Problems posting (internal server error) ... corrected last step in answer below.... ~EP
find f^-1 = y = x+1 solve for x and then switch x and y's x = y-1 so f^-1 = y = x-1
find g^-1 y = 2x solve for x x = y/2 switch y's and x's y = x/2 = g^-1
Now f(5) (this is the original f(x) function....not thte inverse) f(5) = 5+1 = 6
then g(6) = 2x = 2(6) = 12
then f^-1 (12) = 11
then f^-1 (11) = 10
then g^-1 (10 ) = 5
then f (5) = 6
11,111,111,111,111,111,111- 2,222,222,222 = 11111111108888888889
Multiply the given by (4 - i) / (4- i)
= 8 -2i +4i - i^2 / 16 -4i + 4i - i^2
= 8 + 2i +1 / 16 + 1
9 + 2i / (17)
9/17 + 2i/17
Free time = 40 minutes
charge for extra minutes = 40 cents/min
x = original cost
price = 20x
New cost = x- 1.5
new total price is still 20x
30 * (x-1.5) = 20 x
30 x -45 = 20x
10x = 45
x = 4.5
20 (4.5 ) = 90
I think the answer would be '1' H.L.....and it should read 'a quarter OF a dozen' 'a quarter a dozen' sounds like a price, huh?
Number of 3 digit integers = 999-100 +1 = 900
900/6 = 150 of them are divisible by 6
9 x 10 = 90 of them end in 6
every third one is divisible by 6 so 30 30/90 = 1/3 of the three digit numbers that end in 6 (given) are divisible by 6
You are correct....I originally had 01 in there, but it got deleted while I was formatting my answer before posting.....fixed it...THANX ! ~EP
9^0 01
9^1 09
9^2 81
729
6561
59049
531441
4782969
43046721
387420489
3486784401
31381059609
Lo0ks like if even the number ends in 1
the tens pattern as 0 8 2 6 4 4 6 2 8 0 2004 would be the 5th in the sequence
1 + 6 = 7 My guess