3 years ago

Help with one question please.

1 answer:

4 0

I would say the answer is 26.

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MAVERICK [17]

Answer:

If you center the series at x=1

$T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)$

Where $R(x)$ is the error.

Step-by-step explanation:

From the information given we know that

$f(x) = e^{2x^2}$

$f'(x) = 4x e^{2x^2}$ (This comes from the chain rule )

$f^{(2)}(x) = 4e^{2x^2} (4x^2+1)$ (This comes from the chain rule and the product rule)

$f^{(3)}(x) = 16xe^{2x^2}(4x^2 + 3)$ (This comes from the chain rule and the product rule)

If you center the series at x=1 then

$T_3(x) = e^2 + 4e^2 (x-1)+10(x-1)^2 + \frac{56}{3} e^2(x-1)^3 + R(x)$

Where $R(x)$ is the error.

3 0

3 years ago

MrRa [10]

ab+bc>can is the correct answer

3 0

3 years ago

arlik [135]

3(b+6) = 3*b + 3*6 = 3b + 18

5 0

3 years ago

Black_prince [1.1K]

The answer is 0.36. I don't think u need to explain why, right ;)

8 0

3 years ago

gogolik [260]

Answer:47 inches

Step-by-step explanation:

8 0

3 years ago