Melody

I'll do the first on and you can try the others yourself

\(cos(u)=\frac{\sqrt{15}}{5}\\ cos(2*\frac{u}{2})=\frac{\sqrt{15}}{5}\\ cos^2(\frac{u}{2})-sin^2(\frac{u}{2})=\frac{\sqrt{15}}{5}\\ 1-sin^2(\frac{u}{2})-sin^2(\frac{u}{2})=\frac{\sqrt{15}}{5}\\ 1-2sin^2(\frac{u}{2})=\frac{\sqrt{15}}{5}\\ -2sin^2(\frac{u}{2})=\frac{\sqrt{15}-5}{5}\\ sin^2(\frac{u}{2})=\frac{\sqrt{15}-5}{-10}\\ sin^2(\frac{u}{2})=\frac{5-\sqrt{15}}{10}\\ sin(\frac{u}{2})=\sqrt{\frac{5-\sqrt{15}}{10}}\\ sin(\frac{u}{2})=\frac{\sqrt{(5-\sqrt{15})}}{\sqrt{10}}\times \frac{\sqrt{10}}{\sqrt{10}}\\ sin(\frac{u}{2})=\frac{\sqrt{10(5-\sqrt{15})}}{10}\\ sin(\frac{u}{2})=\frac{\sqrt{50-10\sqrt{15}}}{10}\\ \)

Hi Veteran

I will assume that x is a acute angle, is that ok?

Draw a right angled triangle. Mark one acute angle as x.

The adjacent side is \(2\sqrt6\)

The hypotenuse is  5

So the opposite side is  \(\sqrt{(5^2-2\sqrt6)^2}=\sqrt{(25-24)}=1\)

\(cos(x)=\frac{2\sqrt6}{5}\\ sin(x)=\frac{1}{5}\\ tan(x)=\frac{sinx}{cosx}=\frac{1}{5}\div \frac{2\sqrt6}{5}=\frac{1}{2\sqrt6}\)

\(sin(2x) =2sinxcosx =2*\frac{1}{5}*\frac{2\sqrt6}{5}=\frac{4\sqrt6}{25}\\~\\ cos(2x)=cos^2x-sin^2x=\frac{24}{25}-\frac{1}{25}=\frac{23}{25}\\~\\ tan(2x)\\ =\frac{2tanx}{1-tan^2x}\\ =\frac{2}{2\sqrt6}\div (1-\frac{1}{24)}\\ =\frac{1}{\sqrt6}\div \frac{23}{24}\\ =\frac{24}{23\sqrt6}\\ =\frac{24\sqrt6}{23*6}\\ =\frac{4\sqrt6}{23}\\ \)

How many digits are located to the right of the decimal point when   is expressed as a decimal?

\(\frac{3^6}{6^4\times 625}\\ =\frac{3^6}{2^4\times 3^4\times 625}\\ =\frac{3^2}{2^4\times 5^4}\\ =\frac{3^2}{10^4}\\ =9*10^{-4}\\ \therefore\\ \text{4 places after the point}\)

Hello guest,

You are very polite.  I'd like you to become a member :)

\(y=\frac{1}{2}sin(2x+3\pi)\\ \text{Here is the trick }\\~\\ y=\frac{1}{2}sin[2(x+\frac{3\pi}{2})]\\ amplitude =1/2\\ period=\frac{2\pi}{2}=\pi\\ Phase \;shift=\frac{3\pi}{2}\;\;left\\ \text{Since the period is pi, this is 1.5 wavelengths}\\ \text{So it will still start at (0,0) but it will start going down instead of up.} \)

Here is the graph:

ok thanks Alan but that still means that

\(\sqrt{-\frac{1}{2}} \;\;\;\text{Must be interpreted as }\;\;\sqrt{\frac{-1}{2}}\)

I suppose if you think of it as  \(\sqrt{-0.5}\)

the problem is fixed, but that still means the - sign must be in the numerator. 

Mmmm  still thinking. ............

Can I have some clarification on that please Alan?

Are you saying that  you cannot take the squareroot of a negative number unless the negative sign is in the numerator?  

Because obviously      \(\frac{1}{-1} \quad \text{does equal } \;\;\frac{-1}{1}\)

Hi SaraHann,

The question is worded badly, the meaning is not clear.

If something has to happen then it has a probabiliy of 1       r=1

If something has to happent is has a probablility of 0           r=0

All other possiblilities are between these extremes.    So this would be   0<r<1

If you have no idea what the probable outcome might be then r could be over the whole range of possibilities

\(0\le r \le 1\)

Does this answer your question?

Someone else posted it yesterday anyway. :)

I think it was Einstein's birthday too.

I'd say who did the post but I have forgotten.  Sorry. :(

do you mean

1.      \(log_2(1/3)\\ \)

or

2.    \(log 2^{1/3}\)

I think it is a good question too.

Here are some answers that other people have given to this query :)