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Tema [17]
3 years ago
11

write the equation of the line that passes through the given points (0,-1) and (-5,-7) Simplify your answer

Mathematics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

y = 6/5 x-1  slope intercept form

6x-5y = 5  standard form

Step-by-step explanation:

We have the points  (0,-1) and (-5,-7)

We can find the slope using

m = (y2-y1)/(x2-x1)

 = (-7--1)/(-5-0)

 = (-7+1)/(-5-0)

 = -6/-5

 = 6/5

We also have the y intercept ( when x=0)  It is -1

We can use the slope intercept form of the equation y = mx+b

y = 6/5 x-1

Depending on what you mean when you say simplify your answer

We can put it in standard form  ax+by = C

Multiply both sides by 5

5y =5* 6/5 x-1*5

5y = 6x-5

Subtract 6x from each side

-6x+5y = 6x-6x-5

-6x+5y = -5

Multiply by -1

6x-5y = 5

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(11-8)x3+7+27-3<br> (18÷3)+6+(14-8)x5<br> (11-7)x6+4+32-4
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The results of the arithmetic evaluations as given in the task content are; 40,42 and 62.

<h3>What are the results of the arithmetic evaluations?</h3>

The arithmetic operations above can be evaluated and simplified as follows;

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(18÷3)+6+(14-8)x5 = (6)+6+(6)x5 = 42.

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brainly.com/question/4694189

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Find the area of the shaded region ​
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so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)

A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill

\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}

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2 years ago
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