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
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
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
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
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 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
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
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.
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...)
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!
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.
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.
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.
(x) + (x-1) + (x-2) + (x-3) + (x-4) = 185 5x - 10 = 185 5x = 195 x = 39
Integers that add to zero (like 3 and -3 or 5 and -5) are called additive inverses. The general formula for an additive inverse is x + (-x) = 0, where x and (-x) are additive inverses.
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The integers are: x, x+1, and x+2.x2 + (x+2) = 8x2 + x - 6 = 0(x - 2) (x + 3) = 0x = + 2x = - 3Two solutions:=> The integers are 2, 3, and 4=> The integers are -3, -2, and -1