Let f(x)=3x+2 and g(x)=x+4 solve: f(3) g(-1) 2f(x) f(x)+g(x) f(x)-g(x) f(x)•g(x)

f(x) = 3x + 2; g(x) = x + 4

f(3) = 3(3) + 2 = 9 + 2 = 11

g(-1) = -1 + 4 = 3

2f(x) = 2(3x + 2) = (2)(3x) + (2)(2) = 6x + 4

f(x) + g(x) = (3x + 2) + (x + 4) = (3x + x) + (2 + 4) = 4x + 6

f(x) - g(x) = (3x + 2) - (x + 4) = 3x + 2 - x - 4 = (3x - x) + (2 - 4) = 2x - 2

f(x) · g(x) = (3x + 2)(x + 4) = (3x)(x) + (3x)(4) + (2)(x) + (2)(4)

= 3x² + 12x + 2x + 8 = 3x² + 14x + 8

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

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step-by-step explanation: