Question

please do this question​

Answer 1

Answer:

$\huge\bf\underline{\underline{Answer}}$

1. $\sqrt{~}$

2. 8

3. (17)³ = unit place 7³ = 343 => 3 in last . Hence, digit is at one's place in cube of 17 will be 3.

4. true

5. Prime factors of 100 = 2×2×5×5

Here factor 2 and 5 both do not appear in 3’s group.

Therefore 100 must be multiplied by 2×5= 10 to make it a perfect cube.

6. one member of triplet = 12

let 2m = 12

m = 6

let m² - 1 =

= (6)² - 1

= 36 - 1

= 35

let m² + 1

= ( 6 ) ² + 1

= 36 + 1

= 37

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

$\huge\bf\underline{\underline\color{purple}{Purple~you♡}}$

Answer 2

$\huge{ \red{ \fbox{12,35,37}}}$

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Mark upper one Brainliest $\huge\color{red}♡$