What will be the formula of the sum (81) to the power 5 devided by (3 to the power 2) to the power 5?
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Answered by
1
➡ aⁿ / bⁿ = ( a-b ) ⁿ
➡ (81)^5 ÷ ( 3² )^5
➡ ( 81 )^5 ÷ ( 9 )^5
➡ ( 81/9 ) ^5
➡ ( 9 )^5
➡ 9 * 9 * 9 * 9 * 9
➡ 59049
Hope it will help you !!!
➡ (81)^5 ÷ ( 3² )^5
➡ ( 81 )^5 ÷ ( 9 )^5
➡ ( 81/9 ) ^5
➡ ( 9 )^5
➡ 9 * 9 * 9 * 9 * 9
➡ 59049
Hope it will help you !!!
Answered by
0
Heya ❗❗❗
![= {(81)}^{5} \div [{( {3})^{2} ]}^{5} \\ \\ = [{ ({9})^{2} ]}^{5} \div [{( {3})^{2} ]}^{5} \\ \\ = ({9})^{10} \div ( {9})^{5} \\ \\ = \frac{ ({9)}^{10} }{ ({9)}^{5} } \\ \\ = {(9)}^{10 - 5} \\ \\ = {(9)}^{5} \\ \\ = 9 \times 9 \times 9 \times 9 \times 9 \\ \\ = 59049](https://tex.z-dn.net/?f=+%3D++++%7B%2881%29%7D%5E%7B5%7D++%5Cdiv+%5B%7B%28+%7B3%7D%29%5E%7B2%7D+%5D%7D%5E%7B5%7D++%5C%5C++%5C%5C++%3D+++%5B%7B+%28%7B9%7D%29%5E%7B2%7D+%5D%7D%5E%7B5%7D++%5Cdiv+%5B%7B%28+%7B3%7D%29%5E%7B2%7D+%5D%7D%5E%7B5%7D++++%5C%5C++%5C%5C++%3D++++%28%7B9%7D%29%5E%7B10%7D++%5Cdiv+%28+%7B9%7D%29%5E%7B5%7D++++%5C%5C++%5C%5C++%3D+++%5Cfrac%7B+%28%7B9%29%7D%5E%7B10%7D+%7D%7B+%28%7B9%29%7D%5E%7B5%7D+%7D++%5C%5C++%5C%5C++%3D++++%7B%289%29%7D%5E%7B10+-+5%7D++%5C%5C++%5C%5C++%3D++++%7B%289%29%7D%5E%7B5%7D++%5C%5C++%5C%5C++%3D++9+%5Ctimes+9+%5Ctimes+9+%5Ctimes+9+%5Ctimes+9+%5C%5C++%5C%5C++%3D+59049 "= {(81)}^{5} \div [{( {3})^{2} ]}^{5} \\ \\ = [{ ({9})^{2} ]}^{5} \div [{( {3})^{2} ]}^{5} \\ \\ = ({9})^{10} \div ( {9})^{5} \\ \\ = \frac{ ({9)}^{10} }{ ({9)}^{5} } \\ \\ = {(9)}^{10 - 5} \\ \\ = {(9)}^{5} \\ \\ = 9 \times 9 \times 9 \times 9 \times 9 \\ \\ = 59049")
Hope it helps.
Hope it helps.
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