2.5
Sum of fifteen minus b
2 Numbers are a and b. We have( a>b) a+b=40(1) and a-b=10(2) a=10+b( from 2. Subtracts both side by b) => 10+b+b=40 (from 1. Substitute a with 10+b) => 10+2b =40 => 2b =30 => b =15 =>a=10+15=25 => a=25 b=15.
-6b =15 -6b/-6 = 15/-6 b = -2.5
Let the 2 digit number be 10a +b. Then: a + b = 15 10a + b - 27 = 10b + a => 9a -9b = 27 => a - b = 3 adding the first and last equations gives: 2a = 18 => a = 9 and substituting in the first gives: 9 + b = 15 => b = 6 meaning the original number is 96.
2.5
2.5
2.5
2.5
If the sum of the digits in a two-digit number is 15, and the number is 6 more than 15 times the units digit, the number is 96. Let A and B be the digits. (A is the tens digit and B is the units digit) A + B = 15, therefore A = 15 - B 10A + B = 15B + 6Substituting for A, and solving for B...10(15 - B) + B = 15B + 6150 - 10B + B = 15B + 6150 - 9B = 15B + 6144 - 9B = 15B144 = 24BB = 6 Back substitute B into first equation and solve for A...A + 6 = 15A = 9 Therefore, the digits are 9 and 6, so the number is 96.
(a+b)/6
140625
(a + b)/(xy)
Mean = sum of all numbers divided by number of numbers you summed. Call numbers a, b, c, d, e, f (a+b+c+d+e+f)/6 = mean
Sum of fifteen minus b
if you let b= amount of money then b/20 = 7
2 Numbers are a and b. We have( a>b) a+b=40(1) and a-b=10(2) a=10+b( from 2. Subtracts both side by b) => 10+b+b=40 (from 1. Substitute a with 10+b) => 10+2b =40 => 2b =30 => b =15 =>a=10+15=25 => a=25 b=15.