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nignag [31]
4 years ago
7

Write a java program that determines the position of the hour and minute hands of an analog clock at a moment in time. using a 1

2-hour clock (where the hour hand moves between the integers 1 thru 12 and the minute hand moves between the integers 0 and 59), read the current clock time from the user and then read a time increment to add to or subtract from the current time. for this second time, calculate where the location of the hour and minute hand will be.
Mathematics
1 answer:
Keith_Richards [23]4 years ago
4 0
<span>Using a 12-hour clock (where the hour hand moves between the integers 1 thru 12 and the minute hand moves between the integers 0 and 59), read the current clock time from the user and then read a time increment to add to or subtract from the current time.</span>
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Step-by-step explanation:

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Serhud [2]

Answer:

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

Step-by-step explanation:

Given - The circumference of the ellipse approximated by C = 2\pi \sqrt{\frac{a^{2} + b^{2} }{2} }where 2a and 2b are the lengths of 2 the axes of the ellipse.

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Squaring Both sides, we get

[\frac{C}{2\pi }]^{2}   =  [\sqrt{\frac{a^{2} + b^{2} }{2} }]^{2} \\\frac{C^{2} }{(2\pi)^{2}  }   =  {\frac{a^{2} + b^{2} }{2} }\\2\frac{C^{2} }{4(\pi)^{2}  }   =  {{a^{2} + b^{2} }

\frac{C^{2} }{2(\pi )^{2} }  = a^{2} + b^{2} \\\frac{C^{2} }{2(\pi )^{2} }  -  a^{2} = b^{2} \\\sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}  = b

∴ we get

b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

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BabaBlast [244]

Answer:

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