D3.
10-11-12
If the first of these consecutive integers is x, the second integer would be x + 1, and the third integer would be x + 2.Since the sum of the second and the third integer is 17, we can writex + 1 + x + 2 = 172x + 3 = 172x + 3 - 3 = 17 - 32x = 142x/2 = 14/2x = 7Thus, the consecutive integers are 7, 8, and 9.
1. find the product of the first two 2. multiply that product with the third number
Integers are x, x + 2 and x + 4. 2x + 4 = 5x + 10 - 51 3x = 45 so x = 15 and the other integers are 17 and 19. Check: 15 + 19 = 34, 5 x 17 = 85, 85 - 51 = 34. There you have it.
Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.
10-11-12
It is not possible to have the product of an integer. "product" is a binary operation and that means that it is an operation that combines two numbers to make the product - a third number. So you need two numbers as input, not just one.
The let statement is: let the smallest of the three integers be x.
Positive integers are greater than negative integers. For positive integers: * The integer with more digits is larger. * If two integers have the same length, compare the first digit. If the first digit is the same, compare the second digit, then the third, etc., until you find a difference. In each case, the integer with the larger digit (at the first position where you find a difference) is the larger one.
First, separate the negative and positive integers (put them into two separate groups). If there is a zero, you can put it in its own group - or put it into the same group with the positive integers. Negative integers come first, then zero, then positive integers.For positive integers:An integer with less digits comes before an integer with more digits.For integers with the same number of digits, look at the first digit. The integer with the smaller digit in this position comes first.If the first digit is the same, look at the second digit. If those are equal, look at the third digit, etc.For negative integers, it is the other way round - for example, an integer with MORE digits comes first.
No. A negative integer raised to the third power will yield a negative number that is less than the integer. Only whole numbers (positive integers greater than or equal to 1) have the property where that integer raised to the third power is greater than or equal to the integer.
4 is an integer. 1/3 is not an integer. Only whole numbers (1, 2, 3, -1, 500, 1,999, etc.) can be integers.
The second integer is 64 (128/2), so the first is 62 and the third is 66.OrLet the 1st integer be x and the 3rd integer be y.Since even integers differ by 2, then we have:y = x + 4x + y = 128x + x + 4 = 1282x = 124x = 62So the integers are 62, 64, and 66, and their sum is 192
If the first of these consecutive integers is x, the second integer would be x + 1, and the third integer would be x + 2.Since the sum of the second and the third integer is 17, we can writex + 1 + x + 2 = 172x + 3 = 172x + 3 - 3 = 17 - 32x = 142x/2 = 14/2x = 7Thus, the consecutive integers are 7, 8, and 9.
The first integer must equal 77 - 69 = 8 , since doubling it increases the sum by this amount. Similarly, the second integer must = 91 - 69 = 22. Then the third integer is 69 - 22 - 8 = 39.
1. find the product of the first two 2. multiply that product with the third number
Let the first integer be x. Since the consecutive integers differ by 1, the other consecutive integers would be (x + 1), (x + 2), and (x + 3). So we havex + x + 1 + x + 2 + x + 3 = 904x + 6 = 904x + 6 - 6 = 90 - 64x = 844x/4 = 84/4x = 21Thus, the third integer is 23 (21 + 2 = 23).