Melody

Hi Cupcake,

To answer this question you just need to recognise that 

\(n(cos\theta+isin\theta)=ne^{\theta i}\)

Convert them both to this form,

do the division

 then convert the answer back to the original form.   

pi cannot be written exactly, it is an irrational number. 

EP has given you the approximation.  

Hi guest, 

Should I interpret that as 'thanks'   ?

Thanks Chris :)

Your answers are the same as mine.

I think you are just having fun using your C++ programing    

Thanks Chris   

Find the number of integers n that satisfy    \(7 \sqrt{-n^2 + 22n - 21} \le n + 39.\)

Mmm

firstly:

\(-n^2+22n-21\ge0\\ n^2-22n+21\le0\\ (n-21)(n-1)\le0\\ 1\le n\le 21\)

now

\(7 \sqrt{-n^2 + 22n - 21} \le n + 39\\ 49(-n^2 + 22n - 21) \le n^2+78n + 1521\\ -50n^2 +49*22n-78n - 49*21-1521 \le 0\\ -50n^2 +1000n - 2550 \le 0\\ 50n^2 -1000n + 2550 \ge 0\\ n^2 -20n + 51 \ge 0\\ (n-17)(n-3) \ge 0\\ 0\le n \le 3 \qquad or \qquad n\ge17\)

Take the intersection of these restrictions and I get

n can equal    1, 2, 3, 17, 18, 19, 20 or  21

You need ot check this. 

Thanks Heureka,

Another great answer from you.    

Thanks guest,

I should have remembered to use that identity. 

Tiggsy used it recnently and I spent a lot of time making full sense of his answer. 

Thanks for reminding me of it again!