Melody

Well at least you are honest enough to openly admit that your whole point in being here is that you are cheating.

Most people here give their time for free becasue they are generous people who want to help others to learn.

Yes I find YOU offensive and I find it offensive that some people here want to faciliate cheating!

Sounds like you passed because CPhill gave you the answers.

What a clever boy you are, you are learning to cheat your way through life.  What a lot we can learn from the internet!

Students always blame teachers, the only thing wrong with your teacher is the lazy, cheating student that he /she has.

CPhill has a different attitude to me but I see a list of questions llike this and it seems to me that this is probably a test and you are asking us to take your test or do your assignment for you.  Even if this is not the case you are certainly not asking for anyone to teach you anything!

I would not do any of them and I know many other people here would feel the same way.

You re not asking anyone to teach you anything - you are just asking us to help you cheat!

\(y=a* cos[n(\theta+p) ]+ L\)

Amplitude =a

phase shift = p (units in the NEGATIVE direction - opposite direction to what most people expect)

wave length  \(\lambda   = \frac{2\pi}{n}\)

L is the vertical shift   

SO CONSIDER

\(y=1.8cos(3\theta+\frac{\pi}{2})-1.5\\ rewrite\;\; as \\ y=1.8cos(3[\theta+\frac{\pi}{6}])-1.5\\\)

It has the basic \(y=cos(\theta) \)shape.

wavelength = \(\frac{2\pi}{3}\)

Phase (horizontal) shift =\( \frac{\pi}{6}\;\)units  in the negative direction

Amplitude =1.8

Vertical shift is 1.5 units DOWN

check

Here is the graph.

You can play iwth the circles on the left to see how I 'developed' the graph 

Why do you give guest a thumbs down Jonathon. He /she is right!

You did not even offer an appology on or for your former post! 

Why should you have appologize I hear you ask .. Because you wasted the time of everyone who read it!

Rewrite the expression \( 6j^2 - 4j + 12\) in the form \( c(j + p)^2 + q\), where c, p, and q are constants. What is \(\frac{q}{p}\)

\(​​​​6j^2-4j+12\\ =6(j^2-\frac{2}{3}j)+12\\ =6(j^2-\frac{2}{3}j+\frac{1}{9})+12-\frac{6}{9}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+\frac{34}{3}\\~\\ p=\frac{-1}{3}\qquad q=\frac{34}{3}\\ \frac{q}{p}=\frac{34}{3}\div\frac{-1}{3}\\ \frac{q}{p}=\frac{34}{3}\times\frac{3}{-1}=-34\\ \)

How many integers between 200 and 300 have the sum of their digits equal to 15?

First decide how many combinations of 2 digits make 13

So the only numbers are 

249

294

258

285

267

276

So that makes  6 integers

You know you can (and should) give yourself points EP. :)

So should Hectictar ;))

Sorry, I did not think about the question enough :(

Thanks EP :)

So that is 7 200 000 000 people in 27,878,400 square feet

You want to know how many square feet / person.     (area divided by number of people)

Just use the units to help you.

27,878,400 / 7200000000      (sqrare feet per person)

Now just do the division

27,878,400 / 7200000000 = 0.003872 square feet/person