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July 24

Solve the problem

Solve this problem I was unable to solve it.
"If x³ + 1/x³ = 110, find the value of x + 1/x." --Sunny Singh 14:15, 24 July 2012 (UTC) — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs)

Try (x + 1/x)³ and see where that gets you. --Wrongfilter (talk) 14:30, 24 July 2012 (UTC)[reply]

But after using (x+1/x)³ I don,t get the answer. Please, you solve it.--Sunny Singh 14:40, 24 July 2012 (UTC) — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs)

Er, no. --Wrongfilter (talk) 14:41, 24 July 2012 (UTC)[reply]

Hint: If you are only interested in solutions in the real numbers, then x + 1/x is an integer. Gandalf61 (talk) 15:48, 24 July 2012 (UTC)[reply]

I also want to see steps, please, explain this problem with steps. I tried hard but I couldn't solve it. — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs) 15:58, 24 July 2012 (UTC)[reply]

So what did you get when you expanded (x + 1/x)³ ? Gandalf61 (talk) 16:01, 24 July 2012 (UTC)[reply]

I think it is easier if you multiply through by x³ and consider the result as a quadratic. Dmcq (talk) 17:31, 24 July 2012 (UTC)[reply]

I tried that, but it didn't give a very nice solution. You get the cube-root of a radical expression:

x={\sqrt[{3}]{55\pm 12{\sqrt {21}}}}\,.

However, these solutions can be expressed as

x={\frac {1}{2}}(5\pm {\sqrt {21}})\,.

I think that the methods above lend themselves to finding these latter solutions. Besides, the OP wants

x+1/x

. — Fly by Night (talk) 18:00, 24 July 2012 (UTC)[reply]

$x^{3}+1/x^{3}=\left(x+{\frac {1}{x}}\right)\left(\left(x+{\frac {1}{x}}\right)^{2}-3\right)$ . Now use the prime factorization of 110, along with the assumption that x+1/x is an integer. Sławomir Biały (talk) 18:15, 24 July 2012 (UTC)[reply]

Not assuming it is an integer gives the other solutions

{\frac {-5\pm 3i{\sqrt {7}}}{2}}

. Perhaps the OP can try working that out. Dmcq (talk) 20:58, 24 July 2012 (UTC)[reply]

I just tried out Wolfram Alpha at http://www.wolframalpha.com with "solve x^3=110+3x" and it immediately gives the simplified solution. That's another easy way of getting an answer but it is better in the first instance whilst learning to try out for yourself. Dmcq (talk) 21:08, 24 July 2012 (UTC)[reply]

I think Wrongfilter is arguing that

(x+{\frac {1}{x}})^{3}=x^{3}+3x+3{\frac {1}{x}}+{\frac {1}{x^{3}}}

, we know that x³ + 1/x³ = 110 so we have:

(x+{\frac {1}{x}})^{3}=110+3(x+{\frac {1}{x}})

, so we have a cubic in (x+1/x) compare with

y^{3}-3y-110=0

. This has a solution y=5 as 5^3 = 125, 3*5 = 15 and 125-15 = 110. You need to prove/satisfy yourself this is the only real root - is this any help? 83.100.173.200 (talk) 19:01, 24 July 2012 (UTC)[reply]

Word numerical

Can you solve this problem for me? I was unable to solve it. "A machine gun fires 25g bullets at the rate of 600 bullets per minute with a speed of 200m/s. Calculate the force required to keep the gun in position." — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs) 14:58, 24 July 2012 (UTC)[reply]

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. —Kusma (t·c) 15:15, 24 July 2012 (UTC)[reply]

That's a very puny machine gun, with a muzzle velocity of only 200m/s - 750m/s to 850m/s seem to be more typical. Anyways, as you should know from Isaac Newton, every force has an equal and opposite force. Some force accelerates 600*25g from zero to 200m/s every minute. The reaction to that action would accelerate the machine gun backwards if it weren't held in place. You should be able to take it from there. --Stephan Schulz (talk) 16:38, 24 July 2012 (UTC)[reply]

Somewhat puny, but I think you're mistaking feet per second with meters per second, as some slow-velocity rounds, like the Thompson submachine gun shoots a .45 at about 280 m/s. 200 isn't completely outside the realm of possibility (although slow). Shadowjams (talk) 06:39, 25 July 2012 (UTC)[reply]

That's a submachine gun. Submachine guns typically fire pistol rounds. Real machine guns fire rifle rounds, and that means they should go supersonic. I checked the MG 34 (755 m/s), the M60 machine gun (850 m/s), and the PK machine gun (825 m/s). --Stephan Schulz (talk) 22:32, 25 July 2012 (UTC)[reply]

See the Impulse (physics) page. --CiaPan (talk) 06:46, 25 July 2012 (UTC)[reply]

It's not a matter of checking out velocity of the bullet, it's a matter of solving the problem. Don't waste my (Sunny) and your time on commenting on the speed of bullet. Whatever I have written is correct and please don't make excuse.110.227.128.206 (talk) 11:46, 25 July 2012 (UTC)[reply]

What possible connection is there between the header ("Word numerical") and the question posed? Absolutely zero, as far as I can see. -- ♬ Jack of Oz ♬ ^{[your turn]} 11:53, 25 July 2012 (UTC)[reply]

Where I was growing up, such questions in a schoolbook were termed "word problems" (i.e. mathematical problems expressed as prose words, rather than in short algebraic form). I would surmise that wherever Sunny is from, "word numerical" is the equivalent term used (or perhaps the literal translation of the term used). Though I agree a more specific title should have been used, as "word numerical" is basically equivalent to "question".

All that said, the problem is poorly constructed, though that's typical for such word problems. One issue is that the problem is that force is an instantaneous measure, but the problem is pulling in concepts of rate over time, but only poorly specifying the time frames involved. You'll get very different answers for the forces involved if you assume the acceleration happens in, say 10 ms versus 100 ms, even though both can support a rate of 600 bullets/min. (To see the issue extreme example, imagine if the rate wasn't 600 but 6 bullets/min. It would be inaccurate to say that the acceleration from 0 to 200 m/s happens over the full 10 seconds between rounds. Instead, each bullet would be fired quickly, and there would be a 9+ second pause between each firing.) My guess is they're assuming that the the time of acceleration is the full between bullet time (so the firing pin hits the next bullet the instant the first bullet reaches the 200 m/s). I bet they're also assuming that the force during each bullet's firing is constant from 0 to 200 m/s, rather than the strong initial impulse followed by a slow decrease as the expanding gas looses energy.

As to solving the problem, I would probably realize Newton's third law states that the force on the gun is derivable from one on the bullet, and go with Newton's second law to relate force on the bullet to the acceleration on the bullet. Acceleration to velocity can be a little tricky if you haven't had calculus, but if we're assuming that the acceleration is constant, we can simplify things down to acceleration is just the difference in velocity divided by time. You can figure out the change in velocity of the bullet (remembering it starts at rest), and we discussed the period of time issue above, so you should be able to calculate the acceleration, and through it the force. - Just don't go using it to construct a real gun, as the simplifying assumptions make the result highly inaccurate. -- 71.35.117.7 (talk) 15:45, 25 July 2012 (UTC)[reply]

The simplifying assumptions are not all that bad; it doesn't matter over what period of time the acceleration occurs because the impulse is the same (and doesn't require calculus). You do have to interpret the "force required" as being an average force, but that's hardly a practical stretch for a 10 Hz oscillation. --Tardis (talk) 00:45, 26 July 2012 (UTC)[reply]

Wikigraphs

Could someone check this math?

http://www.brianckeegan.com/2012/07/2012-aurora-shootings/ (just kidding)--Canoe1967 (talk) 20:01, 24 July 2012 (UTC)[reply]