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3 years ago

NEED HELP ASAP THANKS

Mathematics

1 answer:

Oxana [17]3 years ago

4 0

1/4w+4 - 1/2w+3
-1/4+7

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Verizon [17]

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3 years ago

What is the volume of the figure?

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3 years ago

In the drawing below, line n is a transversal that intersects two parallel lines. What is the measure of Angle 4?

4vir4ik [10]

Answer:

140°

Step-by-step explanation:

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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