Posted: 10/2/2009 10:20:08 PM EDT
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Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put:
y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... |
I'm wasted. so, bump I guess?
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Quoted:
Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put: y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... y(x)/z(t)=1-[m(k)/z(t)] |
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Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put: y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... y(x) + m(k) = z(t) y(x) = z(t) - m(k) here if you try to divide out z(t) you get.... y(x)/z(t) = (1 - m(k))/z(t) so... to answer your question... you can get it in those terms... but not only those terms.. you are goug to have a z(t) on both sides if you do this... so I guess the answer you are looking for is no... i think............. my last math class was calc 2 and it was 2.5 years ago 
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It's just notation for a function. Solve it as you would solve any other algebra problem, in this case think of the functions them selves as variables, then when you finish you can substitute the original expressions where they belong. This is actually what I already did. The actual problem has about 6 different variables but I simplified it for posting. I cannot for the life of me get y(x)/z(t) isolated without being in terms of themselves. |
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It's just notation for a function. Solve it as you would solve any other algebra problem, in this case think of the functions them selves as variables, then when you finish you can substitute the original expressions where they belong. This is actually what I already did. The actual problem has about 6 different variables but I simplified it for posting. I cannot for the life of me get y(x)/z(t) isolated without being in terms of themselves. valheru's solution is correct, and they will be in terms of themselves. You are simply rearranging the equation. Post the full question. |
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Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put: y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... y(x)/z(t)=1-[m(k)/z(t)] this |
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Quoted: How are you guys getting this? If you subtract m(k) from both sides, then divide both sides by z(t), you get y(x)/z(t)=[z(t)-m(k)]/z(t). Where is the 1 in the numerator coming from?Quoted: Quoted: Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put: y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... y(x)/z(t)=1-[m(k)/z(t)] this |
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How are you guys getting this? If you subtract m(k) from both sides, then divide both sides by z(t), you get y(x)/z(t)=[z(t)-m(k)]/z(t). Where is the 1 in the numerator coming from?
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Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put: y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... y(x)/z(t)=1-[m(k)/z(t)] this z(t)/z(t)=1 The 1 is not in the numerator. |
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I had spent about 15 minutes trying to figure this out, but the browser ate the response. Lets see if I can remember... y(x) + m(k) = z(t) m(k) = z(t) - y(x) y(x) = z(t) - m(k) –––––– y(x)/z(t) = ??? y(x)/z(t) = y(x)/[y(x) + m(k)] = 1 + y(x)/m(k) 1 + y(x)/m(k) = y(x)/z(t) z(t)/y(x) + [z(t)*y(x)]/[y(x)*m(k)] = 0 [y(x) + m(k)]/y(x) + {[y(x) + m(k)]*y(x)]/[y(x)*m(k)]} = 0 [1 + m(k)/y(x)] + [(y(x)^2 + m(k)*y(x)]/[y(x)*m(k)] = 0 Crap, isn't there some TI-84 program that can solve this? |
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Quoted: Okay I see, I guess I was confused by the brackets. Or the Bud Quoted: Quoted: How are you guys getting this? If you subtract m(k) from both sides, then divide both sides by z(t), you get y(x)/z(t)=[z(t)-m(k)]/z(t). Where is the 1 in the numerator coming from?Quoted: Quoted: Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put: y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... y(x)/z(t)=1-[m(k)/z(t)] this z(t)/z(t)=1 The 1 is not in the numerator. |
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How are you guys getting this? If you subtract m(k) from both sides, then divide both sides by z(t), you get y(x)/z(t)=[z(t)-m(k)]/z(t). Where is the 1 in the numerator coming from?
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Wrapping up the first week of class, I spent about two hours trying to figure this out. I'm not sure if I'm getting old and haven't done math in a while, or had a huge brain fart, but I stumbled across something similar to this I could not figure out. Simply put: y(x) + m(k) = z(t) How would you solve for the quantity y(x)/z(t)? Is it even possible? Thanks for any input... y(x)/z(t)=1-[m(k)/z(t)] this Because of this.... y(x)/z(t)=[z(t)-m(k)]/z(t) equals y(x)/z(t)= (z(t) / z(t)) - (m(k) / z(t)) because when you subtract two things they need a common denominator. In this case that is z(t). Because they have the same denominator in the original expression y(x)/z(t)=[z(t)-m(k)]/z(t) You can separate the fraction. And when you get y(x)/z(t)= (z(t) / z(t)) - (m(k) / z(t)) You should obviously know that (z(t) / z(t)) = 1 So you get..... y(x)/z(t)= 1 - (m(k) / z(t)) Enjoy!
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