Area of a circle for (360degrees) = pi*(r^2)
Area of part of a circle with angle θ=(θ/360)*(pi*)*(r^2)
<span>(θ/360)*pi*(6^2)=12*pi
</span>solving, <span>θ=(12/36)*360
</span><span>θ=120 degress
</span>
Since d x pi = circumference
I use 3.14 as pie
I did 9 x pi = 28.26 I DID NOT divide this by two because there are three half circles and 28.26 counts as two of them
Since there is another half circle I divide 28.26 by 2
28.26 + 14.13 = 42.39
Now plus the bottom length
42.39 + 9 = 51.39
Brainliest answer please?
Answer:
![C=\frac{5F-160}{9}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B5F-160%7D%7B9%7D)
Step-by-step explanation:
Let's solve for C.
![F=\frac{9C}{5}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B9C%7D%7B5%7D)
Step 1: Flip the equation.
![\frac{9}{5} C+32=F](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B5%7D%20C%2B32%3DF)
Step 2: Add
to both sides.
![\frac{9}{5} C+32+-32=F+-32](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B5%7D%20C%2B32%2B-32%3DF%2B-32)
![\frac{9}{5}C=F-32](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B5%7DC%3DF-32)
Step 3: Divide both sides by
.
![\frac{\frac{9}{5}C}{\frac {9} {5}}=\frac{F-32}{\frac{9}{5} }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cfrac%7B9%7D%7B5%7DC%7D%7B%5Cfrac%20%7B9%7D%20%7B5%7D%7D%3D%5Cfrac%7BF-32%7D%7B%5Cfrac%7B9%7D%7B5%7D%20%7D)
![C=\frac{5F-160}{9}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B5F-160%7D%7B9%7D)
Answer:
![C=\frac{F-160}{9}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7BF-160%7D%7B9%7D)
Answer:
Step-by-step explanation: