use the equation $y = sqrt3{27x - 54}+5$. which is an equivalent equation of the form $y = asqrt3{x - h}+k$…

use the equation $y = sqrt3{27x - 54}+5$. which is an equivalent equation of the form $y = asqrt3{x - h}+k$ \n$y=-27sqrt3{x + 2}+5$\n$y=-3sqrt3{x + 2}+5$\n$y = 3sqrt3{x - 2}+5$\n$y = 27sqrt3{x - 2}+5$\ndone

use the equation $y = sqrt3{27x - 54}+5$. which is an equivalent equation of the form $y = asqrt3{x - h}+k$ \n$y=-27sqrt3{x + 2}+5$\n$y=-3sqrt3{x + 2}+5$\n$y = 3sqrt3{x - 2}+5$\n$y = 27sqrt3{x - 2}+5$\ndone

Answer

Answer:

C. $y = 3\sqrt[3]{x - 2}+5$

Explanation:

Step1: Factor out the coefficient inside the cube - root.

First, factor 27 out of $27x-54$ in $y=\sqrt[3]{27x - 54}+5$. We know that $27x-54=27(x - 2)$. So $y=\sqrt[3]{27(x - 2)}+5$.

Step2: Use the property of cube - roots.

Since $\sqrt[3]{ab}=\sqrt[3]{a}\cdot\sqrt[3]{b}$ and $\sqrt[3]{27}=3$, we have $y = 3\sqrt[3]{x - 2}+5$.