Question

4 multiple choice 1 point what is the quotient of $\frac{9 - x^{2}}{3x}$ and $\frac{x^{2}+6x + 9}{3x}$? $\frac{x - 3}{3 - x}$ $\frac{x - 3}{x + 3}$ $\frac{3 - x}{x + 3}$ $\frac{x + 3}{x - 3}$ 5 multiple choice 1 point the area of a rectangle is $\frac{x^{2}-4}{2x}in^{2}$ and its length is $\frac{(x + 2)^{2}}{2}in$. what is the width in inches? $\frac{x(x + 2)}{x - 2}$ $\frac{x + 2}{x(x + 2)}$ $\frac{x - 2}{x(x - 2)}$ $\frac{x - 2}{x(x + 2)}$

Explanation:

Question 4

Step1: Recall division of fractions rule

Dividing by a fraction is multiplying by its reciprocal. So, $\frac{\frac{9 - x^{2}}{3x}}{\frac{x^{2}+6x + 9}{3x}}=\frac{9 - x^{2}}{3x}\times\frac{3x}{x^{2}+6x + 9}$.

Step2: Factor the expressions

Factor $9 - x^{2}=(3 + x)(3 - x)$ (using $a^{2}-b^{2}=(a + b)(a - b)$ with $a = 3$ and $b=x$) and $x^{2}+6x + 9=(x + 3)^{2}$ (using $(a + b)^{2}=a^{2}+2ab + b^{2}$ with $a=x$ and $b = 3$). Then we have $\frac{(3 + x)(3 - x)}{3x}\times\frac{3x}{(x + 3)^{2}}$.

Step3: Simplify the expression

Cancel out the common factors $3x$ and $(x + 3)$ (assuming $x
eq - 3$ and $x
eq0$). We get $\frac{3 - x}{x+3}$.

Step1: Recall the formula for the area of a rectangle

The area of a rectangle $A = lw$, where $A$ is the area, $l$ is the length and $w$ is the width. So, $w=\frac{A}{l}$. Given $A=\frac{x^{2}-4}{2x}$ and $l=\frac{(x + 2)^{2}}{2}$.

Step2: Factor the expressions

Factor $x^{2}-4=(x + 2)(x - 2)$ (using $a^{2}-b^{2}=(a + b)(a - b)$ with $a=x$ and $b = 2$). Then $w=\frac{(x + 2)(x - 2)}{2x}\div\frac{(x + 2)^{2}}{2}$.

Step3: Change division to multiplication

Dividing by a fraction is multiplying by its reciprocal. So $w=\frac{(x + 2)(x - 2)}{2x}\times\frac{2}{(x + 2)^{2}}$.

Step4: Simplify the expression

Cancel out the common factors $2$ and $(x + 2)$ (assuming $x
eq - 2$ and $x
eq0$). We get $w=\frac{x - 2}{x(x + 2)}$.

Answer:

$\frac{3 - x}{x + 3}$

Question 5