Thanks, GA....I didn't think about going the "log route"...that's certainly more efficient !!!
Oh well.....I guess we can teach some pigs to fall off a "log"...others???.....Mmmm.....
Original volume.... = pi * r^2 * h = pi * (20)^2 * (20) = 8000pi cm^3
Adding 1500xm^3...we have that
[8000pi + 1500] = pi * (20)^2 * h
[8000pi + 1500] / [ pi *(20)^2 ] = h ≈ 21.19 cm
The water rises ≈ 1.2 cm
[5 - (2 - 2(6 - 3) + 6)] + 2 =
[5 - (2 - 2(3) + 6)] + 2 =
[5 - (2 - (6) + 6)] + 2 =
[ 5 - (2 - 0) ] + 2 =
[ 5 - 2] + 2 =
3 + 2 =
5
3:5 means that there are 8 equal parts to the suppplementary angle....and the smaller one is 3/8 of this....so we have
(3/8)(180) = 67.5°
[8/3] [1/4] = 8 / 12 = 2 / 3 cups of sugar in the cake
[2/3] [3/10] = [ 2 / 10] [ 3 / 3] = 2 /10 * 1 = 2/10 = 1 / 5 cups eaten
x^2-4x=-3
x^2 - 4x + 3 = 0
[ -b ± √ [b^2 - 4ac ] ] / 2 a = 1 b = - 4 c = 3
x = [ 4 ± √ [(-4)^2 - 4*1*3 ] ] / 2
[ 4 ± √ [16 - 12 ] ] / 2
[ 4 ± √ [4 ] ] / 2
[ 4 ± 2 ] / 2 = [4 + 2] / 2 = 6 / 2 = 3 or [4 - 2] / 2 = 2 / 2 = 1
Factor 85184 = 2^6 * 11^3 = [ 2^2 *11] [2^2 *11] [2^2 *11] = [ 2^2 * 11]^3
Factor 1936 = 2^4 * 11^2 = [2^2 *11] [2^2 * 11]= [2^2 *11]^2
So
85184 = ( [2^2 *11]^2])^(3/2) = 1936^(3/2)
So we have
[ 1936^3/2] ^ 85184 = 1936^127,776
Opposite side = 70 sin (52°) ≈ 55.16
BTW ...it's "hypotenuse" not "hypothesis"
7^[5x] * 49^[x+5] / 343^[2x]
7^[5x] * (7^2)^[x+5] / (7^3)^[2x]
7^[5x] * 7^[2x + 10] /(7)^[6x]
7^[ 7x + 10] / 7^[6x]
7^ [7x - 6x + 10]
7^[x + 10]
216^[2n+1] * 36^[5-4n] = 1296^[11n+17]
(6^3)^{2n + 1] * (6^2)^[5-4n] = (6^4)^{11n + 17]
6^[6n + 3] * (6)^[10 - 8n] = (6)^[44n + 68]
6^[ 13 - 2n] = 6^[44n + 68] solve for the exponents
13 - 2n = 44n + 68 rearrange
-55 = 46n → n = -55 / 46
