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rewona [7]
4 years ago
15

A rectangle has a length that is 4 more than twice the width. If the perimeter is 128 yards, find the rectangle’s dimensions.

Mathematics
1 answer:
S_A_V [24]4 years ago
4 0

The length of rectangle is 44 yards and width of rectangle is 20 yards

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

Let the length of rectangle be "a"

Let the width of rectangle be "b"

<em><u>Rectangle has a length that is 4 more than twice the width</u></em>

Length = 4 + 2(width)

Length = 4 + 2b

The perimeter is 128 yards

<em><u>The perimeter of rectangle is given by:</u></em>

Perimeter = 2(length + width)

128 = 2(4 + 2b + b)

128 = 2(4 + 3b)

Divide both sides by 2

64 = 4 + 3b

3b = 60

b = 20

<em><u>Substitute b = 20 in length</u></em>

Length = 4 + 2(20)

Length = 4 + 40 = 44

Thus length of rectangle is 44 yards and width of rectangle is 20 yards

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6 0
3 years ago
Please help me!<br> Find WX.
Fynjy0 [20]
So WX is 7y +4 and XY is 21, and when you add those lines you get the whole thing, which is 12y so write that like an equation
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3 years ago
Help please!<br> 6x^2+2x-9=0<br> A 6.4, -8.4<br> B 1.1, -1.4<br> C -2.2, 2.8<br> D -1.1, -1.4
Yakvenalex [24]
6x^2+2x-9=0\\\\a=6;\ b=2;\ c=-9\\\\\Delta=b^2-4ac\to\Delta=2^2-4\cdot6\cdot(-9)=220\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-2-\sqrt{220}}{2\cdot6}=\dfrac{-2-\sqrt{220}}{12}\approx-1.4\\\\x_2=\dfrac{-2+\sqrt{220}}{2\cdot6}=\dfrac{-2+\sqrt{220}}{12}\approx1.1

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6 0
3 years ago
In a business, if A can earn $7500 in 2.5 years, find the unit rate of his earning per month
andrew11 [14]

2.5 years = 30 months

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x$ in 1 month

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x = (7500*1)/30 = 250

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5 0
4 years ago
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ikadub [295]

Answer:

2

Step-by-step explanation:

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3 0
4 years ago
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