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Pie
3 years ago
12

Fritz attended band practice for 5/6 hour. Then he went home and practiced for 2/5 as long as band practice. How many minutes di

d he practice at home?
PLZ HELP I NEED TO FINISH THIS PROBLEM!!!...
Mathematics
2 answers:
iren2701 [21]3 years ago
4 0
The answer is 1/3 SmartUnicorn101
exis [7]3 years ago
4 0
5/6 of an hour is 50 minutes, so we know that band practice ran 50 minutes. 
If he went home and practiced for 2/5 of 50 minutes, that means that he practiced for 20 minutes. 
Fritz practiced for 20 minutes at home. 
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dezoksy [38]
Volume = area of base *height 

20 = pir^2 *h

20 = 3,14*4*h

h = 20/12,56 = 1,59 rounded 1,6

- using the triangle formed by radius,the height of cup and the slant height of cup so we can writing that

tan ß = h/r so tan ß = 1,6/2 = 0,8

tan ß = 0,8 so ß = arctan 0,8 = 38,6 degrees so 39 degrees 

hope helped 
5 0
3 years ago
To write a polynomial as a product is to _ _ _ _ _ _ it.
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Answer:

factor is the answer

Step-by-step explanation:

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4 years ago
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How many inches are there a in 13 feet
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156

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4 0
3 years ago
Why is it necessary to apply inverse operations on both sides of the equals l sign when solving an equations
liraira [26]
Take this example (cause it's easy to explain)

10+5x=15x

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8 0
4 years ago
The part of a cylindrical soup can that is covered by the label is 8.5 cm tall and has a diameter of 6.5 cm. What is the area of
julia-pushkina [17]

Answer:

180.4~cm^2

Step-by-step explanation:

<u>Surface Area</u>

The surface area of a cylinder of height h and radius r is given by:

A=2\pi rh

It only covers the lateral side of the cylinder. If both the top and the bottom sides are to be included, then:

A=2\pi rh+2\pi r^2

The label will cover only the lateral side of the soup can that has a height of h=8.5 cm and a diameter of 6.5 cm. We need to calculate the radius which is half of the diameter r=6.5 cm / 2 = 3.25 cm.

Now we calculate the side area of the can:

A=2\pi (3.25)(8.5)

A=173.6~cm^2

We need to add the 0.8 cm overlap to the total area already calculated. This overlap has 0.8 cm of width and 8.5 cm of height, so this overlap area is:

A_o= 0.8*8.5=6.8~cm^2

The total area of the label is:

A=173.6~cm^2+6.8~cm^2=180.4~cm^2

The area of the label is \mathbf{180.4~cm^2}

5 0
3 years ago
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