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3 years ago

The sum of six consecutive integers is −9. What are the integers?

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

4 0

Answer:

4, -3, -2, -1, 0, 1

Step-by-step explanation:

since there are 6 consecutive integers, we can say create an equation:

n + (n+1) + (n+2) + (n+3)+ (n+4) + (n+5) = -9

add like terms to get:

6n + 15 = -9

subtract 15 from both sides:

6n = -24

divide both sides by 6 and you get n:

n = -4

plug -4 into n + (n+1) + (n+2) + (n+3)+ (n+4) + (n+5) to find all six integers:

-4 + (-4 + 1) + (-4 + 2) + (-4 + 3)+ (-4 + 4) + (-4 + 5)

simplify the parenthesis:

-4, -3, -2, -1, 0, 1

the sum of all of these number are -9

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(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

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Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

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

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

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

where a = ∞

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

where d(t) = δ(t)

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

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

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

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

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

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

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

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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