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use f(x)=3x-4 and g(x)=2x^2 to evaluate each of the following (a)(f+g)(3) (b)(f-g)(-2) (c)(fg)(7) (d)(f/g)(0)
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(a)(f+g)(3)=(3x-4)+(2x^2) =(3(3)-4)+(2(3)^2) =(9-4)+(2*9) =5+18 = 23
thank u
(b)(f-g)(-2)=(3x-4)-(2x^2) =(3(-2)-4)-(2(-2)^2) =-10-8 =-18
thank you so much I am so lost in this that I did not know there were two answers.
(c)(fg)(7)=(3(7)-4)*(2(7)^2) =(21-4)(2*49) = 17*98 =1.666
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(d)(f/g)(0)=\[\frac{3x-4}{2x^2}=\frac{3(0)-4}{2(0)^2}=\frac{0-4}{0}=0\]
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