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lutik1710 [3]
4 years ago
7

Find the value of the following expression:

Mathematics
1 answer:
FinnZ [79.3K]4 years ago
7 0

\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (2^8\cdot 3^{-5}\cdot 6^0)^{-2}\cdot \left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\implies \stackrel{\textit{distributing the exponent}}{(2^{8\cdot -2}\cdot 3^{-5\cdot -2}\cdot 6^{0\cdot -2})\cdot \left( \cfrac{3^{-2\cdot 4}}{2^{3\cdot 4}} \right)\cdot 2^{28}}

\bf (2^{-16}\cdot 3^{10}\cdot 6^0)\cdot \left( \cfrac{3^{-8}}{2^{12}} \right)\cdot 2^{28}\implies (2^{-16}\cdot 3^{10}\cdot 1)\cdot ( 3^{-8}\cdot 2^{-12})\cdot 2^{28} \\\\\\ 2^{-16}\cdot 2^{-12}\cdot 2^{28}\cdot 3^{10}\cdot 3^{-8}\implies 2^{-16-12+28}\cdot 3^{10-8}\implies 2^{0}\cdot 3^{2} \\\\\\ 1\cdot 9\implies 9

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Step-by-step explanation:

Given:

A catering service offers 10 appetizers, 4 main courses, and 6 desserts.

A costumer is to select 9 appetizers, 2 main courses, and 5 desserts for a banquet.

Question asked:

In how many ways can this be done ?

Solution:

<u>By applying combination's formula:-</u>

^{n} C_{r}=\frac{n!}{(n-r)!\ r!}

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^{10} C_{9}=\frac{10!}{(10-9)!\ 9!}=\frac{10\times9!}{1!\times9!} ,\ 9!\ canceled\ by\ 9! =10\ ways

A costumer can choose 2 main courses out of 4 in =  

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^{6} C_{5}=\frac{6!}{(6-5)!\ 5!}=\frac{6\times5!}{1!\times5!} =\frac{6}{1} =6\ ways

Total number of ways = 10\times12\times6=720\ ways

Therefore, A costumer can select 9 appetizers, 2 main courses, and 5 desserts for a banquet in 720 ways.

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