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Ber [7]
3 years ago
11

Graph x<2, need help on this one please.

Mathematics
1 answer:
pogonyaev3 years ago
5 0
Okay. So, the line shaded would have to be left to 2, because it's less than 2. In that case, A and C are eliminated, because they shade to the right, not left. B and D do this, but wait. The dotted line has to represent the less than, because the one that's fully filled in would mean less than or equal to. B would make sense, because the line is dotted and that makes an inequality symbol of less than, not less than or equal to. The answer is B.

Note: When the sign is < or >, the line on the graph is dotted. If the sign is <= or >=, then the line on the graph is solid.
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An Interesting Equation of Order n: xy" - (x + n)y' + ny = 0: u(x) = e^x. (a) Prove that the given solution is indeed a solution
Igoryamba

Solution:

The given differential equation is

x y" -(x+n)y'+ny=0------------(1)

Let, y'=t

y"=t'

\frac{dy}{dx}=t\\\\y=t x

Substituting the value of , y', y'' and y in equation (1)

→x t' -(x+n)t+n t x=0

→ x t' = x t+ n t- nt x

→ x t'=t(x+n-nx)

\rightarrow \frac{t'}{t}=\frac{x+n-nx}{x}\\\\\rightarrow \frac{dt}{t}=(1-n+\frac{n}{x}) dx\\\\ \text{Integrating both sides}}\\\\ \int{\frac{dt}{t}}=\int {(1-n+\frac{n}{x}) dx}\\\\ \log t=x - nx+n \log x+\log K\\\\ \log t -\log x^n-\log K=x(1-n)\\\\\log \frac{t}{Kx^n}=x(1-n)\\\\t=Kx^n \times e^{x(1-n)}\\\\y'=Kx^n \times e^{x(1-n)}

which is a solution of Differential equation.

(b)

\frac{dy}{dx}=Kx^n\times e^{x(1-n)}\\\\ dy=Kx^n\times e^{x(1-n)} dx

Integrating both sides

y=\frac{Kx^n\times e^{x(1-n)}}{1-n}-\frac{Kn}{1-n}\int{x^{n-1}e^{x(1-n)} dx

required linear independent solution of Differential equation.

7 0
3 years ago
Find the amplitude of y = -4 sin x.
Kisachek [45]

Answer: The amplitude is 4.

Step-by-step explanation:

Given the form of the sine function:

y=Asin(Bx)+k

Where |A| is the amplitude, \frac{2\pi }{B} is the period and "k" is the vertical shift of the wave.

You know that, in the sine function given ( y = -4 sin x), you can identify the amplitude and the period. These are:

  •  The amplitude is:

     |A|=|-4|=4

  • And the period is:

      \frac{2\pi }{1}=2\pi

Therefore, the amplitude of  y = -4 sin x is:

|A|=4

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4 years ago
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yan [13]

3(bx-2ab)=b(x-7a)+3ab\\3bx-6ab=bx-7ab+3ab\\3bx-6ab=bx-4ab\\3bx-bx=-4ab+6ab\\2bx=2ab\\bx=ab\\x=a

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