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)
What’s da problem that u have?
Answer: 4a16b4c12
Step-by-step explanation:
Answer:
I might be wrong but I think its the last answer choice.
Step-by-step explanation:
I'm sorry if its wrong :(
The consecutive positive integers would be: x and (x+1),
We would have to solve the following equation to find these numbers:
x(x+1)-[x+(x+1)]=29
x²+x-2x-1=29
x²-x-30=0
x=[1⁺₋√(1+120)]/2
x=(1⁺₋11)/2
We have two possible solutions:
x₁=(1-11)/2=-5 then: (x+1)=-5+1=-4 This is not the solution.
x₂=(1+11)/2=6 then: (x+1)=6+1=7 This solution is right.
Answer: the numbers would be 6 and 7.
