Question

Can someone please help me?

Answer 1

Question 4:

2x + 3y = 17 -- equation 1

5x + 6y = 32 -- equation 2

(equation 1) * 2

4x + 6y = 34 -- equation 3

(equation 2) - (equation 3)

x = -2 -- equation 4

plug (equation 4)'s value of x into (equation 1)

2(-2) + 3y = 17

3y - 4 = 17

3y = 21

y = 7

Thus the answer is x = -2 and y = 7

Question 5:

The vertex form of a quadrilateral looks like this:

`y = a(x-h)^2 + k`

• a : coefficient of
  `x^2\\`
• (h,k): coordinate of the vertex.

The x-coordinate of the vertex is equal to '-b/2a'. Where b is the

coefficient of x, so:

`h = -(b)/(2a) =-(-8)/(2*1) =4`

There is also another formula to find the y-value of the vertex

coordinate, but the easier way to find it is to plug it into the original

equation:

`k=(4)^2-8(4) + 12= 16 -32 + 12 = -16 + 12 = -4`

Since we know now that h = 4 and k = -4 and that a = 1, lets plug it into

the original equation.

`y = 1*(x-4)^2 - 4 =(x^2-4)-4`

Thus the answer is
`y = (x-4)^2 -4`

Hope that helps!