3+4+1 for each class so it's 8
8 x 6 for all the classrooms so it's 48. The school buys 48 games
Answer: n=8
Step-by-step explanation:
-8(n-6)=-16
-8n+48=-16
-8n=-64
n=8
Answer:
Step-by-step explanation:
To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:
![A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%262%262%5C%5C-7%26-3%265%26-8%5C%5C4%261%261%261%5C%5C3%267%26-1%261%5Cend%7Barray%7D%5Cright%5D)
And the vector B is formed with the solution of each equation of the system:![b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]](https://tex.z-dn.net/?f=b%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D3%5C%5C-3%5C%5C6%5C%5C1%5Cend%7Barray%7D%5Cright%5D)
To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called 
.
![A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]](https://tex.z-dn.net/?f=A_%7B2%7D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%262%262%5C%5C-7%26-3%265%26-8%5C%5C4%266%261%261%5C%5C3%261%26-1%261%5Cend%7Barray%7D%5Cright%5D)
The value of y using Cramer's rule is:
Find the value of the determinant of each matrix, and divide:
9514 1404 393
Answer:
- Keiko: 17
- Chang: 24
- Abdul: 68
Step-by-step explanation:
Let c represent the number of messages sent by Chang. Then Keiko sent (c-7) messages, and Abdul sent 4(c-7) messages. The total number sent was ...
c + (c -7) +4(c -7) = 109
6c -35 = 109 . . . . . . . . . . simplify
6c = 144 . . . . . . . . . . . add 35
c = 24 . . . . . . . . . . . divide by 6
c-7 = 17
4(c-7) = 68
Keiko sent 17, Chang sent 24, and Abdul sent 68 text messages.
Answer:
$90
Step-by-step explanation:
First, converting R percent to r a decimal
r = R/100 = 9%/100 = 0.09 per year,
then, solving our equation
I = 2000 × 0.09 × 0.5 = 90
I = $ 90.00
The simple interest accumulated
on a principal of $ 2,000.00
at a rate of 9% per year
for 0.5 years is $ 90.00.
