Two equations with infinite solutions would look the exact same. Example:
y=mx+b
y=mx+b
Example 2
y=2x+5
y=2x+5
For an equation with no solution they would have the same slope but different y intercepts. An equation with same slope and same y intercepts would have infinite solutions.
Answer:
Given:
I usually walk from home to work. This morning, I walked for 10 minutes until I was halfway to work.
I then realized that I would be late if I kept walking.
I ran the rest of the way. I run twice as fast as I walk.
Find:
The number of minutes in total did it take me to get from home to work
Step-by-step explanation:
Had I kept walking, the second half of my trip would have taken 10 more minutes.
By doubling my speed for the second half of my trip,
I halved the amount of time it took me to finish.
So, the second half of my trip took 5 minutes, for a total trip time of 10+5 = 15 minutes.
The number of minutes in total did it take me to get from home to work is 15 minutes.
Answer:
14 square roots of 2 is correct
Step-by-step explanation:
The correct option is (B) yes because all the elements of set R are in set A.
<h3>
What is an element?</h3>
- In mathematics, an element (or member) of a set is any of the distinct things that belong to that set.
Given sets:
- U = {x | x is a real number}
- A = {x | x is an odd integer}
- R = {x | x = 3, 7, 11, 27}
So,
- A = 1, 3, 5, 7, 9, 11... are the elements of set A.
- R ⊂ A can be understood as R being a subset of A, i.e. all of R's elements can be found in A.
- Because all of the elements of R are odd integers and can be found in A, R ⊂ A is TRUE.
Therefore, the correct option is (B) yes because all the elements of set R are in set A.
Know more about sets here:
brainly.com/question/2166579
#SPJ4
The complete question is given below:
Consider the sets below. U = {x | x is a real number} A = {x | x is an odd integer} R = {x | x = 3, 7, 11, 27} Is R ⊂ A?
(A) yes, because all the elements of set A are in set R
(B) yes, because all the elements of set R are in set A
(C) no because each element in set A is not represented in set R
(D) no, because each element in set R is not represented in set A
