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3 years ago

What is The product of two numbers is equal to 14

Mathematics

2 answers:

Liono4ka [1.6K]3 years ago

6 0

You can do 1x14 to get the product of 14.

Another way is doing 2x7 to get 14.

Here are two pairs you can get the product of 14.

tiny-mole [99]3 years ago

4 0

You can multiply 2x7 to get 14

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Find the inverse laplace transform of: (2 s + 4) / (s - 3)^3

Serhud [2]

Answer:

$e^{3t}(2t+5t^{2})$

Step-by-step explanation:

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

Using the Translation theorem to transform the s-3 to s, that means multiplying by and change s to s+3

Translation theorem: $L^{1} [F(s-a)=L^{-1}[F(s)|_{s \to s-a}\\ L^{1} [F(s-a)=e^{at} f(t)$

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

Separate the fraction in a sum:

$e^{3t} L^{-1}[\frac{2s+10}{s^{3}} ]=e^{3t} L^{-1}[\frac{2s}{s^{3}}+\frac{10}{s^{3}} ]=e^{3t} (L^{-1}[\frac{2}{s^{2}}]+ L^{-1}[\frac{10}{s^{3}}])$

The formula for this is:

$L^{-1}[\frac{n!}{s^{n+1}} ]=t^{n}$

Modify the expression to match the formula.

$e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ \frac{10}{2} L^{-1}[\frac{2}{s^{2+1}}])=e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])$

Solve

$e^{3t} (2L^{-1}[\frac{1}{s^{1+1}}]+ 5 L^{-1}[\frac{2}{s^{2+1}}])=e^{3t}(2t+5t^{2} )$

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3 years ago