Maxima and Minima

Choice (3) i.e 30 is divisible by both 15 and 10 and is least positive integer. 15 is not divisible by 10 and 20 is not divisible by 15. Other choices are more than 30 hence can be eliminated

[- 2.6] - [2.6] = - 3 - 2 = - 5

The possible values for (m, n) consistent with the given conditions are (4, 3 ). Then m – n = 1 or 7. So, the maximum value is 7.

The law of diminishing marginal utility states that as consumption increases, total utility rises at a decreasing rate. When marginal utility reaches zero, total utility is at its maximum point - consuming one more unit would not increase satisfaction. Beyond this point, marginal utility becomes negative and total utility declines.

Some instructions require zero operands, such as NOP (No Operation), HLT (Halt), or RET (Return in some architectures). These instructions perform their function without needing to reference any data operands. Unary instructions need 1 operand, binary operations need 2 operands, and some complex instructions may need 3 operands, but 0 is the theoretical and practical minimum.

The regionMatches() method in Java's String class has two overloaded versions - one taking 4 arguments and another taking 5 arguments. The 5-argument version includes an additional boolean parameter to ignore case differences. Therefore, the maximum is 5 arguments.

When the objective function is parallel to a binding constraint edge, multiple optimal solutions exist. All points on that edge (including both endpoints) yield the same optimal objective value.

An unbounded solution occurs when the objective function can increase (maximization) or decrease (minimization) indefinitely within the feasible region. This typically means constraints don't bound the objective in that direction.

In a decision table, each condition can typically have two states (true or false). For n conditions, there are 2^n possible combinations of conditions, resulting in 2^n columns.

2/3 = 70/105 2/7 = 30/105 3/5 = 63/105 1/3 = 35/105Therefore, 2/7 is smallest.

Fact

x = 9a + 3 = 15b + 3 = 21c + 3, a, b and c are all integers. So 9a = 15b = 21c, the least common multiple of 9, 15 and 21 will be 3*3*5*7 = 315. So 9a = 15b = 21c = 315, a = 35, b = 21, c = 15. Hence, x = 315 + 3 = 318.

When TP reaches its maximum level, MP is zero. MP becomes negative after this stage and MP is positive before this stage.

When TP reaches its maximum level, MP is zero. MP becomes negative after this stage and MP is positive before this stage.

A 16-bit signed integer uses one bit for sign, leaving 15 bits for magnitude. The maximum value is 2^15 - 1 = 32767. This is a standard limit in older C implementations with 16-bit ints.