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
Answer:
-2.82842712475
Step-by-step explanation:
x^2+10+25=27
x^2+35=27
x^2+35-27=0
x^2+8=0
8=-x^2
SQR8=-x
-SQR8=x
748 student tickets were sold.
Step-by-step explanation:
Given,
Number of tickets sold = 1360
Revenue generated = $13328
Cost of each student ticket = $8
Cost of each non student ticket = $12
Let,
x be the number of student tickets sold
y be the number of non student tickets sold
According to given statement;
x+y=1360 Eqn 1
8x+12y=13328 Eqn 2
Multiplying Eqn 1 by 12
Subtracting Eqn 2 from Eqn 3
Dividing both sides by 4
748 student tickets were sold.
Keywords: linear equation, elimination method
Learn more about elimination method at:
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90 is 1/10 of 900 Hope this helps :)
Get a common denominator which you can get by mutiplying by k-1
Now foil and distribute
Now combine the fractions and like terms
Now factor out a k+1
you can cancel out the k+1 leaving