9. what is element $a_{23}$ in matrix a? $a = \begin{bmatrix} 3 & 7 & 0 \\ 8 & 4 & -6 \\ 0 & -4 & 6 \end{bmatrix}$ a. 7 b. 8 c. -4 d. -6 10. what is element $a_{12}$ in matrix a? $a = \begin{bmatrix} 2 & 2 & 4 \\ -5 & 3 & 8 \\ 7 & -2 & 9 \end{bmatrix}$ a. 8 b. 2 c. -5 d. -2 11. identify the maximum or minimum value and the domain and range of the graph of the function $y = 2(x + 2)^2 - 3$. a. minimum value: 3 domain: all real numbers $\geq 3$ range: all real numbers b. maximum value: $-3$ domain: all real numbers $\leq -3$ range: all real numbers c. maximum value: 3 domain: all real numbers range: all real numbers $\leq 3$ d. minimum value: $-3$ domain: all real numbers range: all real numbers $\geq -3$ what is the maximum or minimum value of the function? what is the range? 12. $y = 2x^2 + 28x - 8$ a. minimum value: 7 range: $y \geq 7$ b. minimum value: $-7$ range: $y \geq -7$ c. minimum value: $-106$ range: $y \geq -106$ d. minimum value: $-106$ range: $y \geq -7$ 13. what are the interval(s) on which the function $y = x^2 - 2x - 48$ is positive? a. $x 8$ b. $-6

Question

9. what is element $a_{23}$ in matrix a? $a = \begin{bmatrix} 3 & 7 & 0 \\ 8 & 4 & -6 \\ 0 & -4 & 6 \end{bmatrix}$ a. 7 b. 8 c. -4 d. -6 10. what is element $a_{12}$ in matrix a? $a = \begin{bmatrix} 2 & 2 & 4 \\ -5 & 3 & 8 \\ 7 & -2 & 9 \end{bmatrix}$ a. 8 b. 2 c. -5 d. -2 11. identify the maximum or minimum value and the domain and range of the graph of the function $y = 2(x + 2)^2 - 3$. a. minimum value: 3 domain: all real numbers $\geq 3$ range: all real numbers b. maximum value: $-3$ domain: all real numbers $\leq -3$ range: all real numbers c. maximum value: 3 domain: all real numbers range: all real numbers $\leq 3$ d. minimum value: $-3$ domain: all real numbers range: all real numbers $\geq -3$ what is the maximum or minimum value of the function? what is the range? 12. $y = 2x^2 + 28x - 8$ a. minimum value: 7 range: $y \geq 7$ b. minimum value: $-7$ range: $y \geq -7$ c. minimum value: $-106$ range: $y \geq -106$ d. minimum value: $-106$ range: $y \geq -7$ 13. what are the interval(s) on which the function $y = x^2 - 2x - 48$ is positive? a. $x < 6$ and $x > 8$ b. $-6 < x < 8$ c. $x > 6$ and $x < -8$ d. $6 < x < 8$

Accepted Answer

### Question 9 # Explanation: ## Step1: Understand matrix element notation In a matrix, $ a_{ij} $ represents the element in the $ i $-th row and $ j $-th column. For $ a_{23} $ in matrix $ A $, we look at the 2nd row and 3rd column. ## Step2: Locate the element Matrix $ A = \begin{bmatrix} 3 & 7 & 0 \ 8 & 4 & -6 \ 0 & -4 & 6 \end{bmatrix} $. The 2nd row is $ [8, 4, -6] $ and the 3rd column element here is $ -6 $. # Answer: d. $-6$ ### Question 10 # Explanation: ## Step1: Recall matrix element notation $ a_{ij} $ is the element in row $ i $, column $ j $. For $ a_{12} $, we need row 1, column 2. ## Step2: Find the element Matrix $ A = \begin{bmatrix} 2 & 2 & 4 \ -5 & 3 & 8 \ 7 & -2 & 9 \end{bmatrix} $. Row 1 is $ [2, 2, 4] $, column 2 element is $ 2 $. # Answer: b. $2$ ### Question 11 # Explanation: ## Step1: Analyze the quadratic function form The function is $ y = 2(x + 2)^2 - 3 $, which is in vertex form $ y = a(x - h)^2 + k $, where $ (h, k) $ is the vertex. Here, $ a = 2 > 0 $, so the parabola opens upward, meaning it has a minimum at the vertex. ## Step2: Determine vertex, domain, and range The vertex is $ (-2, -3) $, so the minimum value is $ -3 $. The domain of a quadratic function is all real numbers. Since the parabola opens upward, the range is $ y \geq -3 $ (all real numbers greater than or equal to the minimum value). # Answer: d. minimum value: $-3$; domain: all real numbers; range: all real numbers $\geq -3$ ### Question 12 # Explanation: ## Step1: Use the formula for vertex of a quadratic For a quadratic $ y = ax^2 + bx + c $, the x-coordinate of the vertex is $ x = -\frac{b}{2a} $. Here, $ a = 2 $, $ b = 28 $, so $ x = -\frac{28}{2 \times 2} = -7 $. ## Step2: Find the minimum value Substitute $ x = -7 $ into the function: $ y = 2(-7)^2 + 28(-7) - 8 = 2(49) - 196 - 8 = 98 - 196 - 8 = -106 $. Since $ a = 2 > 0 $, it's a minimum. The range is $ y \geq -106 $. # Answer: c. minimum value: $-106$; range: $ y \geq -106 $ ### Question 13 # Explanation: ## Step1: Factor the quadratic function Factor $ y = x^2 - 2x - 48 $. We need two numbers that multiply to $-48$ and add to $-2$. Those numbers are $-8$ and $6$, so $ y = (x - 8)(x + 6) $. ## Step2: Find the roots Set $ y = 0 $: $ (x - 8)(x + 6) = 0 $, so roots are $ x = 8 $ and $ x = -6 $. ## Step3: Determine where the function is positive The quadratic opens upward (since $ a = 1 > 0 $). So it's positive when $ x < -6 $ or $ x > 8 $? Wait, no—wait, factoring: $ (x - 8)(x + 6) $, so the parabola opens upward, so it's positive outside the roots. Wait, the options: let's check again. Wait, maybe I made a mistake. Wait, $ x^2 - 2x - 48 = (x - 8)(x + 6) $, so roots at $ x = 8 $ and $ x = -6 $. For a parabola opening upward, it's positive when $ x < -6 $ or $ x > 8 $. But looking at the options, option a is $ x < 6 $ and $ x > 8 $? Wait, no, maybe a typo? Wait, original options: a. $ x < 6 $ and $ x > 8 $ – no, wait, the correct roots are $ x = -6 $ and $ x = 8 $. So the function is positive when $ x < -6 $ or $ x > 8 $. But the options: option a is $ x < 6 $ and $ x > 8 $ – no, maybe a mistake in the problem, but looking at the options, the closest correct reasoning: wait, maybe I factored wrong? Wait, $ x^2 - 2x - 48 $: sum of roots is $ 2 $, product is $-48$. So roots are $ 8 $ and $-6$ (since $ 8 + (-6) = 2 $, $ 8 \times (-6) = -48 $). So the parabola opens upward, so positive when $ x < -6 $ or $ x > 8 $. But the options: option a is $ x < 6 $ and $ x > 8 $ – no, maybe a typo, but among the options, the correct intervals (if we consider maybe a typo where $-6$ was written as $6$)—but the intended correct answer should be $ x < -6 $ or $ x > 8 $, but the options have a. $ x < 6 $ and $ x > 8 $ – no, maybe the problem has a typo. Wait, looking at the options again: a. $ x < 6 $ and $ x > 8 $ b. $ -6 < x < 8 $ c. $ x > 6 $ and $ x < -8 $ d. $ 6 < x < 8 $ Wait, maybe I factored wrong. Wait, $ x^2 - 2x - 48 $: let's solve $ x^2 - 2x - 48 = 0 $. Using quadratic formula: $ x = \frac{2 \pm \sqrt{4 + 192}}{2} = \frac{2 \pm \sqrt{196}}{2} = \frac{2 \pm 14}{2} $, so $ x = 8 $ or $ x = -6 $. So the parabola opens upward, so it's positive when $ x < -6 $ or $ x > 8 $. But none of the options have $ x < -6 $ or $ x > 8 $. Wait, option a is $ x < 6 $ and $ x > 8 $ – maybe a typo, and $-6$ was written as $6$. So the closest is option a? Wait, no, maybe I made a mistake. Wait, the function is $ y = x^2 - 2x - 48 $, so when $ x < -6 $, say $ x = -7 $: $ (-7)^2 - 2(-7) - 48 = 49 + 14 - 48 = 15 > 0 $. When $ x > 8 $, say $ x = 9 $: $ 81 - 18 - 48 = 15 > 0 $. Between $-6$ and $8$, say $ x = 0 $: $ 0 - 0 - 48 = -48 < 0 $. So the correct intervals are $ x < -6 $ or $ x > 8 $. But the options: option a is $ x < 6 $ and $ x > 8 $ – maybe a typo, but if we assume $6$ is $-6$, then option a would be correct. Alternatively, maybe the problem has a typo. But among the given options, the intended answer is likely a (assuming a typo in the option, $6$ instead of $-6$). # Answer: a. $ x < 6 $ and $ x > 8 $ (Note: There might be a typo, and the intended roots were $-6$ and $8$, so the correct interval should be $ x < -6 $ or $ x > 8 $, but based on the given options, a is the closest if we consider a typo.)

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Question 9

Step1: Understand matrix element notation

Step2: Locate the element

Step1: Recall matrix element notation

Step2: Find the element

Step1: Analyze the quadratic function form

Step2: Determine vertex, domain, and range

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Question 10