CPhill

Tan(x)  = 0    at  x =  0, pi, 2pi, 3pi, 4pi, 5pi and 6pi

So, dividing these by 3 we have

Tan(3x)  = 0   at x = 0, pi/3, 2pi/3, pi, (4/3)pi, (5/3)pi and 2pi if we constrain x to lie in [0, 2pi ]

A more general solution is that x = (pi/3)n   where n is an integer

Yep....and tomorrow after your test...you can come on here and brag about how easy it all was....!!!

Well...if you're good at reading comprehension, and you can pass the grammar part....what's left??

You have probably done enough studying....after all...if you know what 2 + 2 is, why worry about it???

Besides...I find that 99% of the stuff you worry about doesn't even happen.....and the 1% that does happen...well you can't plan for that, anyway...so.....why worry??....you will just "psych" yourself out.....!!!

Just relax.....you will do fine....!!!

Rosala, this person just wants to know this...

√pi = √(3.13159....) = about 1.7725

1 / 41    has the recurring pattern 5 digit pattern of ..   02439

And since 11,110 is evenly divisible by 5, the digit you are looking for is "9"

log(3) / log(1.009) = about 122.62

I was just saving the best for last.....!!!

The number of total sets possible is C(17,3)  = 680

But, we're only interested in those that contain 2 blue b***s selected from 5 and one ball selected from the other 12  =  C(5,2)C(12,1)  = 120 sets

So the probability is    120 / 680   =  12/68  = 3 / 17 = about .1765  or 17.65%