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PIT_PIT [208]
2 years ago
14

A building access security system is generating inaccurate logs because users often share access tags when entering the building

. How can this be solved effectively?
Engineering
1 answer:
wel2 years ago
6 0

The problem of generation of inaccurate logs because users often share access tags when entering the building solved effectively by using by Amazon Recognition.

<h3>How will this Amazon Recognition help?</h3>

Amazon Recognition  will give enablement of  uptake of imagery as well as video for analysis in applications.

And then identification and distinguishing of facial features and activities will be possible.

Learn more about Amazon Recognition at;

brainly.com/question/14364696

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Show that y = '(t - s)f(s)ds is a solution to my" + ky = f(t). Use g' (0) = 1/m and mg" + kg = 0. 6.1) Derive y' 6.2) Using g(0)
Gre4nikov [31]

Answer:

Explanation:

Given that:

y = \int^t_og'(t-s) f(s) ds \  \text{is  solution to } \ my"ky= f(t)

where;

g'(0) = \dfrac{1}{m}     and mg"+kg = 0

\text{Using Leibniz Formula to prove the above equation:}

\dfrac{d}{dt} \int ^{b(t)}_{a(t)} \ f (t,s) \ ds = f(t,b(t) ) * \dfrac{d}{dt}b(t) - f(t,a(t)) *\dfrac{d}{dt}a(t) + \int ^{b(t)}_{a(t)}\dfrac{\partial}{\partial t} f(t,s) \ dt

So, y = \int ^t_0  g' (t-s) f(s) \ ds

\text{By differentiation with respect to t;}

y' = g'(o) f(t) \dfrac{d}{dt}t- 0 + \int^{t}_{0}g'' (t-s) f(s) ds \\ \\  y' = \dfrac{1}{m}f(t) + \int ^t_0 g'' (ts) f(s) \ ds

y'' = \dfrac{1}{m} f'(t) + g"(0) f(t) + \int^t_o g"'(t-s) f(s)ds --- (1)

Since \ \ mg" (t) +kg (t) = 0  \\ \\  \implies g" (t) = -\dfrac{k}{m} g(t) --- (111) \\ \\  put \  t \  =0 \  we  \ get;\\g" (0) = - \dfrac{k}{m } g(0)  \\ \\  g"(0) = 0 \ \ \ \   ( because \  g(0) =0) \\ \\

Now \ differentiating \ equation (111) \ with \ respect \ to \ t  \\ \\  g"'(t) = -\dfrac{k}{m}g(t)  \\ \\  replacing  \ it \ into  \ equation \ (1) \\ \\ y" = \dfrac{1}{m}f' (t) + 0 + \int ^t_o  \dfrac{-k}{m}g' (t-s) f(s) \ ds \\ \\ y" = \dfrac{1}{m}f' (t) - \dfrac{k}{m} \int ^t_o g' (t-s) \ f(s) \ ds \\ \\  y" = \dfrac{1}{m}f'(t) - \dfrac{k}{m}y \\ \\  my" = f'(t)-ky \\ \\ \implies \mathbf{ my" +ky = f'(t)}

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Explanation:

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