Solve.
-If the area of a rectangle is $\left( 10 x ^ { 2 } + 33 x + 20 \right)$ square feet, and its length is $( 2 x + 5 )$ feet, find its width.
A) $( 5 x + 4 ) \mathrm { ft }$
B) $( 2 x + 4 ) \mathrm { ft }$
C) $( 5 x + 5 ) f t$
D) $( 5 x - 4 ) \mathrm { ft }$

Question

Quizplus · Accepted Answer

The width can be found by dividing the area of the rectangle by its length. Given the area is $10x^2 + 33x + 20$ and the length is $2x + 5$, the width is $\frac{10x^2 + 33x + 20}{2x + 5}$. Factoring the numerator or directly dividing, we find the width to be $5x + 4$ feet.

Solve -If the Area of a Rectangle Is (10x2+33x+20)\left( 10 x ^ { 2 } + 33 x + 20 \right)(10x2+33x+20)

Solve
-If the Area of a Rectangle Is $\left( 10 x ^ { 2 } + 33 x + 20 \right)$