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2 years ago

The dimensions of a triangular prism are shown in the diagram.

Mathematics

1 answer:

user100 [1]2 years ago

5 0

Answer:192

Step-by-step explanation:

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The table shown below shows some values for functions f(x) and g(x). What is/are the solutions for f(x) =g(x)

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Julisha poured 0.9 liter of water into an

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2 years ago

Which equation or function is linear?

zepelin [54]

9514 1404 393

Answer:

C. 2y = (2x-1)/4

Step-by-step explanation:

An equation is linear when the exponents of the variables are 1 and the sum of the exponents of the variables in any term is 1.

a) 3xy = 4 . . . . sum of exponents is 1+1=2

b) f(x) = 2/3(1 -x^2) . . . . exponent is 2

c) 2y = (2x -1)/4 . . . . all exponents are 1 (linear)

d) y = 3/(x+1) ⇒ xy +y = 3 . . . . sum of exponents is 1+1 = 2

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3 years ago

Put in order from least to greatest 12, 8, -9, -12, 10, 16 *

Nina [5.8K]

Answer:

-12,-9,8,10,12,16

Step-by-step explanation:

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2 years ago

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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

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

6 0

2 years ago