Anything to the power of 0 is always 1 right??
2 answers:
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Answer:
Step-by-step explanation:
(27)^2/3=∛27²=∛3³*3³=3*3=9
√36³=√36*36*36=√6²*6²*6²=6*6*6=216
(-243)^3/5=5√2-42³=-27
Answer:
There are a total of functions.
Step-by-step explanation:
In order to define an injective monotone function from [3] to [6] we need to select 3 different values fromm {1,2,3,4,5,6} and assign the smallest one of them to 1, asign the intermediate value to 2 and the largest value to 3. That way the function is monotone and it satisfies what the problem asks.
The total way of selecting injective monotone functions is, therefore, the total amount of ways to pick 3 elements from a set of 6. That number is the combinatorial number of 6 with 3, in other words