Answer:
No
Step-by-step explanation:
Do you have to use substitution?
For number 3, it is reciprocal
and for number 4, it is whole number
so, we have two 54x18 rectangles, so their perimeter is simply all those units added together, 54+54+54+54+18+18+18+18 = 288.
we know the circle's diameter is 1.5 times the width, well, the width is 18, so the diameter of the circle must be 1.5*18 = 27.
![\bf \stackrel{\textit{circumference of a circle}}{C=d\pi }~~ \begin{cases} d=diameter\\[-0.5em] \hrulefill\\ d=27 \end{cases}\implies C=27\pi \implies C=\stackrel{\pi =3.14}{84.78} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perimeter of the rectangles}}{288}~~~~+~~~~\stackrel{\textit{perimeter of the circle}}{84.78}~~~~=~~~~372.78](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bcircumference%20of%20a%20circle%7D%7D%7BC%3Dd%5Cpi%20%7D~~%20%5Cbegin%7Bcases%7D%20d%3Ddiameter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20d%3D27%20%5Cend%7Bcases%7D%5Cimplies%20C%3D27%5Cpi%20%5Cimplies%20C%3D%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7B84.78%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20rectangles%7D%7D%7B288%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7Bperimeter%20of%20the%20circle%7D%7D%7B84.78%7D~~~~%3D~~~~372.78)