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4 years ago

Find any 3 irrational number between 2 by 3 and 4 by 5

Mathematics

2 answers:

FrozenT [24]4 years ago

7 0

Answer:

$\sqrt{2} \sqrt{3 }\sqrt{5} \\ irrational \: means \:has \: no \: square \: \\ root \:$

Tatiana [17]4 years ago

4 0

Answer:

π/4 ≈ 0.785
e/4 ≈ 0.680
(√3) -1 ≈ 0.732

Step-by-step explanation:

It looks like you want irrational numbers between 2/3 ≈ 0.667 and 4/5 = 0.800.

Here are some:

π/4 ≈ 0.785

e/4 ≈ 0.680

(√3) -1 ≈ 0.732

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makkiz [27]

Solution :

Given :

2y''' + 15y'' + 24y' + 11y= 0

Let x = independent variable

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If $Q(x) \neq 0$

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The general solution = complementary solution + particular integral

If Q(x) = 0

It is called the homogeneous then the general solution = complementary solution.

2y''' + 15y'' + 24y' + 11y= 0

$(2D^3+15D^2+24D+11)y=0$

Auxiliary equation,

$2m^3+15m^2+24m +11 = 0$

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∴ $2m^2+13m+11=0$

The roots are

$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-13\pm \sqrt{13^2-4(11)(2)}}{2(2)}$

$=\frac{-13\pm9}{4}$

$=-5.5, -1$

So, $m_1, m_2, m_3 = -1, -1, -5.5$

Then the general solution is :

$= (c_1+c_2 x)e^{-x} + c_3 \ e^{-5.5x}$

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densk [106]

Answer:

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Step-by-step explanation:

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