Given J(1, 1), K(3, 1), L(3, -4), and M(1, -4) and that J'(-1, 5), K'(1, 5), L'(1, 0), and M'(-1, 0). What is the rule that tran
anastassius [24]
(x; y) -> (x - 2; y + 4)
J(1; 1) ⇒ J'(1 - 2; 1 + 4) = (-1; 5)
K(3; 1) ⇒ K'(3 - 2; 1 + 4) = (1; 5)
L(3;-4) ⇒ L'(3 - 2; -4 + 4) = (1; 0)
M(1;-4) ⇒ M'(1 - 2;-4 + 4) = (-1; 0)
Answer:
(-4.5, 4.25) The first one.
Step-by-step explanation:
Plug in -4.5 in for x and 4.25 for y in BOTH problems. When you plug them in and they both work out for BOTH problems then its the answer.
Answer:
D
Step-by-step explanation:
Differentiate the function and set it to zero:
f'(x) = -10x + 13
0 = -10x + 13
x = 1.3 seconds
Plug the x value into f(x):
f(1.3) = -5(1.3)^2 + 13(1.3) + 6 = 14.45 meters
