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3 years ago

Which is the simplified form of (9c^-9)^-3

2 answers:

8 0

$\bf ~~~~~~~~~~~~\textit{negative exponents}\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad 
a^n\implies \cfrac{1}{a^{-n}}\\\\
-------------------------------\\\\
(9c^{-9})^{-3}\impliedby \textit{let's first distribute the exponent}
\\\\\\
(9^{-3}c^{-9\cdot -3})\implies 9^{-3}c^{27}\implies \cfrac{1}{9^3}\cdot c^{27}\implies \cfrac{c^{27}}{729}$

Kazeer [188]3 years ago

3 0

Answer:C

Step-by-step explanation:

Just took the test

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3 years ago

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3 years ago

What property is used to change the sum of 18+6 to the equivalent expression 6(3+1)

Fittoniya [83]

The distributive property is used to get the equivalent expression

Step-by-step explanation:

The properties on mathematical expressions are used to simplify or elaborate the expressions.

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7 0

4 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) :

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

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3 years ago

Part A

nikdorinn [45]

A is one correct answer

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4 years ago