(Ill give brainliest if u helppp me!! :D) Solve six and two sixths times one and one half.

A) six and two twelfths
B) eight and six twelfths
C) nine and six twelfths
D) ten and two twelfths

(here is an image if needed) :

idk bring the function back   im not smart

step-by-step explanation:

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

wait this depends wether emily and zach how they are thinking.

yes they are the same.