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If f(x) varies directly with x2, and f(x) = 75 when x = 5, find the value of f(4). A. 100 B. 50 C. 48 D. 25
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is it A?
" f(x) varies directly with x^2", so f(x) = k*x^2 for some constant k
we know that "f(x) = 75 when x = 5", so... f(x) = k*x^2 75 = k*(5)^2 75 = k*25 75/25 = k 3 = k k = 3
ohhhhhh i get it
Which makes the equation turn from f(x) = k*x^2 into f(x) = 3x^2
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now that you know the full function, you can plug in x = 4 and evaluate
so its 100....
no
plug x = 4 into f(x) = 3x^2 to get ???
144
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no
you square before you multiply
exponents come before multiplication in PEMDAS
ohh of course
its 48
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correct
thx
you're welcome
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