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3 years ago

What is the midpoint of the line segment with endpoints (-5.5 -6.1) and (-0.5 9.1)?

Mathematics

1 answer:

Volgvan3 years ago

4 0

Answer:

(-3, 1.5)

Step-by-step explanation:

(-0.5 - 5.5/2, 9.1 - 6.1/2) = (-3, 1.5)

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Answer:

slope is m=2

Step-by-step explanation:

5-(-7/3)=22/3

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I NEED HELP PLEASEE

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$\bf \cfrac{1+cot^2(\theta )}{1+csc(\theta )}=\cfrac{1}{sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\implies \cfrac{1+\frac{cos^2(\theta )}{sin^2(\theta )}}{1+\frac{1}{sin(\theta )}}\implies \cfrac{~~\frac{sin^2(\theta )+cos^2(\theta )}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}\implies \cfrac{~~\frac{1}{sin^2(\theta )}~~}{\frac{sin(\theta )+1}{sin(\theta )}}$

$\bf \cfrac{1}{\underset{sin(\theta )}{~~\begin{matrix} sin^2(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }}\cdot \cfrac{~~\begin{matrix} sin(\theta ) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{sin(\theta )+1}\implies \cfrac{1}{sin^2(\theta )+sin(\theta )} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{1+cot^2(\theta )}{1+csc(\theta )}\ne \cfrac{1}{sin(\theta )}~\hfill$

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3 years ago