So... let's say the smaller regular octagon has sides of "x" long, then the larger octagon will have sides of 5x.
![\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &Sides&Area&Volume\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array} \\\\ -----------------------------\\\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bratio%20relations%7D%0A%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccllll%7D%0A%26Sides%26Area%26Volume%5C%5C%0A%26-----%26-----%26-----%5C%5C%0A%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%26%5Ccfrac%7Bs%7D%7Bs%7D%26%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%26%5Ccfrac%7Bs%5E3%7D%7Bs%5E3%7D%0A%5Cend%7Barray%7D%20%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
It took them 4 days to install 480 chairs which means if they were working at a constant rate they got 120 chairs installed daily. with 360 chairs left that means in 3 days they will finish. so overall, it took the workers 7 days to finish, or a week.
I think that would be a rectangle with sides 13 and 8 Which = 104 ins^2
4/10 multiplied to 2/3 is 4/15.
Hi there!
Let's break this down:
Two times a number = 2n
Increased by 5 = + 5
is 15 - = 15
Total equation:
2n + 5 = 15
Solving:
2n + 5 = 15
2n = 10
n = 5
Hope this helps!! :)
If there's anything else that I can help you with, please let me know!
