Melody

What is tg(a)  ?

Do you mean tan(a) ??

Hi Goo

Show us why you think   2^(m+2) is, or might be,  correct.       

Are you sure you wrote the question down right?

It has already been answered.

You are just devaluing the good answer that has already been given.

All you are doing is encouraging cheating.

You can do this Ticon6  :)

How many cubes does Eva have?

LOL you got me there Ginger !   

The angles could be  65  65  and x

OR

they could be   x, x,  and 65

The angle sume of a triangle is 180.

There are two possible answers. 

So how are you going to work them out?

PLEASE LET ASKER GUEST RESPOND. 

If x is replaced with x-3 then all the points will be moved in the positive x directions 3 places.

So translate the graph 3 units to the right.

What have you done for yourself?

This is a hint on how to start.

[(q+4)(p+3)]  +   {  -1 *    [(q+2)(p+1)]  } =44+2p+2q

Expand the bits in the square brackets first

Then do the braces curly brackets

Then look at what you have.

Try do do that and if you still need help post what you have tried.

Other answerers:  Please let guest asker attempt this. 

Do not do more for the asker unless he/she displays a proper attempt first

Please check the question.

\(\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\\ \frac{4}{27}\left((2^{\frac{3+m}{3}})^3 + 3 (2^{\frac{3+m}{3}})^2 (2^{\frac{3}{m}}) +3(2^{\frac{3+m}{3}}) (2^{\frac{3}{m}})^2 + (2^{\frac{3}{m}})^3 \right)\\ \frac{4}{27}\left((2^{(3+m)} + 3 (2^{(1+\frac{m}{3})})^2 (2^{\frac{3}{m}}) +3(2^{1+\frac{m}{3}}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \frac{4}{27}\left((2^{(3+m)} + 3 (2^{(2+\frac{2m}{3})}) (2^{\frac{3}{m}}) +3(2^{1+\frac{m}{3}}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \frac{2^2}{3^3}\left((2^32^m + 3 (2^2 2^{(\frac{2m}{3})}) (2^{\frac{3}{m}}) +3(2* 2^{(\frac{m}{3})}) (2^{\frac{6}{m}}) + (2^{\frac{9}{m}}) \right)\\ \)

I can keep going but this is not going to fall out nicely.

LaTex:

\frac{4}{27}\left(2^{\frac{3+m}{3}}+2^{\frac{3}{m}}\right)^3\\

\frac{4}{27}\left((2^{\frac{3+m}{3}})^3  +  3 (2^{\frac{3+m}{3}})^2  (2^{\frac{3}{m}})  +3(2^{\frac{3+m}{3}})  (2^{\frac{3}{m}})^2  +   (2^{\frac{3}{m}})^3 \right)\\

\frac{4}{27}\left((2^{(3+m)}  +  3 (2^{(1+\frac{m}{3})})^2  (2^{\frac{3}{m}})  +3(2^{1+\frac{m}{3}})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

\frac{4}{27}\left((2^{(3+m)}  +  3 (2^{(2+\frac{2m}{3})})  (2^{\frac{3}{m}})  +3(2^{1+\frac{m}{3}})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

\frac{2^2}{3^3}\left((2^32^m  +  3 (2^2 2^{(\frac{2m}{3})})  (2^{\frac{3}{m}})  +3(2* 2^{(\frac{m}{3})})  (2^{\frac{6}{m}})  +   (2^{\frac{9}{m}}) \right)\\

There might be an easier way and you need to check my working for careless errors

\(z+\frac{1}{z}=1\\ \left(z+\frac{1}{z}\right)^2=1^2\\ \left(z^2+\frac{1}{z^2}+2\right)=1\\ \left(z^2+\frac{1}{z^2}\right)=-1\\ \left(z^2+\frac{1}{z^2}\right)\left(z+\frac{1}{z}\right)=-1\left(z+\frac{1}{z}\right)\\ \left(z^3+z+\frac{1}{z}+\frac{1}{z^3}\right)=-1*1\\ \left(z^3+\frac{1}{z^3}+1\right)=-1\\ \left(z^3+\frac{1}{z^3}\right)=-2\\ \left(z^3+\frac{1}{z^3}\right)^2=(-2)^2\\ \left(z^6+\frac{1}{z^6}+2\right)=4\\ \left(z^6+\frac{1}{z^6}\right)=2\\ \left(z^6+\frac{1}{z^6}\right)^2=2^2\\ \left(z^{12}+\frac{1}{z^{12}}+2\right)=4\\~\\ z^{12}+\frac{1}{z^{12}}=2\\ \)

LaTex:

z+\frac{1}{z}=1\\
\left(z+\frac{1}{z}\right)^2=1^2\\
\left(z^2+\frac{1}{z^2}+2\right)=1\\
\left(z^2+\frac{1}{z^2}\right)=-1\\
\left(z^2+\frac{1}{z^2}\right)\left(z+\frac{1}{z}\right)=-1\left(z+\frac{1}{z}\right)\\
\left(z^3+z+\frac{1}{z}+\frac{1}{z^3}\right)=-1*1\\
\left(z^3+\frac{1}{z^3}+1\right)=-1\\
\left(z^3+\frac{1}{z^3}\right)=-2\\
\left(z^3+\frac{1}{z^3}\right)^2=(-2)^2\\
\left(z^6+\frac{1}{z^6}+2\right)=4\\
\left(z^6+\frac{1}{z^6}\right)=2\\
\left(z^6+\frac{1}{z^6}\right)^2=2^2\\
\left(z^{12}+\frac{1}{z^{12}}+2\right)=4\\~\\
z^{12}+\frac{1}{z^{12}}=2\\