Solution:
we have been asked to find
The smallest value of n such that the LCM of n and 15 is 45.
The prime factors of 15 are
So in LCM of n and 15 , we must include both the prime factors and the LCM is 45.
Since 3 is one time in the prime factor of 15. But to make 45 we need one more 3 and its possible if we take the value of n as below"
n=9
Hence the LCM will be
Hence the smallest possible value of n is 9.
Answer:
Subtract 11 from both sides
Step-by-step explanation:
The denominator may not equal 0, so you need to find where -2a+6 = 0.
This happens at a = 3, so 3 is the nonpermissible replacement.
-45 to 49 is 80 degrees different
Answer:
x = 15
Step-by-step explanation:
x multiplied by 0.4 = 6
so you divide 0.4 by each side
x = 6/0.4
and plug that into the calculator.