If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a? (1) a^n = 64(2) n=6

a=2

Step-by-step explanation:

Well, first let us find product of the first 8 positive integers, which is 1*2*3*4*5*6*7*8=40320. It can also be written as 40320=*3²*5*7. From the formula above, it can be extracted that if a=2, n can be any number from 3 to 7 (considering that a and n are not equal and greater than 1).

In (1) a^n=64 and in (2) n=6 so it can be written as 2^6=64 so a=2 and n=6. So the value of a is 2. a=2

the correct answer option is (2x-3)(2x+3).

step-by-step explanation:

we are given the following expression and we are to factorize it:

we are to find such factors that when multiplied they give a result of 36 (9 x 4 = 36) and when added they give a result of 0.

the two such factors of 36 are negative six and postive six.

factorizing the given expression to get:

therefore, the correct factorization of is (2x-3)(2x+3).