cos(x)=15/17 so x = arccos(15/17) tan(76) = 13/x so x = 13/tan(76)
dmA/dt = cmB^2/mA where c is a constant mB=m-mA so dmA/dt = c (m-mA)^2/mA = (c/mA)(mA^2-2mmA+m^2) = cmA-2cm+cm^2/mA integrate with respect to t
sin(a+b)/cos(a)cos(b) = = (sin(a)cos(b)+cos(a)sin(b))/cos(a)cos(b) = sin(a)cos(b)/(cos(a)cos(b) + cos(a)sin(b)/(cos(a)cos(b) = sin(a)/cos(a) + sin(b)/cos(b) = tan(a) + tan(b)
It would take 10 minutes.
Moles of solute = 16 g / 60 g/M = = 0.267 M volume of H2O = 39g/1.3 g/ml = 30 ml = 0.03 l molarity = 0.267 M / 0.03 l = 8.89 M/l
the probability that the 1st marble is red is 10/13. So, the probability that both marbles are red is (10/13)^2 or (100/169)
(a) 3y = x - 10 substitute in the 2nd equation 3(x-2) = x - 10 3x - 6 = x - 10 simplifing... 2x = -4 or x = -2 and y = -4 this is a unique solution (b) this system has infinitely many solutions and cannot be solved without more information. 3x + 3y = 15 divide equation by 3 (3...
y = (x-2)^2 y = x^2-4x -4 dy = 2xdx - 4dx so, dy/dx = 2x-4 dx/dt = 2 so, dy/dt = dy/dx dx/dt = 2(2x-4) = 4x-8 for x=1 then dy/dt = 4(1)-8 = -4 units per second
dx/dt = 2 inches/min V = x^3 so, dV/dt = 2^3 = 8 cubic inches/min
Area = A x B Total length = 3000 = A + 2B B = (3000-A)/2 so, Area = A x (3000-A)/2 = = 3000A/2 - (1/2)A^2 take 1st derivative 0 = 1500 - A and solve for A A = 1500 and B = 750
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