Answer:
28.27 or 28 for rounded
Step-by-step explanation:
Pi r^2 pi(3)^2
Let x and y be your two consecutive whole numbers
x < sqrt(142) < y
x^2 < 142 < y
So, we are looking for x and y such that x^2 < 142 and y^2 > 142.
The closest squared number to 142 is 144 = 12^2.
Next is 11^2 = 121.
11 and 12 are consecutive.
11^2 = 121 < 142 < 144 = 12^2
Thus, 11 and 12 are your numbers
Answer:
x = 22.66
Step-by-step explanation:
Answer:
y=-3/5x+16
Step-by-step explanation:
m=(y2-y1)/(x2-x1)
m=(13-19)/(5-(-5))
m=-6/(5+5)
m=-6/10
simplify
m=-3/5
y-y1=m(x-x1)
y-19=-3/5(x-(-5))
y-19=-3/5(x+5)
y=-3/5x-15/5+19
y=-3/5x-3+19
y=-3/5x+16
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:

