Answer:
If m is nonnegative (ie not allowed to be negative), then the answer is m^3
If m is allowed to be negative, then the answer is either |m^3| or |m|^3
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Explanation:
There are two ways to get this answer. The quickest is to simply divide the exponent 6 by 2 to get 6/2 = 3. This value of 3 is the final exponent over the base m. Why do we divide by 2? Because the square root is the same as having an exponent of 1/2 = 0.5, so
sqrt(m^6) = (m^6)^(1/2) = m^(6*1/2) = m^(6/2) = m^3
This assumes that m is nonnegative.
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A slightly longer method is to break up the square root into factors of m^2 each and then apply the rule that sqrt(x^2) = x, where x is nonnegative
sqrt(m^6) = sqrt(m^2*m^2*m^2)
sqrt(m^6) = sqrt(m^2)*sqrt(m^2)*sqrt(m^2)
sqrt(m^6) = m*m*m
sqrt(m^6) = m^3
where m is nonnegative
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If we allow m to be negative, then the final result would be either |m^3| or |m|^3.
The reason for the absolute value is to ensure that the expression m^3 is nonnegative. Keep in mind that m^6 is always nonnegative, so sqrt(m^6) is also always nonnegative. In order for sqrt(m^6) = m^3 to be true, the right side must be nonnegative.
Example: Let's say m = -2
m^6 = (-2)^6 = 64
sqrt(m^6) = sqrt(64) = 8
m^3 = (-2)^3 = -8
Without the absolute value, sqrt(m^6) = m^3 is false when m = -2
Answer:
32
Step-by-step explanation:
If 
is a number that is both divisible by 4 and 5, then
4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that 
is a solution to the system, where 
is any integer. Simply put, any multiple of 20 fits the bill.
Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have 
when 
, so the sum we want to compute is
Answer: 15/24...i'm not sure though *scratches head*
%27
Hope this helps have a good day
