The simple way of calculating a perimeter is to add all of the sides.
For a rectangle this would equal 2 x length + 2 x width
P = 2(6x + 3) + 2(-2x - 5)
= 12x + 6 - 4x - 10
= 8x - 4
At a higher level both the length and width must be greater than zero (= zero is a trivial rectangle)
6x + 3 > 0
6x > -3
x > -0.5
-2x - 5 > 0
2x + 5 < 0 (multiplying by -1 reverses the inequality)
2x < -5
x < -2.5
This rectangle cannot exist as x cannot be < -2.5 and > -0.5 at the same time!
if it has a diameter of 8, that means its radius is half that, or 4.
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=4\\ h=5 \end{cases}\implies V=\cfrac{\pi (4)^2(5)}{3}\implies V=\cfrac{80\pi }{3} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{using~\pi =3.14}{V= 83.7\overline{3}}~\hfill](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ar%3D4%5C%5C%0Ah%3D5%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%284%29%5E2%285%29%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B80%5Cpi%20%7D%7B3%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20%5Cstackrel%7Busing~%5Cpi%20%3D3.14%7D%7BV%3D%2083.7%5Coverline%7B3%7D%7D~%5Chfill%20)
Answer:
Sure
Step-by-step explanation:
3/7 is the answer for this question
Answer:
15 and 20
Step-by-step explanation: