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alina1380 [7]
3 years ago
5

The width of a rectangle is the length minus 3 units. The area of the rectangle is 54 units. What is the width, in units, of a r

ectangle
Mathematics
1 answer:
saul85 [17]3 years ago
4 0

Width of the rectangle is 9 units

Step-by-step explanation:

  • Step 1: Let the width of the rectangle be x. Then the length = x - 3. Find dimensions of the rectangle if its area = 54 sq. units

Area of the rectangle = length × width

54 = x (x - 3)

54 = x² - 3x

x² - 3x - 54 = 0

x² + 6x - 9x - 54 = 0 (Using Product Sum rule to factorize)

x(x + 6) - 9(x + 6) = 0

(x + 6)(x - 9) = 0

x = -6, 9 (negative value is neglected)

x = 9 units

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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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