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(1 point) The expression
where n, the leading coefficient, is:
and r, the exponent of a, is: 3
and S, the exponent of b, is: 10
and finally t, the exponent of c, is: 15
(5b¹c-5)-³(4b⁰a²)-5 equals na'bs ct
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NOTE: The value for n may be given as a fraction or as a just number. All other values must be
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1 point The expression where n the leading coefficient is and r the exponent of a is 3 and S the exponent of b is 10 and finally t the exponent of c is 15 5bc54 class=

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sqdancefan

Answer:

  • n = 1/128000
  • r = -10
  • s = -3
  • t = 15

Step-by-step explanation:

You want the simplified form of (5b^1c^-5)^-3(4b^0a^2)^-5.

Rules of exponents

There are several useful rules of exponents here:

  (ab)^c = (a^c)(b^c)

  (a^b)^c = a^(bc)

  a^-b = 1/(a^b)

Application

Simplifying the top-level exponents first, we have ...

  [tex](5b^1c^{-5})^{-3}(4b^0a^2)^{-5}=(5^{-3}b^{1(-3)}c^{(-5)(-3)})(4^{-5}b^{0(-5)}a^{2(-5)})\\\\=(5^{-3}4^{-5})a^{-10}b^{-3+0}c^{15}=\dfrac{1}{5^34^5}a^{-10}b^{-3}c^{15}=\boxed{\dfrac{1}{128000}a^{-10}b^{-3}c^{15}}[/tex]

In terms of answer requirements:

  n = 1/128000

  r = -10

  s = -3

  t = 15

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