How many zeroes are at the end of 10! = 1∙2 ∙ … ∙ 9∙10?

2.

Step-by-step explanation:

To find the number of trailing zeroes, given n!, we take n and divide it by increasing powers of 5. For example, the number of trailing zeroes in 50! = 50/5 = 10, 50/25 = 2, 10 + 2 = 12 trailing zeroes.

In this case, 10/5 = 2, and since we cannot get an integer from dividing 10 and 25, we are left with this answer.

I hope this helps!

because its not ur topic and its not for is for homework not that.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b