(3,11)
√33√21 = √(3x11)√(3x7) = √3√11√3√7 = (√3)^2√7√11 = 3√7√11 (or 3√77).
To work this out, you first need to put both numbers as their prime factors: 12 = 2x2x3 33 = 3x11 The next step is to identify any common prime factors. Both numbers have a 3 as a prime factor. Therefore the HCF of 12 and 33 is 3.
x3 times table: 3x1=3 3x2=6 3x3=9 3x4=12 3x5=15 3x6=18 3x7=21 3x8=24 3x9=27 3x10=30 3x11=33 3x12=36 there multiples of 3 are uncountable but these are till 12.
To find the HCF of 3 numbers you first need to write them as the product of their prime factors. In this case you get: 11 = 11 33 = 3x11 121 = 11x11 The next step is to identify any common prime factors. In this case, all three numbers have an 11 as a prime factor. Therefore, the HCF of 11, 33 and 121 is 11.
3x11 = 33
2^3x11
(3,11)
3x11 - Nikki Heat :)
3x1=3, in this case 3x11=33.
(3,11) is a factor pair of 33.
33
It is sqrt(33).
3 33 = 3x11 48 = 3x16 54 = 3x18
33ml is a measue of volume.. it imndicates you ave 33 units of one mililitre.
330= 33 x 10 ^ ^ 3x11 2x5 330= 3x11x2x5
what is the prime factorization of 88 using exponents