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coldgirl [10]
3 years ago
5

The area of the rug is 4 15/36 The rug is 2 3/4 meters in height How wide is the rug

Mathematics
1 answer:
sineoko [7]3 years ago
4 0
I assume the rug is rectangle.

The area of rectangle is width (W) multiply by hight (H)

A=4\frac{15}{36}=4\frac{5\cdot3}{12\cdot3}=4\frac5{12}\\\\H=2\frac34\\\\\\A=W\cdot H\\\\4\frac{5}{12}=W\cdot2\frac34\\\\4\frac5{12}:2\frac34=W\\\\ \frac{4\cdot12+5}{12}:\frac{2\cdot4+3}{4}=W\\\\ \frac{53}{12}:\frac{11}4=W\\\\ \frac{53}{12}\cdot\frac4{11}=W\\\\ \frac{53\cdot4}{3\cdot4\cdot11}=W\\\\W=\frac{53}{33}\\\\W=1\frac{20}{33}

The rug is  1\frac{20}{33}  meters wide
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Step-by-step explanation:

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2 years ago
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Sladkaya [172]

Answer:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}

Operating on the numerators:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}

Removing parentheses:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}

Simplifying:

\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}

The expression cannot be further simplified.

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3 years ago
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Answer:

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