Math
4+5+3+3 = 15 blue = 5/15 or 1/3 so 1 - blue = 2/3 so C yes 12 = 2*2*3 9 = 3 *3 so rulers 2*2*3* (3) = 36 paper 3*3* (2*2)= 36 so 3 of rulers and 4 of paper so B
College math
(2/3)24 = 2 * 8 = 16 check 16/24 = 4/6 = 2/3
incomple problem statement
No problem is given to solve.
Math
Which is in order from least to greatest A.1/4,0.33,2/5,0.5,2/3 Yes A is .25 , .33, .40, .50, .67 B.1/3,1/2,0.4,0.25,2/3 C.1/3,1/4,0/4,1/2,0.66 D.2/3,0.5,2/5,0.33,1/4
Math
3 + 2 = 5 5 * 6 = 30 so 6 teams consisting each containing 3 singing and two dancing 6 * 3 singing groups = 18 singing groups check 30 - 18 = 12 18/12 = 9/6 = 3/2 sure enough
pre calc
3 z^2 - 7 z + 2 = 0 (3 z - 1)(z - 2) = 0 z = 1/3 or z = 2 so if sin x = 1/3 x = .34 radians or pi - .34 = 2.8 radians sin x = 2 has no solutions
pre calculus
sin x + (cos^2 x /sin x) = 1/y sin^2 x/sin^2 x + cos^2 x/sin^2 x = 1/y 1/sin^2 x = 1/y y = sin^2 x
algebra
(3x-2y)(3x-2y) = 9 x^2 - 12 x y + 4 y^2 You can use distributive property to get there if you do not know FOIL 3x (3x-2y) - 2y (3x-2y) = 9 x^2 - 6 x y - 6 x y + 4 y^2 = 9 x^2 - 12 x y + 4 y^2
Physics
I assume that the resistor and the coil are in series I assume you are given the voltage, V In that case, after a long time the current through the inductance is CONSTANT and there is NO voltage across the inductor E = L di/dt but di/dt is zero so only the resistor matters for...
Physics
the cart speeded up so I assume the coal came AT the cart from behind cos 38 = .788 so s cos 38 = .788 s = u = horizontal speed of coal 20 u + 150 (3) = 170 (3.95) 20 u + 450 = 671.5 u = 11.08 s = 11.08/cos 38 = 14.05 m/s
Pages: <<Prev | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | Next>>
For Further Reading
