Factor the expression gcf 12w-18
2 answers:
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Answer:
Problem 2):
which agrees with answer C listed.
Problem 3) : D = (-3, 6] and R = [-5, 7]
which agrees with answer D listed
Step-by-step explanation:
Problem 2)
The Domain is the set of real numbers in which the function (given by a graph in this case) is defined. We see from the graph that the line is defined for all x values between 0 and 240. Such set, expressed in "set builder notation" is:
Problem 3)
notice that the function contains information on the end points to specify which end-point should be included and which one should not. The one on the left (for x = -3 is an open dot, indicating that it should not be included in the function's definition, therefor the Domain starts at values of x strictly larger than -3. So we use the "parenthesis" delimiter in the interval notation for this end-point. On the other hand, the end point on the right is a solid dot, indicating that it should be included in the function's definition, then we use the "square bracket notation for that end-point when writing the Domain set in interval notation:
Domain = (-3, 6]
For the Range (the set of all those y-values connected to points in the Domain) we use the interval notation form:
Range = [-5, 7]
since there minimum y-value observed for the function is at -5 , and the maximum is at 7, with a continuum in between.
First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).
So, we can think of our numbers as
In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.
For example, we could arrange our numbers following the pattern
or
We have possible patterns. Suppose for example that we choose the pattern
One possible way of following this pattern would be the arrangement
In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).
This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have
possible ways of following a fixed pattern. Since the number of patterns was 6, we have
possible arrangements.