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3 years ago

A rectangle with an area of 120 in squared has a length of 8 inches longer then two times its width what is the width of the rec

tangle

Mathematics

1 answer:

Vaselesa [24]3 years ago

6 0

$Answer: \\The\: width\: is \: 6 \: {in}^{2} \\ \\ Step\: by\: step\: explanation: \\ Let \: x \: be \: the \: width \: of \: the \: rectangle,\: x > 0\Rightarrow the \: length:2x + 8 \\ We \: have \: an \: equation:x(2x + 8) = 120 \\ \Leftrightarrow 2 {x}^{2} + 8x + 8= 128 \\ \Leftrightarrow {x}^{2} + 4x + 4 = 64 \\ \Leftrightarrow {(x + 2)}^{2} = 64 \\ \Leftrightarrow x + 2 = \pm8 \\ \Leftrightarrow x = 6 \: \vee \: x = - 10 \\ \Rightarrow x = 6 \: (x > 0)$

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natali 33 [55]

Answer: By the slope formula.

Step-by-step explanation:

Given: ABC is a triangle (shown below),

In which A≡(6,8), B≡(2,2) and C≡(8,4)

And, D and E are the mid points of the line segments AB and BC respectively.

Prove: DE║AC and DE = AC/2

Proof:

Since, And, D and E are the mid points of the line segments AB and BC respectively.

Therefore, By mid point theorem,

coordinate of D are $(\frac{2+6}{2} , \frac{2+8}{2} ) = (\frac{8}{2} , \frac{10}{2} )= (4,5)$

Coordinate of E are $(\frac{2+8}{2} , \frac{2+4}{2} ) = (\frac{10}{2} , \frac{6}{2} )= (5,3)$

By the distance formula,

$DE=\sqrt{(5-4)^2+(3-5)^2}=\sqrt{5}$

$AC=\sqrt{(8-6)^2+(4-8)^2}=2\sqrt{5}$

By the slope formula,

Slope of AC = $\frac{4-8}{8-6} = \frac{-4}{2} = -2$

Slope of DE = $\frac{3-5}{5-4} = \frac{-2}{1} = -2$

Statement Reason

1. The coordinate of D are (4,5) and 1. By the midpoint formula

the coordinate of E are (5,3)

2. The length of DE = √5 2. By the Distance formula

The length AC = 2√5 ⇒ Segment DE

is half the length of segment AC

3. The slope of DE = -2 and the 3. By the slope formula

slope of AC = -2

4. DE║AC 4. Slopes of parallel lines are equal

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3 years ago

Read 2 more answers

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