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disa [49]
3 years ago
7

Compare the value of the following expression with zero. 5a^3b^2/a+b, where a<0, b<0

Mathematics
1 answer:
ira [324]3 years ago
3 0

Answer:

The given expression will be greater than zero.

Step-by-step explanation:

If a < 0 and b < 0, then 5a^3 is < 0 (5a^3) is negative;

but b^2 is > 0.

Then (5a^3)*(b^2) < 0

and if we divide this quantity by a + b (which is negative), the end result will be > 0.

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4 years ago
Help help asap please
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Use theorem 7. 4. 2 to evaluate the given laplace transform. do not evaluate the convolution integral before transforming. (writ
irga5000 [103]

With convolution theorem the equation is proved.

According to the statement

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So, According to this theorem the equation becomes the

\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{\frac{s+1}{(s+1)^2+1}}{s} \\\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).

Then after solving, it become and with theorem it says that the

\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .

Hence by this way the given equation with convolution theorem is proved.

So, With convolution theorem the equation is proved.

Learn more about convolution theorem here

brainly.com/question/15409558

#SPJ4

3 0
2 years ago
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