1. Mrs. Gonzalez signed up to bring two dozen cupcakes for the school bake sale. She made one dozen, but her children ate three
1 answer:
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The complete question is
<span>Bradley made a house for his dog, Bowser, out of wood with a cube base and a triangular prism top. The dimensions of the dog house are a = 2 feet, b = 1 foot, and c = 1.4 feet.
</span><span>If Bradley plans to paint the outside of the dog house blue, not including the bottom, how many square feet of paint will he use?
the picture in the attached figure
</span>
we know that
surface area of the figure=surface area of cube+surface area of triangular prism
surface area of cube=4*a²----> 4*2²-----> 16 ft²
surface area of triangular prism=2*[a*c]+2*[a*b/2]---> 2*[2*1.4]+[2*1]
surface area of triangular prism=5.6+2----> 7.6 ft²
surface area of the figure=16 ft²+7.6 ft²----> 23.6 ft²
the answer is
23.6 ft²
<span>Bradley made a house for his dog, Bowser, out of wood with a cube base and a triangular prism top. The dimensions of the dog house are a = 2 feet, b = 1 foot, and c = 1.4 feet.
</span><span>If Bradley plans to paint the outside of the dog house blue, not including the bottom, how many square feet of paint will he use?
the picture in the attached figure
</span>
we know that
surface area of the figure=surface area of cube+surface area of triangular prism
surface area of cube=4*a²----> 4*2²-----> 16 ft²
surface area of triangular prism=2*[a*c]+2*[a*b/2]---> 2*[2*1.4]+[2*1]
surface area of triangular prism=5.6+2----> 7.6 ft²
surface area of the figure=16 ft²+7.6 ft²----> 23.6 ft²
the answer is
23.6 ft²
Answer:
The cost of the material for the full-sized pad is;
d) 222,750.00
Step-by-step explanation:
The parameters of the steel pad and the scale model are;
The volume of the deck = 18 yd³
The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick
The weight of the sample, m₂ = 75 lbs
The cost of the steel, P = $55/lb
The dimensions of the sample in yards are;
6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0. yards
The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0. yd. = (1/3) yd.³
The density of the sample material, ρ = m₂/V₂
∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³
Therefore, the density of the material of the pad, ρ = 225 lbs/yd³
The mass of the steel pad, m₁ = ρ × V₁
∴ The mass of the steel pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs
The cost of the material for the full-sized steel pad, C = m₁ × P
∴ C = 4,050 lbs × $55/lb = $222,750
