 1,999,999,997.
Object
Specification / Limit
Object name length
100 characters
Invalid characters in a Name
. , ; ' ` : / \ * | ? " & % \$ ! + = () [] {} < >
Number of tables per PowerPivot database
(2^31) - 1 = 2,147,483,647
Number of columns and calculated columns per table
(2^31) - 1 = 2,147,483,647
Number of calculated measures in a table
(2^31) - 1 = 2,147,483,647
PowerPivot memory size for saving a workbook
4GB = 4,294,967,296 bytes
Concurrent requests per workbook
6
Local cubes connections
5
Number of distinct values in a column
1,999,999,997
Number of rows in a table
1,999,999,997
String length
536,870,912 bytes (512 MB), equivalent to 268,435,456 Unicode characters (256 mega characters)
Regards,
Charlie Liao
TechNet Community Support
Thursday, September 29, 2016 7:05 AM
• Hello David,
If the data comes from a supported data source, like a SQL Server database, then you can load the data directly into the Power Pivot data model without a Loop way over an Excel sheet; an Power Pivot can load this amount of rows.
Olaf Helper
[ Blog] [ Xing] [ MVP]
Thursday, September 29, 2016 10:51 AM
All replies
• Hi,
Since the issue is more related to PowerPivot, I would move this thread into the following forum,
http://social.msdn.microsoft.com/Forums/en-US/sqlkjpowerpivotforexcel
Sorry for any inconvenience and have a nice day!
Regards,
Celeste
Thursday, September 29, 2016 1:35 AM
• Hello,
Your question is not clear for me, may can you rephrase it, please?
Power Pivot has a highly compression algorithm and kann easily handle a few million fact data, as Long as your Client machine have enough memory and you are using 64 bit MS Excel; see About the Relativity of Large Data Volumes
Olaf Helper
[ Blog] [ Xing] [ MVP]
Thursday, September 29, 2016 6:33 AM
• Hi David,
When creating a Pivot table from a PowerPivot data model, the data was load to the worksheet first, and then you can create table or chart from it.
For the row limitation, the max row numbers in the table is 1,999,999,997.
Object
Specification / Limit
Object name length
100 characters
Invalid characters in a Name
. , ; ' ` : / \ * | ? " & % \$ ! + = () [] {} < >
Number of tables per PowerPivot database
(2^31) - 1 = 2,147,483,647
Number of columns and calculated columns per table
(2^31) - 1 = 2,147,483,647
Number of calculated measures in a table
(2^31) - 1 = 2,147,483,647
PowerPivot memory size for saving a workbook
4GB = 4,294,967,296 bytes
Concurrent requests per workbook
6
Local cubes connections
5
Number of distinct values in a column
1,999,999,997
Number of rows in a table
1,999,999,997
String length
536,870,912 bytes (512 MB), equivalent to 268,435,456 Unicode characters (256 mega characters)
Regards,
Charlie Liao
TechNet Community Support
Thursday, September 29, 2016 7:05 AM
• Thats really why I wanted to know does the data needed to be renderd onto an excel table first as excel tables have limitations
Feature Maximum limit Worksheet size 1,048,576 rows
how can a table of 1,999,999,997 fit on a sheet of 1,048,576 ?
Does that mean that pivot can link directly to the sql query?
Thanks
D
Thursday, September 29, 2016 7:28 AM
• Hello David,
If the data comes from a supported data source, like a SQL Server database, then you can load the data directly into the Power Pivot data model without a Loop way over an Excel sheet; an Power Pivot can load this amount of rows.
Olaf Helper
[ Blog] [ Xing] [ MVP]
Thursday, September 29, 2016 10:51 AM
• That limitation of 1,0458,576 rows applies to excel worksheets. Power Pivot is not bound by that limitation as the storage mechanism is completely different. Although they look similar, an Excel Worksheet is not the same thing as a Power Pivot or Tabular table. I'm not sure of the upper bound but the limitation is really dictated by how much on-board memory you have. You should be able to import 10's of millions of rows into a power pivot table (via direct sql query) quite easily if you have enough memory.
1,048,576 rows
Thursday, September 29, 2016 11:54 AM | 1,530 | 5,750 | {"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-2022-49 | latest | en | 0.877296 |
https://puzzling.stackexchange.com/questions/61901/you-have-15-boxes-each-containing-a-real-number-ordered-least-greatest-how-do | 1,702,243,429,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679102637.84/warc/CC-MAIN-20231210190744-20231210220744-00696.warc.gz | 513,149,053 | 45,808 | # You have 15 boxes, each containing a real number, ordered least-greatest. How do you find the 5 that will yield the greatest product, the quickest?
As the title says, you have 15 boxes, a real number is in each, and they're ordered such that box 1 contains the lowest value, box 2 the second lowest, box 3 third lowest and so on, until box 15, which contains the highest value.
You want to be sure to pick the 5 boxes with the greatest total product, such that $$\text{box} * \text{box} * \text{box} * \text{box} * \text{box} = \text{maximum amount}$$ and you want to do this with as few operations and comparison as possible. An operation is anything like $x * x$ or $x - y$, and a comparison is anything like $a > b$. What do you do?
You want to be able to know that, no matter what set of 15 in-order (least-greatest) boxes you are given, you can follow this simple formula (whatever you come up with) to always come out with the greatest product of 5 of them. And you want to come out with the minimum total of operations + comparisons that will let you always do that.
And numbers are not necessarily distinct! And all 4 of the basic operators ($+, -, *, /$) are allowed, though I am pretty sure you will only need $*$ ;)
• To be honest, I don't see what's the point of this question. If the boxes are ordered (and assuming we know the order) then obviously the 5 biggest numbers will produce the biggest product. If the boxes are not ordered you can't tell the answer without ordering them. Mar 14, 2018 at 7:30
• Wouldn't that just be a matter of picking boxes 11 - 15 since they'll have the 5 largest numbers? Mar 14, 2018 at 7:55
• Real numbers include negative numbers... ;) Mar 14, 2018 at 7:57
• @Phylyp, not if box11 < 0 < box12.
– Bass
Mar 14, 2018 at 7:57
• Agree with phylyp, i think you could edit your question to include all of the extra details you mentioned here (numbers doesnt have to be distinct, what operators are allowed, etc). What i do want to know is whether we can sneak in a negative operator in there somewhere (e.g. box1 * -box2) which would trivialize the issue of negative numbers. Mar 14, 2018 at 8:21
If we can, we must pick an even number of negative numbers. First, get the special case, where we cannot, out of the way:
if $\boxed{15} < 0$, pick $\boxed{11}\times\boxed{12}\times\boxed{13}\times\boxed{14}\times\boxed{15}$
Then we need the largest absolute value available, with the restriction that there must be an even number of negative values. So the answer is one of these:
$\boxed{11}\times\boxed{12}\times\boxed{13}\times\boxed{14}\times\boxed{15}$
$\boxed{1}\times\boxed{2}\times\boxed{13}\times\boxed{14}\times\boxed{15}$
$\boxed{1}\times\boxed{2}\times\boxed{3}\times\boxed{4}\times\boxed{15}$
Then, note that the boxes are ordered, so out of all the possible pairs of two consecutive boxes, at most one contains both a positive and a negative value. Therefore, when comparing products of two such pairs, at least one of the products is guaranteed to be non-negative, so it's safe to multiply the larger one into our solution without further checks.
So, calculate $\boxed{1}\times\boxed{2}$ and $\boxed{13}\times\boxed{14}$ and compare them. Disqualify the solution that doesn't contain the pair with the larger result.
Then, we need to compare the two remaining possibilities. Luckily, they both contain $\boxed{15}$ and the "winner" of the previous phase, and one of the possibilities contains the losing pair from the previous phase (we already know the multiplication result), so we will need to do only one more multiplication and comparison to distinguish between the two remaining cases.
So the total number of operations required for choosing the boxes is
3 multiplications and 3 comparisons.
If we want to know what the actual maximum product is, we can get that by using
two more multiplications, because we already have two of the pairwise products calculated, so we can multiply those by each other, and then multiply the result by $\boxed{15}$ to get the maximum product.
(For completeness's sake, when we detected the "all numbers negative" case, we hadn't done any multiplication yet, so in that case we can get the maximum product by using 4 multiplications.)
• I like the box notation :-) Edit: Also, thank you for a clear explanation readable by non-math folks. Mar 14, 2018 at 10:24
• you do not address [15] = 0 that makes every other combination with even # of negatives greater than any product that uses box 15. Mar 14, 2018 at 13:01
• @Mindwin - how do you get an even number of negatives from any other combination of 5 boxes since we know all other boxes contain a negative? Mar 14, 2018 at 13:10
• @Mindwin, we have to take exactly 5 boxes. If $\boxed{15}$ is 0, then all the numbers that aren't 0 are negative. If you want a product that is bigger than 0, you need to take 2 or 4 negative values, but the rest of the numbers you pick (there will have to be at least one) are going to be zeroes anyway -> no need for a special case.
– Bass
Mar 14, 2018 at 13:24
To find out which boxes to pick I need
6 operations
Use the following strategy:
Let's call $a = x_{1} \times x_{2}, b = x_{3} \times x_{4}, c = x_{11} \times x_{12}, d = x_{13} \times x_{14}, e = x_{15}$.
The best solutions can only be case 1: $c \times d \times e$ or case 2: $a \times d \times e$ or case 3: $a \times b \times e$.
If $x_{15} < 0$ then case 1 (1 comparison needed).
Now if $a < d$ we need to check $a < c$. If true, then case 1, else case 2 (3 multiplications and 2 additional comparisons needed).
Now if $a >= d$ we need to check $b < d$. If true, then case 2, else case 3 (3 multiplications and 2 additional comparisons needed).
• Beat you to it by all of 8 seconds :-)
– Bass
Mar 14, 2018 at 10:43
• @Bass Was busy writing a program to verify (I never trust my own logic).
– w l
Mar 14, 2018 at 10:54
Here is what I could come up with taking into account a few samples.
Let the boxes be $x_1, x_2, ... x_{15}$ then follow the following procedure:
Step 1: if $x_{15}$ is negative then take the boxes $x_{11}, x_{12}, x_{13}, x_{14}$ and $x_{15}$ else continue to next step.
Step 2: if $x_{13}$ is negative then take the boxes $x_1, x_2, x_3, x_4$ and $x_{15}$ else continue to next step.
Step 3: if $x_{11}$ is negative then go to step 4-a else go to step 4-b.
Step 4-a: take $x_1, x_2, x_3, x_4, x_{15}$ if $(x_3*x_4) > (x_{13}*x_{14})$ else take $x_{1}, x_{2}, x_{13}, x_{14}, x_{15}$. Stop.
Step 4-b: take $x_1, x_2$ and go to step 5 if $(x_1*x_2) > (x_{11}*x_{12})$ else take $x_{11}, x_{12}, x_{13}, x_{14}, x_{15}$. Stop.
Step 5: take $x_3, x_4, x_{15}$ if $(x_3*x_4) > (x_{13}*x_{14})$ else take $x_{11}, x_{12}, x_{13}, x_{14}, x_{15}$. Stop.
Case for all negatives is trivial.
Let's see some other examples:
Example 1: All positives
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Step 1 and 2 will not satisfy. In step three, 11 is positive so we go to step 4-b. Now, $(x_1*x_2) < (x_{11}*x_{12})$ so we will take 11, 12, 13, 14, 15 for max product
Example 2: 1 positive all Negatives -14 -13 -12 -11 ... -1 1
Step 1 will not satisfy but Step 2 will as $x_{13}=-2<0$ so we take -14, -13, -12, -11, 1 which will give max product
Example 3: 4 non-negatives and other negatives
-11 -10 ... -1 0 1 2 3
Step 1 and 2 will not satisfy. In step 3, -1 is negative so we go to step 4-1. In step 4-a, we see $(x_3*x_4) > (x_{13}*x_{14})$ so we will take -11, -10, -9, -8, 3 to get the max product.
In my opinion this involves quite less number of computations, if not the least. Including every multiplication and every comparison as one operation, this takes 9 steps in worst case scenario
I have written a code in C++ to test this which can be found at https://ideone.com/pppGuD
• Nice. I thought I saw a problem in step 2, but realized it actually works as expected :-) Mar 14, 2018 at 9:05
• I think we should include the minimum number of operations+comparisons we need in the worst case scenario for your formula - in this case, it's 7 (1 from step 1, 1 from step 2, 3 from step 3, and 2 from the last step (assuming we can store one of the multiplication results from step 3)) Mar 14, 2018 at 9:10
• Don't think this works. For example, in the case -14 -14 -14 -14 1 1 1 1 1 1 1 1 15 15, we end up with -14,-14,1,15,15 when a better solution is -14,-14,-14,-14,15..
– ffao
Mar 14, 2018 at 9:19
• Phylip: :-) @Votbear I have added that now. Actually it would be 9 considering every comparison and multiplication as one operation Mar 14, 2018 at 9:25
• @ffao oh! Thanks for the example. I will check how can I modify my method for properly working. Mar 14, 2018 at 9:27 | 2,617 | 8,658 | {"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} | 3.984375 | 4 | CC-MAIN-2023-50 | longest | en | 0.947099 |
https://www.physicsforums.com/threads/general-application-of-boyles-law.383343/ | 1,590,417,867,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347388758.12/warc/CC-MAIN-20200525130036-20200525160036-00268.warc.gz | 815,555,494 | 15,025 | # General Application of Boyle's Law
Hello,
I am Year 10 student and would like to get help on the question below based on Boyle's Law.
A piston with a block of mass 120 grams placed on it traps 200cm³ of air in a container. When the block is removed, the volume becomes 250cm³. Find the mass of the piston.
In the question above i only know that you have to use the formula P1V1=P2V2, but i don't know how to apply it. Please can anyone help?
Related Introductory Physics Homework Help News on Phys.org
The formula means that if you have a quantity of gas with pressure P1 and volume V1, and you compress it, or let it expand, and afterwards the pressure and volume are P2 and V2, then
P1V1 = P2V2
you know the volume before and after removing the block. What can you say about the pressure, and how is this related to the mass of the block and the piston. | 216 | 862 | {"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-2020-24 | longest | en | 0.938523 |
https://www.brainscape.com/flashcards/300934-leasehold-improvement-3f-15281720/packs/21924342 | 1,726,778,602,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652067.20/warc/CC-MAIN-20240919194038-20240919224038-00459.warc.gz | 616,360,508 | 18,590 | # 300934 Leasehold Improvement 3F Flashcards
1
Q
On January 1, 20X1, Bay Co. acquired a land lease for a 21-year period with no option to renew. The lease required Bay to construct a building in lieu of rent. The building, completed on January 1, 20X2, at a cost of \$840,000, will be depreciated using the straight-line method. At the end of the lease, the building’s estimated market value will be \$420,000. What is the building’s carrying amount in Bay’s December 31, 20X2, balance sheet?
\$819,000
\$798,000
\$820,000
\$800,000
Question #300934
A
\$798,000
This building is treated as a leasehold improvement. Although the land lease is for 21 years, the building will be in use for 20 years; thus, the depreciation period for the building is 20 years:
Cost ÷ Life = \$840,000 ÷ 20 years = \$42,000/year
Building cost − Depreciation = Carrying value of building
\$840,000 − \$42,000 = \$798,000
Note: Leasehold improvements are depreciated fully without regard to salvage value if there is no option to renew (as is stated in this problem) and they are immovable (e.g., a building). In this case, because the land has a 21-year lease, any items permanently affixed to the land will revert to the owner at the end of the lease. Therefore, their market value at lease-end is irrelevant to Bay Co. because Bay Co. will no longer have the rights to the affixed property upon lease termination.
2
Q
2362.04
A
Leasehold improvements: Considerable judgment should be applied when assessing how to account for leasehold improvements. In general, the lessee should:
• separately capitalize leasehold improvements (i.e., exclude leasehold improvement payments from the fixed lease payments if the lessee is not required to make such payments under the lease).
• include leasehold improvement payments in the fixed lease payment if the lessee is required to make such payments under the lease, provided the leasehold improvements are not specialized and could be used by a future tenant (i.e., would likely be considered an asset of the lessor).
• amortize the leasehold improvement depending upon the lease terms:
• If the asset is to be returned to the lessor, the lessee should amortize over the shorter of the useful life of the leasehold improvements or the remaining lease term.
• If the lessee is reasonably certain to exercise a purchase option, the lessee should amortize over the leasehold improvement useful life.
3
Q
FASB ASC 842-20-35-12
A
4
Q
Carrying Amount (Book Value)
A
The carrying amount or book value is the net amount at which an item is reported in the financial statements of the enterprise. For a receivable, the FASB ASC Glossary indicates the carrying amount is the “face amount increased or decreased by applicable accrued interest and applicable unamortized premium, discount, finance charges, or issues costs and also an allowance for uncollectible amounts and other valuation accounts.”
For a payable, the FASB ASC Glossary indicates the carrying amount is the “face amount increased or decreased by applicable accrued interest and applicable unamortized premium, discount, finance charges, or issue costs.”
5
Q
Deferred
A
Deferred is delayed or postponed recognition of revenue or expense until a future period.
6
Q
Depreciation
A
Depreciation is the process of systematic, rational allocation of the cost of operational assets to the accounting periods benefited. Depreciation is not a process of valuation (FASB ASC 360-10-35-4), does not represent a reserve to replace the asset, and does not mean that cash will be available to replace the asset. Depreciation allowed for tax purposes often differs from depreciation allowed for accounting.
Accounting depreciation attempts to match the cost of the asset to the revenues generated over the life of the asset. It represents accrual accounting and has no effect on cash flows (a noncash expense). Depreciation expense must be added back to accounting income when reconciling to cash from operations using the indirect method.
Computation of depreciation requires the following:
• Acquisition cost
• Estimated useful life
• Estimated residual (salvage) value
• Depreciation method (four GAAP alternatives):
1. Straight-line
2. Sum-of-the-years’-digits
3. Double-declining balance
4. Units of production—units of product and machine hours
Factors which cause the need for depreciation include the following:
• Physical factors:
1. Wear and tear
2. Effects of time and other elements
3. Deterioration and decay
• Functional factors:
2. Obsolescence
In the macroeconomic sense, depreciation is the part of business earnings/gross profit that is considered the replacement of capital stock used or worn out during the period. It is not included in net profit and is not a factor payment. (It is not a claim on the value of output by a factor of production.) Depreciation represents replacement investment, the amount that must be reinvested to maintain the existing level of capital stock, and the amount by which capital contributes to current production. It is a component of GNP (approximately 10%) and is computed by the income approach to national income accounting. Depreciation is the difference between gross and net investment.
In the foreign exchange sense, depreciation is the decline in the value of one currency against or in relation to another, in the sense that it now takes more of a particular currency to buy a unit of a foreign currency. Devaluation is the official change in the value of a country’s currency.
Example: Country A has an inflation rate of 5% and Country B has an inflation rate of 10%. The goods of Country A become relatively cheaper because the relative prices have changed, thus:
• increasing the demand in Country B for Country A’s goods,
• increasing the demand for Country A’s currency (to be able to import Country A’s goods), and
• decreasing the demand for (i.e., depreciating) Country B’s currency by approximately 5% (10%–5%).
7
Q
Fair Market Value (FMV)
A
The fair value of an investment is the amount that the asset could reasonably expect to receive for it in a current sale between a willing buyer and a willing seller, that is, other than in a forced or liquidation sale. Fair value shall be measured by the market price if there is an active market for the investment. If there is no active market for the investment but there is a market for similar investments, selling prices in that market may be helpful in estimating fair value. If a market price is not available, a forecast of expected cash flows, discounted at a rate commensurate with the risk involved, may be used to estimate fair value. The fair value of an investment shall be reported net of the brokerage commissions and other costs normally incurred in a sale.
FASB ASC Glossary
For tax purposes, the fair market value is usually referred to as the sale price between a willing seller and a willing buyer when neither is compelled to buy or sell.
8
Q
Gains
A
Gains are increases in equity (net assets) from peripheral or incidental transactions of an entity and from all other transactions and other events and circumstances affecting the entity except those that result from revenues or investments by owners. (SFAC 6.82–.89)
Gains are similar to revenues—the distinction depends on the nature of the entity, its operations, and its other activities. The primary purpose of distinguishing between revenues and gains is presentation and display.
9
Q
Lease
A
A lease is a contract, or part of a contract, that conveys the right to control the use of identified property, plant, or equipment (an identified asset) for a period of time in exchange for consideration.
FASB ASC Glossary
10
Q
Lease Term
A
The lease term is the noncancelable period for which a lessee has the right to use an underlying asset, together with all of the following:
Periods covered by an option to extend the lease if the lessee is reasonably certain to exercise that option
Periods covered by an option to terminate the lease if the lessee is reasonably certain not to exercise that option
Periods covered by an option to extend (or not to terminate) the lease in which exercise of the option is controlled by the lessor
FASB ASC Glossary
11
Q
Straight-Line Method
A
The straight-line method is a depreciation method based on an equal allocation of the cost of operational assets with the passage of time. It assumes that the useful life of the asset is used up evenly over time and charges a fixed amount per period to expense.
The computation for the straight-line (SL) method is:
• Straight-line depreciation expense per year = (Cost - Residual value) ÷ Useful life (years)
The advantage of the straight-line method is that it is simple to compute. The disadvantage is that the passage of time may not be representative of the use of the asset’s benefits and may not match cost to revenue.
• Straight-line rate = 1 ÷ Useful life (years)
The straight-line method is a method of allocation such that a constant dollar amount is recognized as revenue or expense each period. The straight-line method is commonly used for depreciation.
Example: An asset costing \$100,000, with a salvage value of \$5,000, to be depreciated over 10 years by the straight-line method of depreciation would be depreciated at \$9,500 per year ((\$100,000 - \$5,000) ÷ 10). | 2,069 | 9,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-38 | latest | en | 0.934923 |
http://rutwick.com/index.php/ebooks/information-theory-and-statistical-mechanics-i | 1,638,722,003,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00593.warc.gz | 68,094,064 | 7,999 | ### Download Information theory and statistical mechanics I by Jaynes E.T. PDF
• April 20, 2017
• Data Processing
By Jaynes E.T.
The actual evaluate, Vol. 106, No. four
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Additional resources for Information theory and statistical mechanics I
Example text
2. Range operator. 1 to 10, 10 to 1 Results: 1 2 3 4 5 6 7 8 9 10 Parenthesized Expressions Expressions take precedence on one another. For example, addition has a higher precedence than the comma. Parentheses allow you to change precedence. If the parentheses are empty, the empty sequence is produced. 3. Empty sequence. ( 2 + 3 ) * 5, () Results: 25 Arithmetics JSONiq supports the basic four operations, as well integer division and modulo. You should keep in mind that, as is the case in most programming languages, multiplicative operations have precedence over additive operations.
Comparison is only possible between two compatible types, otherwise, an error is raised. 8. Equality comparison. 50 Comparison 1 + 1 eq 2, 1 lt 2 Results: true true null can be compared for equality or inequality to anything - it is only equal to itself so that false is returned when comparing if for equality with any non-null atomic. True is returned when comparing it with non-equality with any non-null atomic. 9. Equality and non-equality comparison with null. 1 eq null, "foo" ne null, null eq null Results: false true true For ordering operators (lt, le, gt, ge), null is considered the smallest possible value (like in JavaScript).
For example, addition has a higher precedence than the comma. Parentheses allow you to change precedence. If the parentheses are empty, the empty sequence is produced. 3. Empty sequence. ( 2 + 3 ) * 5, () Results: 25 Arithmetics JSONiq supports the basic four operations, as well integer division and modulo. You should keep in mind that, as is the case in most programming languages, multiplicative operations have precedence over additive operations. Parentheses can override it, as explained above. | 616 | 2,835 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-49 | latest | en | 0.886138 |
https://mmerevise.co.uk/gcse-maths-revision/speed-distance-time-gcse-revision-and-worksheets | 1,726,891,624,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725701427996.97/warc/CC-MAIN-20240921015054-20240921045054-00047.warc.gz | 365,797,645 | 57,482 | # Speed Distance Time
GCSEKS3Level 4-5AQACambridge iGCSEEdexcelEdexcel iGCSEOCRWJEC
## Speed Distance Time
Speed is a measure of how quickly something moves, so it is calculated by dividing distance by time.
Make sure you are happy with the following topics before continuing.
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Speed Distance Time – Formula
Speed, distance and time are all related by the formula,
$s = \dfrac{d}{t}$
where $s$ is speed, $d$ is distance, and $t$ is time. You can rearrange this formula to find the other two, for example, if we multiply both sides by $t$ and divide both sides by $s$, we get,
$t = \dfrac{d}{s}$
So, we can calculate the time by dividing the distance by the speed.
Typically, we measure distance in metres (m), kilometres (km), or miles, and we measure time in seconds (s), minutes (mins) or hours (h). As a result, the units we use to measure speed are compound units e.g m/s or km/h.
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Speed Distance Time – Formula Triangle
A handy way of remembering how to calculate one of speed, distance and time is to use one of the triangles below.
The horizontal line means divide and the $\times$ symbol means multiply.
We then cover up the one we want to find (represented by a red circle) and complete the calculation using the other two values from the triangle.
$s = \dfrac{d}{t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, d = s \times t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, t = \dfrac{d}{s}$
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Example 1: Calculating Speed
A truck travels $110$ miles in $2$ hours. What is the average speed of the truck?
[2 marks]
We’re looking for speed, so constructing the triangle and covering up the $s$, we find we must divide the distance by the time. So
$\text{Speed } = 110 \div 2 = 55$ mph
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Example 2: Calculating Distance
Jesse throws a ball that moves at an average speed of $35$ metres per second and travels for a total of $4.5$ seconds. Work out the distance travelled by the ball.
[2 marks]
We’re looking for distance, so covering up $d$, we must multiply the speed by the time. So
$\text{Distance }=35 \times 4.5=157.5$ m
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Example 3: Calculating Time
A car travels at an average speed of $50$ mph. How long will it take for the car to travel $12.5$ miles? Give your answer in minutes.
[2 marks]
We’re looking for time, so covering $t$, we find we must divide the distance by the speed. So
$\text{Time } = \dfrac{12.5}{50} = 0.25$ hours
$0.25$ hours needs to be converted into minutes by multiplying by $60$,
$\text{Time }=0.25\times 60=15$ minutes
Level 4-5GCSEKS3AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
## Speed Distance Time Example Questions
Question 1: Skyler runs a new personal best in the $100$-metre sprint. Her average speed over the course of the race is $8.5$ metres per second. How long, to $2$ decimal places, does it take her to run the race?
[2 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
We are calculating the time taken, so by covering up the $t$ we can see from the triangle above that we have to divide distance, $d$, by speed, $s$.
Therefore:
$\text{Time }=\dfrac{d}{s}=\dfrac{100}{8.5}=11.76$ s ($2$ dp).
Gold Standard Education
Question 2: Gustavo is driving a bus along a motorway with a speed limit of $70$ miles per hour. In $30$ minutes, he travels $36$ miles. Is his average speed during this period exceeding the speed limit?
[2 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
We are calculating speed, so by covering up the $s$ we can see from the triangle above that we have to divide distance, $d$, by time, $t$.
However, before we do that, we have to ensure that the units match. The speed limit is in ‘miles per hour’, but the time we have been given is in ‘minutes’. This is easy, as $30$ minutes is half an hour, or $0.5$ hours. So, Gustavo’s average speed can be calculated as follows:
$\text{speed }=\dfrac{d}{t}=\dfrac{36}{0.5}=72$ mph
So, yes, Gustavo is exceeding the speed limit.
Gold Standard Education
Question 3: On the first part of a journey, a motorcyclist travels for $3$ hours at an average speed of $55$ miles per hour.
On the second part of a journey, the same motorcyclist travels for $90$ minutes at an average speed of $48$ miles per hour.
How far does the motorcyclist travel in total?
[4 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
In order to calculate the total distance travelled, we need to rearrange the speed / distance / time formula.
Since
$\text{ speed} = \text{ distance} \div \text{ time}$
then
$\text{ distance} = \text{ speed} \times \text{ time}$
The distance of the first part of the journey can be calculated as follows:
$\text{Distance } = 3$ hours $\times \, 55$ mph $= 165$ miles
For the second part of the journey, we need to convert the units so that they match. The speed has been given in miles per hour whereas the time has been given in minutes. We can either convert the minutes into hours or we can convert the speed form miles per hour to miles per minute. Converting minutes to hours is probably the easier option:
$90$ minutes $= 1\frac{1}{2}$ hours or $1.5$ hours
The distance of the second part of the journey can be calculated as follows:
$\text{Distance } = 1.5$ hours $\times \, 48$ mph $= 72$ miles
Therefore the total distance travelled is
$165 + 72 = 237$ miles
Gold Standard Education
Question 4: The distance from the Sun to Mars is approximately $210$ million kilometres. What is the speed of light in kilometres per second if it takes $11$ minutes and $40$ seconds for light to reach Mars from the Sun?
[3 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
We know that to calculate speed, we need to divide distance by time as per the formula:
$\text{ speed} = \text{ distance} \div \text{ time}$
The only issue we have in this question is that the distance has been written as ‘$210$ million’, which is not that helpful as ‘million’ has been written as a word and not in figures. The first thing we will need to do is convert $210$ million into figures, hopefully remembering that a million has $6$ zeros:
$210$ million $= 210,000,000$
The time taken has been expressed in minutes and seconds which causes an additional problem. We have been asked to give an answer in kilometres per second, so we need to convert the time from minutes and seconds to seconds.
$11$ minutes $= 11 \times 60$ seconds $= 660$ seconds
$11$ minutes and $40$ seconds $= 660$ seconds $+ \, 40$ seconds = $700$ seconds
We are now in a position to calculate the speed of light as follows:
$\text{ Speed of light} = 210,000,000$ km $\div \, 700$ seconds $= 300,000$ km/s
Gold Standard Education
Question 5: Voyager $1$ became the first spacecraft to leave the solar system after $35$ years of space travel. Given that it is travelling at a speed of $17$ km/s, work out an approximation for the distance travelled. Give your answer in standard form to $3$ significant figures.
[3 marks]
Level 4-5GCSEKS3 AQAEdexcelOCRWJECCambridge iGCSEEdexcel iGCSE
We know that to calculate distance, we need to multiply speed by time as per the formula:
$\text{ speed} = \text{ distance} \div \text{ time}$
Hence converting $35$ years to seconds:
$35$ years $=35\times365\times24\times60\times60=1.104\times10^9$ seconds
The calculation becomes:
$\text{ distance} = 17 \times (1.104\times10^9) =1.88\times10^{10}$ km
Gold Standard Education
## Speed Distance Time Worksheet and Example Questions
### (NEW) Speed Distance Time Exam Style Questions - MME
Level 4-5GCSEKS3NewOfficial MME
Level 4-5GCSEKS3
Level 4-5GCSEKS3
## Related Topics
#### Rearranging Formulae
Level 4-5Level 6-7GCSEKS3
Level 6-7GCSE | 2,317 | 7,989 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 86, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.865889 |
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# Select the sentence that properly corrects the error below.I did my homework, I scored well on the test.O Because I did my homework. I scored well on the test.O After I did my homework. I scored well on the test.O i did my homework. Scored well on the test.O i did my homework, so I scored well on the test.
+5
Answers (2)
1. 6 November, 14:19
0
It is the last one yet again!
Explanation:
If you notice, the first answer is incorrect because it uses a period but then continues like the sentences are connected with a comma.
The second is the same mistake, since there is a period, the reader would figure the subject they were talking about, in this case, homework, is done being talked about.
The third one is kind of the same mistake, but with more depth. The "I" at the beginning is not capitalized and the period placement makes it so there is no filler word on why you scored well on your test, for example,
"I did my homework, therefore scored well on my test."
The last one is correct, even though the I" is not capitalized, I'm going to assume that was an accidental mistake and that in the original problem, the "I" is capitalized. The last sentence has a filler word, just like the one above was missing, therefore it is the right answer.
2. 6 November, 15:00
0
I did my homework, so I scored well on the test.
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Get an answer to your question ✅ “Select the sentence that properly corrects the error below. I did my homework, I scored well on the test. O Because I did my homework. I ...” in 📙 English if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. | 406 | 1,699 | {"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} | 2.75 | 3 | CC-MAIN-2024-18 | latest | en | 0.979897 |
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1. This pack includes four engaging resources for secondary math. The included items are engaging but rigorous! These resources are non-religious. Candy Cane Equations: Students solve multi-step equations to reveal world records about candy canes.Winter Order of Operations: Students simplify difficu
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Product Description
Students find missing angles and solve equations using angle relationships with these fun snowflake worksheets! Vertical angles and supplementary angles are drawn on snowflakes with missing measures.
Six sheets are included:
• Vertical Angles
• Supplementary Angles
• Mixed Vertical and Supplementary Angles
• Vertical Angles with Equations
• Supplementary Angles with Equations
• Mixed Angles with Equations
The snowflake sheets can be printed individually for practice worksheets OR use them for math centers. Answer recording sheets are provided for centers or stations.
Thank you for your interest in this resource from Rise over Run.
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\$3.00 | 425 | 1,930 | {"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-2019-18 | latest | en | 0.91217 |
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Your mail-order company advertises that it ships 90% of its orders within three working days. You...
Your mail-order company advertises that it ships 90% of its orders within three working days. You select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals that 86 of these orders were shipped on time. Note: It could be assumed that it satisfies all four conditions of the Binomial experiment.
a) If the company really ships 90% of its orders on time, what is the probability that 86 or fewer in an SRS of
100 orders are shipped on time? (Hint: Consider Normal approximation) [4]
b) A critic says, “Aha! You claim 90%, but in your sample the on-time percent is only 86%. So the 90% claim is wrong.” Explain in simple language why your probability calculation in (a) shows that the result of the sample does not refute the 90% claim. [1] | 216 | 891 | {"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-2024-22 | latest | en | 0.968801 |
http://slideplayer.com/slide/5742580/ | 1,716,274,340,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058385.38/warc/CC-MAIN-20240521060250-20240521090250-00495.warc.gz | 24,948,896 | 18,239 | # Expressas a decimal. Expressas a decimal. Convertto a decimal..
## Presentation on theme: "Expressas a decimal. Expressas a decimal. Convertto a decimal.."— Presentation transcript:
Expressas a decimal.
Expressas a decimal.
Convertto a decimal..
. Convertto a decimal.
Convert decimal to base 10 fraction; simplify
Express 0.375 as a percent Move decimal point 2 places to the right and add the % sign. Convert 2.867 to a percent
Convert decimals to percents Make sure there are at least two digits at the decimal points; add zero where needed (this does not change the value of the number).
Express 56.25 % as a decimal Move decimal point 2 places to the LEFT and remove the % sign.
Express the percents as decimals
Set up an “ out of 100 ” proportion Express as a percent. Check to see how you would convert the first denominator to 100. x 25 Convert numerator
Express as a percent. x 20
Convert fractions to percents. x 2 x 4 14 24 x 10 30
Or you can express fractions as percents by long division Convert decimal to percent by moving 2 decimal places to the right and adding % sign
Express fractions as percents by long division
Express percents to fractions by dividing by 100 | 283 | 1,199 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2024-22 | latest | en | 0.781608 |
http://mathhelpforum.com/discrete-math/162802-induction.html | 1,480,802,702,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541140.30/warc/CC-MAIN-20161202170901-00464-ip-10-31-129-80.ec2.internal.warc.gz | 176,805,225 | 13,892 | 1. ## Induction.
Hello
wondering if anyone is familiar with ex 3 in chapter 6 of
http://www.cs.sfu.ca/CC/882/jim/ProblemSets.pdf
I appreciate any hints, as I am struggling with proving even the base case.
Thanks
2. Originally Posted by tt11
Hello
wondering if anyone is familiar with ex 3 in chapter 6 of
http://www.cs.sfu.ca/CC/882/jim/ProblemSets.pdf
I appreciate any hints, as I am struggling with proving even the base case.
Thanks
Could you possibly just write out the problem?
3. Thanks for the post. The question might contain some special char, but you may refer to it in the second link (http://www.cs.sfu.ca/CC/882/jim/ProblemSets.pdf
)
The first link is for background if required.
4. I see that you are new to this forum, welcome.
But, if you really expect help, we must be able to read the question.
Therefore, why not learn to post in symbols? You can use LaTeX tags.
It does require some effort on your part. But the reward is worth the effort.
5. Thank you .... I will have a look at latex codes
6. We have a ranked db, and it is made up of antecedents and consequents like
a1, a2 --> q1
a2, not(a3) --> q2
….
it is ranked in the sense that there is function f which assigns to atoms numbers such that for every rule, and every i (between 1 and n), f(q) > = f(ai) and f(not(ai)) = 1+f(ai); the conclusion of the rule assigned g higher than all negative atoms.
It is possible to construct a set of atoms that cannot be derived from this db through reasoning, in an iterative fashion. We can construct a set, Sk = {not(a3), not(a4), ..} if we were to add these to the
DB and perform logical reasoning, , the set of atoms we could not derive would be the ones in Sk.
starting with S0={} and then building list gradually, until Sk is equal to Sk+1 (sequence converges).
S0 = {}
Sk+1 = {not(q) |DB union Sk |/= q }
So in the example db above, DB union (not(a3)) would make q2 derivable. Now we define two natural numbers k, r. S(k, r) = {not(q) ∈ Sk | f(q) < r}. It is possible to verify that for any k, ai, where f(p)= r
DB Union Sk |= p iff DB Union S(k, r) |= p., meaning, to prove p, we only need to look at the negative atoms (not(ai)) that have lower f(ai) than p.
The question is that, for a r < k, we should show S(k+1,r) = S(k+2, r)
For base case, I assume k=1, r=0. For induction, I should show if S(k+1,r) = S(k+2, r) then S(k+2,r) = S(k+3, r). After working on this for a while, I have some ideas about the base case, although not very clear. Can someone recommend some hints or reading material on how to begin transforming induction case to goal case?
Thanks (I know it is a very long question)
7. a1, a2 --> q1
a2, not(a3) --> q2
So in the example db above, DB union (not(a3)) would make q2 derivable.
Why is that? To derive q2, you also need a2, but there is no rule with head a2.
The base case is S(2, 0) = S(3, 0), and it is trivial because by the definition of S(k, r), the set S(k, 0) is empty for any k (f(q) < 0 is impossible).
I'll think more about the induction step tomorrow.
8. First of all thank you very much for the reply.
Yes, you are right about a2. I forgot to mention that, or other facts. (like a1, ...) . I will go through the base case explanation that you provided and looking forward to your help for the induction step.
Thanks again
9. Could you explain why f(q) < 0 is impossible?
Sorry I am missing the point
10. I assumed that f maps atoms to natural numbers, though the definition does not say so. Since the DB is probably finite, it does not matter, but if f maps atoms to integers, I believe the induction's base case should deal with k that is the minimum value of f on any atom.
11. OK, let's assume that stratification ranks of atoms are nonnegative integers. I'll write S(k) for $S_k$. Each S(k) is the union of $\emptyset=S(0)\subseteq S(1)\subseteq\dots$. The program DB is also a union of nested sets. Let $DB(i)=\{p\gets\vec{q}\in DB\mid f(p). Then DB is the union of $\emptyset=DB(0)\subseteq DB(1)\subseteq\dots$.
We claim that $S(\cdot,r)$ stabilizes after r steps, i.e., S(r, r) = S(r + 1, r) = S(r + 2, r) = ... Why? Suppose that f(p) = 0 for some p. Then for any k >= 0, (not p) is in S(k + 1), by definition, iff $DB\cup S(k)\not\models p$. What rules from $DB\cup S(k)$ determines if p is implied? Since f(p) = 0, these rules don't have "not" in them, i.e., they are in DB(0). This also means that S(k), which consists of negative literals (i.e., negations of atoms) don't play a role in whether p is implied. Thus, $DB\cup S(k)\not \models p$ iff $DB(0)\not\models p$ and whether (not p) is in S(k + 1) does not depend on k. All in all, $S(k + 1, 1)=\{\mathbf{not}\;p\mid f(p)=0\text{ and }DB(0)\not\models p\}$ for any k >= 0, so S(1, 1) = S(2, 1) = ...
Now, suppose that r >= 1 and it has been shown that S(r, r) = S(r + 1, r) = S(r + 2, r) = ... We would like to show that S(r + 1, r + 1) = S(r + 2, r + 1) = ... Suppose that f(p) < r + 1. By definition, (not p) is in S(r + 1) iff $DB\cup S(r)\not\models p$. Again, concerning the DB part of $DB\cup S(r)$, only its subset DB(r + 1) is responsible for implying p, and if (not q) occurs in DB(r + 1) for some q, then f(q) < r. Therefore, of the S(r) part of $DB\cup S(r)$, only its subset S(r, r) is responsible for implying p. So, we have
(not p) is in S(r + 1) iff, by definition,
$DB\cup S(r)\not\models p$ iff
$DB(r+1)\cup S(r,r)\not\models p$ iff, by assumption that S(r, r) = S(r + 1, r),
$DB(r+1),S(r+1,r)\not\models p$ iff
$DB, S(r+1)\not\models p$ iff
(not p) is in S(r + 2).
The facts that S(r + 2, r + 1) = S(r + 3, r + 1) = ... are shown similarly.
12. You help so much | 1,749 | 5,639 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "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-2016-50 | longest | en | 0.932217 |
http://www.thesis123.com/gate-2017-eeset-1-q64-solution/ | 1,516,123,968,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084886476.31/warc/CC-MAIN-20180116164812-20180116184812-00140.warc.gz | 565,493,053 | 9,141 | # gate 2017 EE(Set-1) q64-solution
As there are 4 people A,G,N,S and 4 colours so without any restriction total ways have to be
$$4\times4 =16$$
$$Now, Arun \rightarrow dislikes Red and$$
$$Shweta \rightarrow dislikes white$$
$$So 16-2=14 ways$$ | 81 | 246 | {"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} | 2.703125 | 3 | CC-MAIN-2018-05 | latest | en | 0.773757 |
https://brainiak.in/1295/1%C2%A0xi%C2%A0for-1%C2%A0for-where-xi%C2%A0contains-elements-each-yi%C2%A0contains-elements | 1,638,492,950,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362571.17/warc/CC-MAIN-20211203000401-20211203030401-00160.warc.gz | 222,492,201 | 5,199 | # Let Ui = 1 Xi for i ∈ 50 = Ui = 1 for i ∈ n Yi = T, where each Xi contains 10 elements and each Yi contains 5 elements.
Let Ui = 1 Xfor i ∈ 50 = Ui = 1 for i ∈ n Yi = T, where each Xcontains 10 elements and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi 's and exactly 6 of sets Yi 's, then n is equal to :
(1) 45
(2) 15
(3) 50
(4) 30
verified | 150 | 401 | {"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-2021-49 | longest | en | 0.724591 |
https://www.physicsforums.com/threads/voltage-through-a-steady-state-capacitor.552012/ | 1,701,482,559,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100309.57/warc/CC-MAIN-20231202010506-20231202040506-00280.warc.gz | 1,051,211,074 | 16,965 | Voltage through a steady state capacitor.
• TimeToShine
In summary, my attempts at solving this homework equation failed because I don't understand what DC steady state means in this context.
The Attempt at a Solution
According to my notes when a circuit is in DC steady state inductors are short circuited and capacitors are open circuited. This leads me to believe that the voltage across the capacitor is basically the voltage looking in through the terminal of the open circuit:
-----------R------o +
|
|
------------------o _
if it's an open circuit then no current flows through R.
(7sin12t - Vc)/R = 0
Vc = 7sin12t
This doesn't seem right to me, can anyone tell me where I'm going wrong?
Last edited:
Your circuit isn't in DC steady state. It's being driven by a sinusoidal source, so it's in AC steady state. The inductor and capacitor will have reactances (impedances) at the given driving frequency. Find expressions for them and then do the voltage divider thing (or find the current first, then the voltage across the cap.).
This is an AC circuit with a supply voltage given by V = 7Sin12t.
This means that ω = 12.
I don't know what 'steady state' means in this context and would assume it means find the voltage (AC) across the capacitor.
I attempted to find the reactance of L and C and got the following:
Xl = ωL = 12 x (2 +β) = (24 + 12β)Ω
Xc =1/ωC = 1/(12 x 3x 10^-3) = 28Ω
The resistance = (12-ε)Ω
The next logical thing is to get an expression for the impedance, Z,
using Z^2 = [(Xl-Xc)]^2 + R^2
Then use this to get an expression for the current and therefore an expression for Vc
BUT I CAN'T SEE HOW TO DO ALL OF THAT !
Sorry ! Hope you get somewhere with it
gneill said:
Your circuit isn't in DC steady state. It's being driven by a sinusoidal source, so it's in AC steady state. The inductor and capacitor will have reactances (impedances) at the given driving frequency. Find expressions for them and then do the voltage divider thing (or find the current first, then the voltage across the cap.).
Thanks!
The phasor voltage of 7sin12t is 7
The impedance of the resistor is the resistance
the impedance of the cap is 1/j(12)(3mF)
impedance of the inductor is j(12)(L)
What happened to the ε and β?
gneill said:
What happened to the ε and β?
i'm assuming they're 0 for now, they are just variables with given values, sorry i didn't clarify that.
technician said:
This is an AC circuit with a supply voltage given by V = 7Sin12t.
This means that ω = 12.
I don't know what 'steady state' means in this context and would assume it means find the voltage (AC) across the capacitor.
I attempted to find the reactance of L and C and got the following:
Xl = ωL = 12 x (2 +β) = (24 + 12β)Ω
Xc =1/ωC = 1/(12 x 3x 10^-3) = 28Ω
The resistance = (12-ε)Ω
The next logical thing is to get an expression for the impedance, Z,
using Z^2 = [(Xl-Xc)]^2 + R^2
Then use this to get an expression for the current and therefore an expression for Vc
BUT I CAN'T SEE HOW TO DO ALL OF THAT !
Sorry ! Hope you get somewhere with it
thanks, sorry i didnt mention that epsilon and beta are variables, my bad.
i have basically the same work as you, i used voltage division to get the voltage through the capacitor and then reverted back from phasor to time domain form.
thanks for all the help
It is much easier now.
I got Vc =15.5V, Vl =13.2V and Vr = 6.6V
technician said:
It is much easier now.
I got Vc =15.5V, Vl =13.2V and Vr = 6.6V
should the answer not be in the form xsin(wt + y)?
for the voltage across the capacitor i got
vc(t) = 9.2186sin(12t + 0.288)
does this look correct?
edit: i used 6 for epsilon and 2 for beta, but the answer should be roughly there or thereabouts i hope.
I thought ε and β were being taken to = 0 ?
You are right, my voltages should be given as
Vc = 15.5 Sin(12t)... Strictly should be 15.5Sin(12t - ∏/2) because Vc is ∏2 behind I
Vl = 13.2 Sin(12t)... strictly speaking should be 13.2 Sin(12t + ∏/2) because Vl is ∏/2 ahead of I
Vr = 6.6 Sin(12t)...this is OK as it is because Vr is inphase with I
A good check is that Vs^2 = Vr^2 + [Vl - Vc]^2
ie 7^2 = 6.6^2 + [13.2 - 15.5]^2
49 = 43.6 + 5.3 = 48.9 (near enough)
technician said:
I thought ε and β were being taken to = 0 ?
You are right, my voltages should be given as
Vc = 15.5 Sin(12t)... Strictly should be 15.5Sin(12t - ∏/2) because Vc is ∏2 behind I
Vl = 13.2 Sin(12t)... strictly speaking should be 13.2 Sin(12t + ∏/2) because Vl is ∏/2 ahead of I
Vr = 6.6 Sin(12t)...this is OK as it is because Vr is inphase with I
A good check is that Vs^2 = Vr^2 + [Vl - Vc]^2
ie 7^2 = 6.6^2 + [13.2 - 15.5]^2
49 = 43.6 + 5.3 = 48.9 (near enough)
thanks. | 1,438 | 4,671 | {"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} | 4.03125 | 4 | CC-MAIN-2023-50 | latest | en | 0.930368 |
https://biyoterapi.com/technology/how-to-grade-tests/ | 1,632,037,529,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056752.16/warc/CC-MAIN-20210919065755-20210919095755-00460.warc.gz | 194,125,189 | 20,197 | Calculating a test grade involves taking the total amount of points you earned dividing it by the total points of the test. Then you multiply that number by 100 to get a percentage. That percentage is then used to determine your letter grade like in the table below. Enter your points scored and total possible points to determine your grade.
## Test Score Calculator
As you can see in the calculator above, calculating your test grade is fairly straight forward in general. What can make this more tricky is when you start having parts of tests that are weighted more than others instead.
For example, lets say you have 5 different sections of a test. All of those sections have 5 questions each worth one point. Sometimes, those sections are broken down based on how important they are and weight as such. For instance, the first 4 sections may be worth 15% of the total grade each and the last section is worth 40% of the grade.
If that’s the case, you must multiply the total number of points of each section by the % they are worth.
For the last section this would look like the following:
5 points * .40 (40%) = 2 weight points.
You would need to do this for all for sections. Then you would need to add all of the total weighted possible points together. Then divide your scored weight points by your possible weight points to get your grade percentage.
For the example above the total weight points would be 4.40 points. Lets say you got a weighted score of 2 points. Then your grade would be as follows:
This weight scoring makes it more complicated to calculate the final grade, but it allows teachers to adjust tests to focus on the most important subjects.
For classroom teachers, grading tests and papers is second nature. However, if you are a homeschooling parent, you may be unsure about the best way to figure percentage grades, letter grades, and grade point average. You may not even be fully convinced that assigning grades is necessary, choosing instead to work to mastery on each assignment.
## How to Calculate Percentage and Letter Grades
If you decide to grade your students’ schoolwork, use these simple steps to determine the percentage and letter grade for any assignment or test.
To calculate a grade, you will need to figure out the percentage of questions that your student answered correctly. All you need to know to find the grade is the total number of questions on the assignment and how many answers are correct. After that, you will just need to plug a simple equation into a calculator and convert the percentage to a letter grade.
Here’s how:
1. Correct the paper.
2. Determine the number of total questions.
3. Count the number of questions answered correctly.
4. Take the number of correct answers and divide by the total number of questions. (Example: 15 correct answers divided by 20 total questions equals 0.75)
5. Multiply this number by 100 to turn it into a percentage. (Example: 0.75 multiplied by 100 equals 75%)
6. Grade ranges often vary among professors and teachers. However, a typical, easy-to-use grade scale is:
• 90-100% = A
• 80-89% = B
• 70-79% = C
• 60-69% = D
• 59% and below = F
Using the examples above, 75% would earn a C letter grade.
## How to Calculate GPA
If you’re homeschooling high school, you will likely need to figure your student’s overall grade point average (GPA) for his high school transcript. Calculate the cumulative GPA by dividing the total number of grade points earned by the number of credit hours attempted.
A typical grade point scale is:
• A = 4.0
• B = 3.0
• C = 2.0
• D = 1.0
There are variances for +/- grades that will vary based on the percentage grade scale you use. For example, if you use the ten points per letter grade scale, a 95% might indicate an A- which would translate to a grade point of 3.5.
Here’s how:
To figure out your student’s cumulative GPA:
1. Determine the total number of grade points earned. For example, if your student received three A’s and one B, his grade point total would be 15 (3×4 = 12; 1×3=3; 12+3=15).
2. Divide the grade point total by the number of credits attempted. In the example above, if each course reflected one credit hour, your student’s GPA would be 3.75 (15 grade points divided by 4 credit hours = 3.75)
## Why Do Homeschoolers Need Grades?
Many homeschooling families choose not to bother with grades since they don’t move on until a child fully understands the concept. Working to mastery means that the student would ultimately never earn less than an A.
Even if your homeschooling family works to mastery, there are a few reasons you may need to assign percentage or letter grades for your students.
Some students find the challenge of getting good grades motivational.
Some kids like the challenge of seeing how many answers they can get correct. These students are motivated by earning high scores. This may be especially true for kids who have been in a traditional school setting or those who homeschool using a more school-at-home approach. They don’t see the point of completing worksheets or tests if they don’t receive a grade for their work.
Grades can provide valuable feedback for these students to understand how they are performing.
Grades provide an objective means of assessing student performance.
Many homeschooling parents find it difficult to strike a balance between being overly critical and overly lax about their student’s academic performance. It can be helpful to create a grading rubric so that both you and your student know what’s expected.
A rubric can help you assess your student’s work objectively and force you to focus on specific issues. For example, if you’re working on teaching him to write a descriptive paragraph, a rubric can help you stay focused on descriptive elements and ignore run-on sentences or grammar errors until another assignment.
High school students may need grades for their transcript.
Even if you prefer not to assign grades in your homeschool, homeschoolers who will be applying for college admission may need them for their high school transcripts.
Some courses may be difficult to assign a percentage grade, particularly more interest-led topics. An alternative is to assign a letter grade based on your student’s understanding of the topic and the effort put forth in doing the work.
For example, a strong understanding and effort might earn an A. Solid knowledge and a decent but not outstanding effort might earn a B. You might assign a C if your student understands the topic well enough to move on without repeating the course and/or you would have liked to have seen more effort applied. Anything less would mean repeating the course.
Some homeschooling laws may require grades.
Your state homeschooling laws may require submitting grades to the county or state school superintendent, umbrella school, or other governing bodies.
Assigning percentage and letter grades doesn’t have to be difficult. These simple steps can make it easy no matter which route you choose.
Ann Logsdon is a school psychologist specializing in helping parents and teachers support students with a range of educational and developmental disabilities.
James Lacy, MLS, is a fact checker and researcher.
Ariel Skelley/Getty Images
An average test score is the sum of all the scores on an assessment divided by the number of test-takers. For example, if three students took a test and received scores of 69, 87, and 92, these numbers would be added together and then divided by three to get an average of 82.6.
Public schools rely on average, below average, or above-average test scores to identify how well a group of students learns. In some cases, students may be compared to their peers in the school district, county or state.
With the rollout of the Common Core State Standards, which established a common set of academic guidelines for states across the country, students may more frequently be compared to their peers nationally.
Other times, school officials distinguish average students from others to see who’s on grade level or how well an individual child performs in school compared to classmates on nationally normed tests.
## What Does Average Mean?
In special education, average test scores are particularly useful in standardized evaluations and in tests designed by teachers. Educators determine the average by adding a set of numbers and dividing the sum by the total number of numerals used in calculating that sum, also known as the mean.
Anyone who’s been graded on a curve likely knows the concept well. Teachers and specialists can use averages to determine the “middle” group of test-takers.
Statistically, about 68% of any large group of students will score within the low average to high average range on most tests. The other 32% will be in either the above average or below average group.
So how do educators proceed once they’ve identified the average?
## How Educators Use Average Scores
Teachers and specialists may use averages to monitor the rate at which the class is learning the material. Teachers also use averages to estimate where an individual student’s scores place them in relation to the rest of the class.
This is especially important for students with learning disabilities. Educators may also use averages to measure how an individual students’ abilities rate on tests used to diagnose learning disabilities.
## Alternative Methods
Sometimes educators and analysts use other methods to identify an average score. Rather than the mean, they may refer to the median, or the 50th percentile, which represents the score in the exact middle of the list of numbers. If your child scores in the 60th percentile, then they have performed better than 60% of test-takers.
You might have learned about finding the mean or median of a set of numbers in math class. They may be used interchangeably with the term “average,” but mean and median can be very different numbers, so take care when calculating.
## Examples of Averages
Want an example of an average? See if you can figure out the mean test score with the following information. Suppose six students scored 72, 75, 78, 82, 84, and 92 on a test.
To calculate the average, add the test scores together and divide the sum (483) by six. The average score would be 80.5. Anyone with basic math skills can determine an average.
If you are trying to find the median of the same set of numbers, you would identify the exact middle score. Since there is an even amount of numbers and thus no exact middle, you average the two middle scores (78 and 82) to arrive at a median of 80. In this case, the mean and median are very close, but that is not always the case.
## Factors That Affect Test Scores
If your child scores below average on a standardized test, don’t panic. A number of factors could have produced this result. As you’ve likely heard for many years, eating well and getting a good night’s sleep before a standardized test can influence scores.
If your child is struggling emotionally in any way, this can be a factor as well. Some children are very bright but have test anxiety, making these tests an inaccurate measure.
If you are concerned this may be a problem, talk to your child’s teacher. There are a number of ways in which you may able to get around test anxiety so that your child’s scores truly reflect her understanding of the content of the test.
### Learning More About Test Scores
If your child is struggling, you may need to understand more about test scores. For example, teachers use what is known as a standard deviation to describe students who fall outside of the 68 percent of scores which are described as below average, average, and above average.
## What to Do If Your Child Needs Help
If your child performs below average on a test or you find out that your child has learning difficulties in a subject area, it’s important to get the child the help needed. Early intervention can prevent your child’s struggles from worsening.
You may wish to begin by talking to your child’s teachers or the school psychologist.
### How to grade a test
There are about as many ways to test students as there are ways to teach them. And while summative assessments can effectively measure student performance at the end of a particular subject or time period, formative assessments can be instrumental in maximizing student performance along the way. The challenge for teachers, however, is figuring out how to grade a test quickly enough for that feedback to be helpful.
Often, teachers use simplified formats, like multiple choice, true/false, and matching tests, that can either be scored by hand using a printed answer key or automatically using an easy grader app, but not all material lends itself to those kinds of answer options. Sometimes, students need to show their work, demonstrate degrees and levels of competency, or articulate ideas and comprehension through writing – all of which require a closer review by the teacher in order to evaluate and score them appropriately.
### Alternative scoring methods
Even when specific content or coursework doesn’t intuitively conform to a standard bubble test format, teachers still need to assess and respond to student work in a timely manner. Unfortunately, responses like spelling, fill-in-the-blank, short answers, graphs, multi-step processes, problem-solving, and essays can be significantly more time-consuming to grade without the assistance of a test grader.
Many teachers even use rubrics as a formative assessment tool, but scoring them quickly, by hand, is not an easy task. The good news is that there is a grading solution – created by teachers, for teachers – that simplifies and streamlines these kinds of assessments, too.
GradeCam is a teacher grader app that offers all of the standard bubble test functionality, like multiple choice, true/false, number grids, etc. But it also does a whole lot more. In fact, the bubble contents themselves are customizable.
GradeCam also allows teachers to score rubrics using teacher-completed bubbles that can be instantly scanned and recorded, and the rubric with capture option will even scan answer content right into the app for grading on-the-go. Perhaps best of all, a revolutionary and proprietary handwriting recognition capability can read and score handwritten numbers, letters, and short answers, as well, making a whole host of new answer types automatically scannable and scorable.
By Angie Nelson
Last Updated November 4, 2020 . Disclosure: We may receive compensation if you sign up for or purchase products linked below. Details on offers may change, and you should confirm them with the company prior to taking action.
One work-from-home job loved by many in my community is contracting as an online test grader. In this position, you are working online grading tests, essays and other papers. In many cases, these essays are submitted as part of a standardized test like the ACT or SAT. In some cases, they are in conjunction with Engish as a Second Language, or ESL, studies.
To qualify for many of these positions, you often need a bachelor’s degree. Some companies allow that degree to be in any subject. Others may be looking for specific degrees like English. Make sure you read through the company’s current job openings for specifics.
Online grading jobs are seasonal in most cases. We see many companies hiring in the fall and spring as this is when most standardized testing is done. That being said, don’t be afraid to check for openings in the off-season as well. And once you are accepted as an online scorer, most companies will invite you back year after year provided you do a good job and meet the workload requirements.
## 8 Legit Online Grading Jobs
### 1. Measurement, Inc.
Measurement, Inc. is one of the most popular scoring jobs with my readers. They require a bachelor’s degree in any subject. Their projects include test items in English Language Arts, mathematics, science, and other areas.
For most remote positions with this company, they do expect you to work Monday through Friday for the duration of your temporary contract. Their busy season is March through June. Available hours may vary by project.
Onsite positions are usually open year-round. Hiring for remote work is typically done November through March. Openings were listed on the site at the time of writing this post.
### 2. Pearson
Pearson is a well-known name in the learning community. They hire tutors, at-home test developers, online scorers and more. The company was named as a Top 100 Company with Remote Jobs for 2019 by FlexJobs and they have a solid review rating on Glassdoor of 3.6 out of 5 stars.
Opportunities vary throughout the year and some may be location-specific. Most positions are temporary and project-based. A bachelor’s degree is required.
One of their most recent scoring openings was for an edTPA Scorer. This particular test makes sure new teachers are ready for the classroom. This was a more specialized scorer job and required not only a bachelor’s degree but also teaching experience or experience working with teachers.
You can often find Pearson’s most recent needs and requirements on FlexJobs.
### 3. ETS
Educational Testing Service, or ETS, provides scoring for a variety of tests from student leader proficiency tests to high school equivalency exams. As such, the requirements can vary greatly for each position. Some positions require a bachelor’s degree. Some require a Master’s. Some require applicants to possess a current teaching certificate while others do not.
As was stated previously, many of the current openings will be filled in spring and fall 2019.
### 4. Write Score
The Write Score site states that orientation for their scoring season will begin in late July to early August though positions are available throughout the year.
Unlike other companies, Write Score only requires a two-year degree and the ability to pass their qualifying test. They do accept resumes for their waitlist and applicants are contacted as needed.
### 5. ACT
ACT hires Readers to read and score student papers for the ACT Writing Test. This is an opportunity to read and score anytime and from any personal computer.
The site states this is performance-based work. Your earnings will depend on the accuracy and number of essays scored. The company does say Readers can earn \$12 per hour or more.
To qualify, you will need a bachelor’s degree or higher, reside in the U.S., and have current teaching experience. Teaching English to high school junior and senior students is preferred.
### 6. Creative English Solutions
Creative English Solutions, or CES, provides evaluations for students wishing to take the TOEFL, Testing of English as a Foreign Language, and TOEIC, Test of English for International Communication. As such, applicants must be a native English speaker and possess a University degree.
Positions at CES offer a lot of flexibility. Though they do ask for a commitment of at least 10 hours per week, Evaluators can accept and decline assignments as they wish.
In addition to hiring Evaluators, CES also hires freelance writers and voice actors.
### 7. Literably
Literably is a little different in that their scorers evaluate oral recordings of elementary students. Transcriptionists listen to student audio recordings and transcribe errors.
This position does not require any experience or special degrees. You simply need to pass their sample tasks. As with most transcription jobs, pay is based on performance, speed and accuracy. The most recent job posting state “\$10-20 per hour.” PayPal is required to receive payment.
This is a flexible position and you can set your own hours.
### 8. ALTA Language Services
If you possess native fluency in another language, ALTA Language Services currently has several testing evaluator positions available. These are typically on demand, part-time positions.
A degree is required, as is a minimum English proficiency level of a 2+/2+ on the ILR proficiency scale. As with most companies in this industry, training is required and paid.
At the end of the day, you aren’t going to get rich as an online test grader. At the time of writing this post, most positions offer around \$10 to \$13 per hour. That being said, it can be flexible, fulfilling work at home for the right person. Most people report being happy with these positions. The biggest complaint is often that the work isn’t available year-round, but many consider it a nice supplemental income source.
Sharing is caring!
Angie Nelson began working from home in 2007 when she took her future into her own hands and found a way to escape the corporate cubicle farm. Today she balances several successful online ventures and loves to share her passion for home business with others.
Angie Nelson began working from home in 2007 when she figured out how to take her future into her own hands and escape the corporate cubicle farm. Angie’s goal is sharing her passion for home business, personal finance, telecommuting, and entrepreneurship, and her work has been featured on Recruiter, FlexJobs and Business News Daily.
by TeachThought Staff
Google Forms isn’t the first thing you think of when you think of assessment resources.
It’s not especially elegant, it won’t wow students, and the learning curve isn’t as mild as it might be. But with a little bit of work on the front-end, Google Forms can return the favor in spades on the back-end in the form of self-grading assessments.
While there isn’t an app (yet) that can uncover the true nuance of understanding, if you’re using multiple-choice assessments–even just as pre and summative assessments–this trick can save you time, allowing the real potential of assessment to shine through consistently extracting data to revise planned instruction.
How To Create A New Quiz In Google Forms
1. In Google Forms, click Plus .
2. At the top right, click Settings .
3. Click QuizzesMake this a quiz.
5. Click Save.
With just a little work, you can make an answer key, assign points, and provide automatic feedback for students. You can also make an answer key for certain types of assessment items and question types, including:
• Multiple choice
• Checkboxes
• Dropdown
• Multiple choice grid
• Checkbox grid
3. In the bottom left of the question, click Answer key.
5. In the top right of the question, choose how many points the question is worth.
7. Optional: To edit question or answer options, click Edit question.
Note: You can assign points and add feedback on all question types.
How To Grade Individual Assessment Items & Responses
If you collect email addresses, you can assign points and leave feedback on individual responses. After you grade each response, be sure to save your changes.
1. In Google Forms, open a quiz.
2. At the top, click Responses.
3. Click Individual.
4. To move between individuals, click Previous or Next .
5. Find the question you want to grade.
• In the top right, enter how many points the response earned.
6. Enter your feedback and click Save.
7. To save your changes, at the bottom, click Save.
How To See Quiz Results
1. In Google Forms, open a quiz.
2. At the top, click Responses.
3. Click Summary.
nights_stay Dark mode automatically enabled! clear
Either you’ve set your system settings to “dark” or it’s pretty late (11pm – 5am). If you’d rather it not be dark, simply select the sun icon in the top right corner to turn it off.
info v1.0.0 is out! clear
18 months later, Final / Test Grade Calculator has finally been updated with all the features promised in the original version. Scroll down to see what’s new (and why it took so long), and thanks for stopping by!
What do you want to find? What’s the difference? *This will also tell you what you need to raise your letter grade up if possible!
For a final or a test/exam? Final Test/Exam Which one do I choose?
Is each test weighted equally? Yes (default) No (less common) Which one do I choose?
What grade do you want to get? %
What did you get on your final? %
How much is your final worth? %
How many points is it worth? * pts
*“It” refers to your upcoming test, exam, or final. With unweighted grading, only the point value matters.
How much are the tests worth? %
How many tests did you take? *
*The test you’re calculating for should not be included, only previous tests.
What is your test average? * %
*You can put either the overall value (ex. 95%) or the weighed value (ex. 45% (out of a 50% weight))
Points in test category: out of pts
Points your test/exam is worth: pts
*“It” refers to your upcoming test, exam, or final. With unweighted grading, only the point value matters.
How high is the top grade? %
What is n (the root)?
Calculate / Tell Me!
error Sorry, it looks like something is not right!
The missing/invalid inputs have been highlighted in red. Please fix them and press “calculate” again.
info_outline Note that letter grades are not as accurate as percentages, and can result in slight inaccuracies.
info_outline It looks like one or more of the inputs above may have a typo.
You’ll need a . % (??) post-curve in order to get a . % (??) in your class.
Your overall grade in the class would be a . % (??).
This results in you needing to get before the curve.
This is equivalent to getting . , or . before the curve .
If you want to keep your current letter grade, you’ll only need to get a . % (??) post-curve.
If you get a perfect score (100%) on your final, you’ll end with a . % (??).
After dropping your lowest test, your test average increased to 0.00% (0.00% weighted)
You’ll get 0.00% back on your lowest test, increasing your test average by 0.00% (unweighted).
info If you have any, feedback is always greatly appreciated! You can leave feedback here or tell me in person (if you know me), thanks!
Questions that aren’t frequently asked but you’re probably wondering anyways:
1) Why use this site? Why not RogerHub?
• A (hopefully) more intuitive and modern design
• Supports more features with fewer steps (ex. unweighted grades, test grades, typo catching, etc)
• A developer that still actively develops this site and listens to feedback (send feedback here)
• Support someone who’s still a HS student where as Roger already appears rather successful
2) What’s new in v1.0.0, and why did it take 18 months?
• Starting a brand new project right before finals week (May 2019) probably wasn’t a great idea
• Since nobody really used it, the project got abandoned afterwards. until it started showing up on Google and people started sending feedback to me. so I decided to finish it now!
• The big new features in v1.0.0 include the following:
• Dark mode (automatic based on time of day and/or system preferences, but can be toggled with the sun/moon icon on the top right corner)
• Unweighted grade support and test/exam support (rather than “final”), as promised in the first version but never implemented until now
• Support for test curving if your teacher does that!
• Smoother and faster overall with minor UI improvements
3) Why does this site exist?
• I was bored and was totally not procrastinating on my homework
• Wanted a tool that supported more grading policies w/o needing to use RogerHub’s “advanced” mode
• ^ even though it would definitely uses a lot more time in the long run
• Make something people actually want and search for (but currently doesn’t exist)
• Get more traffic on my tiny little website and maybe show up on Google
If you would like to send your assessment scores to the Gradebook so that they can be included in the course grade calculation or allow students to see an answer key or other feedback in Tests & Quizzes, see What are the Grading and Feedback options for an assessment?
On the Total Scores screen for an assessment, you can see the total score each student received based on the points from auto-graded questions (such as multiple choice and true false). You may adjust the students’ scores and for students who submitted the assessment, you can add comments.
• Tests & Quizzes will NOT automatically grade students with no submission. If a student who did not submit an assessment should receive a grade of 0, you’ll need to enter a 0 for their grade.
• Comments cannot be added within the Tests & Quizzes tool for students with no submission and they do not have access to feedback for assessments they did not submit. If you’d like to print an answer key to an assessment to provide to students who did not submit, see How do I print an assessment (i.e. test or quiz)?
• Once a grade has been saved for a student in Tests & Quizzes, you cannot delete the grade to remove it from the student’s course grade. If you inadvertently assign a score to a student who should not have received one for an assessment, and the assessment score is being sent to the Gradebook, you may need to override the course grade in the Gradebook.
Go to Tests & Quizzes.
Select the Tests & Quizzes tool from the Tool Menu of your site.
Select the Published Copies tab.
Click on the Published Copies tab to view the assessments that have been released to students in your site.
Go to the assessment submissions.
Select the Scores option from the drop-down menu for the assessment you would like to grade.
Alternately, you may also click on the number of student submissions in the Submitted column to view the submissions.
Display multiple submissions for students. (Optional)
By default, the submission that displays for each student will be based on your assessment’s Recorded Score setting. For example, if you accept the highest score, the highest scoring submission will display. If you accept the average score, the average point value of all the student’s submissions will display.
If your assessment allows multiple submissions, or if a student has submitted an allowed retake, you can choose to view all student submissions by selecting All Submissions from the View drop-down menu. | 6,306 | 30,071 | {"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} | 3.671875 | 4 | CC-MAIN-2021-39 | latest | en | 0.942612 |
http://www.purplemaths.com/modules/plane.htm | 1,477,009,458,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988717959.91/warc/CC-MAIN-20161020183837-00436-ip-10-142-188-19.ec2.internal.warc.gz | 568,374,824 | 8,090 | Introduction to the x,y-Plane
(The "Cartesian" Plane)
(page 1 of 3)
Sections: Introduction to the plane, Plotting points, The four quadrants
Before the end of the European Renaissance, math was cleanly divided into the two separate subjects of geometry and algebra. You didn't use algebraic equations in geometry, and you didn't draw any pictures in algebra. Then, around 1637, a French guy named René Descartes (pronounced "ray-NAY day-CART") came up with a way to put these two subjects together.
To explain Descartes' method, first think about using a street map. If you're trying to find a street that you've never been on before, you look for the street's name in the map's index. Suppose the index says that your target street is located at D12. This means that you go across the top of the map and find "D", and then you go down the side and find "12". You then trace down and across to find the box labelled as "D12", and look inside that box for the street you need. Somebody figured out that this was a handy way to specify the right general area on the map, telling you how far over (the D) and how far up or down (the 12) you needed to go. Descartes did something similar for mathematics.
You learned about the basic (counting) number line back in elementary school:
Later on, you were introduced to zero and negatives, which completed the number line:
Descartes' breakthrough was in taking a second number line, standing it up on its end, and crossing the first number line at zero: Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
The number lines, when drawn like this, are called "axes" (pronounced "ACK-seez"). The horizontal number line is called the "x-axis" ("eks-ACK-siss"); the vertical number line is the y-axis.
The arrows at the ends of the axes indicate the direction in which the numbers are getting larger. Therefore, only the axes should have arrows, and the arrows should be on one end only. Yes, many textbooks, especially in high school, draw the arrows incorrectly, as do many educators. (Don't point this out to them; for some reason, they're extremely touchy on this subject.) In "real life", not all axes point up or to the right; in the sciences, the x-axis can point left; in finance, the y-axis often points down. This information will be lost if you stick with the high-school way of splashing arrow-tips everywhere on every line.
The whole flat expanse, top to bottom, side to side, bursting outside of the box in the animation above and stretching off to infinity in all directions, is called the "plane". When you put the two axes in the plane, it is then called the "Cartesian" ("carr-TEE-zhun") plane. The name "Cartesian" is derived from the name "Descartes", after its creator, Rene Descartes. Please don't ask me to explain that derivation; I think it's Latin, and I can barely handle English.
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Cite this article as: Stapel, Elizabeth. "Introduction to the x,y-Plane (The 'Cartesian' Plane)." Purplemath. Available from http://www.purplemath.com/modules/plane.htm. Accessed [Date] [Month] 2016
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 963 | 2,824 | {"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-2018-22 | latest | en | 0.882996 |
https://essaybureau.com/blog/find-the-probability-that-our-hypothesis-is-true/ | 1,611,570,850,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00472.warc.gz | 329,382,635 | 11,052 | ## find the probability that our hypothesis is true.
Repeat the above example, but this time use the hierarchical clustering approach with complete linkage as a measure of distance between clusters. Create three clusters (i.e., K=3) based on your hierarchical cluster analysis and add a new variable to the data-identifying cluster ids. Using a contingency table, explain how the three identified clusters are related to the three species of flowers.
1. Suppose that the smoking status, X, has Bernoulli(μ) distribution, where the prior distribution for μ is Beta(1,10). We have interviewed a random sample of 50 people and found that 8 of them smoke regularly. Find the posterior distribution of μ given the observed data. Plot the posterior distribution and find the 95% credible interval for μ. What is the point estimate for μ based on this distribution? Compare this point estimate to the sample proportion, which is also commonly used as a point estimate for the population proportion. Suppose that we hypothesize that less than 20% of the population smoke. Use the posterior probability distribution to find the probability that our hypothesis is true.
### Determine the outlet concentration x1 in the water solution and in the trichloroethane solution using an analytical equation.
Predicting Extraction for an Existing Tower with a Given Number of Steps. An existing tower contains 5.0 theoretical steps. It is desired to predict its performance under the following conditions….
### Calculate the minimum solvent that can be used. [Hint: In this case, the tie line through the feed L0 represents the condition for minimum solvent flow rate.
Minimum Solvent and Countercurrent Extraction of Acetone. An aqueous feed solution of 1000 kg/h containing 23.5 wt % acetone and 76.5 wt % water is being extracted in a countercurrent….
### Determine the break-point time, the fraction of total capacity used up to the break point, the length of the unused bed, and the saturation loading capacity of the solid.
Drying of Nitrogen and Scale-Up of a Column. Using molecular sieves, water vapor was removed from nitrogen gas in a packed bed (C1) at 28.3°C. The column height was 0.268…. | 462 | 2,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2021-04 | latest | en | 0.90474 |
http://www.concepts.org/index.php?title=Volume§ion=+Volume+in+differential+geometry | 1,586,210,673,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371660550.75/warc/CC-MAIN-20200406200320-20200406230820-00392.warc.gz | 217,580,676 | 6,146 | ## Volume in differential geometry::Volume
### ::concepts
Volume::volume Center::style Radius::geometry Sphere::''h'' Height::cubic Manifold::integral
Volume in differential geometry {{#invoke:main|main}}
In differential geometry, a branch of mathematics, a volume form on a differentiable manifold is a differential form of top degree (i.e. whose degree is equal to the dimension of the manifold) that is nowhere equal to zero. A manifold has a volume form if and only if it is orientable. An orientable manifold has infinitely many volume forms, since multiplying a volume form by a non-vanishing function yields another volume form. On non-orientable manifolds, one may instead define the weaker notion of a density. Integrating the volume form gives the volume of the manifold according to that form.
Any oriented Riemannian (or pseudo-Riemannian) manifold has a natural volume (or pseudo volume) form. In local coordinates, it can be expressed as
$\omega = \sqrt{|g|} dx^1\wedge \dots \wedge dx^n$
where the $dx^i$ are the 1-forms providing an oriented basis for the cotangent bundle of the n-dimensional manifold. Here, $|g|$ is the absolute value of the determinant of the matrix representation of the metric tensor on the manifold.
Volume sections
Intro Units Related terms Volume in calculus Volume formulas Volume formula derivations Volume in differential geometry Volume in thermodynamics See also References External links
Volume in differential geometry PREVIOUS: Intro NEXT: Units << >> | 365 | 1,550 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-16 | latest | en | 0.820048 |
http://mathoverflow.net/questions/80710/what-are-the-values-of-the-derivative-of-riemanns-zeta-function-at-the-known-non/80783 | 1,368,941,441,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696383263/warc/CC-MAIN-20130516092623-00027-ip-10-60-113-184.ec2.internal.warc.gz | 169,956,150 | 12,775 | ## What are the values of the derivative of Riemann’s zeta function at the known non-trivial zeros?
Specifically I am interested in the quotients $$-\frac{\zeta'(\rho)}{\zeta'(1-\rho)}=2(2\pi)^{\rho-1}\Gamma(1-\rho)\sin(\pi\rho/2).$$ Obviously they are in $\mathbb{T}$ for all known non-trivial zeros. But how often are these number $\pm 1$? I would find some tables of the derivative at the known zeros rather usefull, or even perhaps tables of the quotients above? I would be very grateful if somebody can provide me with a good reference. Thanks!
-
I imagine you might be stuck calculating this yourself, which shouldn't be too hard (depending on your programming ability). In case you don't have it, here's a link to a table of zeros: dtc.umn.edu/~odlyzko/zeta_tables – BR Nov 11 2011 at 22:59
Mathematica has about $10^7$ zeros built in as the function ZetaZero[n]. – Stopple Nov 11 2011 at 23:11
I'd guess the real and imaginary parts are transcendental. After calculating the first 100 examples, none seem particularly close to $\pm1$ – Stopple Nov 11 2011 at 23:29
Thank you, Stopple. I haven't done the calculations because I don't have Mathematica on my laptop at the moment. I had guessed that the imaginary parts would get small quite quickly. What makes you guess the parts would be transcendental? – Kevin Smith Nov 12 2011 at 1:06
To elaborate a little more, here's some Mathematica code:
ListPlot[Table[{rho = ZetaZero[n];z = N[2 (2 Pi)^(rho - 1) Gamma[1 - rho] Sin[Pi rho/2]]; {Re[z],
Im[z]}}, {n, 1, 100}], AspectRatio -> Automatic]
Here's the output:
-
Fantastic, thank you again. – Kevin Smith Nov 12 2011 at 1:08
I am trying to get at whether these numbers tend to $-1$. – Kevin Smith Nov 12 2011 at 1:09
Your plot seems to suggest that they are repelled from $+1$, is there any further evidence of this conjecture? – Kevin Smith Nov 12 2011 at 1:19
No, this phenomenon does not seem to appear looking at the first 1000. I'd guess they are dense on the circle. The imaginary parts of the zeros are conjectured transcendental - why would you conjecture anything else for these numbers? – Stopple Nov 12 2011 at 1:41
I am not conjecturing that the imaginary parts of the zeros are not transcendental (unless I am missing something implicitly). The quotient of the derivatives describes how much faster (or not, as you suggest) the imaginary part of $\zeta(1/2+it)$ goes to zero than the real part. Since the real part tends to be positive, I wondered if this phenomenon is is evident asymptotically as a tendency for the quotient to be near to $-1$. Your response has been very helpful so I accept your answer... – Kevin Smith Nov 12 2011 at 1:58
show 1 more comment
Here is an answer in a few parts.
1. Ghaith Hiary has computed fairly large tables of zeros at large height, which are at
http://sage.math.washington.edu/home/hiaryg/page/index.html
I believe that each table has 10 million zeros starting at t = 10^n, for n from 12 to 28. There are tables of derivatives of Z'(t) at the zeros, but not zeta'(1/2 + it). At at zero, it is not hard to go from one to the other, though.
2. To compute what you want to compute, I think the formula you wrote down is not going to be computationally practical due to the exponential decay of the gamma function and the exponential growth of the sine function. (I think may already be precision issues in the picture in Stopple's answer.) If I calculate correctly (I hope I'm not embarrassing myself), then $$-\frac{\zeta'(1/2 + i\gamma)}{\zeta'(1/2 - i\gamma)} = e^{-2 i \theta(\gamma)}$$ where $\theta$ is the Riemann-Siegel theta function. This is probably a better way to compute the quotient.
To make the above picture using sage, if you have the optional package database_odlyzko_zeta installed, you can use:
import mpmath
mpmath.mp.prec = 200
L = []
for gamma in zeta_zeros()[:100]:
z = exp(-2 * CC.0 * CC(mpmath.siegeltheta(gamma)))
L.append((z.real(), z.imag()))
P = points(L)
P.save('plot.png', figsize=[5,5])
Something similar will work if you want to parse Hiary's files.
1. As for whether there is any tendency for these numbers to be close to -1: I think there is not. This looks like the question of whether there is any asymptotic relation between the zeros of the zeta function and gram points, and I think that the answer is expected to be no, but I don't know if that is a theorem. (Those thoughts should not stop you from checking, though.)
-
Jonathan, thank you, this is very helpful. For starters, I must point out that my original hypothesis should have been $+1$, not $-1$ (one of those silly late-night errors!). In any case, that there is no such tendancy ($\pm 1$) is equivalent to the statement that $u(t)\notin o(v(t))$ and $v(t)\notin o(u(t))$, where $u,v$ are the real and imaginary parts of derivative. In other words, $\liminf|u/v(\rho)|>0$ and $\limsup|u/v(\rho)|<\infty$, for $\beta=1/2$. I suppose the next question is whether Stopple's guess: that the set is dense in $\mathbb{T}$ is correct, which, as you pointed out... – Kevin Smith Nov 13 2011 at 10:51 amounts to knowing if $\theta(\gamma)$ is uniformly distributed in $[-\pi,\pi]$. – Kevin Smith Nov 13 2011 at 10:52 Typo: replace $t$ with $\rho$ in the comment above. – Kevin Smith Nov 13 2011 at 11:02 In fact, uniform distribution would be much stronger, though still a possibility. I will reformulate this and post another question. – Kevin Smith Nov 13 2011 at 14:22 | 1,492 | 5,448 | {"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} | 3.3125 | 3 | CC-MAIN-2013-20 | latest | en | 0.926583 |
https://ncertmcq.com/ml-aggarwal-class-6-solutions-for-icse-maths-chapter-15-ex-15-4/ | 1,713,763,785,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818081.81/warc/CC-MAIN-20240422051258-20240422081258-00749.warc.gz | 380,881,636 | 9,802 | ML Aggarwal Class 6 Solutions Chapter 15 Data Handling Ex 15.4 for ICSE Understanding Mathematics acts as the best resource during your learning and helps you score well in your exams.
## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 15 Data Handling Ex 15.4
Question 1.
Find the mean of the following data:
(i) 40, 30, 30, 0, 26, 60
(ii) 3, 5, 7, 9, 11, 13, 15
Solution:
Question 2.
Find the mean of the first five even whole numbers.
Solution:
Question 3.
A batsman scored the following number of runs in six innings:
36, 35, 50, 46, 60, 55
Calculate the mean runs scored by him in an inning.
Solution:
Question 4.
The enrolment in a school for six consecutive years was as follows :
1555, 1670, 1750, 2013, 2540, 2825
Find the mean enrolment of the school for this period.
Solution:
Question 5.
The marks (out of 100) obtained by a group of students in a science test are :
85, 76, 90, 85, 39, 48, 56, 95, 81, 75
Find the:
(i) highest and lowest marks obtained by the students.
(ii) mean marks obtained by the students.
Solution: | 319 | 1,045 | {"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-2024-18 | latest | en | 0.927413 |
https://www.scribd.com/document/154290628/how-to-do-a-valuation-project | 1,569,164,396,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575515.93/warc/CC-MAIN-20190922135356-20190922161356-00416.warc.gz | 1,030,947,617 | 60,657 | You are on page 1of 10
# VALUATION:
PROJECTS
Aswath Damodaran
## 1. OVERALL VALUATION PROJECT
Runs the En@re Semester Due in Two Parts
Project Descrip@on
This project is designed to apply the valua@on techniques we learn in class on companies in the real world. It is a group project, with each person in the group picking one company to value. The project analysis is due in two parts:
The discounted cash ow valua@ons are due midway
through the semester on April 5 at 5 pm. They will not be graded, but will be reviewed and returned with comments. The en@re project report is due by 5 pm on the last day of class (May 13 at 5 pm).
3!
## Step 1: Pick the companies
Pick a group of companies (one for each person in the group), making sure you have at least
1. 2.
3. 4.
one company which has nega@ve earnings in the most recent nancial period (year-to-date, if available or most recent nancial year (if not)). one company which has high-growth poten@al (Look for companies whose revenues are expected to grow > 25% over the near future; the could be losing money, in which case you could kill two birds with one stone.) ne non-U.S. company (Value the local lis@ng not the ADR) one service company (retail rm, nancial service rm ..)
4!
## Step 2: Discounted Cashow Valua@on
Value the stock in each company using a discounted cash ow model ( You have the discre@on to choose the model that you think is most appropriate for that company) Evaluate how sensi@ve your value es@mates are to changes in your assump@ons. What are the key drivers of value for your company? Parts 1 & 2 are due midway through the semester
5!
## Step 3: Value rela@ve to comparables
Prepare a list of comparable companies, using criteria that you think are appropriate Choose a mul@ple that you will use in comparing rms across the group. (You might have to try a number of mul@ples out before making this choice) Evaluate your company against the comparable rms using the mul@ple that you have chosen for your valua@on.
using simple techniques (do a qualita@ve analysis adjus@ng the average for comparable rms) using a sector regression (a regression across rm in your sector). You can download the data on comparable rms by going to h]p://www.stern.nyu.edu/~adamodar/New_Home_Page/equity.html (Click on comparable rms)
6!
## Step 4: Value rela@ve to the market
Using the latest regression posted for the market on my web site, and the mul@ple you chose in step 3, evaluate whether your rm is under or over valued.
If you have a non-U.S. company which has an ADR listed on
it, you can use the U.S. regression While I will not require it, I will be very impressed if you run a regression of the mul@ple in your foreign market (use the 50 largest rms, if you want to reduce your work load) against the variables that determine that mul@ple.
7!
## Step 5: Using Op@on Pricing Models
If your nega@ve earnings rm has high nancial leverage, value it using the op@on pricing model. If it does not, do not use the op@on pricing model.
8!
## Step 6: Final Value Es@mate and Recommenda@on
Consider the values you have obtained from the discounted cash ow, rela@ve and op@on valua@on models.
How would you reconcile the dierent es@mates of value? Make a nal recommenda@on on the stocks in your group.
9!
2. MYSTERY PROJECT
Topic: TBA Due: Session TBA (some@me between the 10th and 12th weeks of the class) | 872 | 3,396 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-39 | latest | en | 0.930162 |
https://stats.stackexchange.com/questions/481638/generating-random-numbers-that-are-log-normally-distributed | 1,716,421,084,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00837.warc.gz | 469,847,411 | 39,838 | # Generating random numbers that are log-normally distributed
Even though I don't quite understand why and how this works, I appreciate how simple it is to generate a set of numbers which are Poisson distributed:
public static int getPoisson(double lambda) {
double l = Math.exp(-lambda);
double p = 1.0;
int k = 0;
do {
k++;
p *= Math.random();
} while (p > l);
return k - 1;
}
This "generating" function is basically from this book were I also find generating functions for the binomial and the geometric distribution.
But I did not find a simple generating function for the log-normal distribution. Either the author calls this distribution the "logarithmic series distribution" (then the generating function would not be simple), or it is missing. In the latter case I am looking for some pseudo-code.
• You can probably find details about how to create a random number generator for a normal distribution; then you just have to take the exponent of it. Aug 5, 2020 at 14:19
• @jwimberley: do you want to post your comment(s) as an answer? Better to have a short answer than no answer at all. Anyone who has a better answer can post it. Aug 5, 2020 at 14:21
• The logarithmic series distribution is a completely different beast. Aug 5, 2020 at 15:32
## 1 Answer
Since the log-normal distribution is defined as the distribution of exp(X) where X is a random variable following a normal distribution, you need only find details about how to sample a random number from a normal distribution, and take its exponent.
• Sorry, but don't you mean "take its logarithm"? Aug 5, 2020 at 14:33
• No, I do not. If If X is log-normally distributed, then log(X) is normally distributed, which is what you are probably thinking of. But I'm describing generating a normally distributed Y=log(X) directly. You couldn't take the logarithm of a normally distributed value, anyways, since the normal distribution has support on the negative real numbers, outside the domain of the logarithm. Aug 5, 2020 at 14:38
• The term lognormal is in a sense backward as a lognormal distribution is, as said, an exponentiated normal. The lognormal is named for what you can do with it -- take logarithms, and you get a normal -- rather than for what you need to do to get it. Although it would be appealing to start all over again with statistical terminology -- the term normal is itself a source of much confusion and misunderstanding -- in practice we have to muddle through and hope that with slow campaigns some usages will fade away and better terms will take their place. Aug 5, 2020 at 15:14
• Devroye discusses generating from the normal distribution in chapter 9. In the section on normal distributions, an exercise about lognormal distributions (p. 392) mentions this approach, and also hints at a faster/more direct approach to generating lognormal variates. Aug 5, 2020 at 16:32 | 675 | 2,876 | {"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} | 3.5 | 4 | CC-MAIN-2024-22 | latest | en | 0.920471 |
http://betterlesson.com/lesson/resource/3196995/graphs_of_tangent_and_cotangent-png | 1,477,288,092,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719468.5/warc/CC-MAIN-20161020183839-00375-ip-10-171-6-4.ec2.internal.warc.gz | 26,983,111 | 23,958 | Graphs_of_tangent_and_cotangent.png - Section 2: Whole Class Discussion: What's Weird about Tangent?
Graphs_of_tangent_and_cotangent.png
Graphing the Tangent Function
Unit 8: Trigonometry: Periodic Functions
Lesson 4 of 6
Big Idea: By exploring the graph of the tangent function, students get a preview of some important ideas that they may see in future math classes.
Print Lesson
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Positive: 48 %
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TRANSCRIPT
• Slide 1
• Slide 2
• Contents Introduction Matrix Multiplication Partitioned Matrices Powers of a Matrix Transpose of a Matrix Theorems and Proofs Application of Matrix Multiplication Questions Links
• Slide 3
• The Matrix What is a matrix? A matrix is a rectangular array of numeric or algebraic quantities subject to mathematical operations, or a grid of numbers. Row 247 1 -36 Column The size of this matrix is m n, where m represents the number of rows and n represents the number of columns. For example, the matrix shown above is a 23 matrix.
• Slide 4
• Matrix Multiplication Scalar Multiplication Consider the following If s is a scalar and A is a matrix 1 5 s = 2, and A = 4 3 2 1 1 52(1)2(5) 210 sA = 2 * 4 3 = 2(4)2(3)= 8 6 2 12(2)2(1) 4 2 Rule:if s is a scalar and A is a matrix, then the scalar multiple sA is the matrix whose columns are s times the corresponding entries in A. In other words, you take the scalar and multiply it with every number in the matrix. Theorem: If r and s are scalars, then (r + s)A = rA + sA and r(sA) = (rs)A.
• Slide 5
• Matrix Multiplication Before you multiply a matrix by another matrix, you need to check the compatibility of the two matrices. The number of columns (n) in A must equal the number of rows (m) in B in order to carry out the matrix multiplication. Example 1 A = 2 5B = 3 1 7 1 3 8 2 4 m n m n 22 23 You can see that the number of columns of A matches the number of rows of B. (Later on, you will see that the size of matrix AB will be m of A n of B). Example 2 A = 2 5B = 3 1 7 1 3 8 2 4 41 2 22 33 Obviously, n of A m of B and thus they are not compatible for multiplication.
• Slide 6
• Example 3 Find the product AB from the following matrices A =10-2 53-1 B =63 4-3 -72 A is a 23 matrix and B is a 32 matrix, and they are compatible. To get the final product, we will be sequentially multiplying each row in one matrix by the corresponding column in another matrix. In this example, we take the first row of A and first column of B, multiply the first entries together, second entries together, and third entries together, and then add the three products. 1 0 -2*6= 1(6) + 0(4) + (-2)(-7) = 20 4 -7
• Slide 7
• Example 3 cont. This sum is one of the entries in the product matrix AB; in fact, being the product of row 1 of A and column 1 of B, it is the (1,1)-entry in AB. Column 1 Row 1 20# ## Then we continue in like manner; we multiply row 1 of A by column 2 of B. 1 0-2 *3= 1(3) + 0(-3) + (-2)(2) = -1 -3 2 And its position in matrix AB is Column 2 20-1 Row 1 ##
• Slide 8
• Example 3 cont. Multiplying row 2 of A by column 1 of B gives 5 3 1*6= 5(6)+3(4)+(1)(-7) = 35 4 -7 And its position in matrix AB is at row 2 and column 1. You can do the last multiplication of row 2 of A by column 2 of B by yourself. So the final answer is AB = 10-2 *63= 20-1 5314-3 358 -72 Handy trick: As we mentioned before, the size of matrix AB is 22 (A is a 23 matrix and B is a 32 matrix). AB BA (Can you see why?)
• Slide 9
• Partitioned Matrices Matrices can be written as partitioned matrices. Example 30-1-295 A =-6-40317 2-54801 Can also be written as a 23 partitioned matrix A = A 11 A 12 A 13 A 21 A 22 A 23 A 11 =30-1 A 12 =-29A 13 = 5 -6-4031 7 A 21 =2-54 A 22 =80A 23 = 1
• Slide 10
• Multiplication of Partitioned Matrices Example 4-6 51-42-21-2 A =32140B =37 0-371125 -13 A = A 11 A 12 B = B 1 A 21 A 22 B 2 AB = A 11 A 12 * B 1 = A 11 B 1 + A 12 B 2 =15-56 A 21 A 22 B 2 A 21 B 1 + A 22 B 2 255 1963
• Slide 11
• Powers of a Matrix If A is a nn matrix and if K is a positive integer, then K can be raised to the power of A, as in A K and means that you can have K copies of A. Example A 3 = A*A*A If we have A being nonzero, and X is in the set of real numbers, then A K X is like multiplying X by A repeatedly K times. However, if K = 0, then A 0 X should be X itself as A would be interpreted as the identity matrix. In this way, matrix powers end up being very useful in both theoretical and applied means. Transpose of a Matrix If we have a given matrix, A that is if the size mxn, then the transpose of it would be an nxm matrix and is denoted by A T. In other words, you take an initial row in the original matrix and make it a column in the transposed matrix, and vice versa with the rows. Lets look at an example, A =, and if we follow the rules then A T =
• Slide 12
• Theorems and Proofs All right people, I know this is the part where it gets really technical, but in the true sense of achievement and comprehension of the concepts here, we believe this too is necessary. T heorem 1 Let A, B and C be matrices of the same size, and let p and q be scalars. a.A+B = B+A b.(A+B) + C = A + (B + C) c.A + 0 = A d.p(A + B) = pA + pB e.(p+q)A = pA + qA f.p(q A) = (pq) A The proof of all these equalities lies in showing that each of the matrices of the left hand side are equal in size to the matrix on the right hand side, and also by showing that entries in corresponding columns are equal. We already assumed that the matrices are of equal size at the start, so that is taken care of. As for the other condition, that seems to be satisfied with if we follow the analogous properties of vectors. For those of you who are not aware of these, heres an example.
• Slide 13
• If the jth column of A, B, and C are aj, bj, and cj, respectively, then the jth columns of (A+B)+C and A+(B+C) are (aj + bj) + cj and aj + (bj + cj) respectively. Since the two vector sums are equal for each j, property (b) is verified. Moreover, because of the associative property of addition, we can say that A+B+C = (A+B)+C or A+ (B+C), and the same can be applied to the sum of four or more matrices. Theorem 2 If A is an p x q matrix, and of B and C have sizes for which the indicated sums and products are defined, then using some basic laws of arithmetic and some others we can also verify the following statements: a.A (BC) = (AB) C(associative law of multiplication) b.A (B + C) = AB + AC(left distributive law) c. (B+C)A = BA+ CA(right distributive law) d.r(AB) = (rA) B = A(rB) for any scalar r e.I m A = A = AI n (identity for matrix multiplication)
• Slide 14
• Proof Statements a to d are self explanatory and the basic laws of arithmetic stated are what verify them as well. As for statement e, the identity matrix is just and that multiplied by any matrix is the matrix itself. Example = (You should be able to see why that is correct) Theorem 3 Let A and B denote matrices whose sizes are compatible enough for the following sums and products. a.(A T ) T = A b. ( A+B) T = A T + B T c.For any scalar p, (p A) T = p A T d.( AB ) T = B T A T
• Slide 15
• Proof Proofs of A to C are straight forward and thus omitted. For d, lets look at this example. Let A =, B =, then (AB) = and (AB) T = Whereas, B T = and A T =. Moreover, B T A T =. Thus, as both the answers to the left hand side and the right hand side of the equation are the same, then the statement is proved.
• Slide 16
• Theorem 4Column - row expansion of AB If A is an mxn matrix and B is an nxp, then, AB = [col 1 (A) col 2 (A) col n (A)] = col 1 (A) row 1 (B) + + col n (A)row n (B) Proof For each row index i and column index j, the (i, j)- entry in col k (A) row k (B) is the product of a ik from col k (A) and b kj from row k (B). Hence the (i,j)-entry in the sum shown in (1) is a i1 b 1j + a i2 b 2j + + a in b nj (k=1) (k=2) (k=n) This sum is also the (i, j)-entry in AB, by the row-column rule.
• Slide 17
• Row - Column rule for Computing AB: If the product AB is defined, then the entry in row i and column j of AB is the sum of the products of corresponding entries from row i of A and column j of B. If (AB) ij denotes the (i, j)-entry in AB, and if A is an m x n matrix, then (AB) ij = a i1 b 1j + a i2 b 2j + + a in b nj Warnings: Finally, there are a few things that we would like you to keep in mind. 1.In general, AB does not equal BA. 2.The cancellation laws do not hold for matrix multiplication. That is, if AB = AC. The it is not true in general that B = C. 3.If a product AB is the zero matrix, you cannot conclude in general that either A = 0 or B = 0.
• Slide 18
• Application of Matrix Multiplication Now, We are sure you must all be aware that matrix multiplication is not just a random concept that we learn in mathematics. It has many useful applications in everyday life and also in other branches of math. So now, we will spend some time looking over and trying to understand what exactly it is and where it is that matrix multiplication is applicable. The Matrix Application Ax = b As stated before, one of the most useful applications of matrix multiplication in math is in the matrix equation. The fundamental idea behind this equation is that in Linear Algebra we can view a linear combination of vectors as the product of a matrix and vector (a quantity specified by a magnitude and a direction). And it is in this fundamental concept that lies the existence of matrix multiplication. This is because a vector can be written in matrix notation and then multiplied to the other matrix to obtain one side of the equation and then carry out the appropriate operations and functions from there on end.
• Slide 19
• A formal definition stating this relationship is given below. Definition If A is a m x n matrix, with columns a(1),,a(n), and if x is in the collection of all the lists of real numbers, then the product of Ax, is the | 2,834 | 9,562 | {"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} | 4.75 | 5 | CC-MAIN-2022-21 | latest | en | 0.812056 |
http://www.lwfree.cn/shuxue/20190427/32729.html | 1,560,952,667,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998986.11/warc/CC-MAIN-20190619123854-20190619145854-00088.warc.gz | 259,160,152 | 6,571 | ## 变量代换法在微分方程求解中的应用
Application of Variable Substitution Method in Solving the Differential Equations
Abstract:This paper mainly discusses the three kinds of common variable substitution method—the function transform method,the Fourier transform method and the Laplace transform method,using the relevant properties and theorems, analyze the methods and techniques in solving differential equations of different types,and connect with the concrete examples to illustrate.
Key words:Differential equations;Function transform method;Fourier transform method; Laplace transform method
1.基本理论 3
2.函数变换法 5
2.1求解常系数线性微分方程 5
2.2求解变系数线性微分方程 7
3. 变换法 9
3.1求解波动方程初值问题 9
3.2求解热传导方程初值问题 11
4. 变换法 12
4.1求解常微分方程 12
4.2求解偏微分方程 14
1.基本理论
, 在 上连续或分段连续,且至多只有有限个第一类间断点；
在 上至多有有限个极值点.
, ,
------分隔线---------------------------- | 292 | 843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.765625 | 3 | CC-MAIN-2019-26 | latest | en | 0.33271 |
https://www.kidsacademy.mobi/printable-worksheets/second-grade/math/word-problems/learning-skills/counting-skills/ | 1,720,976,709,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514635.58/warc/CC-MAIN-20240714155001-20240714185001-00127.warc.gz | 753,875,581 | 57,325 | # Counting skills Grade 2 Word Problems Worksheets
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• Word Problems
## Assessment 3 Math Worksheet
Ariel needs to buy yarn. Do your kids know what yarn is? If they've seen you knit, they will. Help them solve the two word problems to figure out how many coins Ariel needs to buy it.
Assessment 3 Math Worksheet
Worksheet
## Assessment 2 Math Worksheet
Using fun exercises and pictures, you can get kids excited about math. With practice and reminders of what they've learnt, they'll be solving simple addition and subtraction equations with ease. Read the word problems in the picture to them, then help them check the equation and choose the correct answer.
Assessment 2 Math Worksheet
Worksheet
This tracing sheet is great for kids to work on math skills and have fun too. Read the word problems to them, then help them trace the dotted lines to match the problem and number sentence to a picture. Not only is it educational, it's beautiful too - your kids will love learning about butterflies!
Worksheet
## Counting Spider Eyes Worksheet
Did you know different spiders have different eye numbers? Kids will love learning this fact and solving the spider word problems on this free worksheet! Word problems promote a deeper understanding of the concept, plus they get to add up the spiders’ eyes while solving addition problems with more than one addend. Fun and math all in one!
Counting Spider Eyes Worksheet
Worksheet
## Counting Seedlings Worksheet
Understanding math word problems is key. Multiple steps can prove challenging - this free worksheet provides one-to-one picture representation to help kids solve multi-step addition word problems. Strengthen addition skills by choosing the matching picture to the answer.
Counting Seedlings Worksheet
Worksheet
This free worksheet helps little mathematicians learn to understand word problems. They'll read each one and match it to the corresponding picture. Important info is highlighted, teaching kids to look for key info when solving problems. Then they'll underline the correct number equation that matches the picture. A great way to practice problem-solving skills!
Worksheet
## Going Bananas Worksheet
Marla needs to send 56 bananas in batches of 8. Help your child practice division and learn about fact families with this free worksheet. It'll help boost their multiplication skills and increase speed when solving problems.
Going Bananas Worksheet
Worksheet
## Sea Math Worksheet
Dive into an ocean of addition with this fun math worksheet! Help your little mathematician count and add colorful coral and fish. Point out phrases like "in total" or "all together" as a reminder to use addition. Watch them add to find the answer and check the boxes to finish!
Sea Math Worksheet
Worksheet
## Australian Animal Subtraction Worksheet
Test your child's wildlife knowledge. Ask them to name animals indigenous to Australia or other countries. Show them the animals in this worksheet and have them match each word problem and picture with the correct number sentence.
Australian Animal Subtraction Worksheet
Worksheet
## Addition at the Zoo Worksheet
Visit the zoo with your kids and point out all the different animals. Ask them to name their favorites and identify animals on a worksheet. Challenge them to solve word problems related to the pictures. Ask them for the equation that fits the picture and check the answer.
Worksheet
## Counting with Drawings:Fruits & Flowers Worksheet
At Kids Academy, we know pictures make math come alive and make sense. This worksheet is perfect for early learners to practice word problems. Kids read each problem and count the images that represent the numbers, then solve the problem with addition or subtraction to get the right answer.
Counting with Drawings:Fruits & Flowers Worksheet
Worksheet | 750 | 3,858 | {"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-2024-30 | latest | en | 0.905969 |
https://www.physicsforums.com/threads/extending-the-fundamental-theorem-of-arithmetic-to-the-rationals.1001476/ | 1,717,075,005,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00724.warc.gz | 798,128,705 | 27,269 | # Extending the Fundamental Theorem of Arithmetic to the rationals
• B
• Warp
In summary, the Fundamental Theorem of Arithmetic states that any positive whole number can be written as a unique product of primes, and that no two distinct combinations of primes will produce the same natural number. It is relatively easy to prove that this also applies to positive rational numbers if we allow any integer as an exponent. However, the uniqueness of this extended version is unsure and it is difficult to express succinctly. Additionally, the FTA is not applicable to the rational numbers since there are no irreducible or prime numbers in a field.
Warp
The Fundamental Theorem of Arithmetic essentially states that any positive whole number n can be written as:
##n = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdot \dots##
where ##p_1##, ##p_2##, ##p_3##, etc. are all the primes, and ##a_1##, ##a_2##, ##a_3##, etc. are non-negative integers. (The theorem also essentially states that no two distinct combinations of the latter will produce the same natural number.)
I have been thinking that it's relatively easy to prove that if we lift the restriction of the exponents being non-negative, ie. we allow any integer as exponent, then the statement is true for any positive rational number n. In other words, any positive rational number n can be written as the above type of product (where the exponents are integers).
(The proof is easy to sketch by realizing that in the most-simplified forms of rational numbers the numerator and denominator are relatively prime, ie. they don't share any prime factors, which means the same prime number is never needed for both the numerator and the denominator.)
I am not sure about the uniqueness, however. Could two different combinations of integers produce the same rational number? I think that completeness (ie. any rational number can be written as a finite product like the above) is relatively easy to prove, but I'm not so sure about uniqueness.
Also, I'm not sure how this "extended" version of the theorem could be succinctly expressed. The original theorem can be expressed as "any positive whole number can be written as a unique product of primes", but this extended version is more difficult to express in such a simple manner.
An extension to the rationals does not make sense. The theorem is about primes, and there are no primes in a field like ##\mathbb{Q}##. So if you ask for a decomposition of a rational number ##n\in \mathbb{Q},## then ##n=n## is already a solution.
The idea is the question: How does a decomposition of an integer into irreducible components look like? Now irreducible and prime are the same in that case, so the question becomes: How does a decomposition of an integer into prime components look like? However, there are no irreducible elements in a field, because all numbers are either ##0## or units.
I don't see why not. Every positive rational number can be expressed as the ratio between two unique products of primes which do not share any primes. When the Fundamental Theorem of Arithmetic is expressed in the form I used in my post, it naturally extends to the (positive) rationals by simply lifting the restriction of the exponents being non-negative. The exact same equation applies in both cases.
Sure, you can write a rational number as a primitive quotient, and then apply the FTA to numerator and denominator. But you cannot generalize the theorem itself because the conditions become empty statements. You will always have to distinguish the two sides of the quotient, for otherwise ##10=2\cdot 5 = \dfrac{1}{\dfrac{1}{2}}\cdot\dfrac{1}{\dfrac{1}{5}}## would be allowed. But if you distinguish numerator and denominator, then it is no more then FTA applied twice.
Try to formulate the theorem and you will see that you cannot do it.
Every non-zero rational number can uniquely be written as a product ##\pm \prod p_i^{\alpha_i}## where the ##p##s are distinct prime integers and the ##\alpha##s are non zero integers (unique up to reordering)
2 and 5 are primes, 1/2 and 1/5 are not.
Office_Shredder said:
Every non-zero rational number can uniquely be written as a product ##\pm \prod p_i^{\alpha_i}## where the ##p##s are distinct prime integers and the ##\alpha##s are non zero integers (unique up to reordering)
Alternatively, ##q = \pm \prod p_i^{\alpha_i}## where pi is the sequence of (all) prime numbers and the integers ##\alpha_i## are non-zero for a finite number of elements (i.e. there is an N such that ##\alpha_i=0 \,\forall i>N##). This is a unique representation as well.
mfb said:
2 and 5 are primes, 1/2 and 1/5 are not.
##2## and ##5## aren't either! You cannot formulate a theorem which doesn't have to apply FTA twice in order to get the expression. There is no way to remain in the field without referring to its ring of integers.
fresh_42 said:
##2## and ##5## aren't either! You cannot formulate a theorem which doesn't have to apply FTA twice in order to get the expression. There is no way to remain in the field without referring to its ring of integers.
2 and 5 are prime numbers in the ring of integers.
Do you think what I wrote in #5 is wrong? If your only claim is it's not a very deep theory that tells you interesting new things to try to extend to more abstract objects, that might be true.
Office_Shredder said:
2 and 5 are prime numbers in the ring of integers.
Do you think what I wrote in #5 is wrong?
Of course not.
If your only claim is it's not a very deep theory that tells you interesting new things to try to extend to more abstract objects, that might be true.
My claim is that the theorem makes only sense for rings, not for fields. If you want to apply it to a quotient field, then it is the version of the integer ring applied twice. A formulation which only refers to the rational numbers is not possible, because there are no primes. Hence the OP's suggestion is simply FTA in its ordinary version applied to numerator and denominator. It is no theorem of rational numbers, i.e. nothing new under the sun. And the reason is, that there are no irreducible or prime rational numbers.
FTA over the rationals is a contradiction in itself. (For nitpickers: a tautology.)
mfb said:
2 and 5 are primes, 1/2 and 1/5 are not.
The FTA is not a statement about primes, but a statement about the natural numbers. The FTA doesn't say that 2 and 5 are primes, it just says that 2 and 5 can be expressed as a product of primes. The FTA doesn't define the prime numbers, it simply states that all positive whole numbers can be composed of prime numbers.
My extension proposal simply extends that idea to all positive rational numbers. What I noticed is that the only thing needed for this is lifting the restriction of the prime exponents being positive. The theorem still works, and now covers all positive rational numbers, not just the whole numbers.
Warp said:
The FTA is not a statement about primes, but a statement about the natural numbers.
It is a statement about the ring of integers. It needs the concept of primes to be phrased.
The FTA doesn't say that 2 and 5 are primes, it just says that 2 and 5 can be expressed as a product of primes. The FTA doesn't define the prime numbers, it simply states that all positive whole numbers can be composed of prime numbers.
Right. You said it yourself: you first need to know what a prime number is.
My extension proposal simply extends that idea to all positive rational numbers.
Wrong. It uses FTA and applies it to numerator and denominator. It does not extend anything, it applies a known fact twice. ##\mathbb{Z}\times \mathbb{Z}## as you use it, is fundamentally something different than ##\mathbb{Q}## and is important that you see this! You cannot define a prime number in the field of rational numbers. O.k., you can define them, but there are none!
What I noticed is that the only thing needed for this is lifting the restriction of the prime exponents being positive.
Wrong. You needed the entire theorem before you could even phrase your statement. It is not a generalization, it is an application.
The theorem still works, and now covers all positive rational numbers, not just the whole numbers.
Wrong. The theorem does not exist over the rational numbers! You confuse ##\mathbb{Z}^2## with ##\mathbb{Q}##.
@fresh_42 , I don't think @Warp understands your reasoning. It doesn't look like he has run across the concept of ring and field.
Vanadium 50 said:
@fresh_42 , I don't think @Warp understands your reasoning. It doesn't look like he has run across the concept of ring and field.
Yes, but the question is whether he wants to understand the concept of prime elements or not. It is important to understand that there are no prime rational numbers. And it is important to understand the difference between an application and a generalization. We don't need ring or field as terms for neither of those. But the terms could be a motivation to learn what's behind the theorem. Rings with that property even have a special name.
fresh_42 said:
It is important to understand that there are no prime rational numbers.
For someone who is so nitpicky about exact details, you have been extraordinarily careless in that statement (which is technically false). You might want to rephrase.
The Fundamental Theorem of Arithmetic is simply a statement about a unique representation of integers in terms of the product of powers of primes, where each prime appears only once. It certainly can be applied to rational numbers, since they are represented by ratios of integers. There is a unique representation of any rational number as the ratio of products of powers of prime numbers, where each prime number appears only once.
The OP expresses some doubt about how to express a theorem for the rationals. I think it would be a good exercise to look up the exact statement of the Fundamental Theorem of Arithmetic and see how it could be stated to apply to the rationals.
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FactChecker said:
I think it would be a good exercise to look up the exact statement of the Fundamental Theorem of Arithmetic and see how it could be stated to apply to the rationals.
I agree. I also think where the thread went south was here:
Warp said:
Also, I'm not sure how this "extended" version of the theorem could be succinctly expressed.
A crisp definition of exactly what we are talking about would have helped.
Warp said:
For someone who is so nitpicky about exact details, you have been extraordinarily careless in that statement (which is technically false). You might want to rephrase.
Nope,
It is important to understand that there are no prime rational numbers.
is completely correct. You will not find a prime in ##\mathbb{Q}##.
Edit: It is not nitpicky. It is about understanding. That the others say you are right isn't helpful. You can only define primality for numbers that do not have an inverse. However, all nonzero rational numbers do have an inverse. The point is, that you cannot leave the integers without destroying the meaning of a prime number. Hence you should not phrase FTA over the rationals, simply because you will learn the wrong message. The fact that others refuse to see their misguiding advice doesn't make them right.
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Any rational number can be reduced to lowest terms and then both the numerator and denominator of the reduced fraction can be factored uniquely into their unique prime representations.
fresh_42 said:
You will not find a prime in ##\mathbb{Q}##.
All integers are rational numbers (they belong to ##\mathbb{Q}##). All the prime numbers are integers. Therefore all prime numbers are rational.
By saying that prime numbers are not rational you are saying that they are irrational numbers.
Does your definition of ##\mathbb{Q}## not include the integers?
Warp said:
All integers are rational numbers (they belong to ##\mathbb{Q}##). All the prime numbers are integers. Therefore all prime numbers are rational.
And this is exactly what you got wrong and refuse to learn. Repetition doesn't make it right. The condition to be prime does not make any sense within ##\mathbb{Q}##. It can only be phrased in ##\mathbb{Z}##.
By saying that prime numbers are not rational you are saying that they are irrational numbers.
No. I am saying that the word 'prime' becomes void over the rationals:
"A number ##p## is prime if ##p## is not invertible, and from ##p|(a\cdot b)## it follows that ##p|a## or ##p|b.##"
There is simply no rational number for which this is true. Only integers.
Does your definition of ##\mathbb{Q}## not include the integers?
It includes the integers, it does not include a meaningful way to define the term 'prime'.
@fresh_42: It's clear to everyone that OP means "primes in the integers" throughout the thread.
mfb said:
@fresh_42: It's clear to everyone that OP means "primes in the integers" throughout the thread.
I'm not sure that it was originally clear, and the comments about rationals not having primes were reasonable. But I think that it was appropriate to steer the thread toward prime factorization of numerator and denominator since that has a reasonable answer.
mfb said:
@fresh_42: It's clear to everyone that OP means "primes in the integers" throughout the thread.
No, it is not. Read the first post:
Warp said:
I have been thinking that it's relatively easy to prove that if we lift the restriction of the exponents being non-negative, ie. we allow any integer as exponent, then the statement is true for any positive rational number n.
The theorem reduces to a tautology over rational numbers. The distinction between ##\mathbb{Z}^2## and ##\mathbb{Q}## is crucial! Everything else gives the OP a false impression, which we can see throughout the thread. It makes no sense to confirm a false impression. The question has been answered in the first 3 posts. The rest is an attempt to teach something which is not true. Sorry, that I do not think this is a good idea.
fresh_42 said:
The rest is an attempt to teach something which is not true. Sorry, that I do not think this is a good idea.
Theorems that establish a unique representation can have value in their own right and are very common in mathematics. Rational numbers have a unique simplest reduced form and that form has a unique representation derived from applying the Fundamental Theorem of Arithmetic separately to the numerator and the denominator. That can be rewritten as the product of powers of distinct primes, allowing negative exponents. And this product is unique up to the order of the factors. Your complaints may have merit regarding other issues, but they do not disprove the unique representation.
FactChecker said:
Theorems that establish a unique representation can have value in their own right and are very common in mathematics.
It is misleading. It has nothing to do with algebra and leads to the anywhere false impression of what prime means. It is as if I said: Every complex polynomial of odd degree has a zero. It is correct, but completely irrelevant and misses the target. As long as the OP doesn't plan to study abstract algebra at all, go ahead teaching misconceptions. But if he sometimes will study algebra, then you haven't done him a favor by using prime and field in the same context.
fresh_42 said:
It is misleading. It has nothing to do with algebra and leads to the anywhere false impression of what prime means.
He asked a simple question and the answer is a simple "Yes, there is a similar statement that can be made about the rational numbers." The integers are a well-defined subset of the rationals that have additional properties and the prime numbers are a well-defined subset of the integers that have additional properties. To deny that is misleading.
There are many contexts where it makes sense to talk about (integer) primes in the context of rational numbers. For example, every norm on ##\mathbb{Q}## is equivalent to either the trivial norm, the usual absolute value, or a ##p##-adic norm. This is an important fact about ##\mathbb{Q}## that references primes in ##\mathbb{Z}.## Just because the ##p_i## are not prime elements of ##\mathbb{Q}## doesn't mean that they shouldn't be referenced in any theorem about ##\mathbb{Q}!##
Anyway, the statement in the OP is of course true, even though the ##p_i## are not prime elements of ##\mathbb{Q}.## Uniqueness follows by clearing denominators and using unique factorization for the integers.
FactChecker
Only n, the product, was allowed to be rational. The primes stay untouched.
Warp said:
I am not sure about the uniqueness, however. Could two different combinations of integers produce the same rational number? I think that completeness (ie. any rational number can be written as a finite product like the above) is relatively easy to prove, but I'm not so sure about uniqueness.
I have been thinking that it's probably quite easy to show that the factorization is unique.
The only way that two rational numbers can be equal is if the numerator and denominator of one are multiples of the numerator and denominator of the other, using the same factor. In other words, $$\frac {a}{b} = \frac {c}{d} \Leftrightarrow \frac {a}{b} = \frac {ka}{kb}; a, b, c, d, k \in \mathbb{N}$$
However, if ##k## is not 1, then the fraction is not most-simplified. In other words, no two most-simplified fractions can be the same unless their numerator and denominator are identical.
Warp said:
the statement is true for any positive rational number
If this is amended to include the condition "irreducible", then it seems clear that it is true.
Warp said:
I am not sure about the uniqueness, however. Could two different combinations of integers produce the same rational number?
Of course they can $\frac{1}{2}$ and $\frac{17}{34}$ represent the same rational number.
Rationals do not have a unique representation. Every rational number is a member of an equivalence class of combinations of integers.
Svein said:
Of course they can $\frac{1}{2}$ and $\frac{17}{34}$ represent the same rational number.
Rationals do not have a unique representation.
But they have a unique reduced form to lowest terms, p/q where GCD(p,q)=1. Then both p and q can be factored uniquely into primes where p and q have no common primes.
FactChecker said:
But they have a unique reduced form to lowest terms, p/q where GCD(p,q)=1. Then both p and q can be factored uniquely into primes where p and q have no common primes.
The crucial point is the title. The word extension suggests a form of generalization, which it is not. You can't extend from ring to field. No way.
fresh_42 said:
The crucial point is the title. The word extension suggests a form of generalization, which it is not. You can't extend from ring to field. No way.
There are some formally defined uses of the word "extension", but extending a theorem is not one of those.
In any case, the original question has probably been answered.
mfb
fresh_42 said:
The crucial point is the title. The word extension suggests a form of generalization, which it is not. You can't extend from ring to field. No way.
It is a generalization in the sense that it is a correct statement which includes unique factorization in the integers as a special case, not in the sense that it deals with prime elements in ##\mathbb{Q}.##
<h2>1. What is the Fundamental Theorem of Arithmetic?</h2><p>The Fundamental Theorem of Arithmetic states that every positive integer can be expressed as a unique product of primes. This means that any positive integer can be written as a multiplication of prime numbers in a specific order.</p><h2>2. How does the Fundamental Theorem of Arithmetic relate to the rationals?</h2><p>The Fundamental Theorem of Arithmetic can be extended to the rationals by expressing each rational number as a product of primes in the numerator and denominator. This extension allows for the representation of any rational number as a unique product of primes, similar to the representation of integers.</p><h2>3. Why is extending the Fundamental Theorem of Arithmetic to the rationals important?</h2><p>Extending the Fundamental Theorem of Arithmetic to the rationals allows for a deeper understanding of the properties and relationships between rational numbers. It also provides a more efficient way of representing and manipulating rational numbers.</p><h2>4. Is the extension of the Fundamental Theorem of Arithmetic to the rationals proven?</h2><p>Yes, the extension of the Fundamental Theorem of Arithmetic to the rationals has been proven and is widely accepted by mathematicians. The proof involves using the unique factorization property of integers to extend to the unique factorization of rationals.</p><h2>5. Can the Fundamental Theorem of Arithmetic be extended to other number systems?</h2><p>Yes, the Fundamental Theorem of Arithmetic can be extended to other number systems, such as complex numbers and algebraic numbers. However, the unique factorization property may not hold for all number systems, making the extension more complex and requiring different approaches.</p>
## 1. What is the Fundamental Theorem of Arithmetic?
The Fundamental Theorem of Arithmetic states that every positive integer can be expressed as a unique product of primes. This means that any positive integer can be written as a multiplication of prime numbers in a specific order.
## 2. How does the Fundamental Theorem of Arithmetic relate to the rationals?
The Fundamental Theorem of Arithmetic can be extended to the rationals by expressing each rational number as a product of primes in the numerator and denominator. This extension allows for the representation of any rational number as a unique product of primes, similar to the representation of integers.
## 3. Why is extending the Fundamental Theorem of Arithmetic to the rationals important?
Extending the Fundamental Theorem of Arithmetic to the rationals allows for a deeper understanding of the properties and relationships between rational numbers. It also provides a more efficient way of representing and manipulating rational numbers.
## 4. Is the extension of the Fundamental Theorem of Arithmetic to the rationals proven?
Yes, the extension of the Fundamental Theorem of Arithmetic to the rationals has been proven and is widely accepted by mathematicians. The proof involves using the unique factorization property of integers to extend to the unique factorization of rationals.
## 5. Can the Fundamental Theorem of Arithmetic be extended to other number systems?
Yes, the Fundamental Theorem of Arithmetic can be extended to other number systems, such as complex numbers and algebraic numbers. However, the unique factorization property may not hold for all number systems, making the extension more complex and requiring different approaches.
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https://www.jiskha.com/similar?question=Solve+using+the+elimination+method.+Show+your+work.+If+the+system+has+no+solution+or+an+infinite+number+of+solutions%2C+state+this.+2x+%2B+10y+%3D+58+-14x+%2B+8y+%3D+140&page=585 | 1,571,519,264,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986697760.44/warc/CC-MAIN-20191019191828-20191019215328-00399.warc.gz | 937,515,838 | 24,351 | # Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this. 2x + 10y = 58 -14x + 8y = 140
92,096 questions, page 585
1. ## math helpp!!!
Bernie put in a garden in her back yard. The garden has a perimeter of 265 feet. She wants to know the total surface area of the garden and all she knows is the perimeter and that the width is 20% or (1/5) of the length. (Remember 20% and the fraction 1/5
asked by Kate on December 13, 2012
2. ## physics
A 16.0 g bullet is fired horizontally into a 87 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 168 N/m. The bullet becomes embedded in the block. The bullet block system
asked by Anonymous on October 22, 2012
3. ## maths
1. Convert the ffl octal numbers into denary number a) 1527 b) 21 364 2. Convert the ffl denary numbers into binary numbers a) 578 b) 709 3. Calculate the follwing using the octal system. a) 704 + 147 b) 652 - 406 c) 657 * 4 d) 754/6 Thanks
asked by temise on February 26, 2007
4. ## help asap!!
Bernie put in a garden in her back yard. The garden has a perimeter of 213 feet. She wants to know the total surface area of the garden and all she knows is the perimeter and that the width is 20% or (1/5) of the length. (Remember 20% and the fraction 1/5
asked by lilly on February 22, 2013
A 20.0 g bullet is fired horizontally into a 85 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 156 N/m. The bullet becomes embedded in the block. The bullet block system
asked by Megan on October 18, 2012
6. ## Calc. Help
Simplify. #1. 5(2-i) - 3(4+3i) ..my work: 10-5i-12-9i -2-14i Do you leave this as the answer or can you simplify it further? #2. (2-i√2)²
asked by Hayden on August 26, 2010
7. ## Science
Which two systems work with the muscular systems to allow the body to move
asked by Brana on November 6, 2016
8. ## Math
Hi there! How would I check if I got the area of a trapezoid correct? Is there a way to check my work? Thank you.
asked by Tim on May 5, 2016
9. ## math
i have 3 numbers 10,7,and 6 and need to make them work in 35...but it cant be like 7 into 35 it has to be all 3 numbers and can be used more than once but after another number like 10 + 7 +10 ect
asked by becky on March 7, 2012
10. ## Physical Science
A machine you designed has an efficiency of 60%. What happens to 60% of the work put into the machine, and what happens to the other 40%?
asked by Alex on January 28, 2014
11. ## Science
If I push a book 1.5 m across the table using a constant for of 10 N, how much work do I do on the book?
asked by Cameron on January 13, 2015
12. ## science
Which requires more work - running around the block or walking around the block?
asked by Lenny on December 2, 2009
How can health care providers work together within the continuum of care?
asked by April on September 6, 2012
14. ## grammar
Plural possesive of Mr. Valdez. (i.e., Mr. Valdez's performance at work....)
asked by Linda on October 17, 2012
15. ## College physcial science
How much work is done in raising a 10.0 kg backpack from the floor to a shelf 1.50 m above the floor?
asked by Tiki on April 25, 2009
16. ## physics
A box is pushed a distance of 2 m with a force of 20 N. How much work is done on the box?
asked by don on June 24, 2018
17. ## math
if a 1/2 kilogram of beef costs £1.60 work out the coast of 3/4 of a kilogram.
asked by Kina on October 16, 2013
18. ## trig!!
sin^4 x (cos^2 x)=(1/16)(1-cos 2x)(1-cos4x) work one 1 side only!
asked by Hayden on February 23, 2009
19. ## College Accounting
What changes have you had to make in your life to make the college experience work for you?
asked by Lynne on February 1, 2010
20. ## Algebra
let do some corrctions sqrt9x+81=x+5= answer was x=35+3sqrt61/2,35-3sqrt61/2. please work with me.
asked by Jack on July 2, 2008
21. ## Physics
A 10 g particle is moving to the left at 20 m/s. How much work must be done on the particle to cause it to move to the right at 46 m/s?
asked by Hayden on February 16, 2012
22. ## Chemistry check
H2O2+MnO2 -----> MnO2 + H2O Is my work correct?
asked by Rain on May 25, 2015
23. ## physics
A 33 g particle is moving to the left at 22 m/s. How much work must be done on the particle to cause it to move to the right at 46 m/s?
asked by kollin on October 17, 2010
24. ## communication
we know that relationships develop in satges. how does movement through relationships work
asked by A on December 1, 2012
25. ## Physics-Circuits
How does a circuit breaker work to protect against an overloaded circuit?
asked by Anonymous on May 10, 2017
26. ## science
How much work is done in raising a 16.1 kg backpack from the floor to a shelf 1.82 m above the floor?
asked by jus jo on February 17, 2018
27. ## Physics
A motor does 5,000 of work in 20 seconds what is the power of the motor
asked by Anonymous on December 14, 2010
28. ## Science
All forms of energy can do work. A.true B.false I think this is true..?
asked by Cassie on March 25, 2013
29. ## math
how would i work this out sari has 7/12 yards of material does she have more or less than 8 yards of cloths?
asked by angel on March 31, 2008
30. ## life orientation
what unethical behaviour is taking place in the work place?
asked by Anonymous on March 10, 2013
31. ## Math
here are some fractions : 3/9, 8/9, 7/9, 4/9, 2/9, 5/9, 6/9 If I were to put them in the order of the greatest, what order does that mean and how did you work that out?
asked by Blake on April 30, 2010
32. ## physics
a particle moves under the effect of force f2 from x=0 to x=4 of the work done by force
asked by ashish on August 31, 2014
33. ## question about math (law of syllogism/detachment)
Does anyone know about the law of syllogism and law of detachment. If so, can you check my work.
asked by Candy on September 13, 2009
34. ## Algebra
Factor this: x^2-5x+12 It may be prime. I tried to work it out, and I ca't figure out how, and that is whyI think it may be prime. Thanks!
asked by Katie on March 8, 2017
35. ## Math
Need help solving for x with work shown 2log6(4x)=0 Log2(x)+log2(x-3)=2
asked by Bob on January 27, 2016
36. ## sci 230
The DNA creates new proteins but once the proteins are created, can they work by themselves
asked by meshelle on April 17, 2009
37. ## science
1) How do hormones from the thyroid gland and parathyoid gland work together?
asked by samantha on November 13, 2011
38. ## algebra
How would you explain how the laws of exponents work with rational exponents?
asked by C on July 6, 2008
39. ## physics
A 75 particle is moving to the left at 29 . How much work must be done on the particle to cause it to move to the right at 46 ?
asked by Anonymous on November 28, 2012
40. ## Some questions..
How does Jiskha work? and is being a jiskha tutor an actual job?
asked by New Member on November 22, 2017
41. ## Ap CHem
For how many electrons in Sr. Does A)n=4; l=1 B)n=3;l=2;ml=1 C)n=1;l=1;ml=0 You must know the variations possible. n = 1, 2, 3, ...... l = 0, 1, 2, etc up to n-1. ml=-l to +l (this is -ell to +ell) ms = +/- 1/2 Start with n = 1 and work your way up.
asked by Susie on October 29, 2006
42. ## scince
A girl lifts a 12 N weight 1.5 meters. How much work does she do on the weight?
asked by autum on April 11, 2011
43. ## Algebra
can you help me graph the function y=-5(3^x)please? f(x)=-5(3^x) f(x)=-5*3 f(x)=-5(9) =-15 =-45 Is this the correct way to work this problem out and how to I graph?
asked by Michelle on February 17, 2017
44. ## PHYSICS!!!
If a machine cant multiply the amount of work , what is the advantage of using such a machine?
asked by Carly!!!!! on January 24, 2010
45. ## PHYSICS!!!
If a machine cant multiply the amount of work , what is the advantage of using such a machine?
asked by Carly!!!!! on January 24, 2010
46. ## Physics
Find the work necessary to run a football (1 lb) from the 2-m mark to the 52-m mark.
asked by Mirania on October 23, 2015
47. ## IT
Why is the waterfall model valuable in the management of information technology projects? Refer to "Arguments for the Waterfall Model" in http://en.wikipedia.org/wiki/Waterfall_model The waterfall model is a sequential software development model (a process
asked by Maria on February 28, 2007
48. ## Math
Can someone please help me with an algebra problem? -|4-8b|=12 I really get confused with the absolute value symbols and a negative sign outside of them. Don't worry about it; it does tend to be a bit confusing. I'll explain the difference between inside
asked by Andrea on September 14, 2006
49. ## PHYSICS
A 54.0-cm diameter disk rotates with a constant angular acceleration of 2.0 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this
asked by Kevin on July 13, 2011
50. ## English
Can you check these sentences, please? 1)If you want to reduce this problem (a synonym for reduce?) 2)He thought that in a man there are good and evil parts (there is a good and an evil part) 3)You have to reuse them (instead of use them more times) 4)He
asked by Franco on January 7, 2010
51. ## Physics
A 0.400 kg block of wood hangs from the ceiling by a string, and a 0.0700 kg wad of putty is thrown straight upward, striking the bottom of the block with a speed of 6.40 m/s. The wad of putty sticks to the block. How high does the putty-block system rise
asked by Physics on April 29, 2010
52. ## English
1.Edgar Allan Poe uses foreshadowing in "The cask of Amontillado" to create mystery in which of the following quotes? A." He had a tight-fitting parti-strip dress, and his head was surmounted by the conical cap and bells." B."And as for Luchresi, he cannot
asked by GummyBears on May 7, 2016
53. ## algebra help
The cost of producing a number of items x is given by: C = mx + b In which b is the fixed cost and m is the variable cost (the cost of producing one more item). (a) If the fixed cost is \$40 and the variable cost is \$10, write the cost equation.and then
asked by TOMMY on April 3, 2007
54. ## math
Omega System Services, a member of the Omega Health Foundation operates a free-standing ambulatory care center that averages \$60 in charges per patient. Variable costs are approximately \$10 per patient, and fixed costs are about \$1.2 million per year.
asked by Luke on August 25, 2011
55. ## Psychology - Writing Paper
Multiple sclerosis [MS] is a disease that targets the central nervous system causing dysfunction in the area(s) of the brain it attacks. Use the Multiple Sclerosis Foundation webpage, along with your text reading to learn about the disease and its impact.
asked by Anonymous on October 4, 2009
56. ## Finance
Omega System Services, a member of the Omega Health Foundation operates a free-standing ambulatory care center that averages \$60 in charges per patient. Variable costs are approximately \$10 per patient, and fixed costs are about \$1.2 million per year.
asked by Karam on June 24, 2011
57. ## algebra ll
describe the system 6x minus 2y equals 10 and 9x minus 3y equals 8 as consistent and independent, consistent and dependent , or inconsistent. explain. i put consistent and dependent .
asked by Anonymous on October 6, 2008
58. ## Accounting For Assets
Kenya Sweets Ltd. Made the following purchases of new Amigo Chocolate products during 2019: - 10th January 2019: - 1,000 Units @ Ksh. 28.00 each - 14th February 2019: - 1,500 Units @ Ksh. 28.20 each - 17th March 2019: - 3,000 Units @ Ksh. 28.40 each - 18th
asked by KNN on June 17, 2019
59. ## Psychology
1. Modern psychoanalysis is defined as ____________________. (1 point) A treatment that focuses on observable behavior A treatment method for mild to moderate disorders that uses a process of patient discovery Demonstrating how unconscious factors affect a
asked by Anonymous on May 22, 2018
60. ## chemistry
A 100.0 ml sample of 0.300 M NaOH is mixed with a 100.0 ml sample of 0.300 M HNO3 in a coffee cup calorimeter. Both solutions were initially at 35.0 degrees celcius; the temperature of the resulting solution was recorded at 37.0 degrees celcius. Determine
asked by jen on July 12, 2012
Assume that at the beginning of 2000 Speedy Delivery a Fed Ex competitor purchased a 737 aircraft at a cost of 60,000,000 Speedy Delivery expects the plane to remain useful for 5 years (5 million miles) and to have a residual value of 5,000,000 Speedy
asked by Ronnie on January 5, 2010
62. ## Chemistry
A concept map was to be created on acids and bases. The concept terms that were given to use were: aqueous solution,bases,acids,ions,weak bases, strong bases, strong acids, weak acids,dissociation, weak electrolyte, strong electrolyte,reaction,salt,
asked by Hannah on October 23, 2011
63. ## accounting
My friend Robert Hughes runs hardware business in down town Dallas. Because of competition he relies on neighborhood customers for sales. From inception of operations to December 31, 2010, Robert Hughes Hardware Company provided for uncollectible accounts
64. ## books (Henry VIII)
I know I have posted this question before put under the wrong context I accidentally wrote Hamlet instead of Henry VIII. That's because I was working on both pieces of work. My answers have changed and want to know if they are correct. Please check over my
asked by Suzanne on October 30, 2013
65. ## Science
How can a graduated cylinder marked in milliliters be used to measure the volume of a small wooden block? Hurry up and answer this for me!!!!!! put 100ml water into the cylinder, then put the block into the water and measure how many ml are indicated on
asked by Margaret on September 14, 2006
66. ## economics
1. The rationale for punishment and correction comes from A. the social contract theory. B. RawlsÂ’s theory. C. utilitarianism. D. egoism. 2. The two major justifications for treatment and punishment are A. incapacitation and deterrence. B. just
asked by Anonymous on August 14, 2009
67. ## physics
A 10.0-g bullet is fired horizontally into a 101-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 155 N/m. The bullet becomes embedded in the block. If the bullet-block system
asked by lilly on November 11, 2012
68. ## Physics
A 11.0-g bullet is fired horizontally into a 101-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 160 N/m. The bullet becomes embedded in the block. If the bullet-block system
asked by Kelly on November 11, 2012
69. ## Physics
Two crates, of mass m1 = 60 kg and m2 = 130 kg, are in contact and at rest on a horizontal surface (Fig. 4-54). A 620 N force is exerted on the 60 kg crate. The coefficient of kinetic friction is 0.15 a)Calculate the acceleration of the system. b)
asked by Syl on October 11, 2010
70. ## Physics
An ice cube of mass 100 g and at 0C is dropped into a Styrofoam cup containing 200 g of water at 25C. The heat of fusion of ice is 80 cal/g and the specific heat capacity of water is 1.0 cal/g C°. Assuming the cup doesn't exchange any heat, the final
asked by DariaV on July 28, 2019
71. ## Chemistry - Stoichiometry/gases
When 2.38 g of methane burns in oxygen, 119 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole of methane under these conditions? Your answer will be in kJ/mol. the Molar Mass of methane is 16.04, so I did 2.38g/16.04=0.14838 mol
asked by Jamie on July 23, 2017
72. ## Physics
A vertical spring has one end attached to the ceiling and a 2.50 kg weight attached to the other one. When the system is at rest, the spring is stretched by 20.0 cm. Now let the weight drop from a position in which the spring is not deformed at all.Use the
asked by Timo on June 29, 2011
plz plz help me with my assignment. This is the assignment:- •Your favorite cousin has moved to your town and is looking for a job. •Her previous experiences are working as a cashier and sales clerk at two department stores. •You know she plans to
asked by jinal on November 10, 2011
74. ## Transportation and Cargo Secuirty
Identify an asset such as an airport or a shipping company within the global transportation system. What does this mean? What is an asset of an airport or shipping company?
asked by chris on April 15, 2012
75. ## Math/calculus HELPPP
Graph the system of inequalities, and label each with a vertex with its coordinates. y-2x≥-3 y+3>−x 3y−2x≤12 2y+x
asked by Ashley on November 20, 2017
76. ## algebra II
Select the best possible first step to solving the system by first eliminating the y variable. x + 8y = –4 2x – 4y = –13 Multiply the first equation by 2. Multiply the second equation by 2. Multiply the first equation by –2. Multiply the second
asked by connor on May 19, 2011
77. ## Psychology
What is the realm of mind and consciousness? How does the realm of mind and consciousness relate to the physical body? Is there such a thing as an independent mind that transcends physical functions of the body through the central nervous system? Why or
asked by Maple C. on August 14, 2011
78. ## Physics
For anti-ballistic missile system, the time of flight tf is determined by the initial speed v0 of the missile and the range R. What is the time (in minute) of the flight tf available to kill an incoming intercontinental ballistic missile launched from
asked by Maxx on October 2, 2011
79. ## PHYSICS
A light spring (k= 130 N/m) is compressed 15 cm between two masses (1.2 kg and 2.4 kg) on a frictionless surface. The system is released and the masses fly apart. Which mass will have the larger magnitude momentum? Why? Which mass will have more kinetic
asked by Camille on February 23, 2016
80. ## Chemistry Urgent
95.0mL of H2O is initially at room temperature (22.0∘C ). A chilled steel rod at 2.0∘C is placed in the water. If the final temperature of the system is 21.3∘C , what is the mass of the steel bar? Specific heat of water = 4.18 J/g⋅∘C Specific
asked by Anna on March 11, 2014
81. ## chemistry
100 mL of is initially at room temperature (22)C. A chilled steel rod at 2 C is placed in the water. If the final temperature of the system is 21.2 C, what is the mass of the steel bar? Specific heat of water = 4.18J/g(C) Specific heat of steel =
asked by Lilit on October 20, 2008
82. ## Physics
a metal ball of radius R is placed concentrically inside a hollow metal sphere of inner radius 2R & outer radius 3R. The ballis given a charge +2Q and the hollow sphere a total charge -Q. Calculate. electrostatic potential energy of system.???
asked by Star on May 13, 2018
83. ## physics
If only external forces can cause the center of mass of a system of particles to accelerate, how can a car on level ground ever accelerate? We normally think of the car’s engine as supplying the force needed to accelerate the car, but is this true? Where
asked by Sandhya on February 18, 2010
84. ## engineering
A steel cylinder has a volume of 0.26 L and holds 4 kg of water at 25˚C and 200 kPa. (a) Find the total mass and total volume of the system (cylinder + water). (b) Give the values for two EXTENSIVE properties and two INTENSIVE properties of the water.
asked by nick on August 29, 2012
85. ## u.s. history
Which cash crops were the most successful in the international interdependence system? Cotton, corn, squash, hemp Indigo, cotton, rice, tobacco *** Indigo, corn, potatoes, rum Tobacco, corn, rice, pumpkins
asked by Labbayk on January 8, 2016
86. ## Physics
I was wondering if my answers are correct. Thanks in advance. 1. A boy pushes down on a car rolling horizontally down the road. The boy pushes down vertically with 10N of force as the car rolls 3.0m horizontally. What work is done? my answer: 30 N•m (or
asked by Liz on February 3, 2014
87. ## Math
There were 7,404,000 unemployed American workers in 1995. In 1980, there were 7,637,000 Americans unemployed. a. What was the average change in unemployment for each of these 15 years? b. If this decline continues at the same rate, how many Americans will
asked by Mariyka on September 17, 2006
88. ## math
In a car repair shop there are 50 cars to be repaired . It is estimated that 23 of that cars need new brakes , and 34 of that cars need to repair the exhaust system . a. How many cars are in need of repair at least two? b . What most cars that are not in
asked by Iman on April 7, 2015
89. ## Science
Which distance is the greatest ? A. The distance across the Milky Way Galaxy. B. The distance to the nearest star Cluster. C. The distance across the solar system. D. The distance to the nearest Galaxy.
asked by Natalie on January 30, 2012
90. ## Social Science
Write for me something about 'Role of women in India's struugle for freedom' with summary and some photos based on the above topic.And it should also contain suitable heading.Introduction part is the must in the whole. I need all this to be done because
asked by Manali on June 19, 2006
91. ## physics
Fleas can jump quite high; experiments show that they reach an altitude of 130 times their own height when they jump vertically! Suppose instead that a flea can jump so that it lands a maximum distance of 19.60 cm away. What is the take-off speed for this
92. ## MATH INTEREST
A FACTORY INCREASES THE PRODUCTION OF CARS ANNUALLY BY 8%. IF PRESENTLY IT MANUFACURES 211,000 CARS PER YR, HOW MANY CARS WILL IT MANUFACTURE PER YR AFTER 4 YRS... i BELIEVE THE ANSWER IS 287063 BUT IM UNSURE OF THE FORMULA... COULD YOU PLEASE SHOW FORMULA
asked by paully on November 7, 2010
93. ## Calculus
A curve is defined by the parametric equations: x = t2 – t and y = t3 – 3t Find the coordinates of the point(s) on the curve for which the normal to the curve is parallel to the y-axis. You must use calculus and clearly show your working, including any
asked by Daksh on July 18, 2014
94. ## Physics
A man holds a flower pot out a window 12m from the ground. He tosses the pot upward with a speed of 5m/s. How long does it take the pot to reach the ground? How fast is it moving just before it hits? Please show steps.
asked by Iris on September 25, 2013
95. ## maths
at a fairground, the table shows the prbability of winning an amount on a spinner. what is the minimum amount that can be charged to make profit from the game? amount 5p 10p 20p 50p prbability 0.5 0.25 0.15 0.1 can you show how you worked it out? thanks
asked by ?????? on November 25, 2012
96. ## economics
(2 points) Suppose that the following equations describe an economy (C, I, G, T, and Y are measured in billions of dollars and r is measured in percent; for example, r = 10 means r = 10%)
asked by kojo on August 16, 2010
97. ## physics
A ball is thrown vertically into the air and when it returns after an interval of 2 seconds it is caught. What is the acceleration of the ball after leaving the hand? I have no idea how to solve this!
asked by Jake on September 27, 2009
98. ## Math
Victoria buys two pairs of sunglasses for \$8 each about how much did she pay for both pairs of glasses round to the nearest ten solve the problem using the four step plan.
asked by Samantha on September 29, 2016
99. ## math
If Nefertiti could pronounce her name in six tenths of a second. At that rate, how many times could she pronounce her name in 15 seconds? (write and solve a fraction division problem to answer the question)
asked by Emily's mom on April 27, 2011
100. ## Math
How would I solve for t if y equals 720 minutes of daylight and t=the number of days in 2012 720 = 713.7+152.7 sin [(2pi/366)(t-80.75)]
asked by anna on November 26, 2012 | 6,752 | 23,858 | {"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} | 3.6875 | 4 | CC-MAIN-2019-43 | latest | en | 0.941599 |
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# Following exercises decide whether the statement is
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Unformatted text preview: hat N is a positive integer. 0 and an integer n nx . 1 n2x2 and we shall do so by giving an example of such a number . Now we want to prove that there exists a number N such that We define To prove that there exists an integer n N 1 N2 1 2 1 2 2 . N such that nx 1 n2x2 we make the observation that Nx 1 N2x2 . 2. For every number x 0, 1 and every number 0 there exists a number 0 such that for every number t 0, 1 satisfying |t x | we have |t 2 x 2 | . Solution: We shall prove that this statement is true. Since we want to prove that something happens for every number x 0, 1 , we begin as follows: Suppose that x 0, 1 . Now we want to prove that something happens for every number Suppose that 0 and so we continue: 0. We are now seeking a number 0 such that the inequality |t 2 x 2 | will hold whenever t 0, 1 we have |t x | . We observe first that whenever t 2|t x |. |t 2 x 2 | |t x ||t x | |t x | |t | |x | With this inequality in mind we define /2. We see at once that whenever t 0, 1 and |t t2 | x2 | 0, 1 and x | we have 2|t x| 2 2 . 3. For every number 0 and every number x 0, 1 there exists a number 0 such that for every number t 0, 1 satisfying |t x | we have |t 2 x 2 | . 24 Hint: The statement that appears here is identical to the one in Exercise 2. 4. For every number 0 there exists a number 0 such that for every number x t 0, 1 satisfying |t x | we have |t 2 x 2 | . 0, 1 and every number Solution: We shall prove that this statement is true. Since we want to prove that something happens for every number Suppose that 0 we begin: 0. We are now seeking a number 0 such that the intequality |t 2 x 2 | will hold for all numbers t and x in the interval 0, 1 satisfying the condition |t x | . To give an example of such a number we continue: Define /2. We see at once that whenever t 0, 1 and x t2 x2 | | 0, 1 and |t x| 2 2|t x | we have 2 . 5. For every number 0 and every number x there exists a number 0 such that for every number t satisfying |t x | we have |t 2 x 2 | . Solution: The statement made in this exercise says a little more than the one in Exercise 2 but it is still true. Since we want to prove that something happens for every number 0 we begin: Suppose that 0. Now we want to prove that something happens for every number x and so we continue: Suppose that x is a real number. We are now seeking a number 0 such that for every number t satisfying |t x | we have |t 2 x 2 | . To help us find an example of such a number we observe first that if |t x | 1 x x 1 t x1 then, since |t | |t x x| |t x | |x | 1 |x | we have |t 2 x 2 | |t x ||t x | |t x | |t | |x | |t x | 1 2|x | With this inequality in mind we define to be the smaller of the two numbers 1 and / 1 2|x | . We observe that whenever a number t satisfies the inequality |t x | we have 1 2|x | . |t 2 x 2 | |t x | 1 2|x | 1 2|x | 6. For...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
Ask a homework question - tutors are online | 1,176 | 3,438 | {"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} | 4.25 | 4 | CC-MAIN-2017-09 | longest | en | 0.909434 |
https://dandiapriman.web.app/171.html | 1,713,184,067,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816977.38/warc/CC-MAIN-20240415111434-20240415141434-00142.warc.gz | 173,963,970 | 5,749 | # More contour lines matlab torrent
The number of contour levels and the values of the contour levels are chosen automatically based on the minimum and maximum values of z. Contour map definition of contour map by merriamwebster. My only wish is that it was even more compatible with the contour functionality of matlab. One of the most important properties of the contour lines is to set the height of the contour lines with the help of peaks function in matlab. I want to know the value of the contour where the line crosses the contour and also the coordinates where intersection happens. Further, there may be no room to insert numerical labels for the contours. In my contourf plot i would really like to draw a contour line around areas that include values above a specified threshold value in my case. For more information about referencing vectors and matrices, see georeferenced. Millions of engineers and scientists around the world use matlab idadansoddi and design systems and. The contour lines must be long enough to fit the label, otherwise clabel does not insert a label.
Mathworks matlab r2015a 64bit mathworks matlab r2016a burst recorded team os the mathworks, matlab software provider, announced the release of the latest version of matlab r2016a. Here we look at the ez versions of contour plot commands ezcontour. How can i compute a contour bounded by the combination of. This is a function that should have been in matlab itself since the day they provided trisurf and trimesh. This app is designed to help us find out, 15 may 2018 the test tower at the mirror lab uses. Use this property to control the number of contour labels along the contour lines. Pdf development of software for making contour plot using. The matlab contour command allows to choose the number of contour lines nc.
The caxis function controls the mapping of data values into the colormap. Create contour plot of image data matlab imcontour mathworks. Ill give it a 5 rating because it deserves that rating. How to avoid artefacts in filled contour plots stack overflow. Learn more about surface, contour, 2d plot, mapping toolbox mapping toolbox, matlab. Contour chart appearance and behavior matlab mathworks. Even though ive stored the z values, they are irrelevant to our problem so i think that you can ignore them. This matlab function plots the contour lines of the function z fx,y for constant. This matlab function creates a contour plot of the regular data grid z with. Add legend labels to map contour display matlab clegendm.
Control the resolution of contour lines by using the meshdensity option. The contour calculations use a regularly spaced grid determined by the dimensions of z c contourcz computes the contour matrix from data in matrix z, where z must be at least a 2by2 matrix. Besides, i would like two show both colorbars to reflect the different limits. Recent examples on the web many locals dont put much stock in the models massport uses to create noise contour maps, published in massive reports long after the readings are taken. To allow further modifications, assign the graphics array containing the surface and contour objects to the variable sc. Draw contours in volume slice planes matlab contourslice. I would like to combine both contour plots in one axes, preserving the original color palettes. The contouring algorithm first determines which contour levels to draw.
The first row represents longitude data and the second row represents latitude data. Note, that most colormaps do not have dedicated colors for these by default. The number of contour levels and the values of contour levels are chosen automatically. I also noticed that matlab seems to have additional heuristics in placing labels. My best attempt so far is to use matlabs polyshape function to transform the contour lines into polygons, and then the intersect function which returns the.
Add contour labels to map contour display matlab clabelm. Use only the colors in the center of the hot colormap by setting the color axis scaling to a range much larger than the range of values in matrix z. This matlab function plots the contour lines of symbolic expression fx,y over. Hatching is supported in the postscript, pdf, svg and agg backends only. It does not label all the lines if the space is tight. In part 1, we focus on the three basic commands for doing this. Part 1 of a twopart series on creating contour plots maps in matlab.
Contour matrix, specified as a matrix with two rows. Active contour segmentation file exchange matlab central. Hi all, i am pretty new to matlab and have the following problem. I am reshaping your z matrix into 405x168x22, considering there are 22 strips in the contour plot of z. This matlab function labels the current contour plot with rotated text inserted into.
Add a color bar to the graph to show how the data values in c correspond to the colors in the colormap. Project 2d contour plot of map data matlab contourm mathworks. How to increase number of lines in co ntour plots learn more about contour plots, number of lines in co ntour. Increasing meshdensity can make smoother, more accurate plots, while decreasing it can increase plotting speed create two plots in a 2by1 tiled chart layout. For example, heres the relevant bit of code and the corresponding surface plot that is produced. Contour map definition is a map having contour lines. You must set the showtext property to on for the labelspacing property to have an effect. I have two different contour data sets shown as two different 2d plots. In summary, matlab cannot cope with outputting artefactfree contour or surface. Hey, ive been playing with the surface function and its really cool, but i cant figure out how to add contour lines to my plot. In the first plot, display the contours of sin x sin y. This example shows how to change the colors used in a filled contour plot. Download mathworks matlab r2016a includes crackteam os the mathworks, matlab software vendor, today announced the release of the latest version of matlab r2016a. I am plotting a 3d graph and a contour of the same.
Well create a surface and a filled contour using contourf. If you want to draw a single contour line per slice at a particular level, then specify the vector in the form lvl lvl example. In this edition, the new versions of matlab and simulink, and updates and patches includes all. Contour plot of matrix matlab contour mathworks nordic. The surface plot uses z for height and c for color. Line 124, for central difference of first derivative, is there a. The column and row indices of z are the x and y coordinates in the plane, respectively. Help with adding contour lines to a surface color plot. Contour plot in matlab learn the examples of contour. How to increase number of lines in contour plots learn more about contour plots, number of lines in contour. This matlab function draws a contour plot of the grayscale image i. The second is that to get the picture youve got above, youre going to need the locations of the bins that hist3 used. Annotation tools on the plot edit toolbar on page 32 for more. Last year we explored how surfaces perform interpolation.
Output a version of the figure without colours in pdf format. Contours for triangular grids file exchange matlab central. It handled some troublesome cases i tried when i tried to break it. This edition includes new versions of matlab and simulink and updates and patches for all other products. Increasing meshdensity can make smoother, more accurate plots, while decreasing it can increase plotting speed. Specify the colors using a colormap, which uses single numbers to stand for colors on a spectrum. Create a second contour plot and use zindex to highlight contour lines at every other integer value. If you do not have the contour matrix c, then replace c with. First, hist3 isnt in core matlab, its in the statistics toolbox. Levels, a matlab program which draws a contour plot of a function, using sampling to determine the appropriate contour levels matlabs contour function displays a contour plot of a data, described either as a table zx,y, or as a function fx,y. Skipping several fruitless deadends, it turns out that in hg2 the text labels, lines and fills are stored. The contourc function calculates the contour matrix for the other contour functions.
Follow 399 views last 30 days supreeth on 7 mar 20. Value of contour levels, specified as a numeric vector with length greater than or equal to. Label contour plot elevation matlab clabel mathworks. Mathworks matlab r2015a x86 torrent download rasenracher. Label spacing along the contour lines, specified as a scalar value in points, where one point is 172 inch. The family of contour functions consists contour, contour3, and contourf and a couple of other minor ones. Also the text is not written over the lines, but there is space in the line where the text goes. Filled 2d contour plot matlab contourf mathworks italia. From matlab graphics r2012a, from page 573 to page 576 the contouring algorithm. Pdf there is several commercial software used to make contour plot, such as. You can create a contour plot with emphasis on selected contour lines by splitting the data and creating two overlapping contour plots. Contourisolineplots contour pseudocolor mesh surf surfc filledcontour contour3 stem3 bar3 figure4. Learn more about contour, interp, interp1, interp2, interp3, smooth, lines, plot. This example shows how to label each contour line with its associated value.
How do i draw the contour for efficiency map of the motor in matlab. The values in z determine the heights of the contour lines with respect to a plane. Contour plot under surface plot matlab surfc mathworks. It is a lowlevel function that is not called from the command line. Part 2 of a twopart video series on creating contour plots in matlab. Matlab, simulink, stateflow, handle graphics, realtime workshop, and xpc targetbox are registered. We can choose any scalar value in the input argument to indicate the height of the lines. Today were going to take a look at some closely related functions.
642 320 784 770 62 1326 378 872 600 775 1248 1600 685 121 1049 279 1327 572 1595 185 35 133 1265 1372 868 729 1123 797 421 757 718 1496 676 173 1529 637 556 640 1049 988 57 143 923 213 1363 842 271 | 2,205 | 10,340 | {"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-2024-18 | latest | en | 0.875541 |
https://anchor-chart.com/s-blends-anchor-chart/ | 1,618,499,364,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00158.warc.gz | 203,277,597 | 14,635 | # S Blends Anchor Chart
S Blends Anchor Chart – Anchor graphs are posters-sized visual demonstrations which emphasize key content in a lecture or particular section of a classroom. Typically positioned at highly visible, easily accessible places, they provide students (and even teachers) with visual cues for referencing material (like tables and graphs ) while teaching and learning.
An anchor graph is a chart of letters, numbers, shapes, colors, or patterns that act as markers. An anchor graph is produced by using a collection of anchors on the graph. This will be followed with the info in the graph. This is a really useful tool in helping children learn things.
Anchor graphs can be used for studying mathematics, science, reading, language, spelling, history, arts, crafts, or perhaps physical pursuits. They can be used to guide children through the learning and teach them what’s important. They also help kids recognize learning aims and establish learning goals. After students may participate in the learning process, they’ve a fantastic part in making anchor charts. Anchors are simple shapes such as squares and circles or triangles or lines, but can also be more complicated shapes such as mountains, trees, or even a picture.
An essential part of learning is always to have fun. Kids will need to know that what they’re doing is studying. It’s okay to make mistakes and move forward. They don’t need to feel bad about anything. This may keep them motivated throughout the learning. Kids will need to realize that their efforts are important. Should they feel that they’re not participating enough, they will be more inclined to get frustrated.
Consonant Blends Anchor Charts By Mrs Davidson S Resources – S Blends Anchor Chart
The fundamental definition of an anchor graph is fairly straightforward. An anchor is something which stands for the next section of data on the graph. A good example could include employing a table, graph, or other graphic representation to provide information on a data point on the chart such as age, sex, race, and income. After the chart is created, the anchor is usually positioned at the upper left corner corner.
Bl Blend Anchor Chart Practice Click File Print By – S Blends Anchor Chart
Among the main intention of an anchor chart is to make sure that the student can quickly refer back to previously studied information and not have to stop their course to discuss every single piece of information. A child’s chart may show their birthdate and the time frame they had been born; an elementary student’s graph may present their very first year of college and the subjects they heard; and a college student’s graph may show the path they finished, in addition to any honors earned. Employing an anchor chart, a teacher can present a clear presentation of significant data and concepts while making sure the students know where they ought to focus their attention throughout the remainder of the lesson.
Consonant Blends Anchor Charts By Mrs Davidson S Resources – S Blends Anchor Chart
Bl Blend Anchor Chart Practice Click File Print By – S Blends Anchor Chart
The main reason why it’s so important to make sure students can easily refer back to their charts is because it provides them with the tools to do so. A pupil can utilize an anchor graph for a reminder instrument while they are studying. They can also use the chart to indicate significant points that they would like to consider future research and to make sure that they understand where to concentrate their attention during the course. Anchor graphs may also be utilized in the classroom to extend a sense of arrangement and order to an otherwise disorganized or chaotic lecture room.
Embedding Quotes Anchor Charts QuotesGram – S Blends Anchor Chart
How to earn anchor chart illustrations operate in the classroom is quite simple. A teacher will begin a lecture by introducing a brand new chart and create a poster of the same chart for students to reference at different points throughout the lecture. He/she then provides text links or images to visually represent significant points in the lesson, but leaves the text clean. {the visual cues of charts or tables]. Pupils then click on the right points to access the corresponding data for that particular chart, thereby ensuring that pupils are able to get it at any time.
Blends Anchor Charts By A Cupcake For The Teacher TpT – S Blends Anchor Chart
There are many methods to utilize an anchor chart, the way to earn chart illustrations work from the classroom is just a matter of finding the ideal way for you. Some educators find it best to place the graph on the board right in front of the course so that pupils can clearly see the graphs without having to discontinue their lectures or take their eyes away from the paper or board. Other educators choose to place the chart on an easel where students can observe the graphs and text without blocking the opinion of the board or paper.
Kids have pleasure when they see something that they enjoy and know that their actions or words can make it happen. This permits them to learn at a much faster rate. They will also enjoy helping others. Kids love receiving and giving presents. A simple way to let kids know they are valued is to include them at the creating of their graph. The children will love knowing they have a function in the creation of the graph. This is likely to make their participation worthwhile.
Learning can be a great deal of fun. Sometimes just having the opportunity to learn could be a challenge due to the many things that can fail. This is a great place to make an anchoring chart because children can anticipate the final result.
Learning can be rewarding, enjoyable, and fun for both kids and parents. Learning an whole subject isn’t always about sitting in front of the computer all day learning. Sometimes kids need to participate and help with something. Learning can also be fun when they are learning with a graph to help children learn and understand at a quicker speed. | 1,170 | 6,050 | {"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-2021-17 | latest | en | 0.947356 |
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m07 #27
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Intern
Joined: 03 Oct 2012
Posts: 10
Location: India
Concentration: General Management, Entrepreneurship
Schools: Tepper '16 (S)
GMAT 1: 620 Q50 V25
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WE: Operations (Computer Software)
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15 Aug 2013, 10:47
Every B.E meets every other person!
10(ways to select a B.E)*16(selecting one other person from the whole = 160
Every C meets every B.E
10*7 = 70
Total= 160+70=330
But our answer should be independent of order, i.e., A meeting B, is same as B meeting A.
Hence correct answer is total / 2
Hence
115
---------------------
bipolarbear wrote:
10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?
(A) 144
(B) 131
(C) 115
(D) 90
(E) 45
[Reveal] Spoiler: OA
C
Source: GMAT Club Tests - hardest GMAT questions
[Reveal] Spoiler: OE
Chairmen shake hands 10*7=70 with business executives times. Business executives shake hands with each other 10 C 2 times or 45 times. The total is 115 .
Can someone explain to me why its 10 C 2?
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Joined: 30 May 2013
Posts: 13
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15 Aug 2013, 15:14
hello,
could someone please explain this one thing to me.
When 10 execs shake hand with each other we say 10*9/2 as the order does no matter.
But when we do 7 chairman shaking hands with 10 execs, we are saying 7*10. Why aren't we dividing this by 2?
Intern
Joined: 13 Aug 2013
Posts: 23
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16 Aug 2013, 02:03
Hand Shakes among the business executive and chairmen - 10c1 x 7c1 = 70
Hand Shakes among the executives alone- 10c2 =45
total- 115 hand shakes
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Posts: 23
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16 Aug 2013, 02:12
1
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makhija1 wrote:
hello,
could someone please explain this one thing to me.
When 10 execs shake hand with each other we say 10*9/2 as the order does no matter.
But when we do 7 chairman shaking hands with 10 execs, we are saying 7*10. Why aren't we dividing this by 2?
This is essentially because executives and a chairmen are a different set of people. We are just selecting one from each group (selecting one from executives 10c1 and one from chirmen 7c1) . And hence the cases. 10c1 x 7c1. Whereas executives is a single group and we need to select two people from the same group for the handshake, hence 10c2
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01 Aug 2014, 00:21
10c2(executives among themselves) +10x7 (executives and chairmen)=115
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01 Aug 2014, 02:47
bipolarbear wrote:
10 business executives and 7 chairmen meet at a conference. If each business executive shakes the hand of every other business executive and every chairman once, and each chairman shakes the hand of each of the business executives but not the other chairmen, how many handshakes would take place?
(A) 144
(B) 131
(C) 115
(D) 90
(E) 45
Can someone explain to me why its 10 C 2?
This question deals with the concept of "Combination". The difference between Permutation and Combination is that in Combination "order" does not matter. Say, if A is shaking hands with B, it's equivalent to B shaking hands with A and either of the event shall be counted as One only. Similarly, in case of A, B & C, the no of "orders" possible is 3 C 2 = 3 ( A & B, B & C and A & C). So this type of counting is done using formula nCr, where n is total no of participants and r is no. of participants in one activity.
Accordingly, when there are 10 people, while 2 shake hands at a time, total no of possible is 10 C 2.
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01 Aug 2014, 10:59
gurpreetsingh wrote:
Another method
17 c 2 - 7 c 2 = 115
17 c 2 = total number of hand shakes between 17 people
7 c 2 = total number of hand shakes between chairmen
Nice way to solve the question in less time....
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Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 1,623 | 5,431 | {"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-2017-26 | longest | en | 0.916002 |
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Info on gear, conditioning, and preparation for hiking/climbing.
rking007
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### Kinda math/climbing/training type question...
For some reason the math is escaping me at the moment but we just got a treadmill at home and I wanted to figure out elevation gain based on the incline I set it too... It goes up to 15% and I can set my own program so I want to set it for a 1000 foot gain and then go from there. Can I enlist the math types out there? If I do a mile at a 15%, for example, how many feet elevation gain is that?
- Stand at the crossroads and look; ask for the ancient paths, ask where the good way is, and walk in it, and you will find rest for your souls.
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### Re: Kinda math/climbing/training type question...
Fletch wrote:880 ft. Sound right?
Wouldn't it be 5280 feet * 0.15 = 792 feet ?
Monster5
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### Re: Kinda math/climbing/training type question...
Fletch wrote:880 ft. Sound right?
Percent grade: just take 15% of the run. 0.15*5280 ft = 792 ft
For 1000 ft gain at 15%, 0.15*X = 1000 ft, X = 6,667 ft = 1.3 mi.
"The road to alpine climbing is pocked and poorly marked, ending at an unexpectedly closed gate 5 miles from the trailhead." - MP user Beckerich
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### Re: Kinda math/climbing/training type question...
jomagam wrote:
Fletch wrote:880 ft. Sound right?
Wouldn't it be 5280 feet * 0.15 = 792 feet ?
So 1000 feet of elevation would be 1000/0.15/5280 = 1.26 miles
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### Re: Kinda math/climbing/training type question...
Technically I think the mile would be the hypotenuse in this triangle because that's where the treadmill track is.
If I'm doing my math and understanding treadmills correctly you have fifteen feet up for every one hundred feet horizontal. So taking the arctangent of 15/100 you have an angle of incline of about 8.53 degrees. Take the sine of that angle and multiply it by your hypotenuse, 5280 feet, and you have a vertical gain of 783 feet.
I could be wrong, I don't know much about how they measure inclines on treadmills.
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### Re: Kinda math/climbing/training type question...
the answer is just a tad over 783 feet. woops! Robzilla beat me to it.
Robzilla
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### Re: Kinda math/climbing/training type question...
climbing_rob wrote:the answer is just a tad over 783 feet. woops! Robzilla beat me to it.
There you go, 100% of Robs agree. Can't be wrong.
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### Re: Kinda math/climbing/training type question...
I'm assuming the 15% means that you are going up .15 units for every 1 unit you go horizontally. Then we can construct a right triangle with base x, height .15x, and hypoteneuse 5280. Using the Pythagorean theorem, I calculated .15x to be about 783 ft.
Alternatively, if you want to know how far you need to run in order to do 1k ft vertical, we need to find the hypoteneuse of a right triangle having height 1000 and base 1000/0.15. Using the Pythagorean theorem, I got 6741' or about 1.3 mi.
woops, Robzilla beat me too!
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### Re: Kinda math/climbing/training type question...
Robzilla wrote:Technically I think the mile would be the hypotenuse in this triangle because that's where the treadmill track is.
If I'm doing my math and understanding treadmills correctly you have fifteen feet up for every one hundred feet horizontal. So taking the arctangent of 15/100 you have an angle of incline of about 8.53 degrees. Take the sine of that angle and multiply it by your hypotenuse, 5280 feet, and you have a vertical gain of 783 feet.
I could be wrong, I don't know much about how they measure inclines on treadmills.
I have no experience in how treadmills work, and suspect they're nowhere near accurate enough for this to matter, but defining the gradient in percentages to mean the ratio of the two legs doesn't sounds right to me. IMO a 100% grade should be mean vertical, not 45 degree as you calculate it. By your logic anything steeper than 45 is over 100% grade; that just sounds silly to me.
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### Re: Kinda math/climbing/training type question...
jomagam wrote:IMO a 100% grade should be mean vertical, not 45 degree as you calculate it. By your logic anything steeper than 45 is over 100% grade; that just sounds silly to me.
Nope, opinion or not, "100% grade" means that for every foot of horizontal, you go a foot of vertical. That's the definition of percent grade. Vertical has infinite "percent grade" (1 foot of vertical for 0 feet horizontal--> 1/0 = infinity).
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### Re: Kinda math/climbing/training type question...
Apparently 100% of Robs here are also aerospace engineers. Kooky. Rocket science came in handy again!
jomagam
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### Re: Kinda math/climbing/training type question...
climbing_rob wrote: Nope, opinion or not, "100% grade" means that for every foot of horizontal, you go a foot of vertical. That's the definition of percent grade. Vertical has infinite "percent grade" (1 foot of vertical for 0 feet horizontal--> 1/0 = infinity).
I stand corrected.
### Who is online
Users browsing this forum: No registered users and 11 guests | 1,741 | 6,113 | {"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} | 2.890625 | 3 | CC-MAIN-2017-13 | latest | en | 0.877799 |
http://mathhelpforum.com/calculus/33518-variables-separable-ode.html | 1,481,222,851,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00036-ip-10-31-129-80.ec2.internal.warc.gz | 174,864,306 | 11,039 | 1. ## variables separable ODE
When solving a variables separable ODE such as:
dy/dx=f(x) => intergral dy = integral f(x) dx
what limits does one integrate between??????
2. Originally Posted by johnbarkwith
When solving a variables separable ODE such as:
dy/dx=f(x) => intergral dy = integral f(x) dx
what limits does one integrate between??????
If the boundary condition $f(x_1) = y_1$ is given, then
$y = \int_{x_1}^{x} f(w) \, dw + y_1$.
If the boundary condition $f(x_1) = y_1$ is given and the the value of y when $x = x_2$ is required, then
$y = \int_{x_1}^{x_2} f(w) \, dw + y_1$.
Otherwise there are no limits:
$y = \int f(x) \, dx$
and the right hand side is an indefinite integral. Of course, this approach can also be taken when boundary conditions are given - see the next reply.
3. Usually you don't use limits on the integral. You evaluate the indefinite integral and add in a constant of integration. Then if you have an initial value for your differential equation you use it to find the value of the constant.
4. ## Maybe
this sample example will help $\frac{dy}{dx}=9y^4$...seperating we get $\frac{dy}{y^4}=9dx$ then by integrating we get $\frac{-1}{3y^3}=9x+C$ multiplying both sides by $-3$ we get $\frac{1}{y^3}=-27x+C_1$ and using the rules of ratios we can switch the means to give us $y^3=\frac{1}{-27x+c_1}$ finally by taking the cuberoot of each side we arrive at our answer $y=\frac{1}{[-27x+C_1]^{
\frac{1}{3}}}$ | 451 | 1,455 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 13, "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-2016-50 | longest | en | 0.77187 |
https://developer.apple.com/library/archive/documentation/mac/Legacy/GXEnvironment/GXEnvironment-367.html | 1,660,891,273,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573623.4/warc/CC-MAIN-20220819035957-20220819065957-00055.warc.gz | 215,568,031 | 3,674 | # Legacy Document
Important: The information in this document is obsolete and should not be used for new development.
Inside Macintosh: QuickDraw GX Environment and Utilities /
Chapter 8 - QuickDraw GX Mathematics / Using QuickDraw GX Mathematics
## Generating Random Numbers
You can use the QuickDraw GX random number functions to return a sequence of random numbers. The `RandomBits` function generates random integers in the range of 0 to 2count - 1, where count is the number of bits in the integer to be generated by the random number generator.
The `SetRandomSeed` function allows you to use a seed other than the default seed. If the `SetRandomSeed` function is not used, the initial seed will always be 0. You can use the `GetRandomSeed` function to return the value of the current seed.
Listing 8-8 is a sample function that generates an unsigned random number between zero and the value passed in the `limit` parameter. It uses the `RandomBits` function, and it also uses the `WideMultiply` function, correcting for the fact that `WideMultiply` works with signed long integers whereas this random generator uses unsigned longs.
Listing 8-8 A random number generator
```unsigned long RandomLong(unsigned long limit)
{
wide temp;
unsigned long random = RandomBits(32, 0);
/* This treats random and limit as signed */
WideMultiply(random, limit, &temp);
if ((long)limit < 0)
temp.hi += random; /* correct for the "sign" of limit */
if ((long)random < 0)
temp.hi += limit; /* correct for the "sign" of random */
return temp.hi;
}
```
The general topic of random numbers and the functions you use to generate them generation are discussed in the section "Random Number Generation" beginning on page 8-58. | 390 | 1,723 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2022-33 | latest | en | 0.726568 |
https://authorscast.com/is-the-letter-d-symmetrical | 1,669,577,575,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710417.25/warc/CC-MAIN-20221127173917-20221127203917-00508.warc.gz | 159,120,794 | 16,190 | # Is the letter D symmetrical?
Capital letters A, M, T, U, V, W, and Y are vertically symmetrical; capital letters B, C, D, E, and K are horizontally symmetrical; capital letters H, I, and X are both horizontally and vertically symmetrical; and capital letter O is endlessly symmetrical. Thus, all capital letters are either completely symmetrical or partially symmetrical.
Dashes are used to indicate missing letters or words. For example, a-b-c indicates that letters a, b, and c are missing from the word background. Dashes can be used in alphabetizing names that contain these gaps. They also can be used when spelling out words that contain diacritics (accents and other markings used in languages other than English). Because of their frequency in names, dashes are important items to identify when building databases for name studies.
The number 2 is considered an essential element in names. There must be at least two characters in a name. If there is only one character, it must be a space between two words or parts of a word separated by a hyphen or some other punctuation mark. Names with only spaces or punctuation marks are called "single-word names." These are the most common names and they can be difficult to sort into categories based on meaning or origin.
Names can also be composed of only digits or only letters. These are called "digit names" and "letter names," respectively.
## Which capital letters are symmetrical?
The uppercase letters H, I, O, and X all have symmetrical horizontal and vertical lines. These letters are called "symmetrical." The lowercase letters b, d, f, g, j, l, m, p, q, r, s, t, v, and w are also symmetrical.
Hence, all capitals are symmetrical.
## Does the letter "I" have point symmetry?
The A has a vertical line of symmetry, whereas the B, C, D, and E have a horizontal line. Let's take a closer look at some more letters! J, K, L, N, and P have no symmetry lines. M has one line of symmetry, while H, I, and O all have two. Thus, I can be divided into two identical shapes that reflect each other about a central axis.
I am a symmetrical shape. I have point symmetry.
Symmetry is the property of being equal to itself after any possible rotation, translation, or reflection. In other words, if you rotate, translate, or reflect any object, it will still look like the same object with only one exception: its position in space has been changed. If two objects are mirror images of each other, they are said to have symmetric properties.
A thing is called symmetrical if it is equal to itself when rotated, translated, or reflected about a central point or axis. For example, the letter I is symmetrical because it is identical to itself when turned upside down or mirrored across its center. The number 1 is also symmetrical because it remains the same after being doubled in size or reduced to half its original size.
Asking whether or not something has symmetry is similar to asking whether or not it is a rotational object. If you rotate an object around its central point, it will always return to its original position.
## What letters of the English alphabet have reflection symmetry about a horizontal mirror?
B, C, D, E, H, I, K, O, and X are the letters exhibiting reflectional symmetry about a horizontal mirror. Any letter that is upright and has an even number of lines on its body can be used as a mirror image; for example, Q and Z. Letters that are inverted (such as P and T) or have an odd number of lines on their body do not reflect symmetrically about a horizontal mirror.
In mathematics, physics, and chemistry, a reflection is the reversal of an object's position and direction but with no change in scale. Reflections can be visualized as reflections of an object within another similar object (such as a mirror), or as images formed by light reflected from a smooth surface. In computer science, a reflection is when the state of a system is identical to the original state after some transformation. For example, if you flip a digital switch that controls an electric light bulb, then the light will go out because the system state has been transformed; it is now "off" instead of "on". However, if you record the state of the system (whether the switch is closed or open) then the state has been preserved and this is called a reflection.
## Which of these letters has at least one line of symmetry with EHMR?
Expert Verified X has two lines of symmetry, whereas the other has one. At least one line of symmetry has the following letters: A, B, C, D, E, H, I, M, O, T, U, V, W, X, Y. (you can choose one).
## What are symmetrical letters?
Letters such as B and D have a horizontal line of symmetry, which means that their top and bottom sections match. Some letters, such as X, H, and O, have both vertical and horizontal symmetry lines. Some, like P, R, and N, have no symmetry lines. Symmetry is an important factor in determining how many different shapes a letter can take.
Symmetrical letters look the same from every angle, which allows you to write them even when viewing the page from a slight angle. This is useful for books and magazines, where turning the page would be difficult if the image was upside down.
Asymmetrical letters do not look the same from all angles. They are more flexible than symmetrical letters and can be written even when viewed from a slight angle. This is useful for advertisements and other printed material where being able to write even when standing back would make reading it easier.
It is also important to note that asymmetrical letters are not the same shape on both sides of the paper or screen. They will usually have one part that is wider than the other, which gives the letter its unique appearance.
Symmetrical letters come in two varieties: upright and inverted. Upright symmetrical letters look the same whether they are face up or face down. Inverted symmetrical letters look the same when turned over right-side up or upside down.
## Does the letter D have reflection symmetry?
The (capital) letters A, B, C, D, E, H, I, K, L, M, O, T, U, V, W, X, and Y all have at least one plane of reflection symmetry (depending on how you design it). This means that there is at least one direction in which these letters look the same if they are reflected across a vertical or a horizontal line.
For example, the letter D has a reflection plane along its diagonals. If we reflect a piece of paper with some printed letters on it across this diagonal, then all the letters will still be able to be read even though some of them have been mirrored over.
This shows that the letter D has reflection symmetry along its diagonals. Some letters, such as the G, N, and Z, don't have any reflection planes so they can't have reflection symmetry.
Here are other examples:
The number 2 has reflection symmetry about its center. If you draw a line through the middle of a square cut out from a sheet of paper, then you will find that it passes through exactly two corners. These are called "reflection points".
The letter D has reflection symmetry about its axis. If you look at a capital D from above, you will see that it is symmetrical from right to left.
##### Veronica Brown
Veronica Brown is a freelance writer and editor with over five years of experience in publishing. She has an eye for detail and a love for words. She currently works as an editor on the Creative Writing team at an independent publisher in Chicago, Illinois. | 1,627 | 7,486 | {"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} | 3.140625 | 3 | CC-MAIN-2022-49 | latest | en | 0.916421 |
https://www.wyzant.com/resources/answers/15575/factoring | 1,527,337,275,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867417.71/warc/CC-MAIN-20180526112331-20180526132331-00105.warc.gz | 869,228,299 | 17,109 | 0
# Factoring?
I can not figure this question out..
Solve the equation by factoring:
15x^2 + 11x - 56 = 0
Is this even possible?
### 2 Answers by Expert Tutors
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
0
15x2 + 11x - 56 = 0
Let's break this down...
15x2
(x)(x)=x2
(5)(3)=15 or (15)(1)=15
11x
11x=(11x+0x) or (10x+1x) or (9x+2x), etc
-56
(-8)(7) or (8)(-7)
Let's factor...
15x2 + 11x - 56 = 0
(?x+ ??)(?x- ??)
Because the 11x is positive, the added ??x must be greater than the subtracted ??x.
(3x+7)(5x-8)=0
Let's FOIL...
FIRST.....15x2
OUTER...-24x
INNER.....35x
LAST.......56
15x2-24x+35x+56=0
15x2+11x-56=0
Thank you!
You are SOOOOO welcome.
Keep those questions coming.
Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...
4.8 4.8 (4 lesson ratings) (4)
0
I will show you a trick that solves all this kind of problems.
15X+ 11X - 56
A quadratic like this can only be factorable, if There exist 2 rational numbers whose Sum is 11, and
their product is - 15 * 56
So we rite 15 * 56 as product of its factor
15 * 56 = 3 * 5 * 7 * 8 = 15 * 56 = 35 * 24 =
Now we see that the difference of 35 and 24 is 11
15 X+ 35X - 24X -56 ( break up 11X into 35X- 24X)
5X( 3X +7 ) - 8 ( 3X + 7) , Now 2 terms have common factor of (3X +7 )
( 3X + 7 ) ( 5X - 8 ) This is the final answer. | 584 | 1,470 | {"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} | 4.21875 | 4 | CC-MAIN-2018-22 | latest | en | 0.760782 |
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1 / 30
# Engineering Orientation - PowerPoint PPT Presentation
Engineering Orientation. Engineering Economics. Engineering Economics. Value and Interest Cost of Money Simple and Compound Interest Cash Flow Diagrams Cash Flow Patterns Equivalence of Cash Flow Patterns. Value and Interest.
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### Engineering Orientation
Engineering Economics
Engineering Economics
• Value and Interest
• Cost of Money
• Simple and Compound Interest
• Cash Flow Diagrams
• Cash Flow Patterns
• Equivalence of Cash Flow Patterns
Value and Interest
• “Value” is not synonymous with “amount”. The value of an amount of money depends on when the amount is received or spent.
• First Cost is what you pay for an item when you buy it
Value and Interest
• The difference between the anticipated amount in the future and its current value is called interest.
• At an interest rate of 10% what is the value now of the expectation of receiving \$1 in one year?
Cost of Money
• Interest that could be earned if the amount invested in a business or security was instead invested in government bonds or in time deposit.
Cost of Money
• Buy a car for \$20,000 of your own cash vs. US bonds returning 5%/yr (\$1,000 forever)
• In effect you are paying \$1,000 for ever (even after the car is a certifiable clunker destined for destruction)
Simple and Compound Interest
• You have a business project costing \$100,000
• You get a loan for 7.5% yearly for 5 years at simple interest payable at the end of the loan
• The loan costs \$7,500 for each of five years for a total interest of \$37,500
• Total cost over 5 years = \$137,500
Simple and Compound Interest
• Is the banker really willing to lend you money for 5 years? Isn’t he also lending you
• \$7,500 for four years,
• \$7,500 for three years,
• \$7,500 for two years,
• \$7,500 for one year
Terms and formulae
• P or PV Principal is the amount borrowed
• N # of pay periods
• r Interest rateper period
• F or FV, Future worth, value in the future of what you have to payback
• Formulae:
• Simple interest = P(1 + Nr) ( = \$137,500)
• Compound interest = P(1 + r)N ( = \$143,563)
Pay periods
• Calculate FV
• Assume your loan is compounded quarterly, monthly or daily instead of yearly.
• Student loan of \$25,000 at 8% for
• Annually for two years,
• Quarterly for two years and
• Daily for two years
Study Examples
• Compute the effective annual interest rate ie equivalent to 8% nominal annual interest compounded continuously.
• Calculate the PV
Example
• What amount must be paid in two years to settle a current debt of \$1,000 if the interest rate is 6% Annually?
Example
• A new widget twister, with a life of six years, would save \$2,000 in production costs each year. Using a 12% interest rate, determine the highest price that could be justified for the machine. Although the savings occur continuously throughout each year, follow the usual practice of lumping all amounts at the ends of years.
Example
• A new widget twister, with a life of six years, would save \$2,000 in production costs each year. Using a 12% interest rate, determine the highest price that could be justified for the machine. Although the savings occur continuously throughout each year, follow the usual practice of lumping all amounts at the ends of years.
Example
• How soon does money double if it is invested at 8% interest?
Example 2
• Find the value in 2002 of a bond described as “Acme 8% of 2015” if the rate of return set by the market for similar bonds is 10%.
Example
• Compute the annual equivalent maintenance costs over a 5-year life of a laser printer that is warranted for two years and has estimated maintenance costs of \$100 annually. Use i = 10%.
Return on Investment
• ROI = The ratio of annual return to the cost of the investment
• If an investment of \$500,000 produces an income of \$40,000 per year, its ROI = \$40,000/\$500,000 = 0.08 = 8%.
• Many successful large companies operate with ROI’s of 15% or more
PAYMENTS AT BEGINNINGS OF YEARS
• Using a 10% interest rate, find the future equivalent of:
Cost of losing one semester
• Two students, Frank and Mary start they Engineering Studies on the same date and they make the commitment of retiring thirty years after their forecasted graduation date (the date they would graduate if no delays are introduced). This date will not change if any delays make any of them graduate later.
• Calculate the difference in their earnings if for some reasons Frank is required to graduate one semester later than what was intended.
The Model
• Assumptions:
• We have a constant inflation
• You have a yearly Salary Increase greater than what you loose because of inflation
• Salary increases and inflation are constant
• They work for the same company their entire carrier
Study Examples
• Your perfectly reliable friend, Frank, asks for a loan and promises to pay back \$150 two years from now. If the minimum interest rate you will accept is 8%, what is the maximum amount you will loan him?
• a) \$119 b) \$126 c) \$129 d) \$139
• The annual amount of a series of payments to be made at the end of each of the next twelve years is \$500. What is the present worth of the payments at 8% interest compounded annually?
• a) \$500 b) \$3,768 c) \$6,000 d) \$6,480
Study Examples
• Maintenance expenditures for a structure with a twenty-year life will come as periodic outlays of \$1,000 at the end of the fifth year, \$2,000 at the end of the tenth year, and \$3,500 at the end of the fifteenth year. With interest at 10%, what is the equivalent uniform annual cost of maintenance for the twenty-year period?
• a) \$200 b) \$262 c) \$300 d) \$325
• The purchase price of an instrument is \$1 2,000 and its estimated maintenance costs are \$500 for the first year, \$1 500 for the second and \$2500 for the third year. After three years of use the instrument is replaced; it has no salvage value. Compute the present equivalent cost of the instrument using 1 0% interest.
• a) \$14,070 b) \$15,570 c) \$15,730 d) \$16,500
Study Examples
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• a) \$7,000 b) \$9,400 c) \$9,625 d) \$9,725 | 1,672 | 6,874 | {"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-2017-47 | latest | en | 0.884152 |
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when a certain prescription drug is taken orally by an adult, the amount of the drug (in mg/L) in the bloodstream at t hours is given by the function y =f(t) shown in the graph
Borys_S
School: Purdue University
I wrote the solutions. Take docx or pdf file, they are the same. Ask if something is unclear.
1. What is the concentration of the drug in the bloodstream at t = 8 hours?
To find the answer, we draw a vertical line that goes through the point (8, 0) at the x-axis (shown in
green). It has only one intersection with the graph. Draw a horizontal line through this point of
intersection (red) and read the value at the y-axis. In this case it i...
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http://en.wikipedia.org/wiki/Respirometry | 1,369,217,595,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368701577515/warc/CC-MAIN-20130516105257-00082-ip-10-60-113-184.ec2.internal.warc.gz | 84,142,329 | 19,623 | # Respirometry
Respirometry is a general term that encompasses a number of techniques for obtaining estimates of the rates of metabolism of vertebrates, invertebrates, plants, tissues, cells, or microorganisms via an indirect measure of heat production (calorimetry).
## Whole-animal metabolic rates
The metabolism of an animal is estimated by determining rates of carbon dioxide production (VCO2) and oxygen consumption (VO2) of individual animals, either in a closed or an open-circuit respirometry system. Two measures are typically obtained: standard (SMR) or basal metabolic rate (BMR) and maximal rate (VO2max). SMR is measured while the animal is at rest (but not asleep) under specific laboratory (temperature, hydration) and subject-specific conditions (e.g., size or allometry[1]), age, reproduction status, post-absorptive to avoid thermic effect of food).[2] VO2max is typically determined during aerobic exercise at or near physiological limits.[3] In contrast, field metabolic rate (FMR) refers to the metabolic rate of an unrestrained, active animal in nature.[4] Whole-animal metabolic rates refer to these measures without correction for body mass. If SMR or BMR values are divided by the body mass value for the animal, then the rate is termed mass-specific. It is this mass-specific value that one typically hears in comparisons among species.
### Closed respirometry
Respirometry depends on a "what goes in must come out" principle.[5] Consider a closed system first. Imagine that we place a mouse into an air-tight container. The air sealed in the container initially contains the same composition and proportions of gases that were present in the room: 20.95% O2, 0.03% CO2, water vapor (the exact amount depends on air temperature, see dew point), 78% (approximately) N2, and a variety of trace gases making up the rest (see Earth's atmosphere). As time passes, the mouse in the chamber produces CO2 and water vapor, but extracts O2 from the air in proportion to its metabolic demands. Therefore, as long as we know the volume of the system, the difference between the concentrations of O2 and CO2 at the start when we sealed the mouse into the chamber (the baseline or reference conditions) compared to the amounts present after the mouse has breathed the air at a later time must be the amounts of CO2/O2 produced/consumed by the mouse. In a closed system, the environment will eventually become hypoxic.
### Open respirometry
For an open-system, design constraints include washout characteristics of the animal chamber and sensitivity of the gas analyzers.[6][7] However, the basic principle remains the same: What goes in must come out. The primary distinction between an open and closed system is that the open system flows air through the chamber (i.e., air is pushed or pulled by pump) at a rate that constantly replenishes the O2 depleted by the animal while removing the CO2 and water vapor produced by the animal. The flow rate must be high enough to ensure that the animal never consumes all of the oxygen present in the chamber while at the same time, the rate must be low enough so that the animal consumes enough O2 for detection. For a 20 g mouse, flow rates of about 200 ml/min through 500 ml containers would provide a good balance. At this flow rate, about 40 ml of O2 is brought to the chamber and the entire volume of air in the chamber is exchanged within 5 minutes. For other smaller animals, chamber volumes can be much smaller and flow rates would be adjusted down as well. Note that for warm-blooded or endothermic animals (birds and mammals), chamber sizes and or flow rates would be selected to accommodate their higher metabolic rates.
#### Calculations
Calculating rates of VO2 and/or VCO2 requires knowledge of the flow rates into and out of the chamber, plus fractional concentrations of the gas mixtures into and out of the animal chamber. In general, metabolic rates are calculated from steady-state conditions (i.e., animal's metabolic rate is assumed to be constant[8] rates of oxygen consumed, one needs to know the location of the flow meter relative to the animal chamber (if positioned before the chamber, the flow meter is "upstream," if positioned after the chamber, the flow meter is "downstream"), and whether or not reactive gases are present (e.g., CO2, water, methane, see inert gas).
For an open system with upstream flow meter, water (e.g., anhydrous calcium sulfate) and CO2 (e.g., a carbon dioxide adsorbent like Ascarite II, a registered trademark of the Arthur H. Thomas Co.) removed prior to oxygen analyzer, a suitable equation is
$VO_2 = \frac {FR \cdot (FiO_2 - FeO_2)} {1 - FeO_2}$
For an open system with downstream flow meter, water and CO2 removed prior to oxygen analyzer, a suitable equation is
$VO_2 = \frac {FR \cdot (FiO_2 - FeO_2)} {1 - FiO_2}$
where
• FR is the flow rate adjusted to STP (see Standard conditions for temperature and pressure)
• FiO2 is the fractional amount of oxygen present in the incurrent air stream (the baseline or reference), and
• FeO2 is the fractional amount of oxygen present in the excurrent air stream (what the animal has consumed relative to baseline per unit time).
For example, values for BMR of a 20 g mouse (Mus musculus) might be FR = 200 ml/min, and readings of fractional concentration of O2 from an oxygen analyzer are FiO2 = 0.2095, FeO2 = 0.2072. The calculated rate of oxygen consumption is 0.58 ml/min or 35 ml/hour. Assuming an enthalpy of combustion for O2 of 20.1 joules per milliliter, we would then calculate the heat production (and therefore metabolism) for the mouse as 703.5 J/h.
### Respirometry equipment
For open flow system, the list of equipment and parts is long compared to the components of a closed system, but the chief advantage of the open system is that it permits continuous recording of metabolic rate. The risk of hypoxia is also much less in an open system.
Pumps for air flow
• Vacuum Pump: a pump is needed to push (i.e., upstream location) or pull (i.e., downstream location) air into and through the animal chamber and respirometry flow-through system.
• Subsample pump: To pull air through the analyzers, a small, stable, reliable pump is used.
Flow meter and flow controllers
• Bubble flow meters: A simple, yet highly accurate way to measure flow rates involves timing movement of bubbles of soap film up glass tubes between marks of known volume.[9] The glass tube is connected at the bottom (for push systems) or at the top (for pull systems) to the air stream. A small rubber pipette bulb attached at the base of the tube acts as both a reservoir and delivery system for the soap bubbles. Operation is simple. First, wet the glass surface along the path bubbles travel (e.g., press the bulb so that copious amounts of soap are pushed up the glass by the air flow) to provide a virtually friction-free surface. Second, pinch the bulb so that one clean bubble is produced. With a stopwatch in hand, record the time required for the bubble to travel between marks on the glass. Note the volume recorded on the upper mark (e.g., 125 = 125 ml), divide the volume by the time required to travel between marks and the result is the flow rate (ml/sec). These instruments can be purchased from a variety of sources, but they may also be constructed from appropriate-sized, glass volumetric pipettes.
• Acrylic flow meters : Under some circumstances of high flow rates we may use simple acrylic flow meters (0 - 2.5 liters/min) to control the flow rates through the metabolic chambers. The meters are located upstream from the metabolic chambers. The flow meters are simple to use but should be calibrated twice daily for use in the respirometry system: once before recording begins (but after the animal has been sealed inside the chamber!!) and again at the end of the recording (before the animal is removed from the chamber). Calibration must be done with a bubble flow meter because the calibration marks on the acrylic meters are only approximate. For proper calibration of flow rates remember that both barometric pressure and temperature of the air streaming through the flow meter (which we assume to be equal to room temperature) must be recorded.
• Mass flow meters: The equations required for calculating rates of oxygen consumption or carbon dioxide production assume that the flow rates into and out of the chambers are known exactly. We use mass flow meters which have the advantage of yielding flow rates independent of temperature and air pressure. Therefore, these flow rates can be considered to be corrected to standard conditions (Standard Temperature Pressure). We only measure and control flow at one location—downstream from the chamber. Therefore, we must assume that the inflow and outflow rates are identical. However, during construction of the respirometry system, flow rate must be measured at all steps, across all connections, to verify integrity of flow.
• Needle valves: Mass flow meters may be purchased with mass flow controllers which permit setting flow rates. These are expensive, however. Respirometry research often will attempt to measure more than one animal at a time, which would necessitate one chamber per animal and thus controlled flow through each chamber. An alternative and more cost-effective method to control flow would be via stainless steel or carbon steel needle valves. Needle valves plus mass flow meters provides a cost-effective means to achieve desired flow rates. The valves cost about \$20.
Tubing and chambers
• Tubing and connections : Various kinds of tubing can be used to connect the components of the respirometry system to and from the animal chamber. A variety of kinds of flexible tubing may be used, depending on the characteristics of the system. Acetyl, Bev-A-Line, Kynar, nylon, Tygon tubing and connectors may be used in regions of the system where oxidizing atmospheres are low (e.g., background levels of ozone only); Teflon tubing would be recommended if there is an expectation for appreciable amounts of ozone to be present because it is inert to ozone. Teflon tubes are more costly and lack flexibility.
• Metabolic chambers: Chambers may be glass jars with rubber stoppers for lids; syringe barrels for small animals and insects; or constructed from Plexiglas. Ideally, chambers should be constructed from inert materials; for example, the acrylic plastics can absorb O2 and may be a poor choice for respirometry with very small insects.[10] Chambers need to be constructed in a manner that yields rapid mixing of gases within the chamber. The simplest metabolic chamber for a small vertebrate might be a glass jar with a stopper. The stoppers are fitted with two ports: short extensions of Teflon tubing are provided for line connections. Teflon tube extensions are pushed through the bulkhead and the line connection is finished by attaching a small hose clip to the base of the Teflon tube extension. Additionally, an extension to the inlet port inside the jar should be provided—this ensures that the animal's expiratory gases are not washed away by the in flow stream. The animal is sealed inside and the rubber stopper is held in place with Velcro straps. If an upstream system is used, any metabolic chamber leak will result in loss of animal air and, therefore, an underestimate of the animal's metabolic rate. When you close an animal inside a metabolic chamber, attention must be paid to the seal. To ensure tight seals before closing the lid, firmly work the stopper into the jar and make sure that it is even. Use 1-2 straps (2 are better) and pull tightly. Acrylic (Plexiglas) chambers will be constructed for some uses, but precise engineering will be needed to ensure proper seating; gaskets will help, and judicious use of tight-fitting clamps will minimize leaks.
• Scrubbing tubes: Water before and after the animal chamber must be removed. One arrangement would use a large acrylic column of Drierite (8 mesh (scale), i.e., relatively coarse) upstream (before the push pump, before the animal chamber) to dry incurrent airstream and several tubes with smaller mesh (10-20, i.e., relatively fine) Drierite to remove water after the animal chamber. To prepare a scrubbing tube, make sure there is a small amount of cotton at either end of the tube to prevent dust particles from traveling to the analyzers. Use small amounts of cotton, say around 0.005 g, just enough to keep the dust out of the tubing. Large amounts of cotton will block air flow when/if it gets damp. Pour the Drierite into the tube with a funnel, tap the tube on the bench to pack the grains tightly (to increase surface area - air + water rushes through loose Drierite, requiring frequent changes of scrubbers), and cap off with a small amount of cotton. To remove carbon dioxide] before and after the animal chamber, Ascarite II is used (Ascarite II is a registered trademark of the Arthur H. Thomas Co.). Ascarite II contains NaOH, which is caustic (so don't get any on your skin and keep away from water). A scrubbing tube is prepared by placing a small amount of cotton into the tube end, filling one-third of the way with 10-20 mesh Drierite, adding a small amount of cotton, then an additional third of the tube with the Ascarite II, another layer of cotton, followed by more Drierite and capping the tube off with another small amount of cotton. Tap the tube on the bench as each layer is added to pack the grains. Note: Driereite can be used over and over again (after heating in an oven), although indicating Drierite will lose color with repeated drying; Ascarite II is used once and will be considered a hazardous waste.
Analyzers
• Carbon dioxide analyzer: CO2 analyzers typically use infrared-based detection methods to take advantage of the fact that CO2 will absorb infra-red light and re-emit light at slightly longer wavelengths. The panel meter on the analyzer displays over the entire 0.01 - 10% CO2 range and a voltage output proportional to CO2 concentration is also generated for data recording.
• Oxygen analyzer: Oxygen analyzers suitable for respirometry use a variety of oxygen sensors, including galvanic ("ambient temperature"), paramagnetic, polarographic (Clark-type electrodes), and zirconium ("high temperature") sensors. Galvanic O2 analyzers use a fuel cell containing an acidic electrolyte, a heavy-metal anode and a thin gas-permeable membrane. Since the partial pressure of O2 near the anode is zero, O2 is driven by diffusion to the anode via the membrane at a rate proportional to ambient O2 partial pressure. The fuel cell produces a voltage linearly proportional to the O2 partial pressure at the membrane. As long as cabinet temperature is stable, and provided that air flow across the fuel cell is stable and within range, the response will be 0.01% or better depending on supporting electronics, software, and other considerations.
Finally, a computer data acquisition and control system would be a typical addition to complete the system. Instead of a chart recorder, continuous records of oxygen consumption and or carbon dioxide production are made with the assistance of an analog to digital converter coupled to a computer. Software captures, filters, converts, and displays the signal as appropriate to the experimenter's needs. A variety of companies and individuals service the respirometry community (e.g., Sable Systems, Qubit Systems, see also Warthog Systems).
## Mitochondrial metabolic rates
Inside the body oxygen is delivered to cells and in the cells to mitochondria, where it is consumed in the process generating most of the energy required by the organism. Mitochondrial respirometry measures the consumption of oxygen by the mitochondria without involving an entire living animal and is the main tool to study mitochondrial function.[11] Three different types of samples may be subjected to such respirometric studies: isolated mitochondria (from cell cultures, animals or plants) permeabilized cells (from cell cultures) permeabilized fibers or tissues (from animals) In the latter two cases the cellular membrane is made permeable by the addition of chemicals leaving selectively the mitochondrial membrane intact. Therefore, chemicals that usually would not be able to cross the cell membrane can directly influence the mitochondria. By the permeabilization of the cellular membrane the cell stops to exist a living, defined organism leaving only the mitochondria as still functional structures. Unlike whole-animal respirometry, mitochondrial respirometry takes place in solution, i.e. the sample is suspended in a medium. Today mitochondrial respirometry is mainly performed with a closed-chamber approach.
### Closed-chamber System [11]
The sample suspended in a suitable medium is placed in a hermetically closed metabolic chamber. The mitochondria are brought into defined “states” by the sequential addition of substrates or inhibitors. Since the mitochondria consume oxygen, the oxygen concentration drops. This change of oxygen concentration is recorded by an oxygen sensor in the chamber. From the rate of the oxygen decline (taking into account correction for oxygen diffusion) the respiratory rate of the mitochondria can be computed.
### Applications
#### Basic research
The functioning of mitochondria is studied in the field of bioenergetics.[12] Functional differences between mitochondria from different species are studied by respirometry as an aspect of comparative physiology.[13][14]
#### Applied research
Mitochondrial respirometry is used to study mitochodrial functionality in mitochondrial diseases or diseases with a (suspected) strong link to mitochondria, e.g. diabetes mellitus type 2,[15][16] obesity[17] and cancer.[18] Other fields of application are e.g. sports science and the connection between mitochondrial function and aging.[19]
### Equipment
The usual equipment includes a seal-able metabolic chamber, an oxygen sensor, and devices for data recording, stirring, thermostatisation and a way to introduce chemicals into the chamber. As described above for whole-animal respirometry the choice of materials is very important.[11] Plastic materials are not suitable for the chamber because of their oxygen storage capacity. When plastic materials are unavoidable (e.g. for o-rings, coatings of stirrers, or stoppers) polymers with a very low oxygen permeability (like PVDF as opposed to e.g. PTFE) may be used. Remaining oxygen diffusion into or out of the chamber materials can be handled by correcting the measured oxygen fluxes for the instrumental oxygen background flux. The entire instrument comprising the mentioned components is often called an oxygraph. The companies providing equipment for whole-animal rspirometry mentioned above are usually not involved in mitochondrial respiromety. The community is serviced at widely varying levels of price and sophistication by companies like Oroboros Instruments, Hansatech, Respirometer Systems & Applications, YSI Life Sciences or Strathkelvin Instruments .
## References
1. ^ White, C. R., and R. S. Seymour. 2005. Allometric scaling of mammalian metabolism. Journal of Experimental Biology 208(9):1611-1619.
2. ^ Blaxter, K. 1989. Energy metabolism in animals and man. Cambridge University Press. ISBN 0-521-36931-2
3. ^ Weibel, E. R., and H. Hoppeler. 2005. Exercise-induced maximal metabolic rate scales with muscle aerobic capacity. Journal of Experimental Biology 208(9):1635-1644.
4. ^ Nagy, K. A. 2005. Field metabolic rate and body size. Journal of Experimental Biology 208(9):1621-1625.
5. ^ Frappell, P. B., H. A. Blevin, and R. V. Baudinette. 1989. Understanding respirometry chambers: what goes in must come out. Journal of Theoretical Biology 138(4):479-494. PMID 2593683
6. ^ Withers, P. C. 2001. Design, calibration and calculation for flow-through respirometry systems. Australian Journal of Zoology49:445-461.
7. ^ Lighton, J. R. B. 2008. Measuring metabolic rates: A manual for scientists. Oxford University Press. ISBN 0-19-531061-6.
8. ^ Bartholomew, G. A., D. Vleck, and C. M. Vleck. 1981. Instantaneous measurements of oxygen consumption during pre-flight warm-up and post-flight cooling in Sphingid moths and Saturniid moths. Journal of Experimental Biology90(1):17-32.
9. ^ Levy, A. 1964. The accuracy of the bubble meter method for gas flow measurements. Journal of Scientific Instruments 41(7):449-453.
10. ^ Stevens, E. D. 1992. Use of plastic materials in oxygen-measuring systems. Journal of Applied Physiology 72:801-804
11. ^ a b c Gnaiger, E. 2008. Polarographic oxygen sensors, the oxygraph and high-resolution respirometry to assess mitochondrial function. In: Mitochondrial Dysfunction in Drug-Induced Toxicity (Dykens J.A. and Will Y., eds) John Wiley: 327-352. ISBN 978-0-470-11131-4
12. ^ Gnaiger E, ed (2007) "Mitochondrial Pathways and Respiratory Control". OROBOROS MiPNet Publications, Innsbruck, Electronic 1st edition, ISBN 978-3-9502399-0-4
13. ^ Hildebrandt, T.M. and Grieshaber, M.K., 2008 Three enzymatic activities catalyze the oxidation of sulfide to thiosulfate in mammalian and invertebrate mitochondria. FEBS J. (275): 3352-3361.
14. ^ Nann A. Fangue N.A., Richards J.G., and Schulte1 P.M. 2009. "Do mitochondrial properties explain intraspecific variation in thermal tolerance?". Journal of Experimental Biology 212:514-522.
15. ^ Phielix E., Schrauwen-Hinderling V.B., Mensink M., Lenaers E., Meex R., Hoeks J., Kooi M.E., Moonen-Kornips E., Sels J.P., Hesselink M.K., Schrauwen P., 2008 Lower intrinsic ADP-stimulated mitochondrial respiration underlies in vivo mitochondrial dysfunction in muscle of male type 2 diabetic patients. Diabetes 57(11): 2943-9.
16. ^ Knauf C., Cani P.D., Ait-Belgnaoui A., Benani A., Dray C., Cabou C., Colom A., Uldry M., Rastrelli S., Sabatier E., Godet N., Waget A., Pénicaud L., Valet P., Burcelin R., 2008. Brain glucagon-like peptide 1 signaling controls the onset of high-fat diet-induced insulin resistance and reduces energy expenditure. Endocrinology 149: 4768-4777.
17. ^ Hoeks J., Briedé J.J., de Vogel J., Schaart G., Nabben M., Moonen-Kornips E., Hesselink M.K., Schrauwen P., 2008. Mitochondrial function, content and ROS production in rat skeletal muscle: effect of high-fat feeding. FEBS Lett. 582: 510-516.
18. ^ Increase in mitochondrial biogenesis, oxidative stress, and glycolysis in murine lymphomas Enrique Sampera, E., Morgadob, L., Estradab, J.C., Bernadb, A., Hubbarda, A., Susana Cadenas, S. and Melova S., 2009. Increase in mitochondrial biogenesis, oxidative stress, and glycolysis in murine lymphomas. Free Radical Biology and Medicine 46(3): 387-396.
19. ^ Hutter E., Unterluggauer H., Garedew A., Jansen-Durr P. and Gnaiger E. 2006 High-resolution respirometry - a modern tool in aging research. Exp. Gerontol. 41:103-109. | 5,232 | 22,871 | {"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": 2, "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-2013-20 | latest | en | 0.919803 |
http://www.chegg.com/homework-help/questions-and-answers/carbonic-acid-h2co3-diprotic-acid-ionization-constants-ka1-45-10-7-ka2-47-10-11-calculate--q2415153 | 1,369,285,293,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368702810651/warc/CC-MAIN-20130516111330-00039-ip-10-60-113-184.ec2.internal.warc.gz | 393,960,407 | 7,937 | How to solve this...?
Carbonic acid, H2CO3, is a diprotic acid with ionization constants: Ka1 = 4.5 × 10-7, Ka2 = 4.7 × 10-11. Calculate the pH of a 0.100 molar solution of Na2CO3.
• effective pKa for a diprotic salt is = (pKa1 + pKa2)/2
and the pH of salt of weak acid and strong base is given by
pH = 0.5(pKa(effective) - logC)
therefore,
pH = 0.5((pKa1 + pKa2)/2 - logC)
pH = 0.5 * ((((-log(4.5 * (10^(-7)))) - log(4.7 * (10^(-11)))) / 2) + log(0.1))
pH = 3.66867241 | 204 | 477 | {"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-2013-20 | latest | en | 0.764566 |
https://betterworld2016.org/how-many-oz-is-one-egg/ | 1,660,505,688,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572063.65/warc/CC-MAIN-20220814173832-20220814203832-00063.warc.gz | 143,047,223 | 6,088 | When a recipe calls because that 1 cup of egg whites, simply how numerous eggs space we talking about using? So, how many eggs room in a cup of egg whites? we tackled this and also other egg questions.
You are watching: How many oz is one egg
In cooking, eggs room usually offered in 3 various ways once they room raw: overall eggs, egg whites and also egg yolks. The volume of an egg counts on numerous things consisting of the dimension of the egg. Since most recipes that call for eggs space referring come "Large" dimension eggs, we made decision to usage a normal big grocery keep egg in our test to answer the concern "What is the volume of an egg?".
we looked at exactly how much egg yolk is in one average large egg. Just how much egg white is in an median egg. And what is the total volume of an egg. It may "crack girlfriend up" come realize that 1 egg yolk actions 1 to 1.25 tablespoon and the white is 2 tablespoons. This provides the entirety shelled egg measure about 3 tablespoons. Using these exact same eggs, the yolk sweet .6 ounce (18 grams) and egg white to be 1.05 ounces (30 grams).
Some people also want to understand "How many eggs in a cup?" The prize is there space 4 to 5 typical size big whole eggs in a cup. If you desire 1 cup of egg white skin - man you"ll need 7 to 8 eggs and also for 1 cup of egg yolks that takes 12 come 13 egg to achieve that quantity. Walk you recognize that the shade of the yolk is figured out by what form of food the hen eats? Also, an egg"s shell color doesn"t indicate the high quality or nutritional worth of one egg, however rather the each other of the hen that laid it. An typical hen lays 250 to 300 egg a year and as she grow older she produce larger and also larger eggs. Ever before wonder if an egg you found in her refrigerator is raw or hard-cooked? no a problem, simply spin it! If the egg spins easily, it is cooked however if it wobbles, the is raw!
Now you know what the volume of an egg is and how plenty of eggs space in a cup, your baking and cooking need to go much an ext smoothly. To identify other exactly egg measurements, you can use the converter below. If you have tendency to different eggs often, you might want to invest in one egg separator. This \$5 OXO good Grips 3-in-1 Egg Separator is among the finest we"ve found and also the one we usage at home.
## One Egg Equals
There is ¾ Ounces (22 mls) that Egg Yolk in a EggThere is 1 ounce (30 mls) the Egg White in a EggThere is 1¾ Ounces (52 mls) of raw Egg in a Egg ns need:¼½¾11 ½22 ½33 ½44 ½5678910Teaspoon(s)Tablespoon(s)Fluid Ounce(s) in VolumeCup(s)Pint(s)Quart(s)Gallon(s)Milliliter(s)Liter(s)ofEgg YolkEgg WhiteRaw EggYou need10.9Eggs
## Want a an elaborate Cocktail?
complete Recipe: acting Manhattan Cocktail
## General exactly how to save Egg Info
To avoid eggs from absorbing odors indigenous other foods items in the refrigerator, save them in the initial carton.
Besides straightforward white or brown chicken eggs, other common types include: duck eggs (larger 보다 chicken, v a more powerful flavor), quail eggs (smaller 보다 chicken and also generally supplied in specialty recipes or together hors d"oeuvres), and goose egg (much larger than chicken and much stronger in flavor).
## Short hatchet Egg Storage
For ideal flavor, usage eggs in ~ 1 week of purchase. Egg will store for 5 weeks there is no loss of nutrients or practical properties.
## Egg long Term Storage
To freeze: crack the egg and beat until just blended, then location 1 egg mixture in each muffin tin compartment to freeze them individually. Ar muffin tin right into the freezer; once solid, pop them out and also store in a plastic freezer bag. Come use, remove but you need and simply thaw in the refrigerator. If girlfriend prefer, beat every the eggs together, ar in an airtight plastic container and also freeze them every together; around 2 tablespoons of the slightly mixed egg mixture will certainly equal about one big egg.
## Egg side Notes
In addition to eating eggs for breakfast, eggs are important throughout the cooking process by helping thicken food, acting together an emulsifier, and also binding food together.
See more: How Long Does It Take To Drive To Florida From New York ? How Far Is Florida From Newyork
The color of the egg covering (white or brown) walk not impact its flavor, quality, or nutrients; shade is identified by the breed of the chicken.
## How to make Ramen-Style Eggs with Sous Vide?
Gary Huang request Jason, my challenge is do ramen eggs the peel there is no making a full mess. I tried adhering to Joule"s cooking recipes at 194°F (90°C) for 9 minutes and then soaking in an ice water bath however the covering sticks like glue destroying the eggs. Ns cracked the shells and tried peeling underwater as well but no dice. I"m curious if girlfriend have any kind of advice to help? | 1,169 | 4,870 | {"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} | 3.6875 | 4 | CC-MAIN-2022-33 | latest | en | 0.961015 |
https://nrich.maths.org/public/leg.php?code=72&cl=4&cldcmpid=2374 | 1,566,412,191,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316150.53/warc/CC-MAIN-20190821174152-20190821200152-00485.warc.gz | 583,266,669 | 9,693 | # Search by Topic
#### Resources tagged with Generalising similar to Thousand Words:
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### Rational Roots
##### Age 16 to 18 Challenge Level:
Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Shape and Territory
##### Age 16 to 18 Challenge Level:
If for any triangle ABC tan(A - B) + tan(B - C) + tan(C - A) = 0 what can you say about the triangle?
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Magic Squares II
##### Age 14 to 18
An article which gives an account of some properties of magic squares.
### AMGM
##### Age 14 to 16 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Generally Geometric
##### Age 16 to 18 Challenge Level:
Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.
### More Twisting and Turning
##### Age 11 to 16 Challenge Level:
It would be nice to have a strategy for disentangling any tangled ropes...
### Fractional Calculus III
##### Age 16 to 18
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
### Interpolating Polynomials
##### Age 16 to 18 Challenge Level:
Given a set of points (x,y) with distinct x values, find a polynomial that goes through all of them, then prove some results about the existence and uniqueness of these polynomials.
### Irrational Arithmagons
##### Age 16 to 18 Challenge Level:
Can you work out the irrational numbers that belong in the circles to make the multiplication arithmagon correct?
### Absurdity Again
##### Age 16 to 18 Challenge Level:
What is the value of the integers a and b where sqrt(8-4sqrt3) = sqrt a - sqrt b?
### Fibonacci Factors
##### Age 16 to 18 Challenge Level:
For which values of n is the Fibonacci number fn even? Which Fibonnaci numbers are divisible by 3?
### What's Possible?
##### Age 14 to 16 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### All Tangled Up
##### Age 14 to 18 Challenge Level:
Can you tangle yourself up and reach any fraction?
### Sums of Pairs
##### Age 11 to 16 Challenge Level:
Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”
### Winning Lines
##### Age 7 to 16
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Pair Products
##### Age 14 to 16 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### One, Three, Five, Seven
##### Age 11 to 16 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Games Related to Nim
##### Age 5 to 16
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Lower Bound
##### Age 14 to 16 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Pentanim
##### Age 7 to 16 Challenge Level:
A game for 2 players with similarities to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Steel Cables
##### Age 14 to 16 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Integral Sandwich
##### Age 16 to 18 Challenge Level:
Generalise this inequality involving integrals.
### Square Pizza
##### Age 14 to 16 Challenge Level:
Can you show that you can share a square pizza equally between two people by cutting it four times using vertical, horizontal and diagonal cuts through any point inside the square?
### Cyclic Triangles
##### Age 16 to 18 Challenge Level:
Make and prove a conjecture about the cyclic quadrilateral inscribed in a circle of radius r that has the maximum perimeter and the maximum area.
### Mystic Rose
##### Age 14 to 16 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Pareq Calc
##### Age 14 to 16 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Generating Triples
##### Age 14 to 16 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Partly Painted Cube
##### Age 14 to 16 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Plus Minus
##### Age 14 to 16 Challenge Level:
Can you explain the surprising results Jo found when she calculated the difference between square numbers?
### Jam
##### Age 14 to 16 Challenge Level:
A game for 2 players
### Sliding Puzzle
##### Age 11 to 16 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
### Nim
##### Age 14 to 16 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Nim-like Games
##### Age 7 to 16 Challenge Level:
A collection of games on the NIM theme
### Cuboid Challenge
##### Age 11 to 16 Challenge Level:
What's the largest volume of box you can make from a square of paper?
### Take Three from Five
##### Age 14 to 16 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### Harmonic Triangle
##### Age 14 to 16 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### For Richer for Poorer
##### Age 14 to 16 Challenge Level:
Charlie has moved between countries and the average income of both has increased. How can this be so?
### Converging Means
##### Age 14 to 16 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
### Odd Differences
##### Age 14 to 16 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Pick's Theorem
##### Age 14 to 16 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Loopy
##### Age 14 to 16 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### Chord
##### Age 16 to 18 Challenge Level:
Equal touching circles have centres on a line. From a point of this line on a circle, a tangent is drawn to the farthest circle. Find the lengths of chords where the line cuts the other circles.
### Gnomon Dimensions
##### Age 14 to 16 Challenge Level:
These gnomons appear to have more than a passing connection with the Fibonacci sequence. This problem ask you to investigate some of these connections.
### Incircles
##### Age 16 to 18 Challenge Level:
The incircles of 3, 4, 5 and of 5, 12, 13 right angled triangles have radii 1 and 2 units respectively. What about triangles with an inradius of 3, 4 or 5 or ...?
### Building Gnomons
##### Age 14 to 16 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible. | 2,181 | 9,257 | {"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} | 3.671875 | 4 | CC-MAIN-2019-35 | latest | en | 0.797205 |
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Multiplying Monomials Worksheet and Answer Key
Free worksheet(pdf) and answer key on multiplying monomials . Over 25 scaffolded questions that start relatively easy and end with some real challenges. Plus model problems explained step by step
Multiplying Monomials - Algebra-Class
This first example shows how you use multiplying powers with the same base in order to multiply monomials. Multiplying Monomials. Need Help With Your Homework? Enter your monomial or polynomial expression and click "simplify" to see the correct answer. For further help with monomials, you may want to visit our next page on dividing monomials.[PDF]
7-1 Multiplying Monomials - CVUSD Home
that will help you multiply monomials. Multiplying monomials is often used when comparing a characteristic of several items, such as acidity of different fruits. It is also used when determining the probability of something, like guessing the correct answer on a test or winning the lottery (Chapter 12).[PDF]
I. Model Problems. II. Practice III. Challenge Problems VI
I. Model Problems A monomial is an expression that is a number, variable or product of a number and variables. Examples of monomials: –3, 4x, 5xy, y2 To multiply monomials, multiply all the coefficients and all the variables.
Multiplying Monomials with Polynomials Worksheet (pdf) and
Free worksheet(pdf) and answer key on Multiplying Polynomials with Monomials. 23 scaffolded questions that start relatively easy and end with some real [PDF]
Chapter 7 Resource Masters - d2ct263enury6rdfront
873950 Al g 1 CH07 EP4 Chapter 7 8 Glencoe Algebra 1 Skills Practice Multiplying Monomials NAME _____ DATE_____ PERIOD _____
Adding, Multiplying, and Subtracting Monomials Worksheets
Multiplying Monomials Worksheet 3 - The difficulty level is slightly higher here than the sheets below. Adding & Subtracting Monomials Homework - Why not send them home with a brief reminder of what you worked on today. Answer Keys View Answer Keys- All the answer [PDF]
Multiplying Monomials and Powers of Monomials - MGCCC
Multiplying Monomials and Powers of Monomials Jefferson Davis Learning Center, Sandra Peterson Simplify. Answers 1. ()()x 3x 1. 3x2 2. (−2ab)(−2a2) 2. 4a3b3.
Multiplying Polynomials Answer Key - Course Hero
View Notes - Multiplying Polynomials Answer Key from MATH Algebra 1 at Monroe Township High School, NJ. Kuta Software - Infinite Algebra 1 Name_ Multiplying Polynomials Date_ Period_ Find each[PDF]
Answers (Anticipation Guide and Lesson 7-1)
Study Guide and Intervention Multiplying Monomials Monomials A monomial is a number, a variable, or the product of a numbe r and one or more variables with nonnegative integer exponents. An expression of the form x n is called a power and represents the product you obtain when x is used as a factor n times. To multiply
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