I am not completely sure but Khan Academy might show how to do that. Or it might at least give an example.
Check the picture below.
so the volume will simply be the area of the hexagonal face times the height.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\stackrel{\qquad degrees}{\cot\left( \frac{180}{n} \right)}~~ \begin{cases} n=\stackrel{number~of}{sides}\\ s=\stackrel{length~of}{side}\\[-0.5em] \hrulefill\\ n=6\\ s=12 \end{cases}\implies A=\cfrac{1}{4}(6)(12)^2\cot\left( \frac{180}{6} \right) \\\\\\ A=216\cot(30^o)\implies A=216\sqrt{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the hexagon}}{(216\sqrt{3})}~~\stackrel{height}{(10)}\implies 2160\sqrt{3}~~\approx ~~3741.2~cm^3](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B4%7Dns%5E2%5Cstackrel%7B%5Cqquad%20degrees%7D%7B%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7Bn%7D%20%5Cright%29%7D~~%20%5Cbegin%7Bcases%7D%20n%3D%5Cstackrel%7Bnumber~of%7D%7Bsides%7D%5C%5C%20s%3D%5Cstackrel%7Blength~of%7D%7Bside%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D6%5C%5C%20s%3D12%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%2812%29%5E2%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7B6%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D216%5Ccot%2830%5Eo%29%5Cimplies%20A%3D216%5Csqrt%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagon%7D%7D%7B%28216%5Csqrt%7B3%7D%29%7D~~%5Cstackrel%7Bheight%7D%7B%2810%29%7D%5Cimplies%202160%5Csqrt%7B3%7D~~%5Capprox%20~~3741.2~cm%5E3)
Answer:
40 minutes.
Step-by-step explanation:
- time spent on the thread mill as a function of d(day) = t(d)
= 30 +2(d-1) minutes
[where d represents day number]
(when d is 1 the time spent should be 30
so substitute d=1 and check it will be 30 only.
on first day he will spend 30 minutes so, I added 30 .
and on every additional day( these additional days are excluding first day), he will increase the time by 2 minutes.so, I added 2(d-1) to initial 30 minutes)
- so, T(6), the minutes he will spend on the treadmill on day 6=
=30+2(6-1)
=30+2(5)
=30+10
=40 minutes
What is 40cm 25% of?
The answer is 160cm.
Rachel is 160cm. 160 - 40 = 120.
Lucy is 120cm tall.
Answer:
Increasing if f' >0 and decreasing if f'<0
Step-by-step explanation:
Difference quotient got by getting
will be greater than 0 if function is increasing otherwise negative
Here h is a small positive value.
In other words, we find that whenever first derivative of a function f(x) is positive the function is increasing.
Here given that for x1, x2 where x1<x2, we have
if f(x1) <f(x2) then the function is decreasing.
Or if x1<x2 and if f(x1) >f(x2) for all x1, and x2 in I the open interval we say f(x) is decreasing in I.