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Feliz [49]
2 years ago
9

One analog clock loses 3 minutes every 2 hours and another analog clock gains 3 minutes every 4 hours. At noon, the owner sets t

he correct time on both clocks and leaves on a business trip. When he returns home in less than two weeks, the clocks show the same
time. What time does he return? What time do the clocks show?

Need answer ASAP! No links please!
Mathematics
1 answer:
Nadusha1986 [10]2 years ago
6 0

Answer:

12 noon

Step-by-step explanation:

Clock 1 loses 3 minutes every 2 hours. Now, 2 hours = 120 minutes.

This means, it has speed of 120 - 3 = 117 minutes for every 2 hours.

Thus, in 1 hours, it's speed = 117/2 = 58.5 min/hr

Clock 2 gains 3 minutes every 4 hours

Which means in an hour, it gains ; 4/3 minutes

Thus, it's speed an hour is; 60 + 4/3 = 61.333 min/hr

Now, When he returns home in less than two weeks, the clocks show the same time.

let's say the number of hours he was away is x. For the 2 clocks to show the same time, it means they have to show it by 12 noon at another day since it was at noon that he set the correct times before leaving.

Thus, they will both be showing 12 noon.

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Step-by-step explanation:

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Step-by-step explanation:

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Mary throws her Grandma’s favorite crystal dish out the second floor window, which is 6 meters from the ground. She is really up
wlad13 [49]

we know that

acceleration due to gravity is 9.8m/s^2

so, we get

a(t)=9.8

and acceleration will always be constant

we know that

integral of acceleration is velocity

so, we can integrate both sides

\int \:a\left(t\right)dt=\int \:9.8dt

v(t)=9.8t+C

we are given

v(0)=35m/s

we can use it and find C

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now, we can plug it back

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we know that

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we can integrate it again

\int \:v\left(t\right)dt=\int \:\left(9.8t+35\right)dt

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now, we have

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we can use it and find C

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6=4.9(0)^2+35(0)+C

now, we can plug back C

C=6

and we get

s(t)=4.9t^2+35t+6.............Answer

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