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devlian [24]
3 years ago
11

Is this graph a function? Explain why or why not.

Mathematics
1 answer:
melomori [17]3 years ago
8 0

Answer:

yes

Step-by-step explanation:

becuase it passes the virtical line test

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Write an equation of a line that is perpendicular to the line y=2/3x and passes through origin
sesenic [268]

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above anyway?

\bf y = \cfrac{2}{3}x\implies y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x+0\qquad \impliedby \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} \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}

so we're really looking for the equation of a line whose slope is -3/2 and runs through (0,0).

\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{3}{2} \\\\\\ \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}{-\cfrac{3}{2}}(x-\stackrel{x_1}{0})\implies y=-\cfrac{3}{2}x

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3 years ago
9(3+c+4) use distributive property to simplify the expression
azamat

9(3+c+4)

9(7+c)

9(7)+9c

Answer:

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The picture cannot be dilated to fit the frame. This is because one side of the picture is already 6-inches. So, therefore you cannot dilate the picture to fit the frame.
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If l || m find the value of y
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That means they are parallel so it’s 6
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