Answer:
![\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:
- a column for the values of x in each equation
- a column for the values of y in each equation
- a column for the independent values of each equation
since our system of equations is:

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:
![\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26%26%5C%5C4%26%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:
![\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%26%5C%5C4%26-2%26%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:
![\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%2612%5C%5C4%26-2%2615%5C%5C%5Cend%7Barray%7D%5Cright%5D)
usually there is a line separating the columns for the values of x and y, and the independent values:
![\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%26-5%20%20%26%7C12%5C%5C4%26-2%20%20%26%7C15%5C%5C%5Cend%7Barray%7D%5Cright%5D)
this is the matrix of the system of equations
Answer: 
Step-by-step explanation:
First you put it in to y= mx+b form.

Then you subtract 34 from both sides

Then you divide 8 from both sides
Leaving you with

Answer:
- 60<em><u>÷</u></em><em><u>15</u></em><em><u> </u></em><em><u>=</u></em><em><u>4</u></em><em><u> </u></em>
<em><u>Therefore</u></em><em><u> </u></em><em><u>Martin</u></em><em><u> </u></em><em><u>uses</u></em><em><u> </u></em><em><u>his</u></em><em><u> </u></em><em><u>power</u></em><em><u> </u></em><em><u>saw</u></em><em><u> </u></em><em><u>4</u></em><em><u> </u></em><em><u>times</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>one</u></em><em><u> </u></em><em><u>hour</u></em><em><u> </u></em>
It costs 5,5! Bc 1 costs 0,5
Answer:
1080 m^2 Don't submit m^2 in your answer.
Step-by-step explanation:
Givens
The catch is to find h
To do that, use a^2 + b^2 = c^2
a b and c are in the same 1/2 triangle.
a = 48/2 = 24 m
b = h = ?
c = 51 meters
Solution
a^2 + b^2 = 51^2 Substitute for b^2 = h^2
24^2 + h^2 = 51^2 Expand 24^2 and 51^2
576 + h^2 = 2601 Subtract 576 from both sides
h^2 = 2601 - 576
h^2 = 2025 Take the square root of both sides
h = 45
Area
Area = 1/2 b * h
Area = 1/2 48 * 45
Area = 1080
Remark
Notice that to find h you only use 1/2 of 48 because that is the base of the right triangle.
To find the area, you need to use all of 48 because 48 is the full length of the base.