Melody

x = 2|y|− 3

Range is all real y

Domain is x>=3

So for the inverse:

Domain is all real x     (-infinty, infinty)

Range is y>=3            [-3,infinity)

check by graphing

The green one is the original and the purple one is the inverse  (reflection across y=x which is the red one)

Use the Euclidean algorithm to find integers x and y  such that  164x + 39y = 1.  State your answer as a list with x first and y second, separated by a comma.

\(164x+39y=1\\~\\ 164=4*\boxed{39}+8\\ 39=4*\boxed{8}+7\\ 8=1*\boxed{7}+1\\~\\ \text{Now the extended euclidean algorithm}\\ 1=8-1*\boxed{7}\\ 1=8-1*\left[39-4*\boxed{8}\right]\\ 1=8+1\left[-39+4*\boxed{8}\right]\\ 1=5*\boxed{8}-1*39\\ 1=5*\left[ 164-4*\boxed{39} \right]-1*39\\ 1=5*164-21*39\\ 1=164(5)+39(-21)\\ 1=164(5-39s)+39(-21+164a)\\~\\ x=5-39a\qquad y=-21+164a \qquad \text{Where a is any integer.} \)

You can finish it.

LaTex

164x+39y=1\\~\\
164=4*\boxed{39}+8\\
39=4*\boxed{8}+7\\
8=1*\boxed{7}+1\\~\\
\text{Now the extended euclidean algorithm}\\
1=8-1*\boxed{7}\\
1=8-1*\left[39-4*\boxed{8}\right]\\
1=8+1\left[-39+4*\boxed{8}\right]\\
1=5*\boxed{8}-1*39\\
1=5*\left[  164-4*\boxed{39} \right]-1*39\\
1=5*164-21*39\\
1=164(5)+39(-21)\\
1=164(5-39s)+39(-21+164a)\\~\\
x=5-39a\qquad y=-21+164a \qquad \text{Where a is any integer.}

The surface area would be the area of the 2 circle ends plus the area of the curved side.

So what would be the surface area in tems of  r and h ?

I do not know how many answers there may be to this.

But I said that one answer would be if they were all the same

I mean  x=y=z

then

\(x^2-x=15\\ x^2-x-15=0\\ x=\frac{1\pm \sqrt{61}}{2}\\ xyz=\left[\frac{1\pm \sqrt{61}}{2}\right]^3\\ \)

Expand this out and you get 2 different answers.

(13/52)^4+(4/52)^4

Hint:

Add all the left hand sides = add all the right hand sides

Now it is easy.

Hi Hectictar,

It is really good to see you here  :)

The parabolas defined by the equations y = x^2 + 4x + 6 and y = 2x^2 + 7x - 4 intersect at points and (a,b) and (c,d), where c >= a. What is c-a?

\(y = x^2 + 4x + 6 \qquad and \qquad y = 2x^2 + 7x - 4\)

These are both concave up parabolas

Looking of  x values the points of intersection

\(x^2 + 4x + 6= 2x^2 + 7x - 4\)

solve that and you will get the 2 x values.

I will leave you to play with it now.   

A guest has answered it.  Thanks guest.   I hope you thank guest too.

I probably would not have answered it.  Your 'request' sounds really rude.

Anyway, that question is ancient...

What calculator?  I don't think it is on the web2.0 calc