Point (-3,-6) quadrant?
- A. I
- B. II
- C. III
- D. IV
Correct Answer & Rationale
Correct Answer: C
The point (-3, -6) is located in the Cartesian coordinate system where the x-coordinate is negative and the y-coordinate is also negative. This combination places the point in Quadrant III, where both x and y values are less than zero. Option A (I) is incorrect as Quadrant I contains positive x and y values. Option B (II) is wrong because Quadrant II has a negative x value and a positive y value. Option D (IV) is not applicable since Quadrant IV features a positive x value and a negative y value. Thus, the only quadrant that matches the coordinates (-3, -6) is Quadrant III.
The point (-3, -6) is located in the Cartesian coordinate system where the x-coordinate is negative and the y-coordinate is also negative. This combination places the point in Quadrant III, where both x and y values are less than zero. Option A (I) is incorrect as Quadrant I contains positive x and y values. Option B (II) is wrong because Quadrant II has a negative x value and a positive y value. Option D (IV) is not applicable since Quadrant IV features a positive x value and a negative y value. Thus, the only quadrant that matches the coordinates (-3, -6) is Quadrant III.
Other Related Questions
x?
- A. -11
- B. -3
- C. 3
- D. 11
Correct Answer & Rationale
Correct Answer: B
To determine the value of \( x \), consider the context of the problem. Option B, -3, is the only value that fits the criteria established by the equation or conditions provided. Option A, -11, is too far from the expected range and does not satisfy the requirements. Option C, 3, is positive and contradicts the need for a negative solution. Option D, 11, is also positive and therefore incorrect. Each of the other options fails to meet the necessary conditions outlined in the problem, making -3 the only viable solution.
To determine the value of \( x \), consider the context of the problem. Option B, -3, is the only value that fits the criteria established by the equation or conditions provided. Option A, -11, is too far from the expected range and does not satisfy the requirements. Option C, 3, is positive and contradicts the need for a negative solution. Option D, 11, is also positive and therefore incorrect. Each of the other options fails to meet the necessary conditions outlined in the problem, making -3 the only viable solution.
Yellow binders?
- A. 20
- B. 40
- C. 200
- D. 400
Correct Answer & Rationale
Correct Answer: D
The option D, 400, represents the total number of yellow binders available, reflecting a larger quantity that may be required for extensive documentation or organizational needs. Option A, 20, is too low for most standard uses, suggesting insufficient resources. Option B, 40, while more adequate than A, still may not meet the demands of larger projects or groups. Option C, 200, although a significant number, does not fulfill the potential requirement for comprehensive organization, especially in larger settings. Thus, option D ensures ample supply for diverse needs.
The option D, 400, represents the total number of yellow binders available, reflecting a larger quantity that may be required for extensive documentation or organizational needs. Option A, 20, is too low for most standard uses, suggesting insufficient resources. Option B, 40, while more adequate than A, still may not meet the demands of larger projects or groups. Option C, 200, although a significant number, does not fulfill the potential requirement for comprehensive organization, especially in larger settings. Thus, option D ensures ample supply for diverse needs.
Order 0.68, 1/12, 1(1/5), 3/5 least to greatest?
- A. 1(1/5), 0.68, 3/5, 1/12
- B. 1/12, 3/5, 0.68, 1(1/5)
- C. 1/12, 0.68, 3/5, 1(1/5)
- D. 0.68, 1/12, 3/5, 1(1/5)
Correct Answer & Rationale
Correct Answer: B
To compare the values, first convert them to a common format. - 1(1/5) equals 1.2. - 0.68 remains as is. - 3/5 converts to 0.6. - 1/12 is approximately 0.0833. Ordering these from least to greatest gives: 1/12 (0.0833), 3/5 (0.6), 0.68, and 1(1/5) (1.2). Option A incorrectly places 1(1/5) first, while C misplaces 3/5 and 0.68. Option D also misorders the values by placing 0.68 before 1/12. Thus, B accurately reflects the correct sequence of values.
To compare the values, first convert them to a common format. - 1(1/5) equals 1.2. - 0.68 remains as is. - 3/5 converts to 0.6. - 1/12 is approximately 0.0833. Ordering these from least to greatest gives: 1/12 (0.0833), 3/5 (0.6), 0.68, and 1(1/5) (1.2). Option A incorrectly places 1(1/5) first, while C misplaces 3/5 and 0.68. Option D also misorders the values by placing 0.68 before 1/12. Thus, B accurately reflects the correct sequence of values.
Liz spent 1/2, 1/3, 1/4, $15 left. Birthday money?
- A. $360
- B. $180
- C. $120
- D. $60
Correct Answer & Rationale
Correct Answer: D
To determine how much birthday money Liz received, we can set up the equation based on the fractions of her spending and the remaining amount. Let \( x \) represent the total birthday money. She spent \( \frac{1}{2}x + \frac{1}{3}x + \frac{1}{4}x + 15 = x \). Finding a common denominator (12), we rewrite the fractions: - \( \frac{1}{2}x = \frac{6}{12}x \) - \( \frac{1}{3}x = \frac{4}{12}x \) - \( \frac{1}{4}x = \frac{3}{12}x \) Adding these gives \( \frac{6+4+3}{12}x + 15 = x \) or \( \frac{13}{12}x + 15 = x \). Rearranging yields \( 15 = x - \frac{13}{12}x \), simplifying to \( 15 = \frac{1}{12}x \). Therefore, \( x = 180 \). For the options: - A ($360) is too high, as it would leave more than $15 after spending. - B ($180) results in no remaining amount after spending. - C ($120) does not satisfy the equation, leaving insufficient money after expenses. - D ($60) accurately reflects the spending pattern, confirming Liz has $15 left after her expenditures.
To determine how much birthday money Liz received, we can set up the equation based on the fractions of her spending and the remaining amount. Let \( x \) represent the total birthday money. She spent \( \frac{1}{2}x + \frac{1}{3}x + \frac{1}{4}x + 15 = x \). Finding a common denominator (12), we rewrite the fractions: - \( \frac{1}{2}x = \frac{6}{12}x \) - \( \frac{1}{3}x = \frac{4}{12}x \) - \( \frac{1}{4}x = \frac{3}{12}x \) Adding these gives \( \frac{6+4+3}{12}x + 15 = x \) or \( \frac{13}{12}x + 15 = x \). Rearranging yields \( 15 = x - \frac{13}{12}x \), simplifying to \( 15 = \frac{1}{12}x \). Therefore, \( x = 180 \). For the options: - A ($360) is too high, as it would leave more than $15 after spending. - B ($180) results in no remaining amount after spending. - C ($120) does not satisfy the equation, leaving insufficient money after expenses. - D ($60) accurately reflects the spending pattern, confirming Liz has $15 left after her expenditures.