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seropon [69]
3 years ago
10

Find area !! GIVING BRAINLIEST

Mathematics
1 answer:
Anton [14]3 years ago
3 0

Answer:

221

Step-by-step explanation:

area= length times width

13 times 17= 221

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Which letter on the number line represents 4.55? <br> F<br><br> E<br><br> D<br><br> C
REY [17]

Answer:

F

Step-by-step explanation:

4 0
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In a group of 60 children there are twice as many girls than boys how many girls are there
Galina-37 [17]
2x + x = 60 that is your equation. Simplify it and you get 3x = 60. 60/3 equals 20 so you have 20 boys but multiply it by 2 since you have twice as many girls and you get 40 girls total.
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Imma need some help chief
nata0808 [166]

Answer:

B

Step-by-step explanation:

in parallel lines, corresponding angles on each line must be the same.

You can also tell that 2 obtuse angles aren't supplementary as they don't add up to 180 degrees.

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2 years ago
Write the linear equation 2x + 3y = 12 in slope intercept form .
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3 years ago
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The region in the first quadrant bounded by the x-axis, the line x = ln(π), and the curve y = sin(e^x) is rotated about the x-ax
charle [14.2K]
First, it would be good to know that the area bounded by the curve and the x-axis is convergent to begin with.

\displaystyle\int_{-\infty}^{\ln\pi}\sin(e^x)\,\mathrm dx

Let u=e^x, so that \mathrm dx=\dfrac{\mathrm du}u, and the integral is equivalent to

\displaystyle\int_{u=0}^{u=\pi}\frac{\sin u}u\,\mathrm du

The integrand is continuous everywhere except u=0, but that's okay because we have \lim\limits_{u\to0^+}\frac{\sin u}u=1. This means the integral is convergent - great! (Moreover, there's a special function designed to handle this sort of integral, aptly named the "sine integral function".)

Now, to compute the volume. Via the disk method, we have a volume given by the integral

\displaystyle\pi\int_{-\infty}^{\ln\pi}\sin^2(e^x)\,\mathrm dx

By the same substitution as before, we can write this as

\displaystyle\pi\int_0^\pi\frac{\sin^2u}u\,\mathrm du

The half-angle identity for sine allows us to rewrite as

\displaystyle\pi\int_0^\pi\frac{1-\cos2u}{2u}\,\mathrm du

and replacing v=2u, \dfrac{\mathrm dv}2=\mathrm du, we have

\displaystyle\frac\pi2\int_0^{2\pi}\frac{1-\cos v}v\,\mathrm dv

Like the previous, this require a special function in order to express it in a closed form. You would find that its value is

\dfrac\pi2(\gamma-\mbox{Ci}(2\pi)+\ln(2\pi))

where \gamma is the Euler-Mascheroni constant and \mbox{Ci} denotes the cosine integral function.
5 0
4 years ago
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