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Rajeev Malhotra 7 years, 11 months ago

let the three numbers are x, x+1 and x+2.

so (x)2+(x+1)2+(x+2)2 = 77

x2+x2+2x+1+x2+4x+4= 77

3x2+6x+5 = 77

3x2+6x -72 =0

3x2+18x-12x-72 =0

3x(x+6)-12(x+6)=0

(3x-12)(x+6)=0

3x-12=0 or x+6 =0

3x=12 or x= -6

x=12/3=4 nor x=-6

x =4 , x=-6 is not possible because a natural number can not be negative.

so numbers are 4,5 and 6.

 

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Gaurav Bansal 7 years, 11 months ago

Yes,this is true when the events A & B are mutually exclusive events.
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Gaurav Bansal 7 years, 11 months ago

Domain={-3,-2,-1,0,1,2,3} Range={4,3,2,1,0}

Ishan Garg 7 years, 11 months ago

Please tell the answer.plz plz
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Kavya Verma 7 years, 11 months ago

AP=1,3,5,7-------2001 a=1,d=2,an=2001 find sn=? as, an=a+(n-1)d 2001=1+(n-1)2 2001=1+2n-2 2001+1=2n 2002=2n n=1001 now, sn=n/2[2a+(n-1)d] =1001/2[2×1+(1001-1)2] 1001/2×2[1+1001-1] 1001×1001 1002001
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Kavya Verma 7 years, 11 months ago

sets and functions ,algebra = 29+37=66
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Kavya Verma 7 years, 11 months ago

sinacosb+cosasinb

Anurag Shakya 7 years, 11 months ago

sinacosb+cosasinb
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Gaurav Bansal 7 years, 11 months ago

Sec^2x
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