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11111nata11111 [884]
3 years ago
12

Given that ABCD is a rectangle, find the perimeter of the figure below. Round your answer to the nearest tenth. *

Mathematics
1 answer:
BARSIC [14]3 years ago
6 0

Answer:

<em><u>ANSWER</u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>OPTION</u></em><em><u> </u></em><em><u>B</u></em><em><u> </u></em>

<em><u>19.7FT</u></em>

PERIMETER OF FIG.=2 (L+B)-B + PI*R

=2 (6+3)-3 + 3.14*1.5

=18-3+4.71

=15+4.71

=19.71

=19.7 ft

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A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11
mixer [17]
Let A(t) denote the amount of salt in the tank at time t. We're given that the tank initially holds A(0)=100 lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}
\implies A'(t)+\dfrac{11}{200+33t}A(t)=484

Find the integrating factor:

\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}

Distribute \mu(t) along both sides of the ODE:

(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}
\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}
A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt
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Since A(0)=100, we get

100=22(200)^{2/3}+C\implies C\approx-652.39

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200+(44-11)t=500\implies 33t=300\implies t\approx9.09

at which point the amount of salt in the solution would be

A(9.09)\approx733.47\text{ lb}
4 0
3 years ago
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