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3 years ago

What is 66/55 in simplest form?

Mathematics

2 answers:

emmainna [20.7K]3 years ago

8 0

66/55 in simplest form is 6/5. Because... 66÷ 11= 6 55÷ 11= 5

liberstina [14]3 years ago

6 0

They both are divisible by 11, so divide both numerator and denominator from 11,
66/11 / 55/11
= 6/5

In short, Your Answer would be 6/5

Hope this helps!

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Goshia [24]

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Step-by-step explanation:

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Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

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Using the characteristic equation:

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Where:

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Therefore the zeros must come from the polynomial:

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Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

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Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

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Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

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4 years ago

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3 years ago

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wlad13 [49]

Answer:

Step-by-step explanation:

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= 6

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3 years ago