<span>The input of days results in the output of total collected books. On day 1, 8 books were collected. Both inputs for days 2 and 3 have the same output of 13, meaning that no books were collected for day 3. After 5 days, 21 books were collected. The most books were collected on day 5.</span>
Answer:

The solution of the system can be given as (5,-8).
Step-by-step explanation:
Given system of equations:


To solve the given system of equations:
Solution:
We will use substitution to solve the given system.
We will substitute the
values in terms of
from the equation
into the other equation to solve for 
Substituting
into
.
We have:

<em>Using distribution.</em>

<em>Combining like terms.</em>

Adding 6 both sides.


Dividing both sides by 3.

∴ 
<em>Plugging in
into
.</em>
We have,

∴ 
The solution of the system can be given as (5,-8).
Answer: 24 possible ways
A set of 4 tires can be fixed in any of the four possible positions in a car as it is mentioned that all four tires are interchangeable. Therefore, the 4 tires can be fixed in 4! ways. The four interchangeable tires can be put on a car in 24 possible ways.
Step-by-step explanation:
looked it up
Answer:

Step-by-step explanation:
