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3 years ago

12

Please help!! Will mark first answerer brainiest!! Thank you so much!

2 answers:

Mrac [35]3 years ago

8 0

The answer is C i believe, always remember to add alike terms.

Ksenya-84 [330]3 years ago

5 0

First distribute -5 to everything in the parentheses:

-15m^3 -35n^2

And simplify like terms:
-35n^2 + 8n^2 = -27n^2

Answer:
-15m^3 - 27n^2

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3 years ago

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Read 2 more answers

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:

<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:

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

3 0

3 years ago