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enyata [817]
2 years ago
6

Prove that given an unlimited supply of 6-cent coins, 10-cent coins, and 15-cent coins, one can make any amount of change larger

than 29 cents.
Mathematics
1 answer:
Andre45 [30]2 years ago
4 0

30¢ = 2×15¢

31¢ = 15¢ +10¢ +6¢

32¢ = 2×10¢ +2×6¢

33¢ = 15¢ +3×6¢

34¢ = 10¢ +4×6¢

35¢ = 15¢ +2×10¢

From this 6¢ range, any subsequent values can be reached by adding multiples of 6¢ to these numbers. For example, 36¢ = 2×15¢ +6¢.

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When a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a ra
Sergeu [11.5K]

Answer:

r=\sqrt{10}\text{ m}

Step-by-step explanation:

We have been given that when a cylindrical tank is filled with water at a rate of 22 cubic meters per hour, the level of water in the tank rises at a rate of 0.7 meters per hour. We are asked to find the approximate radius of tank in meters.

We will use volume of cylinder formula to solve our given problem as:

V=\pi r^2h, where,

r = Radius,

h = Height of cylinder.

Since the level of water in the tank rises at a rate of 0.7 meters per hour, so height of cylinder would be h = 0.7 meters at V=22\text{ m}^3.

Upon substituting these values in above formula, we will get:

22\text{ m}^3=\frac{22}{7}\cdot r^2(0.7\text{ m})

22\text{ m}^3=\frac{22}{10}\cdot r^2\text{ m}

\frac{10}{22}\cdot\frac{22\text{ m}^3}{\text{m}}=r^2

r^2=\frac{10}{22}\cdot\frac{22\text{ m}^3}{\text{m}}

r^2=10\text{ m}^2

Now, we will take positive square root of both sides as radius cannot be negative.

\sqrt{r^2}=\sqrt{10\text{ m}^2}

r=\sqrt{10}\text{ m}

Therefore, radius of tank would be approximately square root of 10 m.

5 0
2 years ago
What is 0.95 rounded off to the nearest whole number
mihalych1998 [28]

1.00 would be your answer because 95 rounded up is 100. So, 0.95 is closest to 1.00

8 0
3 years ago
Two mechanics worked on a car. The first mechanic charged $65 per hour, and the second mechanic charged $90 per hour. The mechan
Katena32 [7]

Answer:

First mechanic worked 20 hours and second mechanic worked 5 hours

Step-by-step explanation:

Let the number of hours the first mechanic worked = a

Let the number of hours the second mechanic worked = b

Therefore, we can write 2 equations and then solve them simultaneously:

a+b=25

65a+90b=1750

Rearranging the first equation: a=25-b

and substituting into the second equation to find b:

65(25-b)+90b=1750

1625-65b+90b=1750\\
25b=125\\
b=5

Now sub b=5 into the first equation to find a

a+5=25\\
a=20

Therefore, first mechanic (a) worked 20 hours and second mechanic (b) worked 5 hours

4 0
2 years ago
Which compound inequality can be represented by the graph below?
Lerok [7]

Answer:

where is the picture of the graph beautiful

8 0
2 years ago
There are 3 in. of snow on the ground when it begins to snow 0.5 in./h . Which linear equation represents the total depth of the
dexar [7]

There are 3 in. of snow on the ground when it begins to snow 0.5 in./h.

Initial depth of snow = 3 in.

it begins to snow 0.5 in./h. The constant rate of snow is 0.5. So slope = 0.5

Let x be the number of hours

y be the total depth of the snow

To frame linear equation we use y=mx+b

where m is the slope and b is the y intercept (initial depth of snow)

We know m=0.5  and b=3

Replace it in the equation

y = 0.5x + 3

The linear equation that represents the total depth of the snow(y), in inches, after x hours

is y= 0.5x + 3


7 0
3 years ago
Read 2 more answers
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