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Oliga [24]
3 years ago
9

Which table corresponds with the function f(x) = x2 + 4?

Mathematics
1 answer:
DerKrebs [107]3 years ago
4 0
X:-2,-1,0,1,2 
<span>y: 7,4,1,-2,-5</span>
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Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. If​po
steposvetlana [31]

The two curves y=x and y=x^{1/15} intersect at x=0 and x=1, with x^{1/15}\ge x over 0\le x\le1.

  • Washer method

\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}

  • Shell method

We have y=x^{1/15}\implies x=y^{15}. The curves x=y and x=y^{15} intersect at y=0 and y=1, with y\ge y^{15} over 0\le y\le1.

\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}

6 0
3 years ago
Find the product <br> -1/2 y(2y3-8)
MissTica
0-(( \frac{1}{2} *y)*2y^3-8) \\ \\ then \ we \ would \ simplify \ 1/2 \\ \\ we \ factor y^3 - 4 \\ \\ 4 \ would \ not \ be \ a \ perfect \ cube. \\ \\ therefore, \ your \ answer \ would \ then \ be \ the \ following: \\ \\ \boxed{ -y * (y^3 - 4)}
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3 years ago
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