The following is a list of triangles: I. Right triangles, II. Isosceles triangles, III. Equilateral triangles. A pair of triangles from which of these groups must be similar to each other?
- A. I only
- B. II only
- C. III only
- D. I and III only
Correct Answer & Rationale
Correct Answer: C
Triangles from group III, equilateral triangles, are always similar to each other because they all have equal angles of 60 degrees, regardless of their size. Group I, right triangles, can vary significantly in angle measures beyond the right angle, so not all right triangles are similar. Similarly, group II, isosceles triangles, can have different base angles, leading to non-similar triangles. Thus, while right and isosceles triangles can share properties, only equilateral triangles guarantee similarity across the group. Therefore, option C accurately identifies the group with universally similar triangles.
Triangles from group III, equilateral triangles, are always similar to each other because they all have equal angles of 60 degrees, regardless of their size. Group I, right triangles, can vary significantly in angle measures beyond the right angle, so not all right triangles are similar. Similarly, group II, isosceles triangles, can have different base angles, leading to non-similar triangles. Thus, while right and isosceles triangles can share properties, only equilateral triangles guarantee similarity across the group. Therefore, option C accurately identifies the group with universally similar triangles.
Other Related Questions
What are the solutions to (x-2)(x+4) = 0?
- A. -4 and 2
- B. -3 and 1
- C. -2 and 4
- D. -1 and 1
- E. -1 and 3
Correct Answer & Rationale
Correct Answer: A
To solve the equation (x-2)(x+4) = 0, we apply the zero product property, which states that if a product of factors equals zero, at least one of the factors must equal zero. Setting each factor to zero gives us the equations x - 2 = 0 and x + 4 = 0. Solving these yields x = 2 and x = -4, confirming that the solutions are -4 and 2. Options B, C, D, and E provide incorrect pairs of solutions that do not satisfy the original equation when substituted back in. Each of these pairs results in non-zero products for the factors, thus failing to meet the requirement of the equation.
To solve the equation (x-2)(x+4) = 0, we apply the zero product property, which states that if a product of factors equals zero, at least one of the factors must equal zero. Setting each factor to zero gives us the equations x - 2 = 0 and x + 4 = 0. Solving these yields x = 2 and x = -4, confirming that the solutions are -4 and 2. Options B, C, D, and E provide incorrect pairs of solutions that do not satisfy the original equation when substituted back in. Each of these pairs results in non-zero products for the factors, thus failing to meet the requirement of the equation.
One online movie-streaming service costs $8 per month and charges $1.50 per movie. A second service costs $2 per month and charges $2 per movie. For what number of movies per month is the monthly cost of both services the same?
- A. 3
- B. 6
- C. 5
- D. 12
- E. 20
Correct Answer & Rationale
Correct Answer: D
To determine when the costs of both services are equal, we can set up equations based on the monthly fees and per-movie charges. For the first service: Cost = $8 + $1.50 * number of movies (m) Cost = $8 + 1.5m For the second service: Cost = $2 + $2 * number of movies (m) Cost = $2 + 2m Setting the two equations equal gives us: $8 + 1.5m = $2 + 2m Rearranging leads to: $6 = 0.5m m = 12 Thus, when 12 movies are rented, the costs are equal. Options A (3), B (6), and C (5) yield different costs, as they do not satisfy the equation. Option E (20) results in a higher cost for the second service, confirming that 12 is the only solution where both services cost the same.
To determine when the costs of both services are equal, we can set up equations based on the monthly fees and per-movie charges. For the first service: Cost = $8 + $1.50 * number of movies (m) Cost = $8 + 1.5m For the second service: Cost = $2 + $2 * number of movies (m) Cost = $2 + 2m Setting the two equations equal gives us: $8 + 1.5m = $2 + 2m Rearranging leads to: $6 = 0.5m m = 12 Thus, when 12 movies are rented, the costs are equal. Options A (3), B (6), and C (5) yield different costs, as they do not satisfy the equation. Option E (20) results in a higher cost for the second service, confirming that 12 is the only solution where both services cost the same.
Through which pair of points could a line of best fit be drawn for the data on the scatterplot?
- A. (0, 36) and (11, 74)
- B. (1, 39) and (6, 60)
- C. (5, 50) and (6, 60)
- D. (6, 60) and (8, 60)
- E. (8, 60) and (11, 74)
Correct Answer & Rationale
Correct Answer: A
Option A, with points (0, 36) and (11, 74), shows a significant range in both x and y values, indicating a strong upward trend that aligns well with the overall direction of the data. Option B, while showing an upward trend, has a narrower range and may not represent the overall data as effectively. Option C features two points that are too close together, limiting their ability to define a clear line of best fit. Option D includes points with the same y-value, suggesting a horizontal line that does not capture the data's trend. Option E, like A, has a valid upward trend but does not span the data range as effectively as A.
Option A, with points (0, 36) and (11, 74), shows a significant range in both x and y values, indicating a strong upward trend that aligns well with the overall direction of the data. Option B, while showing an upward trend, has a narrower range and may not represent the overall data as effectively. Option C features two points that are too close together, limiting their ability to define a clear line of best fit. Option D includes points with the same y-value, suggesting a horizontal line that does not capture the data's trend. Option E, like A, has a valid upward trend but does not span the data range as effectively as A.
A campground rents canoes for either $20 per day or $4 per hour. For what number or numbers of hours, h, is it more expensive to rent a canoe at the daily rate than at the hourly rate?
- A. h = 5
- B. h >= 25
- C. h > 5
- D. h < 5
- E. h ≤ 5
Correct Answer & Rationale
Correct Answer: C
To determine when renting a canoe at the daily rate exceeds the hourly rate, we compare the costs. The daily rate is $20, while the hourly rate is $4 per hour. Setting up the inequality, we have: \[ 20 > 4h \] Dividing both sides by 4 gives: \[ 5 > h \] This means that renting for more than 5 hours makes the daily rate more economical. Option A (h = 5) is incorrect since at 5 hours, both rates are equal. Option B (h ≥ 25) is incorrect because it's not relevant to the threshold we calculated. Option D (h < 5) suggests a scenario where the daily rate is not more expensive, which contradicts our findings. Option E (h ≤ 5) includes values where the rates are equal or less, which doesn't satisfy the condition.
To determine when renting a canoe at the daily rate exceeds the hourly rate, we compare the costs. The daily rate is $20, while the hourly rate is $4 per hour. Setting up the inequality, we have: \[ 20 > 4h \] Dividing both sides by 4 gives: \[ 5 > h \] This means that renting for more than 5 hours makes the daily rate more economical. Option A (h = 5) is incorrect since at 5 hours, both rates are equal. Option B (h ≥ 25) is incorrect because it's not relevant to the threshold we calculated. Option D (h < 5) suggests a scenario where the daily rate is not more expensive, which contradicts our findings. Option E (h ≤ 5) includes values where the rates are equal or less, which doesn't satisfy the condition.