Friday
December 6, 2013

# Posts by FredR

Total # Posts: 108

calculus
180+arctan(12/-5) = 112.6

rotations
a = angular acceleration I = rotational inertia w = angular velocity v = linear velocity Torque = RxFT = Ia so, a = RxFT/I = (R/I)(4t-0.10t^2) w = integral(a) with respect to t v= wR

Algebra 2
It looks right to me. The easiest way to check is to plug the number back into the formula: e^.6931 = 2 2*2-3 = 4-3 = 1

algebra
Permutations of 9 objects taken 5 at a time = 9!/(9-5)! where 4! = 4x3x2x1 and 9! = 9x8x7x6x5x4x3x2x1

ALGEBRA
yes, it could be. You just need to change the sign, so you would have 8-y over y-4 PLUS 7y-9 over y-4

college
dist from station to trains = D ta = time in hours from when A passes station to when trains A & B meet. D = 80ta D = 88(ta-15/60) solve for ta Time at catch-up = 1:20 AM + ta

psychology
NO. 500 = A - A x 0.28 = A (1-.28) = .72A so, A = 500/0.72 = 694

trig 30
csc=1/sin cot=cos/sin so, csc^2-1/cotcsc = ((1/sin^2)-1)sin/(cos/sin) = sin((sin/sin^2)-sin)/(cos) = (sin^2/sin^2-sin^2)/cos = (1-sin^2)/cos = cos^2/cos = cos A ((sin^2/sin^2)-sin/(sin^2 cos)

Pre-Cal
1) it is not correct. The second term would be log3(-2) and you can't take the log of a negative number. 2) it is not correct. The base angle would be 45 degrees, so the cotangent would be 1.

trig
9cos^2(t)+3cos(t)=0 subtract 3cos(t) from both sides 9cos^2(t)=-3cos(t) divide both sides by 3cos(t) 3cos(t)=-1 cos(t)=-1/3 t=arccos(-1/3)

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