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Serggg [28]
4 years ago
14

Use a known maclaurin series to obtain the maclaurin series for the given function. f(x = 8x2 tan?1(7x3

Mathematics
1 answer:
11Alexandr11 [23.1K]4 years ago
6 0
Going out on a limb here and guessing that the function is

f(x)=8x^2\tan^{-1}(7x^3)

Please correct me if this isn't the case.

Recall that

\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^nx^{2n+1}}{2n+1}

which converges for |x|.

It follows that

8x^2\tan^{-1}(7x^3)=8x^2\displaystyle\sum_{n\ge0}\frac{(-1)^n(7x^3)^{2n+1}}{2n+1}
=\displaystyle\sum_{n\ge0}\frac{56(-49)^nx^{6n+3}}{2n+1}
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3 years ago
Which values are solutions to the inequality square root of x is less than 8? A. 55 B. 8 C. -8 D. 4 E. 65 F. -64
Dima020 [189]

9514 1404 393

Answer:

  A.  55

  B. 8

Step-by-step explanation:

We want ...

  √x < 8

  x < 8² . . . . . square both sides

  x < 64

In order for the inequality to make any sense, we must also have ...

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So, the solution space is ...

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Only answer choices A (55) and B (8) are in that range.

4 0
3 years ago
Why is -a² is always negative (a ≠ 0) and why is (-a)² is always positive?
hjlf
-a^2=(-1)\cdot a\cdot a

Regardless of the sign of a, we have a\cdot a=a^2\ge0 (never negative). But multiplying by -1 makes it negative.

On the other hand,

(-a)^2=((-1)\cdot a)^2=(-1)^2\cdot a^2=1\cdot a^2=a^2

which can never be negative for real a.
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3 years ago
Help please!(picture)
UkoKoshka [18]

Answer:

7a - 2

Step-by-step explanation:

A perimeter is just the sum of all sides added together, so just put all the expressions together into a big equation.

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Combine like terms

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