Check the picture below.
so then, the perimeter of that hexagon will just be the sum of all its 6 sides, or namely 3⅖ + 3⅖ + 3⅖ + 3⅖ + 3⅖ + 3⅖, or just 6( 3⅖ ).
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=3\\ p=6\left(3\frac{2}{5} \right) \end{cases}\implies A=\cfrac{1}{2}(3)\left[ 6\left(3\frac{2}{5} \right) \right]\implies A=\cfrac{1}{2}(3)\left[ 6\left(\cfrac{17}{5} \right) \right] \\\\\\ A=\cfrac{1}{2}(3)\left(\cfrac{102}{5} \right)\implies A=\cfrac{1}{2}\left( \cfrac{306}{5} \right)\implies A=\cfrac{153}{5}\implies A=30\frac{3}{5}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D3%5C%5C%20p%3D6%5Cleft%283%5Cfrac%7B2%7D%7B5%7D%20%5Cright%29%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283%29%5Cleft%5B%206%5Cleft%283%5Cfrac%7B2%7D%7B5%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283%29%5Cleft%5B%206%5Cleft%28%5Ccfrac%7B17%7D%7B5%7D%20%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7D%283%29%5Cleft%28%5Ccfrac%7B102%7D%7B5%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%5Cleft%28%20%5Ccfrac%7B306%7D%7B5%7D%20%5Cright%29%5Cimplies%20A%3D%5Ccfrac%7B153%7D%7B5%7D%5Cimplies%20A%3D30%5Cfrac%7B3%7D%7B5%7D)
Answer:
2.25
Step-by-step explanation:
The old width of 4 inches became a new width of 9 inches.
Old: 4 in.
New 9 in.
The old length of 6 inches became a new length of 13.5 inches.
Divide the new size by the corresponding old size.
9/4 = 2.25
13.5/6 = 2.25
You see in both cases, the scale factor is 2.25
The scale factor is 2.25
Answer: 1 + y ≤ -13
Step-by-step explanation:
Answer:
constant
Step-by-step explanation:
