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.
3 times as much as 40
40 x 3
40 + 40 + 40
40
40
40
----
120
120 is the answer
Sum of 2 angles = 80°
Difference = 20°
If we take away 20° from the sum, both angles are equal
⇒80 - 20 = 60°
Divide by 2 to find the smaller angle:
60 ÷ 2 = 30
Add 20° to find the bigger angle:
20 + 30 = 50°
<u>One of the angle is 30° and the other angle is 50°</u>
Given that the sum of the two angles is 80°
⇒ Third angle = 180 - 80 = 100°
Answer: The 3 angles are 30° , 50° and 100°
Answer:
a tenth of it i think is what you're asking.
Step-by-step explanation:
