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2 years ago

7

A pentagon is shown. the pentagon is translated 5 units to the left and then reflected over the x-axis. Use the Connect Line to

draw the pentagon After its transformations

1 answer:

Sever21 [200]2 years ago

4 0

Answer:

soundproofing pendulum Priddy

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3 years ago

A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions

salantis [7]

Dimensions are length 20 meter and width 14 meter

<em><u>Solution:</u></em>

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

<em><u>A rectangle has width that is 6 meters less than the length</u></em>

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

<em><u>The area of the rectangle is given by formula:</u></em>

$Area = length \times width$

<em><u>Substituting the values we get,</u></em>

$Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0$

<em><u>Solve the above equation by quadratic formula</u></em>

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}$

$a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14$

Since, length cannot be negative, ignore a = -14

<em><u>Thus solution of length is a = 20</u></em>

Therefore,

width = length - 6

width = 20 - 6 = 14

Thus dimensions are length 20 meter and width 14 meter

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