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3 years ago

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Two-fifths of one less than a number is less than three-fifths of one more than that number. What numbers are in the solution se

t of this problem? mc026-1.jpg mc026-2.jpg mc026-3.jpg mc026-4.jpg

2 answers:

USPshnik [31]3 years ago

7 0

The answer is x>-5 i took the test this is the correct answer.

adoni [48]3 years ago

3 0

Let

x-------> the number

we know that

The expression " Two-fifths of one less than a number is less than three-fifths of one more than that number" represent the equation

$\frac{2}{5}(x-1) < \frac{3}{5}(x+1)$

Multiply by $5$ both sides

$2(x-1) < 3(x+1)$

$2x-2 < 3x+3$

$-3-2< 3x-2x$

$-5 < x$ --------> $x >-5$

The solution set is the interval -------> (-5,∞) the number $-5$ is not included

therefore

<u>the answer is</u>

The solution set is the interval (-5,∞)

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Aloiza [94]

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(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

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$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

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=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

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Integrating, we have:

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Inputting the boundary conditions t = a = ∞, t = 0:

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$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

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Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

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