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Nikolay [14]
4 years ago
8

Lukas graphed the system of equations shown. 2x + 3y = 2 y = x + 3 What is the solution to the system of equations? (–2, 2) (0,

0.67) (0, 3) (1, 0)

Mathematics
2 answers:
Nat2105 [25]4 years ago
4 0

Answer:

Step-by-step explanation:

GIven that Lukas graphed the system of equations shown

2x+3y =2

y=x+3

x=-1.4: y = 1.6

Hitman42 [59]4 years ago
3 0
The closest answer that I could find would be (-2, 2). The answer to the system is (-1.4, 1.6)

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A scale on a blue print drawing of a house shows that 66 centimeters represents 33 meters. What number of centimeters on the blu
aksik [14]
To find the ratio of centimeters to meters you need to divide 66 by 33 to get 2.
Then multiply 2727 by 2 to get 5454.
5 0
3 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}

3 0
3 years ago
What is the value of x in the equation (-35= 1? Explain.
s344n2d4d5 [400]

Answer:

x=1/35

Step-by-step explanation:

8 0
3 years ago
The domain (Graphically)?
gladu [14]

Answer:

Domain: (-infinite, -8) U (-8, 2) U (2, +infinite)

Step-by-step explanation:

The graphs never intersect with the lines x=-8 and x=2 so -8 and 2 cannot be included in the domain.

Use "[]" for included and "()" for excluded.

The domain will go from far left to -8 => (-infinite, -8).

Then, the domain will go from -8 to 2 => (-8, 2).

Last, the domain will go from 2 to +infinite => (2, +infinite).

8 0
3 years ago
PLS HELP ME WITH MY GEOMETRY
saveliy_v [14]
1/5

9/45=1/5
8/40=1/5

I think this is right
6 0
4 years ago
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