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lara31 [8.8K]
4 years ago
9

Find the laplace transform of f(t) = cosh kt = (e kt + e −kt)/2

Mathematics
1 answer:
iren2701 [21]4 years ago
3 0
Hello there, hope I can help!

I assume you mean L\left\{\frac{ekt+e-kt}{2}\right\}
With that, let's begin

\frac{ekt+e-kt}{2}=\frac{ekt}{2}+\frac{e}{2}-\frac{kt}{2} \ \textgreater \  L\left\{\frac{ekt}{2}-\frac{kt}{2}+\frac{e}{2}\right\}

\mathrm{Use\:the\:linearity\:property\:of\:Laplace\:Transform}
\mathrm{For\:functions\:}f\left(t\right),\:g\left(t\right)\mathrm{\:and\:constants\:}a,\:b
L\left\{a\cdot f\left(t\right)+b\cdot g\left(t\right)\right\}=a\cdot L\left\{f\left(t\right)\right\}+b\cdot L\left\{g\left(t\right)\right\}
\frac{ek}{2}L\left\{t\right\}+L\left\{\frac{e}{2}\right\}-\frac{k}{2}L\left\{t\right\}

L\left\{t\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{t\right\}=\frac{1}{s^2} \ \textgreater \  L\left\{t\right\}=\frac{1}{s^2}

L\left\{\frac{e}{2}\right\} \ \textgreater \  \mathrm{Use\:Laplace\:Transform\:table}: \:L\left\{a\right\}=\frac{a}{s} \ \textgreater \  L\left\{\frac{e}{2}\right\}=\frac{\frac{e}{2}}{s} \ \textgreater \  \frac{e}{2s}

\frac{ek}{2}\cdot \frac{1}{s^2}+\frac{e}{2s}-\frac{k}{2}\cdot \frac{1}{s^2}

\frac{ek}{2}\cdot \frac{1}{s^2}  \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{ek\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{ek}{2s^2}

\frac{k}{2}\cdot \frac{1}{s^2} \ \textgreater \  \mathrm{Multiply\:fractions}: \frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d} \ \textgreater \  \frac{k\cdot \:1}{2s^2} \ \textgreater \  \mathrm{Apply\:rule}\:1\cdot \:a=a
\frac{k}{2s^2}

\frac{ek}{2s^2}+\frac{e}{2s}-\frac{k}{2s^2}

Hope this helps!
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Can i pls get help asap i will give brainly if the answer is right
Zanzabum
<h3>Answer:   0.5</h3>

This is equivalent to the fraction 1/2

==============================================================

Explanation:

The distance from A to B is 3 units. We can count out the spaces, or subtract the x coordinates of the two points and apply absolute value.

|A-B| = |-5-(-8)| = |-5+8| = |3| = 3

or

|B-A| = |-8-(-5)| = |-8+5| = |-3| = 3

We can say that segment AB is 3 units long.

--------------------------

The distance from A' to B' is 1.5 units because...

|A'-B'| = |-2.5-(-4)| = |-2.5+4| = |1.5| = 1.5

or

|B'-A'| = |-4-(-2.5)| = |-4+2.5| = |-1.5| = 1.5

The absolute values ensure the distance is never negative.

We can say A'B' = 1.5

---------------------------

Now divide the lengths of A'B' over AB to get the scale factor k

k = (A'B')/(AB)

k = (1.5)/(3)

k = 0.5

0.5 converts to the fraction 1/2.

The smaller rectangle A'B'C'D' has side lengths that are exactly 1/2 as long compared to the side lengths of ABCD.

3 0
3 years ago
Evaluate the function for <br> f(5)=4x+3<br><br> A. 23 <br> B. 22 <br> C. 18<br> D. 7
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Answer:

B. 22

Step-by-step explanation:

Hope this helps! Ask me anything if you have any quistions!

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Help, plz............
Brilliant_brown [7]

Answer:

I think Question (a) is 11748 m and Question (b) is 32364

Step-by-step explanation:

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What is the simplified form of StartRoot StartFraction 72 x Superscript 16 Baseline Over 50 x Superscript 36 Baseline EndFractio
aalyn [17]

Answer:

  StartFraction 6 Over 5 x Superscript 10 Baseline EndFraction

Step-by-step explanation:

Apparently you want to simplify ...

  \sqrt{\dfrac{72x^{16}}{50x^{36}}}

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  1/a^b = a^-b

  (a^b)^c = a^(bc)

__

So the expression simplifies as ...

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3 years ago
One positive number is 3 more than twice another. If their product is 629, find the numbers.
Soloha48 [4]

Answer:

17,37

Step-by-step explanation:

one number = x

The positive number = 2x + 3

x * (2x+3) = 629

x*2x + x *3 = 629

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2x*(x -17) + 37(x - 17) = 0

(x - 17)(2x + 37) = 0

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x = 17

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3 years ago
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