Answer:
Let us say the domain in the first case, has the numbers. And the co-domain has the students, .
Now for a relation to be a function, the input should have exactly one output, which is true in this case because each number is associated (picked up by) with only one student.
The second condition is that no element in the domain should be left without an output. This is taken care by the equal number of students and the cards. 25 cards and 25 students. And they pick exactly one card. So all the cards get picked.
Note that this function is one-one and onto in the sense that each input has different outputs and no element in the co domain is left without an image in the domain. Since this is an one-one onto function inverse should exist. If the inverse exists, then the domain and co domain can be interchanged. i.e., Students become the domain and the cards co-domain, exactly like Mario claimed. So, both are correct!
Answer:
-3
Step-by-step explanation:
If we directly evaluate the function at -1, we get 0/0, meaning we may still have a limit to find.
In this case, factoring the polynomial at the top would be helpful.
The polynomial can be factored to (x+1)(x-2), so the function would now turn out to be (x+1)(x-2)/(x+1)
The (x+1) cancel out, leaving you with (x-2), which you can directly evaluate by plugging in x as -1:
-1-2 = -3
Quick disclaimer: the function is still undefined at -1; it's just that the function gets closer and closer to -3 as you approach -1.
I hope this helped you.
In 12 days, Wendy will do the both again.
Step-by-step explanation:
Given,
Wendy goes to gym every 4 days
Wendy swims every 6 days
We will find the least common multiple of both the numbers in order to determine the number of days in which she will do both.
4 = 4,8,12,16,20,....
6 = 6,12,18,24,.....
Therefore;
The least common multiple of 4 and 6 is 12.
Thus,
In 12 days, Wendy will do the both again.
Keywords: multiples, LCM
Learn more about multiples at:
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