What is the area of the triangle with vertices (-1,-4), (-3, -2), and d (-4,-2)?

Mathematics

1 answer:

mash [69]1 year ago

4 0

Answer:

Sorry I cant help

Step-by-step explanation:

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hoa [83]

Answer:

Step-by-step explanation:

Look at the picture.

Use cosine:

$cosine=\dfrac{adjacent}{hypotenuse}$

We have a measure of the angle and length of the hypotenuse:

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Substitute:

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2 years ago

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irga5000 [103]

With convolution theorem the equation is proved.

According to the statement

we have given that the equation and we have to evaluate with the convolution theorem.

Then for this purpose, we know that the

A convolution integral is an integral that expresses the amount of overlap of one function as it is shifted over another function.

And the given equation is solved with this given integral.

So, According to this theorem the equation becomes the

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