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Algebra Bloody Hard is this right

  1. Apr 3, 2005 #1
    hey everyone i got a series of questions so can somebody check my working out and see if they are right thanks.

    1. Make t the subject of the formula 1/g= 1/(t+1) - 1/(s-1)
    i did all the working out and got at the end
    t= GS-2G-S+1 / S-1+G not sure if i should make it simplier like
    t= G(S-2) - S + 1 / G+s-1

    2. solve the equation 6x^3 + 8 = 23x^2 + 6x
    i can only get the intercepts using my calculator sketching both graphs and finding the intercepts how do you do it manually????? i tried bringing them all to the LHS and let it equal to zero and subing x to get 0 on the RHS but never happened is there a way i dont think so!!! if there is plz tell me

    3. Write in simplest index form:
    sqrt ( x^8 * y^4 * z^2) / x*y*z^2

    i got totally confused do i just do [(x^8*y^4*z^2)^(1/2) ] / xyz^2
    that gives me
    = x^4*y^2*z / xyz^2 = x^3*Y*z^-1
    therefore
    = x^3*Y / Z IS THAT RIGHT PEOPLE???? PLZ HELP THANKS

    THATS ALL PLZ HELP ME EVERY1 HEHE :P IT WILL BE MOST APPRECIATED
     
  2. jcsd
  3. Apr 3, 2005 #2
    1. is right.

    For 2., note that [itex]x=4[/itex] satisfies the equation. Then use long division to get

    [tex] 6x^3 - 23x^2 - 6x + 8 = (x - 4)(6x^2 + x - 2)[/tex]

    so all you need to solve now is

    [tex]6x^2 + x - 2 = 0[/tex]

    which should be easy enough...

    3. is nearly right. Should be

    [tex] \frac{x^3y}{|z|}[/tex]
     
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