Question

Look at the screenshots for the answers. (50 points)

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

(1)

we are given

$f(x)=5x^2-2x$

$g(x)=3x^2+x-4$

$(f+g)(x)=f(x)+g(x)$

now, we can plug

$(f+g)(x)=5x^2-2x+3x^2+x-4$

$(f+g)(x)=8x^2-x-4$...........Answer

(2)

we are given

$f(x)=2x^2-3$

$g(x)=x+4$

$(fg)(x)=f(x) \times g(x)$

now, we can plug

$(fg)(x)=(2x^2-3) \times (x+4)$

$(fg)(x)=2x^3+8x^2-3x-12$...........Answer

(3)

we are given

$f(x)=3x^2+10x-8$

$g(x)=3x^2-2x$

we have

$(\frac{f}{g} )(x)=\frac{f(x)}{g(x)}$

now, we can plug values

$(\frac{f}{g} )(x)=\frac{3x^2+10x-8}{3x^2 -2x}$

now, we can factor it

$(\frac{f}{g} )(x)=\frac{(3x-2)(x+4)}{x(3x -2)}$

now, we can factor it

$(\frac{f}{g} )(x)=\frac{(x+4)}{x}$

we know that denominator can not be zero

so,

$3x^2-2x\neq 0$

$x\neq 0, x\neq (2/3)$

so, option-C.........Answer

(4)

we are given

$r(x)=x^2+6x+10$

$c(x)=x^2-4x+5$

$(r-c)(x)=r(x)-c(x)$

now, we can plug it

$(r-c)(x)=x^2+6x+10-(x^2-4x+5)$

$(r-c)(x)=10x+5$

now, we can plug x=4

$(r-c)(4)=10*4+5$

$(r-c)(4)=45$

so, option-C..................Answer

(5)

we are given

$c(x)=50+5x$

$p(x)=100-2x$

$(p*c)(x)=p(x)*c(x)$

now, we can plug values

$(p*c)(x)=(50+5x)*(100-2x)$

now, we can plug x=2

$(p*c)(2)=(50+5*2)*(100-2*2)$

$(p*c)(2)=5760$

we know that

price is

$c(x)=50+5x$

we can plug x=2

$c(2)=50+5*2$

$c(2)=60$

so, option-B...............Answer