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Veseljchak [2.6K]

2 years ago

8

In a pair of complementary angles, the measurement of the larger angle is three times that of the smaller angle. Find the measur

ements of the two angles.

1 answer:

gayaneshka [121]2 years ago

8 0

Answer:

idkjrjejdjjdjdjdjfjdjfhjdjdhd

Step-by-step explanation:

idkjsjdjdjdjdjdjdjdjdjdjdjdjjfdjjr

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Wittaler [7]

Answer:

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Step-by-step explanation:

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2 years ago

Inveres laplace transform (3s-14)/s^2-4s+8

Olenka [21]

Complete the square in the denominator.

$s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4$

Rewrite the given transform as

$\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}$

Now take the inverse transform:

$L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}$

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1 year ago