How does $$ g(t)=\frac{1}{9^{t}} $$ change over the interval from $$ t=9 $$ to $$ t=10 $$ ? $$ g(t) $$ decreases by 9 $$ g(t) $$ decreases by a factor of 9 $$ g(t) $$ increases by a factor of 9 $$ g(t) $$ increases by $$ 9 \% $$

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We begin with the function 
$$
g(t)=\frac{1}{9^t}=9^{-t}.
$$
To see how $$ g(t) $$ changes from $$ t=9 $$ to $$ t=10 $$, we compute the value at $$ t=9 $$ and $$ t=10 $$:

1. For $$ t=9 $$:
   $$
   g(9)=9^{-9}.
   $$

2. For $$ t=10 $$:
   $$
   g(10)=9^{-10}.
   $$

Next, we find the ratio of $$ g(10) $$ to $$ g(9) $$:
$$
\frac{g(10)}{g(9)}=\frac{9^{-10}}{9^{-9}}=9^{-10+9}=9^{-1}=\frac{1}{9}.
$$

This ratio tells us that at $$ t=10 $$, the value of $$ g(t) $$ is $$\frac{1}{9}$$ of the value at $$ t=9 $$. In other words, the function $$ g(t) $$ decreases by a factor of 9.

Thus, the correct change over the interval from $$ t=9 $$ to $$ t=10 $$ is:

$$ g(t) $$ decreases by a factor of 9.
$$ g(t) $$ decreases by a factor of 9.

How does change over the interval from to ?
decreases by 9
decreases by a factor of 9
increases by a factor of 9
increases by

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