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(1)we are given [tex]f(x)=5x^2-2x[/tex][tex]g(x)=3x^2+x-4[/tex][tex](f+g)(x)=f(x)+g(x)[/tex]now, we can plug [tex](f+g)(x)=5x^2-2x+3x^2+x-4[/tex][tex](f+g)(x)=8x^2-x-4[/tex]...........Answer(2)we are given [tex]f(x)=2x^2-3[/tex][tex]g(x)=x+4[/tex][tex](fg)(x)=f(x) \times g(x)[/tex]now, we can plug [tex](fg)(x)=(2x^2-3) \times (x+4) [/tex][tex](fg)(x)=2x^3+8x^2-3x-12[/tex]...........Answer(3)we are given [tex]f(x)=3x^2+10x-8[/tex][tex]g(x)=3x^2-2x[/tex]we have [tex](\frac{f}{g} )(x)=\frac{f(x)}{g(x)}[/tex]now, we can plug values[tex](\frac{f}{g} )(x)=\frac{3x^2+10x-8}{3x^2 -2x}[/tex]now, we can factor it [tex](\frac{f}{g} )(x)=\frac{(3x-2)(x+4)}{x(3x -2)}[/tex]now, we can factor it [tex](\frac{f}{g} )(x)=\frac{(x+4)}{x}[/tex]we know that denominator can not be zero so, [tex]3x^2-2x\neq 0[/tex][tex]x\neq 0, x\neq (2/3)[/tex]so, option-C.........Answer(4)we are given [tex]r(x)=x^2+6x+10[/tex][tex]c(x)=x^2-4x+5[/tex][tex](r-c)(x)=r(x)-c(x)[/tex]now, we can plug it [tex](r-c)(x)=x^2+6x+10-(x^2-4x+5)[/tex][tex](r-c)(x)=10x+5[/tex]now, we can plug x=4[tex](r-c)(4)=10*4+5[/tex][tex](r-c)(4)=45[/tex]so, option-C..................Answer(5)we are given [tex]c(x)=50+5x[/tex][tex]p(x)=100-2x[/tex][tex](p*c)(x)=p(x)*c(x)[/tex]now, we can plug values[tex](p*c)(x)=(50+5x)*(100-2x)[/tex]now, we can plug x=2[tex](p*c)(2)=(50+5*2)*(100-2*2)[/tex][tex](p*c)(2)=5760[/tex]we know that price is [tex]c(x)=50+5x[/tex]we can plug x=2[tex]c(2)=50+5*2[/tex][tex]c(2)=60[/tex]so, option-B...............Answer
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general 11 months ago 6523