2 years ago

What is the 110th term of -1,-3,-5,-7

Mathematics

1 answer:

Scilla [17]2 years ago

8 0

Answer:

-219

Step-by-step explanation:

I counted it lol :o

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Pls solve this question

just olya [345]

Answer:

${( \sqrt{ {x}^{ - 3} }) }^{5} = ({( {x}^{ - 3}) }^{ \frac{1}{2} } ) ^{5} \\ \\ = {x}^{ - 3 \times \frac{1}{2} \times 5} = {x}^{ - \frac{15}{2} } = \frac{1}{ {x}^{ \frac{15}{2} } }$

<em><u>Or ;</u></em>

${( \sqrt{ {x}^{ - 3} } )}^{5} = {( \sqrt{ \frac{1}{ {x}^{3} }} })^{5} = ( { \frac{ \sqrt{1} }{ \sqrt{ {x}^{3} } } })^{5} \\ \\ = ( { \frac{1}{ \sqrt{x} \sqrt{ {x}^{2} } } })^{5} = ({ \frac{1}{ x\sqrt{x} }})^{5} \\ \\ = ( \frac{ {(1)}^{5} }{ {x}^{5} ( \sqrt{x} )^{5} } ) = \frac{1}{ {x}^{5} \times {x}^{ \frac{1}{2} \times 5 } } \\ \\ = \frac{1}{ {x}^{5} \times {x}^{ \frac{5}{2} } } = \frac{1}{ {x}^{5} \sqrt{ {x}^{5} } } \\ \\ = \frac{1}{ {x}^{5} \sqrt{ {x}^{4} {x}^{1} } } = \frac{1}{ {x}^{5} \sqrt{x} \sqrt{( { {x}^{2} })^{2} } } \\ \\ = \frac{1}{ {x}^{5} {x}^{2} \sqrt{x} } = \frac{1}{ {x}^{7} \sqrt{x} } = \frac{ \sqrt{x} }{ {x}^{8} }$