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R =(ab)/(a+b) R (a+b) = ab Ra + Rb = ab Rb = ab - Ra Rb = a(b- R ) a = (Rb)/(b- R )

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}