Question

the volume of a rectangular box is the product of its length, width, and height.
write down the expanded form of the expression for the volume: ( 10h^{3}-h^{2}-24h ) (be sto expand the volume expression)
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question 39
the width of a rectangle is 2 feet longer than three times its length, and its area is 33.
if you use ( l ) to represent the length of the rectangle, then an expression representing the width is ( w=)
the length of the rectangle is feet and the width is feet.

Explanation:

Step1: Find the expression for width

The width \(W\) is \(2\) feet longer than three times the length \(L\). So, \(W = 3L+2\).

Step2: Set up the area equation

The area of a rectangle \(A = L\times W\). Given \(A = 33\), substitute \(W = 3L + 2\) into the area formula: \(33=L(3L + 2)\).
Expand the right - hand side: \(33=3L^{2}+2L\).
Rearrange to get a quadratic equation: \(3L^{2}+2L - 33=0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 3\), \(b = 2\), \(c=-33\)), use the quadratic formula \(L=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(2)^{2}-4\times3\times(-33)=4 + 396 = 400\).
Then \(L=\frac{-2\pm\sqrt{400}}{2\times3}=\frac{-2\pm20}{6}\).
We have two solutions for \(L\):
\(L_{1}=\frac{-2 + 20}{6}=\frac{18}{6}=3\) and \(L_{2}=\frac{-2-20}{6}=\frac{-22}{6}=-\frac{11}{3}\).
Since length cannot be negative, \(L = 3\) feet.

Step4: Find the width

Substitute \(L = 3\) into \(W = 3L+2\). Then \(W=3\times3+2=9 + 2=11\) feet.

Answer:

The expression for the width is \(W = 3L + 2\). The length of the rectangle is \(3\) feet and the width is \(11\) feet.