proofs:
<h2>S subset S Union T</h2>
We want to prove 
Let 
. By definition 
is the set that contains the elements of 
and the elements of 
. Then 
must be in . As was arbitrary, we conclude that .
<h2>T Subset S Union T</h2>
This proof is analogous to the previous one. In fact, this result is the same result as the previous one.
<h2>S Intersection T subset S </h2>
We want to prove 
Let 
. By definition of the intersection 
should be in and also in . Then, we already saw that 
. As was arbitrary we can conclude that .
<h2>S Intersection T subset T</h2>
This is the same result as the previous one. There is no need to prove it anymore, but if you wish, you can reply the exact same proof.
First find the slope m. slope formula: 
Let's use (0, -20) for x1 and y1; and (2, 10) for x2 and y2
= 
= 
Now find the y-intercept: the point where the graph crosses the y-axis. The y intercept always has an x value of 0. So -20 is the y-intercept
Use the slope formula 
. Slope=m, y-intercept = b
y = 15x - 20
y is the money in the bank account in dollars
x is the day
That’s the answers. Hope this helps
Answer:
(-3,-4)
Step-by-step explanation:
● (-1,-8) => (0,-11)
Notice that to move from -1 to 0 in the x-axis we should go one unit to the right wich means adding 1.
● -1 + 1 = 0
Notice that to move from -8 to -11 in the y-axis we should go 3 units down wich means substracting 3.
● -8-3 = -11
Let's do the same for the second point.
● (-4,-1) => ?
Add 1 to -4
● -4+1 = -3
We moved 1 unit to the right by adding 1
Substract 3 from -1
● -1-3 = -4
We moved down 3 units by substracting 3
So the image of (-4,-1) is (-3,-4)
