(a) M1, M2, M3, H1, H2 and H3.
(b) possible outcomes are M1, M3, H2.
so probability = these outcomes / all possible outcomes
= 3/6 = 1/2 Answer
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:
x
=
−1
,2
,
3
,
−
4
Step-by-step explanation:
Answer:
If you need it as a fraction then it is 1/18
Step-by-step explanation:
Answer:
Step-by-step explanation:
4. 10x + 30 = 90
10x = 60
x = 6
complementary
5. 4x + 40 + 3x = 180
7x + 40 = 180
7x = 140
x = 20
supplementary
