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Three consecutive odd integers that have a sum of 123?
123/3 = 41 so integers are 39, 41 and 43
Ind three consecutive odd integers that have a sum of 123?
x is lowest of the integers so x + (x + 2) + (x + 4) = 123 ie 3x + 6 = 123 so 3x = 117 and x = 39, so integers are 39, 41 and 43
What is the sum of three consecutive integers is 126 so what are the integers?
:/ We can write it this way: x - 1 + x + x + 1 = 126. 1s cancel out, so we get 3x = 126. 126/3 = 42. Answer: 41, 42, 43.
What are three consecutive odd integers that have a sum of 123?
39, 41, 43 Let x represent the smallest of these numbers. From the problem, we know x+(x+2)+(x+4)=123 Solving for x: 3x+6=123 3x=117 x=117/3=39 So our integers are: 39, 41, 43
How would you reconise a multiple of three?
if the number is a 2 or more diget number, you add up all of the numbers in the number and when you have the sum, if the sum is divisible by 3 , the whole number is divisible by three EX. 123=6, 6 is divisible by 3 so 123 is divisible by three 447=15 is divisible by 3 so 447 is divisible by 3 989=26 is *NOT* divisible by 3 so 989 is not divisible by three this truly does work for every single number
Three consecutive odd integers that have a sum of 123?
123/3 = 41 so integers are 39, 41 and 43
What two numbers have the sum of 123 but the difference of 7?
Suppose the smaller number is x, then the bigger number is x+7 Their sum is 123 ie x + (x+7) = 123 or 2x + 7 = 123 or 2x = 123 - 7 = 116 so that x = 58 and then x+7 = 65 The two numbers are 58 and 65
What is the average of 44 and 123?
E = Sum(i1, i2, .. in) / N So 44 + 123 = 167, N = 2 which make E = 83.5 or "average" of the values.
Ind three consecutive odd integers that have a sum of 123?
x is lowest of the integers so x + (x + 2) + (x + 4) = 123 ie 3x + 6 = 123 so 3x = 117 and x = 39, so integers are 39, 41 and 43
AS 675 is related to 929 in the same way 123 is related to what?
675:929::123:___ SOLUTION : Sum Of Digits in 675 is 6+7+5 = 18 Sum Of Digits in 929 is 9+2+9 = 20 Therefore, difference between them is 2 Sum Of Digits in 123 is 1+2+3 = 6 and sum of digits in 242 is 2+4+2 is 8 Therefore, difference is same So, the answer is 242
What are 5 odd number whose sum is from1 to 19?
There are many answers: if numbers can be repeated-1)all '1s' and all '3s' 2)4'1s' and so on....
How many number 1s has dizzee rascal had?
5 number 1s so far
How do you make hacker in doodle Devil?
You combine life & internet so in a sum formation (Internet + life = cyberspace)
What is the sum of three consecutive integers is 126 so what are the integers?
:/ We can write it this way: x - 1 + x + x + 1 = 126. 1s cancel out, so we get 3x = 126. 126/3 = 42. Answer: 41, 42, 43.
The sum of the first 5 terms of an arithmetic series is 85. The sum of the first 6 terms is 123. Determine the first four terms of the series.?
Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24
One number is 8 more than twice the second number The sum of the numbers is 123 Find the numbers?
Let a be the first number and b the second number then, (1) a = 8 + 2b and (2) a + b =123 From (2) a = 123 - b substituting for a in (1) gives :- 123 - b = 8 + 2b , 3b = 115, b = 115/3 = 38⅓. and substituting for this in (2) gives :- a + 38⅓ = 123, a = 123 - 38⅓ = 84⅔. So, a = 84⅔ and b = 38⅓
What are three consecutive odd integers that have a sum of 123?
39, 41, 43 Let x represent the smallest of these numbers. From the problem, we know x+(x+2)+(x+4)=123 Solving for x: 3x+6=123 3x=117 x=117/3=39 So our integers are: 39, 41, 43
