Question

who is cobe ???,......​

Answer 1

Answer:

please dm yrr place I will not irritated you

Answer 2

Step-by-step explanation:

Answer:

Option C is correct Answer

Why?

First of all let's review table of 9:

$\begin{gathered}\begin{gathered} \boxed{ \begin{array}{c |c} \tt9 \times 1& \tt9 \\ \tt9 \times 2& \tt \red{18} \\ \tt9 \times 3& \tt \purple{27} \\ \tt9 \times 4& \tt36\\ \tt9 \times 5& \tt \orange{45} \\ \tt9 \times 6& \tt54\\ \tt9 \times 7& \tt63 \\ \tt9 \times 8& \tt72 \\ \tt9 \times 9& \tt81 \\ \tt9 \times 10&\tt90\\ \end{array}} \end{gathered}\end{gathered}$

Yep as you can see in 9 table all digits i.e. 18 , 27 and 45 is there,therefore correct option is A.

Now Let's prove why not other options:

Why not option A:

As we know multiples of 5 always end with 5 and 0.

where as in 18 it ends with 8 and in 27 it ends with 7.

Why not option B:

We know that multiples of 6 should also be multiples of both 2 and 3 .

where as 45 is a multiple of 3 but not multiple of 2. .°. it's also not multiple of 6.

Why not option C:

As we know multiple of 12 should also be multiples of both 6 and 3 .

where as 45 is a multiple of 3 but not multiple of 6(we have proved before).°. it's also not multiple of 12.

#CarryOnLearning:D