The grade will round up to around 70 percent the low ones will hold it down
Answer:
$20 off 10
Step-by-step explanation
it really depends on how much there is in your order, but if its more than 10 i would use $20 off 10 because its less stuff so the prices would go lower.. if that makes any sense
Answer:
Notebooks cost $2.75 and pens cost $1.10.
He can also buy 3 notebooks.
Step-by-step explanation:
In order to find this, we need to create two equations given each of the situations.
3n + 2p = 10.45
4n + 6p = 17.60
Now to solve for n, multiply the top equation by -3 and add together.
-9n - 6p = -31.35
4n + 6p = 17.60
----------------------
-5n = -13.75
n = 2.75
Now that we have the value of notebooks, we can find the amount for pens using either equation.
3n + 2p = 10.45
3(2.75) + 2p = 10.45
8.25 + 2p = 10.45
2p = 2.20
p = 1.10
Finally, to find the number of notebooks that he can purchase, find the cost of a notebook with 3 pens.
n + 3p
2.75 + 3(1.10)
2.75 + 3.30
6.05
Now divide 22 by that number
22/6.05 = 3.63
Since we can't have fractional notebooks, we round down to 3.
Answer:
![B)\,\,A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]](https://tex.z-dn.net/?f=B%29%5C%2C%5C%2CA%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20)
Step-by-step explanation:
Expanding with first row
![det(A) = \left\Bigg|\begin{array}{ccc}1&1&1\\3&4&3\\3&3&4\end{array}\right\Bigg|\\\\\\det(A)= (1)\left\Big|\begin{array}{cc}4&3\\3&4\end{array}\right\Big|-(1)\left\Big|\begin{array}{cc}3&3\\3&4\end{array}\right\Big|+(1)\left\Big|\begin{array}{cc}3&4\\3&3\end{array}\right\Big|\\\\det(A)=1[16-9]-1[12-9]+1[9-12]\\\\det(A)=7-3-3\\\\det(A)=1](https://tex.z-dn.net/?f=det%28A%29%20%3D%20%5Cleft%5CBigg%7C%5Cbegin%7Barray%7D%7Bccc%7D1%261%261%5C%5C3%264%263%5C%5C3%263%264%5Cend%7Barray%7D%5Cright%5CBigg%7C%5C%5C%5C%5C%5C%5Cdet%28A%29%3D%20%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D4%263%5C%5C3%264%5Cend%7Barray%7D%5Cright%5CBig%7C-%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D3%263%5C%5C3%264%5Cend%7Barray%7D%5Cright%5CBig%7C%2B%281%29%5Cleft%5CBig%7C%5Cbegin%7Barray%7D%7Bcc%7D3%264%5C%5C3%263%5Cend%7Barray%7D%5Cright%5CBig%7C%5C%5C%5C%5Cdet%28A%29%3D1%5B16-9%5D-1%5B12-9%5D%2B1%5B9-12%5D%5C%5C%5C%5Cdet%28A%29%3D7-3-3%5C%5C%5C%5Cdet%28A%29%3D1)
To find inverse we first find cofactor matrix
Cofactor matrix is
![C=\left[\begin{array}{ccc}7&-3&3\\-1&1&0\\-1&0&1\end{array}\right] \\\\Adj(A)=C^{T}\\\\Adj(A)=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] \\\\\\A^{-1}=\frac{adj(A)}{det(A)}\\\\A^{-1}=\frac{\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right] }{1}\\\\A^{-1}=\left[\begin{array}{ccc}7&-1&-1\\-3&1&0\\-3&0&1\end{array}\right]](https://tex.z-dn.net/?f=C%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-3%263%5C%5C-1%261%260%5C%5C-1%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5CAdj%28A%29%3DC%5E%7BT%7D%5C%5C%5C%5CAdj%28A%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cfrac%7Badj%28A%29%7D%7Bdet%28A%29%7D%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cfrac%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D%20%7D%7B1%7D%5C%5C%5C%5CA%5E%7B-1%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%26-1%26-1%5C%5C-3%261%260%5C%5C-3%260%261%5Cend%7Barray%7D%5Cright%5D)
