Quick question before I answer: Did he have to pay $2,399 when he signed the lease, or was that just how much he had to pay in total over the course of 36 months?

6 0

2 years ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

2 years ago

Solve it............<br>

DedPeter [7]

Answer:

Step-by-step explanation:

1) for '8'; '3'; and '10':

8/2+3+3=10

2) for '2'; '5' and '11':

2/1+5+5=11

3) for '6'; '2' and '?':

6/2+2+2=7

8 0

2 years ago

Can you help me with this please ​

Can you help me with this please