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grigory [225]
3 years ago
13

-8 + 4m = 2. What is m?​

Mathematics
1 answer:
drek231 [11]3 years ago
7 0

●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●

Hi my lil bunny!

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

Let's solve your equation step-by-step.

-8 + 4m = 2

Step 1: Simplify both sides of the equation.

4m - 8 = 2

Step 2: Add 8 to both sides.

4m - 8 + 8 = 2 + 8 \\4m = 10

Step 3: Divide both sides by 4.

\frac{4m}{4} = \frac{10}{4} \\m = \frac{5}{2}

Answer : \boxed {m = \frac{5}{2} }

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●

Have a great day/night!

❀*May*❀

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A quadrilateral has vertices at A(–5, 5), B(1, 8), C(4, 2), D(–2, –2). Use slope to determine if the quadrilateral is a rectangl
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With the properties of rectangle in mind we must perform two verification. Verify to see if opposite sides are parallel and adjacent sides are perpendicular.

We need to determine the slope of each side, using the formula,

m=\frac{y_2-y_1}{x_2-x_1 }

<u>Slope  of AB</u>

m_{AB}=\frac{8-5}{1--5}

\Rightarrow m_{AB}=\frac{8-5}{1+5}

\Rightarrow m_{AB}=\frac{3}{6}

\Rightarrow m_{AB}=\frac{1}{2}


<u>Slope of BC</u>

m_{BC}=\frac{2-8}{4-1}

\Rightarrow m_{BC}=\frac{-6}{3}

\Rightarrow m_{BC}=-2

<u>Slope of CD</u>

m_{CD}=\frac{-2-2}{-2-4}

\Rightarrow m_{CD}=\frac{-2-2}{-2-4}

\Rightarrow m_{CD}=\frac{-4}{-6}

\Rightarrow m_{CD}=\frac{2}{3}

<u>Slope of AD</u>

m_{AD}=\frac{-2-5}{-2--5}

\Rightarrow m_{AD}=\frac{-2-5}{-2+5}

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<u>Verify Parallel sides</u>

If the quadrilateral is a rectangle, then opposite sides should have the same slope. But

m_{AD} = \frac{-7}{3} \neq m_{BC}=-2

m_{AB}=\frac{1}{2} \neq m_{CD}=\frac{2}{3}


<u>Verify Perpendicularity</u>

And also the product of slopes of all sides with a common vertex should be -1. But

m_{AB} \times m_{BC}=\frac{1}{2} \times -2=-1

m_{AB} \times m_{AD}=\frac{1}{2} \times -\frac{7}{3} \neq -1


\Rightarrow m_{CD} \times m_{AD}=\frac{2}{3} \times -\frac{7}{3} \neq -1


\Rightarrow m_{CD} \times m_{BC}=\frac{2}{3} \times -2 \neq -1


Since the quadrilateral fails to satisfy all these conditions, the quadrilateral is not a rectangle




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