First find the gradient of the line
Change in y/change in x
-3–3/-3-3
0/-6
=0 ( so the gradient m is equal to zero)
Y=0x+c
Input the coordinates of one point to find c
-3=(0*3)+c
-3=c
So the equation is
Y= -3
Answer:
8 and 9
Step-by-step explanation:
Answer:
Pretty sure it is C
Step-by-step explanation:

We're given that
and
, and want to find
.
By the chain rule, we have

and


Then

(because the point
corresponds to
)
