A question with unknown variables indicates a good place to plug in.

You need numbers for x and y that will give you a negative product.

Try x = 1 and y = -2.

If you plug these into the statements in the Roman numerals, you find that (I) is false, but (II) and (III) are true.

You can eliminate any answer choice that contains (I).

This leaves (B) and (D).

Now try different numbers to see if you can eliminate another choice.

If you try x = -1 and y = 2, you find that (II) is false and (III) is still true.

This leaves you with (B) as the only correct answer.

Plug in the answers. If you start with (B), the length is 8, and the width is half that, or 4.

Area is length × width. The area of this rectangle is 8 × 4, which is nowhere near 128.

Eliminate (A) and (B), as both are too small.

Try (C): If the length is 16, the width is 8. So, does 128 = 16 × 8? You could write it all out, since you can't use your calculator, but you can also estimate. 16 × 10 = 160, so 16 × 8 would be about 130.

The number in (D) is too large and will give a weird fraction, so (C) is correct.

Alternatively, write an equation. The equation is area = w × 2w. So, 128 = 2w².

Divide by 2 to get 64 = w². Take the square root of both sides to find w = 8.

The length is twice this width, so length = 2 × 8 = 16, and the answer is (C).

The first step to answering this question is to get the equation into the standard form of a quadratic equation by moving all the terms to the left or right side of the equation and setting it equal to zero, like this: rx³ - (¹⁄_s)x - 3 = 0.

Now that you have the equation in standard form, you can begin to solve for the roots.

Since you are given variables instead of numbers, factoring this quadratic would require higher-level math, if it were even possible.

You may have noticed the familiar form of the answer choices. They are in a form similar to the quadratic equation.

Remember that a quadratic in standard form is represented by the equation ax² + bx + c = 0, and the quadratic formula is x = \(\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\).

In this equation, a = r, b = -(¹⁄_s), and c = -3.

Therefore, \(x = \frac{\frac{1}{s} \pm \sqrt{\left ( -\frac{1}{s} \right )^{2} - 4r\left ( -3 \right )}}{2r} = \frac{\frac{1}{s} \pm \sqrt{\frac{1}{s^{2}} + 12r}}{2r}\).

This exact format is not present in the answer choices, but the root part only matches the one in (A), so that is likely the answer.

You will have to do a little more manipulation before you can get the equations to match exactly.

The fractions need to be split up, so rewrite the equation as \(x = \frac{\frac{1}{s}}{2r} \pm \frac{\sqrt{\frac{1}{s^{2}} + 12r}}{2r} \;\; or \;\; x = \frac{1}{2sr} \pm \frac{\sqrt{\frac{1}{s^{2}} + 12r}}{2r}\).

Use FOIL to multiply the two binomials together. The expression becomes 4(¹⁄₂) - 8i + 7i(¹⁄₂) - 14i².

Simplify the result by multiplying through where you can to get 2 - 8i + ⁷⁄₂i - 14i².

To combine the i terms, multiply 8 by ²⁄₂ to get ¹⁶⁄₂.

Now the expression is 2 - ¹⁶⁄₂i + ⁷⁄₂i - 14i², which can be further simplified to 2 - ⁹⁄₂i - 14i².

Substitute -1 for i² and combine like terms: 2 - ⁹⁄₂i + 14 = 16 - ⁹⁄₂i, which is (A).

Since the question gives the value of m, the first step is to plug that value into the original equation to get \(\sqrt{-3x - 5} = x + 3\).

Now square both sides of the equation to remove the square root \(\left(\sqrt{-3x - 5} \right)^{2} = \left( x + 3 \right)^{2}\).

∴ -3x - 5 = x² + 6x + 9.

Now combine like terms. If you combine the terms on the right side of the equation, you can avoid having a negative x² term.

The equation becomes 0 = x² + 9x + 14.

Factor the equation to find the roots: 0 = (x + 2)(x + 7).

The possible solutions to the quadratic are -2 and -7.

Don't forget to plug these numbers back into the original equation to check for extraneous solutions.

Begin by checking x = -2. When you do this, you get \(\sqrt{\left ( -3 \right )\left ( -2 \right ) - 5} = \left ( -2 \right ) + 3\),or \(\sqrt{6 - 5} = 1, \; or \; \sqrt{1} = 1\), which is true.

Now, check x = -7.

Set it up as \(\sqrt{\left ( -3 \right )\left ( -7 \right ) - 5} = \left ( -7 \right ) + 3\) and start simplifying to get \(\sqrt{21 - 5} = -4\).

You can technically stop simplifying here, as there is a negative number on the right-hand side of the equal sign.

Remember, when taking a square root with a radical provided, it will yield the positive root only.

So -7 cannot be part of the solution set.

Start by multiplying the second equation by 2 to clear the fractions.

The equation becomes y + 2x = 2.

To get it into the same form as the other equations, subtract 2x from both sides to get y = -2x + 2.

Set the two x expressions equal to get 3x - 1 = -2x + 2.

Add 2x and 1 to both sides, so the equation becomes 5x = 3, then divide by 5 to find that x = ³⁄₅.

Plug this value into the y = 3x - 1 to get \(y = 3\left(\frac{3}{5} \right) - 1 = \frac{9}{5} - 1 = \frac{9}{5} - \frac{5}{5} = \frac{4}{5}\).

Finally, find the value of \(\frac{x}{y} : \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{5} \times \frac{5}{4} = \frac{3}{4}\) which is (C).

The lowest number that both 8 and 10 are factors of is 40.

Convert the fractions to a denominator of 40: ⁵⁄₄₀ + ⁴⁄₄₀ = ⁹⁄₄₀.

There is no factor that 9 and 40 have in common, so the fraction cannot be reduced.

The number in place of a in ^a⁄_b is 9, so the answer is (B).

Negative exponents mean to take the reciprocal and apply the positive exponent.

So \(9^{-2} = \left(\frac{1}{9} \right)^{2} = \frac{1}{81}\).

Now find what power of ¹⁄₃ equals ¹⁄₈₁.

Because 3⁴ = 81, \(\left(\frac{1}{3} \right)^{4} = \frac{1}{81}\) and x must be 4. The correct answer is (C).

Plug in the number given for a in the expression to find the value: −2 + (−2)² − (−2)³ + (−2)⁴ − (−2)⁵.

Remember PEMDAS, the order of operations:
The first thing to do here is deal with the Exponents, then we can take care of the Addition and Subtraction: −2 + 4 − (−8) + 16 − (−32),

which simplifies to −2 + 4 + 8 + 16 + 32 = 58, (D).

Take it one phrase at a time. The “sum” means you will add two things.

The “squares of x and y” means to square x and square y, or x² and y².

Add these to get x² + y².

Cross out any choice that does not have x² + y² as the first part of the equation.

Only (A) is left.