Reflect the given pre-image over = −1 followed by = −7. Find the new coordinates.

A'(8, -10)

B'(6, -12)

C'(2, -4)

A"(8, -4)

B"(6, -2)

C"(2, -10)

Step-by-step explanation:

The location of the points are at:

A(8, 8) , B(10, 6) and C(2, 2).

Reflection over y = -1:

For point A the y distance between the point and y = -1 line is 9 (8 - (-1)). The reflection of point a over y = -1 would give a point 9 units below y = -1 which is at -10.

The new A coordinate is at A'(8, -10)

For point B the y distance between the point and y = -1 line is 7 (6 - (-1)). The reflection of point a over y = -1 would give a point 7 units below y = -1 which is at -8.

The new B coordinate is at B'(10, -8)

For point C the y distance between the point and y = -1 line is 3 (2 - (-1)). The reflection of point a over y = -1 would give a point 3 units below y = -1 which is at -4.

The new C coordinate is at C'(2, -4)

Reflection over y = -7:

For point A' the y distance between the point and y = -7 line is 3 (-7 - (-10)). The reflection of point a over y = -1 would give a point 3 units above y = -7 which is at -4.

The new A coordinate is at A"(8, -4)

For point B' the y distance between the point and y = -7 line is 1 (-7 - (-8)). The reflection of point a over y = -7 would give a point 1 units above y = -7 which is at -6.

The new B coordinate is at B"(10, -6)

For point C' the y distance between the point and y = -7 line is 3 (4 - (-7)). The reflection of point a over y = -7 would give a point 3 units below y = -7 which is at -10.

The new C coordinate is at C"(2, -10)

the rocket cannot penetrate the ground, in a contextual sense, therefore the roots of this equation should be the domain of this function. i'll let you find the roots of this quadratic equation (hint: use the quadratic formula:  

1- c

2-a

step-by-step explanation:

mark brainliest!

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