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If the average of three distinct integers is 37 and the least integer is 18 .what could be the greatest integer?
Solving for the third integer:(a + b + c) / 3 = 37 (by the definition of "average")(18 + b + c) / 3 = 37 (assuming "a" is the least integer)18 + b + c = 111c = 111 - 18 - b (solving for the third integer)c = 93 - bNow, "b" must be at least 18, in which case:c = 93 - 18 = 75On the other hand, the largest "b" can be is when it is equal to "c" (since I am assuming that "c" is the largest integer):c = 93 - bc = 93 - c2c = 93c = 46.5Adjusting the numbers a bit (since we need integers), in this case we get the numbers 18, 46, 47Thus, the greatest integer can be anything in the range from 47 to 75.
The sum of the digits of n is 11 and 50 is less than n but n is less than 100 When n is divided by 2 the result is a prime integer What is the value of n?
Since 50 < n < 100, n is a 2-digit number. Since the sum of the digits is 11, n could be 56, 65, 74, 83, or 92, but since n is divided (evenly) by 2 and the quotient is a prime integer, then n must be 74 (74÷2 = 37).
The sum of the squares of two consecutive odd integers is 74 find the integers?
Let the first odd integer be x, so the other consecutive odd integer is x + 2. So we have:x^2 + (x + 2)^2 = 74x^2 + x^2 + 4x + 4 = 742x^2 + 4x + 4 - 74 = 74 - 742x^2 + 4x - 70 = 0 divide by 2 to both sidesx^2 + 2x - 35 = 0(x + 7)(x - 5) = 0x + 7 = 0 or x - 5 = 0x + 7 - 7 = 0 -7 or x - 5 + 5 = 0 + 5x = -7 or x = 5Thus, the first odd integer would be -7 or 5, and the second consecutive odd integer would be -5 (-7 + 2) or 7 (5 + 2).Check:
What would be the integer for 0.030?
0.030 is not an integer. So there would be no integer for it.
What is the sum of a positive even integer and the next greater positive integer?
It is an integer.
