11. (5 - 6q - 8q²) - (14q² + 2)
12. (d - 4d² + 8d³) + 3/4(-16d + 32d²)
13. (2/3h - 3h³ + 12) + (2/5h³ - 6)
14. (6 - 3b²) - (5b⁵ - 9b + 2b²) - (2b - b² + 7)
15. write a polynomial expression in standard form for the perimeter of the triangle, where x 2/3.

Question

11. (5 - 6q - 8q²) - (14q² + 2)
12. (d - 4d² + 8d³) + 3/4(-16d + 32d²)
13. (2/3h - 3h³ + 12) + (2/5h³ - 6)
14. (6 - 3b²) - (5b⁵ - 9b + 2b²) - (2b - b² + 7)
15. write a polynomial expression in standard form for the perimeter of the triangle, where x > 2/3.

Accepted Answer

### 11.
# Explanation:
## Step1: Remove parentheses
$$
\begin{align*}
(5 - 6q-8q^{2})-(14q^{2}+2)&=5 - 6q-8q^{2}-14q^{2}-2
\end{align*}
$$
## Step2: Combine like - terms
$$
\begin{align*}
&5 - 2-6q+(-8q^{2}-14q^{2})\
&=3-6q - 22q^{2}
\end{align*}
$$
# Answer:
$-22q^{2}-6q + 3$

### 12.
# Explanation:
## Step1: Distribute the fraction
$$
\begin{align*}
(d-4d^{2}+8d^{3})+\frac{3}{4}(-16d + 32d^{2})&=d-4d^{2}+8d^{3}+\frac{3}{4}\times(-16d)+\frac{3}{4}\times32d^{2}\
&=d-4d^{2}+8d^{3}-12d + 24d^{2}
\end{align*}
$$
## Step2: Combine like - terms
$$
\begin{align*}
&8d^{3}+(-4d^{2}+24d^{2})+(d - 12d)\
&=8d^{3}+20d^{2}-11d
\end{align*}
$$
# Answer:
$8d^{3}+20d^{2}-11d$

### 13.
# Explanation:
## Step1: Remove parentheses
$$
\begin{align*}
(\frac{2}{3}h-3h^{3}+12)+(\frac{2}{5}h^{3}-6)&=\frac{2}{3}h-3h^{3}+12+\frac{2}{5}h^{3}-6
\end{align*}
$$
## Step2: Combine like - terms
$$
\begin{align*}
&-3h^{3}+\frac{2}{5}h^{3}+\frac{2}{3}h+(12 - 6)\
&=(-3+\frac{2}{5})h^{3}+\frac{2}{3}h + 6\
&=(-\frac{15}{5}+\frac{2}{5})h^{3}+\frac{2}{3}h+6\
&=-\frac{13}{5}h^{3}+\frac{2}{3}h + 6
\end{align*}
$$
# Answer:
$-\frac{13}{5}h^{3}+\frac{2}{3}h + 6$

### 14.
# Explanation:
## Step1: Remove parentheses
$$
\begin{align*}
(6-3b^{2})-(5b^{5}-9b + 2b^{2})-(2b - b^{2}+7)&=6-3b^{2}-5b^{5}+9b-2b^{2}-2b + b^{2}-7
\end{align*}
$$
## Step2: Combine like - terms
$$
\begin{align*}
&-5b^{5}+(-3b^{2}-2b^{2}+b^{2})+(9b-2b)+(6 - 7)\
&=-5b^{5}-4b^{2}+7b-1
\end{align*}
$$
# Answer:
$-5b^{5}-4b^{2}+7b - 1$

### 15.
# Explanation:
## Step1: Recall the formula for the perimeter of a triangle
The perimeter $P$ of a triangle with side lengths $a$, $b$, and $c$ is $P=a + b + c$. Here, $a = 3x-2$, $b=4x + 6$, and $c=2x + 8$.
$$
\begin{align*}
P&=(3x-2)+(4x + 6)+(2x + 8)
\end{align*}
$$
## Step2: Combine like - terms
$$
\begin{align*}
P&=(3x+4x+2x)+(-2 + 6+8)\
&=9x+12
\end{align*}
$$
# Answer:
$9x + 12$

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QUESTION IMAGE

Question

Explanation:

11.

Step1: Remove parentheses

Step2: Combine like - terms

Step1: Distribute the fraction

Step2: Combine like - terms

Step1: Remove parentheses

Step2: Combine like - terms

Answer:

12.