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lisabon 2012 [21]
2 years ago
5

What is the constant proportionality of $0.59 per pound​

Mathematics
1 answer:
Tanya [424]2 years ago
6 0

Answer:

The constant of proportionality is just 0.59

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<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ee_1%20%7B%5Cfrac%7B1%7D%7Bx%5Csqrt%7B1-%28logx%29%5E%7B2%7D%20%7D%20%7D%20%7
Zina [86]

For the integral

I=\displaystyle\int_1^3\frac{\mathrm dx}{x\sqrt{1-(\log x)^2}}

(assuming \log x is the natural logarithm with base e) substitute u=\log x and \mathrm du=\frac{\mathrm dx}x. Then the integral is equivalent to

I=\displaystyle\int_{\log1}^{\log e}\frac{\mathrm du}{\sqrt{1-u^2}}=\int_0^1\frac{\mathrm du}{\sqrt{1-u^2}}

Next, substitute u=\sin t and \mathrm du=\cos t\,\mathrm dt:

I=\displaystyle\int_{\sin^{-1}0}^{\sin^{-1}1}\frac{\cos t}{\sqrt{1-\sin^2t}}\,\mathrm dt=\int_0^{\frac\pi2}\frac{\cos t}{\sqrt{\cos^2t}}\,\mathrm dt

We have \sqrt{x^2}=|x| for all x, but in the given integration interval, \cos t\ge0, so

I=\displaystyle\int_0^{\frac\pi2}\frac{\cos t}{\cos t}\,\mathrm dt=\int_0^{\frac\pi2}\mathrm dt=\boxed{\dfrac\pi2}

(Of course, with a little foresight, you could have immediately combined the two substitutions and started off with letting \sin u=\log x.)

5 0
3 years ago
of the 30 students in the six period Math class, eight or also in the same fourth period Science class. which can be used to det
Dmitry_Shevchenko [17]

Answer:

the answer to your question is bbbbb

Step-by-step explanation:

4 0
3 years ago
9(x + 8) + 11(3x + 4)
Sav [38]
Simplify the expression : 42x + 116

Hope this helps!
Have a great day!

5 0
3 years ago
Find the equivalent expressions for (x + 7)(x + 1).
Natali5045456 [20]

Answer:

The answers should be

x^2 + 8x + 7

x(x + 8) +7

Step-by-step explanation:

(x + 7)(x + 1)

x(x) + x(1) + 7(x) + 7(1)

x^2 + x + 7x + 7

x^2 + 8x + 7

5 0
3 years ago
Read 2 more answers
Please help!! i don’t understand how to do this
Arisa [49]

Answer: A is the right angle

Step-by-step explanation:

a. 20^2 + 15^2 = 25^2

400 + 225 = 625

625 = 625

hope this helps :)

8 0
2 years ago
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