The lowest point in a business cycle, which follows a period of economic

Mathematics

2 answers:

Anika [276]3 years ago

7 0

Your answer is b hope this helps

Nonamiya [84]3 years ago

4 0

Answer: it’s a trough

Step-by-step explanation:

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2. Which graph represents the solution to the given system? (1 point)

matrenka [14]

Answer:

1000000000000000million

4 0

2 years ago

Write an equation of a line in slope intercept form that is parallel to y = 3x+6 and passes through the point (-10, 2.5)

TiliK225 [7]

1) Line should be parallel to y = 3x+6, so slope of this line should be =3.

y=mx +b

y=3x +b

Now we have to find b (y-intercept), using the point (-10, 2.5).

2.5 = 3*(-10)+b

b=2.5+30=32.5

y=3x + 32.5

2) The line should be perpendicular to y = -4x -2, so its slope is going to be negative reciprocal m= 1/4.

Now we have to find b (y-intercept), using the point (-16, -11).

y=(1/4)x +b

-11=(1/4)*(-16) + b

-11 = -4 +b

b = -7

y=(1/4)x - 7

3 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$