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Alinara [238K]
3 years ago
6

How can you show that two objects are proportional with an equation

Mathematics
1 answer:
Zina [86]3 years ago
5 0
The equation has to (or be transformed to) contain one object in the left side and the other object at the right side multiplied by a constant, without an independent term.
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Can someone please help me?
natulia [17]

Answer:

#3

Step-by-step explanation:

It implies that Douglass and his siblings nearly forgot about their early lives.

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3 years ago
3 part question. If answer all(brainilest will be given.
docker41 [41]

Answer:

1. 1080

2. It shows the initial debt, i.e., $1080

3.24 months

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3 years ago
What two decimals are equivalent to 3.300​
LenKa [72]

Answer:

3,3 and 3,30

Step-by-step explanation:

No matter how many zeros you wax on or wane off, you will still have equivalent values.

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6 0
3 years ago
Find each unit price. Which is the better by? Explain please.
Luda [366]
1.49/3

0.5 dollars per feet.

This is a fraction equal to 1.49/3.

We want a unit rate where 1 is the denominator, so we divide the top and bottom by 3.

The answer is 0.49666667, which can be rounded to 0.5
5 0
3 years ago
Read 2 more answers
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
3 years ago
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