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Use SUVAT for constant vertical acceleration. Solve for t. S = -10U = 0V = -A = -gT = tS=UT+(AT^2)/2t = sqrt(20/g)

x=0.22222222...10x= 2.222222...10x-x=9x=2x=2/9=0.2222222...
x=2/9

Use substitutionlet u = cos(x)du = - sin(x)dxint(tan(x)) dx = int(sinx/cosx) dx = - int(1/u) du = - ln(u) + c= ln(secx) + c

log2(x + 3) + log2(x - 3) = 4log2 [(x + 3)(x - 3)] = 4x^2 - 9 = 16x^2 = 25x = 5

We will use the cross product of these two vectors to help us find the perpendicular vector. The cross product works by finding a vector which has no effect on the position of a point with respect the the...