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GREYUIT [131]
3 years ago
9

Find the slope-intercept form of the equation of the line through the point (0, -4) and (1, 0).

Mathematics
1 answer:
Arisa [49]3 years ago
4 0

Answer:

The Slope Intercept form of the equation through the point (0, -4) and

(1, 0)  is  y = 4 x - 1

Step-by-step explanation:

Here, the given points are A (0,-4) and B(1,0)

Now, slope of the equation with points (a,b) and (c,d) is given as

m = \frac{d - b}{c -a}

So, here the slope of the line AB is given as:

m = \frac{0 - (-4)}{1-0} = \frac{4}{1}  = 4

⇒ m = 4

Now, by POINT SLOPE  form:

The equation with slope m and point  (x0 , y0) is represented as :

(y-y0)  = m (x-x0)

⇒ The equation of AB is of the form with point (1,0) is

y - 0 = 4 (x -1)

or, y = 4 x - 1

Also, the SLOPE INTERCEPT form of an equation is of the form

y = m x + C ;     m = slope and C =  y -intercept

Hence, the slope intercept form of the equation through the point (0, -4) and (1, 0)  is  y = 4 x - 1

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A colony of bacteria is growing at a rate of 0.2 times its mass. Here time is measured in hours and mass in grams. The mass of t
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Answer:

  • <u>Question 1:</u>      dm/dt=0.2m<u />

<u />

  • <u>Question 2:</u>     m=Ae^{(0.2t)}<u />

<u />

  • <u>Question 3:</u>      m=10e^{(0.2t)}<u />

<u />

  • <u>Question 4:</u>      m=10g<u />

Explanation:

<u>Question 1: Write down the differential equation the mass of the bacteria, m, satisfies: m′= .2m</u>

<u></u>

a) By definition:  m'=dm/dt

b)  Given:  rate=0.2m

c) By substitution:  dm/dt=0.2m

<u>Question 2: Find the general solution of this equation. Use A as a constant of integration.</u>

a) <u>Separate variables</u>

     dm/m=0.2dt

b)<u> Integrate</u>

           \int dm/m=\int 0.2dt

            ln(m)=0.2t+C

c) <u>Antilogarithm</u>

       m=e^{0.2t+C}

       m=e^{0.2t}\cdot e^C

         e^C=A\\\\m=Ae^{(0.2t)}

<u>Question 3. Which particular solution matches the additional information?</u>

<u></u>

Use the measured rate of 4 grams per hour after 3 hours

            t=3hours,dm/dt=4g/h

First, find the mass at t = 3 hours

            dm/dt=0.2m\\\\4=0.2m\\\\m=4/0.2\\\\m=20g

Now substitute in the general solution of the differential equation, to find A:

          m=Ae^{(0.2t)}\\\\20=Ae^{(0.2\times 3)}\\\\A=20/e^{(0.6)}\\\\A=10.976

Round A to 1 significant figure:

  • A = 10.

<u>Particular solution:</u>

           

             m=10e^{(0.2t)}

<u>Question 4. What was the mass of the bacteria at time =0?</u>

Substitute t = 0 in the equation of the particular solution:

         m=10e^{0}\\\\m=10g

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