e^0.6 to the nearest hundredth using the first five terms of the exponential series of eˣ is; e^(0.6) = 1 + 0.6 + 0.6²/2! + 0.6³/3! + 0.6⁴/4....
How to solve Maclaurin Series?
We want to use the first 5 terms of the exponential series of eˣ to approximate e^(0.6)
The type of exponential series we will utilize is called the Maclaurin Series for. Thus, for eˣ, we have;
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4.....
Thus;
e^(0.6) = 1 + 0.6 + 0.6²/2! + 0.6³/3! + 0.6⁴/4....
e^(0.6) = 1 + 0.6 + 0.18 + 0.072 + 0.0324
e^(0.6) = 1.88 to the nearest hundredth
Read more about Maclaurin Series at; https://brainly.com/question/14570303
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