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3 years ago

Algebra 1. Question 13

Mathematics

1 answer:

SIZIF [17.4K]3 years ago

8 0

Hello!

The domain is the set of all possible x-values which will make the function true, and will output real y-values. Filled in points make the function true, and is written as ≤, but open circles, will make the function false and is written as just a less than symbol (<).

In this case, the ordered pair (-5, -2) is part of our domain, while the ordered pair (4, 1) is not because it is an open circle.

Therefore, the domain of the function is -5 ≤ x < 4, [-5, 4), {x | -5 ≤ x < 4}.

The range is the y-values of the function. In this case, the point (1, -4) is part of our range is also the minimum y-value, -4. The maximum y-value is the vertex of the parabola (the curve), and is the ordered pair, (-2, 3).

Therefore, the range of the function is -4 ≤ y ≤ 3, [-4, 3], {y | -4 ≤ y ≤ 3}.

Note: I wrote the domain and range in three ways because it doesn't specify which way to write it.

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Step-by-step explanation:

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3 years ago

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artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

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Using the theorem of the Laplace transform for derivatives, we know that:

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Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

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we deduct from here

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and

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By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

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$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

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Answer:

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Step-by-step explanation:

142=10x+8x+2

Move all numbers with the same variables to each side to separate them.

142-2=10x+8x

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Divide by 18.

x= 140/18

x=7.78

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3 years ago

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