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Mathematics, 26.09.2019 06:30 niki1524

Arectangle has a length of 15x and a width of 2x^3+4-2x^2. find the perimeter of the rectangle when the length is 5 feet

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Answer from: awuaheric

The perimeter of the rectangle is 17.70 feet

Step-by-step explanation:

- The perimeter of a rectangle is → P = 2l + 2w → where l is its length

  and w is its width

- A rectangle has a length 15x and a width 2x³ + 4 - 2x²

- We need to find its perimeter when its length is 5 feet

∵ The length of the rectangle is 15x

∵ The length of the rectangle is 5 feet

- Equate the length's expressions

∴ 15x = 5

- Divide both sides by 15

x=\frac{1}{3}

- Now lets find the width of the rectangle

∵ The width of the rectangle = 2x³ + 4 - 2x²

x=\frac{1}{3}

- Substitute the value of x in the expression of the width

w=2(\frac{1}{3})^{3}+4-2(\frac{1}{3})^{2}

w=2(\frac{1}{27})+4-2(\frac{1}{9})

w=\frac{2}{27}+4-\frac{2}{9})

w=\frac{104}{27} feet

∵ l = 5 feet and w=\frac{104}{27} feet

∵ P = 2l + 2w

∴ P = 2(5) + 2(\frac{104}{27})

∴ P = 10 + \frac{208}{27}

∴ P = 17.70 feet

* The perimeter of the rectangle is 17.70 feet

ansver
Answer from: Quest

4444444444444 i think

ansver
Answer from: Quest

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this

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Arectangle has a length of 15x and a width of 2x^3+4-2x^2. find the perimeter of the rectangle when...