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GaryK [48]
3 years ago
12

X divided by (-2)=8 solve for x

Mathematics
1 answer:
Lunna [17]3 years ago
3 0
The x is equal to -16
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A closed cylindrical can of fixed volume V has radius r.a) Find the surface area, S, as a function of r.b) What happens to the v
andrey2020 [161]

Answer:

Step-by-step explanation:

This question is incomplete; here is the complete question.

A closed cylindrical can of fixed volume V has radius r. (a) Find the surface area, S, as a function of r. (b) What happens to the value of S approaches to infinity? (c) Sketch a graph of S against r, if  V=10 cm³.

A closed cylindrical can of volume V is having radius r and height h.

a). Surface area of a cylinder is given by

S = 2(Area of the circular sides) + Lateral area of the can

S = 2πr² + 2πrh

S = 2πr(r + h)

b). Since surface area is directly proportional to radius of the can

S ∝ r

Therefore, when r approaches to infinity (r → ∞)

c). If V = 10 cm³ Then we have to graph S against r.

From the formula V = πr²h

10 = πr²h

h = \frac{10}{\pi r^{2}}

By placing the value of h in the formula of surface area,

S = 2\pi r(r+\frac{10}{\pi r^{2}})

Now we can get the points to plot the graph,

r       -2             -1         0       1            2

S    -13.72     -13.72     0    26.28    35.13

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Terrence graphs the two equations below on the same coordinate plane.
Vsevolod [243]

Answer:

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6 0
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Which is the graph. or y=1/2x + 1? please hurryyyyyy!
ololo11 [35]

Answer:

Graph 4

Step-by-step explanation:

It crosses at 1 (y-intercept) and has a slope of 1/2

6 0
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Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con
monitta

Let f(x)=e^{-1/x}. Then f'(x)=\frac1{x^2}e^{-1/x}>0 for all x\ge2, so f is strictly increasing. As x\to\infty, e^{-1/x}\to e^0=1, so f is bounded above by 1. This is to say,

e^{-1/x}

and the integral of \frac1{x^2} converges over the same domain, so this integral must also converge by comparison.

We have, by setting y=-\frac1x,

\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}

8 0
3 years ago
Oh lord help :,) please and thank you
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Answer:

The answer is 19.10 $

Step-by-step explanation:

5 0
3 years ago
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