if a five digit number 1b6a3 ,a is the greatest single digit perfect cube and twice of it exceeds b by 7. Then the sum of the number and its cube root is. .
Answers
Answer:
19710
Step-by-step explanation:
5 digit no. is =1b6a3
Greatest single digit perfect cube=8
∴a=8
Given b=2a−7=2×8−7=9
No: 19683
Ans:19683 + 3√19683
= 19683 + 27
= 19710
Answer:
5 digit no. is =1b6a3
5 digit no. is =1b6a3Greatest single digit perfect cube=8
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683Ans:19683+
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683Ans:19683+ 3
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683Ans:19683+ 3
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683Ans:19683+ 3 19683
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683Ans:19683+ 3 19683
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683Ans:19683+ 3 19683 =19683+27
5 digit no. is =1b6a3Greatest single digit perfect cube=8∴a=8Given b=2a−7=2×8−7=9No: 19683Ans:19683+ 3 19683 =19683+27=19710
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