Question 240199
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If *[tex \LARGE f(x)\ =\ 6\ -\ 2x]


Then *[tex \LARGE f(0)\ =\ 6\ -\ 2(0)]


*[tex \LARGE f(-2)\ =\ 6\ -\ 2(-2)]


*[tex \LARGE f(a)\ =\ 6\ -\ 2a]


*[tex \LARGE f\left(\frac{x^2y}{2-pqr}\right)\ =\ 6\ -\ 2\left(\frac{x^2y}{2-pqr}\right)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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