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Answers by Michael F.

[sec(π/4)cos(2π/3)]/[tan(π/6)/csc(3π/4)]. =cos(2π/3)/[cos(π/4)tan(π/6)sin(3π/4)] =-.5/[(√2/2)(√3/3)√2/2)]=-3/√3=-√3 The only tricky parts are finding the reference angles for 2π/3 and 3π/4

D. x=sin2y dx=2sinycosydy √x/√(1-x)=siny/cosy integrand is (siny/cosy)2sinycosydy=2sin2ydy at x=0 y=0 and at x=1/2 y=π/4 ∫2sin2ydy from 0 to π/4

word problems (answer)

Original fraction 5d/d. Adjusted fraction  (5d+12)/(d+12)=2 5d+12=2d+24,  3d=12, d=4 20/4 original fraction, 32/16 when 12 added to both numerator and denominator

Probability (answer)

a.       4C2/52C2=1/221  two aces out of four/choose 2 cards from deck   b.      48C2/52C2=188/221  two cards not aces/choose 2 cards from deck  

Distance= Rate times time and so time=Distance/Rate time=85/√10,800 if you like rational denominators time=85√10,800 /10,800

proportions (answer)

7.5/42=10.9/e where e is the length of the elephant's shadow 7.5e=42×10.9 e=42×10.9/7.5=61.04

5x^2 + 7x = 6 (answer)

The quadratic equation 5x2+7x-6=0 has discriminant 72-4(5)(-6)=49+120=169, it has both its roots real. Since we are so near the solution, the quadratic formula gives x=(-7±13)/10=-2, 6/10 N.B. 5×4+7×(-2)-6=0 as does 5×(3/5)2+7×3/5-6=5×9/25+21/5-6=0  

e=number of ounces of 11%Cu, t=number of ounces of 24% Cu      e+    t=130                 total weight of alloy .11e+.24t=.16×130=20.8 total weight of Cu Rewrite as    ...

x^1/2-5x^1/4+6=0 (answer)

x1/2-5x1/4+6=0 let u=x1/4.  The equation now reads u2-5u+6=0, a quadratic.  Factor it and get (u-3)(u-2)=0 or u=3, u=2 Solving for x we have x1/4=3→x=81 x1/4=2→x=16  

Left side (1+secθ)/secθ = (1+1/cosθ)/(1/cosθ) multiply numerator and denominator by cosθ to get (cosθ+1)/1=1+cosθ   Right side sin2θ/(1-cosθ)=(1-cos2θ)/(1-cosθ)=((1+cosθ)(1-cosθ))/(1-cosθ)=1+cosθ  We are done.

For each x in the equation substitute (x-18). For each y in the equation substitute (y+10).   e.g.  y=x2 is a parabola through (0,0) y+10=(x-18)2 is that same parabola translated to the right 18 units and down 10 units.