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saul85 [17]
3 years ago
14

John wrote 1/5 of a paper in 1/4 of an hour how long did it take him to write an entire paper?\

Mathematics
1 answer:
MatroZZZ [7]3 years ago
7 0

Answer:

1\frac{1}{4}\ hours

Step-by-step explanation:

we know that

using proportion

\frac{(1/4)}{(1/5)}\frac{hour}{paper}=\frac{x}{1}\frac{hour}{paper} \\ \\x=5/4\ hours

covert to mixed number

5/4\ hours=1\frac{1}{4}\ hours

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The law of cause and effect is also called the principle of causality.<br><br> True or false???
kondor19780726 [428]
The answer is falsee
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At the beginning of the week, Stewart's checking account had a balance of $-12.07. On Monday morning he deposited a
MA_775_DIABLO [31]

Answer:

$304.38

Step-by-step explanation:

$-12.07 + $216.45 +$100.00=

216.45+ 100.00= 316.45 - 12.07 = 304.38

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3 years ago
Latisha’s water bottle can hold 0.3 liters. How many milliliters can it hold?
seraphim [82]
I dont know
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3 years ago
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A team of runners ran a relay race. Each runner ran 7 1/2 kilometers. The total length of the race was 45 kilometers. Write and
Travka [436]

Answer:

6 runners

Step-by-step explanation:

Each runner ran 7.5 km.  If there were n runners, then (7.5 km)n = 45 km.

Solve this for n by dividing 7.5 km into 45 km:

n = 6

There were 6 runners.

7 0
3 years ago
A Ferris Wheel 22.0m in diameter rotates once every 12.5s. What is the ratio of a persons apperenet weight to her real weight (a
AnnZ [28]
  <span>Acceleration of a passenger is centripetal acceleration, since the Ferris wheel is assumed at uniform speed: 
a = omega^2*r 

omega and r in terms of given data: 
omega = 2*Pi/T 
r = d/2 

Thus: 
a = 2*Pi^2*d/T^2 

What forces cause this acceleration for the passenger, at either top or bottom? 

At top (acceleration is downward): 
Weight (m*g): downward 
Normal force (Ntop): upward 

Thus Newton's 2nd law reads: 
m*g - Ntop = m*a 

At top (acceleration is upward): 
Weight (m*g): downward 
Normal force (Nbottom): upward 

Thus Newton's 2nd law reads: 
Nbottom - m*g = m*a 

Solve for normal forces in both cases. Normal force is apparent weight, the weight that the passenger thinks is her weight when measuring by any method in the gondola reference frame: 
Ntop = m*(g - a) 
Nbottom = m*(g + a) 


Substitute a: 
Ntop = m*(g - 2*Pi^2*d/T^2) 
Nbottom = m*(g + 2*Pi^2*d/T^2) 

We are interested in the ratio of weight (gondola reference frame weight to weight when on the ground): 
Ntop/(m*g) = m*(g - 2*Pi^2*d/T^2)/(m*g) 
Nbottom/(m*g) = m*(g + 2*Pi^2*d/T^2)/(m*g) 

Simplify: 
Ntop/(m*g) = 1 - 2*Pi^2*d/(g*T^2) 
Nbottom/(m*g) = 1 + 2*Pi^2*d/(g*T^2) 

Data: 
d:=22 m; T:=12.5 sec; g:=9.8 N/kg; 

Results: 
Ntop/(m*g) = 71.64%...she feels "light" 
Nbottom/(m*g) = 128.4%...she feels "heavy"</span>
7 0
3 years ago
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