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)
. . Hmmmmmmmmm, I wonder what the answer could be XD
Answer:
use mathaway or math scanner they will help and provide you with the answer
X=3.2 I believe that's your answer
Answer:
see the explanation
Step-by-step explanation:
First way
we know that
The sum of the interior angles of a triangle must be equal to 180 degrees
so
In this problem
37+97+134 > 180
therefore
At least one of Franklin's measures is incorrect
Second way
we know that
A triangle can only have at most one obtuse internal angle.
In this problem the triangle has two obtuse internal angles
Remember that an obtuse angle is an angle greater than 90 degrees
therefore
At least one of Franklin's measures is incorrect