Answer:
The width of the rectangle is [tex]y^2+6[/tex] meters.
Step-by-step explanation:
Given :
Area of Rectangle = [tex]y^4+11y^2+30 \ m^2[/tex]
Length of Rectangle = [tex]y^2+5\ meters[/tex]
We need to find the width of the rectangle.
Now We know that area of rectangle can be calculated by multiplying length and width.
Area of Rectangle = [tex]length \times width[/tex]
Hence Width can be calculated as;
Width of rectangle = [tex]\frac{\textrm{Area of Rectangle}}{Length}[/tex]
Now Substituting the values we get;
Width of rectangle = [tex]\frac{y^4+11y^2+30}{y^2+5}[/tex]
Now by performing long division we will find the width of rectangle;
Long Division is performed in attachment;
Explanation of Long Division is Given below;
Step 1 : we have dividend [tex]y^4+11y^2+30[/tex] and divisor [tex]y^2+5[/tex] we will first multiply with [tex]y^2[/tex] so the Quotient is [tex]y^2[/tex] and Remainder is [tex]6y^2+30[/tex]
Step 2: Now we have dividend [tex]6y^2+30[/tex] and divisor [tex]y^2+5[/tex] we will fmultiply with [tex]6[/tex] so the Quotient is [tex]y^2+6[/tex] and Remainder is 0.
Hence The width of the rectangle is [tex]y^2+6[/tex] meters.
