When no K is divided by 27, 30, or 45, the remainder is 3. Find the smallest possible value of K. Can someone send me the calcul
1 answer:
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Answer:
g(x)<j(x)<k(x)<f(x)<m(x)<h(x)
Step-by-step explanation:
1.
The denominator of f is defined for all real values of x
Therefore, the function is continuous on the set of real numbers
3.
It cannot be factorize .
Therefore, it has no real values for which it is not defined .
Hence, function h is defined for all real values.
2.
The denominator of g is defined for all real values of x.
Therefore, the function g is continuous on the set of real numbers
4.
x-9=0
x=9
The function i is not defined for x=9
Therefore, the function i is not continuous on the set of real numbers.
5.
The denominator of j is defined for all real values of x.
Therefore, the function j is continuous on the set of real numbers.
6.
It cannot be factorize .
Therefore, it has no real values for which it is not defined .
Hence, function k is defined for all real values.
7.
The function is not defined for x=8 and x=1
Hence, function l is not defined for all real values.
8.
The denominator of m is defined for all real values of x.
Therefore, the function m is continuous on the set of real numbers.
g(x)<j(x)<k(x)<f(x)<m(x)<h(x)