Dot plot help? i need help dot plotting

Mathematics

1 answer:

Svetlanka [38]3 years ago

3 0

I got you drop the promblem

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Find the equation of a line parallel to y=4x−4 that contains the point (−2,0). Write the equation in slope-intercept form.

Allushta [10]

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

$y=\stackrel{\stackrel{m}{\downarrow }}{4} x-4\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so we're really looking for the equation of a line whose slope is 4 and passes through (-2 , 0)

$(\stackrel{x_1}{-2}~,~\stackrel{y_1}{0})\qquad \qquad \stackrel{slope}{m}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{4}(x-\stackrel{x_1}{(-2)}) \\\\\\ y=4(x+2)\implies y=4x+8$

8 0

2 years ago

What is the solution to ln (x2 - 16) = 0?

bogdanovich [222]

ANSWER

$x = \pm \sqrt{17}$
EXPLANATION

The given equation is

$ln( {x}^{2} - 16) = 0$

Take antilogarithm of both sides to base e.

${e}^{ ln( {x}^{2} - 16) } = {e}^{0}$

This will simplify to

${x}^{2} - 16 = 1$

Group like terms to obtain,

${x}^{2} = 1 + 16$

${x}^{2} = 17$

Take square root of both sides to get,

$x = \pm \sqrt{17}$