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nata0808 [166]
4 years ago
7

An ice cream package is done shaped. It has a radius of 5.5 centimeters and a height of 17 centimeters. What is the volume of th

e ice cream cone?
Mathematics
1 answer:
kicyunya [14]4 years ago
8 0
You would do 17 times 5.5 = 93.5, so the volume of the ice cream cone would be 93.5 centimeters. I’m assuming that’s the right answer but you might want to get a second opinion.
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Pls look at this pic and help like i don't understand i will mark u as brainliest
dimulka [17.4K]

Answer:

48 cubic yard

Step-by-step explanation:

volume \: of \: pyramid \\  = area \: of \: base \times height \\  =  {4}^{2}  \times 3 \\  = 16 \times 3 \\  = 48 \:  {yard}^{3}  \\

3 0
3 years ago
Kendrick takes a sheet of paper and makes a cut from corner to corner the cut he makes is 9 inches long the length of the paper
NNADVOKAT [17]

Answer:

Step-by-step explanation:

765

3 0
3 years ago
An Native American tepee is a conical tent. Find the number of cubic feet of air in a teepee 10 ft. in diameter and 12 ft. high.
marta [7]
For this case what you should know is that by definition the volume of the cone is given by:
 V = ((pi) * (r ^ 2) * (h)) / (3)
 Remembering that:
 r = d / 2
 Substituting the values:
 V = ((pi) * ((d / 2) ^ 2) * (h)) / (3)
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 Answer:
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 The nearest option is:
 300 cu. ft
7 0
3 years ago
Read 2 more answers
An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.
Shtirlitz [24]

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.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 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

6 0
3 years ago
8.57×10*-4 in standard form​
Stells [14]

Answer:

342.8

Step-by-step explanation:

8 0
3 years ago
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