Verify given statements by solving both sides —-
1, ( – 7 )power32 ÷ ( – 7)power32 = 1​

Question

Verify given statements by solving both sides —-
1, ( – 7 )power32 ÷ ( – 7)power32 = 1​

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Bella 3 months 2021-07-21T02:25:58+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-21T02:27:10+00:00

      \qquad \bigstar \ \large \overline{ \underline{ \frak{given}}}

    • (-7)³² ÷ (-7)³² = 1

      \qquad \bigstar \ \large \overline{ \underline{ \frak{to \: find}}}

    • Verify the given statement bu Solving both the side

      \qquad \bigstar \ \large \overline{ \underline{ \frak{solution}}}

     \sf \ ∴\:  \:  \:  L.H.S. =  {( - 7)}^{32}  \div  {( - 7)}^{32}  \\  \\  \qquad \sf \:  =  \frac{ {( - 7)}^{32} }{ {( - 7)}^{32} }  \\  \\  \qquad \sf \:  =  \frac{ {( - 1)}^{32}  {(7)}^{32} }{ {( - 1)}^{32}  {(7)}^{32} }  \\  \\

     \qquad  \qquad\sf \:  =  \frac{1. {(7)}^{32} }{1. {(7)}^{32} }  \\  \\

      \:  \sf\qquad\qquad =  \frac{ {(7)}^{32} }{ {(7)}^{32} }  \:   \:  \:  \:  \:  \:  \:  \:  \:  \bigg\{∵ \:  \frac{ {a}^{m} }{ {a}^{n} }   =  {a}^{m - n}  \bigg\} \\  \\

     \sf \qquad \qquad =  {(7)}^{32 - 32}  \\  \\  \sf \qquad =  {(7)}^{0}  \\  \\\sf \qquad = 1 \\\\

     \sf \qquad \qquad \: = 1 = R. H. S.  \\  \\  \qquad \sf \: So, L. H. S = R. H. S \qquad \qquad \qquad \qquad \frak{ \red{proved}}

    0
    2021-07-21T02:27:41+00:00

    Answer:

    (-7)^{32}\:\div(-7)^{32}\:=\:1

    Solving L.H.S :

    (-7)^{32}\div(-7)^{32}

    = (-7)^{32-32}

    = (-7)^0

    = 1

    Hence, L.H.S = R.H.S .

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