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Math Help - Comparing two formulas...

  1. #1
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    Comparing two formulas...

    I have two formulas for a velocity displayed below

    Vn = 175 / ( p^0.43 ) eqn 1

    and

    Ve = 135 / ( p^0.5 ) eqn 2

    I know that for any value of p i choose, Equation 2 will give me a lower velocity.

    I know this from a simple trial and error method, by substituting different values for p.

    Besides this trial and error method, is there any way that i can prove that Ve<Vn all the time?
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  2. #2
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    Quote Originally Posted by Rush View Post
    I have two formulas for a velocity displayed below

    Vn = 175 / ( p^0.43 ) eqn 1

    and

    Ve = 135 / ( p^0.5 ) eqn 2

    I know that for any value of p i choose, Equation 2 will give me a lower velocity.

    I know this from a simple trial and error method, by substituting different values for p.

    Besides this trial and error method, is there any way that i can prove that Ve<Vn all the time?
    If Vn> Ve, then \frac{175}{p^{0.43}}> \frac{135}{p^{0.5}}

    Assuming p is positive, that is the same as 175p^{0.5}> 135p^{0.43} which leads to [tex]\frac{p^{0.5}{p^.43}= p^{.07}> \frac{135}{175}= \frac{4}{5}[tex] so that p> \left(\frac{4}{5}\)^{25}= .0038, approximately. So this is NOT true for [b]all[b] p but is true for p> .004.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    If Vn> Ve, then \frac{175}{p^{0.43}}> \frac{135}{p^{0.5}}

    Assuming p is positive, that is the same as 175p^{0.5}> 135p^{0.43} which leads to [tex]\frac{p^{0.5}{p^.43}= p^{.07}> \frac{135}{175}= \frac{4}{5}[tex] so that p> \left(\frac{4}{5}\)^{25}= .0038, approximately. So this is NOT true for all p but is true for p> .004.

    Sorry, but i cannot read ur equations properly,

    can u retype it for me?
    (esp the below line)


    [b][b][b][b][tex]\frac{p^{0.5}{p^.43}= p^{.07}> \frac{135}{175}= \frac{4}{5}[tex] so that p> \left(\frac{4}{5}\)^{25}= .0038, approximately. So this is NOT true for all p but is true for p> .004.
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  4. #4
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    i specifically cannot understand how u got, p^0.5 x p^0.43 = p^0.07
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