Pls help ASAP i don’t have time!!!!!

Mathematics

1 answer:

Ulleksa [173]3 years ago

6 0

Answer:

i dont have time either >:(

Step-by-step explanation:

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Kerri has a germen shepherd that has a mass of 30,000 grams. how many kilograms is 30,000 grams

wolverine [178]

Answer: 30 kilograms

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8 0

2 years ago

Multiplying desimals

kolezko [41]

Answer:

Line up the numbers on the right - do not align the decimal points.

Starting on the right, multiply each digit in the top number by each digit in the bottom number, just as with whole numbers.

Add the products.

Step-by-step explanation:

7 0

2 years ago

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

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