Algebraic Expressions

Factor out 3 from the second expression: 6x - 3y = 3(2x - y). Since we're given that 2x - y = 4, substituting gives 3 × 4 = 12.

We are given $x = y = 2z$, which means $y = x$ and $z = x/2$. Substituting these into $xyz = 256$ gives $x \times x \times (x/2) = 256$, which simplifies to $x^3 = 512$. Taking the cube root of both sides yields $x = 8$. Distractors represent incorrect algebraic substitutions.

Option C matches the given expression (x+y)^2=x^2+xy2+y^2. The notation xy2 appears to represent the middle term 2xy in the binomial expansion. The standard form of (x+y)^2 is x^2 + 2xy + y^2, but this question uses compressed notation where xy2 serves as the middle term.

Simplify the left side: 3x + x = 4x. So 4x = 2x + 2y. Subtract 2x from both sides: 2x = 2y. Divide by 2: x = y. Option D is correct.

Given 2X - Y = 4, we need 6X - 3Y. Factor out 3: 3(2X - Y) = 3 × 4 = 12. This tests recognition of common factors. Option A (15) would come from incorrectly distributing, option C (18) from multiplying only one term, option D (10) from random calculation.

Given X = Y = 2Z, so Y = 2Z and X = 2Z as well. Then XYZ = (2Z)(2Z)(Z) = 4Z³ = 256, so Z³ = 64 and Z = 4. Therefore X = 2Z = 8. Option B (3), C (5), and D (6) don't satisfy the equation when substituted back.

In REXX, the % operator performs integer division (truncating the fractional part). 12 divided by 4 equals exactly 3, so z=3. Options 0, 4, and 12 are incorrect calculations. The % operator specifically returns the quotient without remainder, unlike / which preserves decimal results.

The SAS intnx function advances dates by intervals. intnx('year', '21oct2009'd, 3, 'mid') adds 3 years using 'mid' alignment, landing on July 1, 2012, which has SAS date value 19175.

In SAS DATA step logic, the assignment statements execute sequentially. When x=4, the condition 'if x=4' is true (using single = for comparison in SAS), so y='four'. The ELSE IF never executes. Later, x is reassigned to 7, but y remains 'four' - the IF/ELSE doesn't re-evaluate. Final values: x=7, y='four'.

The prefix increment ++y makes y equal 2 immediately. The expression evaluates to 0 - 2 + z++ which is -2 + 5 = 3. Post-increment z++ increases z to 6 after evaluation. Thus, x becomes 3, y becomes 2, and z becomes 6.

The post-decrement operator (y--) returns the CURRENT value of y, then decrements y. So if y=5, then x=y-- means x gets 5, and y becomes 4. After execution, x=5 and y=4, so x>y is true. Option A is correct. Option B would mean x

Evaluating the expression: x++ is post-increment (use 5, then x becomes 6), --y is pre-decrement (y becomes -11), so the expression is 5 - (-11 * 2 / 4) = 5 - (-22/4) = 5 - (-5) = 10. In integer division, -22/4 = -5 (truncates toward zero). Final result: z = 10, x = 6, y = -11. Option B is correct.

To solve this question, the user needs to know the order of operations in C programming and the difference between pre-increment and post-increment operators.

First, let's break down the expression x = 0 - (++y) + z++:

1. ++y is a pre-increment operator, which means it increments the value of y by 1 before the value is used in the expression. So, y becomes 2.
2. z++ is a post-increment operator, which means it increments the value of z by 1 after the value is used in the expression. So, z remains 5 in this expression.
3. The expression inside the parentheses is evaluated first, so ++y is evaluated to 2.
4. Next, the multiplication and division are evaluated from left to right, but there are none in this expression.
5. Finally, the addition and subtraction are evaluated from left to right.

So, 0 - (++y) evaluates to -2 and z++ evaluates to 5, therefore the whole expression evaluates to -2 + 5 = 3.

Therefore, the correct answer is:

The Answer is: B. x = 3, y = 2, z = 6

Pre-increment (++y) increments before use: y becomes 2. Post-increment (z++) uses old value then increments: z remains 5 for the expression. x = 0 - 2 + 5 = 3. After expression, z increments to 6. Final: x=3, y=2, z=6.

The Len function counts all characters including spaces. 'TIB Ltd' has 7 characters: T, I, B, space, L, t, d. The space between words is counted.