Question

which is the slope of a line perpendicular to the line $y - 3 = \frac{3}{2}(x + 5)$
$\frac{3}{2}$
$-\frac{3}{4}$
$\frac{2}{15}$
$\frac{15}{2}$
$-\frac{3}{2}$
$\frac{1}{5}$
$\frac{2}{3}$
$-\frac{1}{5}$
$-\frac{2}{15}$
$-\frac{15}{2}$
23 numeric 1 point
simplify $x^0 = $
answer
24 multiple choice 0 points
solve for $f(x) = 0$ for $f(x) = (x - 8)(x + 3)$.
$x = -8$ and $x = -3$
$x = 8$ and $x = 3$

Explanation:

Step1: Identify given line slope

The line is in point-slope form $y-y_1=m(x-x_1)$, so $m=\frac{2}{15}$.

Step2: Find perpendicular slope

Perpendicular slope is negative reciprocal: $m_{\perp}=-\frac{1}{m}$
$m_{\perp}=-\frac{15}{2}$
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Step1: Apply zero exponent rule

For any non-zero $x$, $x^0=1$
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Step1: Set factors equal to 0

Set $(x-8)=0$ and $(x+3)=0$

Step2: Solve for x

$x-8=0 \implies x=8$; $x+3=0 \implies x=-3$

Answer:

1. $-\frac{15}{2}$
2. $1$
3. $x=8$ and $x=-3$