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2 years ago

A jar contains quarters and dimes. There are 35 coins

Mathematics

1 answer:

Tanzania [10]2 years ago

8 0

24 quarters and 2 dimes

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A plane cuts through a cone perpendicular to the base of the cone. What shape is formed by this intersection

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2 years ago

Simplify (12y^7/18y^-3) assume Y=0

iren2701 [21]

$\dfrac{12y^7}{18y^{-3}} = \dfrac{2y^{7+3}}{3} = \dfrac{2y^{10}}{3} = \dfrac{2}{3}y^{10}$

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Find the volume <br> Please help have to turn in a hour from now

pantera1 [17]

Answer:

V ≈ 226.72 cm³

Step-by-step explanation:

Use the formula for volume of a cone using slant height and radius.

V = 1/3 π r² √ l² - r²

Then, plug in dimensions and solve.

V = 1/3 π (10)² √ 10² - 5²

V ≈ 226.72 cm³

Hope I could help!!! Good luck!!!

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2 years ago

A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from th

jeyben [28]

Consider a circle with radius $r$ centered at some point $(R+r,0)$ on the $x$ -axis. This circle has equation

$(x-(R+r))^2+y^2=r^2$

Revolve the region bounded by this circle across the $y$ -axis to get a torus. Using the shell method, the volume of the resulting torus is

$\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx$

where $2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}$ .

So the volume is

$\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx$

Substitute

$x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt$

and the integral becomes

$\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt$

Notice that $\sin t\cos^2t$ is an odd function, so the integral over $\left[-\frac\pi2,\frac\pi2\right]$ is 0. This leaves us with

$\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt$

Write

$\cos^2t=\dfrac{1+\cos(2t)}2$

so the volume is

$\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}$

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2 years ago