There are 4 of them which are 2*3*3*3 = 54
As a product of its prime factors: 2*3*3*3 = 54
2 x 3 x 3 x 3
As a product of its prime factors: 2*3*3*3 = 54
gcf(54, 75, 81) = 3 54 = 2 x 3^3 75 = 3 x 5^2 81 = 3^4 gcf = 3
hcf(18, 54, 81) = 9. Using prime factorization: 18 = 2 x 3^2 54 = 2 x 3^3 81 = 3^4 hcf = 3^2 = 9
There are 4 of them which are 2*3*3*3 = 54
As a product of its prime factors: 2*3*3*3 = 54
2 x 3 x 3 x 3
As a product of its prime factors: 2*3*3*3 = 54
gcf(54, 75, 81) = 3 54 = 2 x 3^3 75 = 3 x 5^2 81 = 3^4 gcf = 3
(2√24) / √54 = (2√4√6) / (√9√6) = (4√6)/(3√6) = 4/3
The LCM of 48 and 54 is 432. Prime factors of 48 are: 2^4 and 3 Prime factors of 54 are: 2 and 3^3 Multiplying the highest prime factors: 2^4 times 3^3 = 432 which is the LCM
178 is the answer By using the formula a{n} = n³ + n² + 5n + 3 For e.g. A{2} =2 ³ +2²+5*2+3=25 =8+4+10+3=25 And so on.    = 10, 25, 54, 103, 178
3/4 * 54 = 3(27)/2 = 81/2 = 40.5.
The factors of 54 are 1, 2, 6, 9, 27, and 54 The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108
t(1) = a = 54 t(4) = a*r^3 = 2 t(4)/t(1) = r^3 = 2/54 = 1/27 and so r = 1/3 Then sum to infinity = a/(1 - r) = 54/(1 - 1/3) = 54/(2/3) = 81.