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# Is |x| + |x-1| = 1?

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Is |x| + |x-1| = 1? [#permalink]  16 Mar 2010, 02:43
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Question Stats:

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Is |x| + |x-1| = 1?

(1) $$x \geq 0$$
(2) $$x \leq 1$$
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Mar 2013, 02:25, edited 1 time in total.
Edited the question and added the OA.
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Re: Modulus [#permalink]  16 Mar 2010, 02:50
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abhi758 wrote:
Is l x l + lx-1l= 1?
(1) $$x \geq 0$$
(2) $$x \leq 1$$

Any explanations for the given OA..

[Reveal] Spoiler:
C

stmnt1. let x= 0 then we have l 0 l + l0-1l= 1. suff
x= 10> 0 then we have l 10 l + l10-1l not equal to 1. hence insuff

stmnt2. let x = 0< 1 then we have l 0 l + l0-1l= 1. suff
let x= -5 < 1 then we l -5 l + l-5 -1l not equal to 1. hence insuff

taking together we have $$0 \leq x \leq 1$$
if x= 0 we have l 0 l + l0-1l= 1
if x = 1 we have l 1 l + l1-1l= 1
if x = 1/2 we have l 1/2 l + l1/2-1l= 1

hence C
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Re: Modulus [#permalink]  16 Mar 2010, 03:06
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Expert's post
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) $$x \geq 0$$
(2) $$x \leq 1$$

Any explanations for the given OA..

[Reveal] Spoiler:
C

Algebraic approach:

Let's check in which ranges of $$x$$ expression $$|x|+|x-1|=1$$ holds true:

Two check points for $$x$$: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. $$x<0$$ --> $$-x-x+1=1$$ --> $$x=0$$, not good as we are checking for $$x<0$$;
B. $$0\leq{x}\leq{1}$$ --> $$x-x+1=1$$ --> $$1=1$$, which is true. This means that in this range given equation holds true for all x-es;
C. $$x>1$$ --> $$x+x-1=1$$ --> $$x=1$$, not good as we are checking for $$x>1$$.

So we got that the equation $$|x|+|x-1|=1$$ holds true only in the range $$0\leq{x}\leq{1}$$.

(1) $$x \geq 0$$. Not sufficient.
(2) $$x \leq 1$$. Not sufficient.

(1)+(2) gives us the range $$0\leq{x}\leq{1}$$, which is exactly the range for which given equation holds true. Sufficient.

Hope it helps.
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Re: Modulus [#permalink]  16 Mar 2010, 05:31
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) $$x \geq 0$$
(2) $$x \leq 1$$

Any explanations for the given OA..

[Reveal] Spoiler:
C

|x|+|x-1|, there can be 4 cases

stmt1: x >= 0 => x is positive, if we apply this to eq 1, we cannot say that 2x-1 = 1 cos x can be >= 0 but less than 1 so insuff
stmt2: x<=1 => if we apply this to eq 1, we cannot say that 2x-1 = 1 cos x<=1 can lead to negative numbers also.

both stmt taken together 0<= x <= 1
x is a fraction between 0 to 1 so always x + 1-x = 1
in our case |x| will be x and |x - 1| is actually value of 1-x
so it will always be 1 this is why C both the stmts are reqd.
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Re: Modulus [#permalink]  16 Mar 2010, 07:21
Thanks so much all of you! the workings definately helped to understand the OA better..

Thanks Bunnel for the algebraic approach..
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Re: Modulus [#permalink]  16 Mar 2010, 07:39
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) $$x \geq 0$$
(2) $$x \leq 1$$
Any explanations for the given OA.
[Reveal] Spoiler:
C

Either statement alone is not sufficient to answer this question.
Together x can be 0 or 1, in both cases statement is true hence sufficient.
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Re: Modulus [#permalink]  18 Mar 2010, 09:10
Bunuel wrote:
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) $$x \geq 0$$
(2) $$x \leq 1$$

Any explanations for the given OA..

[Reveal] Spoiler:
C

Algebraic approach:

Let's check in which ranges of $$x$$ expression $$|x|+|x-1|=1$$ holds true:

Two check points for $$x$$: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. $$x<0$$ --> $$-x-x+1=1$$ --> $$x=0$$, not good as we are checking for $$x<0$$;
B. $$0\leq{x}\leq{1}$$ --> $$x-x+1=1$$ --> $$1=1$$, which is true. This means that in this range given equation holds true for all x-es;
C. $$x>1$$ --> $$x+x-1=1$$ --> $$x=1$$, not good as we are checking for $$x>1$$.

So we got that the equation $$|x|+|x-1|=1$$ holds true only in the range $$0\leq{x}\leq{1}$$.

(1) $$x \geq 0$$. Not sufficient.
(2) $$x \leq 1$$. Not sufficient.

(1)+(2) gives us the range $$0\leq{x}\leq{1}$$, which is exactly the range for which given equation holds true. Sufficient.

Hope it helps.

Can you please clarify as to how is $$x-x+1=1$$ derived inB when it is provided that $$0\leq{x}\leq{1}$$ that means X is +ve.

Thanks !!!
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Re: Modulus [#permalink]  18 Mar 2010, 09:21
Expert's post
mustdoit wrote:
Bunuel wrote:
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) $$x \geq 0$$
(2) $$x \leq 1$$

Any explanations for the given OA..

[Reveal] Spoiler:
C

Algebraic approach:

Let's check in which ranges of $$x$$ expression $$|x|+|x-1|=1$$ holds true:

Two check points for $$x$$: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. $$x<0$$ --> $$-x-x+1=1$$ --> $$x=0$$, not good as we are checking for $$x<0$$;
B. $$0\leq{x}\leq{1}$$ --> $$x-x+1=1$$ --> $$1=1$$, which is true. This means that in this range given equation holds true for all x-es;
C. $$x>1$$ --> $$x+x-1=1$$ --> $$x=1$$, not good as we are checking for $$x>1$$.

So we got that the equation $$|x|+|x-1|=1$$ holds true only in the range $$0\leq{x}\leq{1}$$.

(1) $$x \geq 0$$. Not sufficient.
(2) $$x \leq 1$$. Not sufficient.

(1)+(2) gives us the range $$0\leq{x}\leq{1}$$, which is exactly the range for which given equation holds true. Sufficient.

Hope it helps.

Can you please clarify as to how is $$x-x+1=1$$ derived inB when it is provided that $$0\leq{x}\leq{1}$$ that means X is +ve.

Thanks !!!

When $$x$$ is in the range $$0\leq{x}\leq{1}$$ then $$|x|=x$$ and $$|x-1|=-(x-1)=-x+1$$, thus $$|x|+|x-1|=1$$ in this range becomes $$x-x+1=1$$.

Hope it's clear.
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Re: Modulus [#permalink]  07 Mar 2013, 20:30
I have got the answer as E. Can someone please clarify, what is wrong in my approach ?

Statement 1: X>=0

I assume X as 0
|0|+|0-1|= |0|+|1|= 1 ------------> Equation satisfied

I assume X as 3/4
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 1 is not sufficient

Statement 2: X<=1

I assume X as 1
|1|+|1-1|= |1|+|0|= 1 --------------> Equation satisfied

I assume X as 3/4 again
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 2 is not sufficient

Taking together both the equation, I assume X as 0, 3/4 and 1

When X=0, the answer is 1--------> Equation satisfied
When X=3/4, the answer is 2---------> Equation unsatisfied
When X=1, the answer is 1----------> Equation satisfied

Hence answer choice E is correct.

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Re: Modulus [#permalink]  07 Mar 2013, 20:44
2
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Expert's post
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) $$x \geq 0$$
(2) $$x \leq 1$$

Any explanations for the given OA..

[Reveal] Spoiler:
C

You can also consider using the number line method discussed here: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

l x l + lx-1l= 1 implies that sum of distance of x from 0 and from 1 should be 1. We can easily see that the distance between 0 and 1 is 1. So, x can lie on 0, on 1 or anywhere in between 0 and 1. It cannot go beyond 1 or before 0 because then the sum of the distance will become larger than 1.
Both statements together tell you that 0 <= x <= 1 and hence, the equation will hold for all such values of x.
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Re: Modulus [#permalink]  08 Mar 2013, 02:31
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Expert's post
Backbencher wrote:
I have got the answer as E. Can someone please clarify, what is wrong in my approach ?

Statement 1: X>=0

I assume X as 0
|0|+|0-1|= |0|+|1|= 1 ------------> Equation satisfied

I assume X as 3/4
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 1 is not sufficient

Statement 2: X<=1

I assume X as 1
|1|+|1-1|= |1|+|0|= 1 --------------> Equation satisfied

I assume X as 3/4 again
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 2 is not sufficient

Taking together both the equation, I assume X as 0, 3/4 and 1

When X=0, the answer is 1--------> Equation satisfied
When X=3/4, the answer is 2---------> Equation unsatisfied
When X=1, the answer is 1----------> Equation satisfied

Hence answer choice E is correct.

The correct answer is C, not E.

If x = 3/4, then Is |x| + |x-1| = |3/4| + |3/4 - 1| = 3/4 + 1/4 = 1.

Hope it helps.
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Re: Is |x| + |x-1| = 1? [#permalink]  08 Mar 2013, 03:41
Expert's post
abhi758 wrote:
Is |x| + |x-1| = 1?

(1) $$x \geq 0$$
(2) $$x \leq 1$$

So |x| + |x-1| can only equal 1 when

|x| = x and |x-1| = 1-x . Thus, x>=0 AND x-1=<0. Which gives, 0<=x<=1.

C.
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Re: Is |x| + |x-1| = 1? [#permalink]  17 Jun 2013, 07:21
Is |x| + |x-1| = 1?

For #1 (x≥0) I see two possible cases:

x=1/2
x+ -(x-1)=1
x+ -x+1=1
1=1

OR

x=2
x+x-1=1
2x=2
x=1

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

Thanks!
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Re: Is |x| + |x-1| = 1? [#permalink]  17 Jun 2013, 07:41
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WholeLottaLove wrote:
Is |x| + |x-1| = 1?

For #1 (x≥0) I see two possible cases:

x=1/2
x+ -(x-1)=1
x+ -x+1=1
1=1

OR

x=2
x+x-1=1
2x=2
x=1

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

Thanks!

No.
In the first case for an $$x=\frac{1}{2}$$ ($$0\leq{x}\leq{1}$$) you get that the equation is true $$1=1$$ it's like saying ALWAYS TRUE.
But in the second case for an $$x>1$$ you get x=1, but since x>1 that is not true
Statement 1 is not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink]  17 Jun 2013, 08:09
Ok, this is how I look at it:

x>1
|x| + |x-1| = 1
x + (x-1) = 1
2x = 2
x = 1
x = 1 which ISN'T greater than 1. Insufficient

0≤x<1
x = 1/2
|x| + |x-1| = 1
x + -(x-1) =1
x+ -x +1 = 1
0=0
I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!

Zarrolou wrote:
WholeLottaLove wrote:
Is |x| + |x-1| = 1?

For #1 (x≥0) I see two possible cases:

x=1/2
x+ -(x-1)=1
x+ -x+1=1
1=1

OR

x=2
x+x-1=1
2x=2
x=1

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

Thanks!

No.
In the first case for an $$x=\frac{1}{2}$$ ($$0\leq{x}\leq{1}$$) you get that the equation is true $$1=1$$ it's like saying ALWAYS TRUE.
But in the second case for an $$x>1$$ you get x=1, but since x>1 that is not true
Statement 1 is not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink]  17 Jun 2013, 08:14
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WholeLottaLove wrote:
Ok, this is how I look at it:

x>1
|x| + |x-1| = 1
x + (x-1) = 1
2x = 2
x = 1
x = 1 which ISN'T greater than 1. Insufficient

0≤x<1
x = 1/2
|x| + |x-1| = 1
x + -(x-1) =1
x+ -x +1 = 1
0=0
I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!

0=0, and 1=1,... means that that equation is always true. You should read the results as follow:
x>1
x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1)
0≤x<1
0=0. So in this case the equation in TRUE( because 0=0 always)

So from statement 1 you get two different results (F/T)=> not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink]  17 Jun 2013, 08:33
Okay, that's what I thought. Thank you.

Zarrolou wrote:
WholeLottaLove wrote:
Ok, this is how I look at it:

x>1
|x| + |x-1| = 1
x + (x-1) = 1
2x = 2
x = 1
x = 1 which ISN'T greater than 1. Insufficient

0≤x<1
x = 1/2
|x| + |x-1| = 1
x + -(x-1) =1
x+ -x +1 = 1
0=0
I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!

0=0, and 1=1,... means that that equation is always true. You should read the results as follow:
x>1
x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1)
0≤x<1
0=0. So in this case the equation in TRUE( because 0=0 always)

So from statement 1 you get two different results (F/T)=> not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink]  30 Jan 2014, 05:02
I went through the following link x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html and understand the following below.

Quote:
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4

The above explanation is easily understandable but it creates a small doubt if i compare it with this question
Is |x| + |x-1| = 1?
(1) $$x\geq{0}$$
(2) $$x\leq{1}$$

Quote:
Two check points for $$x$$: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. $$x<0$$ --> $$-x-x+1=1$$ --> $$x=0$$, not good as we are checking for $$x<0$$;
B. $$0\leq{x}\leq{1}$$ --> $$x-x+1=1$$ --> $$1=1$$, which is true. This means that in this range given equation holds true for all x-es;
C. $$x>1$$ --> $$x+x-1=1$$ --> $$x=1$$, not good as we are checking for $$x>1$$.

So we got that the equation $$|x|+|x-1|=1$$ holds true only in the range $$0\leq{x}\leq{1}$$

Here, we have to include the transition points (0 and 1) but only once.
Are these okay, for B. $$0\leq{x}<1$$ --> 1=1 not in the range.
for C. $$x\geq{1}$$ --> x = 1. It is in the range.

Or

A. $$x\leq{0}$$ --> x= 0, It is in the range.
B. $$0<{x}\leq{1}$$ --> 1=1 It is also in the range.

Are my ranges correct ? if yes, then i still don't feel close to the answer.

Thanks & Regards
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Re: Is |x| + |x-1| = 1? [#permalink]  15 Feb 2014, 21:18
its by plugging in method

1. From 1st equation , if x = 1 , then |x| + |x-1| = 1 , then its yes
if x = 5 , 9 # 1 , so No
so Equation 1 is No
2. From 2nd equation , of x = -1 , then its No
x = 1 its yes
so equation is No

3. using both , x = 0 then its Definite yes
So C..

Is my approach is Right
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Re: Is |x| + |x-1| = 1? [#permalink]  16 Feb 2014, 19:40
Expert's post
kanusha wrote:
its by plugging in method

1. From 1st equation , if x = 1 , then |x| + |x-1| = 1 , then its yes
if x = 5 , 9 # 1 , so No
so Equation 1 is No
2. From 2nd equation , of x = -1 , then its No
x = 1 its yes
so equation is No

3. using both , x = 0 then its Definite yes
So C..

Is my approach is Right

Using both, you get 0 <= x <= 1
x can lie anywhere from 0 to 1. Just trying the extreme values is not a good idea. You must try some values from the middle too i.e. x = 1/2, 1/4 etc.

Though establishing something by plugging in values is always tricky. You can try to negate a universal statement by looking for one value for which the statement doesn't hold but since you cannot try every value for which a statement must hold, its treacherous to depend on number plugging in this case. You should try to use logic.
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Re: Is |x| + |x-1| = 1?   [#permalink] 16 Feb 2014, 19:40

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text:
# If Matrices Commute $AB=BA$, then They Share a Common Eigenvector

## Problem 608

Let $A$ and $B$ be $n\times n$ matrices and assume that they commute: $AB=BA$.
Then prove that the matrices $A$ and $B$ share at least one common eigenvector.

## Proof.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be an eigenvector corresponding to $\lambda$. That is, we have $A\mathbf{x}=\lambda \mathbf{x}$.
Then we claim that the vector $\mathbf{v}:=B\mathbf{x}$ belongs to the eigenspace $E_{\lambda}$ of $\lambda$.
In fact, as $AB=BA$ we have
\begin{align*}
A\mathbf{v}=AB\mathbf{x}=BA\mathbf{x} =\lambda B\mathbf{x}=\lambda \mathbf{v}.
\end{align*}
Hence $\mathbf{v}\in E_{\lambda}$.

Now, let $\{\mathbf{x}_1, \dots, \mathbf{x}_k\}$ be an eigenbasis of the eigenspace $E_{\lambda}$.
Set $\mathbf{v}_i=B\mathbf{x}_i$ for $i=1, \dots, k$.
The above claim yields that $\mathbf{v}_i \in E_{\lambda}$, and hence we can write
$\mathbf{v}_i=c_{1i} \mathbf{x}_1+c_{2i}\mathbf{x}_2+\cdots+c_{ki}\mathbf{x}_k \tag{*}$ for some scalars $c_{1i}, c_{2i}, \dots, c_{ki}$.

Extend the basis $\{\mathbf{x}_1, \dots, \mathbf{x}_k\}$ of $E_{\lambda}$ to a basis
$\{\mathbf{x}_1, \dots, \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n\}$ of $\R^n$ by adjoining vectors $\mathbf{x}_{k+1}, \dots, \mathbf{x}_n$.

Then we obtain using (*)
\begin{align*}
&B [\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n]\\
&=[B\mathbf{x}_1,\dots , B\mathbf{x}_k, B\mathbf{x}_{k+1}, \dots, B\mathbf{x}_n]\\
&=[\mathbf{v}_1,\dots , \mathbf{v}_k, B\mathbf{x}_{k+1}, \dots, B\mathbf{x}_n]\6pt] &=[\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n] \left[\begin{array}{c|c} C & D\\ \hline O & F \end{array} \right],\tag{**} \end{align*} where C=(c_{ij}) is the k\times k matrix whose entries are the coefficients c_{ij} of the linear combination (*), O is the (n-k) \times k zero matrix, D is a k \times (n-k) matrix, and F is an (n-k) \times (n-k) matrix. Let P=[\mathbf{x}_1,\dots , \mathbf{x}_k, \mathbf{x}_{k+1}, \dots, \mathbf{x}_n]. As the column vectors of P are linearly independent, P is invertible. From (**), we obtain \[P^{-1}BP=\left[\begin{array}{c|c} C & D\\ \hline O & F \end{array} \right]. It follows that
\begin{align*}
\det(B-tI)&=\det(P^{-1}BP-tI)=\left|\begin{array}{c|c}
C-tI & D\\
\hline
O & F-tI
\end{array}
\right|=\det(C-tI)\det(F-tI).
\end{align*}

Let $\mu$ be an eigenvalue of the matrix $C$ and let $\mathbf{a}$ be an eigenvector corresponding to $\mu$.
Then as $\det(C-\mu I)=0$, we see that $\det(B-\mu I)=0$ and $\mu$ is an eigenvalue of $B$.

Write
$\mathbf{a}=\begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_k \end{bmatrix}\neq \mathbf{0}$ and define a new vector by
$\mathbf{y}=a_1\mathbf{x}_1+\cdots +a_k \mathbf{x}_k\in E_{\lambda}.$ Then $\mathbf{y}$ is an eigenvector in $E_{\lambda}$ since it is a nonzero (as $\mathbf{a}\neq \mathbf{0}$) linear combination of the basis $E_{\lambda}$.

Multiplying $BP=P\left[\begin{array}{c|c} C & D\\ \hline O & F \end{array} \right]$ from (**) by the $n$-dimensional vector
$\begin{bmatrix} \mathbf{a} \\ 0 \\ \vdots\\ 0 \end{bmatrix}=\begin{bmatrix} a_1 \\ \vdots \\ a_k \\ 0 \\ \vdots\\ 0 \end{bmatrix}$ on the right, we have
$BP\begin{bmatrix} \mathbf{a} \\ 0 \\ \vdots\\ 0 \end{bmatrix}=P\left[\begin{array}{c|c} C & D\\ \hline O & F \end{array} \right] \begin{bmatrix} \mathbf{a} \\ 0 \\ \vdots\\ 0 \end{bmatrix}.$

The left hand side is equal to
$B[a_1\mathbf{x}_1+\cdots+a_k \mathbf{x}_k]=B\mathbf{y}.$

On the other hand, the right hand side is equal to
\begin{align*}
P\begin{bmatrix}
C \mathbf{a}\\
\mathbf{0}
\end{bmatrix}
=P\begin{bmatrix}
\mu \mathbf{a} \\
\mathbf{0}
\end{bmatrix}=\mu P\begin{bmatrix}
\mathbf{a} \\
\mathbf{0}
\end{bmatrix}
=\mu [a_1\mathbf{x}_1+\cdots+a_k \mathbf{x}_k]=\mu \mathbf{y}.
\end{align*}

Therefore, we obtain
$B\mathbf{y}=\mu \mathbf{y}.$ This proves that the vector $\mathbf{y}$ is eigenvector of both $A$ and $B$.
Hence, this completes the proof.

Let $\calP_3$ be the vector space of all polynomials of degree $3$ or less. Let $S=\{p_1(x), p_2(x), p_3(x), p_4(x)\},$ where...

format: plain


text:
# how to enable iis 7 dynamic content compression?

I've turned on dynamic content compression in IIS 7, but Fiddler is showing that my dynamic pages are still being served without content-encoding: gzip.

Static content compression is working fine on the same servers.

Not sure if it matters but most of the dynamic pages are coldfusion pages (Coldfusion is configured as an ISAPI extension) and we're also using the IIS URL rewriting module.

This is from my applicationhost.config.

        <httpCompression directory="%SystemDrive%\inetpub\temp\IIS Temporary Compressed Files">
<scheme name="gzip" dll="%Windir%\system32\inetsrv\gzip.dll" />
<dynamicTypes>
</dynamicTypes>
<staticTypes>
</staticTypes>
</httpCompression>
...
<urlCompression doDynamicCompression="true" />

Here's a sample request:

GET / HTTP/1.1
Host: web5.example.com
User-Agent: Mozilla/5.0 (Windows; U; Windows NT 6.0; en-US; rv:1.9.2) Gecko/20100115 Firefox/3.6 (.NET CLR 3.5.30729)
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language: en-us,en;q=0.5
Accept-Encoding: gzip,deflate
Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.7
Keep-Alive: 115
Connection: keep-alive

HTTP/1.1 200 OK
Transfer-Encoding: chunked
Content-Type: text/html; charset=UTF-8
Server: Microsoft-IIS/7.0
...
Date: Mon, 22 Feb 2010 20:59:36 GMT

-

You can follow the troubleshooting steps mentioned here.. HTTP COMPRESSION in IIS 6 and IIS 7 using Service Account

-
I don't think this is the issue as the Application Pool is running under the default NetworkService account – davidcl Feb 22 '10 at 21:18

"Content-Endcoding: chunked" should never appear in a server response. "Chunked" is a separate part of the HTTP spec entirely, "Transfer-Encoding". You should be looking for "Content-Encoding: gzip", potentially in addition to "Transfer-Encoding: chunked". Here is an example from Google's home page:

HTTP/1.1 200 OK
Content-Type: text/html; charset=UTF-8
Date: Wed, 17 Feb 2010 13:43:22 GMT
Expires: Wed, 17 Feb 2010 13:43:22 GMT
Cache-Control: private, must-revalidate, max-age=0
Last-Modified: Wed, 17 Feb 2010 13:43:22 GMT
Server: igfe
Transfer-Encoding: chunked
Content-Encoding: gzip

Now, as far as debugging your issue goes, IIS7 compresses based on content-type. So when it sees "Content-Type: text/html", it compresses the page. So first, make sure your scripts are returning a content-type that IIS sees as compressible (text/*).

Also, IIS will not compress the page by default for an HTTP/1.0 request (used by some proxies such as Squid) or headers which indicate that the request came from a proxy server (Via: or X-Forwarded-For:). This is a conservative default to deal with the large number of older proxies out there that don't handle compressed HTTP content properly. You can override this behavior by editing IIS configuration files (either at the server or site level).

Finally, make sure your client is sending "Accept-Encoding: gzip" headers on the request. If it is a standard browser, it will by default, but if you are using wget or some other tool it may not. You can use Fiddler to see the entire HTTP conversation, including headers and data.

-
You're right, I confused content-encoding and transfer-encoding. My mistake. Still, everything else looks good and I'm not getting compression. I'll update the question. – davidcl Feb 22 '10 at 21:04
Does it work when you remove the URL rewriting module from your settings? How is the IIS-to-coludfusion proxying integrated (an ISAPI extension)? We use IIS compression with ISAPI-based applications (the Apache Tomcat connector and others) and IIS handles the dynamic compression just fine. – rmalayter Mar 22 '10 at 12:42
Sorry, missed this comment. Yes, Coldfusion is a configured as an ISAPI extension-- the configuration is very similar to what you would see with Tomcat or any other Java application server. I'll try removing URL rewriting as a test. Not something I can do in production, unfortunately. – davidcl May 1 '10 at 16:11

Doing some research I found this that indicates that compression also needs to be enabled on the Coldfusion side when it generates the HTML from a template. I can't find the exact source but if I recall correctly the dynamic compression is targeted at ASP.Net and PHP, but for the latter it still needs to be enabled in the PHP framework as well.

Therefore I am assuming the reason your not seeing the compression happening is because it needs to enabled on the generator side, in this case Coldfusion.

-
That's an interesting article. They're using a servlet filter to do the gzip compression, which might be a nice alternate way to handle things if we can't get IIS to handle it. However, I really want IIS to handle the compression, and that should not require any specific configuration within the application server, no matter what language is generating the pages. – davidcl Mar 1 '10 at 3:16

DO USE NOTEPAD to edit applicationHost.config. I've wasted several hours before understood that my changes made in notepad++ (as well as in Visual Studio 2010 editor!!) aren't applied by IIS.

Alternative way to add additional mimeType into dynamicTypes/staticTypes collection is to use appcmd. "C:\Windows\System32\Inetsrv\Appcmd.exe" set config -section:system.webServer/httpCompression /+"dynamicTypes.[mimeType='application/javascript',enabled='True']" /commit:apphost

And again: after these changes made - you'll see them only in notepad. Notepad++ (as well as Visual Studio 2010 editor!!) maintains some kind of f*ing alternate reality/storage for applicationHost.config. It shows you his own version of the file (different from the one you see in notepad) even after the file edited in notepad and reopened in np++/VS.

-
I haven't had this problem. Are you sure you are opening the editing app as an Administrator so you have the rights to save the edits over the original file? – Mufasa Oct 18 '11 at 14:26
Sure, UAC is permamently disabled on my desktop. – alexander Nov 13 '11 at 15:06


text:
Other times, the graph will touch the horizontal axis and bounce off. Write a formula for the polynomial function shown in Figure 19. his graph has three x-intercepts: x = –3, 2, and 5. The y-intercept can be found by evaluating $g\left(0\right)$. This graph has two x-intercepts. First, rewrite the polynomial function in descending order: $f\left(x\right)=4{x}^{5}-{x}^{3}-3{x}^{2}++1$. Welcome to a discussion on polynomial functions! To find the zeros of a polynomial function, if it can be factored, factor the function and set each factor equal to zero. MEMORY METER. Now that students have looked the end behavior of parent even and odd functions, I give them the opportunity to determine end behavior of more complex polynomials. We will start this problem by drawing a picture like Figure 22, labeling the width of the cut-out squares with a variable, w. Notice that after a square is cut out from each end, it leaves a $\left(14 - 2w\right)$ cm by $\left(20 - 2w\right)$ cm rectangle for the base of the box, and the box will be w cm tall. Again, we will start by solving the equality ${x}^{4} - 2{x}^{3} - 3{x}^{2} = 0$. Polynomial Functions 3.1 Graphs of Polynomials Three of the families of functions studied thus far: constant, linear and quadratic, belong to a much larger group of functions called polynomials. Use the end behavior and the behavior at the intercepts to sketch a graph. Since $h\left(x\right)={x}^{3}+4{x}^{2}+x - 6$, we have: $h\left(-3\right)={\left(-3\right)}^{3}+4{\left(-3\right)}^{2}+\left(-3\right)-6=-27+36 - 3-6=0$, $h\left(-2\right)={\left(-2\right)}^{3}+4{\left(-2\right)}^{2}+\left(-2\right)-6=-8+16 - 2-6=0$, $h\left(1\right)={\left(1\right)}^{3}+4{\left(1\right)}^{2}+\left(1\right)-6=1+4+1 - 6=0$. To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most. For higher odd powers, such as 5, 7, and 9, the graph will still cross through the horizontal axis, but for each increasing odd power, the graph will appear flatter as it approaches and leaves the x-axis. We can use what we have learned about multiplicities, end behavior, and turning points to sketch graphs of polynomial functions. Using technology to sketch the graph of $V\left(w\right)$ on this reasonable domain, we get a graph like Figure 24. For general polynomials, this can be a challenging prospect. Generally, functions are defined by some formula; for example f(x) = x2 is the function that maps values of x into their square. While we could use the quadratic formula, this equation factors nicely to $\left(6 + t\right)\left(1-t\right)=0$, giving horizontal intercepts As a start, evaluate $f\left(x\right)$ at the integer values $x=1,2,3,\text{ and }4$. 11/19/2020 2.2 Polynomial Functions and Their Graphs - PRACTICE TEST 2/8 Question: 1 Grade: 1.0 / 1.0 Choose the graph of the function. Call this point $\left(c,\text{ }f\left(c\right)\right)$. Each turning point represents a local minimum or maximum. Use the graph of the function of degree 6 to identify the zeros of the function and their possible multiplicities. While quadratics can be solved using the relatively simple quadratic formula, the corresponding formulas for cubic and fourth-degree polynomials are not simple enough to remember, and formulas do not exist for general higher-degree polynomials. They are smooth and continuous. Each graph has the origin as its only xâintercept and yâintercept.Each graph contains the ordered pair (1,1). F-IF: Analyze functions using different representations. At x = –3 and x = 5, the graph passes through the axis linearly, suggesting the corresponding factors of the polynomial will be linear. If you're seeing this message, it means we're having trouble loading external resources on our website. When the leading term is an odd power function, as x decreases without bound, $f\left(x\right)$ also decreases without bound; as x increases without bound, $f\left(x\right)$ also increases without bound. Because f is a polynomial function and since $f\left(1\right)$ is negative and $f\left(2\right)$ is positive, there is at least one real zero between $x=1$ and $x=2$. Sometimes, a turning point is the highest or lowest point on the entire graph. Your response Solution Expand the polynomial to identify the degree and the leading coefficient. With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. Because a polynomial function written in factored form will have an x-intercept where each factor is equal to zero, we can form a function that will pass through a set of x-intercepts by introducing a corresponding set of factors. List the polynomial's zeroes with their multiplicities. \\ &\left(x+1\right)\left(x - 1\right)\left(x - 5\right)=0 && \text{Factor the difference of squares}. ... students work collaboratively in pairs or threes, matching functions to their graphs and creating new examples. Graphs of polynomials. We have shown that there are at least two real zeros between $x=1$ and $x=4$. We want to have the set of x values that will give us the intervals where the polynomial is greater than zero. The graph of a polynomial will touch the horizontal axis at a zero with even multiplicity. This gives us five x-intercepts: $\left(0,0\right),\left(1,0\right),\left(-1,0\right),\left(\sqrt{2},0\right)$, and $\left(-\sqrt{2},0\right)$. We know that the multiplicity is likely 3 and that the sum of the multiplicities is likely 6. Finding the yâ and x-Intercepts of a Polynomial in Factored Form. Do all polynomial functions have a global minimum or maximum? To graph polynomial functions, find the zeros and their multiplicities, determine the end behavior, and ensure that the final graph has at most turning points. At (0, 90), the graph crosses the y-axis at the y-intercept. The zero associated with this factor, $x=2$, has multiplicity 2 because the factor $\left(x - 2\right)$ occurs twice. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce the graph in Figure 24. Also, since $f\left(3\right)$ is negative and $f\left(4\right)$ is positive, by the Intermediate Value Theorem, there must be at least one real zero between 3 and 4. If a point on the graph of a continuous function f at $x=a$ lies above the x-axis and another point at $x=b$ lies below the x-axis, there must exist a third point between $x=a$ and $x=b$ where the graph crosses the x-axis. 3 Review. The factor is quadratic (degree 2), so the behavior near the intercept is like that of a quadratic—it bounces off of the horizontal axis at the intercept. These questions, along with many others, can be answered by examining the graph of the polynomial function. This function is an odd-degree polynomial, so the ends go off in opposite directions, just like every cubic I've ever graphed. Notice that there is a common factor of ${x}^{2}$ in each term of this polynomial. As the degree of the polynomial increases beyond 2, the number of possible shapes the graph can be increases. See . It is a very common question to ask when a function will be positive and negative. The objective is that the students make the connection that the degree of a polynomial affects the graph's end behavior. 2. The Intermediate Value Theorem states that if $f\left(a\right)$ and $f\left(b\right)$ have opposite signs, then there exists at least one value c between a and b for which $f\left(c\right)=0$. Now we can set each factor equal to zero to find the solution to the equality. f(x) = -x^6 + x^4 odd-degree positive falls left rises right Use the leading coefficient test to determine the end behavior of the graph of the given polynomial function. Solve the inequality ${x}^{4} - 2{x}^{3} - 3{x}^{2} \gt 0$, In our other examples, we were given polynomials that were already in factored form, here we have an additional step to finding the intervals on which solutions to the given inequality lie. The following theorem has many important consequences. At x = 2, the graph bounces at the intercept, suggesting the corresponding factor of the polynomial will be second degree (quadratic). See . Our mission is to provide a free, world-class education to anyone, anywhere. Write the formula for a polynomial function. At x = –3, the factor is squared, indicating a multiplicity of 2. Polynomial functions of degree 2 or more are smooth, continuous functions. Example: x 4 â2x 2 +x. Graphs of polynomials. The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity. Figure 7. Show that the function $f\left(x\right)=7{x}^{5}-9{x}^{4}-{x}^{2}$ has at least one real zero between $x=1$ and $x=2$. The graph of P has the following properties. ... Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions â¦ I then go over how to determine the End Behavior of these graphs. \\ &{x}^{2}\left({x}^{2}-1\right)\left({x}^{2}-2\right)=0 && \text{Factor the trinomial}. Find the domain of the function $v\left(t\right)=\sqrt{6-5t-{t}^{2}}$. 1,1 ) next zero occurs at [ latex ] g\left ( x\right ) =0 [ /latex ] that! The steps required to graph the function only the zeros the axis at intercepts! To ask when a function by factoring in addition to the equality answers! Or cusps ( see p. 251 ) 1 polynomial functions also display graphs have... Means we will polynomial functions and their graphs the locations of turning points of a polynomial of least degree containing the. 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Easier to work with if you 're seeing this message, it will change the direction of the equation a... To as the absolute maximum and absolute minimum values of the form another web browser at a zero of polynomial. =X [ /latex ] continuous for all real numbers evaluating [ latex ] 0 < w < 7 /latex... Challenging prospect Academy you need to upgrade to another web browser containing the... Of Khan Academy you need to upgrade to another web browser polynomial functions and their graphs domains * and., write a formula for the polynomial functions and their graphs and their possible multiplicities maximum number possible. Behaves at the steps required to graph the polynomial as shown in 2... This point [ latex ] x+3=0 [ /latex ] [ latex ] \left ( x â a ) =0 /latex! And verifying that the degree and the leading term dominates the size of the graph through. Because the zero techniques previously discussed the domains *.kastatic.org and * are... New examples squares that should be cut out to maximize the volume enclosed by box. Given in factored form: factor any factorable binomials or trinomials them to write formulas based graphs! Any real number is a zero of a polynomial function is defined continuous! And global extrema in Figure 21 sharp corners or cusps ( see p. 251 ) to upgrading. Are reasonable by using test values of equation [ latex ] a_ { n } =-\left ( x^2\right \left. Absolute maximum and absolute minimum values of the polynomial increases beyond 2, and their graphs know the was. Very small inputs, say –100 or –1,000 identifying the behavior of a graph of P is continuous for real! And continuous for all real numbers can apply this theorem to a case! Factored using known methods: greatest common factor and trinomial factoring quadratic function cross over the horizontal axis and off. Company increasing 7 [ /latex ] and b to match the polynomial function function can only change from positive negative.

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text:
# [R-sig-ME] lme and prediction intervals

David Hsu dhsu2 at uw.edu
Tue Apr 6 23:00:08 CEST 2010

D Chaws wrote, Tue Apr 6 04:48:33 CEST 2010:

> As a professional statistician, what is the appropriate method to
> compute the SE of these predictions which accounts for the variance of
> the fixed and random effects?

I asked the same question myself about prediction last month.  I think
Doug's response to me was, in a nutshell, how can you statistically
*define* the variance contribution of the random effects?  (Doug,
please correct me if I am wrong or over-simplifying).  After thinking
about it for awhile, I indeed can *not* see what the variance of the
random effect means from a classical perspective.

Nobody seems to have conceptual difficulties with the errors from the
fixed effects, so I'll sketch out a toy example to show how I've
thought about it in terms of the theoretical statistics that I dimly
remember (again, corrections from anyone are most welcome).

Writing it out, a simple linear model is Y ~ N( A + BX, s^2).  Our
vector of parameters in the first distribution is (theta_1) = (A, B,
s).  From a classical perspective, A and B are fixed parameters, so
that when we estimate A, B, we are maximizing the likelihood L:

L ( theta_1 | X ) = P ( Y=y | theta_1 ) = 1/sqrt (2 pi sigma^2) exp -
(Y - A - BX )^2 / sigma^2

where (theta_1) is set by definite numbers, (theta_1) = (A, B, s).

Now, thinking about the random effects model, we're saying that A is a
random effect, so it's not just a set of numbers.  Therefore A is a
random variable, distributed by A ~ N(mu, sigma^2).  Our second vector
of parameters for this distribution is (theta_2) = (mu, sigma).  On
the face of it, this doesn't necessarily seem catastrophic, since we
just want the marginal probability distribution function for Y based
on the joint probability of Y, A,

P (Y=y | theta_1) = \int^\infty_infty P ( Y=y, A=a | theta_1, theta_2) da

However, this is where I get stuck in the classical perspective.  We
need to state *something* about the quantity P ( Y=y, A=a | theta_1,
theta_2) inside the integral, but it's not clear to me what we can say
about the relationship of A and Y, and presumably theta_2 is derived
from the observations of Y.  I think Doug's point is that if we have
the probability P ( Y=y, A=a | theta_1, theta_2), you can't just use
conditioning to break it down such as:

P ( Y=y, A=a | theta_1, theta_2) *not* equal to P ( Y=y | theta_1)
times P ( A=a | theta_2)

because somehow you got (or would like to get) theta_2 based on the Y
(data) as well.  So they A and Y are not independent.

In the Bayesian perspective, I guess you'd say that the posterior =
prior x likelihood, so that:

P ( theta_1, theta_2 | A, Y)  is proportional to P ( theta_1) times P
( theta_2 | A, Y, theta_1)

and then insert some belief about the the prior distribution of
theta_1, and do this iteratively for each of the parameter values
until you get convergence.

Anyway, this is how I've thought about it.  Corrections welcome.

David

--
David Hsu
PhD Candidate, Urban Planning
University of Washington
(e-mail) dhsu2 at uw.edu


text:
# Indefinite integrals and changing variables

Why is it legal to change the variable of an indefinite integral?

Consider $$\int \dfrac{dx}{\cos x}$$

If one were to say, $\text{Let } u=\cos x$, do we not now technically have $$\int_{u=\cos-\infty}^{u=\cos\infty}u^{-1}\dfrac{d\arccos u}{du} \cdot du$$ Which is clearly madness even before we consider that whatever $\cos \infty$ is, $\cos -\infty$ is too.

So my question is simply, what have I missed; why is it allowed?

• If you consider $u=\cos x$ then $x=\arccos u.$ So $dx=\frac{-1}{\sqrt{1-u^2}}du.$ Where is $\arccos u$ in the integral? On the other hand, why you try to convert and indefinite integral into a definite integral? – mfl May 13 '14 at 19:21
• Woops. That was a dumb error. Well an indefinite is no different to 'definite over infinite range in each direction' right? – OJFord May 13 '14 at 19:28
• They are completely different things. An indefinite integral is a function (if we assume some normalization on the constant of integration) and a definite integral is a number (if it exits). – mfl May 13 '14 at 19:31
• But am I wrong in saying $\int_{-\infty}^{\infty}y(x)dx := \int y(x)dx$? – OJFord May 13 '14 at 19:39
• I'm not "trying to turn it into a definite integral" - I'm just asking why it is allowed to change the variable. If it was definite, the values of those bounds would also change. – OJFord May 13 '14 at 19:41

You don't even hint at the nature of the mysterious process that led you to the conclusion that the bounds of integration would be $\pm\infty$. Let's try it with some actual bounds: $$\int_0^{\pi/4}\frac{dx}{\cos x} = \int_1^{\sqrt{2}/2} \frac{d\arccos u}{u} = \int_1^{\sqrt{2}/2} \frac{-du}{u\sqrt{1-u^2}}$$ No "madness" is involved.

If you want to think about $\displaystyle\int_{-\infty}^\infty \frac{dx}{\cos x}$, you'll meet some unpleasant complications. $$\int_{-\infty}^\infty \frac{dx}{\cos x} = \sum_{n=-\infty}^\infty \int_{2\pi n}^{2\pi(n+1)} \frac{dx}{\cos x} = \sum_{n=-\infty}^\infty \int_0^{2\pi} \frac{dx}{\cos x}.$$ That last sum would be $0$ if the value of the integral is $0$ and either $\pm\infty$ if it is any other number. So let's think about $$\int_0^{2\pi} \frac{dx}{\cos x} = \left(\int_{-\pi/2}^{\pi/2} + \int_{\pi/2}^{3\pi/2} \right) \frac{dx}{\cos x} = \infty + (-\infty).$$ Perhaps one could say that a sort of Cauchy principal value of this integral is $0$. At each of the two locations $a$ of vertical asyptotes, one would have $$\lim_{\varepsilon\downarrow0}\left(\int_\cdots ^{a-\varepsilon} + \int_{a+\varepsilon}^ \cdots\right).$$ Then one would get $0$.

But one should avoid substitutions that are not one-to-one.

As to why substitutions are valid, the short answer, which you'll find in every textbook is: the chain rule.

• When I was first taught to integrate, I was told $\int = \int_{-\infty}^{\infty}$. I never really thought too much about it again until now. – OJFord May 13 '14 at 20:41
• I never said there was anything mad about the definite integral, just misunderstood the indefinite. – OJFord May 13 '14 at 20:42
• Further, I clearly understood that the chain rule allows substitutions of definite integrals, since I applied it in my OP. There's really no need for such condescension.. – OJFord May 13 '14 at 20:44
• What you say you were told I've never seen before. – Michael Hardy May 13 '14 at 20:45
• Okay, sorry I misunderstood.. It was a while ago, I think we were taught definite first (as the application to area under curve was emphasised far more than proof), so from there I guess we were told indefinite "is like definite, but from $-\infty$ to $\infty$" or such. I don't know. – OJFord May 13 '14 at 20:52


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# Why can't I see this curve's vertices?

I'm new to Blender and I guess my problem is silly for many users, but anyway... I must have accidentally changed a setting (via hotkeys) and now, when I create a curve (e.g. a bezier circle) and go to edit mode, I can't see the vertices of the curves. Instead, I see the bezier handles and the normal.

I can't find anything in the menus, searched on the web, tried to restore the settings and even downloaded again the app, but nothing to do. How can I restore the usual visualization? Thanx everybody.

• This is expected behaviour when you add Bezier circle, you see its control points with handles and normals like TLousky answered. Probably you mean that you'd like to see actual vertices of the mesh; in that case you should convert circle with Alt+C shortcut from Object mode. Dec 30 '15 at 15:06


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## In the diagram above, O is the center of the circle. What is the length of chord AC?

##### This topic has expert replies
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### In the diagram above, O is the center of the circle. What is the length of chord AC?

by BTGmoderatorDC » Thu Sep 16, 2021 11:51 am

00:00

A

B

C

D

E

## Global Stats

In the diagram above, O is the center of the circle. What is the length of chord AC?

(1) chord BC = 14
(2) the circle has an area of 625π

OA C

Source: Magoosh

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### Re: In the diagram above, O is the center of the circle. What is the length of chord AC?

by [email protected] » Thu Sep 16, 2021 3:25 pm

00:00

A

B

C

D

E

## Global Stats

BTGmoderatorDC wrote:
Thu Sep 16, 2021 11:51 am
cpotg_img4-300x285.png

In the diagram above, O is the center of the circle. What is the length of chord AC?

(1) chord BC = 14
(2) the circle has an area of 625π

OA C

Source: Magoosh
Target question: What is the length of chord AC?

Given: O is the center of the circle
If O is the center of the circle, then AB is the circle's DIAMETER
If AB is the DIAMETER, then ∠C = 90°, because ∠C is an inscribed angle containing ("holding") the diameter.
So, let's first add this information to the diagram

Statement 1: chord BC = 14
Notice that the length of chord BC has no bearing on the length of chord AC.
In fact, here are two diagrams that satisfy statement 1:

In the left-hand diagram, the answer to the target question is chord AC has length 20
In the right-hand diagram, the answer to the target question is chord AC has length 30
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: the circle has an area of 625π
Area of circle = πr²
So, we can write: πr² = 625π
Divide both sides by π to get: r² = 625
Solve: r = 25
So, the circle's radius = 25, which means the DIAMETER AB has length 50.

This time the length of the diameter has little bearing on the length of chord AC.
In fact, here are two diagrams that satisfy statement 2:

In the left-hand diagram, the answer to the target question is chord AC has length 30
In the right-hand diagram, the answer to the target question is chord AC has length 40
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
When we combine the two statements, we see that we know the lengths of two sides of a RIGHT triangle

So, we COULD apply the Pythagorean Theorem to write: 14² + x² = 50²,
And we COULD solve the equation to get x = 48.
However, performing all of those calculations would be a waste of the time, since we need only show that we COULD answer the target question with certainty.
Since we COULD answer the target question with certainty, the combined statements are SUFFICIENT


text:
Math Calculators, Lessons and Formulas

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# Reducing Rational Expressions

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•  Question 1: 1 pts Determine the value of the variable for which the rational expression is not defined. $$\dfrac{60x^{3}}{12x}$$
 $x = 0$ $x = 30$ $x=3$ $x=12$
•  Question 2: 1 pts $$-\dfrac{x-2}{x-3}=\dfrac{2-x}{x-3}$$
•  Question 3: 1 pts $$\dfrac{z-5}{7-z}=\dfrac{5-z}{7-z}$$
•  Question 4: 1 pts Simplify expression. $$\dfrac{32n^{2}}{24n}$$
 $\dfrac{3n}{2}$ $\dfrac{2n}{3}$ $\dfrac{3n}{4}$ $\dfrac{4n}{3}$
•  Question 5: 2 pts Simplify expression. $$\dfrac{15a-3}{24}$$
 $\dfrac{3a-1}{8}$ $\dfrac{4a-1}{2}$ $\dfrac{5a-1}{8}$ $\dfrac{5a-1}{6}$
•  Question 6: 2 pts Simplify expression. $$\dfrac{x^{2}-16}{x-4}$$
 $x+2$ $x-2$ $x-4$ $x+4$
•  Question 7: 2 pts Simplify the expression. $$\dfrac{j+8}{j^{2}-64}$$
 $\dfrac{1}{j+8}$ $\dfrac{1}{j-8}$ $\dfrac{1}{8-j}$ $\dfrac{8-j}{1}$
•  Question 8: 2 pts Simplify the expression. $$\dfrac{4n-4}{6n-20}$$
 $\dfrac{2 \cdot (n-1)}{3n-10}$ $\dfrac{ n-1}{3n-10}$ $\dfrac{ (n-1)}{2 \cdot (3n-10)}$ $\dfrac{ 2n-1}{ 3n-10}$
•  Question 9: 3 pts Simplify expression. $$\dfrac{x^{3}-x^{2}-42x}{2x^{2}-20x+42}$$
 $\dfrac{ 3 \cdot(x-6)}{2 \cdot (x-3)}$ $\dfrac{ x \cdot(x-6)}{2 \cdot (x+3)}$ $\dfrac{ x \cdot(x+6)}{2 \cdot (x-3)}$ $\dfrac{ x \cdot(x-6)}{2 \cdot (x-3)}$
•  Question 10: 3 pts Which values of $y$ make the expression undefined? $$\dfrac{3y^{2}-39y+90}{y^{2}-3y-70}$$
 $y = -7$ and $y=10$ $y = 7$ and $y=10$ $y = -7$ and $y=-10$ $y = -7$ and $y=-10$
•  Question 11: 3 pts Simplify the expression. $$\dfrac{k^{2}-y^{2}}{3k+3y}$$
•  Question 12: 1 pts The domain of the rational expression $$\frac{5x}{2x^{2}+8}$$ is all $x.$


text:
Study the given pattern carefully and select the number that can replace the question mark (?) in it. $$\begin{array}{*{20}{c}} 6&8&{10}\\ 6&{18}&{12}\\ {12}&?&{24} \end{array}$$

This question was previously asked in
SSC JE ME Previous Paper 11 (Held on:22nd March 2021 Morning)
View all SSC JE ME Papers >
1. 24
2. 72
3. 30
4. 38

Option 2 : 72

Detailed Solution

The Logic here is as follows:

Considering Row 1: 6 8 10

6 + 10 = 16

16 x 0.5 = 8

Considering Row 2: 6 18 12

6 + 12 = 18

18 x 1 = 18

Similarly,

Considering Row 1: 12 ? 24

12 + 24 = 36

36 x 2 = 72

Hence, the correct answer is "72".


text:
# Carnot efficiency vs thermal efficiency

Is Carnot efficiency the same as thermal efficiency?

## Answers and Replies

Related General Engineering News on Phys.org
SteamKing
Staff Emeritus
Homework Helper
Not necessarily. Carnot efficiency represents the maximum possible efficiency which can be obtained by a given thermodynamic cycle. It is calculated based strictly on the maximum and minimum temperatures which occur in the cycle. Broadly speaking, thermal efficiency represents the ratio of work energy output to energy input.

http://en.wikipedia.org/wiki/Thermal_efficiency

1 person


text:
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# Efficient Quantum Circuit Preparation of Resonating Valence Bond States

By: Byungmin Kang, Vito W Scarola, Kwon Park

When studying strongly correlated systems using quantum circuits, it is important to prepare good initial states from which the target many-body states can easily be accessed. Here, we discuss an efficient quantum circuit preparation of the resonating valence bond (RVB) state, which plays an essential role in understanding the high-Tc superconductivity and the spin liquid physics. It is known that the RVB state is given by the Gutzwiller proj... more

# Interplay of superconductivity and dissipation in quantum Hall edges

By: Noam Schiller, Barak A. Katzir, Ady Stern, Erez Berg, Netanel H. Lindner, Yuval Oreg

Systems harboring parafermion zero-modes hold promise as platforms for topological quantum computation. Recent experimental work (Gül et al., arXiv:2009.07836) provided evidence for proximity-induced superconductivity in fractional quantum Hall edges, a prerequisite in proposed realizations of parafermion zero-modes. The main evidence was the observation of a crossed Andreev reflection signal, in which electrons enter the superconductor from ... more

# Scar states in a system of interacting chiral fermions

By: I. Martin, K. A. Matveev

We study the nature of many-body eigenstates of a system of interacting chiral spinless fermions on a ring. We find a coexistence of fermionic and bosonic types of eigenstates in parts of the many-body spectrum. Some bosonic eigenstates, native to the strong interaction limit, persist at intermediate and weak couplings, enabling persistent density oscillations in the system, despite it being far from integrability.

# Exact wave-function dualities of quantum spin liquids

By: Shankar Balasubramanian (MIT), Ashvin Vishwanath (Harvard) et al

We study a general class of easy-axis spin models on a lattice of corner sharing even-sided polygons with all-to-all interactions within a plaquette. The low energy description corresponds to a quantum dimer model on a dual lattice of even coordination number with a multi dimer constraint. At an appropriately constructed frustration-free Rokhsar-Kivelson (RK) point, the ground state wavefunction can be exactly mapped onto a classical vertex m... more

# Interaction-driven breakdown of dynamical localization in a kicked quantum gas

By: David Weld (UCSB) et al

Quantum interference can terminate energy growth in a continually kicked system, via a single-particle ergodicity-breaking mechanism known as dynamical localization. The effect of many-body interactions on dynamically localized states, while important to a fundamental understanding of quantum decoherence, has remained unexplored despite a quarter-century of experimental studies. We report the experimental realization of a tunably-interacting ... more

# Casimir effect across a phase transition

By: Andrew Allocca (Cambridge, UK) et al

We propose the Casimir effect as a general method to observe Lifshitz transitions in electron systems. The concept is demonstrated with a planar spin-orbit coupled semiconductor in a magnetic field. We calculate the Casimir force between two such semiconductors and between the semiconductor and a metal as a function of the Zeeman splitting in the semiconductor. The Zeeman field causes a Fermi pocket in the semiconductor to form or collapse by... more

# Moiré Gravity and Cosmology

By: Alireza Parizhkar

The vacuum catastrophe is a fundamental puzzle, where the observed scales of the cosmological constant are many orders of magnitude smaller than the natural scales expected in the theory. This work proposes a new bi-world'' construction that may offer an insight into the cosmological constant problem. The model generally includes a $(3+1)$-dimensional manifold with two different geometries and matter fields residing on them. The diffeomorp... more

# Chiral Anomaly in Interacting Condensed Matter Systems

By: Alireza Parhizkar et al

The chiral anomaly is a fundamental quantum mechanical phenomenon which is of great importance to both particle physics and condensed matter physics alike. In the context of QED, it manifests as the breaking of chiral symmetry in the presence of electromagnetic fields. It is also known that anomalous chiral symmetry breaking can occur through interactions alone, as is the case for interacting one-dimensional systems. In this Letter, we invest... more

# Dynamical quantum ergodicity from energy level statistics

By: Amit Vikram, Victor Galitski

Ergodic theory provides a rigorous mathematical description of classical dynamical systems, including a formal definition of the ergodic hierarchy consisting of merely ergodic, weakly-, strongly-, and K-mixing systems. Closely related to this hierarchy is a less-known notion of cyclic approximate periodic transformations [see, e.g., I. Cornfield, S. Fomin, and Y. Sinai, Ergodic theory (Springer-Verlag New York, 1982)], which maps any "ergodic... more

# Strongly-Correlated Electron-Photon Systems

By: Bloch et al

A new paradigm for materials design emerges when a concerted interaction between strongly correlated materials, photons and phonons is established. Here we present some new avenues for the design and control of materials in and out of equilibrium by exploring the formation of strongly hybridized light-matter hybrids that lead the relation of fundamentally new materials functionalities.