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user100 [1]
3 years ago
9

Plz help me with this G(x) F(x)

Mathematics
2 answers:
Ann [662]3 years ago
8 0

Answer:

The domains are the same expect for the point x = -3 where f(x) is undefined.

Step-by-step explanation:

Tatiana [17]3 years ago
6 0

Answer: f(x) D: x > -3

              g(x) D: x ≥ -3

<u>Step-by-step explanation:</u>

Graph: f(x) has an asymptote at x = -3, so the domain is: x > -3

Equation: g(x) = √(x + 3)  + 1   has a domain of x + 3 ≥ 0  --> x ≥ -3

<em>I am not sure what the dropbox options are so I will provide 2 possible answers:</em>

The domain of the function<u> f(x) </u>is greater than the domain of the function <u>g(x)</u>

                                                  or

The domain of the function<u> g(x) </u>is less than the domain of the function <u>f(x)</u>

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\large\boxed{\dfrac{\boxed{8}}{81}}

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\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}

\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}

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