What is the value of $k$ if $-\frac23(k-6) = \frac32(k+6)$?

answer: c

substitute in the values.

for a:

f(0) = 0 + 9

f(0) = 9

a is a solution.

for b:

f(7) = 14 + 9

f(7) = 23

b is a solution.

for c:

f (3) = 6 + 9

f(3) = 15

c is not a solution. it's supposed to be 14.

for d:

f(2) = 4 + 9

f(2)= 13

d is a solution.

isn't it a pentagon? sense [3 (triangle) + 4 (rectangle)] - 2 (triangle and rectangle connected) = 5 (pentagon)