Melody

To be honest,  I am not really sure what this question means ..but

\(\frac{e^{ix}}{e^{iy}}\\ =\frac{cos(x)+isin(x)}{cos(y)+isin(y)}\\ =\frac{cos(x)+isin(x)}{cos(y)+isin(y)}*\frac{cos(y)-isin(y)}{cos(y)-isin(y)}\\~\\ =\frac{cos(x)cos(y)-icos(x)sin(y)+isin(x)cos(y) +sin(x)sin(y) }{cos^2(y)+sin^2(y)}\\~\\ =\frac{cos(x)cos(y) +sin(x)sin(y) +i[-cos(x)sin(y)+sin(x)cos(y) ] }{1}\\~\\ =cos(x-y)+isin(x-y)\)

ALSO

\(\frac{e^{ix}}{e^{iy}}\\ =e^{ (ix-iy) }\\ =e^{ i(x-y) }\\ =cos(x-y)+isin(x-y)\)

I must be almost finished but i don't get what the question actually wants.   

LaTex

\frac{e^{ix}}{e^{iy}}\\
=\frac{cos(x)+isin(x)}{cos(y)+isin(y)}\\

=\frac{cos(x)+isin(x)}{cos(y)+isin(y)}*\frac{cos(y)-isin(y)}{cos(y)-isin(y)}\\~\\

=\frac{cos(x)cos(y)-icos(x)sin(y)+isin(x)cos(y)    +sin(x)sin(y) }{cos^2(y)+sin^2(y)}\\~\\

=\frac{cos(x)cos(y)  +sin(x)sin(y)  +i[-cos(x)sin(y)+sin(x)cos(y) ]    }{1}\\~\\

=cos(x-y)+sin(x-y)

\frac{e^{ix}}{e^{iy}}\\
=e^{ (ix-iy) }\\
=e^{ i(x-y)   }\\
=cos(x-y)+isin(x-y)

one post = one question

I do not know what intersects perpendicularly means.  Perpendicular to what?

Here are your options:

Courtesy of this site:              http://www.ontrack-media.net/gateway/algebra2/g_a2m3l15s1.html    

66^2/12 = 363R0     (1)

363/12=30R3     It has a remainder so it will have a non-zero number in the 12 place

So 1

I think that method is correct

Hi MPS

I think you might do better if you report this with the report button at the bottom of the screen

I think admin will be much more likely to see it then.

I get 630

Rotations and reflections are classed as different.

If you posted this earlier, please post a link to the original.

I have commented on 2 of your questions before.

You ignored me, or at least did not respond (yet) both times.

So will I think about this newest question? 

Maybe I cannot answer it anyway...  Maybe you will never find out.

Most likely someone else will answer it for you.

Please do not post chunks of your homework here and expect people to do it for you.

One question = one post.