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otez555 [7]
3 years ago
15

50 pointssssssss Simplify the following. Please leave it in exponential form.3^8 × 1/3^1

Mathematics
2 answers:
nikitadnepr [17]3 years ago
7 0

Answer:

2187

Step-by-step explanation:

skelet666 [1.2K]3 years ago
5 0

Answer:

3^7

Step-by-step explanation:

We know that 3^8=3\cdot3^7and since it's multiplied by \frac{1}{3} which is the same thing as \frac{1}{3}^1we know that 3\cdot\frac{1}{3}\cdot3^7=3^7.

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3 years ago
Find maclaurin series
Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

The first 3 terms of the series are

\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}

7 0
2 years ago
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