<span>-2i + (9 − 3i) − (6 − 10i)
-2i - 3i + 10i + 9 - 6
5i + 3
3 + 5i</span>
Answer: 4 units to the right, 3 units up
Hope this helped! :)
Answer:
The series is absolutely convergent.
Step-by-step explanation:
By ratio test, we find the limit as n approaches infinity of
|[a_(n+1)]/a_n|
a_n = (-1)^(n - 1).(3^n)/(2^n.n^3)
a_(n+1) = (-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)
[a_(n+1)]/a_n = [(-1)^n.3^(n+1)/(2^(n+1).(n+1)^3)] × [(2^n.n^3)/(-1)^(n - 1).(3^n)]
= |-3n³/2(n+1)³|
= 3n³/2(n+1)³
= (3/2)[1/(1 + 1/n)³]
Now, we take the limit of (3/2)[1/(1 + 1/n)³] as n approaches infinity
= (3/2)limit of [1/(1 + 1/n)³] as n approaches infinity
= 3/2 × 1
= 3/2
The series is therefore, absolutely convergent, and the limit is 3/2
Answer:
Kelvin wrote 60 the wrong way.
He has to write $30.65 - 60%, not 0.60.
Here is the final correct equation:
$30.65 - 60% = 12.26
667.302 would be the correct answer, I did it in a calculator
