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3 years ago

Having trouble does anybody know thisssssssssssss

Mathematics

1 answer:

Korolek [52]3 years ago

3 0

C is the right answer i think

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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

3 years ago

Please help me what is 1462 ÷ 58 as a mixed number then put it in simplest form

cestrela7 [59]

25.20, that might help you hopefully!

5 0

3 years ago

Simplify each expression(OOF PLEASE HURRY FGHJHGFGHJ):

Svetach [21]

(1) -(5y - 2)= -5y+2

(2) -5 (3n + 1)= -15n - 5

(3) too long to do

6 0

4 years ago

What is the measure of angle of 1?

Ludmilka [50]

Answer:

m<1 = 118

Step-by-step explanation:

The remote angles theorem states that when one extends one of the sides of a triangle, the sum of the two non-adjacent angles is equal to the measure of the angle between the extension of the side and a side of a triangle. One can apply this theorem here by stating that

(28) + (90) = m<1

Remember, a box around an angle signifies that its measure is (90) degrees.

Solve this problem by performing the operation,

118 = m<1

6 0

3 years ago

Polynomial division<br>(4x^2 +7x +6) ÷(x+ 2)

jolli1 [7]

Answer:

4x2 + 7x - 6

————————————

x - 2

Step-by-step explanation:

8 0

3 years ago

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