think back to your polar coordinates. Same thing. (a+bi) = r cisθ where tanθ = y/x r^2 = a^2 + b^2 -5+5i r^2 = 5^2+5^2, so r=5√2 tanθ = -5/5 = -1 Now, cosθ=x/r sinθ=y/r since x<0 and y>0, θ is in II, so θ = 3π/4 -5 = -5+0i = ...
that would be 18/100
Extrema-Check my work please?
Ah. I see the error. We have a function of x and t, not just t. So, df/dx = f(x,x) + ∫[a,x] ∂/∂x (t^2-4)/(1+cos(x)^2) dt = x^2-4/(1+cos(x)^2) + ∫[0,x] (t^2-4)sin2x/(1+cos(x)^2)^2 dt = x^2-4/(1+cos(x)^2) + (1/3 t^3 - 4t)sin2x/(1+cos(x)^2)^2 [0,x] = x^2-4...
read n sum=0 for (i=1 to n) sum += i
w(w+2)+8 = w(w+4) w^2+2w+8 = w^2+4w 2w = 8 w = 4 The rectangles are 4x6 and 4x8 The areas are 24 and 32, which differ by 8.
just plug it in: y = 4cos(2(30)+90)-3 = 4cos(150)-3 = 4(-cos(30))-3 = 4(-√3/2)-3 = -2√3 - 3
f(x) has two roots it touches the x-axis at x = 1/3, so that's one turning point. It turns again so it can cross the x-axis at x=2 so, 2 turning points
your line has slope -3 perp line has slope 1/3 So, now you have a point and a slope, so the line is y-5 = 1/3 (x+1)
you have a point and a slope, so the line is y-2 = -1/2 (x-4)
6.98/1.86 = 3.75 you are correct
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