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Makovka662 [10]
3 years ago
12

In the following Laplace transform, how do you get rid of the cos(2at)*sin(at)?

Mathematics
1 answer:
Novay_Z [31]3 years ago
7 0
<span>sin A cos B=1/2[sin(A-B)+sin(A+B)]
sin(at)*cos(2at)=1/2[sin(3at)-sin(at)]</span>
<span>
</span><span>
</span>

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Answer:

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What is the product? (9 t minus 4)(negative 9 t minus 4) negative 81 t squared minus 16 negative 81 t squared + 16 negative 81 t
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The product is negative 81 t squared + 16 ⇒ 2nd answer

Step-by-step explanation:

The product of two binomials (ax + b)(cx + d), where a, b, c, and d are constant

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  • Multiply (ax) by (d) and (b) by (cx) ⇒ ext-reams and nears
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Let us find the product of (9 t - 4) and (-9 t - 4)

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Add the like terms

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Multiply the 2nd two terms

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Write the answer

∴ (9 t - 4)(-9 t - 4) = -81 t² + 0 + 16

∴ (9 t - 4)(-9 t - 4) = -81 t² + 16

The product is  -81 t² + 16

Learn more:

You can learn more about the product of algebraic expressions in brainly.com/question/1617787

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