5.2 The Right Height: True or Not?

Mathematics

2 answers:

Anna71 [15]3 years ago

8 0

The answer is G I think soooooo

Serhud [2]3 years ago

7 0

Answer:

Step-by-step explanation:

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3 years ago

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Find the value of y.<br> m<1 = 3y + 6

Rufina [12.5K]

Step-by-step explanation:

$m > 1 = 3y + 6\\ \frac{m > -1 - 6}{3} = \frac{3y}{3} \\ y = \frac{m > - 5}{3}$

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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

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