Alan

Good try Melody, but this shows why it isn't correct:

Will now have a look at Chris's result (which still doesn't match what I'm expecting!)

Ah! If Chris could do adds and take aways correctly he would have the right answer!

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This can be factored as (x - 9)(x - 11) = 0 so the solutions are x = 9 and x = 11

Do it as follows

\(\frac{o+p}{2}=1.48\\\\\frac{p+1.63}{2}=1.52\)

So p = 2*1.52 + 1.63 so p = ...

and o = 2*1.48 + p  so o = ...

I'm sure you can now complete the calculations.

As follows:

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Both Melody and Omi have produced the correct result for f'(pi/2)  (Omi's 6pi^2/16 can be simplifird to 3pi^2/8)

(but Omi, you should have written f'(pi/2) for your last three f's, not f'(x)) 

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First. draw a picture:

h is the height.  From the diagram we have

tan(30°) = h/x                    ...(1)

tan(60°) = h/(3 - x)            ...(2)

We also know that tan(30°) =  sqrt(3)/3 and tan(60°) = sqrt(3)

Equations (1) and (2) are two equations in two unknowns (h and x).  Can you take it from here?

You can write 15 as 3*5 so both 3a and 15 have the number 3 in common.  This can be factored out

3a - 15 → 3a - 3*5 → 3(a - 5)

That's as far as it goes.

Presumably the motor is driving something (imagine, for example, there are belts over the discs going off to work some machinery).  The discs are applying a force on whatever they are connected to (e.g. they are driving the belts), so, from Newton's third law, whatever is being driven is supplying a reaction force on the discs, opposing the drive.

Hi Omi

I agree, but you forgot the log₈(x) term.  

ln(x-1)/ln(8) - ln(x)/ln(8) = 1 

See Guest's development in #4 above.

Hmmm

\(\log_8{(x-1)}-\log_8{x}=1 \rightarrow\log_8{(\frac{x-1}{x})}=1\rightarrow\frac{x-1}{x}=8^1\rightarrow\frac{x-1}{x}=8\\\\ \rightarrow x-1=8x\rightarrow7x=-1\rightarrow x=-\frac{1}{7}\)

Slightly odd, as the original individual logarithms would only have complex values, though the combined log would be real (the imaginary parts cancel out).