Answer:
The easiest way to do this is to reflect Figure I over the Y axis. Then move it up 3 and left 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
A
The volume (V) of a pyramid is found using the formula
V = 
× area of base × height(h)
area of base = 9² = 81 ← area of a square, hence
324 = × 81 × h = 27h ( divide both sides by 27 )
h = 
= 12 → A
Answer:
D.
Step-by-step explanation:
Lines EG and CD are cut by transversal CF.
By construction, ∠FEG=∠FCD. These two angles are corresponding angles.
Since two corresponding angles are congruent, then lines EG and CD are parallel (by converse of the corresponding angles postulate).
Converse of the Corresponding Angles Postulate: If the corresponding angles formed by two lines and a transversal are congruent, then lines are parallel.
