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The sum of two consecutive odd integers is -8 What are the integers?
The integers are -5 and -3: Where the first integer is x (and the second is x+2) x+(x+2) = -8 x+x+2 = -8 2x + 2 = -8 2x = -10 x = -5 and x+2 = -3
What is the sum of the integers from 5 to 500?
The formula to sum a series of numbers is: sum = 1/2 x number_of_numbers x (first_number + last_number) So to sum the integers 5 to 500: From 5 to 500 there are 500 - 5 + 1 = 496 integers, so sum = 1/2 x 496 x (5 + 500) = 125240
Twice the smallest of three consecutive integers is five more than the largestfind the integers?
7, 8, 9 Let x be the smallest of the three integers; thus, the integers are x, x+1, x+2. From the problem, we get: 2x=(x+2)+5=x+7 2x-x=7 x=7
What are two integers whose sum is 2 and whose difference is 8?
x+y = 2 x-y = 8 +y and -y gets canceled that gives, 2x = 10 x = 5 substituting the value of x x+y = 2 5+y = 2 y =-3 hence the two integers are: x = 5 and y = -3
The product of two positive integers is 1000 If neither of the numbers is a multiple of 10 what is their sum?
Prime factorization of 1000 is2 x 2 x 2 x 5 x 5 x 5The only way to get 2 positive integers out of these factors that aren't multiples of 10 would be{2 x 2 x 2} and {5 x 5 x 5}or 8 and 125So the sum would be 133.(If we multiplied any other combination of factors, you would get a multiple of 10 due to the 5 x 2 term...)
The sum of two consecutive odd integers is -8 What are the integers?
The integers are -5 and -3: Where the first integer is x (and the second is x+2) x+(x+2) = -8 x+x+2 = -8 2x + 2 = -8 2x = -10 x = -5 and x+2 = -3
What is the sum of the integers from 5 to 500?
The formula to sum a series of numbers is: sum = 1/2 x number_of_numbers x (first_number + last_number) So to sum the integers 5 to 500: From 5 to 500 there are 500 - 5 + 1 = 496 integers, so sum = 1/2 x 496 x (5 + 500) = 125240
What integers are greater than -5?
Integers greater than -5 include -4, -3, -2, -1, 0, 1, 2, 3, 4, and so on. In set notation, this can be represented as {x | x > -5 and x is an integer}. These integers form an infinite set that extends to positive infinity.
Twice the smallest of three consecutive integers is five more than the largestfind the integers?
7, 8, 9 Let x be the smallest of the three integers; thus, the integers are x, x+1, x+2. From the problem, we get: 2x=(x+2)+5=x+7 2x-x=7 x=7
How do you multiply integers with different signs?
To multiply two integers with different signs, multiply them together as though both were positive and then make the result negative.Examples:-2 x 5 → 2 x 5 = 10, made negative → -10⇒ -2 x 5 = -103 x -4 → 3 x 4 = 12, made negative → -12⇒ 3 x -4 = -12To multiply two integers of the same sign together, ignore the sings and just multiply them as positive numbers.-2 x -5 → 2 x 5 = 10 ⇒ -2 x -5 = 103 x 4 = 12These rules apply to all numbers, not just integers.
What are two integers whose sum is 2 and whose difference is 8?
x+y = 2 x-y = 8 +y and -y gets canceled that gives, 2x = 10 x = 5 substituting the value of x x+y = 2 5+y = 2 y =-3 hence the two integers are: x = 5 and y = -3
The product of two positive integers is 1000 If neither of the numbers is a multiple of 10 what is their sum?
Prime factorization of 1000 is2 x 2 x 2 x 5 x 5 x 5The only way to get 2 positive integers out of these factors that aren't multiples of 10 would be{2 x 2 x 2} and {5 x 5 x 5}or 8 and 125So the sum would be 133.(If we multiplied any other combination of factors, you would get a multiple of 10 due to the 5 x 2 term...)
The sum of three consecutive odd integers is the same as five times the greatest of these integers find the smallest and the largest of these integers?
x + (x + 2) + (x + 4) = 5 (x + 4) so 3x + 6 = 5x + 20 ie -2x = 14 so x = -7 and integers are -7, -5 and -3 Sneaky!
How do you find three consecutive integers that have a sum of -12?
X + (X + 1) + (X + 2) = - 12 3X + 3 = - 12 3X = - 15 X = - 5 ------------ - 5 + ( - 5 + 1) + (- 5 + 2) = - 12 - 5 + ( - 4) + ( - 3) = -12 - 12 = -12 ----------------so, ( - 3, - 4, - 5 ) are the three consecutive integers
The larger of 2 positive integers is 5 more than twice the smaller integer The product of the integers is 52 find the integers?
y=2x+5 xy=52 x(2x+5)=52 2x2+5x-52=0 (2x+13)(x-4)=0 x=-6.5 or x=4 Since you specified positive integers, x=4. y=2*4+5=13 So, the two numbers are 13 and 4.
What are the two consecutive odd integers such that 3 times the first integer is 5 more than 2 times the second?
Let the two consecutive odd integers be x and x + 2. Given condition can be written in an algebraic form as follows: 3 x = 2 (x +2) + 5 or 3x = 2x + 4 + 5 or x = 9 Thus the integers are 9 and 11!! The question could also be asked by altering the condition as "3 times the second integer is 5 more than 2 times the first". In this case the equation would be as follows. 3 (x+2) = 2 x + 5; which can be readily solved to get -1 and 1 as the required integers.
Sum of 5 consecutive numbers equals 200 what is the smallest integer?
Let your middle integer be x. Then your 5 consecutive integers will be:x-2, x-1, x, x+1, and x+2Sum of all 5 integers:(x-2) + (x-1) + (x) + (x+1) + (x+2)= x -2 + x - 1 + x + x + 1 + x + 2= 5xSince the sum equals 200, we have:5x = 200x = 40Your smallest integer in terms of x is x - 2, which is:40 - 2 = 38.
