The answer would be: 12 bytes^3
Answer:
A
Step-by-step explanation:
If you were to substitute 2 or -3 into equation A, the denominator would be zero and you would have to divide by zero. Thus there are asymptotes for this function at -3 and 2, matching the graph
Let <em>a</em> be the first term in the sequence. If <em>r</em> is the ratio between consecutive terms, then the second term is <em>ar</em>, the third term is <em>ar </em>^2, the fourth is <em>ar</em> ^3, and so on, up to the <em>n</em>-th term <em>ar</em> ^(<em>n</em> - 1).
So the third, fourth, and fifth terms are such that
<em>ar</em> ^2 = 18
<em>ar</em> ^3 = 27
<em>ar</em> ^4 = 81/2
Solve for <em>a</em> and <em>r</em> :
(<em>ar</em> ^3) / (<em>ar</em> ^2) = 27/18 => <em>r</em> = 3/2
<em>ar</em> ^2 = <em>a</em> (3/2)^2 = 9/4 <em>a</em> = 18 => <em>a</em> = 8
Then the <em>n</em>-th term in the sequence is
<em>ar</em> ^(<em>n</em> - 1) = 9 (3/2)^(<em>n</em> - 1)
You can rewrite this by first rewriting 9 = 3^2, then
9 (3/2)^(<em>n</em> - 1) = 3^2 * 3^(<em>n</em> - 1) / 2^(<em>n</em> - 1) = 3^(<em>n</em> + 1)/2^(<em>n</em> - 1)
