Can you guys help me with this?

Mathematics

1 answer:

Ghella [55]4 years ago

7 0

F(x)=2x^2-x-6
Factoring:
f(x)=2(2x^2-x-6)/2=(2^2x^2-2x-12)/2=[(2x)^2-(2x)-12]/2
f(x)=(2x-4)(2x+3)/2=(2x/2-4/2)(2x+3)→f(x)=(x-2)(2x+3)

g(x)=x^2-4
Factoring
g(x)=[sqrt(x^2)-sqrt(4)][sqrt(x^2)+sqrt(4)]
g(x)=(x-2)(x+2)

f(x)/g(x)=[(x-2)(2x+3)] / [(x-2)(x+2)
Simplifying:
f(x)/g(x)=(2x+3)/(x+2)

Answer: Third Option (2x+3)/(x+2)

You might be interested in

Which statements are true about the circle shown below? Select two options.

AfilCa [17]

Answer:

C,D

Step-by-step explanation:

E d g e n u i t y

5 0

4 years ago

Read 2 more answers

Please help I don't understand 5i√-54

Arada [10]

Answer:

Simplify the radical by breaking the radicand up into a product of known factors.

Exact Form:

− 15 √ 6

Decimal Form:

− 36.74234614

Step-by-step explanation:

thats all i got.. if you could please answer my most recent

8 0

3 years ago

An investment of $450 increases at a rate of 6.5% per year. what is the growth factor, b?

Lera25 [3.4K]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$450\\ r=rate\to 6.5\%\to \frac{6.5}{100}\dotfill &0.065\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{per year, thus once} \end{array}\dotfill &1\\ t=years \end{cases}$