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tekilochka [14]
3 years ago
7

Graph a line that contains the point (-5,-3) and has a slope of -2

Mathematics
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

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Step-by-step explanation:

Graph (-5,-3) and use the slope, -2, to find another point. -2 can also be written as -\frac{2}{1}

\frac{rise}{run} =\frac{-2}{1} and  \frac{rise}{run} =\frac{2}{-1} **

Use both. This works because either way, they will lie on the same line with the same slope. So, starting at (-5,-3), go down two points (-)* and to the left 1 (+)*, then, starting at (-5,-3) again, go up two points (+)* then to the right 1 point (-)*.

* If the number is negative, you either go down or to the right. If the number is positive, you go up and to the left.

**Rise over run refers to the change in the y-axis and the change in the x-axis: \frac{rise}{run} =\frac{y-axis}{x-axis}. Only one number is negative because, if both were negative, that would make a positive number, but the slope is -2, not 2.

Another way you could do this is by finding points first. All you need to do is turn the slope into a "point". Remember, 2 is the y and 1 is the x:

-2=-\frac{2}{1} = (1,-2) and (-1,2)

Then, using (-5,-3) and your new points, add them separately:

(-5,-3)+(1,-2)\\(-5+1,-3+(-2))\\(-5+1,-3-2)\\(-4,-5)

(-4,-5) is one point

(-5,-3)+(-1,2)\\(-5+(-1),-3+2)\\(-5-1,-3+2)\\(-6,-1)

(-6,-1) is another point. Connect the three points. You can check your work on the graph:

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Answer:

we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

Step-by-step explanation:

Given the expression

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}

=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}

=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}

\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}

=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}

cancel the common factor: 2

=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}

cancel the common factor: p

=\frac{2p+1}{p^2\left(4p^2-1\right)}

=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}

cancel the common factor: 2p+1

=\frac{1}{p^2\left(2p-1\right)}

Expanding

=\frac{1}{2p^3-p^2}

Thus, we conclude that:

\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}

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