Write in standard form a polynomial function with zeros 2, -2, and 1 has a leading coefficient of 1
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Answer:
B the range, the x- and y-intercept
Step-by-step explanation:
the domain stays the same : all values of x are possible out of the interval (-infinity, +infinity).
but the range changes, as for the original function y could only have positive values - even for negative x.
the new function has a first term (with b) that can get very small for negative x, and then a subtraction of 2 makes the result negative.
the y-intercept (x=0) of the original function is simply y=1, as b⁰=1.
the y-intercept of the new function is definitely different, because the first term 3×(b¹) is larger than 3, because b is larger than 1. and a subtraction of 2 leads to a result larger than 1, which is different to 1.
the original function has no x-intercept (y=0), as this would happen only for x = -infinity. and that is not a valid value.
the new function has an x-intercept, because the y-values (range) go from negative to positive numbers. any continuous function like this must therefore have an x-intercept (again, y = the function result = 0)
Answer:
Step-by-step explanation:
Lets divide it in cases, then sum everything
Case (1): All 5 numbers are different
In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.
The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.
Case (2): 4 numbers are different
We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4
We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.
The total cardinality of this case is
Case (3): 3 numbers are different
As we did before, we pick 3 elements from a set of n. The amount of possibilities is
Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of
Case (4): 2 numbers are different
We pick 2 numbers from a set of n, with a total of possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.
The total amount of possibilities for this case is
Case (5): All numbers are the same
This is easy, he have as many possibilities as numbers the set has. In other words, n
Conclussion
By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is
I hope that works for you!