Graph with vertices r(6, 6), s(3, –6), and t(0, 3) and its image after a reflection over the y-axis.

T'(0, 3) R'(-6,6) and S'(-3,-6)

Step-by-step explanation:

The given vertices are

R( 6, 6), S(3, -6), and T(0, 3)

We have to find the image of these vertices

Since T(0, 3) lies on y axis so its image over the y axis will be same T'(0,3)

since R(6, 6) lies in 1st quadrant so image over the y axis will lie in second quadrant and the point will be R'(-6, 6)

Similarly S(3, -6)  lies in 4th quadrant of which image will lie in 3rd quadrant having co-ordinates as (-3, -6)

Therefore, T'(0, 3) R'(-6,6) and S'(-3,-6) is the answer.

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

⭐⭐

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.