Question

Algebra 1, (Up Hill is not apart of the problem it’s just what the section of my homework is called)

Answer 1

`10in,13in`

Step-by-step explanation

The tabletop is rectangular in shape. The area of a rectangle is the product of its length and width:

`A=L\cdot W`

The area of the rectangular tabletop is given as:

`A=(x^2+5x+4)in^2`

Let us write this area as a product of two terms, just like the general formula for the area of a rectangle.

To do this, factorize the quadratic expression above:

`\begin{gathered} A=x^2+4x+x+4 \\ \Rightarrow A=x(x+4)+1(x+4) \\ \Rightarrow A=(x+1)(x+4) \end{gathered}`

By comparing this to the general formula for the area of a rectangle, we can say that the length and width of the table are:

`\begin{gathered} L=(x+1)in \\ W=(x+4)in \end{gathered}`

The distance around the table is 46 inches. This represents the perimeter of the tabletop.

The perimeter of a rectangle is:

`P=2(L+W)`

Hence, we can substitute the length, width, and perimeter of the tabletop into the equation:

`46=2\lbrack(x+1)+(x+4)\rbrack`

Solve for x in the equation above:

`\begin{gathered} 46=2(2x+5)=4x+10 \\ \Rightarrow4x=46-10=36 \\ x=(36)/(4) \\ x=9 \end{gathered}`

Therefore, substituting the value of x into the equation for the length and width, the actual side lengths of the table are:

`\begin{gathered} \Rightarrow L=9+1=10in \\ \Rightarrow W=9+4=13in \end{gathered}`

That is the answer.