Answer:

Step-by-step explanation:
![\sqrt{8} = \sqrt{4*2} = \sqrt{4} * \sqrt{2} = 2*\sqrt{2} = 2\sqrt{2} \\\\\sqrt[3]{8} = \sqrt{2*2*2} = 2](https://tex.z-dn.net/?f=%5Csqrt%7B8%7D%20%3D%20%5Csqrt%7B4%2A2%7D%20%3D%20%5Csqrt%7B4%7D%20%2A%20%5Csqrt%7B2%7D%20%3D%202%2A%5Csqrt%7B2%7D%20%20%3D%202%5Csqrt%7B2%7D%20%5C%5C%5C%5C%5Csqrt%5B3%5D%7B8%7D%20%3D%20%5Csqrt%7B2%2A2%2A2%7D%20%20%3D%202)
P would equal negative three.
Find the area of the CD. (I used P for pi, too lazy @
[email protected])
A=Pr² - Pr²
radius of the small hole in the middle, 0.75cm
radius of the CD, 6cm
A=3.14*6²-3.14*0.75²
A=111.27375cm²
The Approximate area of the CD is 111.27cm²