MATH SOLVE

4 months ago

Q:
# The rectangle below has an area of 6n4 + 20n3 + 14n?.The width of the rectangle is equal to the greatest common monomial factor of 6n", 20n", and 14n?What is the length and width of the rectangle?LengthWidth201314n2WidthLength =

Accepted Solution

A:

Answer:The width of the rectangle is [tex]2n^2[/tex].The length of the rectangle is [tex]3n^2+10n+7[/tex].Step-by-step explanation:Consider the provided information.The area of rectangle is [tex]6n^4 + 20n^3 + 14n^2[/tex]It is given that the width of the rectangle is the greatest common factor of [tex]6n^4, 20n^3\ and\ 14n^2[/tex].First find the greatest common factor of [tex]6n^4, 20n^3\ and\ 14n[/tex]:[tex]6n^4=2\times3\times n\times n\times n\times n[/tex][tex]20n^3=2\times2\times 5\times n\times n\times n[/tex][tex]14n^2=2\times7\times n\times n[/tex]The greatest common factor is: [tex]2n^2[/tex]Therefore, the width of the rectangle is [tex]2n^2[/tex].Area of rectangle is: A=LWSubstitute the value of A and W in above formula.[tex]6n^4 + 20n^3 + 14n=2n^2L\\\\L=\frac{6n^4 + 20n^3 + 14n}{2n^2}\\\\L=3n^2+10n+7[/tex]Hence, the length of the rectangle is [tex]3n^2+10n+7[/tex]