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Math Help - can someone double check this for me?

  1. #1
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    can someone double check this for me?

    Hi, can someone double check this for me?

    Also sorry about the badly written problem. What programs do you all use? I tried word but i couldn't import it over to here.

    can someone double check this for me?-fsdfsdfds.jpg


    Also, when you do the power, you do that keep that sign first and then incorporate either the - or + sign next right?

    So lets say x=-2, for (x^5)-(X^3)=
    you do -2*-2*-2*-2*-2 which is -32 and then you do -2*-2*-2 which is -8. Then the last step is to subtract -32- (-8) but since there is two neg. signs, -8 becomes a positive 8 right? so it is -32+8

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  2. #2
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    Quote Originally Posted by soyeahiknow View Post
    Hi, can someone double check this for me?

    Also sorry about the badly written problem. What programs do you all use? I tried word but i couldn't import it over to here.

    Click image for larger version. 

Name:	fsdfsdfds.JPG 
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ID:	18229


    Also, when you do the power, you do that keep that sign first and then incorporate either the - or + sign next right?

    So lets say x=-2, for (x^5)-(X^3)=
    you do -2*-2*-2*-2*-2 which is -32 and then you do -2*-2*-2 which is -8. Then the last step is to subtract -32- (-8) but since there is two neg. signs, -8 becomes a positive 8 right? so it is -32+8

    Thanks
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  3. #3
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    I did that but I got -53/3.

    I know it is 0-(64/6)-(16/4)+(-8/3), but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?
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  4. #4
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    Quote Originally Posted by soyeahiknow View Post
    Also sorry about the badly written problem. What programs do you all use? I tried word but i couldn't import it over to here.
    LaTeX help subforum

    Quote Originally Posted by soyeahiknow View Post
    ..image here ..
    You wrote a plus sign where it should have been minus sign, right before the 8/3.

    Quote Originally Posted by soyeahiknow View Post
    Also, when you do the power, you do that keep that sign first and then incorporate either the - or + sign next right?
    I don't follow..

    Quote Originally Posted by soyeahiknow View Post
    So lets say x=-2, for (x^5)-(X^3)=
    you do -2*-2*-2*-2*-2 which is -32 and then you do -2*-2*-2 which is -8. Then the last step is to subtract -32- (-8) but since there is two neg. signs, -8 becomes a positive 8 right? so it is -32+8
    That is correct. If you have a function g(x) = x^5 - x^3 then g(-2) = -24.

    Note that you know in advance that (-2)^5 is negative because the exponent is odd. So it's just -2^5. (Note the lack of parentheses. Meaning take 2^5 and then negate it.)

    Quote Originally Posted by soyeahiknow View Post
    I did that but I got -53/3.

    I know it is 0-(64/6)-(16/4)+(-8/3), but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?
    You seem to have typed the wrong thing into Wolfram Alpha. The answer is -4. See here.
    Last edited by undefined; July 18th 2010 at 12:27 PM. Reason: details
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  5. #5
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    yeah sorry, I messed up that sign.

    So the full problem written out is 0- {(64/6)-(16/4)+(-8/3)}, but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?
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  6. #6
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    Quote Originally Posted by soyeahiknow View Post
    yeah sorry, I messed up that sign.

    So the full problem written out is 0- {(64/6)-(16/4)+(-8/3)}, but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?
     0- \left(\left(\dfrac{64}{6}\right)-\left(\dfrac{16}{4}\right)+\left(-\,\dfrac{8}{3}\right)\right) = -4

    So this is right, just make sure you get -4 as final answer.

    Edit: I thought you were just evaluating the thing inside parentheses first, and then would remember to negate it afterwards. It looks like you were making the mistake pointed out in Archie Meade's post below mine though.
    Last edited by undefined; July 18th 2010 at 02:00 PM.
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  7. #7
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    Quote Originally Posted by soyeahiknow View Post
    yeah sorry, I messed up that sign.

    So the full problem written out is 0- {(64/6)-(16/4)+(-8/3)}, but I thought I just didn't worry about the zero and just have (64/6)-(16/4)+(-8/3). Am I wrong?
    You wouldn't have to worry about the zero if you were adding it to the three values.
    Unfortunately, you are subtracting all of them from zero.

    0-\left(\frac{64}{6}-4-\frac{8}{3}\right)

    is very different to

    0-\frac{64}{6}-4-\frac{8}{3}

    which has a redundant zero.

    0-\left(\frac{64}{6}-4-\frac{8}{3}\right)=-\frac{64}{6}-(-4)-\left(-\frac{8}{3}\right)
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