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3 years ago

2. - 4x + 9y = 14 12x – 10y = -8 ; multiplication

Mathematics

2 answers:

tankabanditka [31]3 years ago

7 0

Answer:

-4x + 9y = 14 answer is x = $\frac{9}{4}y+ \frac{-7}{2}$

12x - 10y = -8 answer is x = $\frac{5}{6}y +\frac{-2}{3}$

Step-by-step explanation:

solve for x

-4x + 9y = 14

add -9y to both sides

-4x + 9y + -9y = 14 + -9y

-4x = -9y + 14

divide both sides by -4

-4x/-4 = -9y + 14/-4

Anwser

x = $\frac{9}{4}y + \frac{-7}{2}$

solve for x

12x - 10y = -8

add 10y to both sides

12x - 10y + 10y = -8 + 10y

12x = 10y -8

Divide both sides by 12

12x/12 = 10y - 8/12

Anwser

x = $\frac{5}{6}y +\frac{-2}{3}$

NeTakaya3 years ago

3 0

Answer:

Step-by-step explanation: (Equation 2) = 12(-12+9y/4) - 10y = -8

-144+108y/4 - 10y = -8

-144 + 108y - 40y/4 = -8

-144 + 108y - 40y = -32

108y - 40y = -32+144

68y = 112

Y= 1.64

2-4x+9y=14

12x-10y= -8

2-14+9y = 4x

-12 + 9y = 4x ( equation 3)

(Put in equation 2)

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2 years ago

(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

Aloiza [94]

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:

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

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

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

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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