At time t=0 hours, a tank contains 3000 litres of water. Water leaks from the tank. At the end of every hour there is x% less water in the tank than at the start of the hour. The volume of water, in litres, in the tank at time t hours is Vt. Given that Vt=KtV0 V2=2881.2 work out the value of k and the value of x.

x = 2; K = 0.98

Step-by-step explanation:

At time, t = 0, Volume of Water = 3,000 litres

After t hours , the Volume of water in the tank = Vt

At the end of every hour there is a x% less water in the tank then at the start of the hour

After 1 hour, volume of water left Vt = V1

V1 = 3,000 - 3000 × (x/100)

V1 = 3,000 - 30x

After 2 hours, Volume of water left is V2

V2 = (3,000 - 30 x) - (3,000 - 30 x) × (x/100)

V2 = (3,000 - 30 x) - (3000x - 30x²)/100

V2 = {100(3000 - 30x) - x(3000 - 30x)} /100

V2 = {(3000 - 30x)(100 - x)}/100

Recall, V2 = 2881.2

Therefore

{(3000 - 30x)(100 - x)}/100 = 2881.2

(3000 - 30x)(100 - x) = 288120

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

30 x ² - 6000 x + 11880 = 0

Dividing both sides by 30

x² - 200 x + 396=0

Factorizing:

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

x = 198 or x = 2

Since the percentage by which the water reduces cannot be greater than 100, x = 2 is chosen.

Step 2: Solving for k

Given that Vt=KtV0

K = Vt/V0

At t = 1

V1 = 3000 - 30x

V1 = 3000 - 30 × 2

V1 = 3000 - 60

V1 = 2940

K = 2940/3000

K = 0.98

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

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answer: d.) the total number of protons and neutrons is 15

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