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)
57. 8
58. 8
59. 24
60. 11.25 or 11
61. 20.52 or 21
62. 144
63. 7.37 or 7
64. 50%
65. 25%
66. 10%
67. 20%
68. 25%
hope this helped :)
Answer:
What equation will model the situation?
✔ x/3 = 4
How many servings did the original recipe make?
✔ 12
Step-by-step explanation:
Hope this helps!
Add the five to each side.
then square it to get rid of the square root
2x+13=(5+x)^2
2x+13=x^2+10x+25
2x=x^2+10x+12
x^2+8x+12=0
(x+6)(x+2)
x+6=0
x+2=0
answers are -2 and -6, however, when pluging -6 in it doesnt work, so it is only -2.
Answer:
C. 32
Step-by-step explanation:
There are 16 lines of symmetry passing through the vertex's and another 16 passing through the mid points of the sides
