Answer:
Can you give me the choices for each of these, I'm not sure how they might phrase it.
Step 1
<u>Find the value of TS</u>
we know that
if PQ is parallel to RS. then triangles TRS and TPQ are similar
so

solve for TS

we have


substitute

Step 2
<u>Find the value of SQ</u>
we know that

we have


substitute

therefore
<u>the answer is</u>
the value of SQ is 
Using the law os cosines formula b^2 = a^2 + c^2 - 2*a*c*cos(B)
a = 17, b = 8, c = 16
8^2 = 17^2 + 16^2 - 2*17*16* cos(B)
64 = 289 + 256 - 544 * cos(B)
544*cos(B) = 289 + 256 - 64
544 * cos(B) = 481
cos (B) = 481/544
B = arccos(481/544)
B = 27.8 degrees
Answer:
-4
Step-by-step explanation:
2/5 times -10 = -4
Number 3 it is a factor because if u expand you will see that x+6 will be the answer