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)
Hey there! :D
Substitute y for 5x-1.
3x+2y=11
y=5x-1
3x+2(5x-1)=11
3x+10x-2=11
13x-2=11
13x=13
x=1
Now, plug that in to find y.
y=5x-1
y=5(1)-1
y=5-1
y=4
(1,4) <== the solution
I hope this helps!
~kaikers
Answer:
10x^3 + 6x^2 + 5x + 3
Step-by-step explanation:
F(x)=(5x+3)(2x^2+1)
= 5x*2x^2 + 5x*1 + 3*2x^2 + 3*1
= 10x^3 + 5x + 6x^2 + 3
= 10x^3 + 6x^2 + 5x + 3
I hope it workssssssssssssssssssssss.
c had the same ? then got it right
