-7,1,3
Domain is the x
Range is the y
C=10 & P=5
3C+4P=50
C+P=15
Solve
3C+4P=50
3C+3P=45
Subtract
P=5
Substitute in equation 2
C+P=15
C+5=15
C=10
It's out of B, D. I did this last year, but it's definitely out of those two choices.
Triangle JKL has vertices J(−2, 2) , K(−3, −4) , and L(1, −2) .
Rule: (x, y)→(x + 8, y + 1 )
J’ (-2, 2) → (-2 + 8, 2 + 1 ) → (6, 3 )
K’ (-3, -4) → (-3 + 8, -4 + 1 ) → (5, -3 )
L’ (1, -2) → (1 + 8, -2 + 1 ) → (9, -1)
J’ (6,3)
K’ (5,-3)
L’ (9,-1)
Hope this helps!