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pantera1 [17]
3 years ago
5

What is the y intercept of the liner function that contains the points (4,1) (-3, -20)

Mathematics
1 answer:
myrzilka [38]3 years ago
8 0

Answer:

<h3>-11</h3>

Step-by-step explanation:

To get the y-intercept, we need to get the slope first

Slope = y2-y1/x2-x1

Slope = -20-1/-3-4

slope = -21/-7

Slope = 3

Get the y-intercept. Substitute m = 3 and any point say (4, 1) into the expression y = mx+c

c is the y-intercept

1 = 3(4) + c

1 = 12 + c

c =1-12

c = -11

Hence the y-intercept is -11

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Step-by-step explanation:

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Answer:

a

   y(t) = y_o e^{\beta t}

b

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c

      \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

Step-by-step explanation:

From the question we are told that

    \frac{dy}{y} =  -\beta dt

Now integrating both sides

     ln y  =  \beta t + c

Now taking the exponent of both sides

       y(t) =  e^{\beta t + c}

=>     y(t) =  e^{\beta t} e^c

Let  e^c =  C

So

      y(t) = C e^{\beta t}

Now  from the question we are told that

      y(0) =  y_o

Hence

        y(0) = y_o  = Ce^{\beta * 0}

=>     y_o = C

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        y(t) = y_o e^{\beta t}

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      \frac{dx}{dt}  = -\alpha xy

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      \frac{dx}{dt}  = - \alpha x(y_o e^{-\beta t })

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Now integrating both sides

         lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c

Now taking the exponent of both sides

        x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}

=>     x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c

Let  e^c  =  A

=>  x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }

Now  from the question we are told that

      x(0) =  x_o

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      x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }

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divide both side  by    (K * x_o)

=>    K = x_o e^{\frac {\alpha y_o  }{\beta } }

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so

    \lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }

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