Solutions
To solve the equation we have to isolate 12y
im sure it would be 2:4. if it was to be simplified it would be 1:2
Answer:
The weight of the water in the pool is approximately 60,000 lb·f
Step-by-step explanation:
The details of the swimming pool are;
The dimensions of the rectangular cross-section of the swimming pool = 10 feet × 20 feet
The depth of the pool = 5 feet
The density of the water in the pool = 60 pounds per cubic foot
From the question, we have;
The weight of the water in Pound force = W = The volume of water in the pool given in ft.³ × The density of water in the pool given in lb/ft.³ × Acceleration due to gravity, g
The volume of water in the pool = Cross-sectional area × Depth
∴ The volume of water in the pool = 10 ft. × 20 ft. × 5 ft. = 1,000 ft.³
Acceleration due to gravity, g ≈ 32.09 ft./s²
∴ W = 1,000 ft.³ × 60 lb/ft.³ × 32.09 ft./s² = 266,196.089 N
266,196.089 N ≈ 60,000 lb·f
The weight of the water in the pool ≈ 60,000 lb·f
The amount of sales before growth according to the article is $153.85 million
<h3>How to find sales before growth?</h3>
let
- Sales before growth = x
- Percentage growth = 30%
- New sales = $200 million
200 = x + (30% × x)
200 = x + (0.3 × x)
200 = x + 0.3x
200 = 1.3x
x = 200/1.3
x = 153.846153846153
Approximately,
x = $153.85 million
Learn more about percentage:
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