Suppose that in standard factored form a = pe1 1 pe2 2 · · · pek k , where k is a positive integer; p1, p2, . . . , pk are prime numbers; and e1, e2, . . . , ek are positive integers. a. What is the standard factored form for a2? b. Find the least positive integer n such that 25 ·3·52 ·73 ·n is a perfect square. Write the resulting product as a perfect square.

a . standard factored form for a2= (pe11pe22...pekk)2

b. n = 810540900

Step-by-step explanation:

25 .3 .52 .73=5square .3. (2square.13).73.n,

Making the power of every prime numbers even, we get

n = 3. 13. 73

The resulting number will be

2square. 3square. 5square. 13square.73square=2×2×3×3×5×5×13×13×73×73 = 28470square = 810540900

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

jwjwkwkwkkkdjebd b jdjejdndj he jenej