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Q: The sum of 15 divided by b and 6?

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2.5

2.5

2.5

2.5

(a+b)/6

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)/(xy)

140625

Sum of fifteen minus b

if you let b= amount of money then b/20 = 7

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

Depends on the digit that "A" represents. E.G. 15 = B and 3 = A 15 divided by 3 = 5 Can use any numbers that work for the equation.

You add 2 fractions with the same denominator [c], so the sum is the sum of the numerators divided by the denominator: a/c + b/c = (a+b)/c

66.6667

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.

B/5 = 6 means that 6 x 5 = B 6 x 5 = 30 Therefore, B = 30

5

-6b =15 -6b/-6 = 15/-6 b = -2.5

(a+b)/2 - 10 a+b divided by 2 and then minus 10

15

B is -42

15 and 3--a + b = 18a = 18 - ba * b = 45(1-b)*b = 4518b-b^2 = 45b^2-18b = -45b^2-18b+81 = 36(b-9)^2 = 6^2b-9 = +/-6b = 9+/-6 = 3, 15a = (18 -b ) = 18 - 9 +/- 6a = 15, 3

a = 6b = 9a + b = 6 + 9 = 15

Let A = rolling a double Let B = sum is 11 P(A)=6/36=1/6 P(B)=2/36=1/18 since (5,6) and (6,5) produce a sum of 11. We want to find P(A/B)= P(A & B) / P(B) = 0 / P(B)=0 P(A & B) represent the event getting a double and the sum being 11.

No. is it not a, b,c, or d.

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