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2 years ago

Write the following numbers in order from smallest to largest. 27.56, 27.509, 27.5998, 27.599

Mathematics

2 answers:

Nookie1986 [14]2 years ago

8 0

Answer: 27.509, 27.56, 27.599, 27.5998

Step-by-step explanation:

Simora [160]2 years ago

3 0

27.509, 27.56, 27.599, 27.5998

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The answer for this question is 100

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3 years ago

Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

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3 years ago

A squared minus 1/4 is 0. solve for a

Ludmilka [50]

Answer:

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Step-by-step explanation:

first we need to isolate a squared. so we add 1/4 to both sides

then you take the square root of a square to get 1/2

to check, 1/2 * 1/2 = 1/4

1/4-1/4 = 0

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4 years ago

What is 34827 x 7 = Jasper over here..

kirill [66]

Answer:

243789

Step-by-step explanation:

(34827) 7 = 243789

Hope this helps!

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2 years ago

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