62.8 = pi * (2)^2 * h divide by pi * (2)^2
62.8 / [ pi * (2)^2] = h ≈ 5 cm
I'm assuming all positive integers....to see why this is true.....note.....
5! = 1 * 2 * 3 * 4 * 5 = 120 ........this minimizes the product of the 1st five integers
So...... 5! * 1000 = 120000 [ any integer > 1000 will produce a product > 120000 ]
This is known as the "pigeonhole" principle
To absolutely guarantee that you have at least 2 left-pointing hands, you would need to select - at a maximum - 22 of the 30 hands......of course.....you might not have to select that many.....but selecting 22 will guaranee at least two.....
1,000,000 / 20 =
10^6 / [2 *10] =
10^5 / 2 =
[2 * 5 ]^5 / 2 =
2^4 * 5^5 =
16 * 3125 =
50000 bags
BTW....10^6 and 20 are not "co-prime"......LOL!!!!!
ln(|8|)-ln(|1|) =
ln(8) - ln(1) =
ln(8) - 0 =
ln (8) ≈ 2.079
2.5 [ .28 ]
.112
25 [ 2.80
2 50
30
25
50
0
5 / 5^5 =
5^1 / 5^5 =
5^(1 - 5) =
5^-4 =
1/ 5^4 =
1 / 3125
Parallel lines have equal slopes.....so slope of w = -5 / 7
8 3/4% = .0875
So
.0875 * 575 ≈ $50.31
4u^3(2u^2+5u) = 8u^5 + 20u^4
4mn^2(3m^2 - 7mn + n^3) = 12m^3n^2 - 28m^2n^3 + 4mn^5
