Answer:What is the greatest number of identical groups that can be made with a number of mums and carnations in each group?
Step-by-step explanation:
EZ
Here , we are provided with a table which shows 5 consecutive terms of an arithmetic sequence . But before solving further , let's recall that ;
The n'th term of a Arithmetic Sequence let's say it be
is given by ;
Where , <u>d</u> is the common difference
Now , here we are given with ;
We have to find the 2nd , 3rd and 4th term respectively ,
Now , by using the above formula , 5th term can be written as ;
Putting the values and transposing 1st term to RHS , we have ;
Now , as we got the common difference , so we can find out the missing terms now ;



Now



Also ,



Now , The given table can be written as ;

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Answer:
![\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}9&9\\-3&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%269%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%260%5C%5C-8%264%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D9%269%5C%5C-3%262%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
To add matrices, we add the corresponding components.
The given matrices is
![\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%269%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%260%5C%5C-8%264%5Cend%7Barray%7D%5Cright%5D)
We add the corresponding components to get;
![\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}3+6&9+0\\5+-8&-2+4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%269%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%260%5C%5C-8%264%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%2B6%269%2B0%5C%5C5%2B-8%26-2%2B4%5Cend%7Barray%7D%5Cright%5D)
We simplify to get:
![\left[\begin{array}{cc}3&9\\5&-2\end{array}\right] +\left[\begin{array}{cc}6&0\\-8&4\end{array}\right]=\left[\begin{array}{cc}9&9\\-3&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%269%5C%5C5%26-2%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D6%260%5C%5C-8%264%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D9%269%5C%5C-3%262%5Cend%7Barray%7D%5Cright%5D)
Answer:
a) 1/2
b) 250
Step-by-step explanation:
The start of the question doesn't matter entirely, although is interesting to read. What we are trying to do is find the value for
such that
is maximized. Once we have that
, we can easily find the answer to part b.
Finding the value that maximizes
is the same as finding the value that maximizes
, just on a smaller scale. So, we really want to maximize
. To do this, we will do a trick called completing the square.
.
Because there is a negative sign in front of the big squared term, combined with the fact that a square is always positive, means we need to find the value of
such that the inner part of the square term is equal to
.
.
So, the answer to part a is
.
We can then plug
into the equation for p to find the answer to part b.
.
So, the answer to part b is
.
And we're done!