Melody

Thanks Alan ...  but I do not know how to calculate that sum to get 3 either.  

Thanks very much Alan,

I am never quite confident about expected values.  :)

Alan's post is being blocked here but available on the answer page  for some weird reason

so I have taken a pit of it and will try to display it here.

Let the sides be a, b and sqrt (a^2+b^2)

So the radii are       \(\frac{a}{2},\quad \frac{b}{2},\quad \frac{\sqrt{a^2+b^2}}{2}\)

Area of triangle =  ab/2

Area of big semicircle = \(0.5 \pi*\frac{(a^2+b^2)}{4}=\frac{(a^2+b^2)\pi}{8}\)

Sum of little white segments = \(\frac{(a^2+b^2)\pi}{8}-\frac{ab}{2}=\frac{(a^2+b^2)\pi-4ab}{8}\)

Sum of the 2 smaller semicircle =       \(0.5*\pi((\frac{a}{2})^2+(\frac{b}{2})^2)= \pi(\frac{a^2+b^2}{8})\\ \)

You can finish it

You are right, I do not, and could not mean cross multiply.

I just mean add.

like how do you do

1/2  +  1/3   ?

You have to get a common denominator.

just turn  1/x+1/y into an entire fraction and it will all fall out.

Given

\(secx+tanx=\frac{4}{3}\\ \frac{1}{cosx}+\frac{sinx}{cosx}=\frac{4}{3}\\ 3+3sinx=4cosx\\ (3+3sinx)^2=(4cosx)^2\\ 9+9sin^2x+18sinx=16cos^2x\\ 9+9sin^2x+18sinx=16(1-sin^2x)\\ 9sin^2x+18sinx+9=16-16sin^2x\\ 25sin^2x+18sinx-7=0\\ \text{solve using the quadratic formula and you get}\\ sinx=-1,\frac{7}{25}\\ \text{Using substitution -1 can be ruled out}\\ sinx=\frac{7}{25} \)

you have to solve for x first.

\(log_{25}(x-4)=0.5\\ \text{can be rearranged as }\\ x-4=25^{0.5}\)

you can take it from there.

Just remember that the LOG of something is a POWER.

I just remember this and then I do my other rearrangements the same way

 \(10^2=100\)    

  \(log_{10}100=2\)

I think this is a trick question.

\(a_{23}=a_{22+1}\\ so \\n=22\)

p=23/2*22=253 though

I work it out with the help of an AP sum formula.