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lapo4ka [179]
2 years ago
10

How to do i solve it

Mathematics
1 answer:
Leona [35]2 years ago
3 0
Use ur brain faqg lololo
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(D. 12/3

Step-by-step explanation:

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Can you answer the attached question which includes the options to answer?
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2 and 4 are both correct statements.
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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 wa
Margaret [11]

Answer:

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

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3 years ago
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y

=

x

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x

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Step-by-step explanation:

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