Alan

You can find cot(3.5°) using the calculator on the home page here.   cot(3.5°) ≈ 16.35

.

Just noticed you seem to have cot (3.5°) given (incorrectly!) on your page!!!

I assume this is cos x*(2[cos x]^2 + 1) = 0

The term in brackets can never be zero, so we must have cos x = 0

That is x = pi/2, 3pi/2 in the range between 0 and 2pi.

.

Provide a specific example of what you are trying to do.

.

The numerator can be factored as \((x+5)(x-4)\) so you have;

\(\frac{x^2+x-20}{x-4}\rightarrow\frac{(x+5)(x-4)}{x-4}\rightarrow x+5\)

.

A = 1, M = 0, S = 6, H = 4, T = 7, O = 9

20264

*       3

--------

60792

32₅,  23₇,  (17₁₀)

The velocity is circumference/period.

circumference = 2pi*r  where r is radius 

r = 29000yr*365days/yr*24hr/day*3*10⁸*m/sec*3600secs/hr*0.001km/m ≈ 2.744*10¹⁷km

period = 290*10⁶yr*365days/yr*24hr/day ≈ 2.54*10¹²hr

velocity ≈ 2*pi*2.744*10¹⁷/(2.54*10¹²) km/hr ≈ 6.8*10⁵ km/hr

.

The general equation for the frequency is:

\(f=\frac{1}{2L}\sqrt{\frac{T*L}{m}}\)

where L is length (0.42m), T is tension (500N), m is mass (0.00075kg) and f is frequency (s^-1)

.

I'll leave you to plug in the numbers.

.

I assume it's as follows:

\(\frac{1}{2}(435+BB)=362-36.44\)

This results in BB = 216.12kJ/mol

.

Don't try to use the "Equation" section of the calculator. Type solve(equation of interest) directly in the "Scientific" section.

eg to solve x²-6=0 type: solve(x²-6) and press enter or =

.