Three numbers are in the ratio 2: 3: 5 and the sum of the squares of the numbers is 608. find the smallest of the three numbers.
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I m sorry...I was mistaken...I made d solution for d sun of digits but few time later I was that it was for d sum of squares...so I have erased my solution....there is another solution above given by some1....
Botlu:
I did...dere4 I erased d solution bcoz I was unable to delete the solution
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Let the numbers are 2x,3x,5x
So their squares are 4x²,9x²,25x²
According to the Q.
4x²+9x²+25x²=608
So, 38x²=608
Or, x²=16
Or, x=±4
So the smallest number is (-4×5)=-20
Or if you take the positive number so the smallest number is (4×2)=8
So their squares are 4x²,9x²,25x²
According to the Q.
4x²+9x²+25x²=608
So, 38x²=608
Or, x²=16
Or, x=±4
So the smallest number is (-4×5)=-20
Or if you take the positive number so the smallest number is (4×2)=8
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