duvide the sum of 65/12and 8/3 by their difference​

Question

duvide the sum of 65/12and 8/3 by their difference​

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Sadie 8 months 2021-10-03T07:35:54+00:00 2 Answers 0 views 0

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    0
    2021-10-03T07:37:04+00:00

    Answer:

    Let A=\frac{65}{12}\: and \:</p><p>B = \frac{8}{3}


    i ) Sum =A+B = \frac{65}{12}+\frac{8}{3}


    =\frac{65+32}{12}


    =\frac{97}{12} —(1)


    ii) Difference = A-B=\frac{65}{12}-\frac{8}{3}


    =\frac{65-32}{12}


    =\frac{33}{12} —(2)


    According to the problem given,


    \frac{A+B}{A-B}


    = \frac{\frac{97}{12}}{\frac{33}{12}}


    = \frac{97}{33}


    Therefore,


    Divide the sum of 65/12 and 8/3 by their difference =\frac{97}{33}



    0
    2021-10-03T07:37:12+00:00

    Answer:

    Let A=\frac{65}{12}\: and \:</p><p>B = \frac{8}{3}

    i ) Sum =A+B = \frac{65}{12}+\frac{8}{3}

    =\frac{65+32}{12}

    =\frac{97}{12} —(1)

    ii) Difference = A-B=\frac{65}{12}-\frac{8}{3}

    =\frac{65-32}{12}

    =\frac{33}{12} —(2)

    According to the problem given,

    \frac{A+B}{A-B}

    = \frac{\frac{97}{12}}{\frac{33}{12}}

    = \frac{97}{33}

    Therefore,

    Divide the sum of 65/12 and 8/3 by their difference =\frac{97}{33}

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