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Doss [256]
3 years ago
14

Simply (m^3np^-3)^3(mn^4p^2)

Mathematics
1 answer:
Tamiku [17]3 years ago
4 0

Answer:

=\frac{m^{10}n^7}{p^7}

Step-by-step explanation:

(m^3np^-^3)^3(mn^4p^2)\\\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=\left(m^3np^{-3}\right)^3mn^4p^2\\\\\left(m^3np^{-3}\right)^3\quad :\quad m^9n^3p^{-9}\\\\=m^9n^3p^{-9}mn^4p^2\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\times\:a^c=a^{b+c}\\\\m^9m=\:m^{9+1}\\\\=n^3p^{-9}m^{9+1}n^4p^2\\n^3n^4=\:n^{3+4}\\\\=n^3p^{-9}m^{10}n^4p^2\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\times\:a^c=a^{b+c}\\=p^{-9}m^{10}n^{3+4}p^2\\\\=p^{-9}m^{10}n^7p^2\\\\=m^{10}n^7p^{-9+2}

=m^{10}n^7p^{-7}\\\\=m^{10}n^7\frac{1}{p^7}\\\\\frac{1\times\:m^{10}n^7}{p^7}\\\\=\frac{m^{10}n^7}{p^7}

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Answer:

A. A reflection over the x-axis and a vertical stretch.

Step-by-step explanation:

Let h(x) = x^{2}, to obtain k(x) = -4\cdot x^{2}, we need to use the following two operations:

(i) Vertical stretch

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(ii) Reflection over the x-axis

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Let prove both transformations in h(x):

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Hence, correct answer is A.

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