CPhill

If he had committed suicide.....the tape player would have still been running

The first one is....... the letter "e"

 The second one is....the bus driver was walking......

Aziz......I brought my old (2003) desktop back from the dreaded BSOD twice by using my recovery disks.....after awhile though, these old computers just aren't worth the trouble.....this new laptop I have is miles better than that old HP......

Nah...I'm not sure, DS.....

Sorry!!

25 are left...he still has the dead one.....

But the riddle is supposed to be.......If a farmer had 20 sick sheep and one dies, how many are left?? Answer 25......The joke is that "20 sick sheep" sounds like "26 sheep"

OK....it's corny.......

By never using big words like "cacography"

f(x+h) = [(x + h)^2 - 2(x + h)] = [x^2 + 2xh + h^2 - 2x - 2h]

So we have

[f(x+h) - f(x)]/h =

[(x^2 + 2xh + h^2 - 2x - 2h) - x^2 + 2x]/ h =

(2xh + h^2 - 2h) / h  =

h (2x + h - 2) / h =

(2x + h - 2)

And if we let h → 0, then we have

2x - 2

And this is the "derivative" of f(x) = x^2 - 2x

(5+9z^3-3z^4)-(7z^4+4)-(6z^3+8z^2)    simplify, fitst

5 + 9z^3 -3z^4 - 7z^4 - 4 -6z^3 - 8z^2   collect like terms

 -3z^4 -7z^4 + 9z^3 - 6z^3  -8z^2 + 5 - 4   simplify again

-10z^4 + 3z^3 -8z^2 +1

Yes, xvxvxv, e is correct for the second question.

As to your hypothesis...let's see if this it is true that cos(n*pi + x) = cos(x)

Using a trig identity, we have

cos(n*pi + x) =

cos(n*pi)cos(x) - sin(n*pi)sinx  =

cos(n*pi)cos(x)

Notice that this equals cos(x) only when n = 0, 2, 4, 6......

2du/dt=u^2, u(0)=1

Separating the variables, we have

1/(u^2) du  = (1/2)dt       Now integrate

∫u^-2 du = ∫ (1/2) dt

-1/u = (1/2)t + C    so we have

1/u = -((1/2)t + C) = -(t + 2C)/2

u(t) = -2/(t + 2C)

And

u(0) = 1

So

-2/(0 + 2C)  = 1

-1 = C

So we have

u = -2/(t - 2)   or

u = 2/(2 - t)