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Q: What is the formula for finding three consecutive integers such that the sum of the largest and 5 times the smallest is -244?

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For x, which is the largest integer of nconsecutive positive integers of which the smallest is m:x = m + n - 1

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There is no difference, only your outcome. The formula for both is x+2

Find two consecuitive integers whose sum is 89. To solve this problem, let x be the smaller of these integers. What is the larger of these two consecutive integers? In terms of x, write a formula that represents the sum of these two consecutive integers.

well its process is science of eratosthenes as i heard,,,,,, science of eratosthenes is the process of crossing out all multiples of 2,3,5&7 ERATOSTHENES is the one who discovered the formula for finding the sum of all integers from 1 to 100

One way to find out is write a formula. Let N and N+1 be the two integers, then N(N+1) = 182 N^2 + N - 182 = 0 This is a quadratic equation. If the factors are not obvious, (N -13)(N + 14) , then use the quadratic formula to find N. The factors tell you there are two possible solutions for N; 13 and -14. Now add 1 to these to get the two consecutive integers. 13 & 14 will work and -14 & -13 will work.

59 + 60 + 61 = 180 ---- To work out consecutive integers, use this formula: (x+1) + (x+2) + (x+3) = 180 ---> 3x + 6 = 180 ---> 3x = 174 ---> 174/3 = x = 58 so (58+1) = 59, (58 + 2) = 60, (58 + 3) = 61

The three integers, since they are consecutive, can be listed as a, a+1, and a+2. Twice the first is 2a. Three times the third is 3(a+2) = 3a+6. First make a formula of the information given: 2a+(3a+6)= -24 Next, solve the formula: 5a + 6 = -24 Subtract 6 from each side. 5a = -30 Divide each side by 5. a = -6 The three consecutive numbers are -6, -5, and -4.

There isn't a formula for finding joules. It is a way for finding a force or giving an example.

A quadratic sequence is when the difference between two terms changes each step. To find the formula for a quadratic sequence, one must first find the difference between the consecutive terms. Then a second difference must be found by finding the difference between the first consecutive differences.

Actually, there are plenty of those. The oldest might be this well-known formula by Euler: n^2 - n + 41 gives a prime for all positive integers smaller than 41.

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