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4 years ago

How to solve 14/15- ?= 1/3

Mathematics

1 answer:

Sophie [7]4 years ago

5 0

Answer:

? = 3/5

Step-by-step explanation:

14/15–x = 1/3

14/15–1/3 = x

14/15–5/15 = x

9/15 = x

3/5 = x

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Answer the Follow: P (7 4) P (6 3)

mote1985 [20]

Answer:

Step-by-step explanation:

Using closest Wolfram|Alpha interpretation: P (7 4) P (6 3)

Assuming

"P"

is referring to prime numbers

Use as

instead

Input interpretation:

primes | between 28 and P(18)

Values:

2 | 3 | 5 | 7 | 11 | 13 | 17 | 19 | 23 (9 primes)

5 0

3 years ago

If anyone can help me with this question thanks!

Vedmedyk [2.9K]

What is it asking for?

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3 years ago

B-2-11. Find the inverse Laplace transform of s + 1/s(s^2 + s +1)

Aleksandr-060686 [28]

Answer:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=1-e^{-t/2}cos(\frac{\sqrt{3} }{2}t )+\frac{e^{-t/2}}{\sqrt{3} }sin(\frac{\sqrt{3} }{2}t)$

Step-by-step explanation:

let's start by separating the fraction into two new smaller fractions

First, s(s^2+s+1) must be factorized the most, and it is already. Every factor will become the denominator of a new fraction.

$\frac{s+1}{s(s^{2} + s +1)}=\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}$

Where A, B and C are unknown constants. The numerator of s is a constant A, because s is linear, the numerator of s^2+s+1 is a linear expression Bs+C because s^2+s+1 is a quadratic expression.

Multiply both sides by the complete denominator:

$[{s(s^{2} + s +1)]\frac{s+1}{s(s^{2} + s +1)}=[\frac{A}{s}+\frac{Bs+C}{s^{2}+s+1}][{s(s^{2} + s +1)]$

Simplify, reorganize and compare every coefficient both sides:

$s+1=A(s^2 + s +1)+(Bs+C)(s)\\\\s+1=As^{2}+As+A+Bs^{2}+Cs\\\\0s^{2}+1s^{1}+1s^{0}=(A+B)s^{2}+(A+C)s^{1}+As^{0}\\\\0=A+B\\1=A+C\\1=A$

Solving the system, we find A=1, B=-1, C=0. Now:

$\frac{s+1}{s(s^{2} + s +1)}=\frac{1}{s}+\frac{-1s+0}{s^{2}+s+1}=\frac{1}{s}-\frac{s}{s^{2}+s+1}$

Then, we can solve the inverse Laplace transform with simplified expressions:

$\mathcal{L}^{-1}\{\frac{s+1}{s(s^{2} + s +1)}\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{\frac{1}{s}\}-\mathcal{L}^{-1}\{\frac{s}{s^{2}+s+1}\}$

The first inverse Laplace transform has the formula:

$\mathcal{L}^{-1}\{\frac{A}{s}\}=A\\ \\\mathcal{L}^{-1}\{\frac{1}{s}\}=1\\$

For:

$\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}$

We have the formulas:

$\mathcal{L}^{-1}\{\frac{s-a}{(s-a)^{2}+b^{2}}\}=e^{at}cos(bt)\\\\\mathcal{L}^{-1}\{\frac{b}{(s-a)^{2}+b^{2}}\}=e^{at}sin(bt)$

We have to factorize the denominator:

$-\frac{s}{s^{2}+s+1}=-\frac{s+1/2-1/2}{(s+1/2)^{2}+3/4}=-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}$

It means that:

$\mathcal{L}^{-1}\{-\frac{s}{s^{2}+s+1}\}=\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}+\frac{1/2}{(s+1/2)^{2}+3/4}\}$

$\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\mathcal{L}^{-1}\{\frac{1/2}{(s+1/2)^{2}+3/4}\}\\\\\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}+\frac{1}{2} \mathcal{L}^{-1}\{\frac{1}{(s+1/2)^{2}+3/4}\}$

So a=-1/2 and b=(√3)/2. Then:

$\mathcal{L}^{-1}\{-\frac{s+1/2}{(s+1/2)^{2}+3/4}\}=e^{-\frac{t}{2}}[cos\frac{\sqrt{3}t }{2}]\\\\\\\frac{1}{2}[\frac{2}{\sqrt{3} } ]\mathcal{L}^{-1}\{\frac{\sqrt{3}/2 }{(s+1/2)^{2}+3/4}\}=\frac{1}{\sqrt{3} } e^{-\frac{t}{2}}[sin\frac{\sqrt{3}t }{2}]$