The definition of the set E gives you a natural choice for the limits in the integral:
![\displaystyle \iiint_E y \, dV = \int_0^6 \int_0^x \int_{x-y}^{x+y} y \, dz \, dy \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciiint_E%20y%20%5C%2C%20dV%20%3D%20%5Cint_0%5E6%20%5Cint_0%5Ex%20%5Cint_%7Bx-y%7D%5E%7Bx%2By%7D%20y%20%5C%2C%20dz%20%5C%2C%20dy%20%5C%2C%20dx)
Computing the integral, we get
![\displaystyle \iiint_E y \, dV = \int_0^6 \int_0^x y ((x+y)-(x-y)) \, dy \, dx = 2 \int_0^6 \int_0^x y^2 \, dy \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciiint_E%20y%20%5C%2C%20dV%20%3D%20%5Cint_0%5E6%20%5Cint_0%5Ex%20y%20%28%28x%2By%29-%28x-y%29%29%20%5C%2C%20dy%20%5C%2C%20dx%20%3D%202%20%5Cint_0%5E6%20%5Cint_0%5Ex%20y%5E2%20%5C%2C%20dy%20%5C%2C%20dx)
![\displaystyle \iiint_E y \, dV = 2 \int_0^6 \frac13 (x^3 - 0^3) \, dx = \frac23 \int_0^6 x^3 \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciiint_E%20y%20%5C%2C%20dV%20%3D%202%20%5Cint_0%5E6%20%5Cfrac13%20%28x%5E3%20-%200%5E3%29%20%5C%2C%20dx%20%3D%20%5Cfrac23%20%5Cint_0%5E6%20x%5E3%20%5C%2C%20dx)
![\displaystyle \iiint_E y \, dV = \frac23 \cdot \frac14 (6^4 - 0^4) = \boxed{216}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciiint_E%20y%20%5C%2C%20dV%20%3D%20%5Cfrac23%20%5Ccdot%20%5Cfrac14%20%286%5E4%20-%200%5E4%29%20%3D%20%5Cboxed%7B216%7D)
The answer is 0.20 times 30 = 6
Answer:
Option B
Step-by-step explanation:
From the graph attached,
x-intercepts of the graph are x = -1, 2 and 5
y-intercept of the graph is y = 10
Since, x intercepts of a graph define the zeros of the function (y = 0)
Therefore, (x + 1), (x - 2) and (x - 5) are the zero factors of the given graphs.
And the equation of the function will be,
f(x) = a(x + 1)(x - 2)(x - 5)
Since, -intercept of the function is y = 10
Therefore, for x = 0,
y = a(0 + 1)(0 - 2)(0 - 5) = 10
a(10) = 10
a = 1
So the function representing the graph is,
f(x) = (x - 1)(x + 2)(x - 5)
Option B will be the answer.
Answer:
The y-intercept is 7
Step-by-step explanation:
y = slope(x) + y-intercept
The amount of space not occupied by the rubber balls is given by:
Volume=(volume of the container)-(volume of the rubber balls)
volume of the container is given by:
V=πr²h
V=π*(5/2)²(20)
V=392.70 cm³
Volume of each ball is:
V=4/3πr³
V=4/3π(2.5)³=65.45 cm³
volume of four balls
65.45×4=261.8 cm³
The volume of the container that is not occupied by the balls will be:
V=392.70-261.8
V=130.9 cm³