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Variable y varies directly with x^2, and y = 72 when x = 6. What is the value of y when x = 4? Y = ___
\(y = kx^2\) Plug in what we know: \(72 = k(6)^2\) Simplify the exponent: \(72 = 36k\) Divide 36 to both sides: \(k = 2\) Now we now that the constant is 2, we can plug this in with x = 4: \(y = kx^2\) \(y = (2)(4)^2\) Simplify and you get 'y'.
one sec. brb. moms calling
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4^2 = 16
2 x 16 = 32
32 = answer?
Yep
You got it.
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thanks :)
No problem.
gonna tag you to another one
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