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Verdich [7]
2 years ago
9

20+20+20+20+20+90000000+?​

Mathematics
2 answers:
Scorpion4ik [409]2 years ago
5 0

Answer:

90000100

Step-by-step explanation:

20 + 20 = 40

20 + 20 = 40

40 + 40 = 80

80 + 20 = 100

90000000 + 100 = 90000100

e-lub [12.9K]2 years ago
4 0

Answer: 90000100

Hope that helped! :D

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r = interest rate
t = term/time

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faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

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Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

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In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

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\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

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There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

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\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

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