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"
Answer:
Number 5. = 69
Number 8. = 19
Step-by-step explanation:
Answer:
9000 pounds 100%
Step-by-step explanation:
dave is fat
The probability of it happening once is 1/5, since 10/50 simplifies to 1/5. Because the ticket goes back in the bag, the probability stays constant at 1/5.
For it to happen twice, you multiply the probability of it happening once, twice.
1/5 • 1/5 = 1/25 probability of it being Gary's name twice in a row.
Answer: SAS
Step-by-step explanation:
There is an angle B or D and the sides being the 26, 21, and 32! (Sorry if I get it wrong)
