Energy is released in the sun’s core through Response area. It then is transferred by Response area through the sun’s radiative zone to the conductive zone, where it is transferred to the sun’s surface by Response area. From the surface, energy from the sun is transferred to Earth by

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)

the answer for the second equation is (4,0)

i believe your answer is a