Graph this function: f(x)=x2-4x-5
1 answer:
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<h3>Answer: 120 different ways</h3>
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Explanation:
There are...
- 6 ways to select the first place winner
- 5 ways to pick the second place winner
- 4 ways to pick the third place winner
We start with 6, and count down by 1 each time we fill up a slot. We stop once the third slot is filled or accounted for. The countdown is to ensure that we don't pick the same person twice. From here, multiply those values: 6*5*4 = 30*4 = 120
Interestingly, this is equal to 5! = 5*4*3*2*1 = 120 because the 3*2 becomes 6 and that *1 at the end doesn't affect things. Though usually results of permutation problems don't always end up like this. The order matters because a result like ABC is different from BAC, where A,B,C,D,E,F are the six school organizations.
As a slightly longer way to do the problem, you can use the nPr formula which is where n = 6 and r = 3 in this case. The exclamation marks indicate factorial. If you go this route, you should find that one of the steps will involve 6*5*4.
Question:
Write out the first three terms and the last term. Then use the formula for the sum of the first n terms of an arithmetic sequence to find the indicated sum.
30
∑ (−3i+5)
i=1
Answer:
The first three terms are : 2, -1 and -4
The last term is: -85
The sum of the sequence is: -1245
Step-by-step explanation:
Given;
==================================
30
∑ (−3i+5) -------------------(i)
i=1
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Where the ith term aₙ is given by;
=
-------------------(ii)
(a) Therefore, to get the first three terms (), we substitute i=1,2 and 3 into equation (ii) as follows;
=
= 2
Since the sum expression in equation (i) goes from i=1 to 30, then the last term of the sequence is when i = 30. This is given by;
(b) The sum of an arithmetic sequence is given by;
-----------------(iii)
Where;
n = number of terms in the sequence = 30
= first term = 2
= last term = -85
Substitute the corresponding values of n, and
into equation (iii) as follows;
= 15[-83]
= -1245