Question

Do as directed
1. Sign of square root is.
2. V64 +2=
3.
digit is at one's place in cube of 17.
4. State whether this statement is true or false: Last three digit of cube of 500 is 0.
5. Find the smallest number which we get by multiplying it with 100 to get perfect
cube.
6. Write a Pythagorean triplet whose one number is 12​

Answer 1

Step-by-step explanation:

1. √ is the sign of square root

2. 8 + 2 = 10

3. cube of 17 is 4913 , digit at ones place is 3.

4. True, the cube of 500 is 125000000.

5. the number is 10. when we multiply it with 100 we get 1000 which is the perfect cube of 10.

6.(2m, m²-1, m²+1)

Here, 2m = 12, m =12/2 = 6.

Hence, the triplet is [ 2 (6), 62-1,62+1]

Therefore, The Pythagorean triplet whose one number is 12 is 12,35,37.

Hope it will help you