CPhill

Melody has given the dot product - a scalar - of these two vectors...

However....if the cross-product was wanted, we have

(-1,-1,3) x (-1,0,2) = (u1, u2, u3) x (v1, v2, v3)  = (s1, s2, s3)

s1 = [-1*2] - [3*0] = -2 - 0  = -2

s2 = [3*-1] - [-1*2]  =  -3 - (-2)  = -3 + 2 = -1

s3 = [-1*0] - [-1* -1 ] = 0 - 1  = -1

The cross-product produces a vector in three-space that is perpendicular to the given vectors, namely..... (-2, -1, -1)

Nice "double-substitution,"  Alan.

P.S -  How did it occur to you to let cos(x/3)  = cos(2Θ)   and sin(x/ 2) = sin(3Θ)  ???

Thanks....Nauseated.....I'll have to plow through that....

Now......

What if N distinguishable b***s were rolled along all the possible "right and up" paths of a 6 x 6 grid  ???

.....eh, never mind....

 We'll let some other SOB figure that one out......

Well...you got most of the difficult part done with the partial fraction stuff....!!!!

That's worth at least 3 points.........

P.S.  - I don't want to appear as though I'm a genius.....I went to the integral tables to find the proper form for that last part.....LOL!!!

Melody....for that last integral.....let u = x , a = √3,   du = dx,  and we have the form.....

-0.75 ∫ 1 / [ u^2 + a^2] du  =

-0.75 [ tan^-1 (u/a)] / a  + C  =

-.0.75 [tan ^-1 (x / √3)] / √3 + C

Thanks, Melody....

Here's the graphical solution that confirms Anonymous' answers

Think about this......

10 =(1/2)bh    multiply both sides by 2

20 = bh    

Note that infinite values work for (b, h)

(5, 4)  (4, 5)  (10, 2) (2, 10), etc......

Thanks, Melody.....!!!!

.01 * 2^52 = 2^52 / 100   = 4,503,599,627,370,496 

In the US, this is about 4.5 "quadrillion"