QuestionNov 16 moodle.straighterline.com My Account Topic 3: Systems of Linear Equations in Two Variables \begin{tabular}{r|l|} \hline rted on & Friday, November 15, 2024, 6:32 PM \\ \hline State & Finished \\ eted on & Friday, November $15,2024,7: 06 \mathrm{PM}$ \\ \hline e taken & 34 mins 2 secs \\ \hline Grade & 90.00 out of $125.00(72 \%)$ \end{tabular}
Simplify the expression $-27+9(y+8)$.
Select one: A. $9 y+45$ B. $y+45$ C. $9 y-19$ D. $72 y-27$
Simplify the expression $\sqrt[5]{\frac{-32}{243}}$. Select one: 2

STEP 1

1. We need to simplify two separate expressions.
2. The first expression is a linear expression involving distribution.
3. The second expression involves simplifying a radical expression with a fractional base.
_HIGH_LEVEL_APPROACH_ for the first expression:
1. Distribute the constant in the expression $-27 + 9(y + 8)$.
2. Simplify the resulting expression.
_HIGH_LEVEL_APPROACH_ for the second expression:
1. Simplify the radical expression $\sqrt[5]{\frac{-32}{243}}$.
2. Determine if the fraction can be simplified further.
**First Expression:**

STEP 2

STEP 3

Distribute the 9 into the terms inside the parentheses:
$$9(y + 8) = 9y + 72$$

STEP 4

Combine the distributed terms with the constant $-27$:
$$-27 + 9y + 72$$

STEP 5

Simplify the expression by combining the constant terms:
$$9y + (72 - 27) = 9y + 45$$
The simplified expression is:
$$\boxed{9y + 45}$$
**Second Expression:**
STEP_1: Simplify the radical expression $\sqrt[5]{\frac{-32}{243}}$.
Recognize that $-32 = (-2)^5$ and $243 = 3^5$, so:
$$\sqrt[5]{\frac{-32}{243}} = \frac{\sqrt[5]{(-2)^5}}{\sqrt[5]{3^5}}$$
STEP_2: Simplify the expression by taking the fifth root of both the numerator and the denominator:
$$\frac{\sqrt[5]{(-2)^5}}{\sqrt[5]{3^5}} = \frac{-2}{3}$$
The simplified expression is:
$$\boxed{\frac{-2}{3}}$$