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Plus predicaments of error modeling
Cropped from Bacon Sandwich source
Sir David Spiegelhalter is a British statistician. He is a strong voice for the public understanding of statistics. His work extends to all walks of life, including risk, coincidences, murder, and sex.
Today we talk about extending one of his inventions.
His invention has to do with grading the performance of people and models that make predictions. A scoring rule grades how often predictions are right. But it may not tell how difficult the situations are. It is easy to look good with predictions when they start with a high chance of success. A weather forecaster predicting sunny-versus-rainy will be right more often in Las Vegas than in Boston. Quoting this FiveThirtyEight item:
If you want to have an easy life as a weather forecaster, you should get a job in Las Vegas, Phoenix or Los Angeles. Predict that it won’t rain in one of those cities, and you’ll be right about 90 percent of the time.
In a 1986 paper, for a particular scoring rule defined by Glenn Brier in 1950, Spiegelhalter worked out how to equalize the forecaster grading. He applied his Z-test not to weather as Brier was concerned with but to medical prognoses and clinical trials.
What I am doing with a small group of graduate students in Buffalo is trying to turn Spiegelhalter’s kind of Z-test around once more. If a forecaster fares poorly, we will try to flag not the model but the behavior of the subjects being modeled. In weather we would want to tell when Mother Nature, not the models, has gone off the rails. Well, we are actually looking for ways to tell when a human being has left the bounds of human predictability for reasons that are inhuman—such as cheating with a computer at chess. And maybe it can shed more light on whether our computers can possibly “cheat” with quantum mechanics.
## Prediction Scores
Let’s consider situations ${t}$ in which the number ${\ell = \ell_t}$ is usually more than ${2}$, that is, usually more than “rain” or “no rain.” The forecaster lays down projections ${\vec{q} = \vec{q}_t = (q_1,\dots,q_\ell)}$ for the chance of each outcome. If outcome ${r}$ happens, then the Brier score for that forecast is
$\displaystyle B^{\vec{q}}(r) = (1 - q_r)^2 + \sum_{j \neq r} q_j^2. \ \ \ \ \ (1)$
If the forecaster was certain that ${r}$ would happen and so put ${q_r = 1}$, all other ${q_j = 0}$, then the score would be zero. Thus lower is better for the Brier score.
If you put probability ${q_r < 1}$ on the outcome that happened, then you get penalized both for the difference and for the remaining probability which you put on outcomes that did not happen. It is possible to decompose the score in another way that changes the emphasis:
$\displaystyle B^{\vec{q}}(r) = 1 + Q - 2q_r \qquad\text{where}\qquad Q = \sum_{j=1}^\ell q_j^2.$
Then ${Q}$ is a fixed measure of how you spread your forecasts around, while all the variability in your score comes from how much stock you placed in the outcome that happened. The worst case is having put ${q_r = 0}$, whereupon your Brier penalty is ${2}$.
We would like our forecasts always to be perfect, but reality gives us situations that are inherently nondeterministic—with unknown “true probabilities” ${\vec{p}_t = (p_1,\dots,p_\ell)}$. The vital point is that the forecaster should not try to hit ${r = r_t}$ on the nose at every time ${t}$ but rather to match the true probabilities. Once we postulate ${\vec{p}}$, the expected Brier score is
$\displaystyle \begin{array}{rcl} \mathsf{E}_{\vec{p}}[B^{\vec{q}}] &=& \sum_{i=1}^\ell p_i B^{\vec{q}}(i)\\ &=& \sum_{i=1}^\ell p_i (1 - 2q_i + Q)\\ &=& 1 + Q - 2\sum_{i=1}^\ell p_i q_i. \end{array}$
This is uniquely minimized by setting ${q_i = p_i}$ for each ${i}$, which defines ${B}$ as a strictly proper scoring rule. Without the second term ${\sum_{j \neq r} q_j^2}$ in (1) the rule would not be proper for ${\ell > 2}$. When ${\vec{q} = \vec{p}}$, ${Q}$ becomes equal to ${P = \sum_{j=1}^\ell p_j^2}$. Thus ${P}$ represents an unavoidable prediction penalty from the intrinsic variance. If all ${p_i}$ are equal, ${p_i = \frac{1}{\ell}}$, then the expected score cannot be less than ${1 - \frac{1}{\ell}}$.
A second example, the log-likelihood prediction scoring rule, is in the original longer draft of this post.
## Spiegelhalter’s Z
Spiegelhalter’s ${z}$-score neatly drops out the unavoidable penalty term by taking the difference of the score with the expectation. Schematically it is defined as
$\displaystyle \mathsf{Z}[B] = \frac{B - \mathsf{E}[B]}{\sqrt{\mathsf{Var}[B]}},$
where ${\mathsf{Var}[B]}$ means the projected variance ${\mathsf{E}_{\vec{p}}[B^2] - (\mathsf{E}_{\vec{p}}[B])^2}$. However, here is where it is important to notate the whole series of forecasting situations ${t = 1,\dots,T}$ with outcomes ${r_t}$ for each ${t}$. The actual statistic is
$\displaystyle \mathsf{Z}_{\vec{p}}[B^{\vec{q}}] = \frac{\sum_{t=1}^T B^{\vec{q}_t}(r_t) - \mathsf{E}_{\vec{p}_t}[B^{\vec{q}_t}]}{\sqrt{\sum_{t=1}^T \mathsf{Var}_{\vec{p}_t}[B^{\vec{q}_t}]}}. \ \ \ \ \ (2)$
The denominator presumes that the forecast situations are independent so that the variances add. The numerator expands to be
$\displaystyle \sum_{t=1}^T \left(2\sum_{i=1}^{\ell_t} p_{i,t} q_{i,t}\right) - 2q_{r,t}.$
The original application is a confidence test of the “null hypothesis” that the projections ${\vec{q}}$ are good. Thus we plug in ${p_{i,t} = q_{i,t}}$ for all ${t}$ and ${i}$ so that we test
$\displaystyle \mathsf{Z}_{\vec{q}}[B^{\vec{q}}] = 2\frac{\sum_{t=1}^T \left(\sum_{i=1}^{\ell_t} q_{i,t}^2 \right) - q_{r,t}}{\sqrt{\sum_{t=1}^T \mathsf{Var}_{\vec{q}_t}[B^{\vec{q}_t}]}}.$
To illustrate, suppose we do ten independent trials of an event with four outcomes whose true probabilities are ${(0.1,0.2,0.3,0.4)}$. The sum in parentheses is ${10(0.01 + 0.04 + 0.09 + 0.16) = 3}$. If the outcomes conform exactly to these probabilities then ${q_{r,t}}$ equals ${0.1}$ once, ${0.2}$ twice, ${0.3}$ three times, and ${0.4}$ four times. This exactly cancels the ${3}$, so ${\vec{q} = \vec{p}}$ makes ${\mathsf{Z}_{\vec{q}}[B^{\vec{q}}] = 0}$, as expected. Most trials will give a nonzero numerator, but in the long run, the numerator divided by ${T}$ tends toward zero and the denominator scales to match it, thus keeping the ${Z}$-statistic normally distributed.
A high ${Z}$, on the other hand—highly positive or highly negative—indicates that the forecasting is way off. That (2) is an aggregate statistic over independent trials justifies treating the ${Z}$-values as standard scores. This applies also to ${Z}$-tests made similarly from other scoring rules besides the Brier score. The test thus becomes a verdict on the model. High ${Z}$-values on certain subsets of the data may reveal biases.
Our idea is the opposite. Suppose we know that the forecasts are true, or suppose they have biases that are known and correctable over moderately large data sets. We may then be able to fit ${\mathsf{Z}[B]}$ as an unbiased estimator (of zero) over large training sets. Then it can become a judgment of whether the data has become unnatural.
## Why This Z?
As I have detailed in numerous posts on this blog, my system for detecting cheating with computers at chess already provides several statistical ${z}$-scores. Why would I want another one?
The motive involves the presence of multiple strong chess-playing programs, each with its own quirks and distribution of values for moves. They are used in two different ways:
1. As inputs telling the relative values ${v_i}$ of moves ${m_i}$, which my model converts into its probability projections ${q_i}$.
2. As output predicates telling how often the player chose the move recommended by a specific program and/or quantifying the magnitude of error for different played moves.
Having multiple engines helps point 1. My intent to blend the values ${v_i}$ from different engines has been blunted by issues I discussed here. Thus I now have to train my model separately (and expensively) for each (new version of each) program. I can then blend the ${q_i}$, but point 2 still remains at issue: My tests measure concordance with a specific program. Originally the program Rybka 3 was primary and Houdini 4B secondary. Now Stockfish 7 is primary and Komodo 10.0 secondary—until I update to their latest versions. The second engine is supposed to confirm a positive result from the first one. This already means that my model is not trying to detect exactly which program was used.
Nevertheless, my results often vary between testing engines. The engines compete against each other and may be crafted to disagree on certain kinds of moves. They agree with each other barely 75–80% in my tests. I would like to factor these differences out.
The Spiegelhalter ${Z}$-test appeals because its reference is not to a particular chess program, but to the prediction quality of my model itself—which per point 1 can be informed by many programs in concert. It gives a way to predicate predictivity. A high value will attest that the sequence of played moves falls outside the range of predictability for human players of the same rated skill level.
## The Method
To harness ${\mathsf{Z}[F]}$ for some scoring rule ${F}$, we need to quantify the nature of my model’s ${q_i}$ projections. In fact, my model has a clear bias toward conservatism in judging the frequency of particular non-optimal moves. This is discussed in my August post on my model upgrade and shown graphically in an appended note on why the conservative setting of a “gradient” parameter is needed to preserve dynamical stability. The fitting offsets this in a way that creates an opposite bias elsewhere. I hope to correct both biases at the same stroke by a specific means of modeling how the ${q_i}$ err with respect to the postulated true probabilities ${p_i}$.
We postulate an original source of error terms ${\epsilon_i}$ all i.i.d. as ${\mathcal{N}(0,\delta^2)}$, where ${\delta}$ governs the magnitude of Gaussian noise. This noise can be transformed and related in various ways, e.g.:
1. ${q_i = p_i \pm \epsilon_i}$,
2. ${q_i = p_i(1 \pm \epsilon_i)}$,
3. ${\frac{1}{q_i} = \frac{1}{p_i} \pm \epsilon_i}$,
4. ${\log(\frac{1}{q_i}) = \log(\frac{1}{p_i}) \pm \epsilon_i}$,
5. ${\log(\frac{1}{q_i}) = \log(\frac{1}{p_i})(1 \pm \epsilon_i)}$,
6. ${\ln(\frac{q_i}{1 - q_i}) = \ln(\frac{p_i}{1 - p_i}) \pm \epsilon_i}$.
There are further forms to consider and it is not yet clear from data within my model which one most applies. We would be interested in examples where these representations have been employed and in observations about their natures.
Given the error terms, we can write each ${p_i}$ as a function of ${q_i}$ and ${\epsilon_i}$. One issue is having at most ${\ell-1}$ degrees of freedom among ${\epsilon_1,\dots,\epsilon_\ell}$, owing to the constraint that the ${q_i}$ as well as ${p_i}$ sum to ${1}$. We handle this by choosing some fixed ${k}$ as the “pivot” and using the constraints to eliminate ${p_k}$ and ${q_k}$, leaving the other error terms free. In all cases, the proposed method of defining what we notate as ${\mathsf{Z}_{\vec{q},\vec{\epsilon}}[F]}$ is:
• Substitute the terms with ${q_i,\epsilon_i}$ for each free ${p_i}$ into ${\mathsf{Z}_{\vec{p}}[F^{\vec{q}}]}$.
• Compute the expectation over ${\epsilon_i \sim \mathcal{N}(0,\delta^2)}$ for the numerator and denominator of (2), separately.
• Holding the other previously-fitted model parameters in place, fit ${\delta}$ so that ${\mathsf{Z}_{\vec{q},\vec{\epsilon}}[F]}$ is zero over the training set (or sets, for each level of Elo rating ${R}$, so ${\delta}$ becomes a function of ${R}$).
If the resulting ${Z}$-scores parameterized by ${\delta_R}$ make sense, the last step will be adjusting them to conform to normal distribution, via the resampling process mentioned recently here and earlier here. We are not there yet. But observations from Spiegelhalter tests with ${\vec{q} = \vec{p}}$ (equivalently, with ${\delta_R}$ fixed to zero) suggest that the resulting single, authoritative, “pure” predictivity test may rival the sharpness of my current tests involving specific chess programs.
## Error Quirks and Queries
To see a key wrinkle, consider the first error form. It is symmetrical: ${p_i = q_i \pm \epsilon_i}$. When we substitute ${q_i + \epsilon_i}$ for ${p_i}$ and take ${\mathsf{E}_{\epsilon_i \sim \mathcal{N}(0,\delta^2)}[\cdots]}$, the symmetry of ${\epsilon_i}$ around ${0}$ makes it drop out of the numerator of (2), and out of everything in the denominator except one place where ${p_i^2}$ becomes ${(q_i^2 + 2\epsilon q_i + \epsilon_i^2)}$. There is hence nothing for ${\delta}$ to fit and we are basically left with the original Spiegelhalter ${Z}$.
In the second form, however, we get ${p_i = q_i \cdot \frac{1}{1 + \epsilon_i}}$. If we presume ${\delta}$ small enough to make the distribution of ${\mathcal{N}(0,\delta^2)}$ outside ${(-1,1)}$ negligible, then we can use the series expansion to approximate
$\displaystyle p_i \approx q_i(1 - \epsilon_i + \epsilon_i^2 - \epsilon_i^3 + \epsilon_i^4).$
Under normal expectation, the odd-power terms drop out (so their signs don’t matter) and we get
$\displaystyle \mathsf{E}_{\epsilon_i \sim \mathcal{N}(0,\delta^2)}[q_i(1 - \epsilon_i + \epsilon_i^2 - \epsilon_i^3 + \epsilon_i^4)] = q_i(1 + \delta^2 + 3\delta^4).$
This credits ${p_i}$ as being greater than ${q_i}$. Provided the projections for the substituted indices ${i}$ were generally slightly conservative, this has hope of correcting them.
Already, however, we have traipsed over some pitfalls of methodology. One is that the normal expectation
$\displaystyle \mathsf{E}_{\epsilon \sim \mathcal{N}(0,\delta^2)}[\frac{1}{1+\epsilon}] = +\infty,$
regardless of how small ${\delta}$ is. For any ${\delta}$, regions around the pole ${\epsilon = -1}$ get some fixed finite probability. Another is the simple paradox of our second form saying:
${q_i}$ is an unbiased estimator of ${p_i}$, but ${p_i}$ is not an unbiased (or even finite) estimator of ${q_i}$.
A third curiosity comes from the fourth error form. It gives ${q_i = p_i e^{\epsilon_i}}$, so ${p_i = q_i e^{-\epsilon_i}}$. We have
$\displaystyle \mathsf{E}_{\epsilon \sim \mathcal{N}(0,\delta^2)}[e^{b\epsilon}] = e^{0.5b^2 \delta^2}$
exactly, without approximation. Again the sign of ${\epsilon_i}$ does not matter. So we get
$\displaystyle \mathsf{E}_{\vec{\epsilon}}[p_i] = q_i e^{0.5\delta^2} > q_i.$
But by the original fourth equation we get
$\displaystyle \mathsf{E}_{\vec{\epsilon}}[q_i] = p_i e^{0.5\delta^2} > p_i.$
So we have ${\mathsf{E}[q_i] > p_i}$ and ${\mathsf{E}[p_i] > q_i}$, with both expectations being over the same noise terms. This is like the famous Lake Wobegon syndrome. What it indicates is the need for care in where and how to apply these error representations.
## Open Problems
Have you seen this idea of directly testing (un)predictability in the literature? Might it improve the currently much-debated statistical tests for quantum supremacy?
Which error model seems most likely to apply? Where have the paradoxes in our last section been noted?
[some wording tweaks]
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# John spends 30% of his income on his children's education
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John spends 30% of his income on his children's education, 20% on recreation and 10% on healthcare. The corresponding spending by Mike are 40%, 25% and 13%.
Who spends more on children education?
1). John spends more on recreation than Mike does.
2). Mike spends more on healthcare than John does.
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John spends 30% of his income on his children's education, 20% on recreation and 10% on healthcare. The corresponding spending by Mike are 40%, 25% and 13%.
Who spends more on children education?
1). John spends more on recreation than Mike does.
2). Mike spends more on healthcare than John does.
Discussed here: https://gmatclub.com/forum/alex-spends- ... 02814.html
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Re: John spends 30% of his income on his children's education &nbs [#permalink] 02 Dec 2018, 00:24
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# John spends 30% of his income on his children's education
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# Funcion y = 1 / x
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### Funcion y = 1 / x
1. 1. <ul>Función del tipo y = 1/x </ul><ul>Jorge A. </ul>
2. 2. <ul>x </ul><ul>y </ul><ul>3 </ul><ul>2 </ul><ul>1 </ul><ul>1/2 </ul><ul>1/3 </ul><ul>-1/3 </ul><ul>-1/2 </ul><ul>-1 </ul><ul>-2 </ul><ul>-3 </ul><ul>Para graficarla, completamos la tabla calculando los valores de y. </ul>
3. 3. <ul>x </ul><ul>y = 1 / x </ul><ul>3 </ul><ul>Y = 1 / 3 </ul><ul>2 </ul><ul>Y = 1 / 2 </ul><ul>1 </ul><ul>Y = 1 </ul><ul>1/2 </ul><ul>Y = 2 </ul><ul>1/3 </ul><ul>Y = 3 </ul><ul>-1/3 </ul><ul>Y = -3 </ul><ul>-1/2 </ul><ul>Y = -2 </ul><ul>-1 </ul><ul>Y = -1 </ul><ul>-2 </ul><ul>Y = 1 / -2 </ul><ul>-3 </ul><ul>Y = 1 / -3 </ul>
4. 4. <ul>x </ul><ul>y = 1 / x </ul><ul>3 </ul><ul>1 / 3 </ul><ul>2 </ul><ul>1 / 2 </ul><ul>1 </ul><ul>1 </ul><ul>1/2 </ul><ul>2 </ul><ul>1/3 </ul><ul>3 </ul><ul>Graficando en el primer cuadrante obtenemos: </ul>
5. 5. <ul>x </ul><ul>y = 1 / x </ul><ul>-1/3 </ul><ul>-3 </ul><ul>-1/2 </ul><ul>-2 </ul><ul>-1 </ul><ul>-1 </ul><ul>-2 </ul><ul>-1/2 </ul><ul>-3 </ul><ul>-1/3 </ul><ul>Graficando en el tercer cuadrante obtenemos: </ul>
6. 6. <ul>La gráfica de la función y = 1 / x se llama HIPÉRBOLA. </ul>
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Sangeetha Pulapaka
1
The method:
Determine the number of days between 1/1/100 and 2/2/200, then divide that number by 4. If the remainder is
0 --> Sunday
1 --> Monday
2 --> Tuesday
3 --> Wednesday
The approach: Step 1: how many days in a year?
1 + 2 + 3 + 4 + ... + 9 + 10
This sum = 55
So in 100 years, you have 55 * 100 = 5500 days. So 1/1/200 is 5500 days from 1/1/100.
Step 2: Count up from 1/1/200 to 2/2/200
1/1/200 0
2/1/200 1
2/2/200 2
So it's 5502 days from 1/1/100 to 2/2/200. Now do the division I mentioned at the beginning.
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I found an answer from english.stackexchange.com
meaning - What date range is being referred to when someone says ...
Due to this ambiguity, this phrase is normally only used when giving rough ... date sound sooner, in which case it will mean the first 7 days of the month. ... If April 1 is on Sunday, then you can pick up your car on Monday April ... the "first week of April" is the week containing the 1st, 2nd, 3rd, and 4th of April.
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I found an answer from www.reddit.com
A tribe in Africa follows a special calendar. They have only 4 days in ...
Sunday, what day will be 02/02/20? ... 1+2+3+…+9+10=55. 55 divided by 4(since 4 days a week) has a remainder of 3. That means that if a ...
For more information, see A tribe in Africa follows a special calendar. They have only 4 days in ...
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I found an answer from stackoverflow.com
Get Day name from Weekday int - Stack Overflow
(If Monday is not the first day of the week, use setfirstweekday to change it). Using the calendar module has the advantage of being location aware: .... "Thursday", " Friday", "Saturday", "Sunday"] return days[weekday] if 0 ... calendar.weekday( year, month, day) - Returns the day of the week (0 is Monday) for year (1970–…) ...
For more information, see Get Day name from Weekday int - Stack Overflow
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I found an answer from workplace.stackexchange.com
salary - 31 days to be paid or 29 days, If joining is from first day of ...
If there are 20 working days in the month and you work the full 20 days, ... on the third working day of the month, then you have worked 18 days of the 20 ... So in your case, if the normal working week is Monday to Friday to ... For example, I worked for a company whose salary date was 10th of every month.
For more information, see salary - 31 days to be paid or 29 days, If joining is from first day of ...
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I found an answer from stackoverflow.com
Get the first Monday of a month - Stack Overflow
getFirstDayOfWeek() returns which day is used as the start for the current ... private static Calendar cacheCalendar; public static int getFirstMonday(int year, int month) .... The Joda-Time library offers a class, LocalDate , for when you need only a ... number of days from Thursday to Monday) plus 1 because days of the month ...
For more information, see Get the first Monday of a month - Stack Overflow
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I found an answer from www.quora.com
Which day can be the first as well as the last day of a century? - Quora
Let assign 0-sunday 1-monday 2-tuesday 3-wednesday 4-thursday 5-friday 6- saturday Last day of the century will go in the order 0,5,3,1 and it repeats ... If we have Monday in the first day of the century which means the previous century ... If u add both u get 29494 days its not a multiple of 7 so obviously the first day and last ...
For more information, see Which day can be the first as well as the last day of a century? - Quora
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I found an answer from parenting.stackexchange.com
newborn - Do you use number of weeks or day of the month to ...
So in non-leap years your birthday is Feb 28th, not March 1st. ... Counting weeks is only relevant for infants, when you see the pediatrician. ... We always did it by day of the month, since months don't all have the same number of days we went ... 2. I do it by day of the month not by number of weeks. 4 weeks ...
For more information, see newborn - Do you use number of weeks or day of the month to ...
Qalaxia Knowlege Bot
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I found an answer from en.wikipedia.org
February - Wikipedia
February is the second and shortest month of the year in the Julian and Gregorian calendar with 28 days in common years and 29 days in leap years, with the quadrennial 29th day being called the leap day. It is the first of five months to have a length of less than 31 days (the other ... In the Southern Hemisphere, February is the third and last month of summer ...
For more information, see February - Wikipedia
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I found an answer from en.wikipedia.org
seven-day week
A week is a time unit equal to seven days. It is the standard time period used for cycles of rest ... 1 Gregorian calendar year = 52 weeks + 1 day (2 days in a leap year); 1 week = 1600⁄6957 ≈ 22.9984% of an average Gregorian month ... The ordering of the weekday names is not the classical order of the planets (by ...
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# Comparing two real nubmers
I have been struggling for 2 days now trying to solve these Math problems. I hope someone could solve these in the next few minutes or so.
Subject: Order on R
1) a and b being two non zero numbers and x>0, compare:
a/b and (a+x)/(b+x)
2) a and b being two strictly positive numbers, compare:
Note by Oussama Jaber
6 years, 11 months ago
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$2$ is simple. For positive numbers $a$ and $b$, $\frac{a+b}{2}\geq \sqrt{ab}$. This is the two variable case of the Arithmetic Mean-Geometric Mean inequality. Here's a simple proof of the two variable case:
We know that $(a-b)^2\geq 0$
$\Rightarrow (a+b)^2 -4ab\geq 0$
And after a little bit of work, $\frac{a+b}{2}\geq \sqrt{ab}$.
For $1$, each one of the following can be true:
$\frac{a}{b}=\frac{a+x}{b+x}$ [happens when $x=0$ or $a=b$]
$\frac{a}{b}<\frac{a+x}{b+x}$ [happens when $x(a-b)<0$]
$\frac{a}{b}>\frac{a+x}{b+x}$ [happens when $x(a-b)>0$]
- 6 years, 11 months ago
Thank you for your reply, We did not take about the Arithmetic Mean- Geometric Mean inequality. The only method we took when comparing two numbers are:
Compare their difference, compare their squares, compare their radicals ...ect
I managed to to solve 2 by comparing their difference. Here is how I solved it.................................................... a and b are strictly positive, therefore a >0 and b>0
$\frac{a+b}{2}-\sqrt{ab}=\frac{a+b-2\sqrt{ab}}{2}=\frac{(\sqrt{a}-\sqrt{b})^2}{2}\geq0$
therefore, $\frac{a+b}{2}\geq\sqrt{ab}$
But for the problem 1, I knew that there is more than one answer because I plugged in numbers to check for comparison and get different comparison each time I plug in a different number.
But how can use the methods we learned to prove these different comparisons?
- 6 years, 11 months ago
sorry I missed one condition from problem 1.... x>0
- 6 years, 11 months ago
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Math Calculators, Lessons and Formulas
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• Question 1: 1 pts Simplify.$$\log_{b}x+\log_b{y}$$
$\log_{b}(x-y)$ $\log_{b}(x^{y})$ $\log_{b}(xy)$ $\log_{b}(x+y)$
• Question 2: 1 pts Simplify.$$\log_{13}6+\log_{13}4$$
• Question 3: 1 pts Simplify.$$\log_{17}30-\log_{17}6$$
$\log_{24}17$ $\log_{17}5$ $\log_{17}24$ $\log_{17}36$
• Question 4: 1 pts Use the power property to simplify the expression $\log_{5}\sqrt[7]{25}.$
$\dfrac{5}{2}$ $\dfrac{7}{2}$ $\dfrac{2}{7}$ $\dfrac{5}{7}$
• Question 5: 2 pts Simplify.$$2\log a+3\log b+4\log c$$
$\log a^{2}+b^{3}c^{4}$ $\log a^{2}b^{3}+c^{4}$ $\log( a^{2}+b^{3}+c^{4})$ $\log a^{2}b^{3}c^{4}$
• Question 6: 2 pts Simplify. $$2\log a-3\log(a^{2}+b^{2})$$
$\log \dfrac{1}{(b^{2})^{3}}$ $\log \dfrac{a^{2}}{(a^{6}+b^{6})}$ $\log \dfrac{a^{2}}{(a^{2}+b^{2})^{3}}$ $\log \dfrac{a^{2}}{(3a^{2}+3b^{2})}$
• Question 7: 2 pts $$\dfrac{1}{3}\log a+\log b=\log\sqrt[3]{ab}$$
• Question 8: 2 pts Simplify. $$\log_{8}\log_{4}\log_{2}16$$
$0$ $2$ $4$ Can't be simplified.
• Question 9: 3 pts Simplify. $$\log_{\frac{1}{9}}\left(\log_{2}\dfrac{1}{2}\cdot\log_{\frac{1}{2}}8\right)$$
$1$ $\dfrac{1}{2}$ $-1$ $-\dfrac{1}{2}$
• Question 10: 3 pts Simplify. $$\log x\cdot \log 12$$
• Question 11: 3 pts Simplify. $$36^{1-\log_{6}3}+25^{-\log_{5}6}$$
$\dfrac{145}{36}$ $\dfrac{125}{36}$ $\dfrac{105}{36}$ $\dfrac{75}{36}$
• Question 12: 3 pts Simplify. $\dfrac{2}{3}\log a+3\log b-\dfrac{2}{5}\log c$
$\log \dfrac{\sqrt[3]{a^{2}}+ b^{3}}{\sqrt[5]{c^{2}}}$ $\log \dfrac{\sqrt[3]{a^{2}}\cdot b^{3}}{\sqrt[5]{c^{2}}}$ $\log \dfrac{\sqrt[3]{a^{2}}}{b^{3}\cdot{\sqrt[5]{c^{2}}}}$ $\log \dfrac{\sqrt[3]{a^{2}}}{b^{3}+{\sqrt[5]{c^{2}}}}$
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# Sets + maximum - math problems
#### Number of problems found: 10
• Butterflies and bees
There were 35 flowers in the garden. The butterflies were on 17 flowers. There were butterflies and bees on 8 flowers. Can you find out on how many flowers were only bees? Note. there was a maximum of one butterfly and a maximum of one bee on each flower.
• Camp
In a class are 26 children. During the holidays 16 children were in the camps and 14 children on holiday with their parents. Determine the minimum and maximum number of children that may have been in the camp and on holiday with their parents at the same
• Trousers
In the class was 12 students. Nine students wearing trousers and turtleneck eight. How many students worn trousers with a turtleneck?
• Glasses
Imagine a set of students in your class (number of students: 19), who wears glasses. How much minimum and maximum element may contain this set.
• Skoda cars
There were 16 passenger cars in the parking. It was the 10 blue and 10 Skoda cars. How many are blue Skoda cars in the parking?
• Ten boys
Ten boys chose to go to the supermarket. Six boys bought gum and nine boys bought a lollipop. How many boys bought gum and a lollipop?
• Deficiencies
During the hygienic inspection in 2000 mass caterers, deficiencies were found in 300 establishments. What is the probability that deficiencies in a maximum of 3 devices will be found during the inspection of 10 devices?
• Left handed
It is known that 25% of the population is left-handed. What is the probability that there is a maximum of three left-handers at a seminar where there are 30 participants?
• Math logic
There are 20 children in the group, each two children have a different name. Alena and John are among them. How many ways can we choose 8 children to be among the selected A) was John B) was John and Alena C) at least one was Alena, John D) maximum one wa
In the box were total of 200 cookies. These products have sugar and chocolate topping. Chocolate topping is used on 157 cookies. Sugar topping is used on 100 cakes. How many of these cookies has two frosting?
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# Probability Distributions with Excel
Last Update: September 15, 2020
Probability distributions consist of all possible values that a discrete or continuous random variable can have and their associated probability of being observed. Classical or a priori probability distribution is theoretical while empirical or a posteriori probability distribution is experimental.
This topic is part of Business Statistics with Excel course. Feel free to take a look at Course Curriculum.
This tutorial has an educational and informational purpose and doesn’t constitute any type of forecasting, business, trading or investment advice. All content, including spreadsheet calculations and data, is presented for personal educational use exclusively and with no guarantee of exactness of completeness. Past performance doesn’t guarantee future results. Please read full Disclaimer.
An example of probability distributions is normal probability distribution which consists of all possible values that a continuous random variable can have and their associated probability of being observed. Its probability distribution function is bell-shaped, symmetric from its mean and mesokurtic. Standard normal probability distribution consists of normal probability distribution with mean zero and unit variance. Any normal probability distribution can be converted into a standard normal probability distribution through continuous random variable standardization.
1. Formula notation.
$pdf\left&space;(&space;x|\mu,\sigma^{2}&space;\right&space;)=\frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{\left&space;(&space;x-\mu&space;\right&space;)^{2}}{2\sigma^{2}}}$
$\mu=\mu,\;&space;\mu=0$
$\sigma^2=\sigma^2,\;\sigma^2=1$
$s=0$
$ek=0$
$z=\frac{x-\mu}{\sigma}$
Where $pdf\left&space;(&space;x|\mu,\sigma^{2}&space;\right&space;)$ = normal probability density function, $x$ = continuous random variable, $\mu=\mu$ = normal probability distribution mean, $\mu=0$ = standard normal probability distribution mean, $\sigma^2=\sigma^2$ = normal probability distribution variance, $\sigma^2=1$ = standard normal probability distribution variance, $s=0$ = normal and standard normal probability distributions skewness, $ek=0$ = normal and standard normal probability distributions excess kurtosis, $z$ = continuous random variable standardization, $\mu$ = continuous random variable mean, $\sigma$ = continuous random variable standard deviation.
2. Excel example.
2.1. Probability distributions data, daily returns and standardized daily returns calculation.
• Data: S&P 500® index replicating ETF (ticker symbol: SPY) daily adjusted close prices (2007-2015).
• Data daily arithmetic returns and standardized daily arithmetic returns calculation.
• Initial daily arithmetic return assumption not fixed and only included for educational purposes.
2.2. Probability distributions data, daily returns and standardized daily returns calculation formulas.
2.3. Standardized daily returns descriptive statistics calculation.
2.4. Standardized daily returns descriptive statistics calculation formulas.
2.5. Standardized daily returns density histogram and standard normal probability distribution curve overlay calculation.
• Standardized daily returns absolute frequency calculation done using Microsoft Excel Analysis Toolpak® Add-In Histogram Analysis Tool.
2.6. Standardized daily returns density histogram and standard normal probability distribution curve overlay calculation formulas.
2.7. Standardized daily returns density histogram and standard normal probability distribution curve overlay chart.
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Proving bounded monotonic sequences must converge
Homework Statement
I'm approaching this problem from a different method than conventially shown.
Homework Equations
if lim=infinity for all M>0, there exists a N such that n>N => {s(n)}>=M
The Attempt at a Solution
this can be rewritten as:
{s(n)} is a sequence. If {s(n)} is bounded and monotonic , then {s(n)} converges.
the contrapositive is,
{s(n)} is a sequence. If {s(n)} doesn't converge then it is either not monotonic or not bounded.
Hence, if I just show that it is not monotonic, then the proof works.
PROOF:
{s(n)} doesn't converge. So if {s(n)}=+infinity, then that means that for all M>0, there exists an N such for all n>N, M>={s(n)}.
Then {s(n)} is not bounded.
Thus all monotonic sequences must converge. Q.E.D
Does this proof work?
1)I'm a bit worried because I don't even know if I can apply contrapostives this way.
2)I don't know if I have to show {s(n)} is not bounded for all cases of {s(n)} diverging(i.e +infinity, -infinity, DNE). It seems plausible that if I can show that {s(n)} is not bounded when {s(n)} diverges for even just one of the cases, then the theorem works.
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A177459 The maximal positive integer m for which the exponents of 2 and prime(n) in the prime power factorization of m! are both powers of 2. 4
19, 131, 34, 19, 35, 35, 35, 67, 259, 575, 67, 67, 67, 131, 259, 515, 1027, 131, 131, 131, 131, 131, 259, 259, 259, 514, 515, 515, 515, 8195 (list; graph; refs; listen; history; text; internal format)
OFFSET 2,1 COMMENTS Or a(n) is the maximal m for which the Fermi-Dirac representation of m! (see comment in A050376) contains single power of 2 and single power of prime(n). LINKS V. Shevelev, Compact integers and factorials, Acta Arithmetica 126 (2007), no. 3, 195-236. FORMULA a(2)=19, a(3)=131; if prime(n) has the form (2^(4k+1)+3)/5 for k>=1,then a(n)=5*prime(n)-1; if prime(n)>=17 is Fermat prime, then a(n)=2*prime(n)+1; if prime(n) has the form 2^k+3 for k>=3, then a(n)=2*prime(n)-3; otherwise, if prime(n) is in interval [2^(k-1)+5, 2^k) for k>=4, then a(n)=3+2^(k+floor(log_2((p_n-5)/(2^k-prime(n)))). In any case, a(n)<=(1/2)*(prime(n)+1)^2+3. Equality holds for Mersenne primes>=31. EXAMPLE For n=31, prime(n)=127 is Mersenne primes. Thus a(31)=(1/2)*128^2+3=8195. CROSSREFS Cf. A000142, A177436, A177378, A177355, A177349, A177458, A177498, A050376, A169655, A169661. Sequence in context: A078851 A202125 A169727 * A142649 A020867 A022679 Adjacent sequences: A177456 A177457 A177458 * A177460 A177461 A177462 KEYWORD nonn AUTHOR Vladimir Shevelev, May 09 2010 STATUS approved
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Homework Help: Measurement help
1. Jun 22, 2007
kakab00
1. The problem statement, all variables and given/known data
http://img487.imageshack.us/img487/5702/untitledna1.png [Broken]
2. Relevant equations
v square = u square + 2as
3. The attempt at a solution
a) Let v square - u square be x
delta x/x = 2(delta v/v) + 2(delta u/u)
delta x/300 = 0.03
delta x = 9
b) i) a = (v square - u square)/2s
delta a/a = 2(delta v/v) + 2(delta u/u) + deltas/s
delta a/3 = 0.03 + 0.002
delta a = 0.096 = 0.1 (round off to 1 sig fig)
ans : 3.0 plus minus 0.1
1. The problem statement, all variables and given/known data
I think I did it correctly, but my answers seem to be wrong , because this is a question on my school website and I have to key in the answers and it'll tell me whether it is right or wrong. I thought maybe I did something wrong, so maybe somebody could help please?
Last edited by a moderator: May 2, 2017
2. Jun 22, 2007
malawi_glenn
a) well if we make a function:
$$f(u,v) = v^{2} - u^{2}$$
then, from error propagation:
$$\delta f = \sqrt{\left( \dfrac{\partial f}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial f}{\partial v} \delta v\right)^{2}}$$
So:
$$\delta f = \sqrt{\left( -2*10 * 0.1 \right)^{2} + \left( 2*20 * 0.1 \right)^{2}} = 4.4721$$
the forumula you used is when f(u,v) is a product and quoitent only.
Last edited: Jun 22, 2007
3. Jun 22, 2007
malawi_glenn
b) the expression for the acceleration is:
$$a(v,y,s) = \dfrac{v^{2} - u^{2} }{2s}$$
then the uncertanty in a is:
$$\delta a = \sqrt{\left( \dfrac{\partial a}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial a}{\partial v} \delta v\right)^{2} + \left( \dfrac{\partial a}{\partial s} \delta s\right)^{2}}$$
Last edited: Jun 22, 2007
4. Jun 22, 2007
kakab00
I don't get your equations, wouldnt the delta u and delta u or other symbols cancel each other out?
5. Jun 22, 2007
malawi_glenn
it is not deltas, it is the partial derivative of f with respect to u, and so on..
what is the answers for these problems?
http://www.rit.edu/~uphysics/uncertainties/Uncertaintiespart2.html#mixtures [Broken]
see section 5(f) , formulas Eq. 4a and Eq. 4b
Last edited by a moderator: May 2, 2017
6. Jun 22, 2007
kakab00
I don't know the exact answers, I only know whether the answers would be right or wrong because we have to key in them inside a system after we found them.
I've not learnt partial derivatives yet though, so I think there should be another easier way to do it using delta/fractional uncertainty?
7. Jun 22, 2007
malawi_glenn
partial derivatives are very easy.. you just do usual derivatives and treat the rest as constans. for example: if f(x,y) = 2xy. the f 'x =2y and f 'y = 2x
According to my knowledges in science of errors, the functions we deal with here we must use error propagation as the formulas I have given.
you can read alot of it on the web page I gave you.
8. Jun 22, 2007
kakab00
it's not easy for me XD
for the 2nd part I tried using your method:
so delta a = square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square + (0.1 x 50 x 2)square)] = root(150) which would give you a value of 12.2, much bigger than a=3 calculated. doesnt make sense, or did I do something wrongly?
9. Jun 22, 2007
malawi_glenn
well doing the partial derivative is just doing the usual treating the rest as constants.
Well you did not to the derivatie for a vs s right.. try again. Remeber to trea't v and u just as constant (values)
10. Jun 22, 2007
kakab00
I have no idea how to do it..
upon further thought, even without finding the derivative for a vs s, the answer already seems to be wrong from the 1st and 2nd part of the equation,
square root[(0.1 x 10 x 2)square + (0.1 x 20 x 2)square] would give you a value of around 4 which is still greater than a=3.
11. Jun 22, 2007
malawi_glenn
are you telling me that you can not differentiate functions like:
f(x) = 1 / x ???
if you wait a minute I can do the function for you and show how you do this..
Last edited: Jun 22, 2007
12. Jun 22, 2007
malawi_glenn
$$\delta a = \sqrt{\left( \dfrac{\partial a}{\partial u} \delta u\right)^{2} + \left( \dfrac{\partial a}{\partial v} \delta v\right)^{2} + \left( \dfrac{\partial a}{\partial s} \delta s\right)^{2}}$$
$$a(v,y,s) = \dfrac{v^{2} - u^{2} }{2s}$$
$$\dfrac{\partial a}{\partial v} = \dfrac{2v-0}{2s} = \dfrac{v}{s}$$
$$\dfrac{\partial a}{\partial u} = \dfrac{0-2u}{2s} = \dfrac{-u}{s}$$
$$\dfrac{\partial a}{\partial s} = \dfrac{-(v^{2} - u^{2})}{2s^{2}}$$
so we get:
$$\delta a = \sqrt{\left( \dfrac{-u}{s} \delta u\right)^{2} + \left( \dfrac{v}{s} \delta v\right)^{2} + \left( \dfrac{(u^{2} - v^{2})}{2s^{2}} \delta s\right)^{2}}$$
Last edited: Jun 22, 2007
13. Jun 22, 2007
kakab00
The answers are still wrong though, I used
a) 4 (round off to one significant figure)
b) 3.0 + - 0.2 (this was the value I calculated from the equation you gave me)
14. Jun 22, 2007
malawi_glenn
Well I got (b) -> error in a = 0.045211
are you sure you have the right formula for the acceleration?
And what kinda sucky school has an electronic answer machine lol
it is also the way you get to the answer that is important, not the result itself. So you can basically try every number from 0 to 20 til it gets right :)
Last edited: Jun 22, 2007
15. Jun 22, 2007
kakab00
hehe
but the answers are still wrong :grumpy: I don't really get your method either, because I haven't learnt them yet. I think the school wouldn't have set something they hadn't taught yet, so I believe there's still another (easier( way ) of doing, maybe some other members might know?
Thanks for your help and time anyway! at least I learnt something from you
16. Jun 22, 2007
malawi_glenn
then you show me the simpler way to do it. It is not the right one I will promise you.
we have functions that are both multiplcation, addition and qouitent. And the errors are independent of each other, hence we must use the error propgation formula.
17. Jun 22, 2007
kakab00
If I knew the easier method I wouldnt be here asking :shy:
Maybe I calculated something wrongly up there? Or maybe the school system just sucks :tongue2:
18. Jun 22, 2007
malawi_glenn
I bet your school sucks hehe :)
CanĀ“t you write donw all that we have discussed here, and maybe print out the relevant section from the www-page i showed you. Then go to the professor and ask "what is wrong?" :P
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Theory of Computations, Final Exam for the courseCS 5315, Spring 2011
1-2. Prove, from scratch, that the binary remainder function a % 2 (or, in mathematical notations, a mod 2) is primitive recursive. Start with the definition of a primitive recursive function, and use only this definition in your proof -- do not use results that we proved in class. Prove also that this function is μ-recursive.
3. Design a Turing machine that, given a binary number a, computes the remainder f(a) = a % 2. For example, for a binary number 11012 (which is equivalent to 13), this Turing machine should return 1. Use the general algorithm for generating a Turing machine for the composition to design a Turing machine that computes the function f(f(a)).
4-5. A student produced a program that claim to provide a new way of computing a % 2. Is there a general way to check that this program halts? If yes, explain how, if no, prove that this is not possible. Suppose now that for this particular proram, we have proven that it always halts. Is there a general way of checking whether this program always produced the correct result? If yes, explain how, if no, prove that this is not possible.
6. Is the intersection of two recursively enumerable (r.e.) sets always r.e.? If yes, prove; if no, provide a counter-example and prove that this counter-example is not r.e.
7-8. Define what is P, what is NP, what is NP-hard, and what is NP-complete; no need to give a detailed definition of reduction. For each of these four classes, indicate whether this class contain the clique problem and whether this class contains a problem of computing the remainder a % 2.
9-13. Reduce the satisfiability problem for the formula (x1 \/ x2) & (~x1 \/ ~x3) & (~x1 \/ x2 \/ x3) to:
• 3-coloring;
• clique;
• subset sum problem;
• interval computations.
Use a greedy algorithm to find a solution to the original satisfiability problem.
14-17. A straightforward algorithm for adding two matrices takes time O(n^2) to compute the sum of two square matrices of size n x n.
• If we do not take into account communication time, how fast can we solve this problem? Is this problem in the class NC?
• If we parallelize and take into account communication time, what is the fastest that we get with such parallel algorithms? Explain your answer.
• Where are similar arguments used in the proof that satisfiability is NP-hard?
• Explain what is the physics behind these arguments and how using non-Euclidean physics can potentially lead to faster computations -- such as solving NP-hard problems in polynomial time.
18-19. What is the Kolmogorov complexity of a sequence 101101 ... 101 (repeated 100,000 times)? Prove that Kolmogorov complexity is not computable. For extra credit: Explain Gell-Mann's potential physical scheme for using equations explicitly depending on Kolmogorov complexity for computing hard problems.
20. Which class of the polynomial hierarchy corresponds to winning a game in two moves? Explain your answer
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# Let $F= \Bbb{Q}(\zeta_9)$ with $\zeta_9 = e^{\frac{2i\pi}{9}}$. Find the Galois group and the intermediate fields.
Let $$F= \Bbb{Q}(\zeta_9)$$ with $$\zeta_9 = e^{\frac{2i\pi}{9}}$$.
a) What is the Galois group of $$F$$ over $$\Bbb{Q}$$?
b) Find all intermediate fields between $$\Bbb{Q}$$ and $$F$$. (Write each in the form $$\Bbb{Q}(\alpha)$$ for some specific $$\alpha \in F$$.)
c) For each intermediate field $$E$$ above, give the Galois group of $$E$$ over $$\Bbb{Q}$$.
a) Since $$\Bbb{Q}(\zeta_9)$$ is the splitting field of a cyclotomic polynomial, $$Gal(F/\Bbb{Q})=\Bbb{Z}^{\times}_9$$.
b) Since $$Gal(F/\Bbb{Q})=\Bbb{Z}^{\times}_9$$, we need to find the subgroups of $$\Bbb{Z}^{\times}_9$$. But $$|\Bbb{Z}^{\times}_9|=6$$ and there are only 2 groups of order 6 up to isomorphism: $$\Bbb{Z}_6$$ and $$D_6$$. Since $$\Bbb{Z}^{\times}_9$$ is abelian, we know that $$\Bbb{Z}^{\times}_9 \cong \Bbb{Z}_6$$. So we can use either of these diagrams, since they are both isomorphic
or
Where $$\langle 4 \rangle = \{1, 4, 7\}$$ and $$\langle 8 \rangle \{1, 8\}$$.
So the corresponding diagram for the intermediate fields is
We need to find $$\alpha$$ and $$\beta$$, where $$[\Bbb{Q}(\alpha):\Bbb{Q}]=2$$ and $$[\Bbb{Q}(\beta):\Bbb{Q}]=3$$.
We know that $$(\zeta^3_9)^3 = 1 \implies (\zeta^3_9)^9-1=0$$. So the minimal polynomial of $$\zeta^3_9$$ divides $$x^3-1$$. We know that $$x^3-1=\Phi_1(x)\Phi_3(x)=(x-1)(x^2+x+1)$$, where $$\Phi_k(x)$$ is the kth cyclotomic polynomial. Since $$1 \in \Bbb{Q}$$, we only need to check if $$x^2+x+1$$ is reducible over $$\Bbb{Q}$$. But since it is a cyclotomic polynomial, it must be. So the minimal polynomial of $$\zeta_9^3$$ over $$\Bbb{Q}$$ has degree $$2 \implies \alpha=\zeta^3_9$$.
Now we need to find $$\beta$$. Using Gerry Myerson's answer, we choose $$\zeta_9 \in \Bbb{Q}(\zeta_9)$$. Since we are looking at a the Galois group of degree $$3$$, we need to find a $$\sigma$$ such that $$\sigma^3(\zeta_9 )=\zeta_9$$. So we can choose $$\sigma(\zeta_9)=\zeta_9^4$$.
Now we need to find the minimal polynomial of $$\zeta_9 + \zeta_9^4 + \zeta_9^7$$. We see that $$(\zeta_9 + \zeta_9^4 + \zeta_9^7)^3 = 9\zeta_9^3 + 9\zeta_9^6 + 9$$
But $$\cos(6\pi/9) + i\sin(6\pi/9) + \cos(12\pi/9) + i\sin(12\pi/9)$$
$$= 2\cos(6\pi/9) = 2\cos(2\pi/3)=2(-1/2) = -1$$
since $$2\pi6/9 - 2\pi = -6\pi/9$$. It follows that $$(\zeta_9 + \zeta_9^4 + \zeta_9^7)^3 = 9(-1+1)=0$$. So we know that the min polynomial for $$\zeta_9 + \zeta_9^4 + \zeta_9^7$$ must divide $$x^3$$.
So we need to check if $$(\zeta_9 + \zeta_9^4 + \zeta_9^7)^2=0$$. We see that $$(\zeta_9 + \zeta_9^4 + \zeta_9^7)^2 = 3\zeta_2^7 + 3\zeta_9^5 + 3\zeta_9^8$$. But we can see from the unit circle, that the sum of the three vectors ends up $$50^{\circ}$$ above the horizontal axis; so it is not zero $$\implies$$ the minimal polynomial for $$\zeta_9 + \zeta_9^4 + \zeta_9^7$$ is of degree $$3$$ $$\implies$$ $$\zeta_9 + \zeta_9^4 + \zeta_9^7$$ is a possible choice for $$\beta$$.
c) The Galois group of $$\Bbb{Q}(\alpha)$$ is isomorphic to $$\Bbb{Z}_3$$, and that of $$\Bbb{Q}(\beta)$$ is isomorphic to $$\Bbb{Z}_2$$.
I have two questions:
2. For part b), we know that $$\Bbb{Z}^{\times}_9 \cong \Bbb{Z}_6$$, so finding the lattice of the subgroups is easy. But what if it was not isomorphic to a cyclic group? Would it be ok to find only $$1$$ subgroup for each possible order, or do we need to find all subgroups of a certain order? For example, we know that $$\Bbb{Z}^{\times}_9$$ has subgroups of orders $$2$$ and $$3$$. So after finding $$\langle 4 \rangle$$ of order $$3$$, and $$\langle 8 \rangle$$ of order $$2$$, can I just stop? Or do I need to search for more groups of orders $$2$$ and $$3$$? The reason why I'm asking this is because we know that for any cyclic group, there is at most 1 subgroup of a given order. So we can just stop when we find $$\Bbb{Z}_2$$ and $$\Bbb{Z}_3$$ in $$\Bbb{Z}_6$$. But we can't guarantee that with other groups, right?
• Is it $\zeta_9=\exp(2\mathrm i\pi/9)$?
– Did
Aug 16, 2013 at 21:44
• Won't you correct your post?
– Did
Aug 16, 2013 at 22:41
• By the way, could it be that (c) asks for $\operatorname{Aut}(E/\mathbb{Q})$ whereas you give $\operatorname{Gal}(F/E)$? Aug 17, 2013 at 0:47
• @Artus: Here, $\operatorname{Aut}(E/\mathbb{Q})=\operatorname{Gal}(E/\mathbb{Q})$ because $E/\mathbb{Q}$ is Galois. Look at $E=\mathbb{Q}(\beta)$ over $\mathbb{Q}$: The extension has degree $3$, and there are $3$ automorphisms of $\mathbb{Q}(\beta)$ that fix $\mathbb{Q}$, namely those generated by the $\sigma$ taking $\beta$ to $\beta^2-2$ (the one that would take $\zeta_9$ to $\zeta_9^{-2}$ in $F$). So $\operatorname{Gal}(E/\mathbb{Q})$ is cyclic of order $3$. You give order $2$ which applies to $\operatorname{Gal}(F/E)$. Aug 17, 2013 at 9:28
• (With $\beta=\zeta_9+\zeta_9^{-1}$, as in my answer.) Aug 17, 2013 at 9:38
As to question (1): Alas, $\zeta_9+\zeta_9^4+\zeta_9^7=\zeta_9(1+\zeta_3+\zeta_3^2)=0$. That is obviously no useful $\beta$.
You need a $\beta$ of degree $3$ over $\mathbb{Q}$, so in order to apply Gerry Myerson's trick, the $\sigma$ you look for should have order $6:3=2$. Pick $\sigma(\zeta_9)=\zeta_9^8=\zeta_9^{-1}$. Then $\beta=\zeta_9+\zeta_9^{-1}$ and you will find $\beta^3=3\beta-1$ which yields a monic, irreducible, hence minimal, polynomial for $\beta$ over $\mathbb{Q}$.
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# Pascal's principle hydraulic lift with torque involved
fordy314
Pascal's principle hydraulic lift with torque involved(need help with lever)
## Homework Statement
A hydraulic jack is used to support a mass which has a mass of 30.7 kg.
The mass is supported by piston 2 which has a mass of 0.250 kg and an area of 40.2 cm2 . A force is applied by the lever on piston 1 which has a mass of 0.085 kg and an area of 14.2 cm2.
The distance between the hinge and piston 1 is 4.47 cm. A force is applied a further 16.4 cm from piston 1.
What is the magnitude of the force F which must be applied to support the mass? You may neglect the mass of the fluid. You may also wish to TORQUE to someone about this problem.
## Homework Equations
ρ=m/v
P=F/A
P=Po+ρgh
Any relevant torque equations (not that far yet)
## The Attempt at a Solution
I'm pretty lost here, but I know that I need to find the force exerted by piston1.
The force required to lift the mass and piston2 is:
30.95kg*9.81m/s^2 = 303.6N
P=F/A=303.6N/(pi*.402^2)=598(k?)Pa
This is where I get lost. normally I would take P=Po+ρgh to find the pressure on piston 1 to find the force needed to push against it, but I don't have the height. Am I missing something, or do you think that I'm supposed to assume that they're at the same height?
Thanks.
Last edited:
Nessdude14
It looks like you're supposed to assume they're at the same height. The pressure from piston 1 needs to equal the pressure from piston 2. Then you can find the force that the lever needs to apply to the piston to create that pressure. Using the law of the lever, you can then find the force that needs to be applied to the end of the lever.
fordy314
Ok, then FPiston1=PA1=37.9N
37.9N-(.085kg*9.81m/s^2)=37.1
Now do I want to use the equation F1*d1=F2*d2?
That would be (37.1*.0447m)/.164m=10.1N, but my answer comes up wrong. Any suggestions?
Nessdude14
Ok, then FPiston1=PA1=37.9N
37.9N-(.085kg*9.81m/s^2)=37.1
Now do I want to use the equation F1*d1=F2*d2?
That would be (37.1*.0447m)/.164m=10.1N, but my answer comes up wrong. Any suggestions?
For the area of piston 2, you used:
A=pi*.402^2
When the problem states that the area is .402 meters (not the radius as you undoubtedly thought).
and yes, you've got the right lever equation.
Also if you're still wondering, the pressures are pascals (not kPA).
fordy314
For the area of piston 2, you used:
A=pi*.402^2
When the problem states that the area is .402 meters (not the radius as you undoubtedly thought).
and yes, you've got the right lever equation.
Also if you're still wondering, the pressures are pascals (not kPA).
Thanks so much. Of course its always that units that'll get me.
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Tải bản đầy đủ - 0 (trang)
Chapter 5. Inner Product Spaces, Orthogonal Projection, Least Squares, and Singular Value Decomposition
# Chapter 5. Inner Product Spaces, Orthogonal Projection, Least Squares, and Singular Value Decomposition
Tải bản đầy đủ - 0trang
5-2
Handbook of Linear Algebra
A Hermitian matrix A is positive definite if x∗ Ax > 0 for all nonzero x ∈ Cn . (See Chapter 8 for more
information on positive definite matrices.)
Facts:
All the following facts except those with a specific reference can be found in [Rom92, pp. 157–164].
1. The vector space Rn is an inner product space under the standard inner product, or dot product,
defined by
n
u, v = uT v =
ui v i .
i =1
This inner product space is often called n–dimensional Euclidean space.
2. The vector space Cn is an inner product space under the standard inner product, defined by
n
u, v = v∗ u =
ui v¯ i .
i =1
This inner product space is often called n-dimensional unitary space.
3. [HJ85, p. 410] In Rn , a function ·, · : Rn × Rn → R is an inner product if and only if there exists
a real symmetric positive definite matrix G such that u, v = uT G v, for all u, v ∈ Rn .
4. [HJ85, p. 410] In Cn , a function ·, · : Cn × Cn → C is an inner product if and only if there exists
a Hermitian positive definite matrix H such that u, v = v∗ Hu, for all u, v ∈ Cn .
5. Let l 2 be the vector space of all infinite complex sequences v = (v n ) with the property that
|v n |2 < ∞. Then l 2 is an inner product space under the inner product
n=1
un v¯ n .
u, v =
n=1
6. The vector space C [a, b] of all continuous real-valued functions on the closed interval [a, b] is an
inner product space under the inner product
f, g =
b
f (x)g (x)d x.
a
7. If V is an inner product space and u, w = v, w for all w ∈ V , then u = v.
8. The inner product on an inner product space V , when restricted to vectors in a subspace S of V ,
is an inner product on S.
9. Let V be an inner product space. Then the norm function · on V has the following basic
properties for all u, v ∈ V :
r
v ≥ 0 and v = 0 if and only if v = 0.
r
av = |a| v , for all a ∈ F .
r (The triangle inequality) u + v
a ∈ F.
≤ u + v with equality if and only if v = au, for some
r (The Cauchy–Schwarz inequality) | u, v | ≤ u
a ∈ F.
r | u − v |≤ u−v .
v with equality if and only if v = au, for some
5-3
Inner Product Spaces, Orthogonal Projection, Least Squares
r (The parallelogram law) u + v 2 + u − v 2 = 2 u 2 + 2 v 2 .
r (Polarization identities)
4 u, v =
⎨ u+v
2
− u − v 2 , if F = R.
⎩ u+v
2
− u−v
2
+ i u + iv
2
− i u − i v 2 , if F = C.
Examples:
1. Let R4 be the Euclidean space with the inner product u, v = uT v. Let x = [1, 2, 3, 4]T ∈ R4 and
y = [3, −1, 0, 2]T ∈ R4 be two vectors. Then
r x, y = 9, x = 30, and y = 14.
r The distance between x and y is d(x, y) = x − y = 26.
9
r The angle between x and y is θ = arccos √ 9√
= arccos √
30 14
2 105
2. u, v = u1 v 1 + 2u1 v 2 + 2u2 v 1 + 6u2 v 2 = uT
G=
1
2
2
v is an inner product on R2 , as the matrix
6
1 2
is symmetric positive definite.
2 6
3. Let C [−1, 1] be the vector space with the inner product f, g =
and g (x) = x 2 be two functions in C [−1, 1]. Then f, g =
and g , g =
1
−1
1
−1
1
−1
f (x)g (x)d x and let f (x) = 1
x 2 d x = 2/3, f, f =
1
−1
1d x = 2,
x 4 d x = 2/5. The angle between f and g is arccos( 5/3) ≈ 0.730 radians.
4. [Mey00, p. 286] A, B = tr(AB ∗ ) is an inner product on Cm×n .
5.2
Orthogonality
Definitions:
Let V be an inner product space. Two vectors u, v ∈ V are orthogonal if u, v = 0, and this is denoted
by u ⊥ v.
A subset S of an inner product space V is an orthogonal set if u ⊥ v, for all u, v ∈ S such that u = v.
A subset S of an inner product space V is an orthonormal set if S is an orthogonal set and each v ∈ S
is a unit vector.
Two subsets S and W of an inner product space V are orthogonal if u ⊥ v, for all u ∈ S and v ∈ W,
and this is denoted by S ⊥ W.
The orthogonal complement of a subset S of an inner product space V is S ⊥ = {w ∈ V | w, v =
0 for all v ∈ S}.
A complete orthonormal set M in an inner product space V is an orthonormal set of vectors in V such
that for v ∈ V , v ⊥ M implies that v = 0.
An orthogonal basis for an inner product space V is an orthogonal set that is also a basis for V .
An orthonormal basis for V is an orthonormal set that is also a basis for V .
A matrix U is unitary if U ∗ U = I .
A real matrix Q is orthogonal if Q T Q = I .
5-4
Handbook of Linear Algebra
Facts:
1. [Mey00, p. 298] An orthogonal set of nonzero vectors is linearly independent. An orthonormal set
of vectors is linearly independent.
2. [Rom92, p. 164] If S is a subset of an inner product space V , then S ⊥ is a subspace of V . Moreover,
if S is a subspace of V , then S ∩ S ⊥ = {0}.
3. [Mey00, p. 409] In an inner product space V , {0}⊥ = V and V ⊥ = {0}.
4. [Rom92, p. 168] If S is a finite dimensional subspace of an inner product space V , then for any
v ∈ V,
r There are unique vectors s ∈ S and t ∈ S ⊥ such that v = s + t. This implies V = S ⊕ S ⊥ .
r There is a unique linear operator P such that P (v) = s.
5. [Mey00, p. 404] If S is a subspace of an n−dimensional inner product space V , then
r (S ⊥ )⊥ = S.
r dim(S ⊥ ) = n − dim(S).
6. [Rom92, p. 174] If S is a subspace of an infinite dimensional inner product space, then S ⊆ (S ⊥ )⊥ ,
but the two sets need not be equal.
7. [Rom92, p. 166] An orthonormal basis is a complete orthonormal set.
8. [Rom92, p. 166] In a finite-dimensional inner product space, a complete orthonormal set is a basis.
9. [Rom92, p. 165] In an infinite-dimensional inner product space, a complete orthonormal set may
not be a basis.
10. [Rom92, p. 166] Every finite-dimensional inner product space has an orthonormal basis.
11. [Mey00, p. 299] Let B = {u1 , u2 , . . . , un } be an orthonormal basis for V . Every vector v ∈ V can
be uniquely expressed as
n
v, ui ui .
v=
i =1
The expression on the right is called the Fourier expansion of v with respect to B and the scalars
v, ui are called the Fourier coefficients.
12. [Mey00, p. 305] (Pythagorean Theorem) If {vi }ik=1 is an orthogonal set of vectors in V , then
k
2
= ik=1 vi 2 .
i =1 vi
13. [Rom92, p. 167] (Bessel’s Inequality) If {ui }ik=1 is an orthonormal set of vectors in V , then
v 2 ≥ ik=1 | v, ui |2 .
14. [Mey00, p. 305] (Parseval’s Identity) Let B = {u1 , u2 , . . . , un } be an orthonormal basis for V . Then
n
for each v ∈ V , v
2
=
| v, ui |2 .
i =1
m×n
15. [Mey00, p. 405] Let A ∈ F
, where F = R or C. Then
r ker(A)⊥ = range(A∗ ), range(A)⊥ = ker(A∗ ).
r F m = range(A) ⊕ range(A)⊥ = range(A) ⊕ ker(A∗ ).
r F n = ker(A) ⊕ ker(A)⊥ = ker(A) ⊕ range(A∗ ).
16. [Mey00, p. 321] (See also Section 7.1.) The following statements for a real matrix Q ∈ Rn×n are
equivalent:
r Q is orthogonal.
r Q has orthonormal columns.
r Q has orthonormal rows.
r Q Q T = I , where I is the identity matrix of order n.
r For all v ∈ Rn , Qv = v .
5-5
Inner Product Spaces, Orthogonal Projection, Least Squares
17. [Mey00, p. 321] (See also Section 7.1.) The following statements for a complex matrix U ∈ Cn×n
are equivalent:
r U is unitary.
r U has orthonormal columns.
r U has orthonormal rows.
r U U ∗ = I , where I is the identity matrix of order n.
r For all v ∈ Cn , U v = v .
Examples:
1. Let C [−1, 1] be the vector space with the inner product f, g =
and g (x) = x be two functions in C [−1, 1]. Then f, g =
1
−1
1
−1
f (x)g (x)d x and let f (x) = 1
xd x = 0. Thus, f ⊥ g .
2. The standard basis {e1 , e2 , . . . , en } is an orthonormal basis for the unitary space Cn .
3. If {v1 , v2 , · · · , vn } is an orthogonal basis for Cn and S = span {v1 , v2 , · · · , vk } (1 ≤ k ≤ n − 1), then
S ⊥ = span {vk+1 , · · · , vn }.
They
4. The vectors v1 = [2, 2, 1]T , v2 = [1, −1, 0]T , and v3 = [−1, −1, 4]T are mutually
√ orthogonal.
T
T
=
v
/
v
=
[2/3,
2/3,
1/3]
,
u
=
v
/
v
=
[1/
2,
−1/
2,
0]
, and
can be normalized to u√
1
1 √1
2
2
2
u3 = v3 / v3 = [− 2/6, − 2/6, 2 2/3]T . The set B = {u1 , u2 , u3 } forms an orthonormal
basis for the Euclidean space R3 .
r If v = [v , v , v ]T ∈ R3 , then v = v, u u + v, u u + v, u u , that is,
1 2 3
1 1
2 2
3 3
v=
2v 1 + 2v 2 + v 3
v1 − v2
−v 1 − v 2 + 4v 3
u3 .
u1 + √ u2 +
3
2
3 2
r The matrix Q = [u , u , u ] ∈ R3×3 is an orthogonal matrix.
1 2 3
5. Let S be the subspace of C3 spanned by the vectors u = [i, 1, 1]T and v = [1, i, 1]T . Then the
orthogonal complement of S is
S ⊥ = {w|w = α[1, 1, −1 + i ]T , where α ∈ C}.
6. Consider the inner product space l 2 from Fact 5 in Section 5.1. Let E = {ei |i = 1, 2, . . .}, where
ei has a 1 on i th place and 0s elsewhere. It is clear that E is an orthonormal set. If v = (v n ) ⊥ E,
then for each n, v n = v, en = 0. This implies v = 0. Therefore, E is a complete orthonormal set.
However, E is not a basis for l 2 as S = span{E} = l 2 . Further, S ⊥ = {0}. Thus, (S ⊥ )⊥ = l 2 ⊆ S
and l 2 = S ⊕ S ⊥ .
5.3
Adjoints of Linear Operators on Inner Product Spaces
Let V be a finite dimensional (real or complex) inner product space and let T be a linear operator on V .
Definitions:
A linear operator T ∗ on V is called the adjoint of T if T (u), v = u, T ∗ (v) for all u, v ∈ V .
The linear operator T is self-adjoint, or Hermitian, if T = T ∗ ; T is unitary if T ∗ T = I V .
5-6
Handbook of Linear Algebra
Facts:
The following facts can be found in [HK71].
1. Let f be a linear functional on V . Then there exists a unique v ∈ V such that f (w) = w, v for
all w ∈ V .
2. The adjoint T ∗ of T exists and is unique.
3. Let B = (u1 , u2 , . . . , un ) be an ordered, orthonormal basis of V . Let A = [T ]B . Then
ai j = T (uj ), ui ,
i, j = 1, 2, . . . , n.
Moreover, [T ∗ ]B = A∗ , the Hermitian adjoint of A.
4. (Properties of the adjoint operator)
(a) (T ∗ )∗ = T for every linear operator T on V .
(b) (aT )∗ = a¯ T ∗ for every linear operator T on V and every a ∈ F .
(c) (T + T1 )∗ = T ∗ + T1 ∗ for every linear operators T, T1 on V .
(d) (T T1 )∗ = T1 ∗ T ∗ for every linear operators T, T1 on V .
5. Let B be an ordered orthonormal basis of V and let A = [T ]B . Then
(a) T is self-adjoint if and only if A is a Hermitian matrix.
(b) T is unitary if and only if A is a unitary matrix.
Examples:
1. Consider the space R3 equipped with the standard inner product and let f (w) = 3w 1 − 2w 3 .
Then with v = [3, 0, −2]T , f (w) = w, v .
⎡ ⎤
x
⎢ ⎥
2. Consider the space R3 equipped with the standard inner product. Let v = ⎣ y ⎦ and T (v) =
z
2x + y
2
⎣ y − 3z ⎦. Then [T ] = ⎣0
x+y+z
1
1
1
1
0
2
−3⎦ , so [T ]∗ = ⎣1
1
0
0
1
−3
1
2x + z
1⎦, and T ∗ (v) = ⎣ x + y + z ⎦.
1
−3y + z
3. Consider the space Cn×n equipped with the inner product in Example 4 of section 5.1. Let A, B ∈
Cn×n and let T be the linear operator on Cn×n defined by T (X) = AX + X B, X ∈ Cn×n . Then
T ∗ (X) = A∗ X + X B ∗ , X ∈ Cn×n .
4. Let V be an inner product space and let T be a linear operator on V . For a fixed u ∈ V , f (w) =
T (w), u is a linear functional. By Fact 1, there is a unique vector v such that f (w) = w, v . Then
T ∗ (u) = v.
5.4
Orthogonal Projection
Definitions:
Let S be a finite-dimensional subspace of an inner product space V . Then according to Fact 4 in Section 5.2,
each v ∈ V can be written uniquely as v = s + t, where s ∈ S and t ∈ S ⊥ . The vector s is called the
orthogonal projection of v onto S and is often written as Proj S v, where the linear operator Proj S is called
the orthogonal projection onto S along S ⊥ . When V = Cn or V = Rn with the standard inner product,
the linear operator Proj S is often identified with its standard matrix [Proj S ] and Proj S is used to denote
both the operator and the matrix.
Inner Product Spaces, Orthogonal Projection, Least Squares
5-7
Facts:
1. An orthogonal projection is a projection (as defined in Section 3.6).
2. [Mey00, p. 433] Suppose that P is a projection. The following statements are equivalent:
r P is an orthogonal projection.
r P∗ = P.
r range(P ) ⊥ ker(P ).
3. [Mey00, p. 430] If S is a subspace of a finite dimensional inner product space V , then
Proj S ⊥ = I − Proj S .
4. [Mey00, p. 430] Let S be a p–dimensional subspace of the standard inner product space Cn , and
let the columns of matrices M ∈ Cn× p and N ∈ Cn×(n− p) be bases for S and S ⊥ , respectively. Then
the orthogonal projections onto S and S ⊥ are
Proj S = M(M ∗ M)−1 M ∗
and Proj S ⊥ = N(N ∗ N)−1 N ∗ .
If M and N contain orthonormal bases for S and S ⊥ , then Proj S = M M ∗ and Proj S ⊥ = N N ∗ .
5. [Lay03, p. 399] If {u1 , . . . , u p } is an orthonormal basis for a subspace S of Cn , then for any v ∈ Cn ,
Proj S v = (u∗1 v)u1 + · · · + (u∗p v)u p .
6. [TB97, p. 46] Let v ∈ Cn be a nonzero vector. Then
r Proj = vv is the orthogonal projection onto the line L = span{v}.
v
v v
r Proj = I − vv is the orthogonal projection onto L ⊥ .
⊥v
v∗ v
7. [Mey00, p. 435] (The Best Approximation Theorem) Let S be a finite dimensional subspace of an
inner product space V and let b be a vector in V . Then Proj S b is the unique vector in S that is
closest to b in the sense that
min b − s = b − Proj S b .
s∈S
The vector Proj S b is called the best approximation to b by the elements of S.
Examples:
1. Generally, an orthogonal projection P ∈ Cn×n is not a unitary matrix.
2. Let {v1 , v2 , · · · , vn } be an orthogonal basis for Rn and let S the subspace of Rn spanned by
{v1 , · · · , vk }, where 1 ≤ k ≤ n − 1. Then w = c 1 v1 + c 2 v2 + · · · + c n vn ∈ Rn can be written as w = s + t, where s = c 1 v1 + · · · + c k vk ∈ S and t = c k+1 vk+1 + · · · + c n vn ∈ S ⊥ .
3. Let u1 = [2/3, 2/3, 1/3]T , u2 = [1/3, −2/3, 2/3]T , and x = [2, 3, 5]T . Then {u1 , u2 } is an orthonormal basis for the subspace S = span {u1 , u2 } of R3 .
r The orthogonal projection of x onto S is
Proj S x = u1T xu1 + u2T x u2 = [4, 2, 3]T .
r The orthogonal projection of x onto S ⊥ is y = x − Proj x = [−2, 1, 2]T .
S
r The vector in S that is closest to x is Proj x = [4, 2, 3]T .
S
5-8
Handbook of Linear Algebra
r Let M = [u , u ]. Then the orthogonal projection onto S is
1 2
5
1⎢
Proj S = M M T = ⎣2
9
4
2
4
8 −2⎦ .
−2
5
r The orthogonal projection of any v ∈ R3 onto S can be computed by Proj v = M M T v. In
S
particular, M M T x = [4, 2, 3]T .
4. Let w1 = [1, 1, 0]T and w2 = [1, 0, 1]T . Consider the subspace W = span{w1 , w2 } of R3 . Define
1
the matrix M = [w1 , w2 ] = ⎣1
0
1
2
0⎦. Then M T M =
1
1
1
.
2
r The orthogonal projection onto W is Proj = M(M T M)−1 M T =
W
1
⎣1
0
1
⎥ 2
0⎦
1
1
1
2
−1
1
1
1
0
2
1⎢
0
= ⎣1
1
3
1
1
2
−1
1
−1⎦ .
2
r The orthogonal projection of any v ∈ R3 onto W can be computed by Proj v. For v = [1, 2, 3]T ,
W
ProjW v = ProjW [1, 2, 3]T = [7/3, 2/3, 5/3]T .
5.5
Gram−Schmidt Orthogonalization and QR Factorization
Definitions:
Let {a1 , a2 , . . . , an } be a basis for a subspace S of an inner product space V . An orthonormal basis
{u1 , u2 , . . . , un } for S can be constructed using the following Gram–Schmidt orthogonalization process:
u1 =
a1
a1
and uk =
ak −
ak −
k−1
i =1
k−1
i =1
ak , ui ui
ak , ui ui
,
for k = 2, . . . , n.
ˆ where Qˆ ∈ Cm×n has
A reduced QR factorization of A ∈ Cm×n (m ≥ n) is a factorization A = Qˆ R,
n×n
is an upper triangular matrix.
orthonormal columns and Rˆ ∈ C
A QR factorization of A ∈ Cm×n (m ≥ n) is a factorization A = Q R, where Q ∈ Cm×m is a unitary
matrix and R ∈ Cm×n is an upper triangular matrix with the last m − n rows of R being zero.
Facts:
1. [TB97, p. 51] Each A ∈ Cm×n (m ≥ n) has a full Q R factorization A = Q R. If A ∈ Rm×n , then
both Q and R may be taken to be real.
ˆ If A ∈ Rm×n ,
2. [TB97, p. 52] Each A ∈ Cm×n (m ≥ n) has a reduced Q R factorization A = Qˆ R.
then both Qˆ and Rˆ may be taken to be real.
ˆ
3. [TB97, p. 52] Each A ∈ Cm×n (m ≥ n) of full rank has a unique reduced Q R factorization A = Qˆ R,
where Qˆ ∈ Cm×n and Rˆ ∈ Cn×n with real r ii > 0.
4. [TB97, p. 48] The orthonormal basis {u1 , u2 , . . . , un } generated via the Gram–Schmidt orthogonalization process has the property
Span({u1 , u2 , . . . , uk }) = Span({a1 , a2 , . . . , ak }),
for k = 1, 2, . . . , n.
Inner Product Spaces, Orthogonal Projection, Least Squares
5-9
5. [TB97, p. 51]
Algorithm 1: Classical Gram–Schmidt Orthogonalization:
input: a basis {a1 , a2 , . . . , an } for a subspace S
output: an orthonormal basis {u1 , u2 , . . . , un } for S
for j = 1 : n
u j := a j
for i = 1 : j − 1
r i j := a j , ui
u j := u j − r i j ui
end
r j j := u j
u j := u j /r j j
end
6. [TB97, p. 58]
Algorithm 2: Modified Gram–Schmidt Orthogonalization
input: a basis {a1 , a2 , . . . , an } for a subspace S
output: an orthonormal basis {u1 , u2 , . . . , un } for S
wi := ai , i = 1 : n
for i = 1 : n
r ii := wi
ui := wi /r ii
for j = i + 1 : n
r i j := w j , ui
w j := w j − r i j ui
end
end
7. [Mey00, p. 315] If exact arithmetic is used, then Algorithms 1 and 2 generate the same orthonormal basis {u1 , u2 , . . . , un } and the same r i j , for j ≥ i .
8. [GV96, pp. 230–232] If A = [a1 , a2 , . . . , an ] ∈ Cm×n (m ≥ n) is of full rank n, then the classic
ˆ with Qˆ =
or modified Gram–Schmidt process leads to a reduced QR factorization A = Qˆ R,
[u1 , u2 , . . . , un ] and Rˆ i j = r i j , for j ≥ i , and Rˆ i j = 0, for j < i .
9. [GV96, p. 232] The costs of Algorithm 1 and Algorithm 2 are both 2mn2 flops when applied to
compute a reduced QR factorization of a matrix A ∈ Rm×n .
10. [Mey00, p. 317 and p. 349] For the QR factorization, Algorithm 1 and Algorithm 2 are not numerically stable. However, Algorithm 2 often yields better numerical results than Algorithm 1.
11. [Mey00, p. 349] Algorithm 2 is numerically stable when it is used to solve least squares problems.
12. (Numerically stable algorithms for computing the QR factorization using Householder reflections
and Givens rotations are given in Chapter 38.)
13. [TB97, p. 54] (See also Chapter 38.) If A = Q R is a QR factorization of the rank n matrix A ∈ Cn×n ,
then the linear system Ax = b can be solved as follows:
r Compute the factorization A = Q R.
r Compute the vector c = Q ∗ b.
r Solve Rx = c by performing back substitution.
5-10
Handbook of Linear Algebra
Examples:
1
1. Consider the matrix A = ⎣2
0
2
0 ⎦.
2
r A has a (full) QR factorization A = Q R:
1
2
⎢2
0⎥
⎦=
0
√1
5
⎢ 2
⎢√
⎣ 5
2
0
4
3 5
− 3√2 5
5
3
− 23
⎤ ⎡√
1⎥
⎥⎢
3⎦⎣
2
3
0
√2
5
√6 ⎥ .
5⎦
0
0
5
r A has a reduced QR factorization A = Q
ˆ R:
ˆ
1
2
0
2
⎢2
0⎥
⎦=
√1
⎢ 25
⎢√
⎣ 5
0
4
⎡√
3 5
5
2
− 3√5 ⎥
0
5
3
√2
5⎦
.
√6
5
3
1 −2
1⎥
⎢3 −4
2. Consider the matrix A = ⎢
⎥. Using the classic or modified Gram–Schmidt process
⎣3 −4 −1⎦
3
1
0
gives the following reduced QR factorization:
⎡1
2
3
1 −2
1
⎥ ⎢
3
−4
1
⎥ ⎢2
⎥= 1
⎣3 −4 −1⎦ ⎢
⎣2
3
1
0
1
2
5.6
1
2
− 12
− 12
1
2
− 12 ⎡
1⎥ 6
2⎥⎢
0
1⎥⎣
−2⎦ 0
1
2
−3 −1
5 −1⎦ .
0
2
Singular Value Decomposition
Definitions:
A singular value decomposition (SVD) of a matrix A ∈ Cm×n is a factorization
A = U V ∗,
= diag(σ1 , σ2 , . . . , σ p ) ∈ Rm×n , p = min{m, n},
where σ1 ≥ σ2 ≥ . . . ≥ σ p ≥ 0 and both U = [u1 , u2 , . . . , um ] ∈ Cm×m and V = [v1 , v2 , . . . , vn ] ∈ Cn×n
are unitary. The diagonal entries of are called the singular values of A. The columns of U are called left
singular vectors of A and the columns of V are called right singular vectors of A.
Let A ∈ Cm×n with rank r ≤ p = min{m, n}. A reduced singular value decomposition (reduced SVD)
of A is a factorization
A = Uˆ ˆ Vˆ ∗ , ˆ = diag(σ1 , σ2 , . . . , σr ) ∈ Rr ×r ,
where σ1 ≥ σ2 ≥ . . . ≥ σr > 0 and the columns of Uˆ = [u1 , u2 , . . . , ur ] ∈ Cm×r and the columns of
Vˆ = [v1 , v2 , . . . , vr ] ∈ Cn×r are both orthonormal.
5-11
Inner Product Spaces, Orthogonal Projection, Least Squares
Facts:
All the following facts except those with a specific reference can be found in [TB97, pp. 25–37].
1. Every A ∈ Cm×n has a singular value decomposition A = U V ∗ . If A ∈ Rm×n , then U and V
may be taken to be real.
2. The singular values of a matrix are uniquely determined.
3. If A ∈ Cm×n has a singular value decomposition A = U V ∗ , then
A∗ u j = σ j v j ,
Av j = σ j u j ,
for j = 1, 2, . . . , p = min{m, n}.
4. If U V ∗ is a singular value decomposition of A, then V
of A∗ .
5. If A ∈ Cm×n has r nonzero singular values, then
u∗j Av j = σ j ,
T
U ∗ is a singular value decomposition
r rank(A) = r .
r A=
r
σ j u j v∗j .
j =1
r ker(A) = span{v , . . . , v }.
r +1
n
r range(A) = span{u , . . . , u }.
1
r
6. Any A ∈ Cm×n of rank r ≤ p = min{m, n} has a reduced singular value decomposition,
A = Uˆ ˆ Vˆ ∗ , ˆ = diag(σ1 , σ2 , . . . , σr ) ∈ Rr ×r ,
7.
8.
9.
10.
where σ1 ≥ σ2 ≥ · · · ≥ σr > 0 and the columns of Uˆ = [u1 , u2 , . . . , ur ] ∈ Cm×r and the columns
of Vˆ = [v1 , v2 , . . . , vr ] ∈ Cn×r are both orthonormal. If A ∈ Rm×n , then Uˆ and Vˆ may be taken
to be real.
If rank(A) = r , then A has r nonzero singular values.
The nonzero singular values of A are the square roots of the nonzero eigenvalues of A∗ A or AA∗ .
[HJ85, p. 414] If U V ∗ is a singular value decomposition of A, then the columns of V are
eigenvectors of A∗ A; the columns of U are eigenvectors of AA∗ .
[HJ85, p. 418] Let A ∈ Cm×n and p = min{m, n}. Define
G=
0
A∗
A
∈ C(m+n)×(m+n) .
0
If the singular values of A are σ1 , . . . , σ p , then the eigenvalues of G are σ1 , . . . , σ p , −σ1 , . . . , −σ p
and additional |n − m| zeros.
11. If A ∈ Cn×n is Hermitian with eigenvalues λ1 , λ2 , · · · , λn , then the singular values of A are
|λ1 |, |λ2 |, · · · , |λn |.
12. For A ∈ Cn×n , |det A| = σ1 σ2 · · · σn .
13. [Aut15; Sch07] (Eckart–Young Low Rank Approximation Theorem)
Let A = U V ∗ be an SVD of A ∈ Cm×n and r = rank(A). For k < r , define Ak = kj =1 σ j u j v∗j .
Then
r
r
A − Ak
2
=
A − Ak
F
=
min
A− B
min
A− B
rank(B)≤k
2
= σk+1 ;
r
rank(B)≤k
F
=
σ j2 ,
j =k+1
m
where M
2
= max ||Mx||2 and M
||x||2 =1
F
n
=
mi2j are the 2-norm and Frobenius norm of
i =1 j =1
matrix M, respectively. (See Chapter 37 for more information on matrix norms.)
5-12
Handbook of Linear Algebra
Examples:
1
Consider the matrices A = ⎣2
0
2
1
0⎦ and B = AT =
2
2
1. The eigenvalues of AT A =
5
2
3. u1 =
1
3
Av1 =
and u2 =
u1 , u2 , and e1 produces u3 =
0
.
2
2
are 9 and 4. So, the singular values of A are 3 and 2.
8
2. Normalized eigenvectors for AT A are v1 =
√ ⎤
5
⎢ 23 ⎥
⎢ √ ⎥
⎣3 5⎦
4
3 5
2
0
1
2
Av2 =
√1
5
√2
5
0
and v2 =
⎢ √2 ⎥
⎣ 5 ⎦.
− √15
√2
5
− √15
.
Application of the Gram–Schmidt process to
2
⎢ 31 ⎥
⎢− ⎥ .
⎣ 3⎦
− 23
4. A has the singular value decomposition A = U V T , where
5
0
1 ⎢
U = √ ⎣2
6
3 5 4 −3
√ ⎤
2√5
−√5⎦ ,
−2 5
3
= ⎣0
0
0
1 1
2⎦ , V = √
5 2
0
2
.
−1
5. A has the reduced singular value decomposition A = Uˆ ˆ Vˆ T , where
5
1 ⎢
Uˆ = √ ⎣2
3 5 4
0
3
6⎦ , ˆ =
0
−3
6. B has the singular value decomposition B = U B
1 1
U B = VA = √
5 2
2
,
−1
B
=
3
0
0
2
1 1
0
, Vˆ = √
2
5 2
T
B VB ,
2
.
−1
where
5
1 ⎢
0
, VB = U A = √ ⎣2
0
3 5 4
0
6
−3
√ ⎤
2√5
−√5⎦ .
−2 5
(U A = U and VA = V for A were given in Example 4.)
5.7
Pseudo-Inverse
Definitions:
A Moore–Penrose pseudo-inverse of a matrix A ∈ Cm×n is a matrix A† ∈ Cn×m that satisfies the following
four Penrose conditions:
AA† A = A; A† AA† = A† ; (AA† )∗ = AA† ; (A† A)∗ = A† A.
Facts:
All the following facts except those with a specific reference can be found in [Gra83, pp. 105–141].
1. Every A ∈ Cm×n has a unique pseudo-inverse A† . If A ∈ Rm×n , then A† is real.
### Tài liệu bạn tìm kiếm đã sẵn sàng tải về
Chapter 5. Inner Product Spaces, Orthogonal Projection, Least Squares, and Singular Value Decomposition
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+0
# How many solutions are there to the equation x1 + x2 + x3 + x4 + x5 = 21, where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that a
+45
41
25813
5
How many solutions are there to the equation x1 + x2 + x3 + x4 + x5 = 21, where xi , i = 1, 2, 3, 4, 5, is a nonnegative integer such that a) x1 ≥ 1? b) xi ≥ 2 for i = 1, 2, 3, 4, 5? c) 0 ≤ x1 ≤ 10? d) 0 ≤ x1 ≤ 3, 1 ≤ x2 < 4, and x3 ≥ 15?
May 24, 2014
#4
+2353
+23
Allright so I found some answers online and I'm going to combine them to give you a proper answer;
'How many solutions total are there to x1 + x2 + x3 + x4 + x5 = 21 ?
Imagine you have 21 "objects" which we can represent with the letter o:
o o o o o o o o o o o o o o o o o o o o o
I think that's 21 of them. Now, how do we divide them up into 5 non-negative values? Just split them up into five sections. The way to do that is just like this:
o o o o | o o o o | o o o o o o |o o o o o | o o
That corresponds to the solution x1=4, x2=4, x3=6, x4=5, x5=2
Now how many ways can you do that? There are 20 viable places for the separator marks -- anywhere between the 21 o's. So you pick a place for four of them -- 20C4.
But those are the solutions with positive values -- you want non-negative values.
How can we compensate for that? Well, we can reduce this to a different problem.
If I want to solve things with non-negative values x1,x2,..,x5, I can simply consider some positive number y1 and then let x1=y1-1. That will produce all non-negative possibilities.
So you just solve:
x1 + x2 + x3 + x4 + x5 = 21
(y1 - 1) + (y2 - 1) + (y3 - 1) + (y4 - 1) + (y5 - 1) = 21
y1 + y2 + y3 + y4 + y5 = 26
So now y1,y2,..,y5 are positive solutions -- we've just "transformed" the problem into a problem we know the answer to, just with different parameters. The answer is 25C4 instead of 20C4. '
25C4 = 12650 possibilities
Now for part a) of your question;
Let z1 = x1-1 and zi = xi for i = 2,3,4,5.
Then x1+x2+x3+x4+x5 = 21 => z1+1+z2+z3+z4+z5 =21 => z1+z2+z3+z4+z5 = 20
Similarly this has 24C4= 10626 solutions
part b) can be solved in the same way leading to 15C4 = 1365 solutions
For part c). The number of solutions with x1<=10 is the total number of solutions minus the solutions where x1>=11. Then again it can be solved in a similar matter.
This leads to 25C4-14C4 = 12650 - 1001 = 11649 solutions
'The solution could also be found using combinations with repetitions,
The solution is equal to a combination of putting 21 elements in 5 boxes(consider each variable a box).
Hence there are 21 elements and 4 borders x1 | x2 | x3 | x4 | x5 where the elements and borders could be combined in different ways.
as x3>=15, the solution could be simplified to 6 elements in 4 borders ( 5 boxes) ; for which the solution is C(10,6)
Now for the remaining conditions, subtract the case when x1>3 and when x2>3 , which is 2*C(6,2)
The final step is to subtract the case when x2=0 , which is tricky because when you take x2=0, there are now 6 elements to fill in 3 borders (4 boxes) and you also have to consider the case when x2=0 and x1>3, the value for which was already subtracted previously.
So in essence the final step would be C(9,6) - C(5,2) (All combinations when x2=1 - All combinations when x2=1 and x1>3)
The final solution is the C(10,6)-C(6,2)-C(6,2)-(C(9,6)-C(5,2)) = 106'
I hope this helps
May 24, 2014
#1
+27396
+10
a) constrains x1 to be greater than zero; c) allows it to be zero.
b) constrains x2 to be greater than 1; d) allows it to be 1.
Which constraints apply?
May 24, 2014
#3
+2353
+18
I'm still thinking about the simplest way to solve this, but I think these are seperate questions Alan.Since the values need to be nonnegative integers there's a limited amount of solutions for each case...
May 24, 2014
#4
+2353
+23
Allright so I found some answers online and I'm going to combine them to give you a proper answer;
'How many solutions total are there to x1 + x2 + x3 + x4 + x5 = 21 ?
Imagine you have 21 "objects" which we can represent with the letter o:
o o o o o o o o o o o o o o o o o o o o o
I think that's 21 of them. Now, how do we divide them up into 5 non-negative values? Just split them up into five sections. The way to do that is just like this:
o o o o | o o o o | o o o o o o |o o o o o | o o
That corresponds to the solution x1=4, x2=4, x3=6, x4=5, x5=2
Now how many ways can you do that? There are 20 viable places for the separator marks -- anywhere between the 21 o's. So you pick a place for four of them -- 20C4.
But those are the solutions with positive values -- you want non-negative values.
How can we compensate for that? Well, we can reduce this to a different problem.
If I want to solve things with non-negative values x1,x2,..,x5, I can simply consider some positive number y1 and then let x1=y1-1. That will produce all non-negative possibilities.
So you just solve:
x1 + x2 + x3 + x4 + x5 = 21
(y1 - 1) + (y2 - 1) + (y3 - 1) + (y4 - 1) + (y5 - 1) = 21
y1 + y2 + y3 + y4 + y5 = 26
So now y1,y2,..,y5 are positive solutions -- we've just "transformed" the problem into a problem we know the answer to, just with different parameters. The answer is 25C4 instead of 20C4. '
25C4 = 12650 possibilities
Now for part a) of your question;
Let z1 = x1-1 and zi = xi for i = 2,3,4,5.
Then x1+x2+x3+x4+x5 = 21 => z1+1+z2+z3+z4+z5 =21 => z1+z2+z3+z4+z5 = 20
Similarly this has 24C4= 10626 solutions
part b) can be solved in the same way leading to 15C4 = 1365 solutions
For part c). The number of solutions with x1<=10 is the total number of solutions minus the solutions where x1>=11. Then again it can be solved in a similar matter.
This leads to 25C4-14C4 = 12650 - 1001 = 11649 solutions
'The solution could also be found using combinations with repetitions,
The solution is equal to a combination of putting 21 elements in 5 boxes(consider each variable a box).
Hence there are 21 elements and 4 borders x1 | x2 | x3 | x4 | x5 where the elements and borders could be combined in different ways.
as x3>=15, the solution could be simplified to 6 elements in 4 borders ( 5 boxes) ; for which the solution is C(10,6)
Now for the remaining conditions, subtract the case when x1>3 and when x2>3 , which is 2*C(6,2)
The final step is to subtract the case when x2=0 , which is tricky because when you take x2=0, there are now 6 elements to fill in 3 borders (4 boxes) and you also have to consider the case when x2=0 and x1>3, the value for which was already subtracted previously.
So in essence the final step would be C(9,6) - C(5,2) (All combinations when x2=1 - All combinations when x2=1 and x1>3)
The final solution is the C(10,6)-C(6,2)-C(6,2)-(C(9,6)-C(5,2)) = 106'
I hope this helps
reinout-g May 24, 2014
#5
+27396
+4
Your interpretation of the question makes sense, Reinout. I'm glad I didn't think of that interpretation or it would have been my head that was spinning!
May 25, 2014
#5
0
Allright so I didnt find some answers online and I'm going to combine them to give you a proper answer;
'How many solutions total are there to x1 + x2 + x3 + x4 + x5 = 21 ?
Imagine you have 21 "objects" which we can represent with the letter o:
o o o o o o o o o o o o o o o o o o o o o
I think that's 21 of them. Now, how do we divide them up into 5 non-negative values? Just split them up into five sections. The way to do that is just like this:
o o o o | o o o o | o o o o o o |o o o o o | o o
That corresponds to the solution x1=4, x2=4, x3=6, x4=5, x5=2
Now how many ways can you do that? There are 20 viable places for the separator marks -- anywhere between the 21 o's. So you pick a place for four of them -- 20C4.
But those are the solutions with positive values -- you want non-negative values.
How can we compensate for that? Well, we can reduce this to a different problem.
If I want to solve things with non-negative values x1,x2,..,x5, I can simply consider some positive number y1 and then let x1=y1-1. That will produce all non-negative possibilities.
So you just solve:
x1 + x2 + x3 + x4 + x5 = 21
(y1 - 1) + (y2 - 1) + (y3 - 1) + (y4 - 1) + (y5 - 1) = 21
y1 + y2 + y3 + y4 + y5 = 26
So now y1,y2,..,y5 are positive solutions -- we've just "transformed" the problem into a problem we know the answer to, just with different parameters. The answer is 25C4 instead of 20C4. '
25C4 = 12650 possibilities
Now for part a) of your question;
Let z1 = x1-1 and zi = xi for i = 2,3,4,5.
Then x1+x2+x3+x4+x5 = 21 => z1+1+z2+z3+z4+z5 =21 => z1+z2+z3+z4+z5 = 20
Similarly this has 24C4= 10626 solutions
part b) can be solved in the same way leading to 15C4 = 1365 solutions
For part c). The number of solutions with x1<=10 is the total number of solutions minus the solutions where x1>=11. Then again it can be solved in a similar matter.
This leads to 25C4-14C4 = 12650 - 1001 = 11649 solutions
'The solution could also be found using combinations with repetitions,
The solution is equal to a combination of putting 21 elements in 5 boxes(consider each variable a box).
Hence there are 21 elements and 4 borders x1 | x2 | x3 | x4 | x5 where the elements and borders could be combined in different ways.
as x3>=15, the solution could be simplified to 6 elements in 4 borders ( 5 boxes) ; for which the solution is C(10,6)
Now for the remaining conditions, subtract the case when x1>3 and when x2>3 , which is 2*C(6,2)
The final step is to subtract the case when x2=0 , which is tricky because when you take x2=0, there are now 6 elements to fill in 3 borders (4 boxes) and you also have to consider the case when x2=0 and x1>3, the value for which was already subtracted previously.
So in essence the final step would be C(9,6) - C(5,2) (All combinations when x2=1 - All combinations when x2=1 and x1>3)
The final solution is the C(10,6)-C(6,2)-C(6,2)-(C(9,6)-C(5,2)) = 106'
Guest Jan 24, 2016
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https://www.godswillhustle.com/discover-indices-in-a-sorted-matrix-the-place-a-brand-new-quantity-will-be-changed/
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Sunday, February 26, 2023
HomeSoftware DevelopmentDiscover indices in a sorted Matrix the place a brand new quantity...
# Discover indices in a sorted Matrix the place a brand new quantity will be changed
Given a matrix arr[][] which is sorted by the rising variety of components and a quantity X, the duty is to search out the place the place the enter integer will be changed with an present ingredient with out disturbing the order of the sorted components. The matrix is sorted in such a way that:
• Each row is sorted in rising order of components.
• Each single ingredient within the present row will likely be better than each single ingredient of the earlier row and smaller than each single ingredient of the following row.
Examples:
Enter: arr[][] = { {1, 1, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 15, 16} }, X = Â 2
Output: 0 2
Rationalization: Within the given matrix, X = 2 so 2 will be changed both with {0, 1} or {0, 2} as a result of alternative at these two positions doesn’t break the order of the sorted matrix.
Enter: arr[][] = { {1, 1, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 15, 16} }, X = 14
Output: 2 2
Rationalization: The enter quantity is 14 so it may very well be both changed with 13 or 15 to keep up the sorted order.
Method: This may be solved with the next concept:
The strategy is to use the binary search for the optimum answer.
 Steps concerned within the implementation of code:
• Initialize l(low) as 0 and h(excessive) as (m*n)-1. Apply binary search utilizing a whereas loop the place l < h.
• Initialize mid(center) as (l+h)/2 Â and entry the center ingredient utilizing arr[mid/m][mid%m] .
• As soon as the binary search’s loop is over then return the indexes that are represented by mid.
Under is the implementation of the above strategy:
## C++
```// C++ code of the above strategy
#embrace <bits/stdc++.h>
utilizing namespace std;
// To seek out the index of changed ingredient
vector<int> findIndex(vector<vector<int> > arr, int quantity)
{
int l = 0, m = arr[0].measurement(), n = arr.measurement(), mid;
int h = m * n - 1;
// Whereas loop to do the binary search
whereas (l <= h) {
// Get the mid ingredient
mid = (l + h) / 2;
// If quantity itself is discovered
if (arr[mid / m][mid % m] == quantity) {
return { mid / m, mid % m };
}
else if (arr[mid / m][mid % m] < quantity) {
l = mid + 1;
}
else {
h = mid - 1;
}
}
// Return the index of
// changed ingredient
return { mid / m, mid % m };
}
// Driver code
int fundamental()
{
vector<vector<int> > arr = { { 1, 1, 3, 4, 5 },
{ 6, 7, 8, 9, 10 },
{ 11, 12, 13, 15, 16 } };
// Perform name
vector<int> ans = findIndex(arr, 25);
cout << ans[0] << " " << ans[1];
return 0;
}```
Time Complexity: O(logN)
Auxiliary House: O(1)Â
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###### \${selected_topic_name}
• Notes
Find the area of the surface obtained by rotating the arc $y=x^{3}$ where $0 \leq x \leq 2$ about the x-axis
area $=S=2 \pi \int y d s=2 \pi \int y \sqrt{1+\left(y^{\prime}\right)^{2}} d x$
(1) $y=x^{3} \Rightarrow y^{\prime}=3 x^{2} \Rightarrow\left(y^{\prime}\right)^{2}=\left(3 x^{2}\right)^{2}=9 x^{4} \Rightarrow 1+(y)^{2}=1+9 x^{4}$
$\Rightarrow \sqrt{1+\left(y^{\prime}\right)^{2}}=\sqrt{1+9 x^{4}} \Rightarrow \text { area }=s=2 \pi \int_{0}^{2} x^{3} \sqrt{1+9 x^{4}} d x$
let $t=1+9 x^{4} \Rightarrow d t=36 x^{3} d x$
when $x=0 \Rightarrow t=1$
when $x=2 \Rightarrow t=1+9(2)^{4}=145$
area $=S=2 \pi \int_{1}^{145} x^{3} \sqrt{t} d x \frac{36}{36}=\frac{2 \pi}{36} \int_{1}^{145} \sqrt{t} \cdot {36 x^{3} d x}$
$=\frac{\pi}{18} \int_{1}^{145} t^{1 / 2} d t=\frac{\pi}{18}\left.\frac{t^{3 / 2}}{3 / 2}\right|_{1} ^{145}$
$=S=\frac{\pi}{18} \cdot \frac{2}{3} \cdot\left.t^{3 / 2}\right|_{1} ^{145}=\frac{\pi}{27} t^{3 / 2}|_{1} ^{145}=$
$=\frac{\pi}{27}\left[(145)^{3 / 2}-1\right]$
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Instructions
For the following questions answer them individually
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
## Let E denote the parabola $$y^{2} = 8x$$. Let P = (−2, 4), and let Q and $$Q^{'}$$ be two distinct points on E such that the lines PQ and $$PQ^{'}$$ are tangents to E. Let F be the focus of E. Then which of the following statements is (are) TRUE ?
Instructions
Consider the region $$R = \left\{(𝑥, 𝑦) \epsilon R \times R ∶ x \geq 0 and y^{2} \leq 4 − x \right\}$$. Let F be the family of all circles that are contained in 𝑅 and have centers on the x-axis. Let C be the circle that has largest radius among the circles in F. Let ($$\alpha, \beta$$) be a point where the circle C meets the curve $$y^{2} = 4 − x$$.
Question 7
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Question 8
## The value of $$\alpha$$ is ___ .
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Instructions
Let $$f_{1}:(0, \infty) \rightarrow$$ R and $$f_{2}:(0, \infty) \rightarrow$$ be defined by
$$f_{1}(x): \int_{0}^{x} \prod_{j=1}^{21}(t-1)^{j}$$ dt, $$x> 0$$ and
$$f_{3}(x) = 98(x-1)^{50}-600(x-1)^{49} + 2450, x > 0$$,
where, for any positive integer n and real numbers $$a_{1}, a_{2}, … , a_{n}, \prod_{}{}^{n}i=1 a_{i}$$ denotes the product of $$a_{1}, a_{2}, … , a_{n}$$. Let $$m_{i}$$ and $$n_{i}$$, respectively, denote the number of points of local minima and the number of points of local maxima of function $$f{i}$$, i = 1, 2, in the interval ($$0, \infty$$).
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## First order logic: formulas, models, tautologies
### Exercise 1
Let $${\mathfrak A} =\langle{\mathbb N}, p^{\mathfrak A}, q^{\mathfrak A}\rangle$$, where:
$$\langle a,b\rangle \in p^{\mathfrak A}$$ iff $$a+b\geq 6$$;
$$\langle a,b\rangle \in q^{\mathfrak A}$$ iff $$b=a+2$$.
Check if formulas
1. $$\forall x p(x,y) \to \exists x q(x,y)$$;
2. $$\forall x p(x,y) \to \forall x q(x,y)$$;
3. $$\forall x p(x,y) \to \exists x q(x,z)$$;
are satisfied under the valuation $$v(y) = 7$$, $$v(z) = 1$$ in structure $${\mathfrak A}$$.
### Exercise 2
Let $${\mathfrak A} = \langle {\mathbb Z}, f^{\mathfrak A}, r^{\mathfrak A}\rangle$$ and $${\mathfrak B} = \langle {\mathbb Z}, f^{\mathfrak B}, r^{\mathfrak B}\rangle$$, where $$f^{\mathfrak A}(m,n) = \min(m,n)$$ for $$m,n\in{\mathbb Z}$$, and $$r^{\mathfrak A}$$ is the relation $$\geq$$;
$$f^{\mathfrak B}(m,n) = m^2+n^2$$ for $$m,n\in{\mathbb Z}$$, and $$r^{\mathfrak B}$$ is the relation $$\leq$$.
Check if formulas
1. $$\forall y(\forall x(r(z,f(x,y))\to r(z,y)))$$;
2. $$\forall y(\forall x(r(z,f(x,y)))\to r(z,y))$$,
are satisfied under the valuation $$v(z) =5$$, $$v(y)=7$$ in structures $${\mathfrak A}$$ and $${\mathfrak B}$$.
### Exercise 3
Is formula $$\forall x(\lnot r(x,y)\to\exists z(r(f(x,z),g(y))))$$ satisfied under the valuation $$v(x) =3$$, $$w(x) = 6$$ and $$u(x) = 14$$
1. in the structure $${\mathfrak A} = \langle {\mathbb N}, r^{\mathfrak A}\rangle$$, where $$r^{\mathfrak A}$$ is the divisibility relation?
2. in the structure $${\mathfrak B} = \langle {\mathbb N}, r^{\mathfrak B}\rangle$$, where $$r^{\mathfrak B}$$ is the congruency modulo 7?
### Exercise 4
In which structures is the formula $$\exists y (y\neq x)$$ satisfied?
And the formula $$\exists y (y\neq y)$$ obtained by naive substitution of $$y$$ to $$x$$?
### Exercise 5
Give examples of models and valuations such that the formula
$$p(x,f(x)) \to \forall x\exists y\, p(f(y),x)$$
is:
a) satisfied;
b) not satisfied.
### Exercise 6
Check if the following formulas are tautologies an if they are satisfiable:
1. $$\exists x\forall y(p(x) \vee q(y)) \to \forall y(p(f(y))\vee q(y))$$;
2. $$\forall y(p(f(y))\vee q(y)) \to \exists x\forall y(p(x) \vee q(y))$$;
3. $$\exists x(\forall y q(y)\to p(x))\to \exists x\forall y(q(y)\to p(x))$$;
4. $$\exists x(\forall y q(y)\to p(x)) \to\exists x(q(x)\to p(x))$$.
### Exercise 7
Let $$f$$ be a function symbol of arity two, that is not used in the formula $$\varphi$$.
Show that the formula $$\forall x\exists y \varphi$$ is satisfiable if and only if the formula $$\forall x \varphi(f(x)/y)$$ is satisfiable.
### Exercise 8
Show that the formula $$\forall x\exists y\,p(x,y)\wedge \forall x\neg p(x,x) \wedge \forall x\forall y\forall z(p(x,y)\wedge p(y,z)\to p(x,z))$$ has only infinite models.
### Exercise 9
For each $$n$$ give a closed formula $$\varphi_n$$ such that $${\mathfrak A}\models\varphi_n$$ iff $${\mathfrak A}$$ has exactly $$n$$ elements.
### Exercise 10
Show that for each finite structure $${\mathfrak A}$$ over a finite signature there exists a set of first order formulas $$\Delta$$ such that $${\mathfrak A}\models\Delta$$ and for all structures satisfying $${\mathfrak B}\models\Delta$$ it holds that $${\mathfrak B}\cong{\mathfrak A}.$$
### Exercise 11
Is it true that $${\mathfrak A} \models \exists x\,\varphi$$ implies existence of a term $$t$$ such that $${\mathfrak A} \models \varphi[t/x]$$?
### Exercise 12
Let $$\varphi = \forall x\forall y\,(y=f(g(x))\to(\exists u\,(u=f(x)\land y=g(u))))$$ and $$\psi = \forall x\,[f(g(f(x)))=g(f(f(x)))]$$. Does it hold that
$$\{\psi\}\models\varphi$$?
Hint
In exercises of the form "Show that the set of formulas $$\Delta$$ is independent", one has to prove that for each $$\varphi\in\Delta,$$ $$\Delta\setminus\{\varphi\}\not\models\varphi,$$ by showing a model of $$\Delta\setminus\{\varphi\},$$ which is not a model of $$\varphi.$$
### Exercise 13
Show that the set of axioms of equivalence relation
$$\left\{\begin{array}[]{c}\forall x\forall y(Exy\to Eyx)\\ \forall x\ Exx\\ \forall x\forall y\forall z((Exy\land Eyz)\to Exz)\end{array}\right\}$$
is independent.
### Exercise14
Show that the set of axioms of linear orders
$$\left\{\begin{array}[]{c}\forall x\forall y((x\leq y)\lor(y\leq x))\\ \forall x\forall y((x\leq y\land y\leq x)\to x=y)\\ \forall x\forall y\forall z((x\leq y\land y\leq z)\to x\leq z)\end{array}\right\}$$
is independent.
### Exercise 15
Show that the set of axioms of of group theory (in multiplicative notation, over signature
$$\Sigma^{F}_{{2}}=\{*\},\Sigma^{F}_{0}=\{ 1\}$$)
$$\left\{\begin{array}[]{c}\forall x((1*x=x)\land(x*1=x))\\ \forall x\forall y\forall z((x*y)*z=x*(y*z))\\ \forall x\exists y((x*y=1)\land(y*x=1))\end{array}\right\}$$
is independent.
### Exercise 16
Show that the formula $$(\forall x\exists y\ Exy)\to(\exists x\forall y\ Exy)$$ is not a tautology.
### Exercise 17
Show that the formula
$$(\forall x\forall y((f(x)=f(y))\to(x=y)))\to(\forall x\exists y(f(y)=x))$$
is not a tautology. Does its negation have a finite model?
### Exercise18
Show that the formula $$\exists x\exists y\exists u\exists v((\lnot u=x)\lor(\lnot v=y))\land(f(x,y)=f(u,v))$$ is not a tautology. How many non-isomorphic finite models does this formula have?
### Exercise 19
Show that the following formulas are tautologies:
• $$(\exists y p(y) \to \forall z q(z)) \to \forall y\forall z(p(y)\to q(z))$$;
• $$(\forall x\exists y r(x,y) \to \exists x\forall y r(y,x))\to \exists x\forall y(r(x,y) \to r(y,x))$$;
• $$\forall x\exists y((p(x)\to q(y))\to r(y)) \to ((\forall x p(x)\to \forall y q(y))\to \exists y r(y))$$;
• $$\forall x(p(x)\to \exists y q(y))\to \exists y(\exists x p(x)\to q(y))$$.
### Exercise 20
Does it hold that
$$\{\forall x\underbrace{f\ldots f}_n(x)= x~|~n=2,3,5,7\}\models\forall x \underbrace{f\ldots f}_{11}(x)= x$$?
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# Convergence of $\sum_{n=2}^\infty {1\over(\log n)^{\log n}}$.
I've been working through exercises in Chapter 8 of Apostol's Mathematical Analysis. Exercise 8.15 gives a number series to be tested for convergence. I've gotten most of them but I'm stuck on
$$\sum_{n=1}^\infty{1\over(\log n)^{\log n}}.$$
The root test is in conclusive, and the things I can think of to compare with have the wrong inequality. I thought I had proven divergence by the integral test, but something was wrong because according to Wolfram Alpha, this converges. Hint, anyone?
P.S. Looking to avoid Cauchy condensation test.
• Is it $\log(n^{\log n})$ or $\log(n)^{\log(n)}$? Commented Feb 2, 2015 at 18:07
• The latter, $(\log n)^{\log n}$. Commented Feb 2, 2015 at 18:07
• Why to avoid the wonderful condensation test? It, together with the $\;n$-th root test, pretty easily tell us there's convergence. Commented Feb 2, 2015 at 18:11
• Well, yes you're correct, but Apostol doesn't cover it. So, there must be a way to do the problem without that test. Commented Feb 2, 2015 at 18:14
• You're right that your integral test must be wrong, since it seems to be converging to $\approx 5.7169706$.
– JPhy
Commented Feb 2, 2015 at 18:16
Hint: $$(\log n)^{\log n} = n^{\log\log n}$$ and as soon as $\log\log n > 1$, you can compare your series with a converging one.
• Shouldn't it be $\log\log n>2$ to compare it with $\frac{1}{x^2}$ Commented Feb 3, 2015 at 12:40
• @avz2611:$\sum_{n\geq 1}\frac{1}{n^{1.00000001}}$ is converging, too. Commented Feb 3, 2015 at 13:18
$\sum u(n)$ and $\sum u(a^n) a^n$ converge/diverge together for $a>0$.
Hence, test $\sum \dfrac {1}{(\log a^{n})^{ \log a^n }} \cdot a^n$ for convergence.
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# Geometry Transformation Composition Worksheet Answers
Geometry Transformation Composition Worksheet Answers. Our Transformations Worksheets are free to obtain, easy to make use of, and very versatile. Reflect \(\Delta DEF\) from Question 2 over the \(x\)-axis, adopted by the \(y\)-axis. A composition of reflections over intersecting traces is identical as a rotation . This Transformations Worksheet will produce problems for training translations, rotations, and reflections of objects.
The transformations are the alterations accomplished to a operate by translation, reflection, rotation, and dilation. The original image generally known as the pre-image is altered to get the picture. This quiz and its attached worksheet assist you to quickly test your information of the compositions of reflections theorem. You might want to carry out a number of transformations and explain how they relate to this principle. In each model, college students might need to write the rule given a graph with the picture and pre-image plotted and write the coordinates after following given rules.
This Transformations Worksheet will produce easy issues for training rotations of objects. This Transformations Worksheet will produce simple problems for working towards translations of objects. Perform a glide reflection over the \(y\)-axis and down 5 items. Perform a glide reflection to the best 6 items, then over the \(x\)-axis. Perform a glide reflection over the \(x\)-axis and to the best 6 items. Key-in the coordinates to the father or mother perform following the rules of transformations.
four.Dilationis when the dimensions of an image is elevated or decreased with out altering its shape. 1.Translationhappens once we move the picture without changing something in it. Hence the shape, measurement, and orientation remain the identical.
Quick activity on Composition of Transformations used for 9th grade Geometry.
You will obtain your rating and answers at the end. Dilations and rotations are centered on the origin. Describe a sequence of transformations that was used to carry ΔBUG onto ΔB’U’G’.
First, the triangle was mirrored over the x-axis. Then translated horizontally 6 unit to the right and vertically 2 items up. A glide reflection is a composition of a mirrored image and a translation. The translation is in a direction parallel to the line of reflection. Learn the way to compose transformations of a figure on a coordinate plane, and perceive the order by which to apply them.
Here is a graphic preview for all of the Transformations Worksheets. You can select totally different variables to customise these Transformations Worksheets on your wants. We have translation, rotation, and reflection worksheets on your use.
Use the trapezoid in the graph to the left to answer questions 20-22. Use the graph of the square to the left to answer questions 7-9. Use the graph of the sq. to the left to reply questions 4-6.
Reflection A reflection is a transformation that flips a figure on the coordinate plane across a given line without changing the shape or dimension of the figure. Glide Reflection A reflection adopted by a translation where the road of reflection is parallel to the direction of translation known as a glide reflection or a stroll. Composite Transformation A composite transformation, also called composition of transformation, is a series of multiple transformations carried out one after the other. We stated there are three forms of isometries, translations, reflections and rotations.
A composition of reflections over parallel lines has the identical impact as a translation . A composition of reflections over intersecting strains is similar as a rotation . Term Definition composition When multiple transformation is performed on a determine.
• Based on how we modify a given picture, there are five main transformations.
• You’re looking for the \(y\) values that come out of these functions.
• TheAn Introduction to the Projectvideo supplies slightly more element about my targets for and my activity during this project.
The 4 primary forms of transformations are rotations, reflections, translations, and resizing. In Preview Activity 1 we experimented with the 4 main forms of operate transformations. You no doubt observed that the values of \(C\) and \(D\) shift the mother or father operate and the values of \(A\) and \(B\) stretch the mother or father perform. The composition of two reflections over parallel lines which would possibly be \(h\) units apart is similar as a translation of \(2h\) items . The type of transformation that happens when each level within the form is mirrored over a line known as the reflection. When the points are reflected over a line, the image is at the identical distance from the line as the pre-image however on the other facet of the road.
If level A is 3 items away from the line of reflection to the right of the road, then level A’ shall be three items away from the line of reflection to the left of the line. Thus the line of reflection acts as a perpendicular bisector between the corresponding factors of the picture and the pre-image. Given a geometric determine and a rotation, reflection, or translation, draw the remodeled determine utilizing, e.g., graph paper, tracing paper, or geometry software. Specify a sequence of transformations that can carry a given determine onto another.
Let the excessive school students translate each quadrilateral and graph the picture on the grid. Rotate, reflect and translate every level following the given rules. Grade 7 students should choose the proper picture of the reworked point. In these worksheets identify the image which best describes the transformation of the given determine. Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective units, paper folding, dynamic geometric software program, and so forth.). SWBAT acknowledge composition notation and perform compositions of transformations.
Our Transformations Worksheets are free to obtain, straightforward to use, and very versatile. This GeoGebra activity allows your college students to discover how rigid transformations work collectively to transform a picture with multiple different transformations occuring. Use the graph of the triangle to the left to reply questions 16-18.
After this activity is full, it results in a great class discussion. Index playing cards, instructions, questions, and answers are offered. A transformation modifications the dimensions, shape, or place of a figure and creates a new figure. A geometry transformation is both inflexible or non-rigid; another word for a rigid transformation is “isometry”. An isometry, such as a rotation, translation, or reflection, does not change the dimensions or shape of the figure.
Contents
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7th Class Mathematics Number System and its Operations Number System
Number System
Category : 7th Class
Learning Objectives:
• To understand integers, decimal, rational number and their representation on number line.
• To learn, addition, subtraction, multiplication and division of integers, decimal number and rational numbers.
• To learn how to compare decimal numbers and rational numbers.
• To learn how to convert fraction to decimal and decimal to fraction.
INTEGERS
VARIOUS TYPES OF NUMBERS:
1. Natural numbers: All numbers from 1 to infinite $\infty$ are known as natural numbers. Thus, $1,\text{ }2,\text{ }3....\infty$ are natural numbers.
2. Whole numbers: All natural numbers including zero is known as whole numbers i.e., $0,\text{}1,\text{ }2,\text{ }3....\infty$ are whole numbers 0 + natural numbers = whole numbers
Note: All natural numbers are whole number but zero is the only whole number which is not natural number.
1. Integers: All natural numbers, 0 and negative numbers are called integers. Thus ............. $-\,5,-\,4,-\,3,-1,\,0,\,1,\,2,\,3,$……………..etc, are all integers.
(i) Positive integers: All natural numbers are positive integers such as 1, 2, 3, 4, ............... etc.
(ii) Negative integers: All negative numbers are negative integers such as ...,$-\,4,-3,-2,-1$
(iii) Zero is neither negative nor positive integer.
Note:
1. Both the positive and negative integer are called directed numbers as they indicate direction. These are also known as signed numbers because of the$+$or$-$sign.
2. The sum of any integer and its negative integer is always zero i.e.,$a+\left( -a \right)=0.$
REPRESENTATION OF INTEGERS ON NUMBER LINE:
Every positive integer is greater than the negative integer.
Zero is less than every positive integer but greater than every negative integer.
1. If two positive or two negative integers are added, we add their values without considering their signs and put common sign before the sum.
(i) $36\,\,\,+27$
\begin{matrix}
+ & 36 \\
+ & 27 \\
+ & 63 \\
\end{matrix}
(ii) $-\,36\,\,\,-27$
\begin{matrix}
- & 36 \\
- & 27 \\
- & 63 \\
\end{matrix}
1. To add a positive and a negative integer, we calculate the difference in their numerical values regardless of their signs and put the sign of greater numerical value integer to the value of difference.
(i) $+\,36\,\,\,-27$
\begin{matrix}
+ & 36 \\
- & 27 \\
+ & 9 \\
\end{matrix}
(ii) $-\,\,36\,\,\,+27$
\begin{matrix}
- & 36 \\
+ & 27 \\
- & 9 \\
\end{matrix}
1. Closure property of addition: The sum of two integers is always an integer.
Examples:
(i) $4+3=7,$which is an integer
(ii) $4\,+\left( -\,3 \right)=1,$which is an integer
(iii) $-\,4+3=-1,$which is an integer
(iv) $-\,4+\left( -\,3 \right)=-7,$which is an integer
1. Commutative law of addition: If$x$and$y$ are any two integers, then, $x+y=y+x$
Examples:
(i) $-7+8=1$ and $8+\left( -\,7 \right)=1$
$\therefore$ $-7+8=8+\left( -\,7 \right)$
(ii) $\left( -\,5 \right)+\left( -\,8 \right)=-13$ and $\left( -\,8 \right)+\left( -\,5 \right)=-13$
$\therefore$ $\left( -\,5 \right)+\left( -\,8 \right)=\left( -\,8 \right)+\left( -\,5 \right)$
1. Associative law of addition: If x, y and z are any three integers then$\left( x+y \right)+z=x+\left( y+z \right)$
Example: $\left\{ \left( -\,5 \right)+\left( -\,6 \right) \right\}+7=-11+7=-\,4$
$\left( -\,5 \right)+\left\{ \left( -\,6 \right)+7 \right\}=-\,5+1=-\,\,4$
$\therefore$ $\left\{ \left( -\,5 \right)+\left( -\,6 \right) \right\}+7=-\,5+\left\{ \left( -\,6 \right)+7 \right\}$
1. Existence of additive identity: For any integer x, we have 0 is called the additive identity for integers.
Examples:
(i) $0+9=9+0=9$
(ii) $\left( -\,6 \right)+0=0+\left( -\,6 \right)=-\,6$
1. Existence of additive inverse: For any integer $x,$ we have $x+\left( -\,x \right)=\left( -\,x \right)+x=0$
(i) The opposite or additive inverse of$x$is $\left( -\,x \right)$ and $\left( -\,x \right)$ is $x$
(ii) The sum of an integer and its opposite is 0.
Example: $4+\left( -\,4 \right)=0$and $\left( -\,4 \right)+4=0$
So additive inverse of 4 is$(-\,4)$and$(-\,4)$is 4.
SUBTRACTION OF INTEGERS:
For any integers$x$and$y,$we define.
(i) $x-y=x+$(additive inverse of$y$)$=x+(-y)$
(ii) $x-(-y)=x+$ {additive inverse of $(-y)$}$=x+y$
PROPERTIES OF SUBTRACTION OF INTEGERS:
1. Closure property for subtraction: If x and y are any integer, then$x-y$is always an integer.
Examples:-
(i) $3-5=-\,2,$which is an integer
(ii) $(-\,3)-6=-\,9,$which is an integer
(iii) $3-(-\,6)=9,$which is an integer
(iv) $(-\,3)-(-\,6)=3,$which is an integer
1. Subtraction of an integer is not commutative:
Examples:-
(i) Consider the integers 2 and 4, we have
$(2-4)=2+(-\,4)=-\,2$and $(4-4)=2+(-\,4)=-\,2$
$\therefore \,\,\,(2-4)\ne (4-2)$
(ii) Consider the integers$(-\,5)$and 3 we have
$(-\,5)-3=(-\,5)+(-\,3)=-\,8$ and $3-(-\,5)=8$
$\therefore \,\,\,(-\ 5)-3\ne -\,3-3-(-\,5)$
(iii) Consider the integers$(-\,6)$and $(-\,4),$we have
$(-\,6)-(-\,4)=-\,6+4=-\,2$and$(-\,4)-(-\,6)=-\,4+6=\,2$
$\therefore \,\,(-\,6)-(-\,4)\ne (-\,4)-(-\,6).$
1. Subtraction of integers is not associative:
Consider the integers $4,$ $(-\,5)$ and $(-\,6),$ we have
$\{4-(-\,5)\}-(-\,6)=(4+5)-(-\,6)=9-(-\,6)$
$=9+6=15$
$4-\{(-\,5)-(-\,6)\}=4-\{(-\,5)+6\}=4-1=3$
$\{4-(-\,5)\}-(-\,6)\ne 4-\{(-\,5)-(-\,6)\}$
MULTIPLICATION OF INTEGERS:-
Rule 1: To find the product of two integers with unlike sing, first get their product regardless to their signs, the give minus sign to the product.
Examples:
(i) $(-\,40)\times 9$
$=-\,360$
(ii) $20\times (-\,3)$
$=-\,60$
Rule 2: To find the product of two integers with like signs, first find their product regardless to their signs then put plus sign to product.
Examples:
(i) $3\times 5$
$=+\,\,15$
(ii) $-\,3\times -\,5$
$=+\,\,15$
PROPERTIES OF MULTIPLICATION OF INTEGERS:
1. Closure properties for multiplication: The product of two integers is always an integer.
Examples:
(i) $3\times 2=6,$ which is an integer
(ii) $(-\,3)\times 2=-\,6,$ which is an integer
(iii) $(-\,3)\times (-\,2)=6,$ which is an integer
(iv) $\,3\times (-\,2)=-\,6,$which is an integer
1. Commutative law of multiplication: For any two integers$x$and$y,$we have.
$(x\times y)=(y\times x)$
Examples:
(i) $2\times \left( -\,6 \right)=-12$ and $\left( -\,6 \right)\times 2=-12$
$\therefore \,\,\,2\times \left( -\,6 \right)=\left( -\,6 \right)\times 2$
(ii) $\left( -\,3 \right)\times \left( -\,7 \right)=21$ and $\left( -\,7 \right)\times \left( -\,3 \right)=21$
$\therefore \,\,\,\left( -\,3 \right)\times \left( -\,7 \right)=\left( -\,7 \right)\times \left( -\,3 \right)$
1. Associative law of multiplication: For any integers x, y and z, we have
Examples:
(i) Consider the integers $3,$ $\left( -\,4 \right)$and $\left( -\,5 \right),$ we have
$\left\{ 3\times \left( -\,4 \right) \right\}\times \left( -\,5 \right)=-\,\,12\times \left( -\,5 \right)=60$ and
$3\text{ }x\text{ }\left\{ \left( -4 \right)\times \left( -5 \right) \right\}=3\times 20=60$
$\therefore$ $\left\{ 3\times \left( -\,4 \right) \right\}\times \left( -\,5 \right)=3\times \left\{ \left( -\,4 \right)\times \left( -\,5 \right) \right\}$
(ii) Consider the integers $\left( -\,5 \right),\left( -\,6 \right)$ and $\left( -\,7 \right),$ we have
$\left\{ \left( -\,5 \right)\times \left( -\,6 \right) \right\}\times \left( -\,7 \right)=30\times \left( -\,7 \right)=-\,210$
$\left( -\,5 \right)\times \left\{ \left( -\,6 \right)\times \left( -\,7 \right) \right\}=\left( -\,5 \right)\times 42=-\,210$
$\therefore$ $\left\{ \left( -\,5 \right)\times \left( -\,6 \right) \right\}\times \left( -\,7 \right)=\left( -\,5 \right)\times \left\{ \left( -\,6 \right)\times \left( -\,7 \right) \right\}$
1. Distributive law of multiplication over addition: For any integer x, y and z, we have:
$x\times \left( y+z \right)=\left( x\times y \right)+\left( x\times z \right)$
Examples: (1) Consider the integers $5,\,\left( -\,6 \right)$ and $\left( -\,7 \right),$we have
$5\times \left\{ \left( -\,6 \right)+(-\,7) \right\}=5\times \left( -\,13 \right)=-\,65$ and $\left\{ 5\times \left( -\,6 \right) \right\}+\left\{ 5\times \left( -\,7 \right) \right\}=-\,30+\left( -\,35 \right)=-\,65$
$\therefore$ $5\times \{(-\,6)+(-\,7)\}=\{5\times (-\,6)\}+\{5\times (-7)\}$
(2) Consider the integers (-5), (-6) and (-7), we have $\left( -\,5 \right)\times \left\{ \left( -\,6 \right)+\left( -\,7 \right) \right\}=\left( -\,5 \right)\times \left( -13 \right)=65$ and
$\left\{ \left( -\,5 \right)\times \left( -\,6 \right) \right\}+\left\{ \left( -\,5 \right)\times \left( -\,7 \right) \right\}=\left( -\,30 \right)+\left( -\,35 \right)=65$
$\therefore \,\,\left( -\,5 \right)\times \left\{ \left( -\,6 \right)+\left( -\,7 \right) \right\}=\left\{ \left( -\,5 \right)\times \left( -\,6 \right) \right\}+\left\{ \left( -\,5 \right)\times \left( -\,7 \right) \right\}$
1. Existence of multiplicative identity: For every integer $x$ we have: $(x\times 1)=(1\times x)=x,$ 1 is known as the multiplicative identity for integers.
Examples:
(i) $13\times 1=13$
(ii) $(-12)\times 1=-12$
1. Existence of multiplicative inverse: Multiplicative inverse of a non-zero integer$x$is
$\frac{1}{x}$as $x\left( \frac{1}{x} \right)=\left( \frac{1}{x} \right)\cdot x=1$
Examples:
(i) Multiplicative inverse of $5=\frac{1}{5}$
(ii) Multiplicative inverse of $-\,5=-\,\frac{1}{5}$
1. Property of zero: For every integer, $x$ we have:
$(x\times 0)=(0\times x)=5$
Examples:
(i) $5\times 0=0\times 5\text{ }=0$
(ii) $(-\,5)\times 0=0\times (-\,5)=0$
IMPORTANT RESULTS:
1. $(-\,{{x}_{1}})\times (-\,{{x}_{2}})\times (-\,{{x}_{3}})\times ..........\times (-{{x}_{n}})$ $=-\,({{x}_{1}}\times {{x}_{2}}\times {{x}_{3}}\times .........\times {{x}_{n}})$ when $n$ is odd.
2. $(-\,{{x}_{1}})\times (-\,{{x}_{2}})\times (-\,{{x}_{3}})\times ..........\times (-{{x}_{n}})$$=({{x}_{1}}\times {{x}_{2}}\times {{x}_{3}}\times .........\times {{x}_{n}})$ when $n$ is even.
3. $(-\,x)\times (-\,x)\times (-\,x)\times ..........\times n\,$times$=-\,{{x}^{n}}$ when $n$ is odd.
4. $(-\,x)\times (-\,x)\times (-\,x)\times .....\times n$ times $={{x}^{n}}$ when $n$ is even.
5. $(-1)\times (-1)\times (-1)........\,n$ times $=-\,1$ when $n$ is odd.
6. $(-1)\times (-1)\times (-1)........\,n$ times $=1$ when $n$ is even.
DIVISION OF INTEGERS:
Rule 1: To find the division of two integers with unlike sign, first get their quotient regardless to their signs, then give minus sign to the product.
Example:
(i) $-\,74\div 2$
$=\frac{-\,74}{2}$
$=-\,37$
(ii) $96\div (-\,3)$
$=\frac{96}{-\,3}$
$=-\,32$
Rule 2: To find the division of two integers with like signs, first find their product regardless to their signs then put plus sign to quotient.
Example:
(i) $48\div 6$
$=\frac{48}{6}$
$=8$
(ii) $-\,155\div (-\,5)$
$=\frac{-\,155}{-\,5}$
$=\,\,31$
PROPERTIES OF DIVISION OF INTEGERS:
1. If $x$ and $y$ are integers then $(x\div y)$ is not necessarily an integer
Example:
(i) $12$ and $5$ are both integer but $(12\div 5)$ is not an integer.
(ii) $(-12)$ and $5$are both integer. But $[(-12)\div 5]$ is not an integer.
1. If $x$ is an integer and $x\ne 0,$ then $x\div x=1$
Examples:
(i) $10\div 10=1$ (ii) $(-\,5)\div (-\,5)=1$
1. If $x$ is an integer, then $(x\div 1)=x$
Examples:
(i) $5\div 1=5$ (ii) $(-\,5)\div 1=(-\,5)$
1. If $x$ is an integer and $x\ne 0,$ then $(0\div x)=0$ but $(x\div 0)$ is not meaningful.
Examples:
(i) $0\div 9=0$
(ii) $0\div (-\,9)=0$
(iii) $6\div 0=$ Meaning less
1. If $x,y$ and $z$ are integers, then $(x\div y)\div z$ $\ne x\div (y\div z)$ unless $z=1.$
Thus division on integers is not associative.
Examples: Let $x=-\,6,y=3,\,z=-\,2$ then,
$(x\div y)\div z=\{(-\,6)\div 3\}\div (-\,2)=(-\,2)\div (-\,2)=1$
$x\div \{(y)\div (z)\}=(-\,6)\div \{3\div (-\,2)\}=(-\,6)\div (-1.5)=4$
$\therefore$ $(x\div y)\div z\ne x\div \{(y)\div (z)\}$
If $x=-\,6,y=3$and$z=1$then.
$(x\div y)\div z=\{(-\,6)\div 3\}\div 1=(-\,2)\div 1=-\,2$
$x\div \{(y)\div z\}=(-\,6)\div \{3+1\}=(-\,6)\div 3=-\,2$
So, $(x\div y)\div z=x\div \{y\div z\}$
1. If $x,y,z$are non-zero integers and$x>y$then
(i) $(x\div z)>(y\div z)$ if $z$ is positive
(ii) $(x\div z)<(y\div z)$ if $z$ is negative
Example:-
(i) If$x=27,y=18$and$z=9$
$(x\div z)>(y\div z)$
$(27\div 9)>(18\div 9)$
$3>2$
(ii) If $x=27,y=18$and $z=-\,9$
$(x\div z)<(y\div z)$
$\{27\div (-\,9)\}<\{18\div (-\,9)\}$
$-\,3<-\,2$
Use of Brackets:
We know that the priority order of different mathematical operations is
(1) Division (2) Multiplication
Sometimes in complex expressions, we require a set of operations to be performed prior to the others.
Here we make use of brackets. Brackets which are commonly used are: -
Brackets Name ( ) Parentheses or common brackets { } Braces or curly brackets [ ] Square or Box brackets — Vinculum
• Proper fraction: A proper fraction is a fraction that represents a part of a whole number. Here denominator is greater than the numerator.
• Improper fraction: An improper fraction is a combination of a whole and a proper fraction. Here numerator is greater than the denominator.
• Mixed fraction: An improper fraction can be written as a mixed fraction, e.g., $\frac{7}{4}=1\frac{3}{4}$.
• Like fractions: Fractions having the same denominators are called like fractions.
• Unlike fractions: Fractions with different denominators are called unlike fractions.
• Addition and subtraction of fractions: For addition or subtraction, we first find out the LCM of the denominators, then we convert each fraction into an equivalent fraction whose denominator is equal to the LCM. Finally, we add or subtract the like fractions obtained above.
For examples, $\frac{3}{5}+\frac{7}{2}$
LCM of $5,\,2=10$
Like fraction $\Rightarrow \frac{3}{5}=\frac{6}{10}$and$\frac{7}{2}=\frac{35}{10}$
$\frac{6}{10}+\frac{35}{10}=\frac{41}{10}$
MULTIPLICATION AND SUBTRACTION OF FRACTIONS
• MULTIPLICATION OFAFRACTION BYAWHOLE NUMBER: To multiply a whole number with a proper or an improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator same.
To multiply a mixed fraction by a whole number, first convert it into an improper fraction and then multiply.
FRACTIONAS AN OPERATOR ‘OF’
Observe this figure. The two squares are exactly same.
Each shaded portion represents $\frac{1}{2}$of 1.
So, both the shaded portions together will represent $\frac{1}{2}$ of 2.
Combine the 2 shaded$\frac{1}{2}$parts. It represents 1.
So, we say $\frac{1}{2}$ of 2 is 1. We can also get it as$\frac{1}{2}\times 2=1.$
Thus, $\frac{1}{2}$ of $2=\frac{1}{2}\times 2=1$
$\text{In}\,\text{general}\,\text{product}\,\text{of}\,\text{fractions}$
$\,\text{=}\,\,\frac{\text{Product}\,\text{of}\,\text{Numerators}}{\text{Product}\,\text{of}\,\text{Denominators}}$
• Value of products of 2 fractions: When two proper fractions are multiplied, the product is less than each of the fractions.
The value of the product of two improper fractions is more than each of the two fractions.
DIVISION OF FRACTIONS
• Division of whole number by a fraction: For dividing a whole number by a fraction. First obtain the reciprocal of the fraction and then multiply it with the whole number.
For example, $6\div \frac{2}{3}\Rightarrow 6\times \frac{3}{2}=9$
• Reciprocal of a fraction: When a fraction is inverted its reciprocal is obtained.
The non-zero numbers whose product with each other is 1 are called reciprocals of each other.
• Division of a fraction by another fraction: For dividing a fraction by another fraction reverse one of the fractions and multiply it with the other.
H.C.F. AND L.C.M.
• The highest common factor (HCF) of two or more numbers is the greatest number which divides each number exactly.
• The least common multiple (LCM) of two or more numbers is the smallest number which is exactly divisible by each number separately.
The H.C.F. of given numbers is not greater than any of the given numbers.
The L.C.M. of given numbers is not less than any of the given numbers.
H.C.F. of two numbers always divides their L.C.M.
The H.C.F. of two co-prime numbers is 1.
The L.C.M. of two co-prime numbers is product of the numbers.
The product of the H.C.F. and the L.C.M. of the two given numbers is equal to the product of those numbers, i.e.,
H.C.F.$\times$L.C.M. = Product of the given numbers.
Example: The HCF of two numbers is 9 and their LCM is 270. If the sum of the numbers is 99, their difference is equal to
Solution:
Let the numbers be $x$ and $99-x.$
So $x\times (99-x)=9\times 270$
On solving the above equation, we get; $x=54,$ or$45$
So difference will be$54-45=9.$
FACTOR THEOREM
If $\left( x+a \right)$ is a factor of polynomial$P(x),$then remainder$=0$
$\Rightarrow P(-\,a)=0$
Example: Show that $(x-3)$ is a factor of the polynomial
$P(x)={{x}^{3}}-3{{x}^{2}}+4x-12$
lf $(x-3)$ is a factor of polynomial
$P(x)={{x}^{3}}-3{{x}^{2}}+4x-12,$ then remainder
$P(3)=0$
$P(3)=0$
$P(3)={{3}^{3}}-3\times {{3}^{2}}+4\times 3-12$
$=27-27+12-12=0$
As remainder$P(3)=0$
$\therefore$ $(x-3)$ is a factor of polynomial$P\,(x)$.
PROPERTIES OF DECIMALS
1. Numbers in form of 1.18, 22.36, and 853.4448 are known as decimal numbers.
2. A decimal number have two parts. A whole number part and decimal number part. These two parts are separated by a dot. This dot is known as decimal point. The whole number part is to the left of the decimal point and the decimal part to the right of decimal point.
Thus, in 63.825
1. The number of digits contained in the decimal part shows its decimal places.
Example: 6.438 have three digits in decimal part which is known as three decimal places.
1. Decimals having the same number of decimal places are called like decimals.
Example: 6.238, 38.937, 48.385 are like decimals.
1. Decimals having different number of decimal places are called unlike decimals.
Example: 6.238, 32.73, 65.3984 are unlike decimals.
Remember: On adding zeros after the last digit of the decimal part of any decimal number does not change the value of the decimal.
Example: 6.28 can be 6.280 or 6.2800 but the value of all three are same.
We use this property to convert the unlike decimal into like decimal.
COMPARINGDEOMALS:
Suppose we have to compare two given decimals. We follow the following steps:
1. Convert the given decimals into like decimals.
2. First compare the whole number part. The decimal with greater whole number part is greater.
3. If the whole-number parts are equal, compare the tenths digits. The decimal with bigger digit in tenths place is greater.
4. If the tenths digits are also equal, compare the hundredths digits, and so on.
Example: Compare the following digits.
5.74, 6.03, 0.8, 0.648 and 8.2.
Solution: Converting the given numbers into like decimals we get them as:
5.740, 6.030, 0.800, 0.648 and 8.200
Clearly 0.648 < 0.800 < 5.740 < 6.030 < 8.200
CONVERSION OFDEOMAL INTO FRACTION:
1. Write the given decimal number without decimal point as the numerator of the fraction.
2. In the denominator, write 1 followed by as many zeros as there are decimal places in the given decimal.
3. Reduce the fraction into simplest form.
Example: Convert 2.85 into fraction.
Solution: $\frac{285}{100}=\frac{57}{20}$
CONVERSION OFA FRACTION INTO DECIMAL:
1. Divide the numerator by the denominator till non-zero remainder is obtained.
2. Put a decimal point in the divided as well as in the quotient:
3. Put a zero on the right of the decimal point in the dividend as well as on the right of the remain
4. Divide again just as we do in whole numbers.
5. Repeat steps 3 & 4, till the remainder is zero.
Example: Convert $\frac{27}{4}$ and $2\frac{3}{8}$ into decimal fraction.
Solution: $\frac{27}{4}$
$4\,\underset{\times \,\times }{\mathop{\underline{\underset{\,\,\,\,-\,20}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,20}{\mathop{\overset{6.75}{\mathop{\underline{\underset{-28}{\mathop{\underset{30}{\mathop{\overline{\underline{\underset{-\,24}{\mathop{)\,\,\,27.00}}\,\,\,\,\,}}}}\,}}\,}}}\,}}\,}}\,}}}\,$
$\therefore$ $\frac{27}{4}=6.75$
and $2\frac{3}{8}=\frac{19}{8}$
$8\,\underset{\times \,\times }{\mathop{\underline{\underset{-\,40}{\mathop{\underset{40}{\mathop{\underline{\underset{\,\,\,\,-\,56}{\mathop{\underset{\,\,\,\,\,\,\,\,\,\,\,60}{\mathop{\overset{2.375}{\mathop{\underline{\underset{-24}{\mathop{\underset{30}{\mathop{\overline{\underline{\underset{-\,16}{\mathop{)\,\,19.000}}\,\,\,\,\,}}}}\,}}\,}}}\,}}\,}}\,}}}\,}}\,}}}\,$
$\therefore$ $2\frac{3}{8}=2.375$
Method:
(i) Write the given numbers one below the other with decimal points in vertical line.
(ii) Equal the digits of numbers by adding zeros at right extremes of decimal parts as required.
(iii) Then add or subtract the numbers and put decimal in result directly under the other decimal points.
Example:
(i) Add: 0.872, 3.46, 4.309, and 3.17
Solution:
\underline{\begin{align} & \underline{\begin{align} & \,\,\,\,0.872 \\ & \,\,\,\,3.460 \\ & \,\,\,\,4.309 \\ & +\,3.170 \\ \end{align}} \\ & \,\,11.811 \\ \end{align}}
(ii) Subtract: 17.182 from 360.05
\underline{\begin{align} & \underline{\begin{align} & \,\,\,\,360.050 \\ & -\,017.182 \\ \end{align}} \\ & 342.868 \\ \end{align}}
MULTIPLICATION OF DECIMAL NUMBERS:
(1) To multiply a decimal number by 10 or any power of 10 move the decimal point as many places to the right as there are zeros in the multiplier or we can say that multiplying a decimal number by ${{10}^{n}}$ moves the decimal point$n$places to the right.
Examples:
(2) To multiply a decimal number by a whole number or a decimal number multiply the numbers first as there were no decimal point at all and fix the position by the rule that there are as many decimal places in the product as they are in multiplier and multiplicand put together.
Example:
(1) $\underset{\text{1}\,\,\text{place}}{\mathop{0.6}}\,\times \underset{\text{2}\,\,\text{places}}{\mathop{0.02}}\,=\underset{\text{3}\,\,\text{places}}{\mathop{0.012}}\,$
(2) $\underset{0\,\,\text{place}}{\mathop{7\,}}\,\,\,\,\times \underset{3\,\,\text{places}}{\mathop{\,0.003}}\,\,\,\,\,=\underset{\text{3}\,\,\text{places}}{\mathop{\,0.021\,}}\,$
DIVISION OF DECIMAL NUMBERS
(1) Dividing a whole number or decimal number by $10$ or any higher power of $10\,({{10}^{n}})$ move the decimal point that number of places $(n)$ to the left.
Examples: (1) $53\div 10=5.3$
(2) $53\div 100=0.53$
(3) $53\div 1000=0.053$
(2) To divide a decimal number by a whole number proceed as with whole numbers, but place the decimal point in the quotient directly above or below the decimal point in the dividend.
Example: (1) Divide 15.064 by 28
$\therefore$ $15.064\div 28=0.538$
(2) Divide 24.2 by 55000
$\because \,\,\,24.50\div 55=0.44$
$\because \,\,\,24.20\div 55000=0.0004$
(3) To divide a decimal number by a decimal number move the decimal point of divisor to the right until it becomes a whole number (i.e., multiply it by 10 or a power of 10). Then move the decimal point of the dividend the same number of places to the right, adding zero if necessary.
Example: Divide 49.08 by 0.012
Solution: $49.08\times 1000=49080$
$0.012\times 1000=12$
RATIONAL NUMBER
A number in form of $\frac{a}{b}$ where $a$ and $b$ are integers and $b\ne 0$ is known as rational number.
Examples: $\frac{1}{2},\frac{-\,6}{7},\frac{3}{-\,2},\frac{-\,5}{-\,8}$is a rational number.
Remember:
1. Zero is a rational number, since we can write it as$\frac{0}{1}.$
2. Every natural number is rational number but a rational number need not to be natural number as $\frac{1}{1},\frac{2}{1},\frac{3}{1},\frac{4}{1}$ etc. are natural numbers and rational numbers but $\frac{2}{3},\frac{6}{8},\frac{1}{3}$ etc. are rational number which can't be natural numbers.
3. Every integer is a rational number but a rational number need not to be an integer.
Example: $1=\frac{1}{1},2=\frac{2}{1},3=\frac{3}{1}\,\,.......\,\,\frac{n}{1}$are integers but rational number like $\frac{5}{7},\frac{-\,8}{9},\frac{11}{-13}$ are not integers.
1. Every fraction is a rational number but a rational number need not to be a fraction.
Let $\frac{a}{b}$ is a fraction where $a$ and $b$ are natural numbers. Since every natural number is $a$ integer so $a$ and $b$ are integers so fraction $\frac{a}{b}$ where $b\ne 0$ is a rational number.
Example:
A number like $\frac{5}{-\,6}$ is a rational number is not a fraction because its denominator $-\,6$ is not a natural number.
Positive rational number: A rational number is said to be positive rational number if both the numerator and denominator are positive or negative.
Examples: $\frac{5}{7},\frac{-18}{-\,27},\frac{-16}{-13},\frac{8}{9}$
Negative rational numbers: A rational number is said to be negative if its numerator and denominator are such that one of them is positive and another is negative.
Example: $\frac{-\,3}{4},\frac{8}{-\,6},\frac{-\,28}{11}$
Remember:
1. Every negative integer is a negative rational number.
Example: $-1,-\,2,-\,3.......$ may be written as: $-\frac{1}{1},\frac{-\,2}{1},\frac{-\,3}{1}........$ are all negative rational numbers.
1. The rational number 0 is neither positive nor negative.
TWO IMPORTANT PROPERTDES OF RATIONAL NUMBERS:
Property 1: Equivalent rational numbers $\div$ If $\frac{a}{b}$ is a rational number and $n$ is a non-zero integer then on multiplying the numerator and denominator of rational number by $n$ we will get its equivalent rational number as $\frac{a}{b}=\frac{a\times n}{b\times n}$
Examples: $-\frac{3}{4}=\frac{(-\,3)\times 2}{4\times 2}=\frac{(-\,3)\times 3}{4\times 3}=\frac{(-\,3)\times 4}{4\times 4}=.....$
All these rational numbers are equal to one another and are called equivalent rational numbers.
Property 2: Reducing to simpler form
If $\frac{a}{b}$ is a rational number and $n$ is a common divisor of $a$ and $b$ then $\frac{a}{b}=\frac{a\div n}{b\div n}$
Examples: $-\frac{48}{60}=\frac{-48\div 2}{60\div 2}=\frac{-48\div 3}{60\div 3}=\frac{-48\div 4}{60\div 4}$
$=\frac{-48\div 6}{60\div 6}=\frac{-48\div 12}{60\div 12}$
$\Rightarrow -\frac{48}{60}=\frac{-24}{30}=\frac{-16}{20}=\frac{-12}{15}=\frac{-8}{10}=\frac{-4}{5}$
The rational number $\frac{-\,4}{5}$ is in lowest number.
STANDARD FORM:
A rational number is said to be in its standard form when its denominator is positive and it is in its lowest term.
Rational number can be in its standard form by following steps-
1. Make the denominator of rational number positive.
2. Divide both the numerator and denominator by their HCF.
Example: Convert the following in their standard form.
(i) $2\frac{4}{9}$
$=\frac{22}{9}$
(ii) $-\frac{8}{2}$
$-\frac{8\div 2}{2\div 2}=-\,4$
(iii) $\frac{4}{-11}$
$=\frac{4}{-11}\times \frac{-1}{-1}=\frac{-\,4}{11}$
(iv) $\frac{9}{15}$
HCF of 9 and 15 is 3
so $\frac{9\div 3}{15\div 3}=\frac{3}{5}$
Note: If the denominator of rational number is negative then multiply both the numerator and denominator by $-1$ to make denominator positive.
RATIONAL NUMBER ON NUMBER LINE:
To express rational numbers on appropriate number lines, divide the unit length into the number of equal parts as the denominator of the rational number and then mark the numbers on the line.
Example: $P=\frac{7}{8},\,\,Q=\frac{-\,3}{8},\,\,R=\frac{1}{8},\,\,S=\frac{1}{-\,8}$
EQUALITY OF RATIONAL NUMBERS:
Two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ are equal $\frac{a}{b}=\frac{c}{d}$ if $a\times d=b\times c.$
Example: $\frac{-24}{27}=\frac{8}{-9}$
$-\,24\times -\,9=27\times 8$
$216=216$
COMPARISON OF TWO RATIONAL NUMBERS:
Positive rational numbers are always greater than negative rational numbers. But if we have to compare two positive or two negative rational numbers we compare them by two methods.
First Method:
(a) Express each rational number with its positive denominator.
(b) Find LCM of positive denominators.
(c) Express each given rational numbers with LCM as the common denominator.
(d) The number having greater numerator is greater.
Example: Compare:
(i) $\frac{9}{15}$ and $\frac{11}{6}$
LCM of 15 and $6=3\times 5\times 2=30$
$\frac{9\times 2}{15\times 2}=\frac{18}{30},\frac{11\times 5}{6\times 5}=\frac{55}{30}$
$\frac{55}{30}>\frac{18}{30},$So, $\frac{11}{6}>\frac{9}{15}$
(ii) $\frac{3}{-14}$and$-\frac{5}{21}$
$\frac{3}{-14}$and$-\frac{5}{21}$
LCM of 14 and $21=2\times 7\times 3=42$
$-\frac{3\times 3}{14\times 3}=\frac{-9}{42},\frac{-5\times 2}{21\times 2}=\frac{-10}{42}$
$\frac{-9}{42}>\frac{-10}{42}$
So, $\frac{-3}{14}>\frac{-5}{12}$
Second Method:
To compare two rational numbers $\frac{a}{b}$ and $\frac{c}{d},$ we compare the products $a\times d$ and $b\times c$ define their inequality accordingly.
If $a\times d>b\times c$ If $a\times d<b\times c$
then $\frac{a}{b}>\frac{c}{d}$ then $\frac{a}{b}<\frac{c}{d}$
Example: Compare $\frac{-5}{9}$ and $\frac{11}{-16}$
First $-\frac{5}{9}$ and $-\frac{11}{16}$
Now, $-\,5\times 16$ and $9\times -\,11$
So, $-\,80>-\,99$
So, $-\,\frac{5}{9}>-\,\frac{11}{16}$
ORDER PROPERTIES OF RATIONAL NUMBERS:
Property 1: For each rational number $a,$ exactly one of the following is true.
(i) $a>0$ (ii) $a=0$
(iii) $a<0$
Property 2: For any two rational numbers $a$ and $b$ exactly, one of the following is true.
(i) $a>b$ (ii) $a=0$
(iii) $a<b$
Property 3: lf a, b and c are any three rational numbers such that a > b and b > c, then a > c
ABSOLUTE VALUE OF RATIONAL NUMBERS:
The absolute value of an integer is an integer similarly the absolute value of a rational number is a rational number regardless to their signs.
Thus$\left| \frac{5}{7} \right|=\frac{5}{7};\left| \frac{-5}{7} \right|=\frac{\left| -5 \right|}{\left| 7\, \right|}=\frac{5}{7};\left| \frac{65}{-23} \right|=\frac{\left| 65 \right|}{\left| -23 \right|}=\frac{65}{23}$
OPERATIONS ON RATIONAL NUMBERS:
Case 1. When denominators of given rational numbers are equal
Let $\frac{p}{q}$ and $\frac{r}{q}$ are two rational numbers. Then $\frac{p}{q}+\frac{r}{q}$or $\left( \frac{p+r}{q} \right)$
Example: Add- $\frac{7}{-11}$ and $\frac{3}{11}$
$\frac{-7}{11}+\frac{3}{11}=\frac{-7+3}{11}=\frac{-4}{11}$
Case 2: When denominator of given numbers are unequal.
1. Take the LCM of denominators of the given rational numbers.
2. Express each of the given rational numbers with above LCM as the common denominator.
3. Now add the numbers as case I.
Example: Add: $\frac{7}{-\,27}+\frac{11}{18}-\frac{7}{27}+\frac{11}{18}$
LCM of 27 and $18=3\times 3\times 3\times 2=54$
$-\frac{7\times 20}{27\times 2}=-\frac{14}{54}$and$\frac{11\times 3}{18\times 3}=\frac{33}{54}$
$\therefore \,\,\,\frac{-14}{54}+\frac{33}{34}=\frac{-14+33}{54}=\frac{19}{54}$
2 Subtraction of Rational Number: If $\frac{a}{b}$ and $\frac{c}{d}$ are two rational numbers, then
$\frac{a}{b}-\frac{c}{d}=\frac{a}{b}+\left\{ \text{Additive}\,\text{inverse}\,\text{or}\,\text{negative}\,\text{of}\frac{c}{d} \right\}$ $\frac{a}{b}-\frac{c}{d}=\frac{a}{b}+\left( -\frac{c}{d} \right)$
Example: Subtract (i) $\frac{7}{8}$ from $\frac{5}{12}$
$\frac{5}{12}-\frac{7}{8}=\frac{5}{12}+\frac{-\,7}{8}$
LCM of 12 and $8=2\times 2\times 3\times 2=24$
$=\frac{5\times 2+(-7)\times 3}{24}=\frac{10+(-21)}{24}=\frac{-11}{24}$
(ii) $\frac{-4}{9}$from $\frac{-7}{18}$
$\frac{-7}{18}-\left( \frac{-\,4}{9} \right)$
$\frac{-7}{18}+\left[ -\left( \frac{-\,4}{9} \right) \right]$
$-\frac{7}{18}+\frac{4}{9}$
LCM of 18 and $9=9\times 2=18$
$\frac{(-7)\times 1+4\times 2}{18}=\frac{-7+8}{18}=\frac{1}{18}$
1. Multiplication of Rational Numbers:
$\text{Product of two rational numbers}$
$=\frac{\text{Product}\,\text{of}\,\text{numerators}}{\text{Product}\,\text{of}\,\text{denominator}}$
Thus if $\frac{a}{b}$ and $\frac{c}{d}$ are two rational numbers, then $\frac{a}{b}\times \frac{c}{d}=\frac{a\times c}{b\times d}$
Example: (i) $\frac{-7}{15}\times \frac{5}{-14}=\frac{-7\times 5}{15\times -14}=\frac{-35}{-210}=\frac{1}{6}$
(ii) $\frac{-7}{9}\times \frac{-4}{5}=\frac{28}{45}$
RECIPROCALOR MULTIPLICATIVE INVERSE OF A RATIONAL NUMBER:
The reciprocal of a rational number $\frac{a}{b}$ is $\frac{b}{a}$ or ${{\left( \frac{a}{b} \right)}^{-1}}=\frac{b}{a}$
Note: (i) Reciprocal of 0 does not exist.
(ii) Reciprocal of 1 is 1
(iii) Reciprocal of$-1$ is$-1$.
Example: Write reciprocal of:
(i) $\frac{12}{7}$ (ii) $\frac{-7}{9}$
(iii) $-5$
Solution:
(i) $\frac{12}{7}$
(ii) $\frac{-7}{9}$
reciprocal$=\frac{7}{12}$ $\frac{9}{-7}$
(iii) $-5\,{{(-5)}^{-1}}=\frac{1}{-5}$
DIVISION OF RATIONAL NUMBERS:
If $\frac{a}{b}$ and $\frac{c}{d}$ are two rational numbers such that $\frac{c}{d}$ then,
$\frac{a}{b}-\frac{c}{d}=\frac{a}{b}\times \left( \text{reciorocal}\,\text{of}\,\frac{c}{d} \right)=\frac{a}{b}\times \frac{d}{c}$
Example:
(i) $\frac{7}{15}\div \frac{2}{3}=\frac{7}{15}\times \left( \text{reciorocal}\,\text{of}\,\frac{2}{3} \right)=\frac{7}{15}\times \frac{3}{2}=\frac{7}{10}$
(ii) $\frac{16}{21}\div \frac{-4}{3}=\frac{16}{21}\times \left( \text{reciorocal}\,\text{of}\,\frac{-4}{3} \right)=\frac{16}{21}\times \frac{3}{-4}=\frac{-4}{7}$
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# Analyse and Interpret Graphs
Term Definition
What is the y-intercept? the point at which the graph line crosses the y axis. When x=0.
What is the y intercept from the following equations? 5x +7 7
What is the y intercept from the following equations? 89x -34 -34
What is the y intercept from the following equations? 12x + 45 45
What is direct variation? when the y intercept equals zero, when there is no additional number added or subtracted onto the equation.
What is a partial variation? When the y intercept does NOT equal 0. When there IS AN additional number added or subtracted onto the equation.
Which equations are direct variations? y= -x, y= 0.25 - 1.3, y= 89x y= -x and y= 89x are direct variations.
Which variations are partial variations? y= 0.98x + 78, y= 4x - 67, and y= 90x. y= 0.98x + 78 and y= 4x - 67 are partial variations.
What is the slope of a graph (or the rate of change)? A measure of the STEEPNESS of a line.
How do you tell if an equation is linear? Has NO VISIBLE exponents
How do you tell if a table of values is linear? Has a consistent pattern.
How do you tell if a graph is linear? The points line up in a straight line.
How do you tell if a graph is partial or direct? Direct: will cross over the 0,0 spot, if not, it is partial. If one of the co-ordinates or one of the places in the table of values is 0,0 then it's direct.
Equation a statement that two mathematical expression are equal and have the same value.
Variable a letter that represent an unknown number.
Numerical Coefficient a number that multiplies the variable.
Constant a known value in an equation or an expression.
Opposite Operation operations that "undo" other operations. sometimes called "inverse operations" addition/subtraction
Distributive Property the rule that states a(b+c) = ab
What is the gradient of y = 3x+4 3
Do you need to compete such a difficult assignment? can assist with any of your college or university tasks. Fill in the order form and enjoy your weekend!
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Thinking Mathematically (6th Edition)
The given line has a slope of 2 and a y-intercept of 3. To graph the given line using the slope and y-intercept, perform the following steps: (1) Plot the y-intercept point $(0, 3)$. (2) From $(0, 3)$, move 2 units up (the rise) and move 1 unit to the right (the run). These moves lead to the point $(1, 5)$. (3) Connect $(0, 3)$ and $(1, 5)$ using a line. (Please refer to the attached image in the answer part above for the graph.)
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# 492 US pints in deciliters
## Conversion
492 US pints is equivalent to 2328.02824716 deciliters.[1]
## Conversion formula How to convert 492 US pints to deciliters?
We know (by definition) that: $1\mathrm{uspint}\approx 4.73176473\mathrm{dl}$
We can set up a proportion to solve for the number of deciliters.
$1 uspint 492 uspint ≈ 4.73176473 dl x dl$
Now, we cross multiply to solve for our unknown $x$:
$x\mathrm{dl}\approx \frac{492\mathrm{uspint}}{1\mathrm{uspint}}*4.73176473\mathrm{dl}\to x\mathrm{dl}\approx 2328.02824716\mathrm{dl}$
Conclusion: $492 uspint ≈ 2328.02824716 dl$
## Conversion in the opposite direction
The inverse of the conversion factor is that 1 deciliter is equal to 0.000429548052614875 times 492 US pints.
It can also be expressed as: 492 US pints is equal to $\frac{1}{\mathrm{0.000429548052614875}}$ deciliters.
## Approximation
An approximate numerical result would be: four hundred and ninety-two US pints is about two thousand, three hundred and twenty-eight point zero three deciliters, or alternatively, a deciliter is about zero times four hundred and ninety-two US pints.
## Footnotes
[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).
Results may contain small errors due to the use of floating point arithmetic.
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the question im trying to solve is: What is the value of x3 / (4y - 7), when x = 5 and y = 8?
(x3 is x cubed just so that makes sense.)
5³/(4×8-7)=125/25=5.
by Top Rated User (765k points)
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A New Kind of Science: The NKS Forum > Pure NKS > Pythagorean Triangles Arranged in Order
Author
New Zealand
Registered: Aug 2010
Posts: 1
Pythagorean Triangles Arranged in Order
I haven¡¦t seen this mentioned on any maths-related websites and believe a lot of people would find it interesting.
Most interesting is the third tab of the Excel file (page 3 of the pdf) which extends the original rows and columns upwards and to the left and right.
This gives four sets of triangles, either with three positive sides or with one or both of the legs negative.
To preserve the formulae, the diagonal axes are triangles with one side of zero length.
The second tab of the Excel file (page 2 of the pdf) shows the in-circle radii of the triangles on the first tab.
I was interested to note that summing the rows on the second tab gives the sum of the first n squares.
Triangle notes
Methods of Generating Pythagorean Triangles
Create a column of Pythagorean triangles for the nth odd number, X:
„h
A = n x 4
„h
B = A + X^2
„h
C = X^2 + 2X
The second triangle follows:
„h
D = A + ((n + 1) x 4)
„h
E = D + X^2
„h
F = C + 2X
And the third:
„h
G = D + ((n + 2) x 4)
„h
H = E + X^2
„h
I = F + 2X
And so on; see Excel spreadsheet ¡§tri gen¡¨.
Patterns in Pythagorean Triples
Hypotenuse and Longest side are consecutive
This sequence is Column B of the spreadsheet, where n = 1
The two legs are consecutive -
These triangles appear on a line bisecting the top angle of the triangular spreadsheet; highlighted yellow on the spreadsheet.
More Patterns
Primitive triangles whose longest side and hypotenuse differ by 2; these triangles appear in a line as well ¡V they are the first triangle of each column in the spreadsheet.
The inradii are shown on the second tab. For a column generated for the nth odd number X the inradii are the multiples of X in numerical order.
Attachment: tri gen 23-08.pdf
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Math and Arithmetic
Cooking Measurements
Volume
# What is the Number of drops in a teaspoon?
91011
###### 2009-04-24 04:39:32
What is the number of drops in a teaspoon?
There are 60 metric drops in a 5ml teaspoon. You'll find many variants on this ranging from 50 - 120, but 60 is a good bet. Generally, drops are not a precise measurement.
There are several exact definitions of a "drop". Here are 3 of them: * the "metric" drop, 1/20 ml (50 μL), which equates to 60 to a 5ml teaspoon.
* the medical drop, 1/12 ml (83⅓ μL). * the Imperial drop, 1/36 of a fluid dram (1/288 of an Imperial fluid ounce or 1/1440 of a gill (approximately 99 μl).
Note: There are 360 drops in a U.S. fluid ounce.
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## Related Questions
1 teaspoon is 76 drops. 1/4 teaspoon is 19 drops.
There are 118 drops in a teaspoon (UK). Then one third of a teaspoon would be 39.333 drops.
1 teaspoon (US) is about 99 (98.578431875) drops.
(1/5) teaspoon (US) is about 19.7(19.715686375) drops.
15 drops equals 1/4 teaspoon.(I had the same question, and found this answer on a few different sites.)
Four US teaspoons equates to about 394 drops.
140 drops equates to 1.4 US teaspoons.
90 drops equals about one (0.91298) US teaspoon.
According to US definition, a "drop" is approximately 1/76 of a teaspoon.
there are 300 million of semen drops in one tea spoon
12 dashes = 1 teaspoon 6 dashes = 1/2 teaspoon 3 dashes= 1/4 teaspoon
A measuring cup is 16 tablespoons or 48 teaspoons or 3648 drops. 1 tablespoon is 3 teaspoons. 1 teaspoon is 76 drops.
From looking at original German recipes and their translations, "3 Tr" would mean "3 Tropfen" or "3 drops". Drops can vary in size.Some sites say a drop is about 1/96 of a teaspoon, so 3 drops would be 3/96 or 1/32 of a teaspoon. That would be 1/4 of a 1/8 tsp measure.Other sites say there are around 76 drops in a teaspoon. That would translate to 3 drops being about 1/25 tsp or about 1/3 of a 1/8 tsp measure.It would be pretty difficult to eye the difference between 1/32 and 1/25 of a teaspoon. I'm not so certain your taste buds would pick up the difference in a cookie or cake recipe. I'd probably get a dropper bottle to use for that recipe, but realize that even drops vary in volume.
For one teaspoon crushed red pepper substitute 1/2 to one teaspoon cayenne pepper or 3 to 6 drops of hot sauce.
For one teaspoon crushed red pepper substitute 1/2 to one teaspoon cayenne pepper or 3 to 6 drops of hot sauce.
1 teaspoon is 5 ml. So a millilitre is closer to 4 drops but still larger. I would estimate a single ml to be about 20 drops of water.
The answer depends on whether you are measuring the drops from a slow drip or the number of drops of water in an ocean!
Teaspoon is a volume measurement, and gram is a weight measurement. The number of grams in a teaspoon of anything depends on what it is that you are measuring with the teaspoon. For example, a teaspoon of mercury would weigh more grams than a teaspoon of water.
Depends on whether it's a metric drop, a medical drop, an imperial drop or a U.S. drop. If it's a U.S. drop, then 60 drops = 1 teaspoon.
A drop is defined as 1/60th of a teaspoon. So it would work out to 82.1486932 microliters.
No. 1.4 is a decimal. The number to the left of the decimal point is the whole number, meaning one. If there were a zero on the right side of the decimal point, then it would be 1 teaspoon. However, there is a 4 there. which is equal to 4/10. That is more than a whole teaspoon
###### Cooking MeasurementsMath and ArithmeticFood & CookingUnits of MeasureTechnologyMedication and DrugsRecipes
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Stability analysis of time domain systems
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Stability analysis of time domain systems - PowerPoint PPT Presentation
Stability analysis of time domain systems. S-Plane: Poles and Zeros . A linear system can be represented by the following transfer function: Zeros: z i (the roots of the numerator) Poles: p i (the roots of the denominator or of the system or characterestic equation).
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Presentation Transcript
Stability analysis of time domain systems
S-Plane: Poles and Zeros
• A linear system can be represented by the following transfer function:
• Zeros: zi (the roots of the numerator)
• Poles: pi (the roots of the denominator or of the system or characterestic equation).
• Example: for the TF shown below
The Stability of Linear Feedback Systems
• In designing a control system, we must be able to predict the dynamic behaviour of the system from knowledge of the components.
• A linear time-invariant control system is stable if the output eventually comes back to its equilibrium state when the system is subjected to an initial condition.
• It is marginally stable if oscillations of the output continue forever.
• It is unstable if the output diverges without bound from its equilibrium state when the system is subjected to an initial condition.
Consider the first-order differential equation:
• Time response: y(t) = e-at
• There are three cases:
i) a > 0, y(∞) = 0 → stable
ii) a = 0, y(∞) = 1 → neutral (marginally stable)
iii) a < 0, y(∞) = ∞ → unstable
A system is stable if ALL the closed-loop poles are in the LEFT HALF of the s-plane and have negative real parts, i.e. all the roots of the characteristic equation are in the left-hand s-plane.
• A general closed-loop system transfer function:
• The task of factorising q(s) becomes difficult for n > 2. Do we have an alternative and simple method?
• Alternative 1 is
• A necessary condition for stability of the system is that all the coefficients of the closed-loop characteristic equation ai are positive. This is to say that if at least one of the coefficients of the characteristic equation q(s) is negative, we can conclude that the system is unstable.
• However, it does not mean that if all the coefficients of q(s) are positive, the system is stable.
• For example:
Routh-Hurwitz Stability Criterion
• Routh's stability criterion tells us whether or not there are unstable roots in the closed-loop characteristic equation without actually solving for them. Using this method, we can tell how many closed-loop system poles are in the right half of the s-plane, but not their exact location.
• Consider the closed-loop characteristic equation:
• The Routh table is constructed as follows:
Routh-Hurwitz criterion
• The Routh-Hurwitz criterion states that the number of roots of q(s) with positive real parts (in right half plane) is equal to the number of sign changes in the first column of the Routh table.
• Example: Use the R-H criterion to determine if the closed-loop system described by the following characteristic equation is stable:
• Since all the coefficients in the first column of the R-H table are all positive and there is no sign change, then the system is stable.
Exercise: Use the R-H criterion to determine if the closed-loop system described by the following characteristic equation is stable:
• The R-H criterion can be used to determine the range of the parameter values which maintain a stable system.
• Example: Consider closed-loop system shown below. Using the R-H criterion, determine the range of K over which the system is stable.
• Solution: Closed-loop transfer function:
Routh's table:
• For stability, all the coefficients in the first column of the Routh's table must be positive, i.e. we require:
Exercise: Consider closed-loop system shown below. Using the R-H criterion, determine the range of K over which the system is stable.
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# Practise Combined Operations Using a Calculator
In this worksheet, students use a calculator to carry out more complex calculations.
Key stage: KS 3
Curriculum topic: Number
Curriculum subtopic: Use Calculators/Technology for Accuracy
Difficulty level:
### QUESTION 1 of 10
This worksheet is about carrying out complex calculations on a calculator.
Example
Use your calculator to work out the following to 1 decimal place:
40 × (84.2 - 31)
-√ 102 + 872
Type the following carefully into your calculator:
40 x ( 84.2 - 31 ) ÷ - √ ( 10 ^ 2 + 87 ^ 2 ) =
This gives the answer -24.29977513.
We round this off to 1 decimal place to give the answer -24.3
Use your calculator to work out the following to 1 decimal place:
√ (592 + 95.12)
Use your calculator to work out the following to 1 decimal place:
√ (362 + 27.92)
Use your calculator to work out the following to 1 decimal place:
√ (612 + 34.62)
Use your calculator to work out the following to 1 decimal place:
√ (252 + 28.72)
Use your calculator to work out the following:
√ (72 + 242)
(Answer is a whole number)
Use your calculator to work out the following to 1 decimal place:
31 × (46.1 - 82)
-√ 1.72 + 492
Use your calculator to work out the following to 1 decimal place:
23 × (30.7 - 79)
-√ 73.42 + 342
Use your calculator to work out the following to 1 decimal place:
25 × (99 - 36)
(17.7 + 10)3
Use your calculator to work out the following to 1 decimal place:
42 × (34.7 - 95)
80.82 + 382
Use your calculator to work out the following to 1 decimal place:
60 × (67.3 - 84)
√ (23 + 703)
• Question 1
Use your calculator to work out the following to 1 decimal place:
√ (592 + 95.12)
111.9
EDDIE SAYS
On the calculator, press:
√ ( 59 ∧ 2 + 95.1 ∧ 2 ) =
• Question 2
Use your calculator to work out the following to 1 decimal place:
√ (362 + 27.92)
45.5
EDDIE SAYS
On the calculator, press:
√ ( 36 ∧ 2 + 27.9 ∧ 2 ) =
• Question 3
Use your calculator to work out the following to 1 decimal place:
√ (612 + 34.62)
70.1
EDDIE SAYS
On the calculator, press:
√ ( 61 ∧ 2 + 34.6 ∧ 2 ) =
• Question 4
Use your calculator to work out the following to 1 decimal place:
√ (252 + 28.72)
38.1
EDDIE SAYS
On the calculator, press:
√ ( 25 ∧ 2 + 28.7 ∧ 2 ) =
• Question 5
Use your calculator to work out the following:
√ (72 + 242)
(Answer is a whole number)
25
EDDIE SAYS
On the calculator, press:
√ ( 7 ∧ 2 + 24 ∧ 2 ) =
• Question 6
Use your calculator to work out the following to 1 decimal place:
31 × (46.1 - 82)
-√ 1.72 + 492
22.7
EDDIE SAYS
On the calculator, press:
31 × ( 46.1 - 82 ) ÷ - √ ( 1.7 ∧ 2 + 49 ∧ 2 ) =
• Question 7
Use your calculator to work out the following to 1 decimal place:
23 × (30.7 - 79)
-√ 73.42 + 342
13.7
EDDIE SAYS
On the calculator, press:
23 × ( 30.7 - 79 ) ÷ - √ ( 73.4 ∧ 2 + 34 ∧ 2 ) =
• Question 8
Use your calculator to work out the following to 1 decimal place:
25 × (99 - 36)
(17.7 + 10)3
0.1
EDDIE SAYS
On the calculator, press:
25 × ( 99 - 36 ) ÷ ( 17.7 + 10 ) ∧ 3 =
• Question 9
Use your calculator to work out the following to 1 decimal place:
42 × (34.7 - 95)
80.82 + 382
-0.3
EDDIE SAYS
On the calculator, press:
42 × ( 34.7 - 95 ) ÷ ( 80.8 ∧ 2 + 38 ∧ 2 ) =
• Question 10
Use your calculator to work out the following to 1 decimal place:
60 × (67.3 - 84)
√ (23 + 703)
-1.7
EDDIE SAYS
On the calculator, press:
60 × ( 67.3 - 84 ) ÷ √ ( 23 + 70 ∧ 3 ) =
---- OR ----
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Why does my height not vary with temperature? Has the ruler dilated?
Sep 24, 2015
Are you certain that it does not? On the dilation question... maybe.
Explanation:
Most physical objects change size (at least a little) with temperature. How much they change depends on a property we call the coefficient of thermal expansion. In most cases, objects get a little larger when heated and smaller when cooled.
It's not unusual for a person's height to change by about half an inch over the course of a day for a variety of reasons. Mostly it just has to do with compression of the spine. If the change due to temperature differences is smaller than this, it might be difficult to measure.
The volume of the human body is about 65 liters. Your body can tolerate temperatures in the range of about 95ºF to 105ºF. The former is nearing hypothermia and the latter is a very high fever. Let's calculate how much 65 liters of water would expand when heated from 95ºF to 105ºF.
The expansion coefficient for water near 98.6ºF is $\beta = 0.000385$.
$\Delta V = {V}_{0} \beta \Delta T$
$\Delta V = 65 \cdot 0.000385 \cdot \left(105 - 95\right)$
$\Delta V = 0.25025 l$
So your volume will change by about one quarter of a liter. If your body expands in all directions equally, this is about the volume of water on your skin when you step out of the shower. You might imagine that this changes your height by about a millimeter. This distance is difficult to measure with a ruler and much smaller than the daily variation in your height.
On the question of changes to the ruler: The ruler will have a different coefficient of expansion. The best rulers are made with combinations of materials which do not expand or contract very much with temperature. So the answer may depend on the materials which make up your ruler.
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# When math is useful
Determining Big O(n*n) is relatively straight forward when compared to other forms. Given,
```start = 0;
end = n;
for (i = start; i < end; i++)
for (j = start; j < end; j++)
doSomething();```
It’s clear, for every iteration of the outer loop, the inner loop iterates n times. The outer loop iterates n times as well, so, total number of iterations are n*n. In Big O notation, the runtime complexity may be represented as O(n*n).
`````` start = 0;
end = n;
for (i = start; i < end; i++)
for (j = i; j < end; j++)
doSomething();
``````
Note: The inner loop starts iteration at i
The code above represents a very practical use case, for example, when finding indexOf, substring et al.
When determining the runtime complexity, intuition will suggest O(n*n), which, is correct. But, how do we arrive at that? Because at each iteration, i, of the outer loop, we are iterating n-i times in the inner loop. The number of inner loop iterations are decreasing with each iteration of the outer loop.
Let’s break it down by each iteration –
When,
```i = 0, inner loop iterations = n
i = 1, inner loop iterations = n - 1
i = 2, inner loop iterations = n - 2
i = 3, inner loop iterations = n - 3
...
i = n - 2, inner loop iterations = n = (n - 2) = 2
i = n - 1, inner loop iterations = n - (n - 1) = 1```
Sum total of all iterations –
`n + (n - 1) + (n - 2) + (n - 3) .... 2, 1`
If we reversed the above, it will look like an arithmetic series of positive integers up to n.
`1, 2 ... (n - 3) + (n - 2) + (n - 1) + n`
Which, as we know is equivalent to –
`n(n+1)/2`
The above expands to –
`(n*n)/2 + (n/2)`
In Big O notation (a notation of asymptotic analysis), we drop the lower terms, which, leaves us with –
`O(n*n)`
# Some thoughts on maintainable code
I take pleasure in writing good code – maintainable, industrial strength, and sometimes beautiful. Here are some of the considerations I make to achieve that goal.
But, first, why is maintainable code important? Maintainable code will generally live longer, be more bug resistant and scale to some degree. With ever increasing engineer attrition, it will result in shorter ramp up times for new hires.
Readability is the most important and biggest consideration. Readability starts with ‘naming’ method/function names, variables, constants etc meaningfully. It may seem like a quick and easy task, but, is usually quite the opposite. Names are important because they tell the story of the code. I believe, readable code requires minimal to no documentation. Similar to a good story, the challenge is in finding succinct and yet meaningful names.
UNIFORMITY
Code uniformity starts with setting up code formatting rules (not guidelines, but, rules), such as, line indentation, spacing between language constructs such as (, {, etc and keywords. It is important to develop rules for where things will be found, such as, constants. Very minute details can make or break code uniformity. For example, in a java class, if ‘this’ is explicitly used to refer to an instance of a Java class, then, it should be done consistently throughout the class. Code uniformity directly affects code readability.
MODULARITY/DRY (Do not repeat yourself)
Functions/methods are the essence of programming languages. As a rule of thumb, if the same piece of code is used more than once, then, it should be written as a function. But, often times, it is also good practice to encapsulate code in a function even if the code is used once. It mostly boils down to readability. Any code reader using a laptop will appreciate if a block of code fits the size of their screen. There are other benefits too, such as, creating units of testable code.
CLEVER
Being clever with code is achieving a desired result with fewer (or least) lines of code. But, if clever code reduces or distorts readability, it is better to refrain from being clever. Clever code can leverage good, but, rarely used features of a programming language – can act as a purveyor of knowledge.
PERFORMANCE
Performance should be the most important consideration if it is practically hindering/slowing the experience for the consumer of code or is the value proposition of a business. But, in most cases, the performance degradation that comes at the cost of code readability is negligible.
Writing a piece of code is like writing a short story in english. There are rules for writing in english, which, help ensure the writing is easy to understand and enjoyable. Well written sentences maintain tense, don’t introduce ambiguity, organize different thoughts in paragraphs etc. Maintainable code too, follows principles that help make it clear and easy to read and understand.
# JavaScript – Primitive vs Reference Types
Two areas of confusion around JavaScript primitive vs reference types
1. If every thing in JavaScript is an object
2. Passed by value vs passed by reference
1. If every thing in JavaScript is an object
This is untrue since JavaScript clearly defines the following primitive types –
1. Number
2. Boolean
3. String
The confusion arises because JavaScript allows us to call methods on primitive types. For example, these are all legal statements –
``` 1['toString']() // outputs 1 as a String```
``` 1..toString() // outputs 1 as a String. Note: 1. means 1 followed by decimal true.toString() // outputs true as a String ```
``` "Some Str".toString() // outputs Some Str as a String ```
It is important to understand how this works. The Number 1, the Boolean true and the String “Some Str” are truly primitive types by themselves, but, when a method is called on them, JavaScript creates a transient or temporary object for the primitive type and tries to call the method. For example, in the case of the Number 1, a transient object is created as follows –
``` new Number(1) // Transient object ```
It is also important to understand that the transient object is ‘transient’ and is garbage collected as soon as the method has returned.
From the explanation and example above, we can see, this object like behavior of primitive types is a feature of the language and not of the primitive types themselves.
2. Passed by value vs passed by reference
As a rule of thumb in any programming language, primitive types should be passed by value, whereas, reference types, passed by reference. In JavaScript this is true, except for Strings.
So, Number and Boolean are passed by value. Whereas, everything else, including Array and Function( represented as objects in JS) are passed by Reference. So, what is the deal with Strings? Here is an explanation –
Since a String can be of infinite size, passing it by value will be inefficient, therefore, JavaScript chooses to pass Strings by reference. The good part is, Strings in JavaScript are immutable, meaning once a String object is created it cannot be modified. There is no method available on String that allows us to modify it. For example, Strings do not provide us any such method – insertCharAt or replaceCharAt. This also means, as Javascript developers, we do not have to worry about Strings ever getting modified, even though, they are passed by reference. Also, note, although, Strings are passed by reference for efficiency, they maintain their primitive nature when we compare two Strings. Take the following for example –
``` "abc" === ("ab" + "c") // true, since values are compared ```
In short, the issue of how Strings are passed in JavaScript is really not an issue because Strings are immutable.
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# How do you calculate the energy change of reaction for the following reaction?
## EDIT: This is nearly impossible unless we are talking about the change in enthalpy of the reaction. We would otherwise have to calculate the ground-state energies of each molecule by solving the Schrodinger equation three times. - Truong-Son
Apr 3, 2018
Using bond enthalpies(?)
#### Explanation:
Assuming you meant the ENTHALPY change of the reaction it becomes clearer. As Truong-Son pointed out it would be a hassle to calculate using the Schrodinger equation if we are truly talking about the ENERGY change.
Given that we are talking about Enthalpy changes, we can use bond enthalpies from a table to solve this. I found my bond enthalpies in this booklet, table 11 (Courtesy of Ibchem.com)
We need to determine what bonds are broken and what bonds are formed. Bond breaking is endothermic- we need to put energy into breaking the bond so the value for $\Delta H$ will be positive.
Bond making is exothermic, meaning energy will be released to the surroundings and $\Delta H$ will be negative.
From the diagram's product side, we can see that the Hydrogen gas and the C-O double bond have vanished, so the respective bonds must have been broken in the first step!
Hence:
Breaking a C-O double bond=$\Delta H = + 745 k j m o {l}^{-} 1$*
Breaking an H-H single bond= $\Delta H = + 436 k j m o {l}^{-} 1$
*(Not the value in the booklet as some pointed out that the value in the booklet was too high)
If we wanted to be thorough, we could compare all the bonds on both the product and reactant side, but here we can see that there is no change in the Methyl $\left(- C {H}_{3}\right)$ groups so the "breaking and making" would cancel out, mathematically.
Anyway, on the product side, we now have the central carbon single bonded to a hydrogen, an oxygen and in turn that oxygen is bonded to a hydrogen. We have 3 new bonds that were not present in the reactant step.
We have formed the following bonds:
Forming a C-H single bond= $\Delta H = - 414 k j m o {l}^{-} 1$
Forming an O-H single bond=$\Delta H = - 463 k j m o {l}^{-}$
Forming a C-O single bond= $\Delta H = - 358 k j m o {l}^{-} 1$
So the total enthalpy change should be all of these enthalpy changes summed up.
$745 + 436 + \left(- 414\right) + \left(- 463\right) + \left(- 358\right) = - 54$
$\Delta H = - 54 k j m o {l}^{-} 1$
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Classworks Common Core Transition Guide. South Carolina 4th Grade Mathematics
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1 Classworks Common Core Transition Guide South Carolina 4th Mathematics OFFICIALLY ENDORSED
3 Applying Mathematical Practices with Classworks Projects
4 CC Standard Unit # Project OA Recycling Estimates Applying Mathematical Practices with Classworks Projects 4th Project Description Students develop a spreadsheet that records the refunds generated by recyclables. MD Local Customs Students create a table showing the conversions for customary units of length then explain how to use the table. MD A Walk In the Park MD Babysitting Record MD Is the Price Right? Students track expenses during a trip to an amusement park to practice adding and subtracting decimals. Students track the time spent babysitting and the money earned during that time. Students cut and paste pictures of money along with various toys to represent how much money is needed to buy the toys. MD Savings Account Students determine how much money they will need to save in order to buy something. NBT Sizing Up the Students demonstrate an understanding of States whole number place value while working with a table listing the areas of some states. NBT Translation Students translate expanded forms of numbers Vacation into standard forms. NF Pizza Part y Students manipulate and compare pizza models to demonstrate equivalent fractions and simplest terms. G Holiday Symmetry College and Career Ready Practices Students select things associated with holidays that have more than one line of symmetry. Make sense of problems, persevere Reason abstractly & quant. Construct viable arguments Model with mathematics Use appropriate tools strategically Attend to precision Look for and make use of structure Express regularity in repeated reasoning Curriculum Advantage 2012
6 Common Core Crosswalks
7 Common Core Crosswalks: South Carolina 4th Math OA.4 OA.4.A OA.4.1 OA.4.2 Operations and Algebraic Thinking Use the four operations with whole numbers to solve problems. Interpret a multiplication equation as a comparison, e.g., : Skill Builder Multiplication interpret 35 = 5 x 7 as a statement that 35 is 5 times as Patterns many as 7 and 7 times as many as 5. Represent verbal 1380: Multiplication and Division Fact statements of multiplicative comparisons as Families multiplication equations. Multiply or divide to solve word problems involving multiplicative comparison, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem, distinguishing multiplicative comparison from additive comparison : Solving Problems by Choosing Multiplication or Division Concept Intro Practice OA.4.3 Write numbers from 0 to 20. Represent a number of objects with a written numeral 0 20 (with 0 representing a count of no objects). OA.4.B Gain familiarity with factors and multiples. OA.4.4 Find all factor pairs for a whole number in the range Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range is a multiple of a given one digit number. Determine whether a given whole number in the range is prime or composite : Skill Builder Finding Reasonable Estimated Answers 1347: Guessing and Checking 1350: Solving Multistep Addition and Subtraction Problems 1400: Solving Problems by Choosing Multiplication or Division 1433: Understanding Choosing Addition or Subtraction NEW 1370: Skill Builder Factors and Multiples to : Skill Builder Prime and Composite Numbers Curriculum Advantage
8 Common Core Crosswalks: South Carolina 4th Math OA.4.C OA.4.5 Generate and analyze patterns. Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. For example, given the rule "Add 3" and the starting number 1, generate terms in the resulting sequence and observe that the terms appear to alternate between odd and even numbers. Explain informally why the numbers will continue to alternate in this way : Understanding Finding a Pattern 1346: Skill Builder Recognizing Number Patterns 1348: More Function Tables 1389: Skill Builder Introducing Tesselations Concept Intro Practice NBT.4 NBT.4.A NBT.4.1 Number and Operations in Base Ten Generalize place value understanding for multi digit whole numbers. Recognize that in a multi digit whole number, a digit in NEW one place represents ten times what it represents in the place to its right. For example, recognize that = 10 by applying concepts of place value and division. 1332: Place Value to Hundred Thousands 1331: Place Value to Thousands NBT.4.2 NBT.4.3 Read and write multi digit whole numbers using base ten numerals, number names, and expanded form. Compare two multi digit numbers based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. Use place value understanding to round multi digit whole numbers to any place : Using Number Forms to Hundred Thousands 1330: Comparing and Ordering Numbers to 100, : Place Value to Thousands NEW 1363: Skill Builder More Rounding 1338: Rounding Rules Curriculum Advantage
9 Common Core Crosswalks: South Carolina 4th Math NBT.4.B NBT.4.4 NBT.4.5 Use place value understanding and properties of operations to perform multi digit arithmetic. NEW Fluently add and subtract multi digit whole numbers using the standard algorithm. Multiply a whole number of up to four digits by a onedigit whole number, and multiply two two digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 1351: Subtracting 4 Digit Numbers 1342: Adding 2 and 3 Digit Numbers with Regrouping 1349: Subtracting 2 and 3 Digit Numbers with Regrouping 1343: Adding 4 Digit Numbers 1353: Skill Builder Add/Subtract 6 Digit Whole Numbers : Using 1 Digit Multipliers 1423: Skill Builder More 1 Digit Multiplication 1446: Using 2 Digit Multipliers Concept Intro Practice NBT.4.6 NF.4 NF.4.A NF.4.1 Find whole number quotients and remainders with up to four digit dividends and one digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. Number and Operations Fractions Extend understanding of fraction equivalence and ordering. Explain why a fraction a/b is equivalent to a fraction (n x a) (n x b) by using visual frac on models, with a en on to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions : 2 Digit Divisors and Dividends Up to 4 Digits 1442: Skill Builder Divide 2 or 3 Digits/1 Digit 1397: 1 Digit Divisors and Dividends up to 4 Digits : Finding Equivalent Fractions Using Models 1443: Skill Builder Working with Fractions 1404: Introducing Equivalent Fractions and Simplest Form Curriculum Advantage
10 Common Core Crosswalks: South Carolina 4th Math NF.4.2 Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model. Concept Intro Practice : Comparing and Ordering Fractions NF.4.B NF.4.3 NF.4.3.a NF.4.3.b Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers. Understand a fraction a/b with a > 1 as a sum of fractions 4 1/b. Understand addition and subtraction of fractions as : Subtracting Fractions with Unlike joining and separating parts referring to the same whole. Denominators 1409: Adding Fractions with Unlike Denominators 1408: Adding and Subtracting Fractions with Models Decompose a fraction into a sum of fractions with the : Finding Fractions as Part of a Whole same denominator in more than one way, recording each decomposition by an equation. Justify decompositions, e.g., by using a visual fraction model. Examples: 3/8 = 1/8 + 1/8 + 1/8 ; 3/8 = 1/8 + 2/8 ; 2 1/8 = /8 = 8/8 + 8/8 + 1/8. Curriculum Advantage
11 Common Core Crosswalks: South Carolina 4th Math NF.4.3.c Add and subtract mixed numbers with like denominators, e.g., by replacing each mixed number with an equivalent fraction, and/or by using properties of operations and the relationship between addition and subtraction : Introducing Adding and Subtracting Mixed Numbers Concept Practice Intro 1 2 NF.4.3.d NF.4.4 NF.4.4.a NF.4.4.b NF.4.4.c Solve word problems involving addition and subtraction of fractions referring to the same whole and having like denominators, e.g., by using visual fraction models and equations to represent the problem. Apply and extend previous understandings of multiplication to multiply a fraction by a whole number. Understand a fraction a/b as a multiple of 1/b. For example, use a visual fraction model to represent 5/4 as the product 5 x (1/4), recording the conclusion by the equation 5/4 = 5 x (1/4). Understand a multiple of a/b as a multiple of 1/b, and use this understanding to multiply a fraction by a whole number. For example, use a visual fraction model to express 3 x (2/5) as 6 x (1/5), recognizing this product as 6/5. (In general, n x (a/b) = (n x a)/b.) Solve word problems involving multiplication of a fraction by a whole number, e.g., by using visual fraction models and equations to represent the problem. For example, if each person at a party will eat 3/8 of a pound of roast beef, and there will be 5 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie? : Fraction Word Problems NEW NEW NEW 1557: Introducing Multiplying Fractions NEW 1413: Fraction Word Problems Curriculum Advantage
12 Common Core Crosswalks: South Carolina 4th Math NF.4.C Understand decimal notation for fractions, and compare decimal fractions. NF.4.5 Express a fraction with denominator 10 as an equivalent 4 fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100. For example, express 3/10 as 30/100, and add 3/10 + 4/100 = 34/100. NF.4.6 Use decimal notation for fractions with denominators : Skill Builder Decimals and or 100. For example, rewrite 0.62 as 62/100; describe a Fractions length as 0.62 meters; locate 0.62 on a number line 1407: Understanding Equivalent Decimals diagram. and Fractions NF.4.7 Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions, e.g., by using a visual model : Comparing and Ordering Decimals to Hundredths Concept Intro Practice MD.4 MD.4.A MD.4.1 Measurement and Data Solve problems involving measurement and conversion of measurements from a larger unit to a smaller unit. Know relative sizes of measurement units within one system of units including km, m, cm; kg, g; lb, oz.; l, ml; hr, min, sec. Within a single system of measurement, express measurements in a larger unit in terms of a smaller unit. Record measurement equivalents in a two column table. For example, know that 1 ft is 12 times as long as 1 in. Express the length of a 4 ft snake as 48 in. Generate a conversion table for feet and inches listing the number pairs (1, 12), (2, 24), (3, 36), : Converting Metric Units of Measure 1416: Converting Customary Units of Length Curriculum Advantage
13 Common Core Crosswalks: South Carolina 4th Math MD.4.2 Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale : Using Maps: Distance Between Locations 1372: Skill Builder Problem Solving With Money 1421: Skill Builder Measurement Word Problems 1429: Elapsed Time Concept Practice Intro MD.4.3 Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor. NEW 1431: Skill Builder Word Problems (Perimeter, Area, and Volume) 1436: Skill Builder Concepts of Area and Perimeter 4 MD.4.B MD.4.4 Represent and interpret data. Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using information presented in line plots. For example, from a line plot find and interpret the difference in length between the longest and shortest specimens in an insect collection. NEW Curriculum Advantage
14 Common Core Crosswalks: South Carolina 4th Math MD.4.C MD.4.5 Geometric measurement: understand concepts of angle and measure angles. Recognize angles as geometric shapes that are formed 4 wherever two rays share a common endpoint, and understand concepts of angle measurement: Concept Intro Practice MD.4.5.a An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a "onedegree angle," and can be used to measure angles. 4 MD.4.5.b MD.4.6 MD.4.7 An angle that turns through n one degree angles is said to have an angle measure of n degrees. Measure angles in whole number degrees using a protractor. Sketch angles of specified measure. Recognize angle measure as additive. When an angle is decomposed into non overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure. 4 NEW 1386: Introducing Angles 1 2 NEW Curriculum Advantage
15 Common Core Crosswalks: South Carolina 4th Math G.4 G.4.A G.4.1 G.4.2 Geometry Draw and identify lines and angles, and classify shapes by properties of their lines and angles. Draw points, lines, line segments, rays, angles (right, : Introducing Points, Lines, acute, obtuse), and perpendicular and parallel lines. Segments, and Rays Identify these in two dimensional figures. 1385: Skill Builder Parallel and Perpendicular Lines 1386: Introducing Angles Classify two dimensional figures based on the presence or absence of parallel or perpendicular lines, or the presence or absence of angles of a specified size. Recognize right triangles as a category, and identify right triangles : Skill Builder Parallel and Perpendicular Lines Concept Intro Practice G.4.3 Recognize a line of symmetry for a two dimensional figure as a line across the figure such that the figure can be folded along the line into matching parts. Identify linesymmetric figures and draw lines of symmetry. NEW 1414: Skill Builder Shape Symmetry 3 3 Curriculum Advantage
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# Find the next number in the sequence
What is the next number in this series?
1, 2, 6, 42, 1806?
SherlockHolmes Expert Asked on 4th May 2016 in
Add 1 to the number and multiply the number by it to get the next number
1 * (1+1) = 2
2 * (2+1) = 6
6 * (6+1) = 42
42 * (42+1) = 1806
1806 * (1806+1) = 3263442
Next number will be 3263442
SreeRoopa Sankararaman Genius Answered on 4th May 2016.
Going acc to the seq the no. get multiplied with 1 more than the numbers,hence the next is 1806*1807 =3263442
Amanjonty23 Guru Answered on 4th May 2016.
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501_PartUniversity Physics Solution
# 501_PartUniversity Physics Solution - Sound and Hearing...
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Sound and Hearing 16-9 16.36. IDENTIFY: Destructive interference occurs when the path difference is a half integer number of wavelengths. SET UP: ( ) ( ) 344 m s, so λ 344 m/s 172 Hz 2.00 m. vv f == = = If 8.00 m A r = and B r are the distances of the person from each speaker, the condition for destructive interference is ( ) 1 2 λ , BA rr n −=+ where n is any integer. EXECUTE: Requiring ( ) 1 2 λ 0 =++ > gives ( ) ( ) 1 2 8.00 m 2.00 m 4, A nr λ + >− =− so the smallest value of B r occurs when 4, n and the closest distance to B is ( ) () 1 2 8.00 m 4 2.00 m 1.00 m. B r =+ + = EVALUATE: For 1.00 m, B r = the path difference is 7.00 m. AB rr −= This is 3.5 . 16.37. IDENTIFY: Compare the path difference to the wavelength. SET UP: ( ) 344 m s 860 Hz 0.400 m vf = EXECUTE: The path difference is 13.4 m 12.0 m 1.4 m. = path difference 3.5. = The path difference is a half- integer number of wavelengths, so the interference is destructive. EVALUATE: The interference is destructive at any point where the path difference is a half-integer number of wavelengths. 16.38. IDENTIFY: beat 1 2 . f ff . = SET UP: 344 m/s, v = Let 1 6.50 cm = and 2 6.52 cm. = 21 > so 12 . f f > EXECUTE: 2 22 1 2 11 ( ) ( 3 4 4 m / s ) ( 0 . 0 2 1 0 m ) 16 Hz. (6.50 10 m)(6.52 10 m) v ffv λλ −− ⎛⎞ −× − = = = ⎜⎟ ×× ⎝⎠ There are 16 beats per second. EVALUATE: We could have calculated 1 f and 2 f and subtracted, but doing it this way we would have to be careful to retain enough figures in intermediate calculations to avoid round-off errors. 16.39. IDENTIFY: beat . ab f For a stopped pipe, 1 . 4 v f L = SET UP: 344 m/s. v = Let 1.14 m a L = and 1.16 m. b L = ba L L > so 11 . f f > EXECUTE: 2 ) ( 3 4 4 m / s ) ( 2 . 0 0 1 0 m ) 1.3 Hz. 4 4 4(1.14 m)(1.16 m) a b L L LL L L = = There are 1.3 beats per second. EVALUATE: Increasing the length of the pipe increases the wavelength of the fundamental and decreases the frequency. 16.40. IDENTIFY: beat 0 . f . 2 v f L = Changing the tension changes the wave speed and this alters the frequency. SET UP: FL v m = so 1 , 2 F f mL = where 0 . FF F = Let 0 0 1 . 2 F f mL = We can assume that 0 / FF Δ is very small. Increasing the tension increases the frequency, so beat 0 . f EXECUTE: (a) 1/2 0 beat 0 0 0 0 . 2 2 f F F F mL F mL ⎡⎤ Δ =− = + Δ− = + ⎢⎥ ⎣⎦ 00 2 ΔΔ += + when 0 / Δ is small. This gives that beat 0 0 . 2 F F Δ = (b) beat 2 2(1.5 Hz) 0.68%. 440 Hz Ff Δ = EVALUATE: The fractional change in frequency is one-half the fractional change in tension. 16.41. IDENTIFY: Apply the Doppler shift equation L LS S . f f + = + SET UP: The positive direction is from listener to source. S 1200 Hz. f = L 1240 Hz. f = EXECUTE: L 0. v = S 25.0 m/s. v S v f f = + gives SL (2 5 m / s ) ( 1 2 4 0 H z ) 780 m/s. 1200 Hz 1240 Hz v = EVALUATE: f f > since the source is approaching the listener.
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16-10 Chapter 16 16.42. IDENTIFY: Follow the steps of Example 16.19. SET UP: In the first step, S 20.0 m/s v =+ instead of 30.0 m/s. In the second step, L 20.0 m/s v =− instead of 30.0 m/s. + EXECUTE: WS S 340 m/s (300 Hz) 283 Hz. 340 m/s 20.0 m/s v ff vv ⎛⎞ == = ⎜⎟ ++ ⎝⎠ Then L LW 340 m/s 20.0 m/s (283 Hz) 266 Hz. 340 m/s v +− = EVALUATE: When the car is moving toward the reflecting surface, the received frequency back at the source is higher than the emitted frequency. When the car is moving away from the reflecting surface, as is the case here, the received frequency back at the source is lower than the emitted frequency.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
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# math
posted by on .
How many 3/8 ounce servings are in 3 1/5 ounces?
• math - ,
3 1/5 = 16/5
(16/5) / (3/8)
(16/5) * (8/3) = 128 / 15 = 8 8/15
• math - ,
thx u ms.sue
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Addition and Subtraction of Polynomials
## Combining like terms in polynomial expressions
Levels are CK-12's student achievement levels.
Basic Students matched to this level have a partial mastery of prerequisite knowledge and skills fundamental for proficient work.
At Grade (Proficient) Students matched to this level have demonstrated competency over challenging subject matter, including subject matter knowledge, application of such knowledge to real-world situations, and analytical skills appropriate to subject matter.
Advanced Students matched to this level are ready for material that requires superior performance and mastery.
This lesson covers adding and subtracting polynomials.
1
## Addition and Subtraction of Polynomials
Add and subtract polynomial expressions and simplify by combining like terms.
0
## Addition and Subtraction of Polynomials
Add and subtract polynomial expressions and simplify by combining like terms.
0
## Addition and Subtraction of Polynomials
by EPISD Algebra 1 Team //at grade
0
by EPISD Algebra 2 Team //at grade
This lesson covers adding and subtracting polynomials.
0
by SFDR Algebra I //at grade
0
0
• PLIX
## Add and Subtract Polynomials: Sliders
Addition and Subtraction of Polynomials Interactive
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• Video
## Addition and Subtraction of Polynomials: An Explanation of the Concept
This video provides an explanation of the concept of the addition and subtraction of polynomials.
0
• Video
## Addition and Subtraction of Polynomials: A Sample Application
This video demonstrates a sample use of the addition and subtraction of polynomials.
0
## Addition and Subtraction of Polynomials Quiz
Quiz for Addition and Subtraction of Polynomials.
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## Addition and Subtraction of Polynomials Practice
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• Critical Thinking
## Addition and Subtraction of Polynomials Discussion Questions
A list of student-submitted discussion questions for Addition and Subtraction of Polynomials.
0
Come up with questions about a topic and learn new vocabulary to determine answers using the table
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## Adding and Subtracting Polynomials Stop and Jot
Strengthen ability to analyze word meaning and symbolic or mathematical notation from context and introduce students to unfamiliar vocabulary or notation. Further solidify understanding by defining new vocabulary words and notation as well as generating personal examples of words, concepts, and notation usage.
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• Real World Application
## Telescope
Students will apply their understanding of polynomials and quadratics to better understand the making of a mirror for an astronomer's telescope.
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• Study Guide
## Operations with Polynomials Study Guide
This study guide looks at terminology for monomials and polynomials, operations with polynomials, FOIL, and special products (square of a binomial, sum and difference patterns).
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# Towards Equations
We finally reconvened our maths club after a mid-term break (children not available) combined with the university exams period (me not available). I correctly assumed nobody remembered anything we talked about before. At least not in detail. Also, we had a significant number of new children joining our group, who had no idea about what we have been up to.
In the previous sessions we discovered and invented our notation, which consisted of three combinators (X, C, I) and three operations: composition (“gluing”), tensoring (“stacking”) and dual (“flipping”).
This time we just took this as a given, the starting point of our development. If the previous sessions were analytic, trying to decompose knots into basic constants and operations, this session was synthetic, starting with formulas and drawing them up and seeing what we get.
The first formula we contemplated was C;C*, which everyone immediately identified as a circle. The knot-theoretic name for this is the unknot, word which caused a certain amount of amusement. One student cleverly observed, unprompted, that (C;C*)* would have been the same thing, also an unknot, and we had a short discussion about how symmetry of a knot K boils down to K=K*. That was quite rewarding and fun and prepared the ground, somewhat serendipitously, for the next formula C;X;C*:
The same student rashly pointed out that this shape is also symmetric, i.e. C;X;C*=(C;X;C*)*. Is this true or false, qua knots?
The ensuing discussion was the most interesting and important part of the session. Initially the students were in majority supportive of the validity of the equation, then someone pointed out that X≠X* so, they reckoned, it must be that
Indeed, the shapes are not identical. The opinion of the group swung so they all agreed that the equation is not valid and the two shapes (original and flipped version) are not equal.
But aren’t they? What do you think? We are talking knots here, and two knots should be equal when they are topologically equal, i.e. they can be deformed into each other without any tearing and gluing. So in fact they are equal because:
So here we have a genuinely interesting equation, where two formulae are equal not because the shapes they correspond to are geometrically equal (i.e. “identical”) but because they are topologically equal, i.e. there is a smooth deformation between them. Also note that the deformation must be in 3D by “twisting” the left (or right) side of the unknot — a 2D transformation would make the wire “pinch” which may or may not be allowable, depending on how we precisely set things up (if we were very serious and careful about what we are doing).
The point was quickly grasped and we moved on. It is a subtle point which I think can be damaged by over-explaining if the initial intuitions are in the right place, which they seemed to be.
Next we looked at a much more complicated formula that took a long time to draw correctly: C;(I⨂C⨂I);(X⨂X*);(I⨂C*⨂I);C*.
As you may see, if we are drawing this as a knot and tidy things up a bit we actually have this:
Which is really just two unknots — they don’t interlock so they can be pulled apart with no tearing and gluing, i.e. topologically. The formula for this is the much simpler C;C*;C;C*!
This point was not easy to make, and it seemed difficult for the students to appreciate it. By this time they were quite tired and the drawing of the more complex formulation of the two unknots diagram drained them of energy and patience — they made quite a few errors and had to restart several times. Showing them a much easier way to achieve the same knot diagram almost made them angry!
I hope to use this as a motivation for the future: how to simplify a formula before we draw it. Until now we have looked at equality semantically, using the underlying model. In due course, as students become more used with manipulating knot diagrams using formulas, I will start introducing some general equational properties.
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# Any ideas on this one? Of those person who became teachers
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Any ideas on this one? Of those person who became teachers [#permalink]
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26 Aug 2005, 14:32
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Any ideas on this one?
Of those person who became teachers in 1968 and who later left the profession, 30 percent today earn salaries above $35,000 a year: of those who became teachers in 1968 and have remained in the profession, only 15 percent today earn salaries above$35,000 a year. These figures indicate how underpaid teachers are today.
The argument above depends on which of the following assumptions about the persons for whom statistics are cited?
(A) At least one-third of the group of persons who have remained in teaching would today be earning more than $35,000 a year if they had left teaching. (B) The group of persons who left teaching and the group who did not are comparable in terms of factors that determine how much people outside the teaching profession are paid. (C) Most of those persons who left teaching did so entirely because of the low salaries teachers earn. (D) As a group, those persons who have remained in teaching are abler and more dedicated than the group of persons who left teaching. (E) The group of persons who left teaching and who today earn more than$35,000 a year were more capable teachers than the group who remained in the profession.
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26 Aug 2005, 14:50
C, D, E easily rejected....
A is not necessarily true..
I go with B, to determine if someone is overpaid or underpaid..B should be true...
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26 Aug 2005, 16:52
I would say it is A.
This is because the argument assumes that the reason that a third of the people who left the teaching profession are able to earn a higher salary is because they left an underpaid profession. So a third of all people who leave the profession should be able to earn the high salaries.
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26 Aug 2005, 17:11
B acc to me
C,D,E do not tie the evidence and the concln together. Only B does!
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26 Aug 2005, 18:00
rigger wrote:
Any ideas on this one?
Of those person who became teachers in 1968 and who later left the profession, 30 percent today earn salaries above $35,000 a year: of those who became teachers in 1968 and have remained in the profession, only 15 percent today earn salaries above$35,000 a year. These figures indicate how underpaid teachers are today.
The argument above depends on which of the following assumptions about the persons for whom statistics are cited?
(A) At least one-third of the group of persons who have remained in teaching would today be earning more than $35,000 a year if they had left teaching. (B) The group of persons who left teaching and the group who did not are comparable in terms of factors that determine how much people outside the teaching profession are paid. (C) Most of those persons who left teaching did so entirely because of the low salaries teachers earn. (D) As a group, those persons who have remained in teaching are abler and more dedicated than the group of persons who left teaching. (E) The group of persons who left teaching and who today earn more than$35,000 a year were more capable teachers than the group who remained in the profession.
I think the answer is A. The assumption is that the two groups are similar. In other words: the people in the two groups are capable to make more than US$35,000 if they are not teachers A says that at least 33% of the people who have remained in teaching are capable to make more than US$ 35,000 in other jobs so do is similiar to the 30% of the other group
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27 Aug 2005, 01:12
The correct ans. seems B 2 me
What is the OA
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27 Aug 2005, 11:07
Thanks! The OA is indeed B.
27 Aug 2005, 11:07
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# How do you solve 4x=sqrt(6-4x) and find any extraneous solutions?
Apr 8, 2017
You can detect any extraneous solutions, if you restrict the argument of the square root to be greater than or equal to zero, then you can square both sides of the equation.
#### Explanation:
Given: $4 x = \sqrt{6 - 4 x}$
Restrict the argument of the square root to be greater than or equal to zero:
4x=sqrt(6-4x); 6-4x>=0
Simplify the restriction:
4x=sqrt(6-4x); 6>=4x
4x=sqrt(6-4x); 3/2>=x
4x=sqrt(6-4x); x<=3/2
Square both sides:
16x^2= 6-4x; x<=3/2
8x^2+2x-3=0; x<=3/2
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 a}$
x = (-2+-sqrt(2^2-4(8)(-3)))/(2(8)); x<=3/2
x = (-2+-sqrt(100))/16; x<=3/2
$x = \frac{- 12}{16} \mathmr{and} x = \frac{8}{16}$
$x = - \frac{3}{4} \mathmr{and} x = \frac{1}{2}$
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# 3D interpolation
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August 19, 2004, 09:13 3D interpolation #1 Lipo Wang Guest Posts: n/a Now I want to do parabolic interpolation in 3-D space Cartesian Coordiate. The form of parabolic should be: a0+a1*x+a2*x^2+b0*y+b1*y^2+b2*x*y+c0*z+ c1*z^2+c2*y*z+c3*x*z Totally 10 unknown coefficients. But for a cube, there are totally 8 end points. And if we also consider the information of 1st derivatives, then totally we have 32 known conditions. So the number of known coditions are not equal to that of unknowns. How to to this problem? Thanks
August 19, 2004, 10:01 Re: 3D interpolation #2 Markus Lummer Guest Posts: n/a Try a local tensor product of Lagrange polynomials. Regards, Markus
August 19, 2004, 11:40 Re: 3D interpolation #3 Lipo Wang Guest Posts: n/a Hello,Markus Could you explain in more details? Lagrange polynomials is parobolic? What is the form of tensor you said? Thanks a lot Lipo
August 19, 2004, 12:54 Re: 3D interpolation #4 ag Guest Posts: n/a You may want to consider a least-squares approach.
August 20, 2004, 01:12 Re: 3D interpolation #5 Markus Lummer Guest Posts: n/a 1) 1d case. Let the grid points be x_i, i=0,1,...,nx and the function values f_i = f(x_i) The Langrange polynomials L_p(x) depend only on the x_i (The L_p are of order nx. The definition of L_p should be found in any book about basic numerical mathematics.) and have the property L_p(x_i) = \delta_{ip} The interpolating polynomial then reads f(x) = \sum_{p=0}^nx f_p L_p(x) 2) 3d case. Cartesian grid points (x_i, i=0,1,...,nx), (y_j, j=0,1,...,ny), (z_k, k=0,1,...,nz), and function values f_{ijk} = f(x_i,y_j,z_k) You calculate the Lagrange polynomials L_p(x),L_q(y),L_r(z) and write for the interpolating polynomial f(x,y,z) = \sum_{p=0}^nx \sum_{q=0}^ny \sum_{r=0}^nz f_{pqr} L_p(x)L_q(y)L_r(z) You can apply this formula locally. I.e., e.g. choose nx=ny=nz=2 (quadratic polynomials) and consider the grid points (x_{i-1},x_{i},x_{i+1}) (y_{j-1},y_{j},y_{j+1}) (z_{k-1},z_{k},z_{k+1})
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## Percentage Yield M.3
Andrew Nguyen 2I
Posts: 68
Joined: Fri Sep 29, 2017 7:07 am
### Percentage Yield M.3
So for this problem, it's saying that from the calcium carbonate's decomposition, a certain amount of CO2 is produced. I know that the percentage yield should be the actual yield over the theoretical yield.
But just to confirm, do I assume that the theoretical yield in this case is 42.73 g (100% for this situation?) ; and it has something to do with the mole ratios correct ?
Naveed Zaman 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times
### Re: Percentage Yield M.3
You are partially right. What you have to realize is that they don't actually give you all the information you need. Remember that BOTH the theoretical yield and the actual yield have to be for the same thing (in this case, for CO2). What you have to do with the 42.73g of CaCO3 is use stoichiometry to convert to grams of CO2; this becomes your theoretical yield (you expect to form "x" grams of CO2 from 42.73g of CaCO3). You are already given the actual yield: 17.5g. All that's left is plugging these values in the %yield formula to get the answer. Hope this helps!
Amanda Hagen 1L
Posts: 26
Joined: Fri Sep 29, 2017 7:04 am
### Re: Percentage Yield M.3
I just want clarify the stoichiometric calculations needed to get from 42.73 g of CaCO3 to the grams CO2 in case more explanation was needed:
1. Divide by the given grams of CaCO3 (42.73 g CaCO3) by the molar mass of CaCO3 (100.09 g.mol-1 CaCO3) to get moles of CaCO3.
2. Convert moles of CaCO3 to moles of CO2 by using a mole ratio (they will be the same number because it is a 1:1 ratio)
3. Convert moles CO2 to grams CO2 by multiplying by the molar mass of CO2 (44.01 g.mol-1 CO2).
Now this gives you the theoretical yield of CO2 and you can use it to find the percent yield as described above.
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Come play IF you Dare
1. Jan 10, 2005
121212
Come play IF you Dare!!!
hey i made a little game i need to know if its you guys can calculate it
atleast try ;) :
Code (Text):
You have 5 balls that you have to divide (put the balls in the cups) between a number of cups in all possible ways.
1 cup:
---
|5|
---
1 possibility
Total: 1 possibility
2 cups:
-----
|1|4|
-----
|2|3|
-----
|3|2|
-----
|4|1|
-----
|5|0|
-----
|0|5|
-----
6 possibilities
Total: 7 possibilities
3 cups:
21 possibilities
Total: 28 possibilities
Continue until you have 500 cups. The total of possibilities from 1 to 500 is the answer.
--------------------------------------------------------------------------
and another one by my friend ! :
Code (Text):
N = number of Queens.
You have to put N Queens onto a chess board of N * N squares, so that the Queens don't threaten each other. (horizontally, vertically, diagonally)
How many possible setups are there?
Thus:
How can I put 4 Queens onto a 4*4 board
How can I put 5 Queens onto a 5*5 board
etc...
Calculate all setups for 4 to 10 Queens, take the sum of the results - the sum is the answer
-------------------------------------------------------------------------
And another one !
-------------------------------------------------------------------------
Code (Text):
If you want to buy something for less than 100 Euro you can pay with different 'units' (coins and bills).
At the moment the 6 best existing units to use are: 1, 2, 5, 10, 20 and 50.
If you want to buy something (that is less than 100 Euro) you have to use some of those units to pay for it.
Now let's go shopping with these 6 units :)
You are going to search for products that are 1 euro untill 99 euro, and you are going to buy one of each. Each time you buy something you choose as less units as possible (you have an infinite amount of units).
Let's see how many units we need if we use the units 1, 2, 5, 10, 20 and 50:
1 EURO -> only 1 unit: 1
2 EURO -> 1 unit : 2
3 EURO -> 2 units : 1 + 2
..
..
..
98 EURO -> 6 units : 1 + 2 + 5 + 20 + 20 + 50
99 EURO -> 6 units : 2 + 2 + 5 + 20 + 20 + 50
Now we will take the average of the needed units between 1 and 99, which is 3.4 units.
That's quite alright, but we can do better...
Your job is to find the lowest average with 6 different (integer) units.
Maybe a 3 euro coin is a good idea? And what about a 22 euro bill? Try it out :)
from lowest to highest coin used
i am gonna try to put these on a site and many more > maybe it needs a little programming to ge tthe correct answers
greetz lassie
Last edited: Jan 10, 2005
2. Jan 10, 2005
Galileo
I think the answer to the first is:
$$\frac{(4+N)!}{5!(N-1)!}$$
where N is the number of cups. That is the number of ways to distribute 5 balls over N cups.
So for 500 cups, there are 265,661,562,600 ways.
You shouldn't add up the number of ways for one cup plus the number of ways for 2 cups and so forth.
Last edited: Jan 10, 2005
3. Jan 10, 2005
121212
maybe u need to programe something then if you cant calculate it ;)
but keep ya head up and you have something to do also
good luck
(use excel)
Last edited: Jan 10, 2005
4. Jan 10, 2005
vincentchan
galileo did it WRONG
here is the solution for the first problem:
$$\Sigma_{k=5,500} \Sigma_{i=o,k} (-1)^i \left( \begin{array}{cc}k\\i\end{array} \right) (k-i)^5$$
If you have learned combinatorics, the idea is simple. this has no different with putting n indistinguishable balls into k indistinguishable cells which allowed cells empty, sum k up from 5 to 500 and let n be 5, that's it...... i am not sure could this be simplify...
edit: is from 1 to 500 instead of 5 to 500
Last edited: Jan 10, 2005
5. Jan 10, 2005
vincentchan
the second answer is 1224: 2+10+4+40+92+352+724
don't say my answer is wrong if you can't find the answer......
6. Jan 10, 2005
Galileo
I didn't assume the cups were indistinguishable.
Assume they are distinguishable.
Let's take 7 cups for simplicity. Then consider the following drawing:
$$\bullet \vert \bullet \vert \vert \bullet \bullet \vert \vert \vert \bullet$$
The 5 dots depict the 5 balls and the 6 vertical bars distinguish 7 area's which represent the 7 cups. So in this particular arrangement we have 1 ball in the first cup, 1 on the second, cup 3 is empty, there are 2 in the 4th cup, cups 5 and 6 are emtpy and there is one in the 7th cup.
It's clear that any order of the 5 dots and 7 lines gives a way to put the 5 balls in the 7 cups. Since there are 5+6 symbols, there are (5+6)!=11! ways to arrange them.
But permutation of the dots do not give a different arrangement and neither does a permutation of the lines. So the number is:
$$\frac{(5+6)!}{5!6!}$$
In general, you can do this for k balls and N cups. You'll have k dots and N-1 bars. So the number of arrangements here is:
$$\frac{(k+N-1)!}{k!(N-1)!}$$
The same problem arises in statistical mechanics, when considering the number of ways to put k bosons in N 'energy states'.
Note that I didn't consider the cups indistinguishable. That is, have 5 balls in the first cup is different from having 5 balls in the second cup.
BTW: 121212 didn't consider the cups indistinguishable either.
Last edited: Jan 10, 2005
7. Jan 10, 2005
vincentchan
oh, yeah, you are right, the question said the cups are distinguishable, because the title of the post, i assume this is a very hard problem and didn't read it carefully, and your answer is probably right... and isn't that his question itself is ADD UP ALL THE possibility.... if not.....why did he say we need writing a computer program.... this indeed is an easy problem
8. Jan 10, 2005
HallsofIvy
9. Jan 10, 2005
121212
wow didnt though u would get so far
btw you answer was correct on the chess game :p
i go to sleep now
in that time i got another one for you guys thats also very hard !!1 grrr
--------------------------------------------------------------------------
Code (Text):
If you want to buy something for less than 100 Euro you can pay with different 'units' (coins and bills).
At the moment the 6 best existing units to use are: 1, 2, 5, 10, 20 and 50.
If you want to buy something (that is less than 100 Euro) you have to use some of those units to pay for it.
Now let's go shopping with these 6 units :)
You are going to search for products that are 1 euro untill 99 euro, and you are going to buy one of each. Each time you buy something you choose as less units as possible (you have an infinite amount of units).
Let's see how many units we need if we use the units 1, 2, 5, 10, 20 and 50:
1 EURO -> only 1 unit: 1
2 EURO -> 1 unit : 2
3 EURO -> 2 units : 1 + 2
..
..
..
98 EURO -> 6 units : 1 + 2 + 5 + 20 + 20 + 50
99 EURO -> 6 units : 2 + 2 + 5 + 20 + 20 + 50
Now we will take the average of the needed units between 1 and 99, which is 3.4 units.
That's quite alright, but we can do better...
Your job is to find the lowest average with 6 different (integer) units.
Maybe a 3 euro coin is a good idea? And what about a 22 euro bill? Try it out :)
from lowest to highest coin used
yah yah this is a hard one ;) :P
10. Jan 10, 2005
chroot
Staff Emeritus
Participants:
Please do not hand out answers to homework problems. You should emphasize technique and methodology, but always leave at least some of the work to the student. Generally, the student must show some of his/her work, up to the point where he/she becomes stuck, before anyone should provide help.
121212:
Please do not attempt to use this site as a way out of your homework. Labelling your homework assignments as "games made by a friend" is ridiculous. If you do not at least make an attempt to do your homework before posting it here, we will not help you.
- Warren
11. Jan 10, 2005
121212
this is not really ment as homework cause a friend of a friend made these
and it is a game to relax but also homework cause u have to do it with your learned maths , and you "Learn" stuff
----edit----
(lol i dont want a teacher that gives me this kinda homework whaha :P)
greetz lassie
12. Jan 10, 2005
vincentchan
okay, then do you have the answer and do you know how to solve it, if yes, please don't post it here, if no, we won't give you the solution......
13. Jan 10, 2005
121212
if i didnt solve it would i post it as a game then ?
i just programmed in vb and perl
and i though that maths forum ""grand masters "" would like to solve this game ....
sorry if my attempt on harmony and fun in life is disaccepted here ...
14. Jan 10, 2005
NateTG
No, talking about this stuff is fine, but it's not appropriate for the homework section. If you had posted it in, for example, the brain teasers section, nobody would be complaining.
15. Jan 11, 2005
121212
ok ill do that then ;) thanks anyway :D
greetz
16. Jul 25, 2005
lazyidiot
All the 3 above questions are programming challenges of net-force.nl ... your friend must be trying to solve them without taking any pain ;)
Ive solved first 2 questions a long time before .. to solve the 3rd one, one must brute-force the possible set of answers, and Ive nt done that. So, If any one can solve that .. gimme a buzz.
Last edited: Jul 25, 2005
17. Jul 25, 2005
NateTG
There's the question of how to prove that the answer is optimal without brute force. Perhaps looking at that first will provide some insight?
18. Jul 25, 2005
NateTG
Hmm, it seems to matter whether negative denominations are allowed.
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At a certain stage of a soccer tournament, the score ratio
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01 Dec 2012, 06:28
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At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
1)8
2)10
3)20
4)40
5)80
I tried this as below:
a:b:c = 3:4:5
a:b= 3:4
doubled:
a:b = 2* 3:4 = 3:2
halved
a:c = (1/2)*3:5 = 3:10
a:c/a:b = (3/10)/ (3/2) = 6/30
given c =40
substituting, we get b = 8.
Pls let me know the basic approach on this.
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Re: Need help on this Ratio basics [#permalink]
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01 Dec 2012, 06:44
2
KUDOS
manukumar wrote:
At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
1)8
2)10
3)20
4)40
5)80
I tried this as below:
a:b:c = 3:4:5
a:b= 3:4
doubled:
a:b = 2* 3:4 = 3:2
halved
a:c = (1/2)*3:5 = 3:10
a:c/a:b = (3/10)/ (3/2) = 6/30
given c =40
substituting, we get b = 8.
Pls let me know the basic approach on this.
A to B = 3 : 4
So, on doubling we get 6 : 4
A to C = 3 : 5
So, on halving we get 1.5 : 5 or 3 : 10 or 6 : 20
So final ratio = 6 : 4 : 20.
If 20x = 40
4x = 8
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17 Feb 2017, 11:02
At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
a) 8
b) 10
c) 20
d) 40
e) 80
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17 Feb 2017, 11:26
Ratio of A:B is originally 3:4. Then this ratio is doubled and we have $$\frac{3}{2}$$ * 2 = $$\frac{6}{4}$$
Ratio of A:C is originally 3:5. Then this ratio is halved so we have $$\frac{3}{5}$$ * $$\frac{1}{2}$$ = $$\frac{3}{10}$$.
In order to compare the two ratio we must express the common term (in this case A) in the same quantity. Thus we multiply the second ratio ($$\frac{3}{10}$$) for 2 and we get $$\frac{6}{20}$$
So now the new score ratio of team A, B and C is 6:4:20. If we divide the score of team C by the ratio for team C we get the multiplier (in this case is 2).
Finally by applying the multiplier to B we get 8.
Hope it is clear.
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Re: At a certain stage of a soccer tournament, the score ratio [#permalink]
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17 Feb 2017, 14:06
dnalost wrote:
At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
a) 8
b) 10
c) 20
d) 40
e) 80
Merging topics. Please refer to the discussion above.
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At a certain stage of a soccer tournament, the score ratio [#permalink]
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17 Feb 2017, 21:24
A : B : C Initially:
3 : 4 : 5
6 : 4 (ratio A:B is doubled )
3 : __ : 10 (ratio is A: C is halved)
6 : __ : 20
6 : 4 : 20 Finally,
12 : 8 : 40 (since C has to become 40)
Therefore, B = 8
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Re: At a certain stage of a soccer tournament, the score ratio [#permalink]
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12 Mar 2017, 10:49
I like making it in a table:
A:B:C
3:4:5 then A:B doubled
6:4:? then A:C halved
3:?:10 let's make a common ration for all 3:
6:4:20 and since C earned 40 points:
12:8:40 B is 8
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Re: At a certain stage of a soccer tournament, the score ratio [#permalink]
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15 Mar 2017, 16:52
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manukumar wrote:
At a certain stage of a soccer tournament, the score ratio of teams A, B and C was 3:4:5. Eventually, the score ratio of A to B has doubled while the score ratio of A to C has halved. If the final score of team C was 40, what was the final score of team B?
1)8
2)10
3)20
4)40
5)80
We are given that the original ratio of A : B : C = 3x : 4x : 5x.
After the ratio of A to B doubles and the ratio of A to C is halved, we have:
A/B = 6x/4x
AND
A/C = 3x/10x
Since we need term A to be the same in both ratios, we can multiply our second ratio by 2/2 and we have:
A/C = 6x/20x
Now the ratio of A : B : C = 6x : 4x : 20x
Since team C had a final of 40 points, we have:
20x = 40
x = 2
So, the final score of team B was 4 x 2 = 8.
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Re: At a certain stage of a soccer tournament, the score ratio [#permalink] 15 Mar 2017, 16:52
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MATH
Calc09_2day2
# Calc09_2day2 - 9.2 day 2 Maclaurin Series Liberty Bell...
• Notes
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9.2 day 2 Maclaurin Series Liberty Bell, Philadelphia, PA Greg Kelly, Hanford High School, Richland, Washington Photo by Vickie Kelly, 2003
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There are some Maclaurin series that occur often enough that they should be memorized. They are on your formula sheet, but today we are going to look at where they come from. Maclaurin Series: (generated by f at ) 0 x = ( 29 ( 29 ( 29 ( 29 ( 29 2 3 0 0 0 0 2! 3! f f P x f f x x x ′′ ′′′ = + + + + ⋅⋅⋅
( 29 ( 29 ( 29 ( 29 ( 29 2 3 0 0 0 0 2! 3! f f P x f f x x x ′′ ′′′ = + + + + ⋅⋅⋅ ( 29 1 1 1 1 x x - = - - ( 29 1 1 x - - ( 29 2 1 x - - ( 29 3 2 1 x - - ( 29 4 6 1 x - - ( 29 5 24 1 x - - ( 29 ( 29 n f x List the function and its derivatives.
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List the function and its derivatives. Evaluate column one for x = 0 . ( 29 ( 29 ( 29 ( 29 ( 29 2 3 0 0 0 0 2! 3! f f P x f f x x x ′′ ′′′ = + + + + ⋅⋅⋅ ( 29 1 1 1 1 x x - = - - ( 29 1 1 x - - ( 29 2 1 x - - ( 29 3 2 1 x - - ( 29 4 6 1 x - - ( 29 5 24 1 x - - 1 1 2 6 3! = 24 4! = ( 29 ( 29 0 n f ( 29 ( 29 n f x 2 3 4 2 3 1 1 2! 3! ! 4! 1 ! 1 4 x x x x x = + + + + +⋅⋅⋅ - 2 3 4 1 1 1 x x x x x = + + + + +⋅⋅⋅ - This is a geometric series with a = 1 and r = x .
( 29 1 1 x - 1 1 x - x x + 2 x x - 2 x + 2 x 2 3 x x - 3 x 3 x + +⋅⋅⋅ We could generate this same series for with polynomial long division : 1 1 x -
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Friday
May 6, 2016
# Homework Help: algebra
Posted by Diana on Friday, January 22, 2010 at 5:09pm.
the width of a rectangle is 1 ft less than the length, the area is 6 ft^2, what is the length and width? thank you
• algebra - Damon, Friday, January 22, 2010 at 5:23pm
L * (L-1) = 6
L^2 -L -6 = 0
(L-3)(L+2) = 0
take it from there
• algebra - bobpursley, Friday, January 22, 2010 at 5:24pm
L*W=area
L(L-1)=6
solve for L
L^2-L-6=0
(L-3)(L+2)=9
L=3, W=2
• algebra - diana, Saturday, January 23, 2010 at 5:15pm
thanks bob, for all your help
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https://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+s+%3D+300&term2=3c+%2B+1.5s+%3D+825&pl=Substitution
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## Enter Simultaneous Equations:
<-- Equation 1
<-- Equation 2
c + s = 300
3c + 1.5s = 825
##### Check Format
Equation 1 is in the correct format.
##### Check Format
Equation 2 is in the correct format.
3c + 1.5s = 825
##### Subtract 1.5s from both sides to isolate c:
3c + 1.5s - 1.5s = 825 - 1.5s
3c = 825 - 1.5s
##### Now divide by 3:
3c 3
=
825 - 1.5s 3
##### Revised Equation 2:
c = 825 - 1.5s 3
##### Plug Revised Equation 2 value into c:
1(c) + s = 300
1 * ((825 - 1.5s)/3) + s = 300
((825 - 1.5s)/3) + s = 300
##### Multiply equation 1 through by 3
3 * (((825 - 1.5s)/3) + s = 300)
3 * (((825 - 1.5s)/3) + s = 300)
825 - 1.5s + 3s = 900
##### Group like terms:
-1.5s + 3s = 900 - 825
1.5s = 75
##### Divide each side by 1.5
1.5s 1.5
=
75 1.5
s = 75 1.5
s = 50
1c + 1(50) = 300
1c + 50 = 300
1c = 300 - 50
1c = 250
##### Divide each side by 1
1c 1
=
250 1
c = 250 1
c = 250
c = 250 and s = 50
##### How does the Simultaneous Equations Calculator work?
Free Simultaneous Equations Calculator - Solves a system of simultaneous equations with 2 unknowns using the following 3 methods:
1) Substitution Method (Direct Substitution)
2) Elimination Method
3) Cramers Method or Cramers Rule Pick any 3 of the methods to solve the systems of equations 2 equations 2 unknowns
This calculator has 2 inputs.
### What 1 formula is used for the Simultaneous Equations Calculator?
Δ = a * e - b * d
For more math formulas, check out our Formula Dossier
### What 7 concepts are covered in the Simultaneous Equations Calculator?
cramers rule
an explicit formula for the solution of a system of linear equations with as many equations as unknowns
eliminate
to remove, to get rid of or put an end to
equation
a statement declaring two mathematical expressions are equal
simultaneous equations
two or more algebraic equations that share variables
substitute
to put in the place of another. To replace one value with another
unknown
a number or value we do not know
variable
Alphabetic character representing a number
Math is always fascinating! Have you noticed? On the path to discovering the ultimate mysteries of math, we move forward with small steps.
Many such discoveries of mathematical formulas validate the research efforts of those who came before us. If you want to inspire yourself and those around you, try making your own Custom Enamel Badges of achievement for small accomplishments.
These vivid pin badges can be a constant reminder that numbers aren't exactly cold and that they can always be accompanied by fun and hope.
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## GMAT Strategy: Comb That Perm!
Statistics questions can be some of the most exasperating Quant questions on the GMAT. And among those, Combination and Permutation questions may just be some of the worst. The good news: statistics questions are some of the least frequently tested concepts on GMAT Quant. The bad news: you’re still likely to see at least 1 Comb-Perm question come test day. Because higher scorers will likely see a difficult Comb-Perm question, strategies to tackle them are needed. That being said, don’t let those tricky Comb-Perm questions make you want pull your hair out. Often times, there’s an easier way to smooth over those Comb-Perm cowlicks (sorry, I’ll try to reign in the hair jokes).
Harder questions on the GMAT are all about reasoning ability and critical thinking. For difficult Comb-Perm questions, the key is properly adapting the typical method of solving these questions. To wit, we present Comb-Perm: Front-to-Back or Back-to-Front.
Let’s take a look at an example:
If from a certain class of 10 children, 4 children will be chosen to form a committee, how many different committees can be formed where Adam and Miles are not on the same committee?
(A) 56
(B) 84
(C) 112
(D) 182
(E) 336
Now, let’s take a look at the straight forward, or Front-to-Back Method, to solving this question. Essentially, this method requires identifying the number of possible combinations for the committee for EACH of the variations that would satisfy the criteria (namely, that Adam and Miles can’t be on the committee together).
So, the three ways to satisfy the criteria:
- Adam on the committee but not Miles.
- Miles on the committee but not Adam.
- Neither Adam nor Miles on the committee.
Now that we have the three possible variations that satisfy the criteria, we can determine the number of possible combinations for each:
Thus, the total possible number of combinations = 56 + 56 + 70 = 182.
With this approach, it is imperative that you see all three variations. If you overlook one (or more), you get the incorrect answer. For many, it is difficult to visualize the variations that would allow one to calculate the correct answer on such problems.
However, if you have an alternative approach, your chances of correctly answering difficult Comb-Perm problems (and all problems) goes up!
Enter the alternative, or Back-to-Front, method:
- Find the total number of possible combinations.
- Find the number of combinations that VIOLATE the criteria (i.e. Miles and Adam together).
- Subtract the two.
Thus, the total number of possible combinations: 210 – 28 = 182.
So we can see that we get the same answer with different approaches. Most problems have multiple methods by which you can solve. Here, finding the more easily identifiable combinations allow us to more safely (and effectively) calculate the correct answer.
The GMAT, at base, tests higher order thinking skills. One of those skills is having alternative methods at your disposal to work around roadblocks. The methodology we employed here can be used to similar effect on a wide variety of GMAT questions, from complicated Geometry based on shaded or overlapping figures, to quadratic equations, to complex probabilities. The principle: find the quantities you know, see if you can manipulate those to find the solution you need.
Until next time, good luck with all your GMAT preparation!
—————————————————————————————————————————————–
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https://penelopethemovie.com/how-many-cubic-feet-does-1-gallon-of-water-take-up/
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# How many cubic feet does 1 gallon of water take up?
## How many cubic feet does 1 gallon of water take up?
0.1337 cubic feet
* 1 US gallon = 0.1337 cubic feet = 3.79 liters.
## How do you convert cubic feet of liquid to gallons?
To convert a cubic foot measurement to a gallon measurement, multiply the volume by the conversion ratio. The volume in gallons is equal to the cubic feet multiplied by 7.480519.
## How many sq ft is in a cubic ft?
Cubic feet to Square feet Calculator
1 cubic feet = 1 ft2 1 cubic feet
2 cubic feet = 1.5874 ft2 2.8284 cubic feet
3 cubic feet = 2.0801 ft2 5.1962 cubic feet
4 cubic feet = 2.5198 ft2 8 cubic feet
5 cubic feet = 2.924 ft2 11.1803 cubic feet
## How do you measure cubic feet of water?
Formula:
1. Formula:
2. L x W x D. = Cubic Feet.
3. Cubic ft x 7.47. = Gallons.
## How many cubic feet are in a 5 gallon bucket of dirt?
How many cubic feet are in a 5 gallon bucket of dirt? The obsolete Imperial gallon would equal 0.802718 cu ft. Regards, James.
## How many gallons is 600 square feet?
Multiply the cubic feet by the amount of water per cubic foot, which is 7.48 gallons. In this example, you would multiply 7.48 times 600 to get 4,488 gallons of water.
## What is water’s volume?
One milliliter (1 mL) of water has a volume of 1 cubic centimeter (1cm3). Different atoms have different sizes and masses. Atoms on the periodic table are arranged in order according to the number of protons in the nucleus. Even though an atom may be smaller than another atom, it might have more mass.
## How many gallons of water will fit in a square foot?
To provide your lawn with one inch of water takes a little over half a gallon per square foot (0.623 gallon to be more exact). That means that every 10’x10’ area will require over 62 gallons of water.
## How many cubic feet are in one gallon of water?
One cubic foot of water contains 7.48052 gallons. There are 1728 cubic inches in one cubic foot. The number of gallons of water in a cubic inch is 7.48052 gallons divided by 1728 cubic inches, which equals 0.004329005 gallons per cubic inch of water. There are 6,272,640 cubic inches of water on one acre.
## What is the formula for Converting cubic feet to gallons?
Convert cubic feet to gallons with this simple formula: gallons = cubic feet × 7.480507. Converting a cubic foot fluid-volume measurement to a gallon measurement involves multiplying your fluid-volume by the conversion ratio to find the result.
## How many cu ft are in a gallon?
There are 0.13368055555562 cubic feet in a gallon. 1 Gallon is equal to 0.13368055555562 Cubic Feet. 1 gal = 0.13368055555562 ft³. A gallon is a unit of volume measurement used widely in the US and a few other countries of the world.
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### Home > A2C > Chapter 4 > Lesson 4.2.1 > Problem4-72
4-72.
Consider a line with a slope of $3$ and a y-intercept at $\left(0, 2\right)$.
1. Sketch the graph of this line.
2. Write the equation of the line.
Try using a slope-intercept equation.
$y = mx + b$
Use the slope and y-intercept to make the equation.
$y = 3x + 2$
3. Find the initial term and the next three terms of the sequence $t\left(n\right) = 3n − 1$. Plot the terms on a new set of axes next to your graph from part (a) above.
Substitute 1, 2, 3, and 4 for n to find the terms.
$t\left(1\right) = 3 –1 t\left(2\right) = 6 –1 t\left(3\right) = 9 –1 t\left(4\right) = 12 – 1$
Graph the results.
$t\left(1\right) = 2 t\left(2\right) = 5 t\left(3\right) = 8 t\left(4\right) = 11$
4. Explain the similarities and differences between the graphs and equations in parts (a) through (c). Are both continuous?
Notice that both graphs have the same slope.
The graphs are parallel, but they have different intercepts. The first is continuous and the second is discrete.
Use the eTool below to graph the line.
Click the link at right for the full version of the eTool: A2C 4-72 HW eTool
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'
# Search results
Found 632 matches
Area of a triangle (by the one side and the sines of the triangle's angles)
A triangle is a polygon with three edges and three vertices. In a scalene triangle, all sides are unequal and equivalently all angles are unequal. When the ... more
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The circumscribed circle or circumcircle of a triangle is a circle which passes through all the vertices of the triangle. The circumcenter of a triangle ... more
Law of tangents for the triangles
The law of tangents is a statement about the relationship between the tangents of two angles of a triangle and the lengths of the opposing sides.The law of ... more
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The interior perpendicular bisector of a side of a triangle is the segment, falling entirely on and inside the triangle, of the line that perpendicularly ... more
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In geometry, Napoleon’s theorem states that if equilateral triangles are constructed on the sides of any triangle, either all outward, or all inward, ... more
Length of internal bisector of an angle in triangle in relation to the opposite segments
In geometry, bisection is the division of something into two equal or congruent parts, usually by a line, which is then called a bisector. If the internal ... more
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An angle bisector divides the angle into two angles with equal measures. An angle only has one bisector. Each point of an angle bisector is equidistant ... more
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Law of sines is an equation relating the lengths of the sides of any shaped triangle to the sines of its angles. The law of sines can be used to compute ... more
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Altitude of a triangle is a line segment through a vertex and perpendicular to a line containing the base (the opposite side of the triangle). This line ... more
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In geometry, the Euler line is a line determined from any triangle that is not equilateral. It passes through several important points determined from the ... more
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A222030 Primes and quarter-squares. 2
0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 12, 13, 16, 17, 19, 20, 23, 25, 29, 30, 31, 36, 37, 41, 42, 43, 47, 49, 53, 56, 59, 61, 64, 67, 71, 72, 73, 79, 81, 83, 89, 90, 97, 100, 101, 103, 107, 109, 110, 113, 121, 127, 131, 132, 137, 139, 144, 149, 151, 156, 157, 163, 167, 169 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS Union of A002620 and A000040. It appears that there is always a prime between two consecutive quarter squares, if n >= 2. Therefore in a square spiral, or zig-zag, whose vertices are the quarter-squares, it appears that there is always a prime between two consecutive vertices, if n >= 2. Apparently the above comment is equivalent to the Oppermann's conjecture. - Omar E. Pol, Oct 26 2013 Union of A000040 and A000290 and A002378. - Omar E. Pol, Oct 28 2013 LINKS Wikipedia, Oppermann's conjecture FORMULA a(n) ~ n log n. - Charles R Greathouse IV, Mar 04 2013 MATHEMATICA mx = 13; Union[Prime[Range[PrimePi[mx^2]]], Floor[Range[2*mx]^2/4]] (* Alonso del Arte, Mar 03 2013 *) CROSSREFS Cf. A000040, A002620, A000290, A014085, A220492, A220506, A220508, A220516. Sequence in context: A060863 A063934 A326643 * A327261 A337133 A062490 Adjacent sequences: A222027 A222028 A222029 * A222031 A222032 A222033 KEYWORD nonn,easy AUTHOR Omar E. Pol, Feb 05 2013 STATUS approved
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > divdirap GIF version
Theorem divdirap 8469
Description: Distribution of division over addition. (Contributed by Jim Kingdon, 25-Feb-2020.)
Assertion
Ref Expression
divdirap ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → ((𝐴 + 𝐵) / 𝐶) = ((𝐴 / 𝐶) + (𝐵 / 𝐶)))
Proof of Theorem divdirap
StepHypRef Expression
1 simp1 981 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → 𝐴 ∈ ℂ)
2 simp2 982 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → 𝐵 ∈ ℂ)
3 recclap 8451 . . . 4 ((𝐶 ∈ ℂ ∧ 𝐶 # 0) → (1 / 𝐶) ∈ ℂ)
433ad2ant3 1004 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → (1 / 𝐶) ∈ ℂ)
51, 2, 4adddird 7803 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → ((𝐴 + 𝐵) · (1 / 𝐶)) = ((𝐴 · (1 / 𝐶)) + (𝐵 · (1 / 𝐶))))
61, 2addcld 7797 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → (𝐴 + 𝐵) ∈ ℂ)
7 simp3l 1009 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → 𝐶 ∈ ℂ)
8 simp3r 1010 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → 𝐶 # 0)
9 divrecap 8460 . . 3 (((𝐴 + 𝐵) ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 # 0) → ((𝐴 + 𝐵) / 𝐶) = ((𝐴 + 𝐵) · (1 / 𝐶)))
106, 7, 8, 9syl3anc 1216 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → ((𝐴 + 𝐵) / 𝐶) = ((𝐴 + 𝐵) · (1 / 𝐶)))
11 divrecap 8460 . . . 4 ((𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 # 0) → (𝐴 / 𝐶) = (𝐴 · (1 / 𝐶)))
121, 7, 8, 11syl3anc 1216 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → (𝐴 / 𝐶) = (𝐴 · (1 / 𝐶)))
13 divrecap 8460 . . . 4 ((𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐶 # 0) → (𝐵 / 𝐶) = (𝐵 · (1 / 𝐶)))
142, 7, 8, 13syl3anc 1216 . . 3 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → (𝐵 / 𝐶) = (𝐵 · (1 / 𝐶)))
1512, 14oveq12d 5792 . 2 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → ((𝐴 / 𝐶) + (𝐵 / 𝐶)) = ((𝐴 · (1 / 𝐶)) + (𝐵 · (1 / 𝐶))))
165, 10, 153eqtr4d 2182 1 ((𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ (𝐶 ∈ ℂ ∧ 𝐶 # 0)) → ((𝐴 + 𝐵) / 𝐶) = ((𝐴 / 𝐶) + (𝐵 / 𝐶)))
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 103 ∧ w3a 962 = wceq 1331 ∈ wcel 1480 class class class wbr 3929 (class class class)co 5774 ℂcc 7630 0cc0 7632 1c1 7633 + caddc 7635 · cmul 7637 # cap 8355 / cdiv 8444 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 603 ax-in2 604 ax-io 698 ax-5 1423 ax-7 1424 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-10 1483 ax-11 1484 ax-i12 1485 ax-bndl 1486 ax-4 1487 ax-13 1491 ax-14 1492 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 ax-ext 2121 ax-sep 4046 ax-pow 4098 ax-pr 4131 ax-un 4355 ax-setind 4452 ax-cnex 7723 ax-resscn 7724 ax-1cn 7725 ax-1re 7726 ax-icn 7727 ax-addcl 7728 ax-addrcl 7729 ax-mulcl 7730 ax-mulrcl 7731 ax-addcom 7732 ax-mulcom 7733 ax-addass 7734 ax-mulass 7735 ax-distr 7736 ax-i2m1 7737 ax-0lt1 7738 ax-1rid 7739 ax-0id 7740 ax-rnegex 7741 ax-precex 7742 ax-cnre 7743 ax-pre-ltirr 7744 ax-pre-ltwlin 7745 ax-pre-lttrn 7746 ax-pre-apti 7747 ax-pre-ltadd 7748 ax-pre-mulgt0 7749 ax-pre-mulext 7750 This theorem depends on definitions: df-bi 116 df-3an 964 df-tru 1334 df-fal 1337 df-nf 1437 df-sb 1736 df-eu 2002 df-mo 2003 df-clab 2126 df-cleq 2132 df-clel 2135 df-nfc 2270 df-ne 2309 df-nel 2404 df-ral 2421 df-rex 2422 df-reu 2423 df-rmo 2424 df-rab 2425 df-v 2688 df-sbc 2910 df-dif 3073 df-un 3075 df-in 3077 df-ss 3084 df-pw 3512 df-sn 3533 df-pr 3534 df-op 3536 df-uni 3737 df-br 3930 df-opab 3990 df-id 4215 df-po 4218 df-iso 4219 df-xp 4545 df-rel 4546 df-cnv 4547 df-co 4548 df-dm 4549 df-iota 5088 df-fun 5125 df-fv 5131 df-riota 5730 df-ov 5777 df-oprab 5778 df-mpo 5779 df-pnf 7814 df-mnf 7815 df-xr 7816 df-ltxr 7817 df-le 7818 df-sub 7947 df-neg 7948 df-reap 8349 df-ap 8356 df-div 8445 This theorem is referenced by: muldivdirap 8479 divsubdirap 8480 divadddivap 8499 divdirapzi 8536 divdirapi 8541 divdirapd 8601 2halves 8961 halfaddsub 8966 zdivadd 9152 nneoor 9165 2tnp1ge0ge0 10086 flqdiv 10106 crim 10642 efival 11450 divgcdcoprm0 11793 ptolemy 12927
Copyright terms: Public domain W3C validator
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# Tips for Solving Word Problems
Dec 6, 2022
##### SUBSCRIBE
There may be no other math skill that conjures up more feelings of fear and anxiety for parents and students alike than the dreaded word problem, which can show up everywhere from elementary school homework to college admissions exams like the SAT and ACT. Maybe it’s because word problems are math in its realest and perhaps most challenging form, asking you to translate a story into a solvable math question. Whether you’re looking for a way to make homework time less stressful or to bank more points on the SAT or ACT math sections, we’re here to share our top tips for solving these pesky problems.
Solve Step-by-Step
The best way to solve a word problem is to have a plan: a set of steps to follow to identify all of the relevant information, translate the problem into an equation, solve it, and check your answer.
• Understand what the question is asking. Read a word problem carefully so you know exactly what it’s asking you to solve. Restate the problem in your own words and cross out any irrelevant information. Is it a particular type of word problem (e.g. create an expression, formula, distance = rate x time)? Have you solved a similar problem before?
• Identify keywords. Circle relevant numbers and underline keywords, especially words that denote mathematical operations (e.g. sum, difference), actions (e.g. equal, less than), or other information describing the situation (e.g. distance, time, people).
• Visualize. Sometimes the key to unlocking a word problem, especially one that involves multiple steps, is to create a model or diagram that helps visualize it. These visuals organize what you know, what information is missing, and what you need to solve the problem.
• Translate the problem into an equation. Use your circled numbers, key words, and variables representing unknown values to create an equation that answers the question.
• Solve the equation. Make sure to write out each step and calculations, so you can go back and reference the steps you took and check your work. If it’s a multiple choice question, can you pick a number and plug it into an equation in the question or answer choices?
• Check your answer. Your answer should make sense given the information provided in the question. Do you answer what the question asked? Does it seem reasonable? If you’re working on a multiple choice word problem, can you plug in an answer choice?
These are just some general guidelines for solving word problems. If you’re curious about exploring other approaches, there are a lot of problem-solving strategies to choose from, such as the basic Plan-Solve-Check routine and CUBES. Experiment with the options out there to add tools to your problem-solving toolbox and to find the strategies that work best for you.
Practice!
Mastering a math skill is not that different from fine tuning your baseball fielding skills or perfecting a pirouette. Practice is key. To get better at word problems, pick a reasonable goal and make a plan to incorporate practice into your routine. Don’t forget to review your answers, so you can adjust your study plan (e.g. increasing the difficulty level, focusing more on a specific question type, etc.).
Don’t be afraid of a challenge
It’s tempting to stick within your comfort zone. If you’re a parent helping your child with his math homework, that might mean knowing how to help without hand holding too much. As much as you may want to guide your child to the correct answer, the best way for him to learn and build confidence is to give him a chance to do it himself, even if that means making mistakes. Likewise, if you’re a student who wants to overcome your word problem woes, don’t be afraid to attempt questions that may have put you off in the past. You may find that the more you challenge yourself, the more you’ll be able to find creative solutions to previously daunting problems.
How A+ Can Help
Hopefully, these tips have helped demystify the often inscrutable word problem. If you still have questions, or believe you may benefit from some expert support, A+ Test Prep and Tutoring offers several services that can help you with word problems and other important math skills. For example, our SAT and ACT test prep programs provide students with seven sessions devoted to honing the math skills your student needs to reach his or her score goals. You can also get math support through our Foundations program and academic tutoring services. Our staff and tutors are committed to providing individualized instruction that builds the skills and confidence students need to succeed.
At A+ Test Prep and Tutoring, our practices are based on the latest developments in educational theory and research. We have an excellent team of tutors who can help you with standardized testing, executive functioning, or achievement in any other school subject. If you want to find out more about our services, our Client Service Directors Joelle Faucette can be reached at 215-886-9188 or email us at [email protected].
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419 S 19th Street
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https://socratic.org/questions/how-do-you-find-the-equation-of-a-circle-center-at-the-origin-passes-through-10-#281534
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# How do you find the equation of a circle center at the origin; passes through (10, 10)?
Jun 26, 2016
${x}^{2} + {y}^{2} = 10 \sqrt{2}$
#### Explanation:
You can use Pythagoras for this by treating the x-axis and y-axis as if they were sides of a triangle and a line from the origin to the point $\left(10 , 10\right)$ as though it is the hypotenuse.
Let the hypotenuse be the radius (r) of the circle
Then ${r}^{2} = {x}^{2} + {y}^{2}$
$\implies r = \sqrt{{10}^{2} + {10}^{2}} = \sqrt{200}$
$\implies r = \sqrt{2 \times {10}^{2}} = 10 \sqrt{2}$
So the equation of the circle is:
${x}^{2} + {y}^{2} = 10 \sqrt{2}$
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https://www.geeksforgeeks.org/class-9-rd-sharma-solutions-chapter-10-congruent-triangles-exercise-10-4/?ref=lbp
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# Class 9 RD Sharma Solutions – Chapter 10 Congruent Triangles- Exercise 10.4
### Question 1: In figure, It is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
Solution:
Given:
AB = CD and AD = BC.
To prove:
Consider ΔADC and ΔCBA.
AB = CD {Given}
BC = AD {Given}
And AC = AC {Common side}
So,
By SSS congruence criterion, we have
Hence, proved.
### Question 2: In a ΔPQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.
Solution:
Given:
In Δ PQR, PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively
To prove:
LN = MN
Join L and M, M and N, N and L
We have PL = LQ, QM = MR and RN = NP
[Since, L, M and N are mid-points of PQ, QR and RP respectively]
And also PQ = QR
PL = LQ = QM = MR = PN = LR —->(equation 1)
MN || PQ and MN = PQ { Using mid-point theorem}
2
MN = PL = LQ —->(equation 2)
Similarly, we have
LN || QR and LN = QR
2
LN = QM = MR —->(equation 3)
From (equation 1), (equation 2) and (equation 3),
We have
PL = LQ = QM = MR = MN = LN
This implies, LN = MN
Hence, Proved.
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Next
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http://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition/chapter-1-limits-and-their-properties-review-exercises-page-92/63
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## Calculus 10th Edition
Since $f(x)$ is a polynomial, it is continuous for all values of $x.$ $f(1)=2(1)^3-3=-1\to f(1)\lt0.$ $f(2)=2(2)^3-3=13\to f(2)\gt0.$ Since $f(x)$ is continuous over the interval $[1, 2]$ and the sign of $f(x)$ changes over the interval $[1, 2]$, then the Intermediate Value Theorem guarantees at least one zero in the interval.
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https://www.physicsforums.com/threads/calculate-intensity-of-sound-waves-at-3-0-m-from-source.793643/
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Calculate Intensity of Sound Waves at 3.0 m from Source
• capudang
In summary, the conversation discusses finding the intensity and intensity level of sound waves emitted from a small source at a distance of 3.0m. The intensity is found to be 0.71W/m and the intensity level is found to be 118.51. The conversation also mentions difficulty in finding the distance at which the intensity is one-fourth of the original value, with an attempted solution of 7.173. The expert provides a reminder to check arithmetic and calculations.
capudang
Homework Statement
A small source emits sound waves with a power output 80W.
a) find the intensity at a point 3.0m from the source. (0.71W/m)
b) find the intensity level at the same point as part a.(118.51)
c) at what distance would the intensity be one-fourth as much as of that part (a).
i have no problem finding the answer for a and b, but i did have a bit of problem in c. i have tried to answer it which i get r=7.173, but i don't know is it right or wrong. help me for C.
2. Homework Equations c
B=10log(I/Io)
Ia=P/4pie r^2
The Attempt at a Solution
I=p/4pie R^2
i get 7.173 after calculate it.
Check your arithmetic. How much is pi? How much is R? How much is 4 pi R^2?
What is the formula for calculating the intensity of sound waves at a distance of 3.0 m from the source?
The formula for calculating the intensity of sound waves at a distance of 3.0 m from the source is I = P/(4πr²), where I is the intensity, P is the power of the sound source, and r is the distance from the source in meters.
How does the intensity of sound waves change as you move further away from the source?
The intensity of sound waves decreases as you move further away from the source. This is because the sound waves spread out and become less concentrated as they travel through space.
What factors can affect the intensity of sound waves at a distance of 3.0 m from the source?
The intensity of sound waves at a distance of 3.0 m from the source can be affected by factors such as the power of the source, the direction of the sound waves, and any obstructions or barriers in the path of the sound waves.
How is the intensity of sound waves measured at a distance of 3.0 m from the source?
The intensity of sound waves at a distance of 3.0 m from the source is typically measured using a sound level meter. This device measures the sound pressure level in decibels (dB) and can provide an accurate measurement of the intensity of sound waves.
What are some real-world applications of calculating the intensity of sound waves at a distance of 3.0 m from the source?
Calculating the intensity of sound waves at a distance of 3.0 m from the source can be useful in various real-world applications such as noise pollution measurement, sound engineering, and acoustic design for buildings and public spaces. It can also help in understanding the potential impact of loud noises on human health and the environment.
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https://socratic.org/questions/how-do-you-find-the-magnitude-of-mp-given-m-19-4-and-p-4-0
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# How do you find the magnitude of MP given M(-19,4) and P(4,0)?
Jun 1, 2017
MP $\approx 23.35 \left(2 \mathrm{dp}\right)$
#### Explanation:
$M \left(- 19 , 4\right) \mathmr{and} P \left(4 , 0\right)$
MP = = sqrt((x_1-x_2)^2+(y_1-y_2)^2 or
MP = $= \sqrt{{\left(- 19 - 4\right)}^{2} + {\left(4 - 0\right)}^{2}} = \sqrt{529 + 16}$ or
MP = $\sqrt{545} \approx 23.35 \left(2 \mathrm{dp}\right)$ [Ans]
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# math
posted by .
1. A teacher has 10 boys and 12 girls in her grade school class. In how many ways can she select 8 of the children to be in a play if she must select 5 boys and 3 girls.
2. The manager of the various displays in the Acme Fine Clothing Emporium has a window which will hold 4 dummies. He has 8 dummies, each showing a different style outfit, from which to choose. How many different displays can he arrange in his display window? (If the order in which the dummies appear changes, it is regarded by the manager as ‘a different display.)
3. If you buy a \$300,000 house, paying 5% down, and take out a 25 year, fixed rate of 6.5%, loan, then your monthly payments for principal and interest will be
4. Three coins are tossed. Find the probability that all the coins land heads up
5. Two dice are rolled. Find the probability of getting a sum greater than 10.
6. A person rolls two dice. What are the odds in favor of throwing at least an eight?
7. A lottery has one \$2000 prize, two \$1000 prizes, and ten \$500 prizes. Thirty-six hundred tickets are sold at \$4 each. Find the expected value if a person buys two tickets.
8. A card is drawn from a standard 52 card deck. What is the probability that it is either a red ‘ace or a black king?
9. A card is drawn from a standard 52 card deck. What is the probability that it is either a red’ ace or a black king?
10 How many three symbol codes using the digits 2 through 9 are possible if repetitions are allowed? (243 is an example of a three symbol code.)
11. An urn contains 5 red marbles, 4 blue marbles, and 3 green marbles. A marble is selected at random and then, without replacing the first marble, a second marble is selected at random. What is the probability of selecting a green marble and then a red marble?
12. In a classroom, the students are 20 boys and 28 girls. If one student is selected at random, find the probability that the student is a boy.
• Math - homework dumping -
If you include only a couple of problems in each post, along with your attempts of answers, someone may be able to help you.
If you don't do this, I will assume that you don't want help, just answers.
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# Null space for $AA^{T}$ is the same as Null space for $A^{T}$
$A$ is an $n\times m$ matrix and $AA^{T}$ is a symmetric real matrix. Also, we have: $\operatorname{rank}(AA^{T})=r\stackrel{?}{=}\operatorname{rank}(A)$. Let $Q= \begin{Bmatrix} q_1,...,q_{n-r} \end{Bmatrix}$ be a basis for the Null space of $AA^{T}$. i.e. $AA^{T}q_i=0$, show that $A^{T}q_i=0$. I guess one proof can be that the Null space for $A^{T}$ is a subspace for Null space for $AA^{T}$, then the question would be why $\operatorname{rank}(A)=\operatorname{rank}(AA^{T})$?
-
Let $q_i$ be a null vector of $A A^\top$, i.e. $A A^\top q_i =0$, then $0 = q_i^\top A A^\top q_i = \vert\vert A^\top q_i \vert\vert_2$, and thus $q_i$ is also a null vector of $A^\top$. Thus $\mathrm{rank}(A A^\top) = \mathrm{rank}(A^\top) = \mathrm{rank}(A)$.
-
I like above answer; it says only what it needed. However, I tried the following. It takes a slightly different approach as it does not rely on $||A^T q_i$$||$ properties.
Let $X = Null(A)$, then $\forall x \in X, Ax = 0$. Assume that $Y = Null(A^TA)$. $\forall y \in Y, A^TAy = 0$. This implies,
1. $Ay= 0$; or
2. $A^TAy=0$ and $Ay \not= 0$
If case (1) is true than we are done: $y$ is in $X$. Now
$A^TAy = 0 \implies y^TA^TA = 0^T \implies y^TA^T = 0^T$ ($A$ is non-zero) $\implies Ay = 0$. And this is same as case (1). Thus $X = Y$.
Now one can argue about their ranks.
-
?? I don't think that you can conclude from $y^TA^TA=O^T$ that you also have $y^TA^T=0$ based on $A\neq0$ alone. After all, $y^TA^T$ could be in the (left-sided) null space of $A$. – Jyrki Lahtonen Mar 8 '12 at 8:23
@JyrkiLahtonen Thanks for pointing it out. Seems like there is no way without bringing in $y^TA^TAy = 0^T$ and then arguing about $|| . ||$? – Dilawar Mar 8 '12 at 12:09
I have found this proof on the Web. I find it easy to read, and I have not found anything clearer on the waters.
We wish to prove that $null(A^TA) \subset null(A)$ Take $x\in R^n$ such that $A^T Ax=0$. Then, as $x$ is orthogonal to every row vector of $A^T A$, and since $A^T A$ is symmetric, then x is orthogonal to every column vectors of $A^TA$.
Thus, $x^T A^T Ax=0$ $\implies$ $(Ax)^T Ax=0$. This implies $Ax.Ax=0$ and $Ax=0$. Which finishes the proof.
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# Name ___________________________
Date _________
Solving Systems of Equations by Graphing Using a Graphing Calculator
Graphing Calculator Directions To graph a line: 1) Make sure the line is written in slope-intercept form. 2) Press and type in your first equation. Note: if your slope is a fraction, is typed in:
you must put it in parentheses. Example,
3) Press 4) Press press
and type in your second equation to view your graphs. If you need to change the scale of the axes, .
To find a point of intersection: 1) Follow this sequence:
1) Use a graphing calculator to find the solution (point of intersection).
2 x9 3 4 y x3 3 y
a)
b)
y 3x 6 y 5 x 2
c)
3 y x 8 2 11 y x6 2
d)
1 y x 8 5 11 y x2 5
2) Graph the system of equations:
4 x5 3 4 y x 1 3 y
a) Describe the graph. What do you notice about the slopes and y-intercepts of these equations?
b) How many solutions does this system have? Explain.
3) The graphs of the following 3 lines contain the sides of a triangle. Find the coordinates of the vertices of the triangle. y 2x 1 y 4 x 7 1 y x2 2
4)
System of Equations
Table of Values
Use the table function on your calculator to fill in the table.
Number of Months Total Cost Option 1 Total Cost Option 2
Option 1: Nextel has an initial fee of \$120 and charges \$40 per month Equation:
0 6 12 18
Option 2: AT&T has an initial fee of \$300 and charges \$25 per month Equation:
24 30 36 42
Graph
Use the graph function on your calculator to graph the equations. Write the window settings you used here. Xmin = Xmax = Xscale = Ymin = Ymax = Yscale = Xres = 1
Analysis
1) What is the point of intersection of the two equations?
2) What does the point mean in the context of this situation?
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http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?s=c2f911b5bf0d8db7d40109bf9d75f15f&p=9538496
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Actuarial Outpost Spring 2019 LTAM Progress Thread
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#81
01-30-2019, 12:08 PM
Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,023
Quote:
Originally Posted by Chappy I don't fully understand this problem. It is from 15th Edition ASM MLC manual. 11.13 Tx is the future lifetime random variable for (x). An insurance pays 1000 at the moment of death if the integral part of Tx is odd. Otherwise no benefit is paid. mu=.04 delta=.02. What is the expected present value. I got the geometric series part, but missed the part about the payment being equal to the difference between a k-year and a k+1 year deferred insurance. Is this because the payment is made at the moment of death? If so, I'm still fuzzy on how that works.
Let ODD be a policy that pays if the integer part of Tx is odd and EVEN be a policy that pays if it is even. Then if you have both, you have a regular policy.
After one year, if you survive, ODD and EVEN (effectively) reverse roles.
To wit: let p = probability of death during year 1 (I'll let you do the time-value-of-money discounting.)
EVEN = p(death benefit) + (1-p)(ODD)
ODD = p(0) + (1-p)EVEN
regular policy = EVEN + ODD
Take it from there. (This works because the forces of mortality and interest are constant.)
__________________
"What do you mean I don't have the prerequisites for this class? I've failed it twice before!"
"I think that probably clarifies things pretty good by itself."
#82
01-30-2019, 01:48 PM
Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,702
Quote:
Originally Posted by Chappy I don't fully understand this problem. It is from 15th Edition ASM MLC manual. 11.13 Tx is the future lifetime random variable for (x). An insurance pays 1000 at the moment of death if the integral part of Tx is odd. Otherwise no benefit is paid. mu=.04 delta=.02. What is the expected present value. I got the geometric series part, but missed the part about the payment being equal to the difference between a k-year and a k+1 year deferred insurance. Is this because the payment is made at the moment of death? If so, I'm still fuzzy on how that works.
Regarding the deferred minus deferred. Notice that a 1-yr deferred minus a 2-year deferred only pays in the second year, when K=1. Then a 3-year-deferred minus a 4-year-deferred only pays in the 4th year, when K=3. Et cetera. So you can write the APV as that for
(1-yr-deferred) - (2-yr-deferred) + (3-yr-deferred) - (4-yr-deferred) + (5-yr-deferred) - (6-yr-deferred) + and so on.
Since, under your assumptions, you can easily evaluate the APV of a k-yr-deferred whole-life moment of death insurance on (x) as (v^k)(e^{-k mu})[mu / (mu + delta)], you end up summing an alternating geometric series for the final answer.
__________________
Jim Daniel
Jim Daniel's Actuarial Seminars
www.actuarialseminars.com
[email protected]
#83
01-31-2019, 09:02 AM
RiskyBusiness7 SOA Join Date: Apr 2018 Posts: 18
still no word on what the official pass rate was for the fall sitting, right?
__________________
Prelims: P | FM | C | MFE | LTAM | PA | VEEs | FAP
#84
01-31-2019, 09:09 AM
Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,073
Quote:
Originally Posted by RiskyBusiness7 still no word on what the official pass rate was for the fall sitting, right?
753 of 2023 candidates passed (37.2%)
753 of 1641 effective candidates passed (45.9%). Effective candidates are those who scored above 0, probably the candidates who got at least 10 multiple choice questions correct.
#85
02-05-2019, 09:30 PM
karenyoung SOA Join Date: Aug 2016 Location: Irvine, California Studying for FAP Final Assessment College: UCLA Posts: 13
Hi everyone! I'm willing to give my gently-used LTAM 1st edition 2nd printing ASM manual for free to anybody willing to pick it up. Work is ramping up and it's hard for me to find time/motivation to ship it. Located in Irvine, California. Message me!
__________________
Exams: P | FM | IFM | LTAM | PA | STAM | SRM Credit
VEE: Econ | Finance | Stats
FAP: 1 | 2 | 3 | 4 | 5 | IA | 6 | 7 | 8 | FA (waiting for results)
APC - April 2019
#86
02-07-2019, 03:22 PM
birdman332 SOA Join Date: Jun 2017 College: Hanover College Posts: 7
Quote:
Originally Posted by Gandalf Have you commented to the SOA about the difficulty of reading the tables? Size / color / two-sided? That might help future candidates.
Where can you comment/complain about something like this?
__________________
Prelims: P FM MFE SRM Credit LTAM STAM PA
VEE: Corporate Finance Applied Stats Econ
#87
02-07-2019, 03:30 PM
Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,073
Quote:
Originally Posted by birdman332 Where can you comment/complain about something like this?
email [email protected]
#88
02-12-2019, 11:36 PM
james2actuary Member SOA Join Date: Oct 2017 Posts: 80
Hey guys, Good to have this thread for discussion.
Can anyone analyse me where I stand rightnow? I am scared that I might not make into next attempt.
1. I did survival from ASM including all exercise questions to built strong base, took me 20 days.
2. I moved to CA from Inusrance section and now I stand at Reserves section, markov chain to start in 2 to 3 days.
3. I am doing only assignments + Examples from CA manual without looking at them and doing well at them, may be 80%. It seems they are easy one and exam is gonna be tough. Although, first 10 chapters from ASM helped me to build strong foundation of survival probabilities which seems to be key for this exam as of now.
4. I am planning to complete manual before 10 march so I take 50 days for practice questions as I am not using Adapt rightnow. I will need alot of practice.
Any guide that I should follow to prepare in more efficient way?
#89
02-13-2019, 01:14 PM
Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,702
Seminar hotel cutoff
The hotel where I hold my 14 - 20 March face-to-face LTAM seminar has a cutoff of Friday 15 February for the reduced rate for a sleeping room. If you are planning to register for my seminar, you an save a little money by registering for the seminar today or tomorrow.
__________________
Jim Daniel
Jim Daniel's Actuarial Seminars
www.actuarialseminars.com
[email protected]
#90
02-15-2019, 01:41 AM
tkt Member CAS SOA Join Date: Jun 2011 Location: Des Moines College: Drake University Posts: 509
Quote:
Originally Posted by james2actuary Hey guys, Good to have this thread for discussion. Can anyone analyse me where I stand rightnow? I am scared that I might not make into next attempt. 1. I did survival from ASM including all exercise questions to built strong base, took me 20 days. 2. I moved to CA from Inusrance section and now I stand at Reserves section, markov chain to start in 2 to 3 days. 3. I am doing only assignments + Examples from CA manual without looking at them and doing well at them, may be 80%. It seems they are easy one and exam is gonna be tough. Although, first 10 chapters from ASM helped me to build strong foundation of survival probabilities which seems to be key for this exam as of now. 4. I am planning to complete manual before 10 march so I take 50 days for practice questions as I am not using Adapt rightnow. I will need alot of practice. Any guide that I should follow to prepare in more efficient way?
I think you have a solid plan. Most of the assignment questions are taken from the sample or past year questions; it's good that you are doing well at them! For this sitting, we also included written-answer questions in appropriate sections so that will help expose to written-answer questions sooner. Target to spend the last 45 days just on practicing and reviewing. Pay more attention to the written-answer questions, but also don't neglect the multiple-choice questions as well. Focus on the basic principles. When in doubt, utilize our discussion forums.
All the best!
__________________
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Product Manager, Actuarial
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## Question
### Comprehension
Directions: Study the chart carefully and the questions that follow.
The pie charts show the distribution of the total number of cookies sold by companies A, B, C, and D and the distribution of chocolate flavor cookies sold by companies A, B, C, and D respectively. Also, Total Cookies = Chocolate Cookies + Butterscotch flavor cookies.
Number of Butterscotch flavor cookies sold by A = 110
Number of Butterscotch flavor cookies sold by B = 30
# The total number of Butterscotch cookies sold by all companies together is approximate how much percentage more or less than the total number of chocolate cookies sold by all the companies together?
This question was previously asked in
IBPS PO Mains Memory Based Paper 4th Feb 2021
View all IBPS PO Papers >
1. 50%
2. 33.33%
3. 60%
4. 30%
5. None of these
Option 2 : 33.33%
## Detailed Solution
From the given pie charts, we have:
(N + 5) + N + (N + 20) + 2N = 100
⇒ N = 15
Also, M + 20 + (M + 10) + 40 = 100
⇒ M = 15
Let the total number of cookies sold be x
Also, total number of cookies sold by A = (20/100) × x = 0.2x
Also, the number of Butterscotch cookies sold by A = 110
So, the number of chocolate cookies sold by A = 0.2x – 110
Hence, 0.2x – 110 = (15/100) × total number of chocolate cookies ----(i)
Similarly,
Also, total number of cookies sold by B = (15/100) × x = 0.15x
Also, the number of Butterscotch cookies sold by B = 30
So, the number of chocolate cookies sold by B = 0.15x – 30
Hence, 0.15x – 30 = (20/100) × total number of chocolate cookies ----(ii)
Multiply equation (i) by 3,
⇒ 0.6x – 330 = (45/100) × total number of chocolate cookies ----(iii)
and equation (ii) by 4:
⇒ 0.6x – 120 = (80/100) × total number of chocolate cookies ----(iv)
Subtract equation (iii) from equation (iv)
⇒ -120 + 330 = (0.8 – 0.45) × total number of chocolate cookies
⇒ Total number of chocolate cookies = 210/0.35 = 600
Also, from the pie chart of total cookies,
200 = (20/100) × total cookies
| 621 | 2,028 |
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# Tag Archives: 3DM
## Numerical 3-Dimensional Matching
SP15 is 3-Parttion.
The next problem is one of G&J’s “unpublished results” problems. I tried figuring out an elegant way to doing it, but couldn’t make it happen.
The problem: Numerical 3-Dimensional Matching. This is problem SP16 in the appendix.
The description: Given three sets W, X, and Y, each containing the same amount (m) of elements with positive “sizes” and a positive bound B. Can we create m sets A1 through Am (containing 3 elements each), such that:
• Each Ai has exactly one element from W, X, and Y
• The sum of the sizes of the elements in each Ai is exactly B
• Each element in W, X and Y is in some Ai
Example: Suppose we have the following sets:
• W has elements with sizes {1,2,3,4}
• X has elements with sizes {12,11,7,5}
• Y has elements with sizes {1,1,4,5} (the ability to allow repeat numbers is why we define the sets as elements with sizes rather than sets of integers)
If B=14, then the partition where A1 is the first element in W, X, and Y, A2 is the second element in W, X, and Y, and so on gives each Ai set a sum of 14. Obviously, we don’t need to choose corresponding elements from W, X, and Y to form the sets (for example, rearranging the elements in X to be in increasing order doesn’t change whether the problem can be solved, just the exact composition of the Ai sets)
Reduction: I tried doing a reduction using 3-partition, but got stuck (I’ll show it below, in case you want to try to fix it). G&J refer you to Theorem 4.4 in the book, which is the proof that 3-partition itself is NP-Complete.
We can follow that along and do similar steps to our problem:
• Theorem 4.3 shows how to turn 3DM into 4-partition (a problem like 3-partition but each set in the solution has 4 elements instead of 3). Since the sets that are created in the 4-partition solution come from 4 different places (page 98 calls then a “ui, a wi[·], an xj[·], and a yk[·]). Since these partitioned sets all add to the same total (B) and come from 4 disjoint parent sets, we can see how we could do basically the same reduction and show that the “numerical 4-dimensional matching problem” is NP-Complete.
• Theorem 4,4 shows how to turn a 4-parition problem into a 3-partition problem. The idea is to add enough “pairing” and “filler” elements to the 3-partition instance to make any 4-partition set be split into two 3-partition sets, each consisting of 2 elements from the 4-parittion, plus the “pairing” element of one of the 2 elements chosen. We can do something similar converting numerical 4-dimensional matching to numerical 3-dimentional matching. (The difference is that we are given specifically which sets the elements are coming from) So, if we’re given W, X, Y, and Z in our numerical 4DM instance, we construct W’ to be elements from W and Y, X’ to be elements from X and Z, and Y’ to be the pairing elements of pairs from W’ and X’. We then need to add enough filler elements to our 3 sets in a similar way to the 3-partition proof (again, the difference is that we have to specifically assign them to W’, X’, or Y’. But that can be determined by how the 3-partition proof allocates the items)
Difficulty: If you have gone over the 3-partition reduction, this is probably a 6. Lots of tricky math but doable in a (hard) homework. But keep in mind you’re tacking it on to the difficulty 8 of understanding the 3-partition reduction in the first place.
My reduction I couldn’t get to work: I really want there to be an easier way to do this. I tried reducing from 3-partition directly because the problems are so similar. Here is where I got to:
We’re given a 3-parititon instance S, and an integer B. Our goal is to split S into sets of size 3 that all add up to B.
So, let’s use S to create 3 sets in our numerical 3DM instance:
• W has all of the elements in S
• X has all of the elements in S, but the sizes are each increased by 10B.
• Y has all of the elements in S, but the sizes are each increased by 100B.
This would make B’ be 111B.
S has a 3-parititon, then for each set {si, sj, sk}, we take the three sets {wi, xj, yk}, {wj, xk, yi}, and {wk, xi, yj} This will solve the numerical 3DM instance.
My problem comes showing this in the other direction. If we have a numerical 3DM solution, we can only construct the 3-partition instance if the sets in the 3DM solution arrange themselves nicely like they do above. I need to show that if the 3DM solution has {wi, xj, yk}, then the set in the 3DM solution that contains wj also contains xi (or xk) and yk (or yi). I think you can get there by using the rules about how the bounds of the elements in the 3-partition instance work, but the work you need to do to show that it’s true makes this way of doing things no longer “easier” than the Theorem 4.4 proof I sketched above, so I gave up on it.
I still wish there was a more elegant way to transform this problem, though.
## Partition
First, an administrative note. I wanted to call this site “Annotated NP-Complete Problems”, because the idea is that I’m going through the Garey&Johnson book and adding notes to each problem talking about how to do the reduction and how applicable it is for student assignments. But that name is sort of taken already, and I don’t want to step on any toes or cause any confusion. So I figured that I’d change the title now, before anyone finds out about the site.
And as I’ve been writing, these notes feel more like “discussions” than “short annotations” anyway, so I think this is a better title.
This is the last of the “core six” problems in the G&J book, as defined in Chapter 3. There are several other problems presented in that chapter, with proofs, but since the point of this exercise is to get to the problems without proofs, I’m going to skip over them, and come back to them only if I need to (because they’re the basis for a future reduction, for example).
The problem: Partition (PART)
The definition: Given a set of integers A, can I fid a subset A’⊆A such that the sum of all of the elements in A’ is exactly half the sum of all of the elements in A?
(Alternately, given a set of integers A, can I split all of the elements of A into two subsets B and C, such that the sum of all of the elements in B is equal to the sum of all of the elements in C?)
(Alternately (this is the G&J definition), given any set A, where each element a∈A has a size s(z) that is a positive integer, can we find a subset A’ of A where the sum of the sizes of everything in A’ is exactly equal to the sum of the sizes of everything in A-A’?)
Example: Suppose A was the set {1,2,3,4,6}. Then A’ could be {1,3,4}, forming a partition (A’ and A-A’ both sum to 8). If instead, A was the set {1,2,3,4,5}, then no possible partition exists. This may seem like a silly case (any set A where the sum of the elements is odd has no partition), but even if the sum of the elements of A is even, it’s possible that no partition exists- for example if A is {2,4,100}.
Note: Partition is one of my favorite NP-Complete problems, because the description is so easy, and it seems so simple. It’s probably my go-to example if I want to explain this stuff to non-technical people in under a minute- pretend that the elements of A are weights, and the value of each element is the weight in ounces. The partition problem asks “given this group of weights, and a scale, can you make the scale balance using all of the given weights?”. It’s pretty surprising to most people that the only known way to answer that question basically boils down to “try all possible combinations of arranging things on the scale”
The reduction: From 3DM. G&J provide the reduction on pages 61-62. The basic idea is to have one element in A for each element in M, and to represent the elements in A as binary numbers that have 1’s in positions corresponding to which element from W, X, and Y we get the triple from. The number is set up so that we have a maximum possible value of the sum of all of these elements in A. They then add two “extra” elements to A so that each side of the partition (elements of M that make a matching, and everything else) will add up to the same value
Difficulty: 7. The idea of using binary numbers to keep track of “positions” in a list is tricky, but comes up in lots of places. If students have seen that idea before, this becomes a hard but doable problem. If students haven’t seen that idea before, I’d make this a 9.
## Exact Cover by 3-Sets
This is not one of the “core six”, but it is used a lot in reductions, so it’s worth including since it builds right off of 3DM
Also, I think I will include examples for lots of these problems. Lots of times I have trouble parsing the problem description, so creating a concrete example is helpful.
The problem: Exact Cover by 3-Sets (X3C)
The definition: Given a set X, with |X| = 3q (so, the size of X is a multiple of 3), and a collection C of 3-element subsets of X. Can we find a subset C’ of C where every element of X occurs in exactly one member of C’? (So, C’ is an “exact cover” of X).
Example: Suppose X was {1,2,3,4,5,6}
If C was {{1,2,3},{2,3,4},{1,2,5},{2,5,6}, {1,5,6}} then we could choose C’ to be {{2,3,4},{1,5,6}} as an exact cover because each element in X appears exactly once.
If instead, C was {{1,2,3},{2,4,5},{2,5,6}}, then any C’ we choose will not be an exact cover (we need all 3 subsets to cover all elements in X at least once, but then the element 2 appears three times).
Note that if we do have an exact cover, C’ will contain exactly q elements.
The proof:G&J prove this “by restriction” which basically means that they show how X3C is a more general version of 3DM . If you view an instance of 3DM as a special case of X3C by letting XX3C = W∪X3DM∪Y and C = all q3 triples taking one element from W, one element from X3DM, and one element from Y), then the C’ you get as a solution to X3C is also a matching for 3DM.
(Note that lots of these reductions will be between problems that use the same symbols in both problems. I’ll do my best to disambiguate by using subscripts to mark where the common letter comes from. So, in this case X3DM is the set X that we get from the 3DM problem (one of the 3 input sets), and XX3C is the set we build for the X3C problem (the set we need to cover). Hopefully that doesn’t make things more confusing)
(Also note that like many (most?) Computer Science people, I’m a big fan of nested parentheses. I’m sure all of you can follow along with that. (Right?))
Personally, I don’t like proofs by restriction as a way to teach this stuff to students. It’s very easy to mess up the “this is a special case” argument- you get incorrect arguments like “this is just a special case of SAT where we return true wherever there’s a cover and false when we don’t!”. Also it feels like you’re going backwards from a real reduction, and since getting the direction wrong is probably the #1 most common issue students have in doing reductions, anything that makes their job harder isn’t a great idea.
If I taught this in a class, I’d make them do a proper reduction out of it- start with an instance of 3DM (W, X3DM, and Y), and build an instance of X3C (creating XX3C and C), and going from there.
Difficulty: 3. It’s only not a 2 because I, at least, have trouble understanding what the X3C and 3DM problems are asking. It’s not as straightforward to explain as many other problems
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7-9
a) First derivatives: the number of components of a symmetric 4×4 matrix is 10. There are therefore 10×4 = 40 first derivatives of the metric and 4×10 = 40 second derivtives 2xa/x¢bx¢g since these are symmetric in b and g.
b) Second derivatives: 2g¢ab/x¢gx"d is symmetric in both a and b and in g and d. There are therefore 10×10 = 100 conditions. The number of third derivatives 3xa/x¢bx¢gx¢d is 4×(the number of symmetric combinations of 3 indices) = 80. This leaves 20 second derivatives that cannot be made to vanish.
7-10
a) Substitute the expressions for X, Y, Z, W into X2 + y2 + Z2 + W2 and find, identically, R2. This shows that every value of c,q,f corresponds to a point on the 3-sphere. (It does not show that the coordinates cover all of the three-sphere without singularity - which is, in fact, not so.)
b) To find the metric, work out
dX = R [( sinc sinq cosf) df + . . . ]
and substitute into dS2 = dX2 + dY2 + dZ2 + dW2, with the result
dS2 = R2 [ dc2 + sin2c (dq2 + sin2q df2 )].
8-3
8-8 Note that the problem gives you a Killing vector (does not ask you to show that it is a Killing vector), so you can start knowing that /f in spherical coordinates is a Killing vector. There are then many ways to find its rectangular components, for example by relating the Killing vector to angular momentum operators. One of the more straightforward approaches uses the coordinate transformation equations, x = r sinq sinf etc. and the "chain rule", /f = (xi/f) /xi = -y /x + x /y. Cyclic permutation then gives the other two Killing vectors.
8-12
a) The distance along an x-constant line to a point on the x-axis is ò0dy/y which diverges. Similarly the distance aloing any other curve to a point on the x-axis will diverge.
b) The geodesics can be obtained from the variational principle
d ó
õ
y-1 æ
Ö
.x 2 + .y 2
dt = 0 or d ó
õ
y-2 æ
è
.
x
2
+ .
y
2
ö
ø
dS = 0
where · means d/dt resp d/dS. The first version gives the "geodesic equations" only if supplemented by the normalization, dS/dt = 1. One finds
ddS æç è dxy2 dS ö÷ ø = 0 (1)
d2ydS2 = - 1y æç è dxdS ö÷ ø 2 + 1y æç è dydS ö÷ ø 2 (2)
c) One has two integrals of the motion, Eq (1) above implying
dx/dS = y2/r, r = const, (1a)
and the normalization,
1y2 éê ë æç è dxdS ö÷ ø 2 + æç è dydS ö÷ ø 2 ùú û = 1 (3)
We compute
dxdy = dx/dSdy/dS = y(r2 - y2)½
which integrates to give (x - x0)2 + y2 = r2.
d) Use 1(a) in (3) to eliminate dx/dS, integrate:
y(S) = rcosh(S) , then from (1) x(S) = r tanh(S).
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## Official Guide Explanation:Data Sufficiency #13
Background
This is just one of hundreds of free explanations I've created to the quantitative questions in The Official Guide for GMAT Review (12th ed.). Click the links on the question number, difficulty level, and categories to find explanations for other problems.
These are the same explanations that are featured in my "Guides to the Official Guide" PDF booklets. However, because of the limitations of HTML and cross-browser compatibility, some mathematical concepts, such as fractions and roots, do not display as clearly online.
Question: 13
Page: 273
Difficulty: 3 (Very Easy)
Category 1: Algebra > Linear Equations-One Unk >
Category 2: Word Problems > Other >
Explanation: First, simplify the information given. The first three minutes, at 42 cents per minute, costs a total of \$1.26. The total cost of the call can be given as:
1.26 + 0.18(a)
where a is the number of additional minutes. Alternatively, since the number of "additional minutes" is 3 minutes less than the length of the call, we can use this:
1.26 + 0.18(m - 3)
where m is the total number of minutes of the call.
Statement (1) is sufficient. We know that the first three minutes cost \$1.26. If that's 36 cents less than the cost of the remainder of the call, the remainder of the call cost:
\$1.26 + \$0.36 = \$1.62.
We can divide that number by 18 cents to determine the number of additional minutes.
Statement (2) is also sufficient. Using the formula we derived from the question:
1.26 + 0.18(m - 3) = 2.88
Here, we have only one variable in a linear equation, so we have enough information to solve. At this point you can conclude that (D) is the correct answer. To solve it for good measure:
1.26 + 0.18(m - 3) = 2.88
0.18(m - 3) = 1.62
m - 3 = 9
m = 12
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# If r – s = 3p , is p an integer? (1) r is divisible by 735
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If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
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21 Jan 2012, 15:46
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If r – s = 3p , is p an integer?
(1) r is divisible by 735
(2) r + s is divisible by 3
OA is C. I am struggling to find how. This is how I approaching the question. Can someone please help?
Considering Questions Stem
We have to find whether r-s/3 as p is an integer?
Considering statement 1
r is a factor of 735. That means r is divisible by all factors of 735 and 3 is a factor of 735 [Because 7+3+5=15]. But as the statement doesn't mention anything about s, it's INSUFFICIENT to answer the question.
Considering statement 2
r+s is divisible by 3
Case 1
r=6 s = 3 then r+s and r-s both divisible by 3.
Case 2
r=7 and s =5 then r+s is divisible by 3 and r-s is NOT divisible by 3. Therefore this statement alone is INSUFFICIENT.
Now combining the two statements : I am struggling after this?
[Reveal] Spoiler: OA
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Re: Is p an integer? [#permalink]
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21 Jan 2012, 16:01
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enigma123 wrote:
If r – s = 3p, is p an integer?
(1) r is divisible by 735
(2) r + s is divisible by 3
If r – s = 3p , is p an integer?
Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.
(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.
(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.
(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.
Below might help to understand this concept better.
If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.
If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.
If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.
Hope it's clear.
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Re: If r–s =3p, is p an integer? (1) r is divisible by 735 [#permalink]
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21 Jan 2012, 16:15
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Classic explanation. Thanks very much.
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Re: Is p an integer? [#permalink]
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25 May 2012, 01:18
Bunuel wrote:
enigma123 wrote:
If r – s = 3p, is p an integer?
(1) r is divisible by 735
(2) r + s is divisible by 3
If r – s = 3p , is p an integer?
Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.
(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.
(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.
(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.
Below might help to understand this concept better.
If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.
If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.
If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.
Hope it's clear.
Hi,
I think B should be sufficient. Here is how:
We are given r+s is divisible by 3.
r+s=3a (where a is any integer) -----(1)
r-s+2s = 3a
3p + 2s = 3a (since we are given that r-s=3p)
2s = 3(a-p)
since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.
similarly if we replace r by (-r+2r), we will get that r is also divisible by 3.
hence, if both r and s are divisible by 3, therefore p is an integer.
please correct me if i am wrong somewhere.
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Re: Is p an integer? [#permalink]
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25 May 2012, 01:32
Expert's post
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kunalbh19 wrote:
Bunuel wrote:
enigma123 wrote:
If r – s = 3p, is p an integer?
(1) r is divisible by 735
(2) r + s is divisible by 3
If r – s = 3p , is p an integer?
Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.
(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.
(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.
(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.
Below might help to understand this concept better.
If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.
If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.
If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.
Hope it's clear.
Hi,
I think B should be sufficient. Here is how:
We are given r+s is divisible by 3.
r+s=3a (where a is any integer) -----(1)
r-s+2s = 3a
3p + 2s = 3a (since we are given that r-s=3p)
2s = 3(a-p)
since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.
similarly if we replace r by (-r+2r), we will get that r is also divisible by 3.
hence, if both r and s are divisible by 3, therefore p is an integer.
please correct me if i am wrong somewhere.
OA for this question is C, not B. OA is given in the initial post under the spoiler.
Next, for the second statement there are two examples given in my post which give different answer to the question whether p is an integer.
(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
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06 Nov 2012, 05:23
The above mentioned solution is valid for integers .... What if s is a fraction say 3/4 ... The answer should be E in that case..
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
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06 Nov 2012, 05:32
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himanshuhpr wrote:
The above mentioned solution is valid for integers .... What if s is a fraction say 3/4 ... The answer should be E in that case..
(1) says that r is divisible by 735, which implies that r is an integer. Next, (2) says that r + s is divisible by 3, which implies that r +s is an integer and since r is an integer then so is s. Thus, when we consider the two statements together we know that both r and s are integers.
On GMAT when we are told that $$a$$ is divisible by $$b$$ (or which is the same: "$$a$$ is multiple of $$b$$", or "$$b$$ is a factor of $$a$$"), we can say that:
1. $$a$$ is an integer;
2. $$b$$ is an integer;
3. $$\frac{a}{b}=integer$$.
So the terms "divisible", "multiple", "factor" ("divisor") are used only about integers (at least on GMAT).
Hope it helps.
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
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06 Nov 2012, 05:36
Thank you Bunuel , it's a new and very important thing which I have come to know.
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Re: Is p an integer? [#permalink]
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06 May 2013, 13:32
kunalbh19 wrote:
Bunuel wrote:
enigma123 wrote:
If r – s = 3p, is p an integer?
(1) r is divisible by 735
(2) r + s is divisible by 3
If r – s = 3p , is p an integer?
Question basically asks whether r-s is a multiple of 3 (..., -6, -3, 0, 3, 6, ...), because if it is then p would be an integer.
(1) r is divisible by 735 --> r is a multiple of 3 (as the sum of the digits of 735 is divisible by 3), though this statement is insufficient as no info about s.
(2) r + s is divisible by 3 --> r + s is a multiple of 3. Now, if r=2 and s=1 then r-s=1 and the answer is NO but if r=s=0 then r-s=0 and the answer is YES. Not sufficient.
(1)+(2) From (1) r={multiple of 3} then as from (2) r+s={multiple of 3}+s={multiple of 3} then s is also a multiple of 3 --> r-s={multiple of 3}-{multiple of 3}={multiple of 3}. Sufficient.
Below might help to understand this concept better.
If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.
If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.
If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.
Hope it's clear.
Hi,
I think B should be sufficient. Here is how:
We are given r+s is divisible by 3.
r+s=3a (where a is any integer) -----(1)
r-s+2s = 3a
3p + 2s = 3a (since we are given that r-s=3p)
2s = 3(a-p)
since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.
similarly if we replace r by (-r+2r), we will get that r is also divisible by 3.
hence, if both r and s are divisible by 3, therefore p is an integer.
please correct me if i am wrong somewhere.
I did the same mistake as you and assumed statement 2 by itself would suffice .
After pondering on it for a while , figured where I went wrong.
r+s=3a (where a is any integer) -----(1)
r-s+2s = 3a
3p + 2s = 3a (since we are given that r-s=3p)
2s = 3(a-p)
since 2 and 3 are both primes, we can conclude that a-p is a multiple of 2 and s is a multiple of 3.
Let me try to explain why statement 2 can be insufficient.
2s=3(a-p) . you got it right till here.
But you cannot conclude with just statement 2 , that a-p is a multiple of 2 . Cos all we know untill this point is "a" is an integer. We do not know weather S or/and P is an integer .
To illustrate what I mean above, lemme give you an example -
consider 2s=3(a-p) ===> 2 (0.3) = 3(0.2) , where s=0.3 and a-p=0.2 , a is an integer , lets say 2 , in which case P would be 1.8 . Hence we can prove that p is not an integer.
similarly , we can prove otherwise that P is an integer.
Now if you consider statement 1 , which says S is an integer , we can conclude from the equation 2S=3(a-p) , that P is an integer.
cos 2*integer = 3 * integer , i.e a-p SHOULD be an integer , since a is an integer and S is an integer .
Hope that helps
Jyothi
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
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20 Feb 2014, 10:33
Terrific Bunuel, the way you have explained the solution is just amazing. +1 kudos for the excellent concept explained
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
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01 Jul 2014, 03:42
Clearly p=(r-s)/3
Statement 1
r is divisible by 735,so p=(735x-s)/3=245x-s/3……..not sufficient
Statement 2
r+s=3y↪↪p=(3y-s-s)/3=(3y-2s)/3=y-2s/3…….not sufficient
Combining (1)+(2)
p=(735x-s)/3
p=(735x-3y-r)/3
now r is a multiple of 3,so it is a multiple of 3
therefore,p=(735x-3y-3z)/3=integer…..sufficient
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Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink]
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13 Apr 2017, 01:47
1] r=735k (k is any integer)
We can just say that r is an integer. Nothing known about s.
So not sufficient.
2] r+s=3p
Again, we can just say that r+s is an integer. nothing known independently of r and s.
For example r=7/3, s=2/3, r+s=3 but r-s=5/3
So not sufficient.
Combined, r is an integer and s is also an integer.
Sufficient. Ans C.
Re: If r – s = 3p , is p an integer? (1) r is divisible by 735 [#permalink] 13 Apr 2017, 01:47
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# If r – s = 3p , is p an integer? (1) r is divisible by 735
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## Binary Search & Newton-Raphson Root Finding
### The Two Methods and Their Uses
Binary Search is a technique found in the field of Computer Science that is used to find, or search for, an element in a sorted list/array. It is incredibly efficient, with a complexity of $$O(log\space n)$$, also known as logarithmic run time.
Newton-Raphson is a method for finding the square root of a number, similar to the iterative method of Heron of Alexandria. It involves making an initial guess of the square root, then refining that guess through a specific formula.
So what do these two methods, one an efficient technique to find an element in a list-like structure, and the other a root calculator, have in common?
### Binary Search in-depth (sort of)
Let's first present the Binary Search method, a.k.a the Binary Chop. The technique is actually quite simple to execute. Take a list represented as a line, the list contains the elements from 0 to 10:
Now let's say that the element we want to find is 4. One way to do this would be to go to the beginning of the list and check every element one at a time. If the element we are searching for is right at the beginning we're lucky, if not we could go all the way to the end of the list before finding it.
Binary Search solves this problem by "guessing" where the element will be by adding the least variable it could be (in this case 0) and the largest (10), and getting the average. Thus it's first guess would be 5:
Now it does a check, is 5 the element it was looking for? Nope. Is 5 greater than the element wanted? Yes. So the largest variable (thus far 10), is changed the guess it just made (we know that 5 is larger than the element wanted, so we keep that as the upper bound, lower bound remains unchanged at 0). The rest of the list is thrown away, our new list looks like this:
Again we add the upper bound and lower bound together and get the average ( $$(0 + 5)/2$$), this is the new guess.
Again we check, is the guess greater than the value we're looking for? This time's it's a no... so is it less than the value wanted? Yes, 2.5 is less than 4. In this case, the upper bound is changed to 2.5, our previous guess, while the upper bound remains unchanged. This is the new line:
Again the upper bound and lower bound are added and averaged to produce our new guess. The average is 3.75:
As you can see, as the formula iterates, the list gets smaller and smaller as more of it is thrown away and the guesses get closer and closer. Eventually (in the next two iterations) the code will find the element it was looking for. This method is extremely useful in large data sets, where the lists searched have elements that number in the hundreds if not thousands.
#### Lisp Implementation
We're printing out the computer's guesses so that we can see how well the algorithm performs. The syntax is as follows: chop is followed by the value that we're looking for and the list from which we perform the search.
In the first run, we use quite a small list, only 6 elements, and it gave us the value after 2 guesses. In the second one, we used a list of 1,000 elements and asked it to find the element 21. Here it took 9 guesses... not bad for such a big list. Now how about if we give it a list of 10,000 elements? Will it take 10 times as much (90 guesses) as the list with 1,000 elements?
Well it took more than 9... 4 more to be exact! Nowhere close to 10 times as much, even though the previous list was 10 times smaller. This tells us that this particular algorithm is incredibly efficient and is perfect for dealing with Big Data.
### Newton-Raphson
In it's original form, the Newton-Raphson formula for finding roots is this:
$$X_1 = X_0 - \frac{f(X_0)}{f'(X_0)}$$
This is an interactive formula, where the next value of $$X_1$$ is found by substituting the present value of $$X_1$$ into the algorithm as $$X_0$$.
#### Lisp Implementation and Sample Run
This will print out every guess the program makes until it comes up with a reasonably accurate answer.
We can see that it's very quick and can produce an incredibly accurate answer (in this case the accuracy is hard-coded into the code and will give us an answer that is within 0.01 to the correct answer, this of course can be adjusted).
### Comparing Newton-Raphson and Binary Search for Root Finding
We haven't yet written an example that shows how Binary (or Bisection) Search can be used for root finding...
This is quite crude, but will work for demonstration purposes. We're also printing out each guess the code makes so that we can see how the guesses progress. Let's see what the code can do when given the same root to find as the Newton-Raphson algorithm.
So we can see that it does give a decent answer, the Newton-Raphson code though is much more accurate when set with the same precision test (< 0.01), and much quicker. Our binary search algorithm took 43 guesses while Newton-Raphson came up with the answer after 6 guesses.
Is there a way that the two can be combined to further improve what the Newton-Raphson can algorithm can do? What can the Binary Search technique bring to the table?
### Using Binary Search's Upper and Lower Bounds to Prevent Drift
Binary Search's min and max limit can be used to ensure that if the guess provided by the Newton-Raphson algorithm is outside of the boundaries (since we know that no number's square can be more than half of that number we use that as the initial upper limit), then instead of using that guess for the next iteration, we take the number between the upper and lower limits and use that instead. That way we can avoid unnecessary iterations (although, to be honest, in the few trials I've run I've only ever found the improvement to be one iteration less at best).
Implementation of the combined algorithms:
In the nr-chop function, we first set accuracy to 0.01. If the guess is not good enough we then check to see if $$\text{lower} < \text{guess} < \text{upper}$$, which means that we're making sure that the guess falls within the known boundaries. If it doesn't, then we discard the guess and instead use the midpoint between the boundaries as the next guess. If the guess falls within the boundaries but is greater than the upper boundary when squared, we then set the guess as the new upper boundary. If it is less than the upper boundary when squared, we instead set reset the lower boundary to the guess.
I've also set the initial guess to the midpoint of the starting boundaries (beginning upper boundary is half of the given number), though you can change this around and it will not affect the number of iterations (that's the point of combining the two algorithms).
First we'll compare the algorithms when a fourth of the value entered as it's initial guess.
In action:
Ok so that was expected, after all they're basically using the same underlying formula so they should come up with the same solution in the same number of iterations.
How about if we switch up the initial guess?
We can see that the pure version iterated 6 times when given 99 as it's initial guess. Can the hybrid do better?
As you can see, we changed the initial guess to the value entered minus 1 (in this case 99). However, the algorithm rejected the first guess given by the Newton-Raphson formula, which would have been 50.00505 as when we ran the pure formula, but instead substituted the initial guess with 26 (upper boundary + lower boundary divided by 2). This enabled one less iteration and slightly more accurate guesses (actually the algorithm did iterate the same number of times, it just rejected the initial guess and gave a new one).
### That's the Great Optimization?
If this were the end-all-be-all for the hybrid algorithms involving Newton-Raphson and binary search then it would have a died an ignoble death almost immediately, yet this combination is used in the field (in graphing calculators for example) quite often.
#### Non-convergence
There are actually scenarios where the Newton-Raphson formula cannot provide a clear answer. However, this happens when you try to find the root of a polynomial or a system of equations.
For example, we implemented Newton-Raphson to look for square roots as follows:
$$\text{guess}_1 = {guess}_0 - \frac{\text{number entered} - \text{guess}^2}{\text{guess}^2}$$
This is quite a simple expression and the derivative is stable, ensuring that the formula converges on a point with ever increasing accuracy.
If however, you try to solve for the root of this particular function with Newton-Raphson:
$$f(x) = x^3 - 2x + 2$$
Initial guess is $$0$$
First iteration: $$x_1 = x_0 - \frac{f(x)}{f'(x)} = 0 - \frac{2}{-2} \\ x_1 = 1$$
Second iteration: $$x_2 = 1 - \frac{f(1)}{f'(1)} = 1 - \frac{1}{1} \\ x_2 = 0$$
Uh oh... we an see exactly where this is going... the points oscillate between 0 and 1.
#### Binary Search to the Rescue
How about if we used the binary search method? First thing is to find the upper and lower bounds, with polynomials this means two points that have opposite signs (positive and negative).
$$f(0) = 0^3 - 2(0) + 2 = 2$$
So one point is $$0$$
$$f(-2) = _2^3 - 2(-2) + 2 = -2$$
The other point is $$-2$$
So the root is between $$0$$ and $$-2$$:
$$\frac{0 + -2}{2} = -1$$
Now plug that in to the equation: $$-1^3 - 2(-1) + 2 = 3$$. Since 3 is positive and we have to maintain one positive and one negative result as the upper and lower bounds for the root, $$-1$$ takes the place of $$0$$ as the upper limit.
Iterate
$$\frac{-1 + -2}{2} = -1.5$$.
Plugging in $$-1.5$$ results in $$1.625$$, another positive number, so again $$-1.5$$ replaces $$-1$$ as the upper limit. On the next iteration, $$-1.75$$ is the midpoint and yields $$0.140625$$ when plugged in. Eventually you will end up with $$-1.769$$, which, when plugged into the equation:
$(-1.769)^3 - 2(-1.769) + 2 = 0.0021$
Close enough I think that we can declare $$-1.769$$ as the root for the polynomial.
Whew! So as you can see, what's happened is that due to a "bad" initial guess, the Newton-Raphson formula failed to find a solution, while the binary search method just kept plodding along.
This is precisely the reason that graphing calculators contain a hybrid algorithm for root finding. If the root to be found is a root of a fixed integer or float, then the Newton-Raphson method is brutally effective. Change that to the root of a polynomial and things can become problematic, as the algorithm can start to oscillate between two points without converging (there are a few other monkey wrenches as well).
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# Chapter 6 binary, octal, and hexadecimal in vernacular C ++
Source: Internet
Author: User
Chapter 6 binary, octal, and hexadecimal
6.1 why do we need octal and hexadecimal?
6.2 decimal number conversion to decimal number
6.2.1 convert binary to decimal
6.2.2 convert octal to decimal
6.2.3 expression of the octal number
6.2.4 use of Octal numbers in escape characters
6.2.5 convert the hexadecimal number to the decimal number
6.2.7 use of hexadecimal numbers in escape characters
6.3 decimal number to binary, octal, and hexadecimal number
6.3.1 convert a 10-digit number to a 2-digit number
6.5 original code, reverse code, and supplemental code
6.6 view variable values through debugging
6.7 summary of this Chapter
This is a course titled "no village or no store. The conversion between different numerics is purely a mathematical calculation. However, you don't have to worry about the complexity. It's nothing more than multiplication or Division calculation.
In fact, the counting methods in many places in life are somewhat different in hexadecimal notation.
For example, the most commonly used 10-digit system actually originated from people with 10 fingers. If our ancestors never get rid of this situation, I think we must be using a 20-digit system.
For binary ...... No sock is called 0, and one sock is called 1. But if there are two sock, we often say: 1 pair of sock.
In life, there are also: 7 hexadecimal, such as weeks. Hexadecimal format, such as hour or "A Dozen", 60 bytes, such as minute or angle ......
6.1 why do we need octal and hexadecimal?
In programming, we usually use 10 hexadecimal notation ...... C/C ++ is a high-level language.
For example:
Int A = 100, B = 99;
However, because the representation of data in a computer exists in binary format, sometimes binary can be used to solve the problem more intuitively.
However, the binary number is too long. For example, the int type occupies 4 bytes and 32 bits. For example, if the value is 100, the value expressed in the binary number of the int type is:
0000 0000 0000 0000 0110 0100
No one would like to think about or operate on such a long number. Therefore, C and C ++ do not provide a method to directly write binary numbers in code.
This problem can be solved in hexadecimal or octal. Because,The larger the number, the shorter the expression length.. However, why is it 16 or 8 hexadecimal instead of 9 or 20?
2, 8, and 16 are respectively the power 1 of 2, the Power 3, and the Power 4. This allows the three hexadecimal systems to directly convert each other. In octal or hexadecimal mode, the binary number is shortened, but the expression of binary number is maintained. In the following course about hexadecimal conversion, you can find this point.
6.2 convert binary number to decimal number 6.2.1 convert to decimal number
0th bits in binary are 0 to the power of 2, and 1st bits are 1 to the power of 2 ......
Therefore, there is a binary number: 0110 0100, which is converted to a 10-digit system:
The following is a vertical layout:
0110 0100 to decimal
0th-Bit 0*20 = 0
1st-Bit 0*21 = 0
2nd-bit 1*22 = 4
3rd-Bit 0*23 = 0
4th-Bit 0*24 = 0
5th-bit 1*25 = 32
6th-bit 1*26 = 64
7th-Bit 0*27 = 0 +
---------------------------
100
The horizontal calculation is as follows:
0*20 + 0*21 + 1*22 + 1*23 + 0*24 + 1*25 + 1*26 + 0*27 = 100
The number multiplied by 0 is 0, so we can skip the bit with a value of 0:
1*22 + 1*23 + 1*25 + 1*26 = 100
6.2.2 convert octal to decimal
In October, it means every 8 to 1.
The eight-digit number ranges from 0 ~ 7.
The 0th bits in eight dimensions are 0 to the power of 8, the 1st bits are 1 to the power of 8, and the 2nd bits are 2 to the power of 8 ......
Therefore, there is an octal digit: 1507, which is converted to decimal:
Vertical representation:
1507 to decimal.
0th-bit 7*80 = 7
1st-Bit 0*81 = 0
2nd-bit 5*82 = 320
3rd-bit 1*83 = 512 +
--------------------------
839
Similarly, we can use horizontal calculation:
7*80 + 0*81 + 5*82 + 1*83 = 839
The result is that the number of octal digits is 1507 to 839 in decimal format.
6.2.3 expression of the octal number
In C and C ++, how does one express an octal number? If the number is 876, we can conclude that it is not an octal number, because the octal number cannot contain more than 7 Arabic numbers. However, if the number is 123, 567, or 12345670, it is possible to set the number to octal or decimal.
Therefore, C and C ++ rules,If you want to specify that a number uses an octal value, you must add a value of 0 before it.For example, 123 is in decimal format, but 0123 indicates octal. This is the expression of Octal numbers in C and C ++.
Since neither c nor C ++ provides the expression of binary numbers, we learned octal here. The second method of expressing the value of CTC ++ is used.
Now, for the same number, for example, 100, we can use the normal 10 hexadecimal expression in the Code, for example, during variable initialization:
Int A = 100;
We can also write:
Int A = 0144; // 0144 is 100 of the octal value. We will learn how to convert a 10-digit number into an octal value.
Remember, you cannot lose the first zero when expressing it in octal. Otherwise, the computer is regarded as a 10-digit system. However, when the number of octal characters is used, the value 0 cannot be used, which is the "Escape Character" expression we learned before for expressing characters.
6.2.4 use of Octal numbers in escape characters
We have learned how to use an escape character '/' and a special letter to represent a character. For example, '/N' indicates line ), '/t' indicates the Tab character, and'/'indicates the single quotation mark. Today, we have learned another method to use escape characters: The Escape Character '/' is followed by an octal number to indicate characters whose ASCII code is equal to this value.
For example, check the ASCII code table in Chapter 5th and find the question mark character (?) The ASCII value of is 63, so we can convert it into an octal value: 77, and then use '/77' To Represent '? '. Because it is octal, it should be written as '/077', but because C, C ++ does not allow the use of a slash plus a 10-digit number to represent characters, so here 0 can be left empty.
In fact, we seldom use escape characters and Octal numbers to represent a character in actual programming. Therefore, we only need to understand the content in section 6.2.4.
6.2.5 convert the hexadecimal number to the decimal number
In decimal format, two Arabic numbers are used: 0 and 1;
8 digits: 0, 1, 2, 3, 4, 5, 6, and 7;
10 in hexadecimal notation, with 10 Arabic numerals: 0 to 9;
In hexadecimal notation, sixteen Arabic numerals ...... Wait, the Arab, or the Indian, only invented 10 numbers?
In hexadecimal notation, the number is 16 to 1, but we only have 0 ~ 9 These 10 numbers, so weUse the letters A, B, C, D, E, and F to represent 10, 11, 12, 13, and 15 respectively.. Uppercase letters are not case sensitive.
The 0th bits in hexadecimal notation are 0 to the power of 16, 1st bits are 1 to the power of 16, and 2nd bits are 2 to the power of 16 ......
Therefore, in the Nth (N starts from 0) bit, if it is a number x (x is greater than or equal to 0, and X is less than or equal to 15, that is, F) it indicates the N power of x * 16.
Suppose there is a hexadecimal number 2af5, how can we convert it into a hexadecimal number?
Use vertical calculation:
2af5 to 10:
0th bits: 5*160 = 5
1st bits: F * 161 = 240
2nd bits: A * 162 = 2560
3rd bits: 2*163 = 8192 +
-------------------------------------
10997
Direct calculation is:
5*160 + F * 161 + A * 162 + 2*163 = 10997
(Remember, in the above calculation, a represents 10, and f Represents 15)
Now we can see that the key to converting all hexadecimal values into a decimal value is that their respective weights are different.
Suppose someone asks you, why is the ten-in-one number 1234 one thousand two hundred and thirty-four? You can give him the following formula:
1234 = 1*103 + 2*102 + 3*101 + 4*100
If no special writing format is used, the hexadecimal number will be mixed with the hexadecimal number. A random number: 9876, it cannot be seen that it is in hexadecimal or 10 hexadecimal format.
C, C ++,Hexadecimal number must start with 0x. For example, 0x1 indicates a hexadecimal number. 1 indicates a decimal number. For example, 0xff, 0xff, 0x102a, and so on. X is not case sensitive. (Note: 0 in 0x is the number 0, not the letter O)
Some examples are as follows:
Int A = 0x100f;
Int B = 0x70 +;
So far, we have learned all the hexadecimal expressions: 10 hexadecimal, 8 hexadecimal, and 16 hexadecimal numbers. The last point is very important. In C/C ++, there are positive and negative numbers in hexadecimal notation. For example, 12 indicates positive 12, while-12 indicates negative 12.The octal and hexadecimal values can only be unsigned positive integers.If you are in the Code:-078, or write:-0xf2, C, C ++, it is not treated as a negative number.
6.2.7 use of hexadecimal numbers in escape characters
An escape character can also be expressed as a hexadecimal number. As mentioned in section 6.2.4 '? 'Character, which can be expressed as follows:
'? '// Enter the character Directly
'/77' // use the octal sequence. The value 0 at the beginning can be omitted.
Similarly, this section is only used for understanding. Except for null characters expressed by octal numerals '/0', we rarely use the last two methods to represent a single character.
6.3 decimal number to binary, octal, and hexadecimal number 6.3.1 10 hexadecimal number to binary number
Give you a decimal number, for example, 6. What if I convert it into a binary number?
The 10th hexadecimal number is converted to the binary number, which is a process of consecutive division of 2:
Divide the number to be converted by 2 to obtain the quotient and remainder,
Divide the quotient by 2 until the quotient is 0. Finally, sort all the remainder in reverse order. The result is the conversion result.
Sounds confused? We will illustrate it with examples. For example, convert 6 to the binary number.
"Divide the number to be converted by 2 to get the quotient and remainder ".
So:
The number to be converted is 6, 6 then 2.The quotient is 3, and the remainder is 0.. (Don't tell me you won't calculate 6 then 3 !)
"Divide the merchant by 2 until the merchant is 0 ......"
The current business is 3, not 0, so continue to divide by 2.
So: 3 then 2, getThe quotient is 1, and the remainder is 1..
"Divide the merchant by 2 until the merchant is 0 ......"
The current business is 1, not 0, so continue to divide by 2.
Then: 1 second 2, getThe quotient is 0, and the remainder is 1.(Take a pen and paper to calculate it. 1, 2, is it 0, more than 1 !)
"Divide the merchant by 2 until the merchant is 0 ...... Finally, sort all the remainder in reverse order"
Excellent! The current business is 0.
After three computations, the remainder is 0, 1, and 1 respectively. All the remainder is sorted in reverse order, that is, 110!
6. Convert to binary. The result is 110.
Change the preceding section to a table:
Divisor Computing process Vendors Remainder 6 6/2 3 0 3 3/2 1 1 1 1/2 0 1
(In a computer, tokens are expressed)
During the examination, it would be a little time-consuming to draw such a table. The more common conversion process is the Division:
(Figure: 1)
Compare the graph, table, and text description, and calculate how to convert 6 to binary number.
After talking about it for half a day, are our conversion Results Correct? Is the binary number 110 6? You have learned how to convert a binary number to a 10-digit number. Therefore, calculate whether the value of 110 is 6 if it is 10.
Very happy. The method for converting a 10-digit number to an 8-digit number is similar to the method for converting to a 2-digit number. The only change is that the divisor is changed from 2 to 8.
Let's look at an example of how to convert the decimal number 120 to the octal number.
Table representation:
Divisor Computing process Vendors Remainder 120 120/8 15 0 15 15/8 1 7 1 1/8 0 1
120 is converted to octal, and the result is: 170.
Very happy. The method for converting a 10-digit number to a 16-digit number is similar to the method for converting to a 2-digit number. The only change is that the divisor is changed from 2 to 16.
It is also 120, and The hexadecimal format is:
Divisor Computing process Vendors Remainder 120 120/16 7 8 7 7/16 0 7
120 is converted to hexadecimal with the result of 78.
Take pen paper and use the form (figure: 1) to calculate the process of the above two tables.
The conversion between binary and hexadecimal is important. However, there is no need to calculate the conversion between the two. Every C, C ++ programmer can see the binary number and convert it directly to the hexadecimal number.
The same is true if we finish this section.
First, let's look at a binary number: 1111. What is it?
You may need to calculate as follows: 1*20 + 1*21 + 1*22 + 1*23 = 1*1 + 1*2 + 1*4 + 1*8 = 15.
However, since 1111 is only 4 bits, we must remember the weights of each bits, and remember them from high to low: 8, 4, 2, and 1. That is, the maximum bit is 23 = 8, followed by 22 = 4, 21 = 2, 20 = 1.
Remember 8421. For any 4-bit binary number, we can quickly calculate the corresponding 10-digit value.
The following lists all possible values of the four-digit binary xxxx (skipped in the middle)
A four-digit binary number is used to calculate the hexadecimal value of a decimal number.
1111 = 8 + 4 + 2 + 1 = 15 f
1110 = 8 + 4 + 2 + 0 = 14 E
1101 = 8 + 4 + 0 + 1 = 13 d
1100 = 8 + 4 + 0 + 0 = 12 C
1011 = 8 + 4 + 0 + 1 = 11 B
1010 = 8 + 0 + 2 + 0 = 10
1001 = 8 + 0 + 0 + 1 = 10 9
....
0001 = 0 + 0 + 0 + 1 = 1 1
0000 = 0 + 0 + 0 + 0 = 0 0
To convert a binary number to a hexadecimal value, the binary number is converted to a hexadecimal value in four digits.
For example ):
1111 1101,101 0 0101,100 1 1011
F d, a 5, 9 B
When we see FD, how can we quickly convert it to a binary number?
First convert F:
When we see F, we need to know that it is 15 (maybe you are not familiar with ~ F). How can we use 8421 for 15? It should be 8 + 4 + 2 + 1, so the four digits are all 1: 1111.
Then convert D:
When I see D, how can I use 8421 to get it together? It should be: 8 + 2 + 1, that is: 1011.
Therefore, FD is converted to the binary number, which is 1111 1011.
Because the hexadecimal conversion to binary is quite direct, when we need to convert a decimal number to a binary number, we can also convert it to a hexadecimal number first, then convert it to a binary system.
For example, to convert a decimal number 1234 to a binary number, if you want to divide it by 2 all the time, you need to calculate a large number of times. So we can divide it by 16 to get the hexadecimal number:
Divisor Computing process Vendors Remainder 1234 1234/16 77 2 77 77/16 4 13 (d) 4 4/16 0 4
Result: The hexadecimal value is 0x4d2.
Then we can directly write the binary format 0x4d2: 0100 1011 0010.
The ing relationship is:
0100 -- 4
1011 -- d
0010 -- 2
Similarly, if a binary number is very long, we need to convert it into a 10-digit number, except for the previous method, we can also convert the binary to hexadecimal and then to hexadecimal.
The following is an example of the binary number of the int type:
01101101 11100101 10101111 00011011
We convert the four-digit group to hexadecimal: 6d E5 af 1b
6.5 original code, reverse code, and supplemental code
After the conversion of various hexadecimal formats, let's talk about another topic: source code, reverse code, and complement code.
We already know that all data in a computer is expressed in binary.
We have also learned how to convert a decimal number to a binary number.
However, we still haven't learned how to express a negative number in binary.
For example, suppose there is a number of int type, the value is 5, then we know that it is represented in the computer:
00000000 00000000 00000000 00000101
5 is converted to a binary system of 101, but the number of int types occupies 4 bytes (32 bits), so a bunch of 0 values are filled in front.
Now I want to know how-5 is represented in a computer?
In a computer, a negative number is expressed as a positive complement..
What is a supplemental code? This should start with the original code and the reverse code.
Original code: an integer that is converted into binary numbers based on the absolute value. It is called the original code.
For example, 00000000 00000000 00000000 00000101 is the source code of 5.
Reverse code: returns the bitwise result of the binary number. The new binary number obtained is called the reverse code of the original binary number.
The reverse operation indicates that the original value is 1, and the value is 0. The original value is 0, and the value is 1. (1 to 0; 0 to 1)
For example, if the bitwise of 00000000 00000000 00000000 is reversed, 00000101 11111111 11111111 is obtained.
Said: 11111111 11111111 11111111 11111010 00000000 is an anti-code of 00000000 00000000 00000101.
The anti-code is mutual, so it can also be called:
11111111, 11111111, 11111111, 11111010, 00000000, 00000000, 00000000, 00000101, and are mutually inverse codes.
Complement: the anti-code plus 1 is called a complement.
That is to say, to get a complement of a number, first obtain the reverse code, and then add 1 to the reverse code. The resulting number is called a complement.
For example, the 00000000 00000000 00000000 00000101 anti-code is: 11111111 11111111 11111111 11111010.
Then, the complement code is:
11111111 11111111 11111111 11111010 + 1 = 11111111 11111111 11111111
Therefore,-5 is expressed in the computer as 11111111 11111111 11111111 11111011. Convert to hexadecimal: 0 xfffffffb.
Let's look at how integer-1 is represented in a computer.
Suppose this is also an int type, then:
1. first obtain the original code of 1: 00000000 00000000 00000000 00000001
2. Reverse code: 11111111 11111111 11111111 11111110
3. Supplemental code: 11111111 11111111 11111111 11111111
It can be seen that the Binary Expression of-1 in a computer is full 1. Hexadecimal: 0 xffffff.
Everything is on paper ...... -1 is expressed as 0 xffffff in the computer. Can I see it with my own eyes? Of course. With the debugging function of C ++ builder, we can see the hexadecimal value of each variable.
6.6 view variable values through debugging
Next, let's complete a small experiment. Through debugging, we can observe the value of the variable.
We declare two int variables in the Code and initialize them to 5 and-5 respectively. Then, we can see the decimal and hexadecimal values of the two variables when the program is running through the debugging methods provided by CB.
Create a console project. Add the following simhei part (just one line ):
//---------------------------------------------------------------------------
# Pragma hdrstop
//---------------------------------------------------------------------------
# Pragma argsused
Int main (INT argc, char * argv [])
{
Int AAAA = 5, bbbbb =-5;
Return 0;
}
//---------------------------------------------------------------------------
We are not familiar with the line:
Getchar ();
Therefore, if you run the program at full speed, the DOS window will flash. However, today we will setBreakpointTo use the program to stop where we need it.
Set breakpoints: One of the most common debugging methods. When a program is running, it is paused at a certain code location,
In CB, you can set a breakpoint by pressing F5 on a line of code or clicking the mouse in the first line.
For example:
In, we set a breakpoint on the return 0; line. The row where the breakpoint is located is displayed in red by CB.
Then, run the program (F9) and the program will stop at the breakpoint.
(Note the differences between the two images. The preceding figure is before running, and the following figure is running. the arrow on the left indicates the row to which the operation is running)
When the program stops at the breakpoint, we can observe the visible variables in the current code snippet. There are many ways to observe variables. Here we will learn to use DEBUG inspector to fully observe a variable.
The following describes how to call up the debug Inspector window to observe a variable:
Make sure that the code window is an active window. (Click the code window with the mouse)
Press ctrl and move the mouse over the variable AAAA. You will find that the AAAA in the Code turns blue and the underline appears. The effect is like a hyperlink on the webpage, And the mouse turns into a small hand:
Click the mouse to display the AAAA check window:
(The operating system used by the author is Windows XP, and the window looks different from Win9x)
From this window, I can see:
Aaaa: variable name
INT: Data Type of the Variable
5: This is the value of the variable, that is, AAAA = 5;
0x00000005: it is also the value of the variable, but expressed in hexadecimal notation. Because it is of the int type, it occupies 4 bytes.
First, close the debug Inspector window used to observe the AAAA variable.
Now, we use the same method to observe the variable Bbbb. Its value is-5, and the negative number is represented by a supplementary code in the computer.
As we think, the complement of-5 is: 0 xfffffffb.
Press F9 again, and the program continues to run from the breakpoint and ends.
6.7 summary of this Chapter
A difficult chapter?
Let's take a look at what we have learned:
1) We learned how to convert hexadecimal numbers to decimal numbers.
The three conversion methods are the same, and multiplication is used.
2) We learned how to convert a decimal number to a binary, octal, or hexadecimal number.
The methods are the same. Division is used.
3) We learned how to swap binary numbers and hexadecimal numbers quickly.
The trick is to convert binary numbers into hexadecimal numbers in a group of four digits.
After learning the hexadecimal number, we will use the hexadecimal number in many places to replace the binary number.
4) We learned the original code, reverse code, and complement code.
Change the value of the original code 0 to 1, 1, and 0 to obtain the reverse code. To obtain the complement code, you must first obtain the reverse code and then add 1.
In the past, we only knew how a positive integer is expressed in a computer, and now we know that a negative number is expressed using the complement code of its absolute value in a computer.
For example, how does-5 express in a computer? The answer is: 5.
5) Finally, In the hands-on experiment, we learned how to set breakpoints and how to call up the debug Inspector window to observe variables.
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A polygon having actually eight sides is recognized as an octagon. If every the political parties of one octagon space equal and also angles space the same then the octagon is dubbed a regular octagon. A continuous octagon has a total number of 20 diagonals. The sum of all internal angles the a regular octagon is 1080 degrees. Also, each inner angle is 135 degrees. Theexterior edge of one octagon measures45 degrees and the amount of every exterior angles is 360 degrees. Theoctagon formula is used to calculation its area, perimeter of one octagon. Learn about the octagon formula with couple of examples offered below.
You are watching: What is the exterior angle of a regular octagon
## What Is Octagon Formula?
The octagon formula is provided to calculation the area, perimeter, and diagonals of an octagon. To find the area, perimeter, and diagonals of an octagon we usage the adhering to octagon formulas.
### Formulas because that Octagon:
To uncover the area of an octagon we use the following formula: Area of octagon formula=2× s2× (1 + √2)
To uncover the perimeter of an octagon we use the adhering to formula: Perimeter the octagon= 8s
To uncover the variety of diagonals of one octagon we use the following formula: number of Diagonals = n(n - 3)/2 = 8(8 - 3)/2 = 20
where,
s = next lengthn = number of sides
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## Examples UsingOctagon Formula
Example 1: calculation the perimeter and also area of one octagon having a side same to 4 units using the octagon formula.
Solution:
To Find: Perimeter and also AreaGiven:s= 4 units.Using the octagon formula because that perimeterPerimeter(P) = 8sP = 8 × 4P = 32 unitsUsingthe octagon formula because that areaArea of octagon= 2s2(1 + √2)= 2 × 42(1 + √2)= 77.25483 units2
Answer: Perimeter and area the the octagonare 32 units and also 77.25483 units2.
See more: What Is A Trunk Lift Exercise ? Trunk Lifts
Example 2:An octagonal board has actually a perimeter equal to 24 cm. Uncover its area utilizing the octagon formula.
Solution:
To Find:Area of the octagon.Given: Perimeter = 24 cm.The perimeter the octagon = 8s24 = 8 ss = 3 cm.Usingthe octagon formula for area,Area the octagon = 2s2(1 + √2)= 2 × 32(1 + √2)= 43.45cm2
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+0
# Mensuration
+8
1979
2
+132
A square wall is covered with square tiles. There are 85 tiles altogether along the two diagonals. How many tiles are there on the whole wall.
May 18, 2017
#1
+22290
+3
A square wall is covered with square tiles.
There are 85 tiles altogether along the two diagonals.
How many tiles are there on the whole wall.
$$\left(\dfrac{85+1}{2}\right)^2 = 43^2=1849$$
May 18, 2017
#2
+101149
+3
Let's use a more simple example to figure this out.....
Look at the pic below :
Note that we have a total of 13 points on both diagonals..........add 1 to this = 14 and divide by 2 = 7.....so...letting n = 7 we have a grid with n^2 = 7^2 = 49 points (tiles)
So...in your case....we have 85 tiles (points) on both diagonals......add 1 to this = 86 and divide by 2 = 43.....so, now.....letting n = 43 we have a grid with 43^2 = 1849 points (tiles)......which is exactly what heureka found.....!!!
Another way to see this is that we have :
(n - 2)^2 interior points and 2 [ n + (n - 2) ] "border" points
Letting n = 7 we get (7 - 2)^2 + 2 [ 7 + 5] = 25 + 24 = 49 points (tiles)
And with n = 43 we get (43 - 2)^2 + 2 [ 43 + 41 ] = 1849 points (tiles)
May 18, 2017
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Which function is graphed below?
Students were enquired to answer a question at academics and to declare what is most important for them to succeed. Of the many comments, one that that stood out was practice. People who are successful do not become successful by being born. They work hard and perseverance their lives to succeeding. If you wish to reach your goals, keep this in mind! following some question and answer examples that you could simply utilize to elevate your knowledge and gain insight that will assist you to continue your school studies.
Question:
Which function is graphed below?
The general basic exponential function is of the form,
The function has a intercept of
This point must satisfy equation (1). So let us do the substitution to obtain,
Now our equation 1 becomes
Also note that the graph passes through, , hence must also satisfy its equation. Substituting this point into equation (2), gives us
Equation (2) can be rewritten as;
Now let us substitute the value of into equation (3) to obtain,
Hence the correct answer is option B
From the answer and question examples above, hopefully, they can potentially guide the student deal with the question they had been looking for and keep in mind of everything that stated in the answer above. Then could actually have some sharing in a group discussion and also study with the classmate regarding the topic, so another student also ought to have some enlightenment and still keeps up the school learning.
READ MORE WHich principle established during the Enlightenment does this excerpt reflect?
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# Statistical Analysis, Capital Budgeting, and Risk Analysis...
Statistical Analysis, Capital Budgeting, and Risk Analysis
Introduction:
Learning exercise 1 will serve as an introduction to statistical analysis and introduce the basics of corporate finance including capital budgeting models, bond and stock valuation, and risk analysis. This learning exercise will use data from the Wharton Database to perform specific statistical analysis using Excel.
1) Using the following information determine the NPV, Profitability Index, and IRR of the following project. For simplicity assume that the company only has debt and equity and all of the company’s debt is in 15 year bonds. Would the company move forward with the project? Please explain.
Data: Price of 15-year bond is $1,075, face value of the bond is$1,000, and the coupon rate is 5.35% and the coupon payments are paid annually.
The EPS of the company is $9.76, the dividend payout ratio is 32%, and the stock price is$67.89 (Assume a constant growth model)
Assume the debt equity ratio is .35
The Net Cash Outlay (NCO) of the project is - $18,750, 000 After tax cash flows for year 1 = -$2,278,000
After tax cash flows for year 2 = $4,790,000 After tax cash flows for year 3 =$5,987,000
After tax cash flows for year 4 = $6,350,000 After tax cash flows for year 5 =$8,145,000
Corporate Tax of 23 %.
If you can’t solve for the cost of capital, assume that the cost of capital is 10% so you will be able to move forward with the problem.
B) Please discuss the advantages and disadvantages to the NPV model, IRR model, and the profitability index.
2) A) Using the Gordon constant growth model for stock valuation determine the price of the stock assuming that:
EPS 1 = \$7.45
Ke = 11.56%
Dividend Payout Ratio (DPR) = 28%
What happens to the price of the stock if the DPR is changed to 34%? What happens to the price of the stock if the DPR is 100%? What rate do you think retained earnings should be invested at in the Gordon model? Can growth
Attachments:
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Closed set?
1. Feb 12, 2010
michonamona
1. The problem statement, all variables and given/known data
Is [0, infinity) a closed set?
2. Relevant equations
N/A
3. The attempt at a solution
It's easy to say that its not. But the solution in my textbook suggests otherwise. Why is this so?
Thanks!
M
2. Feb 12, 2010
LCKurtz
If it isn't closed, you should be able to find a limit point that isn't in the interval, right? What point would that be?
3. Feb 12, 2010
vela
Staff Emeritus
Is its complement open?
4. Feb 12, 2010
michonamona
LCKurtz:
Ok, I think I understand it for integers. Say, [1,5), where the limit point 5 is not in the interval. But what confuses me is infinity. How is it precisely defined with respect to actual numbers like integers?
vela:
What is its complement? R\[0, infinity)?
Hmm....
5. Feb 12, 2010
vela
Staff Emeritus
Yes, that's the complement.
Just curious, what's the definition of a closed set that your class uses?
6. Feb 12, 2010
michonamona
If its complement is open, then I know that the interval is closed. But why? why is its complement open?
I appreciate you guys for not giving me the answers directly. I'm really trying hard to understand it.
M
7. Feb 12, 2010
vela
Staff Emeritus
What's the definition of an open set?
8. Feb 12, 2010
michonamona
A set that contains only its interior points....
Its boundary is contained in its complement...
9. Feb 12, 2010
vela
Staff Emeritus
The complement of $[0,\infty)$ is $(-\infty,0)$. Let $x\in(-\infty,0)$. Is there an open interval centered on x that's contained in $(-\infty,0)$?
10. Feb 12, 2010
chr0nicbudd
i think that it needs to be open on both sides.
if it were (0, infinity) it would be open. but since its its [0, infinity), it is a closed interval since you could make x = 0 and have a point, y, that lied to the left of 0, yet still within the Br(0) (ball of radius r about 0).
11. Feb 12, 2010
LCKurtz
Infinity is not part of the real numbers. [0,oo) is just a convenient alternative notation for $x \ge 0$. You might think of the symbol as indicating the values of x are not bounded above.
12. Feb 12, 2010
Hurkyl
Staff Emeritus
Just FYI -- "closed" only makes sense relative to a space. The right question is
Is [0, +infinity) a closed subset of the reals
It seems to be the habit to introduce it as a sort of "useful fiction". The set of nonnegative reals is quite interval-like, and it's useful to introduce a formal interval-like notation to write such things. Thus "[0, +infinity)".
However, if you go on, you should learn about the extended real numbers which are, IMO, a much better way to go about doing things. This number system has two additional numbers that the real numbers don't: -infinity and +infinity. And it turns out that every "useful fiction" that you learn in the introductory classes turns out to port to ordinary things in this more sophisticated approach, despite having the exact same notation. e.g. In the extended real numbers, [0,+infinity) is a perfectly ordinary interval, which consists of the nonnegative real numbers. (And the interval [0,+infinity] would consist of all nonnegative extended real numbers)
13. Feb 13, 2010
bennyska
It's like putting too much air in a balloon!
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09FMAT10_Test01Rev
# 11 a special window in the shape of a rectangle with
• Notes
• 2
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Unformatted text preview: 11. A special window in the shape of a rectangle with semicircles at each end is to be constructed so that the outside dimensions are 100 feet in length. Find the dimensions of the rectangle that maximizes its area. Math 10 – Precalculus RCC – Norco Fall 2009 Jason Rey 12. Beth has 3,000 feet of fencing available to enclose a rectangular field. a. Express the area A of the rectangle as a function of x, where x is the length of the rectangle. b. What is the domain of A? c. For what value of x is the area largest? d. What is the maximum area? 13. The price p (in dollars) and the quantity x sold for a certain product obey the demand equation g G ¡ ¢ £ ¤ ¥ uUU ¦U § ¤ § ¨UU© . a. Express the revenue R as a function of x. b. What is the revenue if 100 units are sold? c. What quantity x maximizes the revenue? What is the maximum revenue? d. What price should the company charge to maximize revenue? 14. Determine the end behavior of the polynomial function a. ( ) 7 6 2 3 +- = x x x g b. ( ) ( ) ( ) 7 4 2 3 2-- = x x x x p c. ( ) ( )( ) 7 2 5 3--- = x x x q 15. Find all zeros of the polynomial function, then create a sign chart and use it to sketch the graph of: a. ( ) 4 4 5 3 2 3-- + = x x x x p b. ( ) x x x x x g 30 40 25 5 2 3 4 +- +- = 16. Sketch the graph of the rational function: a. ( ) 12 4 2 2 2- + + = x x x x x R b. ( ) 16 6 4 2 2 3- + +- = x x x x x R c. ( ) 16 4 3 2-- = x x x f d. ( ) 2 3 6 2 2 + +-- = x x x x x g 17. Solve the inequality: a. 1 3 > +- x x b. 2 4 x x ≥ c. 1 3 2 1 + ≤ + x x...
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## Intermediate Algebra: Connecting Concepts through Application
$(2x-3)(4x^2+6x+9)$; formula applied incorrectly.
Use the formula $a^3-b^3=(a-b)(a^2+ab+b^2)$ to factor. Paul did not apply the formula correctly in his work. $8x^3-27$ =$(2x-3)(4x^2+6x+9)$
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2-8 (9) - Create assignment 00223 Homework 8 Oct 24 at 7:27...
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Create assignment, 00223, Homework 8, Oct 24 at 7:27 pm 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Cable Over a Pulley 10:07, trigonometry, numeric, > 1 min, nor- mal. 001 A cable passes over a pulley. Because the cable grips the pulley and the pulley has non- zero mass, the tension in the cable is not the same on opposite sides of the pulley. The force on one side is 120 N, and the force on the other side is 100 N. Assuming that the pulley is a uniform disk of mass 2 . 1 kg and ra- dius 0 . 81 m, determine the magnitude of its angular acceleration. Correct answer: 23 . 5156 rad / s 2 . Explanation: The resultant torque is given by (120 N)(0 . 81 m) - (100 N)(0 . 81 m) = 16 . 2 N m The moment of inertia is: I = 1 2 m r 2 = 1 2 (2 . 1 kg)(0 . 81 m) 2 = 0 . 688905 kg m 2 . Then, τ = I α gives α = τ I = 16 . 2 N m 0 . 688905 kg m 2 = 23 . 5156 rad / s 2 . Complex Atwood Machine 10:07, trigonometry, numeric, > 1 min, nor- mal. 002 An Atwood machine is constructed using two wheels (with the masses concentrated at the rims). The left wheel has a mass of 2 kg and radius 22 cm. The right wheel has a mass of 2 . 5 kg and radius 30 cm. The acceleration of gravity is 9 . 8 m / s 2 . 3 m 1 m 4 m 2 m The hanging mass on the left is 1 . 5 kg and on the right 1 kg. What is the acceleration of the system? Correct answer: 0 . 7 m / s 2 . Explanation: Basic Concepts: The net acceleration a = is in the direction of the heavier mass m 1 . Each pulley’s mass is concentrated on the rim, so I = m pulley r 2 3 m 1 1 m 4 m 2 2 2 m T 1 T 3 T a a r r For each pulley, τ net = = mr 2 a r · = mra so that for the leftmost pulley, T 1 r 1 - T 2 r 1 = m 1 r 1 a m 1 a = T 1 - T 2 (1) and for the rightmost pulley, T 2 r 2 - T 3 r 2 = m 2 r 2
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# Busn135Ch01.xls
Document Sample
``` Whole Numbers and Decimals
Define:
Whole Numbers
Decimals
Whole Numbers
Decimals
Round:
Whole Numbers
Decimals
Whole Numbers
Decimals
Order of Operations:
By hand
In Excel
Multiply and Divide:
Whole Numbers
Decimals
Omitting zeros when multiplying or dividing
Math Word-Application Problems:
Translating words into math symbols
Steps to solve math word-application problems
Whole Numbers: numbers to the left of the decimal point. Uses
the ten one-place digits: 0,1,2,3,4,5,6,7,8,9. No "parts of the
whole", or no numbers between 1 and 0. Use a comma every
third place.
Trillions Billions Millions Thousands Ones
Hundred Thousands
Hundred Trillions
Hundred Millions
Hundred Billions
Ten Thousands
Ten Trillions
Ten Millions
Ten Billions
Thousands
Trillions
Millions
Billions
4 5 6 7 5 8 4 5 2
4
Fifteen Million, Seven Hundred Eighty Nine Thousand, Four
15,789,458 Hundred Fifty Eight
Whole
Four Hundred Thirty Trillion, Four Hundred Fifty Seven Billion, Eight
Numbers
Hundred Forty Five Million, Eight Hundred Ninety Six Thousand,
430,457,845,896,214 Two Hundred Fourteen
Four Hundred Fifty Six Billion, Seven Hundred Fifty Eight Million,
Four Hundred Fifty Two Thousand, One Hundred Nineteen And
456,758,452,119.155 One Hundred Fifty Five Thousandths
Decimal
Two Hundred Fifty Four Thousand, Four Hundred Nineteen
Numbers
0.254419 Millionths
One Hundred Fifty Four Thousand Five Hundred Eighty Seven
\$154,587.25 Dollars And Twenty Five Cents
Decimals: numbers to the right
decimal point. Uses of the decimal point -
The word representing parts of a whole -
,9. No "parts of the "and" goes "a whole" is the number 1 and
Use a comma every here when the "parts" is a number between
you say the 1 and 0. Define Decimal ==> A
word number written with a decimal
Ones such as 4.987 or .062
Hundred-Thousandths
Decimal Point (and)
Ten-Thousandths
Thousandths
Hundredths
Hundreds
Millionths
Tenths
Ones
Tens
1 1 9 1 5 5
5 6 3 2 5 4 4 1 9
Formatting is like a façade on a house: it is only what is on top. Formatting does not necessarily refle
number
General Number Currency Accounting Date Time
Shows date
Shows "floating" Shows "fixed" format, Shows time
No commas, Shows commas dollar sign, dollar sign, underneath is the format,
shows as many and as many commas and as commas and as number of days underneath is the
decimals as you decimals as you many decimals as many decimals as since Jan. 1, proportion of one
type in tell it to show you tell it to show you tell it to show 1900 24 hour day
12569.45698 12,569.46 \$12,569.46 \$ 12,569.46 7/3/2006 8:00:00 AM
12569.45698 12569.45698 12569.45698 12569.45698 38901 0.333333333
12569.45698 12569.45698 12569.45698 12569.45698 38901 0.333333333
s not necessarily reflect the underlying
Percentage Fraction
Shows
Percentage
format (slides
decimal to right
two positions and
percentage Division
symbol). What is represented with
underneath is a a numerator and
decimal or whole a denominator
number (.06 or 1 and a line
)) between the two
6.00% 1/4
0.06 0.25
0.06 0.25
Rules for Rounding by hand
Select the position to which you would
like to round. We will call this the
1 "selected position".
If you are rounding to a whole number
draw a line under the "selected
position".
If you are rounding to a decimal
position, draw a vertical line after the
"selected position" to indicate that you
will drop the remaining digits.
2 Look one position to the right
If this digit is 5 or more, increase the
digit in the "selected position" by one
If this digit is 4 or less, DO NOT
CHANGE the digit in the "selected
position"
Change all the digits to the right of the
3 "selected position" to zeros.
If you have decimal positions that have
zeros, you can drop them.
by hand
Financial Statements are often rounded
to the Thousands position \$5,698,700,459.25
Income Taxes are rounded to the ones
position \$555.25
Payroll deductions are rounded to the
hundredths position \$98.2569
ROUND function in Excel
To round to the thousands position use "-
3" as the second function argument
=ROUND(C2,-3)
To round to the ones position use "0" as
the second function argument
=ROUND(C3,0)
To Round to the Hundredths Position use
"2" as the second function argument
=ROUND(C4,2)
ROUND function
Financial Statements are often rounded
to the Thousands position \$5,698,700,459.25
Income Taxes are rounded to the ones
position \$555.25
Payroll deductions are rounded to the
hundredths position \$98.26
Increase/Decrease decimals with
formatting
Use the Increase/Decrease Decimal
button on the Formatting toolbar
Income Taxes are rounded to the ones
position \$555.25
Payroll deductions are rounded to the
hundredths position \$98.2569
Name Gross Pay Deduction 1 Net Pay
Suix Chin 5,468.25
Stephanie Walting 7,667.54
Rachel Hensley 6,900.20
Totals
Assumption
Deduction Rate 0.32
Rule for when to use the ROUND function: If you will use the
result of a formula in a subsequent Excel formula and you are
multiplying or dividing decimals where you are required to round
to a certain position, you must use the ROUND function to get a
In the range C4 to C6 (green color) create efficient formulas that
5) Because each deducti
calculate the deduction based on the rate in cell B10 (Gross * subtracted from each pay
deduction rate). Then calculate the net pay in the range D4:D6. function. The Decimal bu
Then add totals in cells C7 and D7. Hint: When we calculate a are Rounded, but they ar
deduction, we must round to the penny.
Name Gross Pay Deduction 1 Net Pay Name Gross Pay
Suix Chin 5,468.25 1,749.84 3,718.41 Suix Chin 5,468.25
Stephanie Walting 7,667.54 2,453.61 5,213.93 Stephanie Walting 7,667.54
Rachel Hensley 6,900.20 2,208.06 4,692.14 Rachel Hensley 6,900.20
Totals 6,411.51 13,624.48 Totals
Assumption Assumption
Deduction Rate 0.32 Deduction Rate 0.32
ROUND function
2) and then typed the numbers she saw into the cells here.
3) When she added them up she got a different number than
you did. Her number is correct because what you see is what
5) Because each deduction is a bunch of pennies that are
subtracted from each paycheck, you must use the ROUND
function. The Decimal button on the toolbar makes it look like they
are Rounded, but they are not.
Deduction 1 Net Pay
1,749.84 3,718.41
2,453.61 5,213.93
2,208.06 4,692.14
6,411.52 13,624.47
1,749.84000
2,453.61280
1749.84 2,208.06400
2453.61
2208.06
6411.51
can see there are some decimals that you originally could not
see, but the SUM function does see them. The SUM function
sees them and used them when it adds. That is why your SUM
shows 6,411.52 (it has more decimals to add and so it gets one
penny more).
Whole Numbers
16
+ 15
62
- 43
Decimals
16.256
+ 15.128
62.07
- 43.29
Whole
Income From Bookkeeping 15,580.00
Income From Tax Preparation 11,504.00
Rent Expense 1,000.00
Insurance Expense 1,300.00
Salary Expense 3,145.00
Supplies Expense 526.00
Depreciation Expense 244.00
subtract Net Income
s
Decimals
Income From Bookkeeping 15,580.58
Income From Tax Preparation 11,504.19
Total Revenue
Rent Expense 1,000.12
Insurance Expense 1,300.26
Salary Expense 3,145.37
Supplies Expense 526.86
Depreciation Expense 244.86
Total Expenses
Net Income
Name Gross Pay Deduction 1 Deduction 2 Deduction 3 Total Deductions
Name 1 881.13 13.06 12.94 29.76
Name 2 690.13 24.94 12.58 20.38
Name 3 1086.01 18.26 26.8 24.8
Name 4 672.43 19.99 25.02 25.56
Name 5 937.85 25.66 21.59 14.08
Name 6 994.07 31.72 14.94 29.51
Name 7 1005.24 16.27 35.35 37.27
Name 8 988.26 24.7 27.42 29.17
Name 9 688.82 37.78 15.61 15.62
Totals 7943.94 212.38 192.25 226.15
Name Gross Pay Deduction 1 Deduction 2 Deduction 3 Net Pay
Name 1 881.13 13.06 12.94 29.76
Name 2 690.13 24.94 12.58 20.38
Name 3 1086.01 18.26 26.8 24.8
Name 4 672.43 19.99 25.02 25.56
Name 5 937.85 25.66 21.59 14.08
Name 6 994.07 31.72 14.94 29.51
Name 7 1005.24 16.27 35.35 37.27
Name 8 988.26 24.7 27.42 29.17
Name 9 688.82 37.78 15.61 15.62
Totals 7943.94 212.38 192.25 226.15
Name Gross Pay
Suix Chin 5,468.25
Stephanie Walting 7,667.54
Rachel Hensley 6,900.20
Totals
Assumption
Deduction Rate 0.32
Net Pay
efficient formula
=B15-SUM(C15:E15)
inefficient formula
=B16-C16-D16-E16
Deduction 1 Net Pay
1,749.84 3,718.41
2,453.61 5,213.93
2,208.06 4,692.14
6,411.51 13,624.48
d67cc0eb-3fd2-449d-8e1c-09d4495967ae.xls - Order of Operations
Arithmetic operation signs in Excel:
( ) represents Parentheses
^ represents Exponents (powers and roots)
* represents Multiplication
/ represents Division
– represents Subtraction
Order of Operations form Algebra class (Each one is left to right):
Please Parenthesis ( )
Excuse Exponents ^ 2^2 = 4 or 4^(1/2) = 2
My Dear Multiplication * and Division /
Aunt Sally Adding (SUM) + and Subtraction -
Excel's Order of Operations:
Parenthesis ( )
Ranges use of colon symbol ":"
Example: =SUM(A1:A4)
Evaluate unions (,)
Example: =SUM(A1:A4,B2:C7)
Negation (-)
Example: =-2^4 16
Example: =-(2^4) -16
Converts % (1% .01)
Exponents (^)
Example 1: 4^(1/2) = 2
Example: 3^2 = 9
Multiplication (*) and division (/), left to right
Adding (+) and division (-), left to right
Text operators (& Concatenation)
Comparative symbols: =, <>, >=, <=, <, >
If anything is still left, then left to right
Page 15 of 29
d67cc0eb-3fd2-449d-8e1c-09d4495967ae.xls - OOO (2)
3+3*2 =D1+E1*F1 3 3 2
(3+3)*2 =(D2+E2)*F2 3 3 2
2+(2+1)*3 =D3+(E3+F3)*G3 2 2 1 3
2-1+(3-1)*2 =D4-E4+(F4-G4)*H4 2 1 3 1 2
Page 16 of 29
Multiplying Whole Numbers
690
* 43 3 4
2 6
690 multiplicand factor 6
x 43 multiplier factor 4
2070 partial product (3 x 69)
+ 27600 partial product (4 x 69) 2
3
5
3
300 300 2 zeros
* 70 x 70 1 zero
21 + 3 zeros
Omitting zeros when multiplying: "remove at
Dividing Whole Numbers
256
4 Numerator = Dividend
16
2 Denominator = Divisor
4 (Numerator) 2 (Denominator) = 2 (Quotient or answer)
2 (Denominator) 4 (Numerator)
108000
900 Zeros in division:
0/4 = 0 because 0 * 4 = 0
however
4/0 = ? = in not defined because what?*0 = 4 ????
1080
Reciprocals: 1/2 and 2/1 are reciprocals
9
Omitting zeroes in division
108000 900 = 1080 9 = 120 876
2
2
3
Multiply Decimals
0.032 Items Boomerangs
* 0.070 Quantity 5.00
Price 22.25
Total
Hours 39.50 Gross -
Wage 15.75 Tax Rate 0.06
Gross Tax Rate
Divide Decimals
17.60
0.25
17.6 .25 = 1760 25 = 70.4
1,760.00
25.00
When the denominator is
15.00 between 1 and 0, the quotient
will always be bigger than the
0.25
Numerator
Different ways to express quotient: Quick way to estimate
20 29548 round up ==>
15 * 0.10 slide decimal to right
1.333333 1R5 1 1/3
Decimal Remainder Fraction slide decimal back to left <==
158 56,986
190 23
197
r) = 2 (Quotient or answer)
ause what?*0 = 4 ????
30000
1
30000
3000
Steps for solving word-application problems:
Read problem carefully – be sure you know what
List all relevant facts. List all information you want
Step 2 to find.
Step 4 Set up the problem and solve.
Write answer in words – go back and use the
Step 6 nouns and verbs from the problem.
Short list of steps for solving word-application problems:
Step 1 Read problem carefully and list all relevant facts
Step 2 Set up problem, solve, check and double check
Write answer in words – go back and use the
Step 3 nouns and verbs from the problem.
Translate words into math symbols
Add Subtract Multiply Division Equal Powers
plus less product divided by is squared
more subtract double divided into the same as raised to
more than subtracted from triple quotient equals to the second power
added to difference times goes into equal to cubed
increased by less than of divide yields to the 12th power
sum fewer twice divided equally results in
total decreased by twice as much per are
sum of loss of
increase of minus
gain of take away
The three invoices each had a total of \$14.26
listed on the bottom line. What is the total that is
owed?
If the total gross earnings for the year were
\$29,013.45 and the first \$15,000 of those gross
earnings were taxed at a rate of .10 and the
remaining amount was taxed at .15. What is the
total tax paid for the year if you round your
1) Setup for Math Problems In Excel
2)
3)
4)
5)
6)
7)
8)
9)
If there are 10 Boomerangs for sale at \$20 each and the cost for
each boom is \$10, what is the total profit if you sell them all?
Summary Exercise
Step 1 Read and list all relevant facts
1 Average wedding costs in USA in 2004
2 Cost of average 2001 wedding = \$20,357.00
Average cost has increased as the average
number of wedding guest has grown to over
3 200
4 Graph shows details
Wedding reception \$9,590.00
Rings (engagement and wedding) \$5,099.00
Photographer \$1,647.00
Misc. expense \$1,583.00
Music \$1,042.00
Bridesmaids' dresses (5) \$992.00
Bouquets and other flowers \$950.00
Rehearsal dinner \$877.00
Men's formal wear \$849.00
Mother's of the Bride apparel \$297.00
Step 2 and 3
a The total Costs shown in the graph =
The increase in the average wedding cost in
b 2004 over the cost in 2001 =
c Our cost for wedding = \$7,000.00
Cost per guest = \$32.00
The number of guests we can invite =
Since we can not invite a part of a guest, we
must round ==>
Total costs for us =
Amount left over =
Number of guests that are invited to wedding
d = 150
Budgeted amount for wedding = \$7,000.00
Amount that can be spent on each person =
Round to the nearest cent =
e Budgeted amount for wedding = \$4,000.00
Cost per guest = \$27.00
The number of guests we can invite =
Since we can not invite a part of a guest, we
must round ==>
Total costs for us =
Amount left over =
f Budgeted amount for wedding = \$1,000.00
Cost per guest = \$15.00
The number of guests we can invite =
Since we can not invite a part of a guest, we
must round ==>
Total costs for us =
Amount left over =
g Number of bridal party bouquets = 5
Cost per bridal party bouquets = \$36.25
Number of boutonnieres = 5
Cost per boutonnieres = \$7.50
Total amount budgeted for flowers = \$863.00
Amount spent for bridal party bouquets and
boutonnieres =
Amount left over
s
#1 Round to the nearest ten
number to round = 844
#2 Round to the nearest hundred
number to round = 21,958
#3 Round to the nearest thousand
number to round = 671,529
#4 Front End Round
number to round = 50,987
#5 Front End Round
number to round = 851,004
#6 Commissions Earned
Monday \$124.00
Tuesday \$88.00
Wednesday \$62.00
Thursday \$137.00
Friday \$195.00
Total = \$606.00
The total
commissions that
Suzanne earned =
#7 Total Costs
Cost per airless sprayer = \$1,540.00
Number of airless sprayers purchased = 3
Total = \$4,620.00
Cost per rototillers = \$695.00
Number of rototillers purchased = 5
Total = \$3,475.00
Number of ladders purchased = 8
Total = \$304.00
Total for all purchases = \$8,399.00
#8 Round to the nearest cent 2
Number 21.0568
#9 Round to the nearest cent 2
Number 364.345
#10 Round to the nearest cent 0
Number 7246.49
8.42
3.715
159.8
Total = 181.535
32.78
426.3
37
Total = 498.795
#13 Subtract 341.4
207.8
Difference = 133.6
#14 Subtract 3.8
0.0053
Difference = 3.7947
#15 Multiply 21.98
0.72
Product = 15.8256
#16 Multiply 218.6
0.037
Product = 8.0882
#17 Division
Numerator (Dividend) 252.008
Denominator (Divisor) 21.8
Quotient = 11.56
#18 Division
Numerator (Dividend) 57.358
Denominator (Divisor) 2.41
Quotient = 23.8
#19 Division
Numerator (Dividend) 79.135
Denominator (Divisor) 18.62
4.25
#20 Total costs
Number of solvent gallons to buy 32.6
Solvent cost per gallon \$13.48
Total = \$439.45
Number of acid gallons to buy 18.5
Acid cost per gallon \$3.56
Total = \$65.86
Total for solvent and acid = \$505.31
#21 Total costs
Roof Materials per 10x10 squares \$84.52
Installation labor per 10x10 squares \$55.75
Supplies per 10x10 squares \$9.65
Total Costs per 10x10 square \$149.92
Number of 10x10 squares 26.3
Total costs \$3,942.90
#23 How much change was received
Number of meters of electrical wire
purchased 16.5
Electrical wire cost per meter \$0.48
Number of meters of brass rod 3
Cost of brass rod per meter \$1.05
Total cost \$11.07
Amount of one fiver \$5.00
Number of fivers 3
Total amount of money given to cashier \$15.00
Amount of change he got from three fivers = \$3.93
#24 Price of bananas per pound
Price of bananas per kilogram \$1.74
Number of pounds in one kilogram 2.2
Price for 2.2 pounds of banana \$1.74
Price per 1 pound of banana \$0.79
#25 Find Number of seedlings
Ounces of concentrated fertilizer needed for
one seedling 0.058
Ounces of concentrated fertilizer available
for use 14.674
Number of seedlings that can be fertilized 253.0000
50,000
900,000
The total commissions that Suzanne earned = \$606.00
```
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https://stats.stackexchange.com/questions/295920/logistic-regression-is-a-convex-problem-but-my-results-show-otherwise
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# Logistic Regression is a Convex Problem but my results show otherwise?
I know that logistic regression is a convex problem. Furthermore, from Lemma 1.17 in these optimization lecture notes, if a function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ is convex, then the function over a restricted domain (as it were) should be convex. For example, if $f(x,y,z)$ is convex, then $g(x,y) = f(x,y,2)$ should be convex (assuming $f$ is defined at $z=2$), right?
However, I have been experimenting with my own implementation of logistic regression and have found that the loss function (ie opposite of log-likelihood) is not convex. All my code is at the bottom
I have created the following fake dataset which can be perfectly classified:
I defined a loss function as the negative log likelihood, scaled by $1/n$:
$$L(\beta) = - \frac{1}{n}\sum_{i=1}^{n} y_i \log(p_i) + (1-y_{i})\log(1-p_i)$$
where
$$p_i = \dfrac{\exp(\beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i})}{1+\exp(\beta_0 + \beta_1 x_{1i} + \beta_2 x_{2i})}$$
My optimization function gives the following solution (which seems to be right because it can perfectly classify the data),
$$\hat{\mathbf{\beta}} = (0, -20, 23)$$
If I plot the loss function at a fixed value of the third parameter $g(a,b) = L( (a, b, 23)$, it does not appear convex. I have tried this for multiple values of the third parameter.
What is going on here?
#
# Create dataset
#
x1 <- rep(seq(1, 10, length = 100),2)
x2 <- c(x1[1:100]+3 , x1[101:200]+5)
# Scale
x1 <- scale(x1)
x2 <- scale(x2)
# Store in data frame
dat <- data.frame(x1 = x1, x2 = x2,
y = c(rep(0, 100), rep(1,100))
)
# Plot
plot(dat$x1, dat$x2, col = dat$y + 1, main = "Logistic Regression Dataset") # # Loss Function # sigmoid <- function(a){ sapply(a, function(arg){ if(arg > 18){return(1)} if(arg < -18){return(0)} return(1/(1+exp(-arg))) }) } # A log function that is zero when its argument is zero, not -Inf log0 <- function(a){ sapply(a, function(arg){ if(arg==0){return(0)} return(log(arg)) }) } loss <- function(y,x,w){ # x a matrix with unity first col, w a vector if(all(x[,1] == 1) == FALSE){ stop("first column of x must be unity") } # Compute vector of dot products sigmoid.dot <- sigmoid(x %*% w) # Compute elementwise loss li <- y*log0(sigmoid.dot)+(1-y)*log0(1-sigmoid.dot) # # Take negative average and return # return(-mean(li)) } # #OMITTING SOLLUTION CODE # PROVE THAT IT WORKS: X <- cbind(rep(1, nrow(dat)), dat$x1, dat$x2) w <- c(0,-20, 23) preds <- ifelse(sigmoid(X %*% w) > 0.5, 1, 0) table(preds, dat$y)
#
# Plot the loss
#
lossPlot <- function(d){
lossf <- function(a,b){
w <- c(a, b, d)
return(loss(dat$y, X, w)) } a <- seq(-5, 5, length = 100) b <- seq(-30, 30, length = 100) z <- matrix(nrow = 100, ncol = 100) for(ii in 1:100){ for(jj in 1:100){ z[ii,jj] <- lossf(a[ii],b[jj]) } } persp(a,b,z, phi = 45, theta = 45, main = paste("Plot of restricted loss function with third coordinate", d) ) } lossPlot(5) lossPlot(23) Thanks! • It looks like you're encountering some numeric issues when the arguments to$exp\$ get very large. Aug 2, 2017 at 23:10
• 1. As others have pointed out, the approximation in R\[-18, 18] leads to some ill condition; 2. log(0) should be -INF, but you returned 0 instead. If you want to smooth this behaviour, you should return a large negative number. Returning 0 corrupted the convexity of the log function around 0, which corrupted the loss of LR subsequently. Mar 31, 2019 at 18:29
The problem with your data set is called Complete separation of the data. The likelihood associated to logistic regression models is concave, provided that there is no complete separation of the data.
The phenomenon of complete separation of the data is defined and discussed in:
Albert, Adelin, and J. A. Anderson. "On the existence of maximum likelihood estimates in logistic regression models." Biometrika 71.1 (1984): 1-10.
It has also been discussed in this site:
How to deal with perfect separation in logistic regression?
• This is weird. According to is-there-any-intuitive-explanation-of-why-logistic-regression-will-not-work-for, the complete separation problem only leads to a degenerated solution (solution goes to infinity) instead of turning the loss surface from concave to non-concave. Mar 31, 2019 at 17:45
• This logistic-model-maximum-likelihood webpage says the log-likelihood of LR is always concave (i.e.: the cost function is convex) as long as the design matrix is of full rank. Convex algebra also tells us the cost function of LR is convex (ln (1 + exp(x_i \beta)) is the log-sum-exp of affine functions) + convex (y_i \beta x_i is an affine function) = convex function. Mar 31, 2019 at 17:56
• I tried 2 experiments: Firstly I made a toy dataset without perfect separation, the non-convex surface still exists. Secondly, I removed the special treatment for |arg| > 18 in the sigmoid function, the loss surface looks convex. Now it's clear: @Commodore's answer is correct. The non-convexity is caused by approximation of sigmoid function in (-inf, -18] and [18, inf). Mar 31, 2019 at 18:13
There are two different things going on here.
1. As you correctly stated, logistic regression is a convex problem. You made your loss function not convex by altering your definition of the sigmoid function in your code when |arg|>18. Your plots include the region where |arg|>18, so it's not surprising that they don't appear convex.
2. Convexity guarantees uniqueness of minima, but not existence of minima. For example, the exponential function e^x is convex, but has no minimum. Similarly, your logistic regression problem is convex, but has no minimum. The minimization algorithm converged at (0,-20,23) because the loss function's first and second derivatives are very close to zero there.
In general, a minimum will not exist when there is complete separation, as described in Hyco's answer above.
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# Continuity of a function between $l^p$ spaces
Let us consider the function defined as $$F: l^4 \rightarrow l^6 \\ (x_1, \dots,x_n, \dots) \mapsto (x_1^{20}, \dots, x_n^{20}, \dots)$$ I am asked to prove whether this function is continuous or not. Let us try proving the continuity, then using the $$\epsilon - \delta$$ definition I need to find, given $$\epsilon > 0$$ a suitable $$\delta$$ such that $$\lvert \lvert (x_1, \dots,x_n, \dots) - (y_1, \dots,y_n, \dots) \rvert \rvert_{l^4} = \left(\sum_{n \in \mathbb{N}}\left| x_n-y_n \right|^{4}\right)^\frac{1}{4} < \delta$$ Implies $$\lvert\lvert F(x_1, \dots,x_n, \dots) - F(y_1, \dots,y_n, \dots)\rvert\rvert_{l^{6}} = \left(\sum_{n \in \mathbb{N}}\left| x_n^{20}-y_n^{20} \right|^{6}\right)^\frac{1}{6} < \epsilon$$ Anyway I am stuck here since I cannot find any good inequalities, so I am starting having doubts about its continuity.
For sequences, $$\|\boldsymbol{x}\|_q\le\|\boldsymbol{x}\|_p$$ whenever $$p\le q$$. Also, $$|x^{20}-y^{20}|=|x-y||x^{19}+\cdots+y^{19}|\le c|x-y|.$$ Hence $$\|(x_n^{20})-(y_n^{20})\|_6\le c\|x-y\|_6\le c\|x-y\|_4
To justify the constant $$c$$, note that \begin{align*}\left|\sum_{i=0}^{19}x_n^iy_n^{19-i}\right|&\le\sum_{i=0}^{19}|x_n|^i|y_n|^{19-i}\\ &\le\sum_{i=0}^{19}\frac{i|x_n|^{19}+(19-i)|y_n|^{19}}{19}\\ &= 20(|x_n|^{19}+|y_n|^{19})\\ &\le 20(\|\boldsymbol{x}\|_\infty^{19}+\|\boldsymbol{y}\|_\infty^{19})=c \end{align*}
• What about the constant c: How do you control it? I think that it depends on $x,y$ hence it shouldn't be uniform for all $x_n, y_n$. Maybe I am missing something. – JCF Jan 31 at 10:52
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# If x^3=-27,then what is x?
Apr 23, 2018
$x = - 3$
#### Explanation:
${x}^{3} = - 27$
${x}^{3} = - {3}^{3}$
$x = - 3$
Apr 23, 2018
$x = - 3$
#### Explanation:
We can take the cube root of both sides, which is the same as raising both sides to the $\frac{1}{3}$ power. Thus, we have:
$x = \sqrt[3]{- 27}$
which evaluates to
$x = - 3$ (the number in the radical is negative, and we have an odd-numbered root, so our answer must be negative).
If you can solve ${x}^{3} = 27$, you can easily find ${x}^{3} = - 27$, because the only thing that'll change is the sign of the answer. It will become negative.
Hope this helps!
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# math
posted by on .
(6) Which equation is equivalent to
4^-x=8?
a. log8 4=x
b. logx = -4
c. log -x 4=8
d. log 4 8= -x
• math - ,
1/4^x = 8
4^x = 1/8
x log4 (4) = log4 (1/8)
but log4 (4) = 1 because b^logb(a) = a
x = log4(1/8) = log4(1) - log4(8)
but log4(1) = 0 for the same reason
so
x = -log4(8) or -x = log4 (8)
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+0
# Adding and subtracting rational expressions
0
439
3
a - 3b + a + 5b
a + b a + b
Jan 11, 2018
#1
+8961
+1
$$\frac{a-3b}{a+b}\,+\,\frac{a+5b}{a+b}$$
These fractions have a common denominator so we can combine them.
$$=\,\frac{(a-3b)+(a+5b)}{a+b} \\~\\ =\,\frac{a-3b+a+5b}{a+b} \\~\\ =\,\frac{a+a+5b-3b}{a+b} \\~\\ =\,\frac{2a+2b}{a+b}$$
Let's factor a 2 out of the numerator.
$$=\,\frac{2(a+b)}{a+b}$$
Now we can reduce the fraction by (a+b) .
$$=\,2$$ This is true for all values of a and b as long as a + b ≠ 0 .
Jan 11, 2018
#1
+8961
+1
$$\frac{a-3b}{a+b}\,+\,\frac{a+5b}{a+b}$$
These fractions have a common denominator so we can combine them.
$$=\,\frac{(a-3b)+(a+5b)}{a+b} \\~\\ =\,\frac{a-3b+a+5b}{a+b} \\~\\ =\,\frac{a+a+5b-3b}{a+b} \\~\\ =\,\frac{2a+2b}{a+b}$$
Let's factor a 2 out of the numerator.
$$=\,\frac{2(a+b)}{a+b}$$
Now we can reduce the fraction by (a+b) .
$$=\,2$$ This is true for all values of a and b as long as a + b ≠ 0 .
hectictar Jan 11, 2018
#2
+1
Thank you!!!
Guest Jan 11, 2018
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# Canonical boundary conditions.
• CPL.Luke
In summary, canonical boundary conditions are constraints applied to a physical system at its boundaries to fully define its behavior. They are important because they allow for accurate predictions and meaningful observations. These conditions are determined through analysis of fundamental principles and can change over time due to various factors. They directly affect the behavior of a system by altering the way energy, mass, and momentum are transferred across the boundaries.
CPL.Luke
how do you prove/show that there really is a vector space defined by certain boundary conditions?
unfortunatly this part of pde's was glossed over in my professor's lecture notes and I don't recall him talking about it in class.
haha actually I think I just realized how to do it, you have to show that any two functions that any combination of funcions that meet the conditions must be a function that also meets the condition, which will be met with any set of boundary conditions where the function =0 at that point.
af(0) +bf'(0)=0+0=0is this correct?
Canonical boundary conditions refer to a set of boundary conditions that fully define a vector space for a given problem in partial differential equations (PDEs). These conditions are typically specified at the boundaries of a domain and are essential for finding a unique solution to the PDE problem. In order to prove or show that there is a vector space defined by certain boundary conditions, we can follow a few steps:
1. Define the problem domain and the PDE that needs to be solved. This includes specifying the boundary conditions at the boundaries of the domain.
2. Identify the type of PDE problem (e.g. elliptic, parabolic, hyperbolic) and the type of boundary conditions (e.g. Dirichlet, Neumann, Robin).
3. Use the properties of the PDE and the given boundary conditions to show that the problem satisfies the fundamental properties of a vector space. These properties include closure under addition and scalar multiplication, existence of a zero vector, and existence of additive and multiplicative inverses.
4. Show that the solution space of the PDE problem is a subset of the vector space defined by the boundary conditions. This can be done by demonstrating that the solution to the PDE satisfies the given boundary conditions.
5. Prove that the solution space of the PDE problem is a vector space by showing that it satisfies the additional properties of a vector space, such as associativity, distributivity, and commutativity.
Overall, proving the existence of a vector space defined by certain boundary conditions involves carefully examining the properties of the PDE problem and the given boundary conditions, and showing that they satisfy the fundamental properties of a vector space. This can be a rigorous process, but it is essential for ensuring the uniqueness and validity of the solution to the PDE problem.
## 1. What are canonical boundary conditions?
Canonical boundary conditions are mathematical constraints that are applied to a physical system to fully define its behavior. These conditions are typically applied at the boundaries of a system, and they provide information about how the system will behave at those boundaries.
## 2. Why are canonical boundary conditions important?
Canonical boundary conditions are important because they allow us to fully define the behavior of a physical system. By applying these constraints, we can accurately predict how the system will behave under different conditions and make meaningful scientific observations.
## 3. How are canonical boundary conditions determined?
Canonical boundary conditions are typically determined by analyzing the fundamental principles and equations that govern the behavior of a system. These conditions can also be experimentally determined by observing the behavior of the system under different conditions.
## 4. Can canonical boundary conditions change over time?
Yes, canonical boundary conditions can change over time. This can happen due to changes in the underlying physical system or as a result of external factors such as temperature, pressure, or other environmental conditions.
## 5. How do canonical boundary conditions affect the behavior of a system?
Canonical boundary conditions directly affect the behavior of a system by constraining its behavior at the boundaries. These conditions can alter the way energy, mass, and momentum are transferred across the boundaries, ultimately impacting the overall behavior of the system.
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GMAT Exam Prep
Book
• Sorry, this book is no longer in print.
Description
• Dimensions: 8-3/8x10-7/8
• Pages: 432
• Edition: 1st
• Book
• ISBN-10: 0-7897-3593-8
• ISBN-13: 978-0-7897-3593-5
The GMAT Exam Prep provides a comprehensive text for studying for the GMAT test. This book takes each topic and provides numerous examples, explanations, strategies, and tips for both understanding the content of knowledge that the test covers, as well as in helping in preparing the student to score higher on the test. In addition to coverage of all topic areas, the book has an expanded topical review coverage on the Quantitative section, due to the fact that many students find they don't understand this topic very well. The book also offers specific strategies for answering various question types, reviews the psychology of testing, and provides a guide for understanding scores. Finally the book has four full practice exams with detailed answers to all questions covering all topics areas.
Sample Content
Online Sample Chapters
Part I: Introduction to the GMAT
Chapter 1 Overview of the GMAT
The GMAC
Format of the GMAT
What’s Next
Chapter 2 Taking the GMAT
Registering for the GMAT
Paying for the GMAT
Taking the GMAT
The Testing Center
What to Bring to the Testing Center
Test Takers with Disabilities
Scoring the GMAT
What Your Scores Mean to Schools
A Note on Scoring the Practice Tests in This Book
Retaking the GMAT
What’s Next
Part II: Preparing for the GMAT
Chapter 3 Diagnostic Test with Answers and Explanations
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Section 3
Section 4
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Chapter 4 GMAT General Testing Strategies
KSA (There Is Some Science Behind This Process)
Focus on the Easy Stuff
Stay “On Point”
Simplify
Anatomy of the GMAT
Guess Wisely
Manage Stress
Relax to Succeed
Specific Relaxation Techniques
A Note on Music
What to Expect on Test Day
What’s Next
Chapter 5 Introduction to GMAT Logic
Arguments
Context
Chain of Reasoning
Assumptions
Validity, Truth, and Soundness
Conditionals
Contrapositive
Sufficiency and Necessity
Compound Conditionals
Negation
Fallacies
Slippery Slope
Correlation—Causation
Equivocation
Appeal to Authority
Appeal to Majority—Quantity-Quality Fallacy
Circular Argument
Begging the Question
Sampling Error
Percent Versus Number
What’s Next
Chapter 6 GMAT Analytical Writing Assessment (AWA)
Scoring the Analytical Writing Sections
What the GMAT Readers Are Looking For
Sample Essay Prompt: Analysis of an Issue
Sample Essay Prompt: Analysis of an Argument
Writing Techniques
Writing Strategies
Practice Writing Prompts
Chapter 7 GMAT Quantitative
Multiple Choice
Anatomy of a Multiple Choice Question
General Strategies for Multiple-Choice Questions
Data Sufficiency
Anatomy of a Data Sufficiency Question
General Strategies for Data Sufficiency Questions
What’s Next
Chapter 8 GMAT Verbal
Anatomy of a GMAT Reading Comprehension Question
General Strategies for Reading Comprehension Questions
Critical Reasoning
Anatomy of a Critical Reasoning Question
General Strategies for Critical Reasoning Questions
Critical Reasoning Question Types
A Note on Question Format
Practice Critical Reasoning Questions
Sentence Correction
Anatomy of a GMAT Sentence Correction Question
General Strategies for Sentence Correction Questions
Practice Sentence Correction Questions
What’s Next
Part III: GMAT Content Area Review
Chapter 9 Quantitative Review
Numbers and Operations
Properties of Integers
Real Numbers
Order of Operations (PEMDAS)
Decimals
Fractions, Rational Numbers, and Scientific Notation
Ratio, Proportion, and Percent
Squares and Square Roots
Arithmetic and Geometric Sequences
Factors and Multiples
Mean, Median, and Mode
Probability and Outcomes
Absolute Value
Algebra and Functions
Factoring
Exponents
Inequalities
Word Problems
Geometry
Coordinate Geometry
Triangles
Other Polygons
Circles
Three-Dimensional Figures
Data Analysis
Internalize the Rules
Chapter 10 Verbal Review
Subject/Verb Agreement
Person
Number
Voice
Verb Tense
Nouns and Pronouns
Personal Pronouns
Relative Pronouns
Indefinite Pronouns
Pronoun Consistency
Verbs and Verb Forms
Simple Past Versus Past Participle
Parallel Construction
Run-on Sentences
Sentence Fragments/Incomplete Sentences
Misplaced Modifiers
Idiom
Rhetoric
Commonly Misused Words
Accept, Except
Affect, Effect
Among, Between
Assure, Insure, Ensure
Because, Since
Compare to, Compare with
Complement, Compliment
Farther, Further
Fewer, Less
Imply, Infer
Its, It’s
Lay, Lie
Like, Such As
Number, Amount
Precede, Proceed
Principal, Principle
Set, Sit
Than, Then
That, Which
There, Their, They’re
To, Too, Two
Whether, If
Your, You’re
Internalize the Rules
Part IV: Practicing for the GMAT
Chapter 11 Practice Test 1 with Answers and Explanations
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Section 3
Section 4
Analytical Writing: Sections 1 and 2
Quantitative: Section 3
Verbal: Section 4
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Chapter 12 Practice Test 2 with Answers and Explanations
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Section 3
Section 4
Analytical Writing: Sections 1 and 2
Quantitative: Section 3
Verbal: Section 4
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Chapter 13 Practice Test 3 with Answers and Explanations
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Section 3
Section 4
Analytical Writing: Sections 1 and 2
Quantitative: Section 3
Verbal: Section 4
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Chapter 14 Practice Test 4 with Answers and Explanations
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Section 3
Section 4
Analytical Writing: Sections 1 and 2
Quantitative: Section 3
Verbal: Section 4
Section 1: Analysis of an Argument
Section 2: Analysis of an Issue
Section 3: Quantitative
Section 4: Verbal
Part V: Appendixes
Appendix A GMAT Vocabulary List
Appendix B Glossary of GMAT Math Terms
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0
# What are the perfect squares 100-225?
Wiki User
2010-05-11 20:05:05
325
Jaylin Konopelski
Lvl 10
2021-02-27 00:47:23
Study guides
20 cards
## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials
➡️
See all cards
3.8
2018 Reviews
Wiki User
2010-05-11 20:05:05
If you include 100 and 225, there are 6:
10 squared = 100
11 squared = 121
12 squared = 144
13 squared = 169
14 squared = 196
15 squared = 225
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edHelper subscribers - Create a new printable
Sample edHelper.com - Number Word Searches Worksheet
Name _____________________________ Date ___________________
Solve.
1 1 x 7 =
2 5 x 1 =
3 1 x 1 =
4 12 x 1 =
5 4 x 1 =
6 1 x 3 =
7 11 x 1 =
8 9 x 1 =
9 2 x 1 =
10 1 x 10 =
11 6 x 1 =
12 8 x 1 =
Name _____________________________ Date ___________________
Find the equations, from the previous page, in the number word search.
6 + 9 1 7 = 3 2 + 6 x 1 = 1 2 4 = 8 = 0 5 9 + 8 2 6 = + 1 0 4 7 + = 3 5 + 0 8 = 5 3 = x 1 6 2 4 + 9 7 6 = 2 + 1 x 1 0 = 1 0 = 5 7 0 + 3 = 1 9 4 + 8 = 3 + 0 6 1 9 = + 4 7 8 1 5 = 2 + 4 0 2 = + 1 9 8 = 8 x 1 = 8 6 7 3 + 5 = + 4 8 6 = 5 9 + 3 7 9 = 7 0 = 2 1 6 3 + 8 7 5 = 1 + 6 5 + 9 2 = 4 0 6 5 = 9 + 3 6 0 = 7 = = 2 + 4 1 + 8 = 0 8 7 1 9 5 + = 6 x = 2 3 + = 4 7 1 + 9 = 0 2 3 = 5 1 1 4 + 6 8 + = 2 9 3 8 1 x 7 = 1 2 6 1 1 5 + = 7 4 = + 1 4 1 0 = + 4 x x 2 5 6 8 + 3 1 = 9 0 + 7 8 1 = 0 7 = = x 6 4 + 2 = 5 3 9 x x + 2 7 1 2 = + 5 1 2 x 1 = 1 2 1 x 1 0 = 1 2 6 8 6 5 = + 1 9 3 4 = 0 6 1 1 4 1 1 + 1 x 1 = 6 7 = 9 2 + 0 = 3 8 + 5 1 2 = 0 + 6 5 7 = 1 3 4 + 8 + = = 9 2 9 9 = 3 + 4 6 5 = 0 8 7 + 1 = + 5 = 9 0 3 6 + = 4 8 1 3 = 2 + 7 1 1 1 4 7 + = = 6 2 0 5 3 = 9 8 + + 6 9 1 1 x 1 = 1 1 5 = 4 1 0 7 + 3 = 8 4 2 + 5 2 = 1 9 + 1 0 = 7 8 + 6 4 = 3 + 2 = 1 3 9 0 7 + 6 3 9 = 5 + 8 4 = x 5 2 4 7 = x 6 + 8 1 0 = 7 9 3 + + 9 5 = 6 x 3 = 1 4 + 8 7 x = 0 2 + 7 5 1 9 + 4 6 1 9 = 3 2 = = 0 + 8 + 1 6 8 = 4 1 7 6 0 3 + 5 = 2 1 = + 9 3 5 7 = 0 2 = + 4 8 6 + 7 1 9 = 8 9 4 = 6 + 0 9 3 = = 2 + 7 5 + 1 = 7 = 2 3 6 = 4 + 4 8 0 = 5 x 1 + 9 5 2 4 = + 3 + 9 = = 3 0 6 8 = 7 + 1 5 1 3 + 6 7 4 = 1 = 2 8 + 1 0 9 1 8 4 = 2 + 6 9 7 = 1 3 x + 5 0 + = 2 8 3 + 1 = 0 9 6 7 4 1 + 5 = 1 7 0 = 8 + 4 + 9 = 5 2 + x = 1 6 3 2 1 = 0 3 + 5 6 + 4 8 = 9 2 x 1 = 2 7 4 + 9 8 = 0 = 1 7 3 + 5 5 + 6 2 2 7 = 3 5 4 8 = + 9 6 0 + = 1 0 4 7 + = 5 + 3 = 2 6 8 1 + 9 6 5 = 9 + 8 2 7 = 1 3 + 0 4 = 6 3 + = 2 8 1 4 0 + 9 = = 7 5 1 + 8 = 6 + 4 9 5 2 + = 7 3 0
1 x 1 0 = 1 0 8 x 1 = 8 6 5 x = 1 1 1 = x x 1 2 x 1 = 1 2 6 5 1 = 9 3 1 = 1 1 x 1 = 1 1 4 1 3 x x 7 x 1 9 = 1 = 7 4 x 1 2 x 1 = 2
1 1 x 7 = 7
2 5 x 1 = 5
3 1 x 1 = 1
4 12 x 1 = 12
5 4 x 1 = 4
6 1 x 3 = 3
7 11 x 1 = 11
8 9 x 1 = 9
9 2 x 1 = 2
10 1 x 10 = 10
11 6 x 1 = 6
12 8 x 1 = 8
edHelper subscribers - Create a new printable
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# CSC445: Theory of Computation, Spring 2013
```Dr. Hasmik Gharibyan
Computer Science Department
California Polytechnic State University
San Luis Obispo, CA, 93407
CSC445: Theory of Computation, Spring 2013
Midterm 2 Study Sheet
Midterm 2 is scheduled on Wednesday, May 22.
This exam covers topics discussed in the following chapters and sections of your textbook:
- Chapter 5 (sections 4-6)
- Chapter 6 (sections 1-6)
Use lecture handouts for guidance. Here is the list of handouts in the chronological order:
- Nondeterministic Finite Automata
- Nondeterministic Finite Automata with -transitions
- Removing nondeterminism
- Finite Automata and Regular Sets
- Regular Grammars and Finite Automata
- Closure Properties of Regular Languages
- The Pumping Lemma For Regular Languages (with one-page handouts for examples)
In midterm 2 you can expect the following type questions:
Part1. Exercises on learned topics (similar to homework exercises):
Construct an NFA (or NFA-) to accept the given language.
Construct an equivalent DFA for the given NFA (or NFA-) using the Algorithm 5.6.3.
Construct a regular expression for the given finite automaton using the Algorithm 6.2.2.
You will be asked to provide the graph after each deleted node (after each loop-iteration).
Construct a finite automaton for the given regular grammar using the algorithm presented
in Theorem 6.3.1.
Construct a regular grammar for the given finite automaton using the algorithm presented
in Theorem 6.3.2.
Prove that the given language is regular.
To prove a language is regular you need to do one of the following:
(i) construct a regular expression to specify that language,
(ii) construct a regular grammar to generate that language,
(iii) construct a finite automaton to accept that language,
(iv) show that it can be obtained from regular languages using closure properties.
Note: you will need to prove the regularity of each participating language – you can
not just say “we know L1 is regular” – you need to prove it.
Using the Pumping Lemma, prove that the given language is not regular.
There will be a question where you will be asked to provide the whole proof. You will
also have additional short-answer questions asking you to identify all strings in a given
list (or give an example of a string) that would be appropriate to take as the z string to
correctly carry out the proof of non-regularity of a given language.
Continued on the back
1
Dr. Hasmik Gharibyan
Part2. Short answer questions on learned concepts, definitions, theorems, results (similar
to quiz questions posted on the course site: www.csc.calpoly.edu/~hghariby/CSC445
I expect you to know the following definitions, concepts, theorems, conclusions:
Chapter 5: Nondeterministic finite automata (NFA and NFA-), -transitions,
acceptance of a string, language, machine configuration, tracing of a computation,
-closure, and t functions, Lemma 5.5.2, Theorem 5.5.3, Theorem 5.6.4.
Chapter 6: Expression graphs (minimal expression graphs and regular expressions for
each one’s language), the relationships between Regular Sets, Regular Grammars
and Finite Automata. Theorem 6.2.3 (Kleene), Theorems 6.4.1-6.4.3 (closure
properties), Theorem 6.6.3 (Pumping Lemma), Theorem 6.6.4, Corollary 6.6.5,
Corollary 6.6.6.
2
```
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+0
# Fraction Question
0
277
3
+524
How would I express (2x - 1) / (x + 2) as A + B / (x+2), where A and B are integers?
I can do 2x / (x + 2) - 1 / (x+2) of course, but that doesn't work...
Any ideas?
EDIT: Sorry if I wasn't clear, but it's NOT:
$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}} = {\mathtt{A}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{B}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$
But instead I need to change:
$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$
Into something that looks like:
$${\mathtt{A}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{B}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$
Thank you!
Will85237 Jan 19, 2015
#2
+91412
+10
I think anon's logic should work but I would do it much more simply.
$$\frac{2x-1 }{ x+2}\\\\ =\frac{2(x+2)-1-4}{x+2}\\\\ =\frac{2(x+2)-5}{x+2}\\\\ =\frac{2(x+2)}{x+2}+\frac{-5}{x+2}\\\\ =2+\frac{-5}{x+2}$$
Melody Jan 19, 2015
Sort:
#1
+5
## Now just solve for B. 🚲
Guest Jan 19, 2015
#2
+91412
+10
I think anon's logic should work but I would do it much more simply.
$$\frac{2x-1 }{ x+2}\\\\ =\frac{2(x+2)-1-4}{x+2}\\\\ =\frac{2(x+2)-5}{x+2}\\\\ =\frac{2(x+2)}{x+2}+\frac{-5}{x+2}\\\\ =2+\frac{-5}{x+2}$$
Melody Jan 19, 2015
#3
+91412
+5
YES anon understood what you wanted and gave you a correct answer!
I shall show you
$$\\\frac{2x-1}{x+2}=A+\frac{B}{x+2}\\\\ multiply both sides by (x+2)\\ 2x-1=A(x+2)+B\\ 2x-1=Ax+2A+B\\ 2x-1=Ax+(2A+B)\\ equating co-efficients\\ 2=A\\ -1=2A+B\\ -1=4+B\\ B=-5\\ so\\ \frac{2x-1}{x+2}=2+\frac{-5}{x+2}\\\\$$
so maybe you owe anon an appology
Melody Jan 19, 2015
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# WEEK 5- CLIC TARGETS
C- Counting- Core Numbers- Decimal Place Value- working our the place value of decimal numbers and ordering a range of decimal numbers from smallest to largest and visa versa.
L- Learn it’s- Times Tables- recalling times table facts with speed in relation to division questions
I- It’s Nothing New- Large Sum Multiplication- using coin multiplication to solve sums and comparing with other methods
C- Calculations- Decimal Multiplication and Division- using a preferred method to solve multiplication and division sums with numbers with decimals
WEEK 4- SPELLING
# PUPIL OF THE WEEK- WEEK 4, TERM 3
🌈⭐️✨CONGRATULATIONS TO ✨⭐️🌈
Archie!
Archie worked really hard in his numeracy CLIC sessions this week. His positive attitude and determination allowed him to learn some tricky maths. He also tried really hard with his poetry writing this week, writing a brilliant poem in the rhyming scheme ABAB.
# WEEK 4- CLIC TARGETS
C- Counting- Core Numbers- Decimal Place Value- working our the place value of decimal numbers and ordering a range of decimal numbers from smallest to largest and visa versa.
L- Learn it’s- Times Tables- recalling times table facts with speed in relation to division questions
I- It’s Nothing New- Large Sum Multiplication- using coin multiplication to solve sums and comparing with other methods
C- Calculations- Decimal Multiplication and Division- using a preferred method to solve multiplication and division sums with numbers with decimals
# TERM 3, WEEK 4- SPELLING (REVISION AND ASSESSMENT WEEK)
This week we are completing assessments and revising previous phonemes!
# SCIENCE- RESPIRATORY SYSTEM LESSON
Primary 7 have been looking at the Respiratory system. In order to investigate and attain a scientific understanding of the Respiratory system, we have been conducting experiments looking at the impact of exercise on our breathing.
We began by assessing how big we could blow the balloon up in one breath. Then we compared it to how we could blow up a balloon after doing star jumps or jogging on the spot. We used a string to measure the difference.
We were all able to make scientific predictions, review these and apply our findings to facts.
We all loved this fun experiment and now know the importance of our lungs, breathing and respiratory system
.
# PUPIL OF THE WEEK- WEEK 3, TERM 3
🌈⭐️✨CONGRATULATIONS TO ✨⭐️🌈
Lee!
Lee has given excellent effort in all of his learning activities this week and has made great choices that have shown his maturity and determination to make great progress. Well done Lee!
# WEEK 3- CLIC TARGETS
C- Counting- Core Numbers- Ordering Numbers- ordering a range of decimal numbers from smallest to largest and visa versa
L- Learn it’s- Times Tables- recalling times table facts with speed in relation to division questions
I- It’s Nothing New- Large Sum Multiplication- using coin multiplication to solve sums and comparing with other methods
C- Calculations- Decimal Multiplication and Division- using a preferred method to solve multiplication and division sums with numbers with decimals
Archimedes have differing targets- these can be found on Google Classroom.
WEEK 3- SPELLING
# PUPIL OF THE WEEK- WEEK 2, TERM 3
🌈⭐️✨CONGRATULATIONS TO ✨⭐️🌈
Maja
Maja has put in fantastic effort in all areas of her learning this week, particularly with her limerick. Maja has shown her unique artistic skills while creating her own Steven Brown inspired McCoo, great job Maja.
## Just another blogs.glowscotland.org.uk – Aberdeen site
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## Sunday, May 25, 2008
### 22. Differentiation - Revision points 1
4. Fundamental rules for differentiation
1. d/dx of constant = 0
2. d/dx of (c.f(x)) = c d/dx of f(x)
3. d/dx of (f(x) ±g(x)) = d/dx of f(x) ± d/dx of (g(x))
4. Product rule d/dx of uv = u*dv/dx + v*dv/dx
5. quotient rule d/dx of u/v = [v*du/dx – u*dv/dx]/v²
5. Relation between dy/dx and dx/dy
dy/dx = 1/{dx/dy)
6. Differentiation of implicit functions
If variables are given as f(x,y) = 0 and if it is not possible to find y as a function of x in the form y = ф(x), then y is said be an implicit function of x.
To find dy/dx in such a case, differentiate both sides of the equation with respect to x, by writing the derivative of g(y) w.r.t. to x as (dg/dy)*(dy/dx).
7. Logarithmic differentiation
To find derivatives of the functions of the form [f(x)] g(x)
Procedure is:
Let y = [f(x)] g(x)
Take logarithms on both sides
Log y = g(x)*log [f(x)]
Differentiate w.r.t. x
(1/y)*dy/dx = g(x)*(1/f(x))*d(f(x))/dx + log [f(x)]*d[g(x)]/dx
Therefore dy/dx = (1/y)*[ g(x)*(1/f(x))*d(f(x))/dx + log [f(x)]*d[g(x)]/dx]
8. Differentiation of parametric form
If x = f(t) and y = g(t) are given and we have to find dy/dx
Then, first find dy/dt and dx/dt
dy/dx will be obtained as (dy/dt)/(dx/dt)
9. Differentiation of a function with respect to another function
u = f(x) and v = g(x) be two functions. To find the derivative of f(x) with respect of g(x) or du/dv use the formula
du/dv = (du/dx)/(dv/dx)
10. Higher order derivatives
Derivative of y w.r.t. x = dy/dx
Derivative of dy/dx w.r.t. x = d²y/dx²
and so on.
The alternative notations of higher order derivatives are
dy/dx, d²y/dx²
y1, y2
y’, y’’
Dy, D²y
f’(x), f’’(x)
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Resources tagged with Factors and multiples similar to Dice Stairs:
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Broad Topics > Numbers and the Number System > Factors and multiples
Scoring with Dice
Age 7 to 11 Challenge Level:
I throw three dice and get 5, 3 and 2. Add the scores on the three dice. What do you get? Now multiply the scores. What do you notice?
Abundant Numbers
Age 7 to 11 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
Sweets in a Box
Age 7 to 11 Challenge Level:
How many different shaped boxes can you design for 36 sweets in one layer? Can you arrange the sweets so that no sweets of the same colour are next to each other in any direction?
Seven Flipped
Age 7 to 11 Challenge Level:
Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time.
It Figures
Age 7 to 11 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
The Moons of Vuvv
Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
Two Primes Make One Square
Age 7 to 11 Challenge Level:
Can you make square numbers by adding two prime numbers together?
Multiplication Squares
Age 7 to 11 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
Zios and Zepts
Age 7 to 11 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
A Mixed-up Clock
Age 7 to 11 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
Tiling
Age 7 to 11 Challenge Level:
An investigation that gives you the opportunity to make and justify predictions.
Neighbours
Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
Being Resilient - Primary Number
Age 5 to 11 Challenge Level:
Number problems at primary level that may require resilience.
Fractions in a Box
Age 7 to 11 Challenge Level:
The discs for this game are kept in a flat square box with a square hole for each. Use the information to find out how many discs of each colour there are in the box.
Three Dice
Age 7 to 11 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
A Square Deal
Age 7 to 11 Challenge Level:
Complete the magic square using the numbers 1 to 25 once each. Each row, column and diagonal adds up to 65.
Sets of Numbers
Age 7 to 11 Challenge Level:
How many different sets of numbers with at least four members can you find in the numbers in this box?
Gran, How Old Are You?
Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
Number Tracks
Age 7 to 11 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
Mystery Matrix
Age 7 to 11 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
Multiply Multiples 2
Age 7 to 11 Challenge Level:
Can you work out some different ways to balance this equation?
Number Detective
Age 5 to 11 Challenge Level:
Follow the clues to find the mystery number.
Pebbles
Age 7 to 11 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
Multiply Multiples 1
Age 7 to 11 Challenge Level:
Can you complete this calculation by filling in the missing numbers? In how many different ways can you do it?
Multiples Grid
Age 7 to 11 Challenge Level:
What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
Being Collaborative - Primary Number
Age 5 to 11 Challenge Level:
Number problems at primary level to work on with others.
Fitted
Age 7 to 11 Challenge Level:
Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
Making Pathways
Age 7 to 11 Challenge Level:
Can you find different ways of creating paths using these paving slabs?
Curious Number
Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
Ip Dip
Age 5 to 11 Challenge Level:
"Ip dip sky blue! Who's 'it'? It's you!" Where would you position yourself so that you are 'it' if there are two players? Three players ...?
Surprising Split
Age 7 to 11 Challenge Level:
Does this 'trick' for calculating multiples of 11 always work? Why or why not?
Round and Round the Circle
Age 7 to 11 Challenge Level:
What happens if you join every second point on this circle? How about every third point? Try with different steps and see if you can predict what will happen.
Multiply Multiples 3
Age 7 to 11 Challenge Level:
Have a go at balancing this equation. Can you find different ways of doing it?
Got it for Two
Age 7 to 14 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
Have You Got It?
Age 11 to 14 Challenge Level:
Can you explain the strategy for winning this game with any target?
Ducking and Dividing
Age 7 to 11 Challenge Level:
Your vessel, the Starship Diophantus, has become damaged in deep space. Can you use your knowledge of times tables and some lightning reflexes to survive?
In the Money
Age 7 to 11 Challenge Level:
One quarter of these coins are heads but when I turn over two coins, one third are heads. How many coins are there?
Path to the Stars
Age 7 to 11 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
What Is Ziffle?
Age 7 to 11 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
What Two ...?
Age 7 to 11 Short Challenge Level:
56 406 is the product of two consecutive numbers. What are these two numbers?
Sets of Four Numbers
Age 7 to 11 Challenge Level:
There are ten children in Becky's group. Can you find a set of numbers for each of them? Are there any other sets?
Crossings
Age 7 to 11 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?
Tom's Number
Age 7 to 11 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
Age 7 to 11 Challenge Level:
If you have only four weights, where could you place them in order to balance this equaliser?
Divide it Out
Age 7 to 11 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
Factor-multiple Chains
Age 7 to 11 Challenge Level:
Can you see how these factor-multiple chains work? Find the chain which contains the smallest possible numbers. How about the largest possible numbers?
Factor Lines
Age 7 to 14 Challenge Level:
Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
What Do You Need?
Age 7 to 11 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
American Billions
Age 11 to 14 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
Which Is Quicker?
Age 7 to 11 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
| 2,092 | 8,698 |
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https://commons-credit-portal.org/how-to-calculate-the-interest-rate-on-a-car-loan/
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# How to Calculate the Interest Rate on a Car Loan
How to Calculate the Interest Rate on a Car Loan
Checkout this video:
## Introduction
Car loans are a common way to finance the purchase of a car. The interest rate on a car loan is determined by a number of factors, including the type of loan, the lender, the borrower’s credit score, and the length of the loan. In general, the interest rate on a car loan is higher than the interest rate on a personal loan or home equity loan.
To calculate the interest rate on a car loan, you will need to know the purchase price of the car, the down payment, the length of the loan, and the annual percentage rate (APR). The APR is the interest rate plus any fees charged by the lender.
Here’s an example:
You are buying a car for \$20,000. You have a down payment of \$4,000 and you will finance the remaining \$16,000 for 60 months at an APR of 8%.
To calculate your monthly payment, you would first need to calculate your interest charge. To do this, you would multiply your remaining balance by your APR (0.08). This gives you an interest charge of \$128 per month (\$16,000 x 0.08 = \$1,280).
To calculate your monthly payment amount, you would add your interest charge to your remaining balance and divide by your number of payments (60). This gives you a monthly payment amount of \$322 (\$1,280 + \$16,000 = \$17,280; \$17,280 / 60 = \$288).
## How to calculate the interest rate on a car loan
In order to calculate the interest rate on a car loan, you will need to know the purchase price of the car, the down payment, the length of the loan, and the APR. You can find all of this information on the car loan calculator on our website.
The first step is finding the advertised rate, which is the annual percentage rate, or APR. This is the stated rate that the lender charges, and it’s the rate that’s used to calculate your monthly payment. You can find the APR in the loan contract or in the ad for the loan.
### Find the true cost
The interest rate on a car loan is the cost of borrowing money from the lender. The annual percentage rate (APR) is the interest rate charged for the entire term of the loan, even if it is for a shorter period of time. Most car loans are during periods of 4 years or less, which means that you will be charged interest for the entire length of the loan. The only way to avoid paying interest on a car loan is to pay cash for the car.
When you’re car shopping, it’s important to know the true cost of ownership, which includes not just the monthly payment but also fuel, insurance, maintenance and repairs, and depreciation. That’s why it’s important to calculate the interest rate on a car loan. Here’s how:
1. Find the sticker price of the car and add any options you plan to purchase. This is the total purchase price.
2. Find the down payment amount. This is typically 20% of the purchase price but may be more or less depending on your credit score and other factors.
3) Add in any taxes and fees associated with the purchase price and down payment amount. These can vary by state and by dealer but may include things like sales tax, registration fees, documentation fees, etc.
4) Calculate your monthly payment by subtracting your down payment from your total purchase price and dividing that number by the number of months in your loan term (often 36 or 48). This is your principal plus interest payment each month.
5) Finally, calculate your interest rate by dividing your monthly interest payments (calculated in Step 4) by your total purchase price plus taxes and fees (calculated in Step 3). This will give you your APR—the true cost of borrowing money to finance your car purchase.
### Find the money factor
To calculate the interest rate on a car loan, you need to know the money factor. The money factor is a number that represents the interest rate. To find the money factor, you need to divide the interest rate by 2400. For example, if the interest rate is 4%, the money factor would be 0.004 / 2400 = 0.0001666%. To calculate the interest rate, you need to multiply the money factor by 2400.
### Find the APR
The interest rate on a car loan is the cost of borrowing money from a lender, expressed as a percentage of your loan amount. The interest rate you pay will affect both the total amount you pay for your car and how much your monthly payments will be. Before you start shopping for a car, it’s important to understand how car loan interest works and use that knowledge to get the best deal on financing.
Most lenders will quote you an Annual Percentage Rate (APR) when you apply for a car loan, which includes the interest rate plus other fees and charges. The APR is the best way to compare different offers, because it shows you the total cost of borrowing money.
To calculate the interest rate on your car loan, divide the APR by the number of days in a year (365). Then, multiply that number by the number of days between when you take out the loan and when it is due (the “term” of the loan).
For example, let’s say you borrow \$20,000 at 4% APR for 60 months (5 years). To calculate your daily interest rate, divide 4 by 365 to get 0.0109%. Then, multiply 0.0109% by 153 (the number of days in 5 years), to get 0.1673%. This means that your daily interest rate is 0.1673%.
Now let’s say you want to calculate your monthly interest rate. To do this, divide 4 by 12 to get 0.3333%. Then, multiply 0.3333% by 60 (the number of months in 5 years), to get 1%. This means that your monthly interest rate is 1%.
## Conclusion
The interest rate on a car loan is important to consider when you are buying a car. The higher the interest rate, the more you will have to pay for the car over time. There are a few things that you can do to try to get a lower interest rate on your loan, such as shopping around for a better deal, or considering a shorter loan term. You can also try to negotiate with the dealership or lender to get a lower interest rate. Ultimately, the interest rate that you get on your loan will depend on your credit score and the market conditions at the time that you apply for the loan.
| 1,409 | 6,139 |
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| 3.71875 | 4 |
CC-MAIN-2024-38
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https://www.proofwiki.org/wiki/Sum_of_Sequence_of_Products_of_3_Consecutive_Integers
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| 1,046,106,966 | 11,881 |
# Sum of Sequence of Products of 3 Consecutive Integers
## Theorem
$\ds \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2}$ $=$ $\ds 1 \times 2 \times 3 + 2 \times 3 \times 4 + \cdots + \paren {n - 1} n \paren {n + 1} + n \paren {n + 1} \paren {n + 2}$ $\ds$ $=$ $\ds \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$
## Proof
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$
### Basis for the Induction
$\map P 1$ is the case:
$\ds \sum_{j \mathop = 1}^1 j \paren {j + 1} \paren {j + 2}$ $=$ $\ds 1 \times 2 \times 3$ $\ds$ $=$ $\ds \dfrac {1 \times 2 \times 3 \times 4} 4$ $\ds$ $=$ $\ds \dfrac {1 \paren {1 + 1} \paren {1 + 2} \paren {1 + 3} } 4$
Thus $\map P 1$ is seen to be true.
This is our basis for the induction.
### Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
$\ds \sum_{j \mathop = 1}^k j \paren {j + 1} \paren {j + 2} = \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4$
Then we need to show:
$\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} \paren {j + 2} = \dfrac {\paren {k + 1} \paren {k + 2} \paren {k + 3} \paren {k + 4} } 4$
### Induction Step
This is our induction step:
$\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} \paren {j + 2}$ $=$ $\ds \sum_{j \mathop = 1}^k j \paren {j + 1} \paren {j + 2} + \paren {k + 1} \paren {k + 2} \paren {k + 3}$ $\ds$ $=$ $\ds \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4 + \paren {k + 1} \paren {k + 2} \paren {k + 3}$ Induction hypothesis $\ds$ $=$ $\ds \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} + 4 \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4$ $\ds$ $=$ $\ds \dfrac {\paren {k + 1} \paren {k + 2} \paren {k + 3} \paren {k + 4} } 4$ algebra
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
$\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$
$\blacksquare$
| 1,014 | 2,262 |
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| 4.625 | 5 |
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https://documen.tv/question/how-many-meters-are-in-5-0-cm-500-0-050-0-0005-0-5-17514595-55/
| 1,632,009,049,000,000,000 |
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| 257,914,398 | 15,584 |
## How many meters are in 5.0 cm? 500 0.050 0.0005 0.5
Question
How many meters are in 5.0 cm?
500
0.050
0.0005
0.5
in progress 0
2 months 2021-07-24T01:55:35+00:00 2 Answers 0 views 0
1. 0.05 meters i’m pretty sure
2. Explanation:
100cm=1m
5 cm= x
cross multiple
x=5/100
=0.05m
| 131 | 288 |
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| 3.6875 | 4 |
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https://www.smore.com/hj5hn-exam-review
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| 1,056,993,571 | 10,791 |
# Exam Review
## Take a look at these videos and examples:
Simplifying Expressions:
https://youtu.be/KL5mZ1gArQE
Solving Equations:
https://youtu.be/oIxxqztQz3Y
Solving equations with fractions:
https://youtu.be/vk4CP10bys0
2 other suggestions for solving equations with fractions:
1) Turn all fractions into decimals and solve like you would with integers.
Subtract .25 from both sides of the equation
.3333x + .25 = .75
- .25 = -.25
.333x = .50
Divide both sides by .333 so x = =1.5
2) Multiply the equation by a number that will get rid of all of the denominators.
Writing the equation of a line given 2 points:
https://youtu.be/4il4haYASys
(3, 4) and (6, 10)
1) Find the slope (m). Change in the Ys over Change in the Xs
2) Pick one of the two points. It does not matter which one. I will chose (3, 4) because I like smaller numbers J
3) You now have a Y and an X and an M so you can substitute into the equation of a line y=mx+b
y = 4 x = 3 and m = 2
4= 2(3) +b
4 = 6 + b
-6 -6
-2 = b
So y=mx + b à y = 2x - 2
Finding the equation of a line given a point and a parallel line:
https://youtu.be/TrONIeOpJHg
Finding the equation of a line given a point and a perpendicular line:
https://youtu.be/QCWVlSr0khQ
| 395 | 1,240 |
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| 4.1875 | 4 |
CC-MAIN-2019-43
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| 0.844597 |
http://www.docstoc.com/docs/656433/Absolute-Value-Equations---PowerPoint
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| 208,176,077 | 14,503 |
# Absolute Value Equations - PowerPoint
Document Sample
``` SOLVING ABSOLUTE-VALUE EQUATIONS
You can solve some absolute-value equations using mental math. For instance, you learned that the equation | x | 8 has two solutions: 8 and 8. To solve absolute-value equations, you can use the fact that the expression inside the absolute value symbols can be either positive or negative.
Solving an Absolute-Value Equation
Solve | x 2 | 5
SOLUTION
The expression x 2 can be equal to 5 or 5.
x 2 IS POSITIVE x 2 IS POSITIVE
|| x 2 || 5 x2 5 x 2 5 x 2 5
x 2 IS NEGATIVE |x2|5 x 2 5 x 3
x7 x7
The equation has two solutions: 7 and –3.
CHECK
|72||5|5
| 3 2 | | 5 | 5
Solving an Absolute-Value Equation
Solve | 2x 7 | 5 4 SOLUTION Isolate the absolute value expression on one side of the equation.
2x 7 IS POSITIVE IS POSITIVE
2x 7 IS NEGATIVE IS NEGATIVE | 2x 7 | 5 4 | 2x 7 | 9 2x 7 9 9 2x 2 TWO SOLUTIONS x 1
| 2x 7 | 5 4
| 2x 7 | 9 2x 7 +9 2x 16
x8
Solving an Absolute-Value Equation
Recall that | xis the distance between xxand 0. IfIf xx x | is the distance between and 0. | | 8, 8, then any number between and 8 8 is a solution of the then any number between 88 andis a solution of the inequality.
8 7 6 5 4 3 2 1 8
0
1
2
3
4
5
6
7
You can use the following properties to solve absolute-value inequalities and equations.
SOLVING ABSOLUTE-VALUE INEQUALITIES
SOLVING ABSOLUTE-VALUE EQUATIONS AND INEQUALITIES
| ax b | c means ax b c and a x b c.
| a x b | an absolute value b less than a number, c. a x is c and a x b the When c means
inequalities are connected by and. When an absolute | a x b | cgreater thana x number, the inequalities c. or a x b are value is means a bc connected by or.
| ax b | c means means ax b c or a x b c.
| ax b | c
ax b c
or
a x b c.
Solving an Absolute-Value Inequality
Solve | x 4 | < 3
x 4 IS POSITIVE |x4|3 x 4 3 x7
x 4 IS NEGATIVE |x4|3 x 4 3 x1
Reverse inequality less than The solution is all real numbers greater than 1 and symbol. 7
This can be written as 1 x 7.
Solving an Absolute-Value Inequality
Solve | 2x 1 | 3 6 and graph the 2x + 1 IS POSITIVE 2x + 1 IS NEGATIVE solution.
2x + 1 IS POSITIVE 2x | 1 | 6 | 2x| 1 3 9 2x 1 9 | 2x 1 |+9
| 2x 1 | 3 6
2x + 1 IS NEGATIVE 2x | 1 | 9 | 2x| 1 3 6
| 2x 1 | 3 6
2x 1 9 | 2x 1 | 9 2x 10 2x 8 2x 1 9 2x 1 +9 x is x 5 The solution 4all real numbers greater10 or 2x than 2x 8 equal x4 x 5 to 4 or less than or equal to 5. This can be written as the compound inequality x 5 or x 4. Reverse 5 inequality symbol. 4.
6 5 4 3 2 1 0 1 2 3 4 5 6
Writing an Absolute-Value Inequality
You work in the quality control department of a manufacturing company. The diameter of a drill bit must be between 0.62 and 0.63 inch. a. Write an absolute-value inequality
to represent this requirement.
b. Does a bit with a diameter of 0.623
inch meet the requirement?
Writing an Absolute-Value Inequality
The diameter of a drill bit must be between 0.62 and 0.63 inch.
a. Write an absolute-value inequality to represent this requirement.
Let d represent the diameter (in inches) of the drill bit.
Write a compound inequality. Find the halfway point. Subtract 0.625 from
0.62 d 0.63 0.625
0.62 0.625 d 0.625 0.63 0.625
each part of the 0.005 d 0.625 0.005 compound inequality. an absolute-value inequality. d 0.625 | 0.005 Rewrite as | This inequality can be read as “the actual diameter must differ from 0.625 inch by no more than 0.005 inch.”
Writing an Absolute-Value Inequality
The diameter of a drill bit must be between 0.62 and 0.63 inch. b. Does a bit with a diameter of 0.623 meet the requirement?
| d 0.625 | 0.005 | 0.623 0.625 | 0.005 | 0.002 | 0.005
0.002 0.005
Because 0.002 0.005, the bit does meet the requirement.
```
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| 4.625 | 5 |
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