Answer:
A two-digit number can be written as:
a*10 + b*1
Where a and b are single-digit numbers, and a ≠ 0.
We know that:
"The sum of a two-digit number and the number obtained by interchanging the digits is 132."
then:
a*10 + b*1 + (b*10 + a*1) = 132
And we also know that the digits differ by 2.
then:
a = b + 2
or
a = b - 2
So let's solve this:
We start with the equation:
a*10 + b*1 + (b*10 + a*1) = 132
(a*10 + a) + (b*10 + b) = 132
a*11 + b*11 = 132
(a + b)*11 = 132
(a + b) = 132/11 = 12
Then:
a + b = 12
And remember that:
a = b + 2
or
a = b - 2
Then if we select the first one, we get:
a + b = 12
(b + 2) + b = 12
2*b + 2 = 12
2*b = 12 -2 = 10
b = 10/2 = 5
b = 5
then a = b + 2= 5 + 2 = 7
The number is 75.
And if we selected:
a = b - 2, we would get the number 57.
Both are valid solutions because we are changing the order of the digits, so is the same:
75 + 57
than
57 + 75.
Answer:
Hope Choose Capital one who is the person I was just thinking about parallel with you and I hope you feel
The answer is 1 2/3+4 2/5=91/15
Answer:
y = -7/4x - 11
Step-by-step explanation:
formula is y = mx + b
m is the slope
b is the y intercept
y = -7/4x -11
which is the first one
Answer:
Step-by-step explanation:
Let us assume that if possible in the group of 101, each had a different number of friends. Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.
Say P has 100 friends then P has all other persons as friends. In this case, there cannot be any one who has 0 friend. So a contradiction. Hence proved
Part ii: EVen if instead of 101, say n people are there same proof follows as
if different number of friends then they would be 0,1,2...n-1
If one person has n-1 friends then there cannot be any one who does not have any friend.
Thus same proof follows.
