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**unique****Member**- Registered: 2006-10-04
- Posts: 419

find the length of the line segment with endpoints (-4,0) and (5,4)

Desi

Raat Key Rani !

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

the length of the line segment defined by two end points is found by the differences in coordinates

(5,4) - (-4,0) = (9,4) , sqrt( 9^2 + 4^2) = sqrt(97) = 9.85....

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The End Of All Things To Come.

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**LQ****Real Member**- Registered: 2006-12-04
- Posts: 1,285

I've heard that gravity has that value in some places. What memories.

I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

If my calculations are right, then you'd need to be around 14km below sea level for gravity to be that high.

Why did the vector cross the road?

It wanted to be normal.

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**er.neerajsrivastava****Member**- Registered: 2006-12-27
- Posts: 9

help me please. what is the fastest way to compute 285714! (i.e. the factorial of 285714)?

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**Devantè****Real Member**- Registered: 2006-07-14
- Posts: 6,400

Use this:

http://www.mathsisfun.com/forum/viewtopic.php?pid=54379#p54379

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

mathsyperson wrote:

If my calculations are right, then you'd need to be around 14km below sea level for gravity to be that high.

What are these calculations?

And the gravity at poles is different than the gravity at the Equatior.

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I searched on Google to find the radius of the Earth and a precise value of g.

Then I used the g = Gm/r² thing to work out the mass of the earth, and then used that to work out at what distance from the centre the gravitional strength would be sqrt97. And then I just took that away from the standard radius to get the distance below sea level that it would have to be.

I didn't include all that in the post because it was off the point of the topic.

Why did the vector cross the road?

It wanted to be normal.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Sorry for the off topic, but thank you for explainig me.

but... what's G? //sorry for the off-topic again

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

G is the gravitational constant (6.67 x 10^-11). The force due to gravity between two objects is given by:

Where m[sub]1[/sub] and m[sub]2[/sub] are the masses of the 2 objects, and r is the distance between them.

And don't worry about going off-topic, the original question has been resolved so it doesn't matter too much.

Why did the vector cross the road?

It wanted to be normal.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Thank you again.

I'm asking, because in school we use 9.8 for g.

IPBLE: Increasing Performance By Lowering Expectations.

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

G is the gravitational constant, it is universal. 9.8 rather, is the value of (G×mass of earth) / (average earth radius)^2

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