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maks197457 [2]
3 years ago
15

At time t=0 hours a tank contains 3000 litres of water .

Mathematics
2 answers:
Degger [83]3 years ago
8 0

Answer:

At t=0, Volume of Water = V_{0} = 3,000 Litre

After t hours , the Volume of water in the tank = V_{1}

As it is also given , At the end of every hour there is a x% less water in the tank then at the start of the hour

V_{1} = Amount of water after an hour = 3,000 - 3,000 \times \frac{x}{100} = 3,000 - 30 x

V_{2} = Amount of water after two hour =(3,000 - 30 x) - (3,000 - 30 x)×\frac{x}{100}= (3,000 - 30 x)(\frac{100 - x}{100})

Also, V_{2} = 2881.2

⇒(3000 - 30 x)(100 -x) = 288120

⇒300000 - 3000 x - 3000 x + 30 x²  = 288120

⇒ 30 x ² - 6,000 x + 11880= 0

Dividing both sides by 30, we get

⇒ x² - 200 x + 396=0

Factorizing the quadratic expression

(x -198)(x-2) = 0

x = 198, x = 2

⇒ Which gives a value of x = 2 as a possible solution of this equation.

Also, V_{1} = k ^1 V_{0}

k = \frac{V_{1}}{V_{0}}

V_{1} = 3,000 - 30 \times 2 = 3,000 - 60= 2,940

k = \frac{2940}{3000}= 0. 999.......(non terminating repeating)

k= 0.999 (approx)



goldfiish [28.3K]3 years ago
7 0

Finding x:

v^2=2881.2

2881.2=3000(1-x)^2

2881.2/3000=(1-x)^2=0.9604

1-x=0.098

x=2%

Finding k:

v^2=k^2*3000

2881.2=k^2*3000

k^2=0.9604

k=0.98

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To calculate the distance between two points, we use the formula d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}.

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