4 square root of 3 over 9
Answer:
4
Step-by-step explanation:
<h3><u>some relevant limit laws</u></h3>
lim C = C where c is a constant.
lim( f(x) + g(x)) =lim f(x) + lim g(x)
lim( f(x)g(x)) =lim f(x) * lim g(x)
lim( cg(x)) =clim g(x)
lim( f(x)/g(x)) =lim f(x) / lim g(x) if lim g(x) is not equal to zero.
lim( f(x))^2 = (lim f(x) )^2
lim square root( f(x)) = square root(lim f(x) )
![\lim_{n \to 3} g(x) = 9\\\\\lim_{n \to 3} f(x) = 6\\\\ \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10} \\\\ = \lim_{n \to 3} \sqrt[3]{f(x)g(x) + 10}\\\\= \sqrt[3]{lim_{n \to 3}f(x) \times lim_{n \to 3}g(x) + 10}\\\\= \sqrt[3]{6 \times 9 + 10}\\\\= \sqrt[3]{64}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%203%7D%20g%28x%29%20%20%3D%209%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%203%7D%20f%28x%29%20%20%3D%206%5C%5C%5C%5C%20%5Clim_%7Bn%20%5Cto%203%7D%20%5Csqrt%5B3%5D%7Bf%28x%29g%28x%29%20%2B%2010%7D%20%5C%5C%5C%5C%20%3D%20%5Clim_%7Bn%20%5Cto%203%7D%20%5Csqrt%5B3%5D%7Bf%28x%29g%28x%29%20%2B%2010%7D%5C%5C%5C%5C%3D%20%5Csqrt%5B3%5D%7Blim_%7Bn%20%5Cto%203%7Df%28x%29%20%5Ctimes%20lim_%7Bn%20%5Cto%203%7Dg%28x%29%20%2B%2010%7D%5C%5C%5C%5C%3D%20%5Csqrt%5B3%5D%7B6%20%5Ctimes%209%20%2B%2010%7D%5C%5C%5C%5C%3D%20%5Csqrt%5B3%5D%7B64%7D)
= 4
The formula for circumference = 2*Pi*r
R= radius. Radius is half of the diameter, and d= 20. This means that radius = 10. Now we have a full formula to work with.2*Pi*10.
Substitute 3.14 for pi.
3.14 * 10 = 31.4.
31.4*2 = 62.8.
62.8 rounded equals 63.
C = 63.