Create an arithmetic sequence with a common difference of -10
1 answer:
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Answer:
a. The critcal points are at
b. Then, is a maximum and
is a minimum
c. The absolute minimum is at and the absolute maximum is at
Step-by-step explanation:
(a)
Remember that you need to find the points where
Therefore you have to solve this equation.
From that equation you can factor out and you would get
And from that you would have , so
.
And you would also have .
You can factor that equation as
Therefore .
So the critcal points are at
b.
Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since
Then, is a maximum and
is a minimum
c.
The absolute minimum is at and the absolute maximum is at
Well, you could assign a letter to each piece of luggage like so...
A, B, C, D, E, F, G
What you could then do is set it against a table (a configuration table to be precise) with the same letters, and repeat the process again. If the order of these pieces of luggage also has to be taken into account, you'll end up with more configurations.
My answer and workings are below...
35 arrangements without order taken into consideration, because there are 35 ways in which to select 3 objects from the 7 objects.
210 arrangements (35 x 6) when order is taken into consideration.
*There are 6 ways to configure 3 letters.
Alternative way to solve the problem...
Produce Pascal's triangle. If you want to know how many ways in which you can choose 3 objects from 7, select (7 3) in Pascal's triangle which is equal to 35. Now, there are 6 ways in which to configure 3 objects if you are concerned about order.
A, B, C, D, E, F, G
What you could then do is set it against a table (a configuration table to be precise) with the same letters, and repeat the process again. If the order of these pieces of luggage also has to be taken into account, you'll end up with more configurations.
My answer and workings are below...
35 arrangements without order taken into consideration, because there are 35 ways in which to select 3 objects from the 7 objects.
210 arrangements (35 x 6) when order is taken into consideration.
*There are 6 ways to configure 3 letters.
Alternative way to solve the problem...
Produce Pascal's triangle. If you want to know how many ways in which you can choose 3 objects from 7, select (7 3) in Pascal's triangle which is equal to 35. Now, there are 6 ways in which to configure 3 objects if you are concerned about order.