Answer:
The reduced row-echelon form of the linear system is ![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%260%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
We will solve the original system of linear equations by performing a sequence of the following elementary row operations on the augmented matrix:
- Interchange two rows
- Multiply one row by a nonzero number
- Add a multiple of one row to a different row
To find the reduced row-echelon form of this augmented matrix
![\left[\begin{array}{cccc}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D2%263%26-1%2614%5C%5C1%262%261%264%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
You need to follow these steps:
- Divide row 1 by 2
![\left[\begin{array}{cccc}1&3/2&-1/2&7\\1&2&1&4\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C1%262%261%264%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 from row 2
![\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\5&9&2&7\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C0%261%2F2%263%2F2%26-3%5C%5C5%269%262%267%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 multiplied by 5 from row 3
![\left[\begin{array}{cccc}1&3/2&-1/2&7\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%263%2F2%26-1%2F2%267%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%263%2F9%269%2F2%26-28%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 3 from row 1
![\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&3/9&9/2&-28\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%263%2F9%269%2F2%26-28%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 3 from row 3
![\left[\begin{array}{cccc}1&0&-5&16\\0&1/2&3/2&-3\\0&0&0&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%261%2F2%263%2F2%26-3%5C%5C0%260%260%26-19%5Cend%7Barray%7D%5Cright%5D)
- Multiply row 2 by 2
![\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&-19\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%262%263%26-6%5C%5C0%260%260%26-19%5Cend%7Barray%7D%5Cright%5D)
- Divide row 3 by −19
![\left[\begin{array}{cccc}1&0&-5&16\\0&2&3&-6\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%2616%5C%5C0%262%263%26-6%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 3 multiplied by 16 from row 1
![\left[\begin{array}{cccc}1&0&-5&0\\0&1&3&-6\\0&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26-5%260%5C%5C0%261%263%26-6%5C%5C0%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Add row 3 multiplied by 6 to row 2
7+x is the awnser
unless you meant to type “A number is more than 7” then it would be x>7
Answer:
Your answer is B
Step-by-step explanation:
First we have to find the Y intercept, which is -2.
Then you have to find the slope. Rise/Run
If you count up 5 and left 2 you get your next point. Meaning that 5/-2 is your slope.
You can find the value of the hypotenuse if you apply the Pythagorean Theorem, which is show below:
h²=a²+ b² ⇒ h=√(a² + b²)
h: hypotenuse (the opposite side of the right angle and the longest side of the triangle).
a and b: legs (the sides that form the right angle).
Then, you have:
h²=a² + b²
h²=12²+12²
h=√ ((12)² + (12)²)
h=12√2
What is the lenght of the hypotenuse?
The answer is: The length of the hypotenuse is 12√2
Answer
Find out the how many pounds of metal are in 1,950 lb of ore .
To proof
let us assume that the pounds of metal are in 1,950 lb of ore be x .
As given
In an ore, 9.8% of its total weight is metal.
ore weight = 1,950 lb
9.8% is written in the decimal form
= 0.098
Than the equation becomes
x = 0.098 × 1950
x = 191.1 pounds
Therefore the 191.1 pounds of metal are in 1,950 lb of ore .
Hence proved
