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klio [65]
2 years ago
12

If k is a nonzero constant, what is the slope of the line containing the points (k,2k) and (3k,k)

Mathematics
1 answer:
ivolga24 [154]2 years ago
3 0

Answer:

The slope is -1/2

Step-by-step explanation:

m=\frac{y_2-y_1}{x_2-x_1}\\ \\m=\frac{k-2k}{3k-k}\\ \\m=\frac{-k}{2k}\\ \\m=-\frac{1}{2}

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4 0
3 years ago
Read 2 more answers
What is the peremeter of this polygon? (With picture)
lyudmila [28]
Check the picture below.

so.. simply, use the distance formula, to get their length an add them up, and that's the perimeter of the polygon.


\bf \textit{distance between 2 points}\\ \quad \\&#10;\begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%  (a,b)&#10;&({{ -1}}\quad ,&{{ 2}})\quad &#10;%  (c,d)&#10;&({{ 2}}\quad ,&{{ 4}})\\&#10;&({{ 2}}\quad ,&{{ 4}})\quad &#10;%  (c,d)&#10;&({{ 3}}\quad ,&{{ -2}})\\&#10;&({{ 3}}\quad ,&{{ -2}})\quad &#10;%  (c,d)&#10;&({{ -2}}\quad ,&{{ -3}})\\&#10;&({{ -2}}\quad ,&{{ -3}})\quad &#10;%  (c,d)&#10;&({{ -1}}\quad ,&{{ 2}})&#10;\end{array}\qquad &#10;%  distance value&#10;d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}

\bf -------------------------------\\\\&#10;d=\sqrt{[2-(-1)]^2+(4-2)^2}\implies d=\sqrt{(2+1)^2+(2)^2}&#10;\\\\\\&#10;d=\sqrt{3^2+2^2}\implies \boxed{d=\sqrt{13}}\\\\&#10;-------------------------------\\\\&#10;d=\sqrt{(3-2)^2+(-2-4)^2}\implies d=\sqrt{1^2+(-6)^2}\implies \boxed{d=\sqrt{37}}\\\\&#10;-------------------------------\\\\&#10;d=\sqrt{(-2-3)^2+[-3-(-2)]^2}\implies d=\sqrt{(-5)^2+(-3+2)^2}&#10;\\\\\\&#10;d=\sqrt{(-5)^2+(-1)^2}\implies \boxed{d=\sqrt{26}}

\\\\&#10;-------------------------------\\\\&#10;d=\sqrt{[-1-(-2)]^2+[2-(-3)]^2}\implies d=\sqrt{(-1+2)^2+(2+3)^2}&#10;\\\\\\&#10;d=\sqrt{(1)^2+(5)^2}\implies \boxed{d=\sqrt{26}}

so, those are their lengths, sum them all up, that's the polygon's perimeter.

4 0
3 years ago
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