Since the rotation is about the y-axis, I'll integrate by dy.
![\displaystyle y=x^3\\x=\sqrt[3]y\\\\V=\pi \int \limits_1^8(2^2-(\sqrt[3]y)^2)\, dy\\V=\pi \Big[4x-\dfrac{3}{5}x^{\tfrac{5}{3}}\Big]_1^8\\V=\pi \left(4\cdot8-\dfrac{3}{5}\cdot8^{\tfrac{5}{3}\right-\left(4\cdot1-\dfrac{3}{5}\cdot1^{\tfrac{5}{3}\right)\right)\\V=\pi \left(32-\dfrac{96}{5}-\left(4-\dfrac{3}{5}\right)\right)\\V=\pi \left(\dfrac{64}{5}-\dfrac{17}{5}\right)\\V=\dfrac{47\pi}{5}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%3Dx%5E3%5C%5Cx%3D%5Csqrt%5B3%5Dy%5C%5C%5C%5CV%3D%5Cpi%20%5Cint%20%5Climits_1%5E8%282%5E2-%28%5Csqrt%5B3%5Dy%29%5E2%29%5C%2C%20dy%5C%5CV%3D%5Cpi%20%5CBig%5B4x-%5Cdfrac%7B3%7D%7B5%7Dx%5E%7B%5Ctfrac%7B5%7D%7B3%7D%7D%5CBig%5D_1%5E8%5C%5CV%3D%5Cpi%20%5Cleft%284%5Ccdot8-%5Cdfrac%7B3%7D%7B5%7D%5Ccdot8%5E%7B%5Ctfrac%7B5%7D%7B3%7D%5Cright-%5Cleft%284%5Ccdot1-%5Cdfrac%7B3%7D%7B5%7D%5Ccdot1%5E%7B%5Ctfrac%7B5%7D%7B3%7D%5Cright%29%5Cright%29%5C%5CV%3D%5Cpi%20%5Cleft%2832-%5Cdfrac%7B96%7D%7B5%7D-%5Cleft%284-%5Cdfrac%7B3%7D%7B5%7D%5Cright%29%5Cright%29%5C%5CV%3D%5Cpi%20%5Cleft%28%5Cdfrac%7B64%7D%7B5%7D-%5Cdfrac%7B17%7D%7B5%7D%5Cright%29%5C%5CV%3D%5Cdfrac%7B47%5Cpi%7D%7B5%7D%20)
The mass of the first shipment at time t is

The mass of the second shipment at time t is

At time t, the ratio of m₁ to m₂ is

Therefore as a percentage,

Answer: B. 81.2%
Answer:
37.68cm
Step-by-step explanation:
C = 2
r
C = circumference
= the constant pi (3.14)
r = radius of the circle (6cm)
d = diameter of the circle (12cm)
d = 2r
C = 2 · 3.14 · 6
C = 37.68cm
6:2 because there are 6 cups of flour and 2 cups of milk if u put the 2 first it means 2 cups to 6 cups and you need to answer it the exact way the it is asked I am in elementary school and I know this
<h3>
Answer: A) 52 square inches</h3>
==========================================================
Explanation:
a = 8, b = 13, and c = 15 are the sides of the triangle
s = (a+b+c)/2 = (8+13+15)/2 = 18 is the semi-perimeter, aka half the perimeter.
Those values are then plugged into Heron's Formula below

The triangular plaque has an area of approximately 52 square inches.