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schepotkina [342]
3 years ago
9

What is the slope of a line that is parallel to the line y = 3/4x + 2?

Mathematics
1 answer:
alexandr402 [8]3 years ago
5 0
3/4
when it’s parallel it has the same slope!
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zepelin [54]

Answer: A/The LINE graph

Step-by-step explanation:

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3 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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3 years ago
Y = -4/3x − 1<br> y = 1/3x + 4
Tom [10]

Answer:

x = -3, y =3

Step-by-step explanation:

y = -4/3x - 1

y = 1/3x + 4 -------> 4y = 4/3x + 16

Adding the first equation and the new second equation together:

5y = 15 ------> y = 3

Plug y = 3 into one of the original equations:

3 = 1/3x + 4

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x = -3

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butalik [34]

Answer: 1,341 3/4

Step-by-step explanation:

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3 years ago
Read 2 more answers
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CaHeK987 [17]

Answer:

Number of large boxes: 55

Number of small boxes:65

Step-by-step explanation:

Let be "l" the number of large boxes and "s" the number of small boxes.

Set up a system of equations:

\left \{ {{l+s=120} \atop {60l+30s=5250}} \right.

Use the methof of elimination. Mulitply the first equation by -60 and add both equations. Then solve for "s":

\left \{ {{-60l-60s=-7200} \atop {60l+30s=5250}} \right.\\.........................\\-30s=-1950\\s=65

Substitute s=65 into any of the original equations and solve for "l":

l+65=120\\l=55

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