Algebra
posted by Taurus .
R varies jointly as r and B^2. When r=3, and B=3 R=135. Find R when r is 1
and B=10.

R = krB^2
135 = k(3)(3^2) = 27k
k = 5
So, R(r,B) = 5rB^2
R(1,10) = 5(1)(100) = 500
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