$\bf f(x)=x-1\qquad \qquad g(x)=3\cdot f(x)\implies g(x)=\underline{3(x-1)} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{ccll} x&g(x)\\ \cline{1-2} 1&0\\2&3\\3&6 \end{array}\qquad \qquad (\stackrel{x_1}{1}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{6})$

$\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-0}{3-1}\implies \cfrac{6}{2}\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-0=3(x-1)\implies y=\underline{3(x-1)}$

7 0

3 years ago

Read 2 more answers

Round this to the nearest hundred. 728

hammer [34]

<span>Round this to the nearest hundred. 728

= 700</span>

6 0

3 years ago

8p + 2q = -16<br> 2p - q=2

yKpoI14uk [10]

Answer:

p = -1 q = -4

Step-by-step explanation:

a system of eq and solve for p and q ??? can do :)

Eq. 1) 8p + 2q = - 16

Eq. 2) 2p - q = 2

use Eq .2 and solve for q

2p - 2 = q

plug into Eq.1 with q

8p +2(2p - 2) = - 16

8p +4p -4 = -16

12p = - 12

p = -1

plug -1 into Eq. 1 for p and solve for q

8(-1) + 2q = - 16

-8 + 2q = - 16

2q = -8

q = -4

8 0

3 years ago