Melody

Thanks for answering BuilderBoi,

I'd like you to think about your presentation.

In my oppinion if you spread your ideas out they would be easier to comprehend.

'Blocks' always look intimidating to me.

Just a suggestion.

Consider the polynomial f(x) = 1 - 12x + 3x^2 - 4x^3 + 5x^4 and g(x) = 3 - 2x - 6x^3 + 19x^4. Find c such that the polynomial f(x) + cg(x) has degree 3.

\(f(x) = 1 - 12x + 3x^2 - 4x^3 + 5x^4 \)

\(g(x) = 3 - 2x - 6x^3 + 19x^4\)

Looking just at the x^4 term

   \(f(x) + cg(x)=..... 5x^4+c*19x^4\\ f(x) + cg(x)=..... (5+19c)x^4\)

if the x^4 term is to disappear then    5+19c=0     etc

Let's see

\(\sqrt{60x}*\sqrt{12x}*\sqrt{63x}*\sqrt{42x}\\ =\sqrt{60}*\sqrt{12}*\sqrt{63}*\sqrt{42}*\sqrt{x^4}\\ =2\sqrt{3*5}*2\sqrt{3}*3\sqrt{7}*\sqrt{7*6}*x^2\\ =2*2*3 *\sqrt{3*5*3*7*7*6}*x^2\\ =12 *3*7*\sqrt{5*6}*x^2\\ =12 *3*7*\sqrt{5*6}*x^2\\ =252\sqrt{30}\;x^2\)

There you go BilderBoi,    we are in full agreement   

Have you drawn the pic yet?

The angle that they have given you is not actually in the triangle.

Depending on which angle of the triangle that you use you will need to use either sin or cos to get the answer.

If you want to get something right you must display it in a form suitable for the media.

Are you incapable of comprehending that?

It says, in Latex, (you haven't bothered to write it in English) "In the digatram below"

There is no diagram, it is just a bunch or unrendered code.

I would not bother with this if I were you.  Stick to 2+2.  You should not try to move on too quickly.

I still don't know what you are talking about.

"REDA REDB is one hand of two cards. "

How is this one hand of two cards.  There are 2 lots of 4 letters here.  It makes no sense.

Thanks for responsing though.  I gave you are point just for responding.

What is the $ sign for?

Your question is also rubbish.

So I suppose it is a  reasonable answer.      Garbage in, garbage out.

What quadrant is 133 degrees in?

Is tan pos or neg in this quadrant? 

Which other quadrant has the same sign?

What is the acute angle equivalent of 133 degrees?

So what is the answer?

Get back to us with what you can do for yourself.

Thanks very much Alan, It is nice to get confirmation that our answers are most likely correct. :))