CPhill

In general,  (d/dx) (a^x)  = (ln a)* a^x * (the derivative of the exponent on "a").....so......

(d/dx) 10^-x =  ln(10)* 10^-x * (-1)  = -(ln10)10^-x     .....and.....

(d/dx) [- ln(10)* 10^-x ] =  -(ln10) * (ln 10)* 10^-x (-1)  =  (ln10)^2* 10^-x

So.......the "nth" derivative would be

(-1)^n * (ln 10)^n * 10^-x

(tan - sin)^2 + (1 - cos)^2 = (1 - sec)^2

tanx^2 - 2tanxsinx + sin^2x + 1 - 2cosx + cos^2x =  1 - 2secx + sec^2x =

tan^2x - 2tanxsinx + 1 + sin^2x + cos^2x - 2cosx  = 1 + sec^2x - 2secx

tan^2x - 2 tanxsinx  - 2cosx + 1 + 1 =  1 +  sec^2x - 2secx     [subtract 1 from each side]

tan^2x + 1 - 2tanxsinx - 2cosx  = sec^2x - 2secx            [tan^2x + 1 = sec^2x]

sec^2x - 2tanxsinx  - 2cosx  = sec^2x - 2sec     [ subtract sec^2x from both sides, add 2cosx to each side]  

-2tanxsinx  = 2cosx - 2secx     [divide both sides by -2]

tanxsinx = secx - cosx

(sinx/cosx) * sinx  =  secx - cosx

(sin^2x)cosx  = (1/cosx) - cosx           [sin^2x = 1 - cos^2x ......get a common denominator on the right]

(1 -cos^2x)/cosx  = (1 -cos^2x)/cosx

a(1+x^2)+2bx+c(1-x^2)  =

a + ax^2 + 2bx + c  - cx^2  =

(a - c)x^2 + 2bx + (a + c) =       (√(a - c)x + √(a + c) ) ( √(a - c)x + √(a + c))

This implies that 2√(a^2 -  c^2)x  =  2bx    →  √(a^2 -  c^2)  = b

So, one solution is  that any "Pythagorean Triple"  for a, b and c   would work   [However, we might - or might not - have "rational" coefficients ]

For example, let a = 5, b= 4 and c = 3

Then we have:

(a - c)x^2 + 2bx + (a + c) =

(5 - 3)x^2 + 2bx + (5 + 3)  =

2x^2 + 8x + 8 =

(√2x + √8) (√2x + √8)  = (√2x + √8)^2

The sum of the first "N" positive odd  integers = N^2

There are 13 odd imtegers between 1 and 26, inclusive.....and there are 4 odd integers between 1 and 8, inclusive

So....the sum of the odd integers between  8 and 26  equals the difference in the above, and we have....

13^2  - 4^2  =   169  - 16  =  153

This would be a difficult one to solve, algebraically.....here's a graph of the solution......

Let x be the number of match boxes that Fred started with....then, Joe had 432 - x

And we know that :

x  - (2/5)x + (1/4)[(432 -x) +(2/5)x] = [ (432 -x) + (2/5)x - (1/4)[(432 - x) +(2/5)x]   simplify

x - (2/5)x + 108 - (1/4)x + (1/10)x = 432 -x + (2/5)x - 108 +(1/4) x - (1/10)x

108 + (9/20)x = 324 - (9/20)x

(18/20)x = 216

(9/10)x = 216       multiply both sides by (10/9)

x = 240    →     this is what Fred started with

Happy " I"  Day, rosala.......are there going to be a lot of festivals???

These graphs are always a nice addition to the answers.........thanks, Melody

Let a, b, c, d and e be the digits in that order...so,we have the following system:  

c + e = 14  

d - b  = 1

2b - a = 1 

b + c = 10

a + b + c + d + e  = 30    (5)

Notice that 

c + e = 14

c + b = 10      subtracting the second equation from the first, we have

e - b = 4    → e = b + 4

And c = 10 - b

And d = 1 + b

And a = 2b - 1

And substituting all these into equation (5), we have

(2b - 1) + b + (10 - b) + (1 + b) + (b + 4) = 30    simplify

4b + 14 = 30     subtract 14 from each side

4b = 16  → b = 4

Then

c = 10 - 4 = 6

d = 1 + 4 = 5

e = 4 + 4  = 8

a = 2(4) - 1 = 7

And the code is:

abcde = 74658

Just as Anonymous found!!!!!

We hate to see you go, jillybean....

Please visit when you can!!!!!

Sincerely, CPhill