I don't why you put -+ but I will go ahead and assume that it is -2.
4-+6
I think it is saying 4 - positive 6 but I'm guessing.
Answer:
lateral area = 2320 m²
Step-by-step explanation:
The question wants us to calculate the lateral area of a square base pyramid. The square base pyramid has a side of 40 meters.The height is 21 meters.
Half of the square base is 40/2 = 20 meters . With the height it forms a right angle triangle. The hypotenuse side is the slant height of the pyramid.
Using Pythagoras's theorem
c² = a² + b²
c² = 20² + 21²
c² = 400 + 441
c² = 841
square root both sides
c = √841
c = 29 meters
The slant height of the pyramid is 29 meters.
The pyramid has four sided triangle. The lateral area is 4 multiply by the area of one triangle.
area of triangle = 1/2 × base × height
base = 40 meters
height = 29 meters
area = 1/2 × 40 × 29
area = 580
area of one triangle = 580 m²
Lateral area = 4(580)
lateral area = 2320 m²
Answer:
B
Step-by-step explanation:
The equation is Y = mx + b so just plug in the m and the b
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
