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mart [117]
1 year ago
10

Pls solve this question ​

Mathematics
1 answer:
just olya [345]1 year ago
3 0

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} }

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Answer:

Step-by-step explanation:

Find the measure of the vertex angle ∠ABD of an isosceles triangle

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Step 2

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<h3>How to depict the information?</h3>

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