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Alexus [3.1K]
2 years ago
6

Use the laplace transform to solve the given initial-value problem. y' + y = (t − 1), y(0) = 3

Mathematics
1 answer:
GaryK [48]2 years ago
5 0

In mathematics, the Laplace transform, named after its discoverer Pierre Simon Laplace, transforms a function of real variables (usually in the time domain) into a function of complex variables (in the time domain). is the integral transform that Complex frequency domain, also called S-area or S-plane).

Learn more about Laplace transform here brainly.com/question/17062586

#SPJ4

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dezoksy [38]

Answer:

Im not so sure about what your asking, but I'm assuming that your asking about the lengths of the circle. The 6 inches would be the diameter and the 3 inches would be the radius and the 3.14 would be the circumference.

Step-by-step explanation:

I hope this helps, if it doesn't then just message me and ill be more than happy to help :)

6 0
2 years ago
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The ratio of the sides of a triangle is 3:4:5 and its perimeter is 48 inches. Find the length of the shortest side.
exis [7]

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12 inches

Step-by-step explanation:

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1. x=-3 y=4 2. x=5/2 y=15/2 3. x=9/5 y=-36/5 4. x=-2 y=1
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3 years ago
Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)
fredd [130]

Answer:

f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3

First tangent line:

y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5

Second tangent line:

y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7


Notice: slope of -1 means that both f'(x_1), \ f'(x_2) are equal to -1, so f'(x_1)=-1 \ and \ f'(x_2)=-1


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3 years ago
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