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3 years ago

6

The graph of a linear equation passes through (0, 5) and (-10, 0). What is the equation?

1 answer:

anzhelika [568]3 years ago

8 0

$(\stackrel{x_1}{0}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{-10}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{-10}-\underset{x_1}{0}}}\implies \cfrac{-5}{-10}\cfrac{1}{2}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{0}) \\\\\\ y-5-\cfrac{1}{2}x\implies y=\cfrac{1}{2}x+5$

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bulgar [2K]

The two inequalities can be rewritten as

$\begin{cases}y-2\end{cases}$

The solution is composed by every point below the graph of the line $y=-4x+2$ , and above the horizontal line $y=-2$

To draw the line $y=-4x+2$ we can build a little table (only 2 points are required):

$\begin{array}{c|c}x&y\\0&2\\1&-2\end{array}$

So, if you draw the points (0,2) and (1, -2) and connect them, you'll have your line.

To draw the line $y=-2$ , simply pick any point with y coordinate -2, for example (0, -2), and draw a horizontal line.

Choose the above/below areas as specified above to solve the system.

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DiKsa [7]

Answer:

Step-by-step explanation:

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4 years ago

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3 years ago

Which figure has the same area as the parallelogram shown below? A. A triangle with a base of 4 mm and a height of 20 mm B. A tr

Arisa [49]

<h2>Answer: A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>

The parallelogram in the figure has an area of $80mm^{2}$ , according to the following formula, which works for all rectangles and parallelograms:

$A_{parallelogram}=(b)(h)$ (1)

Where $b$ is the base and $h$ is the height

The<u> area of a triangle</u> is given by the following formula:

$A_{triangle}=\frac{1}{2}(b)(h)$ (2)

So, for option A:

$A_{triangle}=\frac{1}{2}(4mm)(20mm)=40mm^{2} \neq 80mm^{2}$

Now, the <u>area of a trapezoid </u>is:

$A_{trapezoid}=\frac{1}{2}(b_{1}+ b_{2})(h)$ (3)

For option B:

$A_{trapezoid}=\frac{1}{2}(15mm+25mm)(2 mm)=40mm^{2} \neq 80mm^{2}$

For option C:

$A_{trapezoid}=\frac{1}{2}(6mm+14mm)(8 mm)=80mm^{2}$ >>>>This is the correct option!

For option D:

$A_{rectangle}=(30mm)(8mm)=240mm^{2} \neq 80mm^{2}$

<h2>Therefore the correct option is C</h2>

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