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Math Help - evaluating numerical expression

  1. #1
    Senior Member vaironxxrd's Avatar
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    evaluating numerical expression

    Hello forum, Vaironxxrd here.

    I have a question about exponents. If I have,

    (-9)(-9)^3 Would that be (-9)^4 or (-9^4).

    If possible can you guys provide other examples?
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    MHF Contributor Siron's Avatar
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    Re: evaluating numerical expression

    It's indeed (-9)^4. For example, consider (-2)^4\cdot (-2)^2 which is offcourse (-2)^6=64 and -2^6=-(2^6)=-64 therefore there're not equal.
    So it's important to use brackets!
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    Senior Member vaironxxrd's Avatar
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    Re: evaluating numerical expression

    Quote Originally Posted by Siron View Post
    It's indeed (-9)^4. For example, consider (-2)^4\cdot (-2)^2 which is offcourse (-2)^6=64 and -2^6=-(2^6)=-64 therefore there're not equal.
    So it's important to use brackets!
    Finally, I now understand it (-9)^3 = -729 is going to be something negative since the exponent is odd and I just do the normal exponent.
    9^2 is positive = 81.

    Now what is the difference between (-9^2) and (-9)^2
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    Re: evaluating numerical expression

    Quote Originally Posted by vaironxxrd View Post
    Finally, I now understand it (-9)^3 = -729 is going to be something negative since the exponent is odd and I just do the normal exponent.
    9^2 is positive = 81.

    Now what is the difference between (-9^2) and (-9)^2
    Exponents come before multiplication in the order of operations and a minus sign in front is multiplying by -1.

    However, brackets come before exponents so if the multiplication is done inside the bracket it too is affected by the exponents.

    (-9)^2 = -9 \times -9 = 81 \text{  OR  } (-9)^2 = (-1 \times 9)^2 = (-1)^2 \times 9^2 = 81

    -9^2 = -1 \times 9^2 = -81 - the brackets here are unnecessary and only serve to complicate matters IMO.
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    Senior Member vaironxxrd's Avatar
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    Re: evaluating numerical expression

    Quote Originally Posted by e^(i*pi) View Post
    Exponents come before multiplication in the order of operations and a minus sign in front is multiplying by -1.

    However, brackets come before exponents so if the multiplication is done inside the bracket it too is affected by the exponents.

    (-9)^2 = -9 \times -9 = 81 \text{  OR  } (-9)^2 = (-1 \times 9)^2 = (-1)^2 \times 9^2 = 81

    -9^2 = -1 \times 9^2 = -81 - the brackets here are unnecessary and only serve to complicate matters IMO.
    Thanks allot Pi!
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    Re: evaluating numerical expression

    Quote Originally Posted by vaironxxrd View Post
    Now what is the difference between (-9^2) and (-9)^2
    Remember BIDMAS (Brackets, Indices, Division, Multiplication, Addition, Subtraction).
    The order of operands is very important, and this is where you're having trouble.

    -9 is essentially like saying 0 - 9, you just omit the 0.
    Now, from BIDMAS, we know that everything in brackets is evaluated first, and that indices/multiplication is evaluated before subtraction.

    So (-9^2) is like saying (0 - 9^2). You evaluate the index first, then you subtract the result from 0.
    0 - (9)(9) = 0 - 81 = -81

    Whereas, in (-9)^2 you evaluate everything in brackets first, so (0 - 9)^2 = (-9)(-9) = 81, because multiplying an even number of negatives together results in a positive.

    I hope that was clear enough.
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    Senior Member vaironxxrd's Avatar
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    Re: evaluating numerical expression

    Quote Originally Posted by FrameOfMind View Post
    Remember BIDMAS (Brackets, Indices, Division, Multiplication, Addition, Subtraction).
    The order of operands is very important, and this is where you're having trouble.

    -9 is essentially like saying 0 - 9, you just omit the 0.
    Now, from BIDMAS, we know that everything in brackets is evaluated first, and that indices/multiplication is evaluated before subtraction.

    So (-9^2) is like saying (0 - 9^2). You evaluate the index first, then you subtract the result from 0.
    0 - (9)(9) = 0 - 81 = -81

    Whereas, in (-9)^2 you evaluate everything in brackets first, so (0 - 9)^2 = (-9)(-9) = 81, because multiplying an even number of negatives together results in a positive.

    I hope that was clear enough.
    To not start a new thread can you tell me if I correctly solved this problem?

    \frac{5x^3y^9}{20x^2y^-^2} Solution: \frac{x&y^1^1}{4}

    Is this right?
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    Re: evaluating numerical expression

    Quote Originally Posted by vaironxxrd View Post
    To not start a new thread can you tell me if I correctly solved this problem?

    \frac{5x^3y^9}{20x^2y^-2} Solution: \frac{x&y^1^1}{4}

    Is this right?
    if that is y^{-2} in the denominator, then yes, it's correct.

    In future, go ahead and start a new problem with a new thread.
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    Senior Member vaironxxrd's Avatar
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    Re: evaluating numerical expression

    Yeah sorry for that. Does it matter how many threads I start? I feel like it might be annoying.
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    Re: evaluating numerical expression

    Quote Originally Posted by vaironxxrd View Post
    Yeah sorry for that. Does it matter how many threads I start? I feel like it might be annoying.
    a single thread with multiple problems all strung out is more annoying.
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