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Q: The sum of 3 times m and n?

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4(N-3) is the formula, do the math from there and you get "N-3"

3 (n + 7) = 16

7(n+3)

3*(n + 5) or 3n + 15

6(3n+5)

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sum = 2n + 3m

4(N-3) is the formula, do the math from there and you get "N-3"

10n = -30 n = -3

m and n are 70 and 90

Where n = any number, n(12+3)

3 (n + 7) = 16

7(n+3)

To find the (mean) average, add all the numbers and divide by the number of numbers. The sum of a series of digits (in arithmetic progression, like 13, 14, 15, ... 37) is sum = (first + last) x number_of_digits / 2 So their average is: average = ((first + last) x number_of_digits / 2) / number_of_digits = (first + last) / 2 = average of first and last digits! So the average of the numbers 13, 14, 15, ..., 37 is: average = (13 + 37) / 2 = 50 / 2 = 25 To find the sum of n digits starting with m: Sum = m + (m+1) + ... + (m+n-2) + (m+n-1) Rewrite the sum in reverse order: Sum2 = Sum = (m+n-1) + (m+n-2) + ... + (m+1) + m Add the two sums, term by term: Sum + Sum2 = 2 Sum = (m + (m+n-1)) + (m + (m+n-1)) + ... + (m + (m+n-1)) + (m + (m+n-1)) There are n terms, all (m + (m+n-1)), so: 2 Sum = (m + m+n-1) x n Sum = (m + m+n-1) x n / 2 But n is the number of digits, m is the first number and (m+n-1) is the last, so: Sum = (first + last) x number_of_digits / 2

3*(n + 5) or 3n + 15

4*(n + 1) = 6*n*3

long n= 100, m= 500;printf ("The sum is %ld\n", (m-n+1)*(m+n)/2);

You record the various speed on the same distance and sum it. Then divide that sum with how many times you record it. Like for x meters : 1. a m/s 2. b m/s 3. c m/s . . n. y m/s The average would be : (a+b+c +...y)/n

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