At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
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.
So the amount of chlorine in the tank changes according to

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):


![\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%5Cdfrac%7Bc%28t%29%7D%7B%28200-3t%29%5E%7B5%2F3%7D%7D%5Cright%5D%3D0)


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

![\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7Bc%28t%29%3D%5Cdfrac1%7B200%7D%5Csqrt%5B3%5D%7B%5Cdfrac%7B%28200-3t%29%5E5%7D5%7D%7D)
You add all of those numbers together and get your answer which is 60, so basically like this
25.50 + 14.25 + 18 + 2.25 = 60
Answer:
62
Step-by-step explanation:
The enclosing rectangle is 24 units long and 12 units high. Its area is ...
(24 units)(12 units) = 288 square units
The white circles obviously cover more than half the area, so the only viable answer choice is 62.
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If you want to go to the trouble to actually figure it out, you can find the area of each circle using the area formula ...
A = πr² = (3.14)(6²) = 113.04
Then both circles have an area of 2×113.04 = 226.08, and the shaded area is the difference between that and the rectangle area:
shaded area = 288 -226.08 = 61.92 ≈ 62
Answer:
20
Step-by-step explanation:
Total no = 75
N (P) = 48 , N (H) = 45 , N (T) = 58
N (P∩H) = 28 , N (H∩T) = 37 , N (P∩T) = 40
N (P∩H∩T) = 25
Total no = N (P) + N (H) + N (T) - N (P∩H) - N (H∩T) - N (P∩T) + N (P∩H∩T) + neither
75 = 48 + 45 + 58 - 28 - 37 - 40 + 25 + neither
75 = 71 + neither → neither = 4
N (only P) = N (P) - N (P∩H) - N (P∩T) + N (P∩H∩T) = 48 - 28 - 40 + 25 = 5
N (only H) = N (H) - N (P∩H) - N (H∩T) + N (P∩H∩T) = 45 - 28 - 37 + 25 = 5
N (only T) = N (T) - N (H∩T) - N (P∩T) + N (P∩H∩T) = 58 - 37 - 40 + 25 = 6
So, total liking either one or neither = 4 + 5 + 5 + 6 = 20