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I work in the german forum. I have a trick how to upload images.

I had the same problem, but I found a way out. I am using https://de.free-picload.com/. I upload the image and enter the path to the image in the URL and click ok. Try it.

I hope I could help.

Find all integers X for which there exists an integer Y such that

1/X + 1/Y = 1/7
(In other words, find all ordered pairs of integers (X, Y) that satisfy this equation, then enter just the X's from these pairs.)

Hi guest and Rom,

I didn't think it was that bad either.  LOL       

I gave you the other thumbs up.       

I've been around since 2010

Given vectors \(\mathbf{v}\) and \(\mathbf{w}\) such that \(\|\mathbf{v}\| = 3,\ \|\mathbf{w}\| = 7,\) and \(\mathbf{v} \cdot \mathbf{w} = 10,\) then find \(\|\operatorname{proj}_{\mathbf{w}} \mathbf{v}\|.\)

\(\begin{array}{|rcll|} \hline \|\operatorname{proj}_{\mathbf{w}} \mathbf{v}\| &=& \left\| \dfrac{\mathbf{v} \cdot \mathbf{w}}{\left\|\mathbf{w}\right\|^2}\mathbf{w} \right\| \\\\ &=& \left\| \dfrac{10}{7^2}\mathbf{w} \right\| \\\\ &=& \dfrac{10}{7^2} \| \mathbf{w} \| \\\\ &=& \dfrac{10}{7^2} \cdot 7 \\\\ \|\operatorname{proj}_{\mathbf{w}} \mathbf{v}\| &=& \mathbf{\dfrac{10}{7}} \\ \hline \end{array}\)

omg yeah, i've been trying to register and i tried different devices, different browsers, etc, and i keep getting an internal server error. and this has been going on for like 7-8 weeks now for me. 

\(\text{A quadratic has a common root when it's discriminant is 0}\\ p(x) = a x^2 + bx+c \Rightarrow D = b^2 - 4ac\\ D = b^2 - (4)(1988)(8891) = 0\\ b^2 = 70701232\\ b = \pm 4 \sqrt{4418827}\)

true, true. and it is a pretty good website too :( so it would be a shame if anything happened :(((

\(6 = \dfrac{35}{x} - \dfrac{49}{x^2}\\ 6x^2 = 35x - 49\\ 6x^2 - 35x +49 = 0\\~\\ x = \dfrac{35\pm \sqrt{35^2 - (4)(6)(49)}}{12} = \left\{\dfrac 7 3, \dfrac 7 2\right\} \)

\(\text{The key is that}\\ a^n \mod p = (a \mod p)^n \mod p\)

\(13^{13}+5 \mod 6 = \\ 1^{13} + 5 \mod 6 = \\ 6 \mod 6 = 0\)

I meant z multiplied by the conjugate of w=6+8i. Sorry for the confusion!!

thank you for your help!! i know how to solve them but like idk how to write them in solution set, like do we write it like this for the second one? 
{x: -3/2 \(\)

It is a 42-inch TV. This means the length of the diagonal is 42 inches.

Aspect ratio is the ratio of the width to the height.

So if we let the unknown scale factor be called  s ,

then  the width  =  4s

and  the height  =  3s

Now we can use the Pythagorean Theorem to find what  s  is.

(4s)² + (3s)²  =  42²    This is the equation given in the hint.

(4s)(4s) + (3s)(3s)  =  42²

16s² + 9s²  =  42²

                                   Just like  16 apples + 9 apples = 25 apples, so does  16s² + 9s² = 25s²

25s²  =  42²

                       Let's rewrite  25s²  like this...

(5s)²  =  42²

                       and take the square root of both sides.

5s  =  42

s  =  8.4    (inches)

Now that we know what  s  is, we can find the width and the height.

width  =  4s  =  4(8.4)  =  33.6    (inches)

height  =  3s  =  3(8.4)  =  25.2    (inches)

 2 ^ 17179869184 = 9.2743643701 9009168775 4290625551 9701007064 3102612995 9636058398 4155296225 6148051675 9380057885 370267332  x 10^5171655945

Sorry, Rom... all I remembered was the idea and the punch line.  But "butchered"...? 

I didn't think my improvisation of the set up was that bad.   Thanks for the thumbs up. 

.

You are going in the right direction.....we can split your answer  up into these  functions

         cosΔx - 1                             sinΔx 

(1/2)  _______    +    (√ 3/2)    _________ 

            Δx                                       Δx

We have two identities to remember  :

As   x →  0,     cos Δx  -  1

                       __________     approaches 0......

                                Δx 

So   the limit of the first function  just simplifies to  0

Also    as x   → 0 ,     sin Δx

                                  ______   approaches  1......

                                     Δx

So the limit of the second function just simplifies  to (√ 3/2) 

So  

lim               sin ( pi/6  +  Δx)  - 1/2           

 x→ 0         __________________   =   (√ 3/2) 

                            Δx

See the graph here : https://www.desmos.com/calculator/mnpxjiqitj

Here is the graph :   https://www.desmos.com/calculator/eyxfsgsa6y

The only x intercept  is   x  = 0

y intercepts   →  y  = 0

To find any possible holes in the graph......factor  the numerator  and denominator

x ( x - 3 )                          x

____________  =        ______       a "hole"  occurs when  x - 3  = 0  ⇒   x  =   3 

(2x + 3) (x - 3)              2x - 3

And the y coordinate  of this point  is

   3                       3                   1

______  =         _____   =      ____      so....the "hole"   is  at    (3, 1/3)

2(3) + 3                9                    3

We have a horizontal asymptote at    y  = 1/2

And we have vertical asymptotes at  x  =   -3/2  

I'll give you a thumbs up but you butchered that joke.

It helps one to have been born with an amount of analytical intuition, but even if one hasn't, mathematics is still a learnable skill.  My opinion is that the success one has in learning the skill depends on a few things. 

Here are a couple. 

1) Like math.  Enjoy it.  Have fun with it.  Appreciate its balance, its absolute dependability, and, I would add, its beauty.

2) The second thing, I'll tell an anecdote.  A tourist in New York, sightseeing, didn't know his way around the city very well.  He spotted a policeman and decided to ask directions.  He said, "Can you tell me how to get to Carnegie Hall?"  The policeman replied, "Practice, practice, practice!"  LOL.  Math is like other things, the more you do of it, the better you get at it. 

.

1. I am sort of surprised that the concept of the conjugate did not spring into my mind. Thanks for that small tip. 

2. No, I have not learned L'Hopital's rule. There is a hint near the footnote of the page for this problem; it says to think about trigonometric identities. It seems to me that the sum-difference identity is the proper one to use. Specifically, \(\sin(A+B)=\sin A\cos B+\cos A\sin B\).

\(\lim_{\Delta x \to 0}\frac{\textcolor{red}{\sin\left(\frac{\pi}{6}+\Delta x\right)}-\frac{1}{2}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\textcolor{red}{\sin\left(\frac{\pi}{6}\right)\cos \Delta x+\cos\left(\frac{\pi}{6}\right)\sin \Delta x}-\frac{1}{2}}{\Delta x}\\ \lim_{\Delta x \to 0}\frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\frac{1}{2}\cos \Delta x+\frac{\sqrt{3}}{2}\sin \Delta x-\frac{1}{2}}{\Delta x}\\ \lim_{\Delta x \to 0}\frac{\sin\left(\frac{\pi}{6}+\Delta x\right)-\frac{1}{2}}{\Delta x}=\lim_{\Delta x \to 0}\frac{\cos \Delta x+\sqrt{3}\sin \Delta x-1}{2\Delta x}\\ \)

I still do not see a way to simplify this limit. Although, I may be closer than I was before. The indeterminate form still exists. 

Thank you, CPhill !