Answer:
115 degrees.
Step-by-step explanation:
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
11x-2; SInce you have to add 6x and 5x because they are 'like terms' meaning they share the same variable. You also add the 6 and the -8, which is basically doing 6-8, which equals -2. So, you put these two numbers together to get an expression of 11x-2.
F(x)*g(x)=(2x^2+x-3)(x-1)=(2X^3+x^2-3x)-(2x^2+x-3)=2x^3-x^2-4x-3
so the answer should be C