Let P = the perimeter = 4S
Area of square = S^2
S^2 = 4 (4S )
S^2 = 16S
S^2 - 16 S = 0
S (S - 16) = 0
S = 16
Diameter of 10 = radius of 5
pi ( 20^2 - 5^2) =
375 pi (in^2)
2^3 * 4^5 * 6^7 * 8^9 =
2^3 * (2^2)^5 * 2^7 * 3^7 * (2^3)^9 =
2^3 * 2 ^10 * 2^7 * 2^27 * 3^7 =
2^47 * 3^7
We need 2 * 3 = 6
CORRECTED !!!!
Volume of 8 steel balls =
8 * pi (5)^2 = 200pi
200 pi = pi (r^2)
200 = r^2
10sqrt (2) in = r
Also.....notice that if angle AOB = 60
Then angle AOX = 30 = angle BOX
And angle OAB = Angle BOA = 60
So triangle AOB is equilateral with AB = 20
Thx, BB.....I didn't read this one correctly !!!!
Taking this from
x^2 = 3y^2 + 2xy
x^2 - 2xy - 3y^2 = 0
(x - 3y) ( x + y) = 0
x + y = 0 not possible
x - 3y = 0
x = 3y
y / x = 1 / 3
NVM !!!
(3/4)N -10 = 14
(3/4)N = 24
N = 24 (4/3) = 32
2x > 3x - 3 8x - a > -6 + 2x
3 > 3x - 2x 8x -2x > -6 + a
3 > x 6x > -6 + a
If x = 2 then
6(2) > -6 + a
12 + 6 > a
18 > a
17 positive integers
