Answer:
[tex]W =2 {y}^{3} \: \:meters[/tex]
Step-by-step explanation:
The figure is a rectangle whose area can be calculated from the relation;
[tex]A = L \times W[/tex]
where A , L and W represent the area, length and width respectively.
From the question,
[tex]A = 60 {y}^{4} [/tex]
and L=30y meters
We substitute the values of A and L into the formula to solve for W
[tex] \implies60 {y}^{4} = 30y \times W[/tex]
Dividing through by 30y
[tex]\implies\frac{60 {y}^{4}}{30y}= \frac{30y \times W}{30y} [/tex]
[tex]\implies W =2 {y}^{3} \: \:meters[/tex]
NB: From the law of indices;
[tex] \frac{ {y}^{4} }{y}= {y}^{(4 - 1)}={y}^{3} [/tex]
