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r = 4 cscθ then r sinθ = 4. now, r cosθ = x, while r sinθ = y.... these are the equations needed to transform cartesian to polar ... thus you have y = 4. (a horizontal line) §

Let f: R -> R' be an isomorphism. If R is commutative, and a... of R', then a' = f(a) and b' = f(b) for some a,b in R. Then we have: a'b' = f(a)f(b) = f(ab) = f(ba) = f(b)f(a) = b'...

r^2+r=A/(2ph)? If so, then r^2+r-A/(2ph)=0=> r=[-1+/-sqrt(1+2A/(ph)]/2 => r=[-1+sqrt(1+2A/(ph))]/2 or r=[-1-sqrt(1+2A/(ph))]/2

Use the facts that sin(θ) = y/r and r = √(x^2 + y^2). First, for r = 3sin(θ): r = 3sin(θ) r = 3(y/r... is a circle. In the subject line, I'll assume you mean r = sin(3θ) and not r = sin^3 (θ). You'll have to use a trig identity for...

r = 7(c - x) r/7 = 7(c-x)/7 r/7 = c - x r/7 + x = c

(R+r)(R-r) = A/pi R²-r² = A/pi <--- this is the difference of two squares method R² = A/pi + r² R = √(A/pi+r²)

- ( R-3 ) = 10 -r +3 = 10 -3 -3 --------------- -r=7 ----------- -1 -1 r= -7 (r-3)=10 +3 +3 -------------- r=13 {-7,13}

r = 2 / sin(t) r * sin(t) = 2 y = 2 x = r * cos(t) y = r * sin(t) x^2 + y^2 = r^2