What are the vertices of ΔA'B'C produced by T (ΔABC) = ΔA'B'C? Triangle A B C graphed on a coordinate plane with vertices at A 3 comma 0, B 3 comma negative 2, C 0 comma negative 3.
Group of answer choices

A′(0, 6), B′(0, 4), C′(−3, 3)

A′(0, −6), B′(0, −8), C′(−3, 9)

A′(6, −6), B′(6, −8), C′(3, 9)

A′(6, 6), B′(6, 4), C′(3, 3)

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.