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3 years ago

PLEASEEEEEE HELPPPPPPPPPP MEEEEEEEEEEEEEEEE

Mathematics

1 answer:

polet [3.4K]3 years ago

3 0

Answer:

120.2 or in other words, the answer is the last option, D

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Approximate the square root value of 5 to the nearest Hundredth (50 points and will give most brainiest answer)

Softa [21]

Answer:The approximate square root is 2.23

Step-by-step explanation:√ 5 = 2.23

Thus: 2.23 x 2.23 = 5

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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) :

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

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2 years ago

Explain how to estimate 368+231

alina1380 [7]

Hi there,

368 + 231 = 599

Hope this helps :)

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3 years ago

What is the first quartile, Q1, of the data represented by the box plot?

atroni [7]

In this box plot 12.5 would be your answer

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3 years ago

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True or False. There are values of t so that sin t =.35 and cos t =.6

IceJOKER [234]

It is a completely false statement that there <span>are values of t so that sin t =.35 and cos t =.6. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that you were looking for and the answer has actually come to your desired help.</span>

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3 years ago

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