Alan

Total energy produced is  15*750 BTUs

volume is 38*8.5*6.25 ft³ 

Divide total energy by volume to get the number of BTUs per cubic foot.

I'll leave you to crunch the numbers.

Density = mass/volume

Here the volume is 8³ cm³ or 512 cm³

so the density is 400/512 g/cm³ = 0.78 g/cm³ to the nearest hundredth. 

I assume you mean there to be an infinite number of 4's, in which case:

Let x = 0.44444...

10x = 4.44444...

So 10x - x = 4.44444... - 0.44444...

or 9x = 4

so x = 4/9

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x! Is x*(x-1)*(x-2)*...*2*1

That means that x!/(x-2)! → x*(x-1)*(x-2)*..,*2*1/( (x-2)*(x-3)*...*2*1 ) → x*(x-1) as everything else cancels out.

Similarly (x-1)!/(x-3)! → (x-1)*(x-2)*(x-3)*...*2*1/( (x-3)*...*2*1 ) → (x-1)*(x-2)

Hence:  x!/(x-2)! + (x-1)!/(x-3)! = 8 becomes: x*(x-1) + (x-1)*(x-2) = 8

Expanding and simplifying this we get:  2x² - 4x + 2 = 8 or x² - 2x -3 = 0

This factors as: (x + 1)(x - 3) = 0

Becase the question originally involved factorials we can only accept a positive result, so x = 3

Check:

3!/(3-2)! = 6/1 → 6

(3-1)!/(3-3)! = 2/1 → 2.   (Note that 0! = 1)

These two sum to 8.

Thinking about this further:

Write equation as: n*ln(n) = 18*10^7*ln(2)

Let w = ln(n), so n = e^w, then we have w*e^w = 18*10^7*ln(2) so w = LambertW(18*10^7*ln(2))

and n = e^LambertW(18*10^7*ln(2))

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Yes, I know about the LambertW function.  This is usually written as w*e^w = x (where w is to be found given a value of x), or as w + ln(w) = ln(x).  I didn't think to seek a solution to your problem in terms of LambertW as I suspect most people using this site are not familiar with it. (Also, not immediately obvious how to manipulate your equation to LambertW form!)

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We have  \(z=k\frac{x^2}{\sqrt{y}}\)   where k is a constant

Find k as  \(k=12\frac{\sqrt{4}}{2^2}\rightarrow6\)

So when x = 6 and y = 32 we have \(z=6\frac{6^2}{\sqrt{32}}\rightarrow27\sqrt2\)

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You could also do this by simple iteration.

\(\text{First scale the problem to deal with more manageable numbers. Let }x=\frac{n}{18\times10^7}\\.\\\text{Then the equation becomes: }x\log_2(18\times10^7x)=1 \\.\\\text{Rewrite in the form: }x=\frac{1}{\log_2(18\times10^7x)} \text{ and use a simple iterative approach.}\\.\\ x_0=1\\ x_1=\frac{1}{\log_2(18\times10^7*x_0)}\rightarrow 0.036465\\ x_2=\frac{1}{\log_2(18\times10^7*x_1)}\rightarrow0.044158\\ ...\\ x_8=\frac{1}{\log_2(18\times10^7*x_7)}\rightarrow0.0436571\\.\\ n=0.0436571\times18\times10^7\approx7858278\)

The result you quote isn't consistent with the integral you specify:

\(\int_3^4(x^2-1)e^{x-1}-7ex+11edx=9e^3-4e^2-\frac{27e}{2}\)

However:

\(\int_3^4x^2e^{x-1}-7ex+11edx=10e^3-5e^2-\frac{27e}{2}\)

Note that  -49e/2 + 11e  =  -27e/2

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In an isentropic process we have   \(\frac{P_i}{P_f}=(\frac{V_f}{V_i})^\gamma\).

This becomes  \((\frac{P_i}{P_f})^\frac{1}{\gamma}=\frac{V_f}{V_i}\)    .

Your original would only be valid if   \(\gamma=2\)    

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