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Using suitable identity evaluate 101 multiply …

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Using suitable identity evaluate 101 multiply 102
  • 5 answers

Smriti Gupta 1 year, 7 months ago

10302

Shubh Goyal 1 year, 7 months 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

Shivanand Tiwari 1 year, 7 months 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

Rudra . 1 year, 7 months 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

Rudra . 1 year, 7 months 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
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