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densk [106]
3 years ago
11

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Mathematics
1 answer:
Butoxors [25]3 years ago
5 0

Answer:   \bold{a_n=\dfrac{2^{n-1}}{3^{n+1}}}

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

The explicit rule for a geometric sequence is: a_n=a_1(r)^{n-1}\quad \text{where}\ a_1\ \text{is the first term and r is the common ratio}

Given the sequence {9, 6, 4, \dfrac{8}{3}, ... } we know

  • the first term (a₁) is 9
  • the common ratio (r) is \dfrac{6}{9}=\dfrac{2}{3}\ \text{when simplified}

So, the explicit rule for the given sequence is:

a_n=9\bigg(\dfrac{2}{3}\bigg)^{n-1}\\\\\\.\quad =3\cdot3\dfrac{(2)^{n-1}}{(3)^{n-1}}\\\\\\.\quad =3\dfrac{(2)^{n-1}}{(3)^{n}}\\\\\\.\quad =\dfrac{(2)^{n-1}}{(3)^{n+1}}

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please help! functions and relations. f(x)= square root of x-4. find the inverse of f(x) and it’s domain.
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ANSWER:

\:D.\text{ }f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge-4

STEP-BY-STEP EXPLANATION:

We have the following equation:

f(x)=\sqrt{x}-4

The inverse is the following (we calculate it by replacing f(x) by x and x by f(x)):

\begin{gathered} x=\sqrt{f^{-1}(x)}-4 \\  \\ \sqrt{f^{-1}(x)}=x+4 \\  \\ f^{-1}(x)=(x+4)^2 \end{gathered}

The domain would be the range of the original equation, and it would be the range of values that f(x) could take, which was from -4 to positive infinity, that is, f(x) ≥ -4.

Therefore, the domain is x ≥ -4.

So the correct answer is D.

\:f^{-1}\left(x\right)=\left(x+4\right)^2;x\ge -4

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