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Harlamova29_29 [7]

3 years ago

6

If x/3= x+1/4, What is the value of x

1 answer:

olya-2409 [2.1K]3 years ago

8 0

Well first because in this equation the first value is less than one, we should multiply the whole equation by 3, so
<span>x/3= x+1/4
x=3x+3/4
now we should switch the values
3x+3/4=x
now subtract x from both sides
2x+3/4=0
now subtract 3/4 from both sides
2x=-3/4
now divide both sides by 2
x=3/8
So that's your answer!

</span>

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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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