A figure is formed by shaded squares on a grid. Which figure has a perimeter of 12units and an area of 8 square units?
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A.
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B.
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C.
- D. None of the above
Correct Answer & Rationale
Correct Answer: C
To determine the figure that meets the criteria of having a perimeter of 12 units and an area of 8 square units, we analyze each option. Option C achieves both requirements: it has a perimeter of 12 units, calculated by adding the lengths of all sides, and an area of 8 square units, determined by multiplying its length and width (2 x 4). In contrast, Option A has a perimeter exceeding 12 units, while its area is less than 8 square units. Option B has a perimeter of 10 units and an area of 6 square units, failing both criteria. Option D is not applicable since Option C meets the conditions. Thus, Option C stands out as the only figure that satisfies both the perimeter and area requirements.
To determine the figure that meets the criteria of having a perimeter of 12 units and an area of 8 square units, we analyze each option. Option C achieves both requirements: it has a perimeter of 12 units, calculated by adding the lengths of all sides, and an area of 8 square units, determined by multiplying its length and width (2 x 4). In contrast, Option A has a perimeter exceeding 12 units, while its area is less than 8 square units. Option B has a perimeter of 10 units and an area of 6 square units, failing both criteria. Option D is not applicable since Option C meets the conditions. Thus, Option C stands out as the only figure that satisfies both the perimeter and area requirements.
Other Related Questions
Which graph represents a function?
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A.
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B.
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C.
Correct Answer & Rationale
Correct Answer: B
To determine which graph represents a function, we apply the Vertical Line Test. This test states that if a vertical line intersects the graph at more than one point, the graph does not represent a function. Option A fails this test, as a vertical line can intersect the graph at multiple points, indicating it is not a function. Option C also does not satisfy the criteria, showing multiple intersections with vertical lines. In contrast, Option B passes the Vertical Line Test, as any vertical line drawn will intersect the graph at only one point, confirming it represents a function.
To determine which graph represents a function, we apply the Vertical Line Test. This test states that if a vertical line intersects the graph at more than one point, the graph does not represent a function. Option A fails this test, as a vertical line can intersect the graph at multiple points, indicating it is not a function. Option C also does not satisfy the criteria, showing multiple intersections with vertical lines. In contrast, Option B passes the Vertical Line Test, as any vertical line drawn will intersect the graph at only one point, confirming it represents a function.
What is the area, in square inches, of a circle with diameter 2 inches?
- A. 6.28
- B. 3.14
- C. 1
- D. 12.56
Correct Answer & Rationale
Correct Answer: B
To find the area of a circle, the formula \( A = \pi r^2 \) is used, where \( r \) is the radius. Given a diameter of 2 inches, the radius is 1 inch. Substituting this into the formula yields \( A = \pi (1)^2 = \pi \), which approximates to 3.14. Option A (6.28) incorrectly doubles the area, possibly confusing it with the circumference. Option C (1) neglects the use of \(\pi\), leading to an inaccurate calculation. Option D (12.56) mistakenly uses the formula for circumference, multiplying the diameter by \(\pi\) instead of squaring the radius. Thus, 3.14 accurately represents the area of the circle.
To find the area of a circle, the formula \( A = \pi r^2 \) is used, where \( r \) is the radius. Given a diameter of 2 inches, the radius is 1 inch. Substituting this into the formula yields \( A = \pi (1)^2 = \pi \), which approximates to 3.14. Option A (6.28) incorrectly doubles the area, possibly confusing it with the circumference. Option C (1) neglects the use of \(\pi\), leading to an inaccurate calculation. Option D (12.56) mistakenly uses the formula for circumference, multiplying the diameter by \(\pi\) instead of squaring the radius. Thus, 3.14 accurately represents the area of the circle.
The number line below shows the solution set of an inequality: Which two inequalities represent the graph shown?
- A. -2x>4 and 4x<-8
- B. 3x>-6 and x-4>6
- C. 4x<-8 and x≥6
- D. 4x<-8 and x≥6
Correct Answer & Rationale
Correct Answer: B
The graph indicates a solution set that includes values greater than -2 and less than 6. Option B, with inequalities 3x > -6 and x - 4 > 6, accurately reflects this range. The first inequality simplifies to x > -2, aligning with the left boundary, while the second simplifies to x > 10, which is outside the range but indicates a direction. Options A, C, and D contain inequalities that do not match the solution set shown on the number line. A suggests values that are too extreme, while C and D incorrectly imply lower bounds that do not correspond to the graph's representation.
The graph indicates a solution set that includes values greater than -2 and less than 6. Option B, with inequalities 3x > -6 and x - 4 > 6, accurately reflects this range. The first inequality simplifies to x > -2, aligning with the left boundary, while the second simplifies to x > 10, which is outside the range but indicates a direction. Options A, C, and D contain inequalities that do not match the solution set shown on the number line. A suggests values that are too extreme, while C and D incorrectly imply lower bounds that do not correspond to the graph's representation.
At what point does the function stop decreasing and start increasing?
- A. (1, -4)
- B. (3, 0)
- C. (-4, 1)
- D. (0, -3)
Correct Answer & Rationale
Correct Answer: A
To determine where the function stops decreasing and starts increasing, we look for a local minimum, which occurs where the derivative changes from negative to positive. Option A: (1, -4) indicates a point where the function transitions from decreasing to increasing, making it a local minimum. Option B: (3, 0) does not represent a minimum; the function is still increasing here. Option C: (-4, 1) is not relevant to the transition, as it does not indicate a change in direction. Option D: (0, -3) also does not represent a point of change, as the function continues to decrease. Thus, A is the point where the function stops decreasing and begins to increase.
To determine where the function stops decreasing and starts increasing, we look for a local minimum, which occurs where the derivative changes from negative to positive. Option A: (1, -4) indicates a point where the function transitions from decreasing to increasing, making it a local minimum. Option B: (3, 0) does not represent a minimum; the function is still increasing here. Option C: (-4, 1) is not relevant to the transition, as it does not indicate a change in direction. Option D: (0, -3) also does not represent a point of change, as the function continues to decrease. Thus, A is the point where the function stops decreasing and begins to increase.