Answer:
Step-by-step explanation:
Juan has 20 books to sell. He sells the books for $15 each.
The range of the function is the set of all possible values of the dependent variable. The dependent variable here is the amount of money that is made and this amount depends on
the number of books, x sold
The amount of money Juan makes from selling books is represented by a fucntion. f(x)=15x
The maximum amount that can be made from 20 books at a rate of $15 each would be 20×15 = $300
The minimum amount that fan be made is $0 and this is when no book is sold. Let y = f(x). So the range is
0 lesser than or equal to y lesser than or equal to 300
Answer:
Let
be the time for the
th task.
We know these times have a certain structure:
- Any 3 adjacent tasks will take half as long as the next two tasks.
In the form of an equations we have

- The second task takes 1 second

- The fourth task takes 10 seconds

So, we have the following system of equations:

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.
Here is the augmented matrix for this system.
![\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcccccc%7Cc%7D%201%20%26%201%20%26%201%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%200%20%26%200%20%5C%5C%5C%5C%200%20%26%201%20%26%201%20%26%201%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%200%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5Cend%7Barray%7D%20%5Cright%5D)
b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.
- Subtract row 2 from row 1
. - Subtract row 2 from row 3
. - Add row 3 to row 2
. - Multiply row 3 by −1
. - Add row 4 multiplied by
to row 1
. - Subtract row 4 from row 3
.
Here is the reduced echelon form for the augmented matrix.
![\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccccccc%7D%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20%26%2015%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%201%20%26%200%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-11%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5Cend%7Barray%7D%20%5Cright%5D)
c) The additional rows are

and the augmented matrix is
![\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccccccc%7D%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%20%5Cfrac%7B1%7D%7B2%7D%20%26%2015%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%201%20%26%200%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-%20%5Cfrac%7B1%7D%7B2%7D%20%26%20-11%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%26%2020%20%5C%5C%5C%5C%201%20%26%201%20%26%201%20%26%200%20%26%200%20%26%200%20%26%2050%20%5Cend%7Barray%7D%20%5Cright%5D)
d) To solve the system you must use these row operations.
- Subtract row 1 from row 6
. - Subtract row 2 from row 6
. - Subtract row 3 from row 6
. - Swap rows 5 and 6.
- Add row 5 to row 3
. - Multiply row 5 by 2
. - Subtract row 6 multiplied by 1/2 from row 1
. - Add row 6 multiplied by 1/2 to row 3
.
![\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccccccc%7D%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%200%20%26%205%20%5C%5C%5C%5C%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%5C%5C%5C%5C%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%200%20%26%2044%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%200%20%26%2010%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%200%20%26%201%20%26%200%20%26%2090%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%26%200%20%26%200%20%26%201%20%26%2020%20%5Cend%7Barray%7D%20%5Cright%5D)
The solutions are:
.
Zero is greater than 2, so second one from the left.