Melody

I suspect guest two is the one that is positing many one line incorrect answers.

Maybe he/she thinks it is funny. 

\(\frac{ax^{13}+......}{d(x)}=(bx^{7}+.....)+\frac{cx^4+....}{d(x)}\\ so\\ ax^{13}+...... =[(bx^{7}+.....)*d(x)]\;\;+\;\;cx^4+... \)

For this to work, d(x) would have to have a  degree of 6

because x^7*x^6 = x^13  and that will be the highest power on both sides.

So the remainder has no relevance.

coding:

\frac{ax^{13}+......}{d(x)}=(bx^{7}+.....)+\frac{cx^4+....}{d(x)}\\
so\\
ax^{13}+...... =[(bx^{7}+.....)*d(x)]\;\;+\;\;cx^4+...

Sorry Ginger, I did not even realise there WAS a question 2. 

I had no idea what you were on about.

Your answer is probably perfect.

There is truth to that Chris   ,

            although I strongly believe that a teaching part answer is better than a full answer.

However I was offering a genuine warning.

There is a guest who is putting up incorrect short answers, presumably on purpose.

Why doesn't it?

How do you know?  You should justify such a claim.

Maybe Ginger but I am not convinced.

No, you do not need to.

There are lots of isosceles triangles in there.

You can label all the angles in terms of x.

You can use

1. base angles in isosceles triangles,

2. angle sum of triangle

3. supplementary angles

I thank that is all you really need to use.  

I already answered this a couple of days ago.

And I see EP found an even earlier answer of mine.

Half the infor was missing then.  On scanning this new question, some of the required info is still missing.

You did not even bother to comment on my comprehensive answer!

I guess you are just another brat here who reposts without comment when they do not get the answer that they want.

Here was my second answer

Yes that is correct.

y=sinx    (or y=cosx)   has a wavelength of    2pi

The symbol for wavelength is  lambda.    so        \(\lambda=2\pi\)

y=sinx =sin(1x)      wavelength (lambda)= \(2\pi\)

y=sin(2x)    wavelength = \(\frac{2\pi}{2}=\pi\)

\(y=sin(\frac{x}{4})  =sin(\frac{1}{4}\pi)\)         wavelength       \(\lambda=2\pi\div(\frac{1}{4})=8\pi\)

So, for cos and sin graphs,  the wavelenth will be 2pi divided by the coefficient of x

----------------------------------------------------------------------

For the next two questoins, do a freehand sketch BEFORE you check using Desmos

If

1)   y=asin(kx)   then what  is  (answer in terms of a and k)

a)amplitude

b)midline

c) yintercept

d) wavelength

--------------------------------------------------------------------------

If

2)     \(y=4-3sin(\frac{x}{4})\)   

a) What is the coefficient of x, I mean what number is x multipled by?

b)amplitude

c)midline

d) yintercept

e) wavelength

All give you a good starting point.

Let  angle b = x

Now you can get every angle there in terms of x.

Start with the most obvious and work from there.

It is not all that hard.

eventually you will have 2 or 3 angles all in terms of x that add to give 180 degrees.

Then you will just have a little equation to solve to get x.

When you finish you should check that it works,

Please no one else give a full answer.

Jenny, you can interact with me if you need/want to.