Question

using prime factorisation find if hcf (t , a) =2 and lcm is (6,a)=60
please answer this
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Answer 1

Step-by-step explanation:

hlo mate here's your answer

6= 2× 3=

6= 2× 3= 2 1

6= 2× 3= 2 13 1

6= 2× 3= 2 13 120= 2× 2× 5=

6= 2× 3= 2 13 120= 2× 2× 5= 2 2

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 1

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 12

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 12LCM=Product of greates power of each prime factor

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 12LCM=Product of greates power of each prime factor=

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 12LCM=Product of greates power of each prime factor=2 2

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 12LCM=Product of greates power of each prime factor=2 23 1

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 12LCM=Product of greates power of each prime factor=2 23 15 1

6= 2× 3= 2 13 120= 2× 2× 5= 2 25 1HCF=Product of smallest power of each common prime factor=2 12LCM=Product of greates power of each prime factor=2 23 15 160

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