Melody

Sorry but there is no question here.

cotx/sinx.cosx

 $$cotx\div sinx\times cosx\\\\
=\frac{cosx}{sinx}\div sinx\times cosx\\\\
=\frac{cosx}{sinx}\times \frac{1}{sinx}\times cosx\\\\
=\frac{cos^2x}{sin^2x}\\\\
=Cot^2x$$

I'll just finish it.

In 500 rolls he can expect to get this oucome about  27.77...%* 500 =  139 times.

$${\mathtt{0.277\: \!777\: \!777\: \!77}}{\mathtt{\,\times\,}}{\mathtt{500}} = {\frac{{\mathtt{1\,250}}}{{\mathtt{9}}}} = {\mathtt{138.888\: \!888\: \!885}}$$

It goes from left to right.

In primary school you learned about number lines.  They go from left to right across the page.

X comes before Y in the alphabet.

so the one you learned first is the X axis.

$$\sqrt{\frac{9}{49}}=\frac{3}{7}$$

Your username does not inspire a lot of confidence failed.

Your method is reasonable but it would only be correct if the order of selection mattered and i don't think that it does in this case.

The answer is 12C5 = $${\left({\frac{{\mathtt{12}}{!}}{{\mathtt{5}}{!}{\mathtt{\,\times\,}}({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{5}}){!}}}\right)} = {\mathtt{792}}$$

this is

$$\frac{12*11*10*9*8}{5*4*3*2*1}$$           

So first you select your 5 players just as you did.  

but then you think about how many ways those 5 players can be ordered.  And you divide by this number.

does this make sense?

26*25*24=15600

or

$$^{26}P_3 =$$     $${\left({\frac{{\mathtt{26}}{!}}{({\mathtt{26}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{15\,600}}$$

It is also     $$^8P_8$$

That is, there are 8 different objects.  In how many ways can you select 8 and order them.

$${\left({\frac{{\mathtt{8}}{!}}{({\mathtt{8}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)} = {\mathtt{40\,320}}$$

Alan's method is probably best but if you want to start playing with LaTex you can open it up and type

log_3(x+1)=4

and this will be your output

$$log_3(x+1)=4$$

-6+12=

Use the number line.

Start at -6 then go 12 places in the positive direction.  What is the answer?