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lyudmila [28]
3 years ago
5

3 security guards below whistle at interval of 50 minutes, 60 minutes and 100 minutes if they blow together at 1:00am,at what ti

me will they blow together again​
Mathematics
1 answer:
LekaFEV [45]3 years ago
4 0

Answer:

They will blow together again at 6:00 am

Step-by-step explanation:

Guard 1 whistles every 50 minutes.

Guard 2 whistles every 60 minutes.

Guard 1 whistles every 100 minutes.

Assuming that at the beginning they whistle together, the next time that they will whistle together is after the smallest common multiple of 50, 60, and 100.

Because one of the numbers is 100, it only has multiples like:

100*2 = 200

100*3 = 300

100*4 = 400

etc...

So we need to find a common multiple of 50 and 60 that is also a multiple of 100.

For example, the multiples of 50 that are also multiples of 100 are:

50*2 = 100

50*4 = 200

50*6 = 300

And the multiples of 60 that are also multiples of 100 are:

60*5 = 300

(this is the only one)

Then we can see that a common multiple of 50, 60 and 100 is 300.

Then they will whistle together 300 minutes after the first common whistle.

We know that they blow together at 1:00 am

300 minutes after it is:

First, we need to write 300 minutes in hours.

We know that one hour has 60 minutes.

then 300 minutes = (300 minutes/60 minutes) hours = 5 hours.

300 minutes after 1:00 am is the same as 5 hours after 1:00 am

And 5 hours after 1:00 am is 6:00 am

They will blow together again at 6:00 am

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31600 rounded to the nearest thousand is what
Zigmanuir [339]
32000 is your answer

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4 0
3 years ago
(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
WHAT ARE THESE TWO ILL GIVE BRAINLESS AND SHOW UR WORK
belka [17]
N= 66°
k= 29°

All triangles add up to 180 so that’s how i got those
4 0
3 years ago
How times can 38 go into 7
kaheart [24]
Hmm, this answer is mathematically incorrect. If the question has no errors, 38 can go into 7 0 times, it can't because its bigger than 7. 

However, if it were meant to be the other way round, how many times does 7 go into 38, the answer would be 5. 
6 0
4 years ago
Read 2 more answers
6. The angle of depression from the top of a tower to a boulder on the ground is 38°. If the tower is 25 m high,
gizmo_the_mogwai [7]
Tan38 = 25/x
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31.998 = x
3 0
3 years ago
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