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Thanks again man!! I at least had the right idea in one of my previous attempts but I think I screwed up the order of processes for myself. Anyways, you've helped me out a lot :D 

Yes, but it is not the exact math problem

Hi

Welcome to the forum :)

Midpoint =   ( add x coordinates/2 , add y coordinates/ 2  )  =

( [ - 3 + 4 ] / 2 , [ - 7 + 5 ] / 2 )    =

( 1/2 , -2/2 ) =

(1/2, -1) 

g(x)=2x^2-4x-5

For g(x), write y

y =  2x^2 - 4x - 5   factor out the 2

y = 2 ( x^2 - 2x -  5/2)    divide by 2

(y / 2) =  x^2 - 2x - 5/2      comlplete the square on x

(y /2) = x^2 - 2x + 1  - 5/2 - 1       factor the first three terms andsimplify

(y/2)  = (x - 1)^2  - 7/2   add 7/2 to both sides

[y + 7 ] / 2  =  (x - 1)^2      take both roots

±√  [ ( y + 7) / 2 ]   =  x - 1      add 1 to both sides

±√  [ ( y + 7) / 2 ] + 1   =  x       swap x and y 

±√  [ ( x + 7) / 2 ] + 1   =  y       for y, write  g^-1(x)

±√  [ ( x + 7) / 2 ] + 1  = g^-1 (x)

So   

g^-1 (-2)  =  ±√  [ ( -2 + 7) / 2 ] + 1  =   √ [5/2]  + 1     or   - √ [5/2] + 1

This will be a vertical hyperbola with the following equation

(y - 1)^2     -    x^2

______           ___      =     1

    4                   1

In the following form

(y - k)^2    -   (x - h)^2

______         ______      =  1

   a^2                b^2

The center is  ( h, k)   =   (0, 1)

The slope of asymptote with the positive slope, m, is    a / b   =  2 / 1  = 2

And the equation of this line is

y =  m ( x - h) + k  =     2 (x - 0) + 1     =     2x + 1

Your answer is good :)

I had not worked through it before, sorry :)

p.s. I know it doesn't show (because of someone else) but I did give you a point for your answer :/

5 (5/2)   = 25 / 2 units = length after dilation

-2+(-3)-5+2(3)

(4, 1)    and ( 9, 13)

The coordinates of the point are given by

( 4 + (4/5)(9 - 4) ,  1  + (4/5)(13 - 1)  )   =

( 4 +  4,  1  + 48/5 )  =

(8 , 53/5)

Draw  WY...this will be the hypotenuse of  isosceles right triangle WYX

Its length will be   √2  * leg length  = √2 * 4    =   √32  units

Angle WYX  =  45....so angle  WYZ  =  m [angle ZYX]  - m [ WYX]  =  135 - 45  = 90°

So triangle  WYZ is a right triangle  with WZ  the hypotenuse  and legs  WY  and YZ 

And we can use the Pythagorean Theorem to find  YZ  =   "a"

So

WY^2  + YZ^2  =  WZ^2

(√32)^2  + YZ^2  =  9^2

32  +  YZ^2  =  81

YZ^2  =  81 - 32

YZ^2  =  49       take the square root of both sides

YZ  =  "a"   =   7

This is a perfect square trinomial

We're looking for two integers that sum to 10  and multiply to 25

5    and   5    are just what we need

So.....factor as

(x + 5) ( x+ 5)  =   (x + 5)^2

See my other answer.....this is a vertical hyperbola

The equation is

(y - 1)^2    -    x^2

_______        ___     = 1

    4                  1

The center is   (0, 1)

The   vertices are   at     ( 0,  1  - 2)  =  (0, -1)      and   (0, 1 + 2)  =  (0, 3)

thank you very much!!! it was very helpful

Factor:

\(x2 + 10x + 25\)

First we can rewrite 25 to 5² and we will check the middle term by multiplying 2ab. This is factoring by the perfect square rule.  Now we can check the result of this new problem compared to the original expression.

We now have \(2ab = 2*x*5 = 0\)  

Simplify this to \(2ab = 10x =0\) 

Now we factor by the perfect square trinomial rule  [  \(a2+2ab+b2=(a+b)2a2+2ab+b2=(a+b)2\) where \( a=xa=x \) and \(b=5b=5.\) ]  to get: \(( x + 5 ) 2 = 0\).

Next we can set \(x+5\) equal to 0 and solve for \(x \).

Set the factor equal to 0 and we get: \(x+5=0\) 

Solve that and you get  x = -5

Sorry I did it on a white board, and I erased to work on the net problems.

A =  100/150  = 2 / 3

B = 72/108  =  2 / 3

C = 80/132  = 20 / 33

A and B  are the similar polygons

△ABC∼△EFG .

Given m∠A=39° and m∠F=56° , what is m∠C

Measure of angle  F  =  Measure of angle B

So.......

Measure  of  angle A + Measure of angle B + Measure of angle C  =  180°

39   +   56    +  Measure of angle C  = 180

95      +   Measure of angle C  =   180      subtract 95 from both sides

Measure of angle C  =   180  -  95   =  85°

Is there an image to go along with this question? If so, it would be very helpful if you could post that too .   

:-)

Yeah, forget about this question, it has been edited and is not worth your trust. DO NOT BELIVE THE QUESTION IT IS FULL OF LIES!

-4x^2 + y^2 - 2y  = 3

Rewrite as

y^2 -2y  - 4x^2  = 3     comlplete the square on y

y^2 - 2y + 1  -  4x^2  =  4       factor the frst three terms

(y - 1)^2   - 4x^2  = 4       divide through by 4

(y - 1)^2      -   x^2              

_______        ____         =      1

    4                  1

This is a "vertical" hyperbola.....the center is   (0, 1)

Please do not double post a question. It is spam and will be removed.

Can we see your work and how you tried to solve it?

In the question it specifies that we are looking for the probobility that a triangle can exist with side lengths of 1, x and y.

We do know that x and y can be either 0, 1, 2 or 3. We can list that all out:

1,0,0

1,0,1

1,0,2

1,0,3

1,1,0

1,1,1

1,1,2

1,1,3

1,2,0

1,2,1

1,2,2

1,3,0

1,3,1

1,3,2

1,3,3

This gives us a 50/50 chance or 1/2 chance that a triangle with 1,x and y can exists.

Btw I am looking at the problem now and it says the answer is 1/2.

This is just my attempt to explain this but if you see issues with my solving of this problem please point them out.

If you are just here for the answer it is:  1/2