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Using suitable identity evaluate 101 multiply …

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Using suitable identity evaluate 101 multiply 102

    • Posted by Nikunj Aggarwal 2 years, 1 month ago

    CBSE > Class 09 > Mathematics
    • 5 answers

    Smriti Gupta 2 years, 1 month ago

    10302

    0Thank You

    Shubh Goyal 2 years, 1 month ago

    101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x 2 +(a+b)x+ab, we have (100+1)(100+2)=(100) 2 +(1+2)100+(1)(2) =10000+(3)100+2=10000+300+2 =10302

    1Thank You

    Shivanand Tiwari 2 years, 1 month ago

    101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x 2 +(a+b)x+ab, we have (100+1)(100+2)=(100) 2 +(1+2)100+(1)(2) =10000+(3)100+2=10000+300+2 =10302

    2Thank You

    Rudra . 2 years, 1 month ago

    101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x2+(a+b)x+ab, we have (100+1)(100+2)=(100)2+(1+2)100+(1)(2) =10000+(3)100+2 =10000+300+2 =10302

    2Thank You

    Rudra . 2 years, 1 month ago

    101×102=(100+1)(100+2) Now, using identify (x+a)(x+b)=x 2 +(a+b)x+ab, we have (100+1)(100+2)=(100) 2 +(1+2)100+(1)(2) =10000+(3)100+2=10000+300+2 =10302

    2Thank You

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