Question

find three numbers in the ratio 2:3:5,the sum of whose square is 608​

Answer 1

Answer:

8,12,20

8,12,20(2z)²+(3z)²+(5z)²=608

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38or (z)²= 16

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38or (z)²= 16or (z)² = (4)²

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38or (z)²= 16or (z)² = (4)²or z = 4

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38or (z)²= 16or (z)² = (4)²or z = 4 Hence the three number are

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38or (z)²= 16or (z)² = (4)²or z = 4 Hence the three number are2×z=2×4=8

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38or (z)²= 16or (z)² = (4)²or z = 4 Hence the three number are2×z=2×4=83×z=3×4=12

8,12,20(2z)²+(3z)²+(5z)²=608or (4z)²+(9z)²+(25)²= 608or (38z)²= 608or (z)²= 608÷38or (z)²= 16or (z)² = (4)²or z = 4 Hence the three number are2×z=2×4=83×z=3×4=125×z=5×4=20