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pogonyaev
3 years ago
7

Suppose while at a store, you ask for change for five dollars. You get in return exactly twenty-four coins, all of which are nic

kels and quarters. How many nickels and quarters did you receive?
Mathematics
2 answers:
Galina-37 [17]3 years ago
8 0

Answer:

19 quarters and 5 nickles

Step-by-step explanation:

19 * .25 = 4.75     .05 * 5 = .25     4.75+.25= 5

Readme [11.4K]3 years ago
7 0

Answer: 5 nickels and 19 quarters

Step-by-step explanation:

Let represent the number of nickels and let represent the number of quarters.

From the problem we can write these two equations:

+ = 24; (0.05) + (0.25) = 5.00

Isolate in the first equation: = 24 − .

Substitute 24 − for and solve for :

(24 − )(0.05) + (0.25) = 5.00

24(0.05) − (0.05) + (0.25) = 5.00

1.20 − 0.05 + 0.25 = 5.00

1.20 + 0.20 = 5.00

0.20 = 3.80

= 19

Use = 19 to find :

+ = 24; + 19 = 24; = 5

Answer: 5 nickels, 19 quarters

Hope this helps!

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Bill's speed is constant after 5 seconds and 45 meters into the race.

Correct responses:

a. Bill's maximum speed is 8 m/s

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<h3>Methods used to calculate speed and distance traveled</h3>

Given parameters are;

The distance of the race = 100 meter

Distance at which Bill reaches maximum speed = 45 meters

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Solution:

Distance Bill runs at maximum speed, d = 85 m - 45 m = 40 m

Time at which Bill runs the 40 m at maximum speed, <em>t</em> = 10 s - 5 s = 5 s

Speed = \mathbf{\dfrac{Distance}{Time}}

Therefore;

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b. i. Required:

The distance Bill will run for 3 seconds at the maximum speed;

Solution:

Distance,<em> s</em> = Speed, <em>v</em> × Time, <em>t</em>

<em />

The distance traveled at maximum speed in 3 seconds is therefore;

Distance = 8 m/s × 3 s = 24 m

The distance Bill travels in 3 seconds at the maximum speed is 24 meters.

ii) The distance Bill travels at 8 seconds after start, is given as follows;

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Distance traveled in the next 3 seconds = 24 meters

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c. Bill's distance 6.5 seconds after the start of the race is given as follows;

Distance traveled in the first 5 seconds = 45 meters

Distance traveled in the next 1.5 seconds = 1.5 s × 8 m/s = 12 meters

  • Bill's distance from the starting line 6.5 seconds after start of the race = 45 meters + 12 meters = <u>57 meters</u>

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