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)
Yes because you only need to look at the first decimal for problems like these and the first number you see a seven and a one and seven is greater than one
Answer:
The value now = $207,360
Step-by-step explanation:
Rate of increase = 20% per year = 20/100 = 0.2 per year
value 3 years ago = $120,000
interest at year 1:
20% of 120,000
= 0.2 × 120,000 = 24,000
Value at the end of year 1
= initial value + interest
= 120,000 + 24,000 = $144,000
In year 2:
20% of 144,000
= 0.2 × 144,000 = 28,800
Value at the end of year 2:
144,000 + 28,800 = $172,800
In year 3
0.2 × 172,800 = 34,560
Value at the end of year 3:
34,560 + 172,800 = $207,360
That’s cool. i need p01nts so imma comment
Answer:
median=mode
Step-by-step explanation: