2 3/4
--------------------
You just subtract them. :)
The answer is (A) because the (quantity of a number minus nine) implies that x-9 is inside parenthesis.
Median: 6
Range: 9
Mode: 7, appeared 3 times
Answer:
He made a mistake in the second line.
Step-by-step explanation:
He factored out x out of 6-2x and 5+4x, which is not possible because there is not an x in both 6 and 5.
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)
