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sergey [27]
3 years ago
11

Solve. Which is the solution to 2^x=9^{x-9}

Mathematics
2 answers:
avanturin [10]3 years ago
7 0

Answer:

No solution

Step-by-step explanation:

1=9\left\{\left(-\frac{u-9\ln \left(2\right)}{\ln \left(2\right)}\right)-9\right\}e^{-\ln \left(2\right)\left(-\frac{u-9\ln \left(2\right)}{\ln \left(2\right)}\right)}

Vitek1552 [10]3 years ago
7 0

Answer:

okay, this isn't as helpful but its for sure not A, cause I got it wrong, so it's either B, C, or D.... sorry broskis

Step-by-step explanation:

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(i) Given that

V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}

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In Mathematica, you can first define the velocity function with

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(ii) In full, the volume function is

\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr

Compute the integral:

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V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr

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In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

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