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:
X=7 y=-5
Step-by-step explanation:
2x+y=9
3x-y=16
use the process of elimination finding out what works for one problem and try it on the other until they both work
Answers:
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Explanation:
BC = 78 is parallel to the midsegment DE.
This parallel midsegment is half as long as BC, so DE = BC/2 = 78/2 = 39
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DF = 34 doubles to AB = 2*DF = 2*34 = 68, since the midsegment is half as the parallel side, so we just think in reverse of that process.
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BE = 34 for two reasons
- Quadrilateral EBFD is a parallelogram with opposite sides BE and DF that are congruent.
- E is the midpoint of AB, so BE is half as long as AB
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We don't use the info that AC = 48.
They are counting by twos.
2,4,6,8,10,12,14,16,18,20,etc
I have to do a lot of stuff to do this in the time of time I have to do a lot of work to do this really
