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Title: Mathematics question
Source: .
URL Source: http://.
Published: Mar 31, 2015
Author: .
Post Date: 2015-03-31 12:54:22 by Artisan
Keywords: None
Views: 1611
Comments: 41

Can anyone tell me the formula I would use to solve this question.

If 2001.51 = 70% of X, what is X?

I know what X is, (2859.30) but if I didn't, and only had the 2001.51 number, what is the formula I would use to determine X?

And what area of math is this referred to as, (since I obviously need a course on it ;-) Thanks.

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Begin Trace Mode for Comment # 13.

#6. To: Artisan (#0)

That's a 6th grade math problem.

Abraham  posted on  2015-03-31   13:38:11 ET  Reply   Untrace   Trace   Private Reply  

Thank you for pointing that out, Abraham. What would we do without you?

Are any of you seriously mathematical? If so I have a question that should be simple, but none of my local geniuses can answer it.

NeoconsNailed  posted on  2015-03-31   13:57:11 ET  Reply   Untrace   Trace   Private Reply  

If so I have a question that should be simple, but none of my local geniuses can answer it.

Shoot it here.

Fred Mertz  posted on  2015-03-31   14:00:07 ET  Reply   Untrace   Trace   Private Reply  

Goody! How do you calculate the points at which the hour and minute hands of a clock are in the same exact position (or exactly opposite) other than 12:00?

Mind you, the exact times are of secondary interest at best. I mainly want to know how they'd be arrived at, simply because I'm an endlessly curious person (if a total mathematical failure).

NeoconsNailed  posted on  2015-03-31   15:03:57 ET  Reply   Untrace   Trace   Private Reply  

#18. To: NeoconsNailed (#13) (Edited)

I don't know how to calculate it.

The big hand passes the little hand once an hour, and is opposite it once an hour.

2 times an hour for whatever length of time you want. I don't know why you eliminate 12:00 - same thing.

If this is a trick question and you make me look like a retard, I'm going to knock you over the head with my mouse.

Fred Mertz  posted on  2015-03-31 15:18:46 ET  Reply   Untrace   Trace   Private Reply  

Seems like it would be 120 instances of the same or opposite positions.

Lod  posted on  2015-03-31 15:29:49 ET  Reply   Untrace   Trace   Private Reply  

You can figure out the exact times the hands will be together and opposite, but it is a bit complicated:

Say starting at 12, you wanted to know the next time the hands would be together after 1. The hour hand moves 1/12 as fast as the minute hand. So the minute hand moves all the way around, but the hour hand moves only to one. When the minute hand gets to one the hour hand will be at 1 + 1/12.

Following this progression gives you the time when the two hands converge:

It is the infinite series 1 +1/12 +(1/12)^2 + 1/12^3.......

However this infinite series converges to definite value of 1 +1/(1 - 1/12). That is the next time the two hands will be together after 12. Using that point and the same method of calculation, you can follow the times around.

By simple geometry, it works the same way for when the hands are opposite. You could start with 12;30 and work your way around the clock.

Abraham  posted on  2015-03-31 17:49:41 ET  Reply   Untrace   Trace   Private Reply  

Goody! How do you calculate the points at which the hour and minute hands of a clock are in the same exact position (or exactly opposite) other than 12:00?

Mind you, the exact times are of secondary interest at best. I mainly want to know how they'd be arrived at, simply because I'm an endlessly curious person (if a total mathematical failure).

I'm sure that you could work this to the nanosecond. How close do you want it?

From Noon, the next time that the two hands are equally placed are 65 and a fraction of a minute later, again, pending to what significant figures you want it calculated to.

Since this is simply a curiosity thing primarily if not exclusively, that extra five minutes that the minute hand has to move every hour to get past exactly 65 minutes later, is approximately 25 seconds. 5/60 of the next 5 minutes.

So every 65 minutes and 25 seconds the two hands will be equal.

Simple linear math entails that they're opposite exactly half-way prior to or after (however you'd like to view it since it's identical) the next (prior to) or last (after), which means that every 65 minutes and 25 seconds, exactly halfway out of phase with when they're equally aligned, they are opposite.

This also entails that this occurs 32 minutes and 43 seconds after and before the next identical alignment.

Another way of looking at it is that in one 12 hour period they align, or are opposite, 11 times, either one.

12 hours = 720 minutes.

720 minutes/11 = 65.45

.45 minutes = .45 x 60 = 27 seconds. That extra 2 seconds or so, is from the additional time of movement that the hour hand moves in the five minutes after the minute hand crosses the 1 hour mark.

So in essence, the first time, from Noon, that the hands are in perfect alignment is 65 minutes and 27 seconds later. Then 65 minutes and 27 seconds after that. So on and so forth.

For oppositely aligned, it's the same thing except that it starts approximately 32 minutes and 43 seconds later.

Make sense?

Is that what you wanted?

Time is linear.

Katniss  posted on  2015-03-31 20:54:06 ET  Reply   Untrace   Trace   Private Reply  

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