let's notice something, we have a circle with a radius of 12 and one 90° sector is cut off, so only three 90° sectors of the circle are left shaded, so namely the cone will be using 3/4 of that circle.
think of it as, this shaded area is some piece of paper, and you need to pull it upwards and have the cutoff edges meet, and when that happens, you'll end up with a cone-shaped paper cup, and pour in some punch.
now, once we have pulled up the center of the circle to make our paper cup, there will be a circular base, its diameter not going to be 24, it'll be less, but whatever that base is, we know that is going to have the same circumference as those in the shaded area. Well, what is the circumference of that shaded area?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=12 \end{cases}\implies C=2\pi 12\implies C=24\pi \implies \stackrel{\textit{three quarters of it}}{24\pi \cdot \cfrac{3}{4}} \\\\\\ 6\pi \cdot 3\implies 18\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D12%20%5Cend%7Bcases%7D%5Cimplies%20C%3D2%5Cpi%2012%5Cimplies%20C%3D24%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bthree%20quarters%20of%20it%7D%7D%7B24%5Cpi%20%5Ccdot%20%5Ccfrac%7B3%7D%7B4%7D%7D%20%5C%5C%5C%5C%5C%5C%206%5Cpi%20%5Ccdot%203%5Cimplies%2018%5Cpi)
well then, the circumference of that circle at the bottom will be 18π, so, what is the diameter of a circle with a circumferenc of 18π?
![\bf \textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ C=18\pi \end{cases}\implies 18\pi =2\pi r\implies \cfrac{18\pi }{2\pi }=r\implies 9=r \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{diameter is twice the radius}}{d=18}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bcircumference%20of%20a%20circle%7D%5C%5C%5C%5C%20C%3D2%5Cpi%20r~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20C%3D18%5Cpi%20%5Cend%7Bcases%7D%5Cimplies%2018%5Cpi%20%3D2%5Cpi%20r%5Cimplies%20%5Ccfrac%7B18%5Cpi%20%7D%7B2%5Cpi%20%7D%3Dr%5Cimplies%209%3Dr%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bdiameter%20is%20twice%20the%20radius%7D%7D%7Bd%3D18%7D~%5Chfill)
Answer:
nonlinear increasing
Step-by-step explanation:
Answer:
(0, 75), (75, 0), (-75, 0), (0, -75)
Step-by-step explanation:
I used desmos graphing calculator. c:
Answer:
is it fighet spinner or I am wrong hahahaa fighet spinner
Answer:
Time of murder = 10:39 am
Step-by-step explanation:
Let the equation of exponential function representing the final temperature of the body after time 't' is,
f(t) = 
Here, a = Initial temperature
n = Constant for the change in temperature
t = Duration
At 11:30 am temperature of the body was 91.8°F.
91.8 = 
--------(1)
Time to reach the body to the morgue = 12:30 pm
Duration to reach = 12:30 p.m. - 11:30 a.m.
= 1 hour
Therefore, equation will be,
84.4 = 
eⁿ = 
ln(eⁿ) = ln(0.9194)
n = -0.08403
From equation (1),
91.8 = 
![ln[(e)^{0.08403t}]=ln[\frac{98.6}{91.8}]](https://tex.z-dn.net/?f=ln%5B%28e%29%5E%7B0.08403t%7D%5D%3Dln%5B%5Cfrac%7B98.6%7D%7B91.8%7D%5D)
0.08403t = 0.07146
t = 0.85 hours
t ≈ 51 minutes
Therefore, murder was done 51 minutes before the detectives arrival.
Time of murder = 11:30 - 00:51
= 10:90 - 00:51
= 10:39 am
