text:
# Security of KDF1 and KDF2 (hash based KDF's)

It's still common to come across implementations of KDF1 and KDF2. Basically these are KDF's that simply derive multiple keys from the key seed and a counter:

$K_i = KDF(K_{master}, i) = H(K_{master} | c)$

In this function $|$ means concatenation and $c$ is the encoded value of $i$ in 4 bytes using an unsigned big endian notation. KDF1 and 2 only differ with regards to the starting value of $i$.

The issue with the KDF is that a hash is not necessarily a PRF. Actually, I've only seen MD5, SHA-1 or SHA-256 being deployed.

Are there any particular attack vectors that can be used against this construction? Is there any practical/pressing need to switch to HKDF ora NIST SP 800-108 approved hashing algorithm or are the concerns purely theoretical of nature?

Please note: above only shows KDF1/2 in their least complex form, using only a single output block and with an empty $OtherInfo$.

-
Hopefully it does not matter if $c$ is signed or unsigned, of course :P. Maybe the security of the KDF can be directly mapped to a hash vs HMAC comparison? –  Maarten Bodewes Apr 19 '14 at 16:43
This should work as long as the message being hashed has constant length (or at least is prefix free). Else length extensions might bite you. –  CodesInChaos Apr 21 '14 at 9:15
@CodesInChaos Yeah, I figured as much. Otherwise HMAC did not need to have additional passes. But most (if not all) of the time the input is just a key seed or an actual key, so length extensions are generally not applicable, at least not as far as I can see... –  Maarten Bodewes Apr 21 '14 at 12:23

As far as I know (which, admittedly, might be limited; I do not claim to possess encyclopedic knowledge of attacks on KDFs), there are no known practical attacks against KDF1 or KDF2 (which are also mentioned on this page, following ISO-18033-2) when instantiated with a secure hash function.

Regarding the relative security of these KDFs vs. HMAC-based KDFs like HKDF, it's worth noting that the HMAC security proof is based on the assumption that the compression function of the underlying hash is itself a PRF. Therefore, when used with any hash function to which the standard HMAC security proof applies, it seems that KDF1 / KDF2 should also be provably secure, at least as long as the master key length equals (or is padded to) the input block size of the hash.

-
Your "this page" says, nebulously, that $H(K | c)$ has "a possible security issue" fixed by using $H(c | K)$ instead. The former is used by ISO KDF1 and KDF2, the latter by KDF3 and NIST SP 800-56A. Does it matter? –  Matt Nordhoff May 11 '14 at 12:26
I should, perhaps, have mentioned the optional "OtherInfo" concatenated on, e.g. $H(x | y | OtherInfo)$. –  Matt Nordhoff May 11 '14 at 12:45
@owlstead I don't get it, though. $c|K$ and $K|c$ would be equally vulnerable to a length extension attack as long as $OtherInfo$ comes last, wouldn't they? In other words, KDF3 is equally vulnerable, if that's the "possible security issue". –  Matt Nordhoff May 11 '14 at 13:30


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# Derivative of $\frac{d}{dx}(xy)$

I'm having trouble understanding the solution to this question which asked me to use the product rule. I understand the product rule portion but I don't understand why the last bit is what it is.

$$\frac{d}{dx} (xy)$$

The solution that is shown is: \begin{align} & = \frac{d}{dx}(x)y + x\frac{d}{dx}(y)\\ & = y + x\frac{d}{dx}\\ \end{align}

The main problem I am having with this is the last portion. Why does the $\frac{d}{dy}(y)$ not get reduced to $1$ and we are only left with $y+x$?

Thanks!

EDIT:

Would my understanding be correct, if I said the $y$ is not reduced to $1$ because it is considered a function and $x$ is reduced to $1$ because it is the slope?

-
It is not $\frac{d}{dy}y$ but $\frac{d}{dx}y$ as you first correctly wrote. Otherwise, yes, it would be $1$. –  1015 Feb 22 '13 at 1:44
I updated the question. I understand the product idea but I just don't understand the $y + x \frac{d}{dx}$ –  Jeel Shah Feb 22 '13 at 1:49
@gekkostate: Please see my comment below. :) –  Haskell Curry Feb 22 '13 at 1:57

Things are clearer if you let $y = f(x)$ and consider the function $x \longmapsto x \cdot f(x)$. Then use the Product Rule: \begin{align} \frac{d}{dx} [x \cdot f(x)] &= \frac{d}{dx} [x] \cdot f(x) + x \cdot \frac{d}{dx} [f(x)] \\ &= 1 \cdot f(x) + x \cdot f'(x) \\ &= f(x) + x \cdot f'(x) \\ &= y + x \frac{dy}{dx}. \end{align}
We have $\dfrac{d}{dx} [x] = 1$, because we are interpreting $x$ as a function, and as a linear function, $x$ has slope $1$ everywhere.
Sorry but why did the $\frac {d}{dx} (x)$ become $1$ and not the $\frac{d}{dx} f(x)$? Is it because $y$ is a function and $x$ isn't? –  Jeel Shah Feb 22 '13 at 1:51
The derivative of $y = x$ is the slope of the function $y = x$, which is equal to $1$. When you write $\dfrac{d}{dx} [x]$, you are differentiating $y = x$ and not $y = f(x)$. –  Haskell Curry Feb 22 '13 at 1:52
therefore, when you are differentiating $\frac{d}{dx}(y)$ you are actually differentiating $f(x)$? –  Jeel Shah Feb 22 '13 at 1:59

format: plain


text:
# Neural Network Process Question - Updating weights after each training set

When creating a neural network, do I update the weights after each run of forward then back propogation? Or do I just keep the random weights and update the Delta variables?

I am looking at slide 8 on these notes:https://d396qusza40orc.cloudfront.net/ml/docs/slides/Lecture9.pdf

It says:

For i = m
Set a(1) = x(i)
Perform Forward-Propogation
Compute delta
Compute DELTA
QUESTION: Do I update the Weights that I use in Forward-propogation, or do I
use random weights and just keep updating the accumulator 'DELTA'? And if I
update the weights, do I set them to DELTA?

For pure stochastic gradient descent, you update the weights after each forward-backward.

When you do the backward phase, you compute the gradients w.r.t. the weights.

You then do the following update:

weights = weights + learningRate * gradWeights

I did not mention more complicated update rules for simplicity.

For batch gradient descent, you accumulate your gradients over the whole batch of samples (i.e. you do a forward-backward pass and do gradWeightsBatch = gradWeightBatch + gradWeight after each backward), and then once you are finished processing the batch, you apply the same update rule:

weights = weights + learningRate * gradWeightsBatch

• Exactly. Each training pattern gives you a weight update. You can perform the weight update either immediately (on-line learning), every N samples by adding up the updates (mini-batch), or summing up the updates over the entire epoch and then applying them (batch-learning). The discussion which would be the best is an expansive topic. – PawelP Jun 4 '15 at 10:55
• @smhx I am pretty sure you subtract the learningRate * gradWeights(Batch) – Sebastian Nielsen Oct 19 '18 at 18:37


text:
# Is a constant function periodic? [closed]

Can we regard a constant function "$f(x)=\text{constant}$" to be a periodic function? If yes, what is its period?

## closed as off-topic by Claude Leibovici, BLAZE, Shailesh, user91500, Davide GiraudoAug 7 '17 at 10:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Claude Leibovici, BLAZE, Shailesh, user91500, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.

• Yes, and anything. – user296602 Aug 7 '17 at 7:05
• See Sassatelli last paragraph. A constant function is a somewhat trivial example of a periodic function with any period. – Peter Szilas Aug 7 '17 at 7:34
• You just deleted a similar "question". The question was : what do you want to show really ? Is it the Fourier series vs the Fourier transform ? – reuns Aug 7 '17 at 7:38
• I just had the exactly the same question pop up in the course of my day, and this question (and it's excellent answers) was the first hit on duckduckgo. I am confused to see it was down voted and closed. – Ron Jensen Feb 23 '18 at 23:04

Definition 1: For $P>0$, a function $f:\Bbb R\to\Bbb R$ is periodic of period $P$ (or $P$-periodic) if $$\forall x\in\Bbb R,\ f(x+P)=f(x)$$ A function is commonly said to be periodic if there is some $P>0$ such that $f$ is $P$-periodic.

Definition 2: For a periodic function $f$, the fundamental period of $f$ is, if existing, the least positive real number $T$ such that $f$ is $T$-periodic. Specifically, if we call $$L=\inf\{ P>0\,:\,\forall x\in\Bbb R,\ f(x+P)=f(x)\},$$ then $f$ has a fundamental period if and only if $L>0$; in which case, $L$ is the fundamental period of $f$.

Answer to your question: Constants are periodic functions of any period and, therefore, they do not have a fundamental period.

Nowhere in the definition of a period function is it stated that the function must have a least period.

If $f(x) = c$ then for any $p$ we have $f(x+p) = f(x)$. So $f$ is periodic and $p$ is a period. Obviously any other non-negative value will also be a period.

There is nothing in the definition of periodic function that says that is not allowed.

• Ah... G. Sassatelli beat me by a minute with a good definition. The idea of a least period simply called the "fundamental period". It does not exist for constant functions. – fleablood Aug 7 '17 at 7:17
• For clicks and gigles let $f(x) = 0$ if $x$ is rational. And $f(x) = 1$ if $x$ is irrational. Then for any rational $q$, $f(x+q) = f(x)$ so $f(x)$ is also a periodic function that has no fundamental period. But $f(x) = c$ is the only periodic continuous function that has no fundamental period. FYI. – fleablood Aug 7 '17 at 7:24

Yes, a constant function is a periodic function with any T∈R as its period (as f(x)=f(x+T) always for howsoever small 'T' you can find).

However, the fundamental period of a constant function is not defined for the above reason.


text:
## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - Review Exercises: 76

#### Answer

$${e^{2x}} + C$$

#### Work Step by Step

\eqalign{ & \int {2{e^{2x}}} dx \cr & {\text{take out the constant}} \cr & = 2\int {{e^{2x}}} dx \cr & {\text{use the formula }}\int {{e^{ax}}dx} = \frac{1}{a}{e^{ax}} + C \cr & = 2\left( {\frac{1}{2}{e^{2x}}} \right) + C \cr & {\text{simplify}} \cr & = {e^{2x}} + C \cr}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
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Gottfried Wilhelm Leibniz

(Redirected from Gottfried Leibniz)
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 This page contains too many unsourced statements and needs to be improved. Gottfried Wilhelm Leibniz could use some help. Please research the article's assertions. Whatever is credible should be sourced, and what is not should be removed.

Gottfried Wilhelm Leibniz (1646–1716) was a German diplomat, and polymath with multiple important contributions to mathematics and philosophy.[1]

Career and contributions

Mathematics

Leibniz invented[notes 1] discovered calculus independently of Sir Isaac Newton. As a consequence of his quest for the appropriate symbols to represent ideas, Leibniz introduced the notations (such as $\frac{dy}{dx}$ for differentials and $\int\limits_a^b$ for integrals) that are still used today.[1] Newton's dot notation $\dot{y}$, on the other hand, is more commonly used in physics, for example, to represent the time derivative. In fact, some expressions, such as the Euler-Lagrange equation, are most conveniently expressed using both Newton and Leibniz's notations. Also note that while the stylized s is universally used to represent an integral today, people still use different notations for the derivative. Joseph-Louis Lagrange used the prime notation $y'$ while Leonard Euler used the letter $D$, treating it as a linear operator. For partial derivatives, Gustav Jacob Jacobi employed the stylized d that resemble the number six in reverse $\partial$,[2] which may be thought of as an extension or modification of the Leibniz notation. Unfortunately, however, Newton and Leibniz engaged in a prolonged and bitter but ultimately pointless priority dispute. We now know that Newton formulated his own version of calculus first, but published later as he despised controversies. Leibniz, on the other hand, arrived at his later but published first. We also know from historical records that before the dispute, the two communicated with each other about Newton's work on the binomial theorem. [3] It is intriguing, but ultimately fruitless, to speculate on what might have been if the two cooperated and there were no priority disputes.

While Newton preferred to think of transcendental functions as infinite series to be differentiated or integrated term-by-term, Leibniz treated functions in a much less messy manner. Like Newton, Leibniz began by considering infinite series. But unlike Newton, whose starting materials were Euclidean, geometry, Cartesian geometry, and the work of John Wallis, Leibniz had his entryway via the differential triangle of Pascal.[1] He unearthed the Product and Quotient rules of differentiation. One can generalize the Product Rule using mathematical induction to arrive at the General Leibniz Rule, which gives a higher-order derivative of a product of two function and is most conveniently expressed using the binomial notation.[4] Knowing how to differentiate the products and quotients of functions, Leibniz moved on to composite functions and discovered the all-important Chain Rule. Newton independently discovered it via substitution.[1] Given the fact that he independently came up with the Fundamental Theorem of Calculus, it is no surprise that he also introduced the Leibniz Rule for Differentiating Under an Integral Sign, or simply the Leibniz Integral Rule.

Alongside James Bernoulli, Leibniz showed us how to solve what we now call separable ordinary differential equations (ODEs). His use of differentials definitely paved the way.[1] Taking advantage of the Product Rule, he introduced a technique to solve a slightly more complicated class of ODEs called first-order linear equations known as an integration factor. When James Bernoulli challenged other mathematicians to solve the brachistochrone problem, a classic in the calculus of variations, Leibniz responded with his own correct answer, independently of Newton and a few other mathematicians. His work with infinite series led Leibniz to a beautiful[citation NOT needed] formula for $\pi$ [1] that we today typically derive using the Maclaurin series of the inverse tangent function.[3]

Leibniz also invented the binary system (01010100011001001110110110110000001),[notes 2] was a rationalist, and made contributions to medicine, philosophy, law, geology, and many other subjects.

Metaphysics

The Metaphysics of Leibniz is most famous for his assertion that everything in the material universe consists of monads, which are a physical manifestation of God.[5]

Theology

If we are living in the best of all possible worlds, surely Leibniz cookies are the best of all possible cookies?

Leibniz was also a Lutheran apologist who, in his Essais de Théodicée sur la bonté de Dieu, la liberté de l'homme et l'origine du mal ("Theodical Essays on the Goodness of God, Liberty of the Person, and the Origin of Evil") claimed we are living in the best of all possible worlds, which Voltaire satirized as the character Pangloss in his Candide.[6]


text:
Probability

# Probability of an event

## Keywords:

$P\left(A\right)=\frac{\left|A\right|}{\left|\Omega \right|}$

|A| - number of ways an event can happen

|Ω| - total number of outcomes

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$
 $P\left(\varnothing \right)=0$ Probability of an impossible event $P\left(\Omega \right)=1$ Probability of a sure event
$P\left(\overline{A}\right)=1-P\left(A\right)$
$P\left(A\cap B\right)=P\left(A|B\right)·P\left(B\right)=P\left(B|A\right)·P\left(A\right)$
 $P\left(A\cap B\right)=P\left(A\right)·P\left(B\right)$ In cas events A and B are independent


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### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A star converts all its hydrogen to helium, achieving a 100% helium composition. Next, it converts the helium to carbon via the triple-alpha process, ${}^{{\mathbf{4}}}{\mathbit{H}}{\mathbit{e}}{\mathbf{+}}{}^{{\mathbf{4}}}{\mathbit{H}}{\mathbit{e}}{\mathbf{+}}{}^{{\mathbf{4}}}{\mathbit{H}}{\mathbit{e}}{{\mathbf{\to }}}^{{\mathbf{12}}}{\mathbit{C}}{\mathbf{+}}{\mathbf{7}}{\mathbf{.}}{\mathbf{27}}{\mathbf{}}{\mathbit{M}}{\mathbit{e}}{\mathbit{V}}$The mass of the star is role="math" localid="1661754478822" ${\mathbf{4}}{\mathbf{.}}{\mathbf{6}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{32}}}{\mathbit{k}}{\mathbit{g}}$, and it generates energy at the rate of ${\mathbf{5}}{\mathbf{.}}{\mathbf{3}}{\mathbit{x}}{{\mathbf{10}}}^{{\mathbf{3}}}{\mathbf{}}{\mathbit{W}}$. How long will it take to convert all the helium to carbon at this rate?

The required time is $1.6×{10}^{8}yr$.

See the step by step solution

## Step 1: Describe the expression for the time

Let m be the mass of helium, the number of three-helium atoms in this mass equals the number of moles multiplied by the number of atoms in one mole, where the number of moles equals the mass divided by three the molar mass of the helium.

${{\mathbit{N}}}_{\mathbf{3}\mathbf{H}\mathbf{e}}{\mathbf{=}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}}$

Let be the energy release per fusion, then the total energy released by fusion equals the number of fusions multiplied by ${{\mathbit{N}}}_{\mathbf{3}\mathbf{He}}$.

${{\mathbit{E}}}_{\mathbf{f}\mathbf{u}\mathbf{s}\mathbf{i}\mathbf{o}\mathbf{n}}{\mathbf{=}}{{\mathbit{N}}}_{\mathbf{3}\mathbf{H}\mathbf{e}}{\mathbit{Q}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{=}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}}$

The energy also can be written as follows:

${{\mathbit{E}}}_{\mathbf{f}\mathbf{u}\mathbf{s}\mathbf{i}\mathbf{o}\mathbf{n}}{\mathbf{=}}{\mathbit{P}}{\mathbit{t}}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}}{\mathbf{=}}{\mathbit{P}}{\mathbit{t}}\phantom{\rule{0ex}{0ex}}{\mathbit{t}}{\mathbf{=}}\frac{{\mathbf{m}}_{\mathbf{H}\mathbf{e}}{\mathbf{N}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{3}{\mathbf{M}}_{\mathbf{H}\mathbf{e}}\mathbf{P}}$

## Step 2: Find the time required to convert all the helium to carbon

Substitute all the known values in equation (1).

$t=\frac{\left(4.6×{10}^{35}g\right)\left(6.022×{10}^{23}mo{l}^{-1}\right)\left(7.27MeV\right)\left(1.602×{10}^{-13}J/MeV\right)}{3\left(4.0g/mol\right)\left(5.3×{10}^{30}W\right)}\phantom{\rule{0ex}{0ex}}=5.07×{10}^{15}s\phantom{\rule{0ex}{0ex}}=\frac{5.07×{10}^{15}s}{3.154×{10}^{7}s/yr}\phantom{\rule{0ex}{0ex}}=1.6×{10}^{8}yr$

Therefore, the required time is $1.6×{10}^{8}yr$.


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# An orange drink contains 2 ounces of real orange juice per 100 ounces

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An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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An orange drink contains 2 ounces of real orange juice per 100 ounces of the orange drink. If a bottle contains 32 ounces of the drink, how many ounces of
real orange juice does it contain?

a)2/100
b)2/68
c)2/32
d)32/100
e)64/100

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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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04 Oct 2016, 22:21
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100 ounces of juice have 2 ounce of real juice.

1 ounces of juice have $$\frac{2}{100}$$ ounce of real juice.

32 ounces of juice have $$\frac{(2*32)}{100} = \frac{64}{100}$$ ounce of real juice.

Ans. E
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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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05 Oct 2016, 07:32
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alanforde800Maximus wrote:
An orange drink contains 2 ounces of real orange juice per 100 ounces of the orange drink. If a bottle contains 32 ounces of the drink, how many ounces of
real orange juice does it contain?

a)2/100
b)2/68
c)2/32
d)32/100
e)64/100

The question effectively means that 2% in a given quantity is orange juice (i.e.\frac{2}{100})
Therefore in 32 ounces of the drink 2% is orange juice.
so, $$\frac{2}{100} * 32$$
= \frac{64}{100}
Ans. E
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An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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Updated on: 08 Oct 2016, 07:18
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This post was
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alanforde800Maximus wrote:
An orange drink contains 2 ounces of real orange juice per 100 ounces of the orange drink. If a bottle contains 32 ounces of the drink, how many ounces of
real orange juice does it contain?

a)2/100
b)2/68
c)2/32
d)32/100
e)64/100

the easiest way to solve this problem is using a simple proportion.

Indeed the ratio $$\frac{(RealJuice)}{(total )}$$ has to be the same =$$\frac{2}{100}$$

$$\frac{2}{100}= \frac{x}{32}$$

$$x=\frac{64}{100}$$

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Originally posted by damybox on 05 Oct 2016, 12:25.
Last edited by damybox on 08 Oct 2016, 07:18, edited 1 time in total.
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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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05 Oct 2016, 14:28
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2/100=x/32

cross multiply and solve
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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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08 Oct 2016, 05:37
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damybox wrote:
alanforde800Maximus wrote:
An orange drink contains 2 ounces of real orange juice per 100 ounces of the orange drink. If a bottle contains 32 ounces of the drink, how many ounces of
real orange juice does it contain?

a)2/100
b)2/68
c)2/32
d)32/100
e)64/100

the easiest way to solve this problem is using a simple proportion.

Indeed the ratio $$\frac{(RealJuice)}{(total )}$$ has to be the same =$$\frac{2}{100}$$

$$\frac{2}{100}= \frac{x}{32}$$

$$x=\frac{64}{100}$$

I believe you meant E.
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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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08 Oct 2016, 07:19
abhimahna wrote:
damybox wrote:
alanforde800Maximus wrote:
An orange drink contains 2 ounces of real orange juice per 100 ounces of the orange drink. If a bottle contains 32 ounces of the drink, how many ounces of
real orange juice does it contain?

a)2/100
b)2/68
c)2/32
d)32/100
e)64/100

the easiest way to solve this problem is using a simple proportion.

Indeed the ratio $$\frac{(RealJuice)}{(total )}$$ has to be the same =$$\frac{2}{100}$$

$$\frac{2}{100}= \frac{x}{32}$$

$$x=\frac{64}{100}$$

I believe you meant E.

yes, I meant E!

thanks for the correction
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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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21 Dec 2017, 20:01
Setting this problem up as a ratio of an amount divided by a total makes this really easy to solve. The question tells us that we have 2 ounces of real orange juice for every 100 ounces of drink. Thus, our juice-to-total ratio would look like this:

$$\frac{J}{T} = \frac{2}{100}$$

The problem then tells us that a particular bottle contains 32 ounces total. We can simply plug that 32-ounce amount into our total, $$T$$, and solve for $$J$$. Thus:

$$\frac{J}{32} = \frac{2}{100}$$

$$J = \frac{2*32}{100}=\frac{64}{100}$$

While we could further simplify 64/100 down to 16/25 or even to the decimal 0.64, the answers keep it fractional form, so we can stop there.

Another way to solve this problem would be to look down at the answer choices and realize this problem might be a prime candidate for approximating. While all the answer choices are small, they are very different from each other, ranging from $$\frac{2}{100}$$ to $$\frac{64}{100}$$. The question tells us that there are only 2 ounces of juice in 100 ounces of drink. The "target" amount of drink (32 ounces) is approximately $$\frac{1}{3}$$ of 100 ounces. Thus, the amount of juice should be about $$\frac{1}{3}$$ of 2 ounces. $$\frac{1}{3}$$ of 2 is $$\frac{2}{3}$$, or about .67 ounces. The only answer choice that is even close to .67 is answer choice (E).

Any way you solve it, the answer is still (E).
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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces [#permalink]

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27 Jan 2018, 13:12
Hi All,

The first sentence of this prompt tell us that every 100 ounces of orange drink contains 2 ounces of real orange juice (and by extension, the other 98 ounces are something else). We're meant to assume that this ratio is maintained regardless of the amount of orange drink that we have. Thus, 2/100 = 1/50 of any orange drink will be real orange juice.

We're asked for the number of ounces of real orange juice in 32 ounces of orange drink...

(32)(1/50) = 32/50 = 64/100 ounces of real orange juice

[Reveal] Spoiler:
E

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Re: An orange drink contains 2 ounces of real orange juice per 100 ounces   [#permalink] 27 Jan 2018, 13:12
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Study of jet quenching with Z+jet correlations in PbPb and pp collisions at $\sqrt{s_{NN}}$ = 5.02 TeV

The collaboration
No Journal Information, 2017

Abstract
The production of jets in association with Z bosons, reconstructed via the mu+mu- and e+e- decay channels, is studied in pp and, for the first time, in PbPb collisions. Both data samples were collected by the CMS experiment at the LHC, at a center-of-mass energy of 5.02 TeV. The PbPb collisions were analyzed in the 0-30% centrality range. The back-to-back azimuthal alignment was studied in both pp and PbPb collisions for Z bosons with transverse momentum ptz > 60 GeV/c and a recoiling jet with ptj > 30 GeV/c. The pt imbalance, xjz= ptj/ptz, as well as the average number of jet partners per Z, rjz, were studied in intervals of ptz, in both pp and PbPb collisions. The rjz is found to be smaller in PbPb than in pp collisions, which suggests that in PbPb collisions a larger fraction of partons, associated with the Z bosons, lose energy and fall below the 30 GeV/c ptj threshold.

• #### Table 1

Data from top panel of Fig. 2

10.17182/hepdata.77604.v1/t1

Distribution of the azimuthal angle difference between the Z boson and the jet in pp collisions. The distribution is normalized...

• #### Table 2

Data from top panel of Fig. 2

10.17182/hepdata.77604.v1/t2

Distribution of the azimuthal angle difference between the Z boson and the jet in PbPb collisions. The distribution is normalized...

• #### Table 3

Data from bottom panel of Fig. 2

10.17182/hepdata.77604.v1/t3

Distribution of the transverse momentum ratio between the Z boson and the jet in pp collisions. The distribution is normalized...

• #### Table 4

Data from bottom panel of Fig. 2

10.17182/hepdata.77604.v1/t4

Distribution of the transverse momentum ratio between the Z boson and the jet in PbPb collisions. The distribution is normalized...

• #### Table 5

Data from top panel of Fig. 3

10.17182/hepdata.77604.v1/t5

The mean value of the X_JZ distribution in pp collisions. The resolutions of the measured jet energy and azimuthal angle...

• #### Table 6

Data from top panel of Fig. 3

10.17182/hepdata.77604.v1/t6

The mean value of the X_JZ distribution in PbPb collisions.

• #### Table 7

Data from bottom panel of Fig. 3

10.17182/hepdata.77604.v1/t7

The average number of jet partners per Z boson in pp collisions. The resolutions of the measured jet energy and...

• #### Table 8

Data from bottom panel of Fig. 3

10.17182/hepdata.77604.v1/t8

The average number of jet partners per Z boson in PbPb collisions.

• #### Table 9

Data from top panel of Fig. 4

10.17182/hepdata.77604.v1/t9

Distribution of the transverse momentum ratio between the Z boson and the jet in pp collisions. The distribution is normalized...


text:
1. Sep 4, 2008

### CrazyAmerican

1. The problem statement, all variables and given/known data

10 moles of an ideal gas at 100 atm pressure are expanded isothermally against a constant pressure of 10 atm, to 10 at. calculate Q W deltaU and delta H (Take temp to be T)

2. Relevant equations

PV = nRT
DeltaH = Q (for constant P)
DeltaU = Q-W
U = H+PV
W = Integral(PdV)

3. The attempt at a solution

I wish i knew where to start...
Isothermal means a constant pressure, in this case 10 atm. ...i just have no idea.

PV = nRT
V1 = .8314T
V2 = 8.314T

W= Integral(PdV)
W = PIntegral(dv)
W = P(v2-v1)
W = 74.826

Got IT!

Last edited: Sep 4, 2008


text:
# What is the derivative of this function y= (1+sin x)/(1-sin x)?

Nov 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cos x}{1 - \sin x} ^ 2$

#### Explanation:

If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $y = \frac{1 + \sin x}{1 - \sin x}$ Then

$\left\{\begin{matrix}\text{Let "u=1+sinx & => & (du)/dx=cosx \\ "And } v = 1 - \sin x & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = - \cos x\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \sin x\right) \left(\cos x\right) - \left(1 + \sin x\right) \left(- \cos x\right)}{1 - \sin x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos x - \sin x \cos x + \cos x + \sin x \cos x}{1 - \sin x} ^ 2$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \cos x}{1 - \sin x} ^ 2$