Answer:
Therefore the tank will be half empty after 10.10 days.
Therefore there will be 2735.53 liters water in the tank after 4 days.
Step-by-step explanation:
Given that,the square root of the volume of remaining water in the tank is proportional to the rate of water leakage.
Let V be volume of water at any instant time t.
![\therefore \frac{dV}{dt} \propto \sqrt V](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cfrac%7BdV%7D%7Bdt%7D%20%5Cpropto%20%20%5Csqrt%20V)
where k is constant of proportionality.
![\Rightarrow \frac{dV}{\sqrt V}= k \ dt](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7BdV%7D%7B%5Csqrt%20V%7D%3D%20k%20%5C%20dt)
Integrating both sides
![\Rightarrow \int \frac{dV}{\sqrt V}= \int k \ dt](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cint%20%5Cfrac%7BdV%7D%7B%5Csqrt%20V%7D%3D%20%5Cint%20k%20%5C%20dt)
[ C is integrating constant]
At t=0, the volume of water is 350 liters
![2\sqrt{350}=k.0+C](https://tex.z-dn.net/?f=2%5Csqrt%7B350%7D%3Dk.0%2BC)
![\Rightarrow C=2\sqrt{350}](https://tex.z-dn.net/?f=%5CRightarrow%20C%3D2%5Csqrt%7B350%7D)
The equation becomes
![\Rightarrow 2\sqrt V=kt+2\sqrt{350}](https://tex.z-dn.net/?f=%5CRightarrow%202%5Csqrt%20V%3Dkt%2B2%5Csqrt%7B350%7D)
Again at t=1, the volume of water is(350-20)liters=330 liters
![2\sqrt {330}=k.1+2\sqrt{350}](https://tex.z-dn.net/?f=2%5Csqrt%20%7B330%7D%3Dk.1%2B2%5Csqrt%7B350%7D)
![\Rightarrow k=2\sqrt {330}-2\sqrt{350}](https://tex.z-dn.net/?f=%5CRightarrow%20k%3D2%5Csqrt%20%7B330%7D-2%5Csqrt%7B350%7D)
The equation becomes
![2\sqrt V=2(\sqrt{330}-\sqrt{350} ) t+2\sqrt{350}](https://tex.z-dn.net/?f=2%5Csqrt%20V%3D2%28%5Csqrt%7B330%7D-%5Csqrt%7B350%7D%20%29%20t%2B2%5Csqrt%7B350%7D)
![\Rightarrow \sqrt V=(\sqrt{330}-\sqrt{350} ) t+\sqrt{350}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%20V%3D%28%5Csqrt%7B330%7D-%5Csqrt%7B350%7D%20%29%20t%2B%5Csqrt%7B350%7D)
Now when the tank is half empty,then V= (350÷2) liters = 175 liters
![\sqrt {175}=(\sqrt{330}-\sqrt{350} ) t+\sqrt{350}](https://tex.z-dn.net/?f=%5Csqrt%20%7B175%7D%3D%28%5Csqrt%7B330%7D-%5Csqrt%7B350%7D%20%29%20t%2B%5Csqrt%7B350%7D)
![\Rightarrow (\sqrt{330}-\sqrt{350} ) t=\sqrt{175}-\sqrt{350}](https://tex.z-dn.net/?f=%5CRightarrow%20%28%5Csqrt%7B330%7D-%5Csqrt%7B350%7D%20%29%20t%3D%5Csqrt%7B175%7D-%5Csqrt%7B350%7D)
![\Rightarrow t=\frac{\sqrt{175}-\sqrt{350}}{ (\sqrt{330}-\sqrt{350} )}](https://tex.z-dn.net/?f=%5CRightarrow%20t%3D%5Cfrac%7B%5Csqrt%7B175%7D-%5Csqrt%7B350%7D%7D%7B%20%28%5Csqrt%7B330%7D-%5Csqrt%7B350%7D%20%29%7D)
days
Therefore the tank will be half empty after 10.10 days.
After 4 days,
![\sqrt V=(\sqrt{330}-\sqrt{350} ) (4)+\sqrt{350}](https://tex.z-dn.net/?f=%5Csqrt%20V%3D%28%5Csqrt%7B330%7D-%5Csqrt%7B350%7D%20%29%20%284%29%2B%5Csqrt%7B350%7D)
![\Rightarrow \sqrt V= 16.54](https://tex.z-dn.net/?f=%5CRightarrow%20%5Csqrt%20V%3D%2016.54)
liters
Therefore there will be 2735.53 liters water in the tank after 4 days.
<span>since h=2t+3
i.e. at h=13 we find
13= 2t+3
i.e. 13-3= 2t
10=2t
i.e. t=10/2
i.e. t=5 days
similarly at h=27
we find
27= 2t+3
i.e. 27-3= 2t
24=2t
i.e.
t=24/2
i.e.
t=12 days
hence in 5 and 12 days the height of the tomato plant will be between 13 and 27 inches respectively.</span>
Answer: $25639.04
plz mark me as brainliest took long time to type
Step-by-step explanation:
i am doing the method in which u find the simple interest of first year then second year
SI FOR 1ST YEAR= P X R X T / 100
SI = 23600 X 2 X 4 / 100
SI = 1888
SI FOR SECOND YEAR =
P = 23600 + 1888= 25488
SI = 25488 X 2 X 4 / 100
SI= 2039.04
COMPOUND INTEREST (CI) = PRINCIPLE + SI OF 2ND YEAR
CI =$25639.04
or u can solve be the method
CI = amount - principle
Amount= principle x (change in ratio) raised to time
The answer is 6r + 21
Can you mark it brainliest?