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muminat
3 years ago
7

Helppppppppppppp meeeeee

Mathematics
1 answer:
lora16 [44]3 years ago
7 0

Answer D and E

Step-by-step explanation:

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Trava [24]

2,1 I think I hope this helps

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3 years ago
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))

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5 0
4 years ago
Please find the area of each figure.
salantis [7]

Answer:

The first answer is 3456

Step-by-step explanation:

Area=L*W

First shape~ 36*36=1296

Second shape~ 36*60=2160

1296+2160=3456

8 0
3 years ago
Read 2 more answers
Find the solutions of 6(2-7x)+x(7x-2=0)
azamat
6(2-7x)+x(7x-2)=0

Use distributive property: a(b - c) = ab - ac

(6)(2)-(6)(7x)+(x)(7x)-(x)(2)=0\\\\12-42x+7x^2-2x=0\\\\7x^2-44x+12=0\\\\7x^2-42x-2x+12=0\\\\7x(x-6)-2(x-6)=0\\\\(x-6)(7x-2)=0\iff x-6=0\ \vee\ 7x-2=0\\\\x-6=0\ \ \ |+6\\\\x=6\\\\7x-2=0\ \ \ |+2\\\\7x=2\ \ \ |:7\\\\x=\dfrac{2}{7}

Answer:\ x=6\ \vee\ x=\dfrac{2}{7}
8 0
3 years ago
. Solve the inequality, graph the solution. 9+q≤15
KATRIN_1 [288]

Answer:

q=6 creates a verticle line at x=6 shaded to the left and encludes the line.

Step-by-step explanation:

9 + q <= 15

q <= 15 - 9 = 6

5 0
3 years ago
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