8. What is a counterexample of the following conjecture? Conjecture: Any number that is divisible by 3 is also divisible by 6.
027
O 36
O 23
O 18

Any odd multiple of 3 would be the counterexample of the given conjecture.

That is 3(2n - 1), for any natural number n, will be divisible by 3 but not divisible by 6 as 2 is not a factor.

Examples:

3(3) = 9

3(5) = 15

3(7) = 21 are some examples.

Step-by-step explanation:

none of the option is correct.

step-by-step explanation:

it says to find which expression equals 9^3 sqrt 10 = ?

9^3 sqrt 10 = 729 sqrt 10.

checking option a. 5 sqrt 10 + 4 sqrt 10

5 sqrt 10 + 4 sqrt 10

= 9 sqrt 10.

checking option b. 5^3 sqrt 10 + 4^3 sqrt 10

5^3 sqrt 10 + 4^3 sqrt 10

= 125 sqrt 10 + 64 sqrt 10

= 189 sqrt 10.

checking option c. 5 sqrt 10 + 4^3 sqrt 10

5 sqrt 10 + 4^3 sqrt 10

= 5 sqrt 10 + 64 sqrt 10

= 69 sqrt 10.

checking option d. 5^3 sqrt 10 + 4 sqrt 10

5^3 sqrt 10 + 4 sqrt 10

= 125 sqrt 10 + 4 sqrt 10

= 129 sqrt 10.

hence, none of the option is correct.