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bagirrra123 [75]
3 years ago
13

One of the factors of 1331x^3-8y^3

Mathematics
1 answer:
pav-90 [236]3 years ago
3 0
Difference of 2 perfect squares

remember
a^3-b^3=(a-b)(a^2+ab+b^2)
so
(11x)^3-(2y)^3=(11x-2y)(121x^2+22xy+4y^2)
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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
Perform the indicated operation and write in standard form 5i+(-9-i)
antoniya [11.8K]
Since the sign before the parentheses is a + , we can just remove the parentheses.

<span>5i + (-9 - i) = 5i - 9 - i = -<u>9 + 4i</u></span>


3 0
3 years ago
Read 2 more answers
Danny brought a car for £10000 the value of the car depreciated by 20% in the first year then the value of the car depreciated b
Levart [38]

£7200

In first year depreciates by 20%, that is it is worth 80% of it's original value

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value after 1 year = 0.8 × £10000 = £8000

In the second year it depreciates by 10% of it's value, that is it is worth 90% of it's value at the end of the first year.

90% = \frac{90}{100} = 0.9

value after 2 years = 0.9 × £8000 = £7200


6 0
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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