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tigry1 [53]
3 years ago
10

Ashleigh rode her bicycle 26.5 miles in 4 hours.which gives best estimate of how far Ashleigh rode in 1 hour

Mathematics
1 answer:
Firdavs [7]3 years ago
4 0
Ashleigh rode her bicycle about 6 miles in 1 hour 
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Pls I’m about to cry...
marissa [1.9K]

FORMULA:

  • Volume of cylinder = πr²h
  • Volume of sphere = 4/3πr³

ANSWER:

Volume of capsule = 4/3πr³ + πr²h

πr²(4/3r + h)

  • 3.14 × 2²(8/3 + 5.6)
  • 3.14 × 4(2.6 + 5.6)
  • 12.56(8.2)
  • 102.99 or 103 rounded.

Now, Density = 0.7mg/mm³

  • 0.7/103
  • 0.00679mg

7 0
2 years ago
-|-2-3(6)| evaluation
Fiesta28 [93]

Answer:

-20

Step-by-step explanation:

5 0
3 years ago
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Use the table to determine the location with the greatest amount of rainfall and the location with the least amount of rainfall.
Kryger [21]

Step-by-step explanation:

I think it is hilo hawii 480 cm it has the longest cm

7 0
2 years ago
Help please. (12 points)
Sphinxa [80]
 because the symbol is a less than sign < the blue area needs to be below the red line and does not include the solution to Y ( if it included Y it would be a ≤ symbol.
 When the solution isn't included, it is a dashed line.
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6 0
3 years ago
A person is standing 50 ft from a statue. The person looks up at an angle of elevation of 8 degrees when staring at the top of t
borishaifa [10]

The height of the statue is 19.5 feet.

Why?

We can solve the problem using trigonometric formulas. In this case, we are going to use the trigonometric formula of the tangent.

We know that the person is standing 50ft from the statue, so, it will be the base of the two triangles formed by both angles (elevation and depression)

Using the trignometric formula, we have:

First triangle:

tg(\alpha )=\frac{h_{1}}{base} \\\\tg(8\°)=\frac{h_{1}}{50ft}\\\\0.14*50ft=h_1\\\\h_1=7ft

Second triangle:

tg(\alpha )=\frac{h_{2}}{base} \\\\tg(14\°)=\frac{h_{1}}{50ft}\\\\0.25*50ft=h_1\\\\h_1=12.5ft

Now, the total height of the statue will be:

TotalHeight=h_1+h_2=7ft+12.5ft\\\\TotalHeight=19.5ft

Have a nice day!

8 0
3 years ago
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