The total sum
No. Two odd numbers added together always give an even number. Two even numbers added together always give an even number. An odd and an even number added together always give an odd number. So, if we have 5 odd numbers - a, b, c, d and e a + b will be even c + d will be even So if (a + b) + (c + d) is even, adding e to that will have to be an odd number - but 50 is even, so it cannot be done.
I just did a very similar problem where you have to find a set of numbers so that when multiplied or added together you get 7. So i just plugged in your 7.11 into my formula and got an answer. Its a bit complicated how I got there. Essentially I used the formula a/b+(a-b)/b+b^2=(whatever number you want). And the formula a/b*(a-b)/b*b^2=(the number you want). Then I plugged in 7.11, combined the two formulas to solve for b. Used my calculator to find the solution (which should be in terms of fourth roots and square roots) and got these numbers. (they are rounded) a=3.121604582 b=.84392981. Notice that each of those equations uses 3 terms (a/b, (a-b)/b, b^2). Those are the three numbers. So finally, the numbers are... 3.698891237, 2.698891237, .7122175253. If you want whole numbers though... I cant help.
|a + b| ≤ |a| + |b|
-10
You cannot swap two numbers using call by value, because the called function does not have access to the original copy of the numbers.Swap with call by reference... This routine uses exclusive or swap without temporary variable.void swap (int *a, int *b) {*a ^= *b;*b ^= *a;*a ^= *b;return;}
Yes, the mean can be a decimal because the mean is a+b+c+d+(the numbers)....=e(the sum of the numbers), then e/(the quantity of numbers added together to get e)=f(the mean). Sometimes the sum may not go into the quantity in a whole number, which gives you a decimal.
The associative property of multiplication states that for any three numbers a, b and c, (a * b) * c = a * (b * c) and so we can write either as a * b * c without ambiguity. ie, when multiplying three numbers together, you can multiply the first two together and then multiply the result of that by the third, or multiply the second two numbers together and multiply that result by the first, and you will get the same answer.
void swap(int& a, int& b ) { a^=b^=a^=b; }
The numbers are (1/2)(5 + i root(15)) and (1/2)(5 - i root(15))Note: You can get this from the Wolfram Alpha site, using the input: a+b=5, ab=10
Mean = sum of all numbers divided by number of numbers you summed. Call numbers a, b, c, d, e, f (a+b+c+d+e+f)/6 = mean
Let's call the numbers A and B, A being the bigger number. So, A - B = 3 and A + B =47 That makes, (A - B) + (A + B) = 2A = 50 Thus, A = 25 and B = 22.
Let the numbers be a and b. Then a + b = 56 And a - b = 18.........adding these two equations together to eliminate b gives : 2a + b - b = 56 + 18 : 2a = 74 : a = 37 Substituting for a in the first equation gives : 37 + b = 56 : b = 56 - 37 = 19 The two numbers are 37 and 19.