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liubo4ka [24]
3 years ago
8

The formula for the area of a rectangle is A-ly, where A is the area, I is the length, and w is the width. If a rectangle has an

area of 1800 square feet, and the length is twice the distance of the width, determine the width. ​
Mathematics
2 answers:
Colt1911 [192]3 years ago
7 0

Answer:

idk

Step-by-step explanation:

Galina-37 [17]3 years ago
4 0

Answer:

w = 30

Step-by-step explanation:

a = lw

l = 2w

a = 2w × w

1800 = 2w × w

1800 ÷ 2 = w × w

900 = w × w

square root of 900 = square root of w^2

(I don't have the symbol for square root)

30 = w

check work:

a = lw

l = 2w

l = 2(30)

l = 60

1800 = 60 × 30

1800 = 1800

correct answer

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3 years ago
Solve for x<br><br> THANK YOU
Anestetic [448]

Answer:

x=\dfrac{-17+ \sqrt{451} }{18}, \quad x=\dfrac{-17- \sqrt{451} }{18}

Step-by-step explanation:

\textsf{Given equation}:

\dfrac{2}{3}x+\dfrac{3x}{x}=\dfrac{4x}{2}(3x+6)

\textsf{Cancel the common factor } x \textsf{ in }\dfrac{3x}{x}:

\implies \dfrac{2}{3}x+3=\dfrac{4x}{2}(3x+6)

\textsf{Simplify }\dfrac{4x}{2}\textsf{ by dividing the numbers}:

\implies \dfrac{2}{3}x+3=2x(3x+6)

\textsf{Apply the distributive law}\quad \:a\left(b+c\right)=ab+ac:

\implies \dfrac{2}{3}x+3=6x^2+12x

\textsf{Multiply both sides by 3 to cancel the denominator of }\dfrac{2}{3}:

\implies \dfrac{2}{3}x\cdot 3+3\cdot 3=6x^2\cdot 3+12x \cdot 3

\implies 2x+9=18x^2+36x

\textsf{Switch sides}:

\implies 18x^2+36x=2x+9

\textsf{Subtract }2x \textsf{ from both sides}:

\implies 18x^2+36x-2x=2x+9-2x

\implies 18x^2+34x=9

\textsf{Subtract 9 from both sides}:

\implies 18x^2+34x-9=9-9

\implies 18x^2+34x-9=0

Solve using the <u>Quadratic Formula:</u>

x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0

Define the variables:

\implies a=18, \quad b=34, \quad c=-9

Substitute the defined variables into the quadratic formula and solve for x:

\implies x=\dfrac{-34 \pm \sqrt{34^2-4(18)(-9)} }{2(18)}

\implies x=\dfrac{-34 \pm \sqrt{1156+648} }{36}

\implies x=\dfrac{-34 \pm \sqrt{1804} }{36}

\implies x=\dfrac{-34 \pm \sqrt{4 \cdot 451} }{36}

\implies x=\dfrac{-34 \pm \sqrt{4}\sqrt{451} }{36}

\implies x=\dfrac{-34 \pm 2\sqrt{451} }{36}

\implies x=\dfrac{-17 \pm \sqrt{451} }{18}

Therefore, the solutions to the given equation are:

x=\dfrac{-17+ \sqrt{451} }{18}, \quad x=\dfrac{-17- \sqrt{451} }{18}

Learn more about the Quadratic Formula here:

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