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olga_2 [115]
3 years ago
12

ANSWER NOW PRONTO STACKED PLSSSSSS 20 POINTS

Mathematics
1 answer:
skad [1K]3 years ago
6 0

Answer:

6.7 I think

Step-by-step explanation:

hope this helps

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What value of c makes the polynomial below a perfect square? 4x2 + 12x + c?
Tema [17]
This is what we know:4x^2 + 12x + c = (2x + k)^2
<span> 4x^2 + 12x + c = 4x^2 + 2kx + 2kx + k^2 
c=9
</span>
6 0
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Read 2 more answers
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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2 years ago
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Step-by-step explanation:

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Is (10,4),(-2,4),(-1,1),(5,6) a function?
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