Inequality form: u>6
Interval Form: (6,0)
Let x and y be the two integers.
The sum of the integers is x+y while the difference is x-y assuming x is larger than y.
If x+y > x-y, then
x+y > x-y
x+y-x > x-y-x
y > -y
y+y > -y+y
2y > 0
2y/2 > 0/2
y > 0
So as long as y is positive, this makes the sum greater than the difference
For example, if x = 10 and y = 2, then
x+y = 10+2 = 12
x-y = 10-2 = 8
clearly 12 > 8 is true
If y is some negative number (say y = -4), then
x+y = 10+(-4) = 10-4 = 6
x-y = 10-(-4) = 10+4 = 16
and things flip around
Saying a blanket statement "the sum of two integers is always greater than their difference" is false overall. If you require y to be positive, then it works but as that last example shows, it doesn't always work.
So to summarize things up, I'd say the answer is "no, the statement isn't true overall"
Step-by-step explanation:
there it is
and if u need more help tell me
Answer:
((2a+b)/2 , a/2)
Step-by-step explanation:
The midpoint is the mean/average of both the x and y values.
Our x values are a+b and a
Our y values are a-b and b
The midpoint x value is (a+b + a ) /2 = (2a + b)/2
The midpoint y value is (a-b + b) /2 = (a)/2
The midpoint combines these two values to get ((2a+b)/2 , a/2) as our midpoint
