<h3><u>Question</u><u>:</u></h3>
<u>The difference between a 2-digit number and the number formed by reversing its digits is 45. If the sum of the digits of the original number is 13, then find the number. </u>
<h3><u>Statement</u><u>:</u></h3>
<u>The difference between a 2-digit number and the number formed by reversing its digits is 45. </u><u>T</u><u>he sum of the digits of the original number is 13</u><u>.</u>
<h3><u>Solution:</u></h3>
- Let one of the digit of the original number be x.
- So, the other digit = (13-x)
- Therefore, the two digit number = 10(13-x) + x = 130-10x+x = 130-9x
- The number obtained after interchanging the digits is 10x+(13-x) =9x+13
- Therefore, by the problem
130-9x-(9x+13) = 45
or, 130-9x- 9x-13 = 45
or, -18x = 45-130+13
or, -18x= -72
or, x = 72/18 = 4
or, x = 4
- So, the original number = 130-9x = 130 -9(4) = 130 - 36 = 94
<h3>Answer:</h3>
The number is 94.
I think the answer you have given isn't right. The answer should be 94.
A^2 + b^2 = c^2....this is the pythagoream theroum (sorry, probably spelled it wrong)....it is used for right triangles.
a and b are ur legs of the triangle and c is the hypotenuse. Therefore, the value of a^2 is the value of one of the legs, squared.
OR. if u mean this..
a^2 + b^2 = c^2
a^2 = c^2 - b^2
Answer:
1. 4y-12
2. c
For #1 you distribute the 4 to the y and the 3. 4 times 3 equals 12.
For #2 you factor out the 3 and you get 3(5x+1)
You just need to factor a to get
