Answer:
1
Step-by-step explanation:
There is one whole 3 inside of 4 1/2 or 4.5.
Multiples of 3 are 3x1=3 3x2=6....
Only a single whole 3 can be inside of 4.5.
A. Linear y= -x - 1 would be the answer
Answer:
A the sybol is facing the wrong way so it means greater than or equal not less than or equal to
B the gross value is accuratly represented and the symbol is facing the correct way
I think
Answer:
![\displaystyle \large{\sum_{n=1}^{42}(6n-1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7B%5Csum_%7Bn%3D1%7D%5E%7B42%7D%286n-1%29)
Step-by-step explanation:
Given:
To find:
- Summation notation of the given series
Summation Notation:
![\displaystyle \large{\sum_{k=1}^n a_k}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7B%5Csum_%7Bk%3D1%7D%5En%20a_k%7D)
Where n is the number of terms and
is general term.
First, determine what kind of series it is, there are two main series that everyone should know:
A series that has common difference.
A series that has common ratio.
If you notice and keep subtracting the next term with previous term:
Two common difference, we can in fact say that the series is arithmetic one. Since we know the type of series, we have to find the number of terms.
Now that brings us to arithmetic sequence, we know that first term is 5 and last term is 251, we’ll be finding both general term and number of term using arithmetic sequence:
<u>Arithmetic Sequence</u>
![\displaystyle \large{a_n=a_1+(n-1)d}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Ba_n%3Da_1%2B%28n-1%29d%7D)
Where
is the nth term,
is the first term and
is the common difference:
So for our general term:
![\displaystyle \large{a_n=5+(n-1)6}\\\displaystyle \large{a_n=5+6n-6}\\\displaystyle \large{a_n=6n-1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7Ba_n%3D5%2B%28n-1%296%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7Ba_n%3D5%2B6n-6%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7Ba_n%3D6n-1%7D)
And for number of terms, substitute
= 251 and solve for n:
![\displaystyle \large{251=6n-1}\\\displaystyle \large{252=6n}\\\displaystyle \large{n=42}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7B251%3D6n-1%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7B252%3D6n%7D%5C%5C%5Cdisplaystyle%20%5Clarge%7Bn%3D42%7D)
Now we can convert the series to summation notation as given the formula above, substitute as we get:
![\displaystyle \large{\sum_{n=1}^{42}(6n-1)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clarge%7B%5Csum_%7Bn%3D1%7D%5E%7B42%7D%286n-1%29)
5,040 I think because 7•6•5•4•3•2•1=5040