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3 years ago

[ f(x) = 3x + 2 g(x) = x2 Find gf(5) Find fg(x)

Mathematics

1 answer:

marusya05 [52]3 years ago

8 0

Answer:

gf(5) = 289.

fg(x) = 3x^2 + 2.

Step-by-step explanation:

f(5) = 3(5) + 2 = 17

So gf(5) = 17^2 = 289.

fg(x) = 3(x^2) + 2 (we replace the x in f(x) by g(x)).

= 3x^2 + 2.

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Afina-wow [57]

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It's x = -1.5

Step-by-step explanation:

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2 years ago

Consider the function f(x) = x^3 - 7x^2 - 2x + 14

mars1129 [50]

Answer:

There are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:x^3\:-\:7x^2\:-\:2x\:+\:14$

To get the zeros of $f(x)$ , set $y$ or $f(x) =0$

$0=x^3-7x^2-2x+14$

$x^3\:-\:7x^2\:-\:2x\:+\:14=\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)$

$\left(x-7\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0$

Using the zero factor principle:

$\mathrm{ \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$

$x-7=0\quad \mathrm{or}\quad \:x+\sqrt{2}=0\quad \mathrm{or}\quad \:x-\sqrt{2}=0$

solving

$x-7=0$

$x=7$

$x+\sqrt{2}=0$

$x=-\sqrt{2}$

$x-\sqrt{2}=0$

$x=\sqrt{2}$

Therefore, there are 3 zeros, which are:

$x=7,\:x=-\sqrt{2},\:x=\sqrt{2}$

8 0

2 years ago

What is the solution to this system of equations?<br> x = 12 − y<br> 2x + 3y = 29

Elenna [48]

Let’s arrange the fist equation
x+y=12
2x+3y=29

Multiply equation 1 by 2
So it will be

2x+2y=24

Subtract eq 1 from eq 2

2x+3y=29
2x+2y=24
—————-
Y =5

Put value of y in eq 1

2x+2y=24
2x +10= 24
2x= 24-10
2x= 14
x=14/2
x=7

S.S= (7,5)

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2 years ago

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