Question

y varies inversely with the square of x, y also varies with x+1 direclty.What is the ratio of the constant \[\frac{k_1}{k_2}\] if x=1

Answer 1

\[\frac{k_1}{k_2}\]

Answer 2

i have been looking at alot of proportion problems on OS heres my own

Answer 3

It's easier than you think

Answer 4

\(y = \frac{k_1}{x^2}\)
\(y = k_2(x + 1)\)

x = 1

\(y = \frac{k_1}{1^2} = \frac{k_1}{1}\)
\(y + k_2(1 + 1) = 2k_2\)

y = y

\(\frac{k_1}{1} = 2k_2\)
\(\frac{k_1}{k_2} = 2\)

Answer 5

@Jonask

Answer 6

you win

Answer 7

lol, okay :D

Answer 8

good training for such questions ....