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SSSSS [86.1K]
3 years ago
15

What are two ways the debt- to- GDP ratio can in crease

Mathematics
1 answer:
yulyashka [42]3 years ago
7 0

Answer:

Governments should reduce spending, and encourage growth through production and exportation, or increase tax revenues.

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PLSS helpp <br><br> btw have a nice day
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It’s the first one! The table shows X going up and Y going down, so the slope would be negative!
8 0
3 years ago
Write the recurring decimal 0.5 as a fraction in its simplest form.​
sergejj [24]

Answer:

0.5×2/2=1/2 is the required fraction.

8 0
3 years ago
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NEED HELP, ASAP! !!! 2 QUESTIONS.
Maksim231197 [3]

Answer:

1.we know the perimeter of the triangle=x-3+x+6+x=3x+3

so,answer is (b)3x+3

2.area of the triangle=1/2*b*h

=1/2*x+2*2x+6=x+2*2x+6/2=2x^2+6x+4x+12/2=2x^2+10x+12/2=2(x^2+5x+6)/2=x^2+5x+6

so,answer is(b)x^2+5x+6

8 0
2 years ago
Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

3 0
3 years ago
What is the solution of the equation when solved over the complex numbers?<br><br> x^2+27=0
Luba_88 [7]

Answer:

(0,27)

Step-by-step explanation:

x2 + 27 = 0

Solutions based on quadratic formula:

x1  

= −0 − √ 02 − 4×1×27   2×1 = 0 − 6 × √ 3 i   2 ≈ −5.196152i

x2  

= −0 + √ 02 − 4×1×27   2×1 = 0 + 6 × √ 3 i  2 ≈ 5.196152i

Extrema:

Min = (0, 27)

8 0
3 years ago
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