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:
Step-by-step explanation:
Given that,
A sequence 352,345,338.
First term = 352
Common difference = 345-352 = -7
We need to find the 37th term of the sequence.
The nth term of an AP is given by :
So, the 37th term of the sequence is 100.
Answer:
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Step-by-step explanation:
We know that Step 1 is correct, because it is just a restatement of the equation. Therefore, we can eliminate Step 1:
2(5y – 2) = 12 + 6y
In Step 2, the student tried using the Distributive Property. The Distributive Property can be written as one of the two following formulas:
a(b + c) = ab + ac
a(b – c) = ab – ac
In this case, we'll use the second formula. Substitute any known values into the equation above and simplify:
2(5y – 2) = 2(5y) – 2(2)
2(5y – 2) = 10y – 4
In Step 2, the student calculated 2(5y – 2) to equal 7y – 4. However, we have just proven that 2(5y – 2) is equal to 10y – 4.
The student first made an error in Step 2, and the correct step is:
Step 2: 10y – 4 = 12 + 6y
I hope this helps!
