Write the system in augmented-matrix form:
![\left[\begin{array}{ccc|c}2&2&4&16\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D2%262%264%2616%5C%5C5%26-2%263%26-1%5C%5C1%262%26-3%26-9%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 1 by 1/2:
![\left[\begin{array}{ccc|c}1&1&2&8\\5&-2&3&-1\\1&2&-3&-9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C5%26-2%263%26-1%5C%5C1%262%26-3%26-9%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 1) to row 3, and add -5(row 1) to row 2:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&-7&-7&-41\\0&1&-5&-17\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%26-7%26-7%26-41%5C%5C0%261%26-5%26-17%5Cend%7Barray%7D%5Cright%5D)
Swap rows 2 and 3:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&-7&-7&-41\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%26-5%26-17%5C%5C0%26-7%26-7%26-41%5Cend%7Barray%7D%5Cright%5D)
Add -7(row 2) to row 3:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&-42&-160\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%26-5%26-17%5C%5C0%260%26-42%26-160%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by -1/42:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&-5&-17\\0&0&1&\frac{80}{21}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%26-5%26-17%5C%5C0%260%261%26%5Cfrac%7B80%7D%7B21%7D%5Cend%7Barray%7D%5Cright%5D)
Add 5(row 3) to row 2:
![\left[\begin{array}{ccc|c}1&1&2&8\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%261%262%268%5C%5C0%261%260%26%5Cfrac%7B43%7D%7B21%7D%5C%5C0%260%261%26%5Cfrac%7B80%7D%7B21%7D%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) and -2(row3) to row 1:
![\left[\begin{array}{ccc|c}1&0&0&-\frac53\\0&1&0&\frac{43}{21}\\0&0&1&\frac{80}{21}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%26-%5Cfrac53%5C%5C0%261%260%26%5Cfrac%7B43%7D%7B21%7D%5C%5C0%260%261%26%5Cfrac%7B80%7D%7B21%7D%5Cend%7Barray%7D%5Cright%5D)
So the solution to the system is
If there were 4 pieces eaten, and 3/7 left, then there must have been 7 pices altogether and 3 are left.
To find the median cancel out numbers on both sides, until one is left in the middle and if there are two in the middle add them up and divide by two.
So in this case the median is
53+78
131 / 2
65.5
Answer:
y= 3
Step-by-step explanation:
The triangle on the left is a right triangle, so we can use trig functions to solve for the missing side, y.
First, find the length of the side that both triangles are connected to.
tan30=x/9
Isolate the x variable and solve.
9(tan30)=x
x= 5.19615
Now use the Pythagorean Theorem to solve for the side length, y.
a^2+ b^2 = c^2
(5.19615)^2+ b^2= 6^2
c= 3
