MATH SOLVE

4 months ago

Q:
# The area of a rectangular room is 750 square feet. The width of the room is 5 feet less than the length of the room. The area of a rectangular room is 750 square feet. The width of the room is 5 feet less than the length of the room. Which equations can be used to solve for y, the length of the room? Select three options. y(y + 5) = 750 y2 – 5y = 750 750 – y(y – 5) = 0 y(y – 5) + 750 = 0 (y + 25)(y – 30) = 0

Accepted Solution

A:

Answer:[tex]y^{2}-5y=750[/tex][tex]750-y(y-5)=0[/tex][tex](y + 25)(y -30) = 0[/tex]Step-by-step explanation:GivensThe area of a rectangular room is 750 square feet.The width of the room is 5 feet less than the length of the room.Let's call [tex]w[/tex] the width and [tex]l[/tex] the length. According to the problem they are related as follows[tex]w=l-5[/tex], because the width is 5 feet less than the lenght.We know that the area of the room is defined as[tex]A=w\times l[/tex]Where [tex]A=750 ft^{2}[/tex]Replacing the given area and the expression, we have[tex]750=(l-5)l\\750=l^{2}-5l \\l^{2}-5l-750=0[/tex]We need to find two number which product is 750 and which difference is 5, those numbers are 30 and 25.[tex]l^{2}-5l-750=(l-30)(l+25)[/tex]Using the zero property, we have[tex]l=30\\l=-25[/tex]Where only the positive number makes sense to the problem because a negative length doesn't make any sense.Therefore, the length of the room is 30 feet.Also, the right answers are the second choice where [tex]y=l[/tex], the third choice and the last choice.