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4 years ago

Order -3.25, -3 1/8, -3 1/4 , and -3.1 from least to greatest . Explain

Mathematics

2 answers:

Andrei [34K]4 years ago

6 0

Answer:

YES THATS VERY COOL DAD

Step-by-step explanation:HEHEHEHEHHEHEHEHEHEHHEHEHEHEHE

noname [10]4 years ago

5 0

Make all of them either a decimal or all as fractions. the least is the furthest from 0 if it on the left side of 0 on a number line. Negatives are to the left.
-3 and 1/8 = -3.125
-3 and 1/4 = -3.25
so your numbers are now
-3.25, -3.125, -3.25, -3.1
Least to greatest
-3.25, -3 1/4, -3 1/8, -3.1

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3 years ago

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masha68 [24]

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4 years ago

Read 2 more answers

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seropon [69]

The answer is The second one because I took this test before and I got it right

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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