For floating point numbers, do #include <cmath> and use std::abs(). For integers do #include<cstdlib> and use std::abs().
If you are a beginner, I recommend Absolute C++ by Savitch. If you are experienced, I suggest buying the K & R.
-8c+1 = -3 -8c = -3-1 -8c = -4 Divide both sides by -8 in order to find the value of c remembering that a minus number divided into minus number is equal to a plus number: c = 1/2
any number
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#define biggest (a) > (b) && (a) > (c) ? (a) : (b) > (c) ? (b) : (c)
Yes.
The answer depends on absolute deviation from what: the mean, median or some other measure. Suppose you have n observations, x1, x2, ... xn and you wish to calculate the sum of the absolute deviation of these observations from some fixed number c. The deviation of x1 from c is (x1 - c). The absolute deviation of x1 from c is |x1 - c|. This is the non-negative value of (x1 - c). That is, if (x1 - c) ≤ 0 then |x1 - c| = (x1 - c) while if (x1 - c) < 0 then |(x1 - c)| = - (x1 - c). Then the sum of absolute deviations is the above values, summed over x1, x2, ... xn.
If the sum of squares of digits of a number equals to the number itself, then that number is called an aram strong number.
If for the two integers a + b = c, a is larger than zero and b is larger than the absolute value of the first number, then c is always positive. If a is smaller than zero, and b is larger than the absolute value of the second number, then c is positive. If a is smaller than zero, and b is smaller than the absolute value of the second number, then c is negative.
the correct answer is the 15 number in the alphabet which is o
negative
See related link. It's in C rather than C++, but conversion to C++ is fairly simple.