CPhill

Let s+1 = u

Then s = u -1

And ds = 1 du

So we have

4pi∫(u - 1)/u³ du =

4pi [∫1/u^-2 du - ∫ u^-3 du ] + C =

4pi [ (-1)u^-1 - (-1/2)u^-2 ]  + C =

4pi [ 1/(2u²) - 1/u ] + C =

2pi [ 1- 2u]/ u² + C =   [back substitute ....( u = s+ 1) ]

2pi [ 1 -2(s + 1) ]/ (s + 1)² + C

2pi [ -(2s + 1) ] / (s + 1)² + C =

-2pi (2s + 1) / (s + 1)² + C

The probability of at least one red one is...... 1 - P(no red selected). In essence, we're just counting the number of possible sets of green marbles only divided by the number of total possible sets and subtracting this result from 1.

So we have......

1 - C(6,3) / C(10, 3)  = 5/6 or about 83.333%

Thanks for that answer, Alvarez......nicely presented......!!!

ND !!!!!!!....glad you dropped in........we've been REALLY missing you around these parts  !!!!

(We also had a visit by Golden Leaf just a while ago.......it's like a "class reunion")

If we are not using rotational matrices, we can solve this using basic trig. Note that the angle between the x axis and a line drawn from the origin to the point (2, 5) is given by:

tan^-1(5/2) = about 68.2°

And the length of this line (i.e., r) is given by √(2² + 5²) = √(4 + 25) = √29

And rotating this line by 45° gives us ( 68.2° +  45°) = 113.2°

So the coordinates of the new point are given by [r * cos(113.2°), r * sin(113.2° ] = [√29*cos(113.2°), √29*sin(113.2° ] = 

(-2.12, 4.95)

And that's it......!!!!!

Assuming that we know one of the sides (other than the hypoteneuse, of course), we can use this....

h = s* √2    where h is the hypoteneuse and s is the side length of one of the legs

Notice that the "divide and average" method is very simple.......the only problem with the method - and some others, as well - is that the numbers generated are difficult to work with........unless you have a calculator......!!!!

Take the log of both sides.....so we have

log x^0.4 = log 4.065   ...... and by a log property...... logx^a = (al)ogx      ......so we have

0.4log x = log 4.065       divide both sides by 0.4

log x  = log 4.065/0.4       and, in exponential form, we have

10 ^{log 4.065/0.4}  =   x   = about 33.316

All the possible rational real zeroes are given by all the factors of p divided by all the factors of q where p is 3 and q is 2.

So we have  ±3  ±1  ±3/2  ±1  ±1/2  

And, as Alan found, the only rational real is 1/2.

Notice that the Rational Zeroes Theorem doesn't guarantee any real rational roots. It just says that, if there are any, they will come from the p/q list.