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Posted: 5/10/2004 5:12:43 PM EDT
[#1]
Uh, what's the "2" above the equation for? Need to go eat right now, but I'll work on it for ya and come back and see if anyone else figured it out yet.
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Posted: 5/10/2004 5:13:02 PM EDT
[#2]
the two with the variables need to be subtracted -63 and divided by -119.
it comes out funny...but then again, what the hell is that 2 up there? |
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Posted: 5/10/2004 5:13:08 PM EDT
[#3]
ok lets try this again.
9h^2 -72h + 119 = 0 (3h - 7) * (3h - 17) = 0 3h-7 = 0 3h - 17 = 0 3h = 7 3h = 17 h = 7/3 h = 17/3 |
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Posted: 5/10/2004 5:14:46 PM EDT
[#4]
The 2 is the first h squared.
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Posted: 5/10/2004 5:15:23 PM EDT
[#5]
thats what i got... but that 2 has me thinking. is that the number of the problem in your book?! if it is dude, its fucking everything up. |
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Posted: 5/10/2004 5:17:58 PM EDT
[#6]
It should have been written 9(h^2) then. |
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Posted: 5/10/2004 5:25:57 PM EDT
[#7]
solve for zero:
9h^2-72h+119=0 I do not know how to factor that, no common multiples for 9 and 119. |
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Posted: 5/10/2004 5:29:25 PM EDT
[#8]
You need to use the quadratic equation. I got h=2.3333333 or h=5.666666667 |
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Posted: 5/10/2004 5:36:46 PM EDT
[#9]
If I'm not mistaken: -b +- (b^2-4ac) ------------------- 2a Where the quadratic is : a^2x + bx + c = 0 Edit to say: Note the "+-" ...this is supposed to be plus or minus. You have to do both, there are usually two answers to the equation. Also, quadratic is no fun. When possible, just factor them. |
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Posted: 5/10/2004 5:40:12 PM EDT
[#10]
9h^2 - 72h= -119
9h^2-72h+119=0 (3h-17)(3h-7)=0 3h-17=0 3h-7=0 3h=17 3h=7 h=17/3 h=7/3 |
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Posted: 5/10/2004 5:42:12 PM EDT
[#11]
Yep, solve for zero.
Either factor it or use the quadratic formula. 9h^2 - 72h + 119 = 0 AH HAH! It factors (3h - 7) (3h - 17) = 0 3h - 7=0 or 3h - 17 = 0 3h=7 3h=17 h=7/3 h=17/3 I hope I did all that right... I know that is the answer with factoring. Let me try the quadratic formula. 72 +/- square root of 72-4(9)(119) divided by 2(9) 72 +/- squre root of 900 divided by 18 72 +/- 30... divided by 18 102/18 or 42/18 reduced to 17/3 or 7/3 So those are your two answers 17/3 and 7/3. Hope you can understand my notes. |
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Posted: 5/10/2004 5:50:20 PM EDT
[#12]
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Posted: 5/10/2004 5:50:56 PM EDT
[#13]
h=2 1/3 & h=5 2/3
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Posted: 5/10/2004 5:52:24 PM EDT
[#14]
You can solve it by factoring it into 2 binomials or you can use the quadratic formula.
Dang, Cypher214 nailed it. See what happens when you talk on the phone. |
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Posted: 5/10/2004 6:04:01 PM EDT
[#15]
Looks like you got the help you needed. Besides, before you edited your post I assumed the equation was 2/9h-72h=119 so I have the wrong answer.
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Posted: 5/10/2004 6:11:55 PM EDT
[#16]
Thanks guys, I'm trying to help my little brother out on his homework and I'll be damned if he stumped me on that one. Thanks again!!!!!
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Posted: 5/10/2004 7:45:14 PM EDT
[#17]
A more interesting problem is solving:
AX^3 +BX^2 + CX +D = 0 or AX^4 + BX^3 + CX^2 + DX + E = 0 In attempting to find a general solution to the 5 order polynomials, Galois invented a who new field of mathematics...(and in the process proved that a general solution to the 5th order polynomial is impossible) |
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