Melody

If it takes Mike and Joshua 2 hours to do a certain job, and it takes Mike and Kelvin 3 hours to do the same job, how long would it take for Mike, Joshua and Kelvin to do the job if all 3 of them worked together?

I got in a mess trying to solve this alsgebraically so I decided to takle it with an example.

Let mike and Josh both do  1/4 of the job in an hours so together they do 1/2 the job in an hour which is 1 job in 2hours.

Now Mike and Kevin together do 1/3 of a job in an hour.   so 1/4 + K = 1/3       K=1/12 of a job in an hour

so together 

M+J+K = 1/4 +1/4 + 1/12 = 7/12 jobs in an hour   = 12/7 hours for a job = aprrox  1 hour and 43 minutes

2nd  then   tan

or just type in     atan

Please would be a big improvement on Quick!  

You are lucky I have answered!

Suppose that 6 kids are put into a queue. However, we know that two of them are especially naughty and they must be separated. In how many ways can the queue be arranged?

How many ways can 6 children be arranged?     6!

I'm going to tie the two naughty ones together, Now how amny ways are there.   5!  But either one could be first so that is 2*5!

so the number of ways that they are not together is    6! - 2*5!  

6!-2*5! = 480 ways

33177600000000 ^ 2376000

\(let\\ y=33177600000000 ^ {2376000}\\ logy=log33177600000000 ^ {2376000}\\ logy=2376000*log33177600000000\\\)

2376000*log(33177600000000) = 32125527.641227103424

\(logy=32125527.641227103424\\ 10^{logy}=10^{32125527.641227103424}\\ y=10^{32125527}\times 10^{0.641227103424}\\ y=10^{0.641227103424}\times 10^{32125527}\\ y=4.377509562\times 10^{32125527}\\ 33177600000000 ^ {2376000}=4.377509562\times 10^{32125527}\\\)

WE need more information.

Prove that cos(7 1/2 ) degree= squareroot6+sq. root 4+sq.root 3+squ. root2

\(Prove\;\; that \\\\\:cos\left(7\frac{1}{2}\right)^{\circ }=\:\sqrt{6}+\sqrt{4}+\sqrt{3}+\sqrt{2}\\\)

I can see straight off that these are NOT equal because the right hand side is obviously more than 1 and the left hand side cannot be more than one!

mmm ok...

thanks for responding :)