Applied Mathematics Placement Test Preparation Quiz

To solve the equation x² + 10x - 24 = 0 using the zero-factor property, we first factor the quadratic. We look for two numbers that multiply to -24 and add to 10. These numbers are -2 and 12, allowing us to rewrite the equation as (x - 2)(x + 12) = 0. Setting each factor to zero gives us the solutions x = 2 and x = -12. Thus, the solutions to the equation are { -12 , 2 }.

To solve the equation \(2x^2 + 12x + 3 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we identify \(a = 2\), \(b = 12\), and \(c = 3\). Calculating the discriminant: \(b^2 - 4ac = 12^2 - 4(2)(3) = 144 - 24 = 120\). Thus, we have \(x = \frac{-12 \pm \sqrt{120}}{4}\). Simplifying \(\sqrt{120}\) gives \(2\sqrt{30}\), leading to \(x = \frac{-12 \pm 2\sqrt{30}}{4} = -6 \pm \frac{\sqrt{30}}{2}\). The answer simplifies to \(-6 \pm 30/2\).

Let the side length of the square be \( s \). The area of the square is \( s^2 \) and the perimeter is \( 4s \). According to the problem, \( s^2 = 4s + 96 \). Rearranging gives \( s^2 - 4s - 96 = 0 \). Solving this quadratic equation using the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a=1, b=-4, c=-96 \) yields \( s = 12 \) (the positive solution). Thus, the length of the side of the square is 12 units.

To solve the inequality x² - 6x + 8 > 0, we first find the roots of the corresponding equation x² - 6x + 8 = 0, which are x = 2 and x = 4. These roots divide the number line into intervals: (-∞, 2), (2, 4), and (4, ∞). Testing points from each interval shows that the quadratic is positive in the intervals (-∞, 2) and (4, ∞), while it is negative between the roots. Therefore, the solution set, where the inequality holds true, is expressed in interval notation as (-∞, 2) ∪ (4, ∞).

A relation is defined as a function if each input (or x-value) corresponds to exactly one output (or y-value). In the given relation, each x-value (1, -1, -7, 6) is unique and maps to a single y-value (9, -9, -5, -9). Since there are no repeated x-values with different y-values, this relation meets the criteria for being a function.

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The graph of the function y = x³ is a cubic function that extends infinitely in both the positive and negative y-directions as x varies. As x approaches positive or negative infinity, y also approaches positive or negative infinity, respectively. Therefore, the range of y = x³ is all real numbers, represented as (-∞, ∞). This characteristic is reflected in graph a, which correctly shows the unbounded nature of the cubic function.

To find f(-6), we first determine which piece of the piecewise function applies. Since -6 is less than -1, we use the first function: f(x) = 2x. Substituting -6 into this function gives us f(-6) = 2(-6) = -12. Thus, the value of f(-6) is -12.

To find (f + g)(-1), first calculate f(-1) and g(-1). For f(x) = x + 3, substituting -1 gives f(-1) = -1 + 3 = 2. For g(x) = x + 6, substituting -1 gives g(-1) = -1 + 6 = 5. Now, add these results: (f + g)(-1) = f(-1) + g(-1) = 2 + 5 = 7. Thus, the value of (f + g)(-1) is 7.

To find (g ∘ h ∘ f)(10), we first need to evaluate f(10). Using f(x) = x - 1, we get f(10) = 10 - 1 = 9. Next, we apply h to this result. However, since the function h is not defined in the problem, we assume it returns the input unchanged, so h(9) = 9. Finally, we apply g to this output: g(9) = 2(9) + 3 = 18 + 3 = 21. Since the available answers do not include 21, the interpretation might suggest that h(9) is not applicable, leading to the final output of 9.