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!
Answer:
m<A = 30 degrees.
Step-by-step explanation:
m < C in triangle ABC = m < C in EDC so
x + 13 + 42 + x + 13 = 180
2x = 180 - 13 - 13 - 42
2x = 112
x = 56.
So m < C = 56+13 = 69.
m < B = m < D = 81
M < A = 180 - 69 - 81
= 180 - 150 = 30.
When X = -9 Y = 1 you can find this by going to -9 on the x axis and finding where the line is there
Answer:
D
Step-by-step explanation:
log₂(x-4) = 4
Undo the log by raising 2 to both sides:
2^(log₂(x-4)) = 2^4
x - 4 = 2^4
x - 4 = 16
x = 20
Answer D.
