Question 1:
Slope = 1/5
y = mx + c
y = 1/5 x + c
at point (5, -1)
-1 = 1/5 (5) + c
- 1= 1 + c
c = - 2
y = 1/5x - 2
5y = x - 10
Question 2:
slope = (9-5)/(3-1)
Slope = 2
y = mx + c
y = 2x + c
at point (1, 5)
5 = 2(1) + c
c = 5 - 2
c = 3
y = 2x + 3
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:
times 5
Step-by-step explanation:
√11 √5 = √(11 * 5) = √55 = approx 7.42 (rounded)
Answer:
x = ± 2i
Step-by-step explanation:
note that 
= i
x² = - 12 ( take square root of both sides )
x = ± 
= ± 
= ± 2i
