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-   -   Evaluate formulas (https://www.askmehelpdesk.com/showthread.php?t=11435)

  • Jul 31, 2005, 05:49 AM
    vermonstervt
    evaluate formulas
    evaluate:
    x^2+y^3 if x=2 y=-3
    4+-27=-23
    --------------------
    pm^2-z^3 if p=1 m=-4 z=-2
    (1)(-4^2)-(-2)^3=(1)(16)-(-8)=16+8=24
    -------------------------------------
    -x^2-y^3 if x=-3 y=-2
    -3^2-2^3
    -9-8=-17
    -------------------
    a^2-b^2a if a=-2 b=3
    -2^2-3^2
    -4-9-2=-13-2=-15
    If not correct please explain step by step what I did wrong so I can get the hang of this thanks :)
  • Jul 31, 2005, 08:14 AM
    wzartv
    Hello and welcome,

    Number 1 looks to be correct.

    Number 2, however, I noticed an error.
    (-4^2) comes out to -16, not 16.
    That is the only error in that problem. Be sure to look over the problem and see where you made that mistake. Try working it out again and the answer is -8.

    Number 3, I noticed an error.
    The variables in the problem (x and y) have a negative sign in front of them. Also, the numbers that are replacing the variables are negative. That's a double negative, therefore making them a positive number. Look at the problem, see what I mean, and the answer comes out to 17, not -17.

    Number 4, could you please explain the problem. Is there an "a" after -b^2, are there parenthesis present, was it a typo? You have written a^2-b^2a - is that the correct problem?

    All right, I double checked my answers and I am quite sure that they are correct. If anyone sees any errors, please point them out. Let me know about the fourth problem.
  • Jul 31, 2005, 12:43 PM
    CroCivic91
    Quote:

    Originally Posted by wzartv
    Number 2, however, I noticed an error.
    (-4^2) comes out to -16, not 16.
    That is the only error in that problem. Be sure to look over the problem and see where you made that mistake. Try working it out again and the answer is -8.

    Let me just add to this...
    (-4^2) = -(4^2) = -16
    but...
    ((-4)^2) = 16
    Just watch out how the braces are placed.

    Quote:

    Originally Posted by wzartv
    Number 3, I noticed an error.
    The variables in the problem (x and y) have a negative sign in front of them. Also, the numbers that are replacing the variables are negative. That's a double negative, therefore making them a positive number. Look at the problem, see what I mean, and the answer comes out to 17, not -17.

    I have to dissagree here.

    Let's look at the braces again...
    -x^2-y^3 if x=-3 y=-2
    that means...
    -(x^2)-(y^3) = - ((-3)^2) - ((-2)^3) = -9 - (-8) = -9 + 8 = -1

    You have to be careful with the braces.

    About the 4th problem...

    a^2-b^2a if a=-2 b=3
    Let's imagine that you misstyped "-b^2a" and that it should be "-b^2".
    a^2 - (b^2) = (-2)^2 - (3^2) = 4 - 9 = -5
    Let's imagine that you didn't misstype that...
    You have multiple options... does that mean
    - (b^(2a))... (case 1)
    or does it mean
    - (b^2)*a... (case 2)

    Case 1:
    -(b^(2a)) = - (3^(2*(-2))) = - (3^(-4)) = - ( 1 / (3^4) ) = - ( 1 / 81 )

    Case 2:
    - (b^2)*a = - (3^2) * (-2) = (-9) * (-2) = 9 * 2 = 18

    So the whole result in case 1 would be "4 - ( 1 / 82 )" and in case 2 "4 + 18 = 22"
  • Jul 31, 2005, 12:50 PM
    wzartv
    All right thanks for the posts... I guess I have to go to school again... hahaha
  • Jul 31, 2005, 12:51 PM
    wzartv
    Good point on number three... didn't think of that
  • Aug 1, 2005, 02:53 AM
    CroCivic91
    It's all right, everybody makes mistakes :)
  • Aug 1, 2005, 07:48 AM
    wzartv
    vermonstervt - I left you a response to your e-mail... the subject will come up as "Re: Thanks"

    Check it out.

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