Algebra Math Problems Help?

Could you solve these and explain?

(3x^5 y^0)^ -1

(2x^2)(2x^-4)^2

(-2p^3)(3p^5)

Comments

  • (3x^5 y^0)^ -1

    =1/(3x^5 y^0)

    =1/(3x^5 )

    =1/3 (x^-5)

    (2x^2)(2x^-4)^2

    =(2x^2)(4x^-8)

    =8x^-6

    (-2p^3)(3p^5)

    = -6 p^8

  • 1).

    (3x^5 y^0)^-1

    = 1/3x^5

    2).

    (2x^2)(2x^-4)^2

    = 8/x^6

    3).

    (-2p^3)(3p^5)

    = -6p^8

  • 1) 1/(3x^5)

    2) (2x^2)(4x^-8)

    8/(x^6)

    3) -6p^8

    I hope this information was very helpful.

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