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3 years ago

Which of the following numbers is greater than 190,090?

Mathematics

1 answer:

zaharov [31]3 years ago

4 0

A 190,900
(after the comma there is a 9 which makes it one hundred ninety thousand nine hundred 900 is greater than 90)

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0/16

Step-by-step explanation:

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Answer:

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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

11Alexandr11 [23.1K]

Answer:

Question 1: $dm/dt=0.2m$

Question 2: $m=Ae^{(0.2t)}$

Question 3: $m=10e^{(0.2t)}$

Question 4: $m=10g$

Explanation:

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

a) By definition: $m'=dm/dt$

b) Given: $rate=0.2m$

c) By substitution: $dm/dt=0.2m$

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

a) Separate variables

$dm/m=0.2dt$

b) Integrate

$\int dm/m=\int 0.2dt$

$ln(m)=0.2t+C$

c) Antilogarithm

$m=e^{0.2t+C}$

$m=e^{0.2t}\cdot e^C$

$e^C=A\\\\m=Ae^{(0.2t)}$

Question 3. Which particular solution matches the additional information?

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.

Particular solution:

$m=10e^{(0.2t)}$

Question 4. What was the mass of the bacteria at time =0?

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

$m=10e^{0}\\\\m=10g$

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Answer:

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Step-by-step explanation:

3150 ÷150

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