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Wikipedia:Reference desk/Archives/Mathematics/2014 November 27

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November 27

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properties of parallelogram

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If the area is 96cm2 and LN=8 cm, what is KM? — Preceding unsigned comment added by 121.54.58.233 (talk) 10:35, 27 November 2014 (UTC)[reply]

KM...? --CiaPan (talk) 11:03, 27 November 2014 (UTC)[reply]
Are LN and KM four points? Does it look similar to this:

        ________________________
       /|                      /
      / |                     /
     /  |                    /
    /   |                   /
   /    |                  /
  /     |                 /
 /      |                /
/       |               /
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
K       N              M

If it looks like that then see Parallelogram#Area formula. If N is the fourth corner then there is not enough information. PrimeHunter (talk) 12:52, 27 November 2014 (UTC)[reply]

I think it is more likely that the OP was referring to a parallelogram with vertices labelled K, L, M and N, so LN is one diagonal, and KM is the other. The fact required to answer the question is that the area is half the product of the diagonals, a fact that is missing from our article. The calculation is left for the OP since it is probably homework. (corrected below) Dbfirs 12:57, 27 November 2014 (UTC)[reply]
Are you sure about that? See the image above and imagine the shorter edges rotate clockwise about their lower endpoints—they will eventually fall onto the KM line, thus making the height zero and area zero, while both diagonals will stay non-zero (although different than in original figure)... --CiaPan (talk) 13:18, 27 November 2014 (UTC)[reply]
This would be true if (and only if) the parallelogram is a rhombus. Otherwise half the product of the diagonals is strictly greater than the area. Sławomir Biały (talk) 13:56, 27 November 2014 (UTC)[reply]
Oops! Yes, I was trying to make sense of the OP's question and thinking of a rhombus (which the diagram almost is). I've struck my error above. More information is needed to be able to answer the question. You need to multiply half the product of the diagonals by the sine of the angle at which they intersect. Dbfirs 14:04, 27 November 2014 (UTC)[reply]
→Parallelogram#Area formula, you need to calculate the angle α. --Hans Haase (talk) 13:01, 27 November 2014 (UTC)[reply]

What is it "5mIU/ml."?

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5.28.173.111 (talk) 14:56, 27 November 2014 (UTC)[reply]

mIU means one thousandth of an International unit so 5mIU/ml is the same as 5 International units per litre. There is no general correspondance between International units and mass or volume because it will be different for each particular substance. Dbfirs 16:51, 27 November 2014 (UTC)[reply]
Thanks. Btw, what does the letter m stand for? 5.28.173.111 (talk) 23:26, 27 November 2014 (UTC)[reply]
See m. -- ToE 01:11, 28 November 2014 (UTC)[reply]

What does it mean " 0.5974 kg/m3"?

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Does it mean 0.5974 of 1 kg per 1 cubic centimeter? 5.28.173.111 (talk) 20:32, 27 November 2014 (UTC)[reply]

There is no c for centi. If you copied it correctly except it was formatted 0.5974 kg/m3 then it means 0.5974 kg per cubic meter. It's a measure of density where water is around 1000 kg/m3 (originally by definition of kilogram). PrimeHunter (talk) 20:47, 27 November 2014 (UTC)[reply]
Yes. 0.5974 kg/m3 would be about half the density of air under typical sea-level conditions. --65.94.50.4 (talk) 10:32, 28 November 2014 (UTC)[reply]
(EC) The density of air is approximately 1.225 kg/m3 at sea level at 15 °C. Your number, about half that, is equal to the density of air at the pressure and temperature found at an altitude of about 6,900 m according to our graph at density of air#Altitude. (ρ, the Greek letter rho, is typically used as the symbol for density. In many fonts it is unfortunately similar to our letter p which is typically used for pressure.) Your 0.5974 kg/m3 is similar to the density of methane under standard conditions. (ρ_CH4 = 0.656 kg/m3 at 25 °C, 1 atm, so to match your density exactly, you would have to warm it up another 30 °C while maintaining 1 atm. See ideal gas law.)
Your initial guess of 0.5974 kg/cm3 = 597.4 g/cm3 is exceedingly dense, much denser than anything naturally occurring on earth. Water is only 1.0 g/cm3, mercury is 13.5 g/cm3, gold is 17.3 g/cm3 , and osmium, the densest naturally occurring element, is 22.6 g/cm3. Even the material at the core of the sun is only 150 g/cm3, about one fourth of your number. Densities as great or greater than your number can be achieved for very short periods of time in the laboratory, such as with inertial confinement fusion, and of course, neutron stars have much greater densities in the rage of 3.7×1017 to 5.9×1017 kg/m3. -- ToE 11:03, 28 November 2014 (UTC)[reply]
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