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zheka24 [161]
3 years ago
5

The value of which of these expressions is closest to e?

Mathematics
2 answers:
irinina [24]3 years ago
6 0

<u>Answer:</u>

The correct answer option is (1+\frac{1}{26} )^{26}.

<u>Step-by-step explanation:</u>

We are given a few mathematical expressions and we are to figure out which one among them has the value closest to that of <em>e</em><em>.</em>

Using a calculator, we can check that <em>e </em>= 2.718281828.

So let's evaluate each one of our given expressions to check which one of them is the closest to <em>e</em>:

(1+\frac{1}{26} )^{26}=2.667784966

(1+\frac{1}{25} )^{25}=2.665836331

(1+\frac{1}{23} )^{23}=2.661450119

(1+\frac{1}{24} )^{24}=2.663731258

Therefore, by looking at the above calculated values, we can conclude that the value of (1+\frac{1}{26} )^{26} is the closest to the value of <em>e</em>.

Mazyrski [523]3 years ago
5 0

Answer:

A. (1 + 1/26)²⁶

Step-by-step explanation:

e = \lim_{n \to \infty} (1+\frac{1}{n})^{n}

and e ≈ 2.71828

The greater the value of n,the closer is the value of the expression (1+ 1/n)ⁿ to the value of e.

Of the given options, the greatest value of n = 26.

Hence (1 + 1/26)²⁶ is closer to the value of e.

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You buy 3 bags of sand and 5 bags of pea gravel for $26. In a second purchase (at the same prices), you buy 4 bags of sand for $
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2z and 4y

Step-by-step explanation:

We first have to use the second purchase to figure this out, if 4 bags of sand is $8. We have to divide 8 by 4, which we get $2. We use this to find the bags of pea gravel. Since you bought 3 bags, 3 times 2 is 6. We subtract 6 from 26 leading to 20 dollars. We also buy 5 bags of pea gravel so we have two numbers, 20 and 5. We divide 20 and 5 and we get 4 dollars for 1 bag of pea gravel. If this is not what you meant I apologise.

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Find the form of the general solution of y^(4)(x) - n^2y''(x)=g(x)
Dennis_Churaev [7]

The differential equation

y^{(4)}-n^2y'' = g(x)

has characteristic equation

<em>r</em> ⁴ - <em>n </em>² <em>r</em> ² = <em>r</em> ² (<em>r</em> ² - <em>n </em>²) = <em>r</em> ² (<em>r</em> - <em>n</em>) (<em>r</em> + <em>n</em>) = 0

with roots <em>r</em> = 0 (multiplicity 2), <em>r</em> = -1, and <em>r</em> = 1, so the characteristic solution is

y_c=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}

For the non-homogeneous equation, reduce the order by substituting <em>u(x)</em> = <em>y''(x)</em>, so that <em>u''(x)</em> is the 4th derivative of <em>y</em>, and

u''-n^2u = g(x)

Solve for <em>u</em> by using the method of variation of parameters. Note that the characteristic equation now only admits the two exponential solutions found earlier; I denote them by <em>u₁ </em>and <em>u₂</em>. Now we look for a particular solution of the form

u_p = u_1z_1 + u_2z_2

where

\displaystyle z_1(x) = -\int\frac{u_2(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx

\displaystyle z_2(x) = \int\frac{u_1(x)g(x)}{W(u_1(x),u_2(x))}\,\mathrm dx

where <em>W</em> (<em>u₁</em>, <em>u₂</em>) is the Wronskian of <em>u₁ </em>and <em>u₂</em>. We have

W(u_1(x),u_2(x)) = \begin{vmatrix}e^{-nx}&e^{nx}\\-ne^{-nx}&ne^{nx}\end{vmatrix} = 2n

and so

\displaystyle z_1(x) = -\frac1{2n}\int e^{nx}g(x)\,\mathrm dx

\displaystyle z_2(x) = \frac1{2n}\int e^{-nx}g(x)\,\mathrm dx

So we have

\displaystyle u_p = -\frac1{2n}e^{-nx}\int_0^x e^{n\xi}g(\xi)\,\mathrm d\xi + \frac1{2n}e^{nx}\int_0^xe^{-n\xi}g(\xi)\,\mathrm d\xi

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u(x)=C_1e^{-nx}+C_2e^{nx}+u_p(x)

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\displaystyle y(x)=C_1+C_2x+C_3e^{-nx}+C_4e^{nx}+\int_0^x\int_0^\omega u_p(\xi)\,\mathrm d\xi\,\mathrm d\omega

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