Floor 80 sq ft, 4 sq tiles, 5 rect tiles. 4S+5R=80
Cover floor? Select ALL.
- A. 15s4r
- B. 8s10r
- C. 5s12r
Correct Answer & Rationale
Correct Answer: A,C
To determine which options cover the floor effectively, we analyze the dimensions given. Option A (15s4r) indicates a larger area, suggesting it can cover more floor space due to its higher values. This makes it suitable for extensive coverage. Option B (8s10r) has moderate dimensions but does not provide sufficient area to cover larger floors, making it less effective compared to A and C. Option C (5s12r) also presents a viable coverage area, complementing A's larger dimensions. Thus, A and C collectively ensure adequate floor coverage, while B falls short.
To determine which options cover the floor effectively, we analyze the dimensions given. Option A (15s4r) indicates a larger area, suggesting it can cover more floor space due to its higher values. This makes it suitable for extensive coverage. Option B (8s10r) has moderate dimensions but does not provide sufficient area to cover larger floors, making it less effective compared to A and C. Option C (5s12r) also presents a viable coverage area, complementing A's larger dimensions. Thus, A and C collectively ensure adequate floor coverage, while B falls short.
Other Related Questions
Joe’s age 4 more than 3x Amy’s. Equation?
- A. A=J/3+4
- B. A=3J+4
- C. J=3A+4
- D. J=3(A+4)
Correct Answer & Rationale
Correct Answer: C
To find the equation representing Joe's age in relation to Amy's, we start with the statement: Joe's age (J) is 4 more than 3 times Amy's age (A). This can be expressed mathematically as J = 3A + 4, which aligns with option C. Option A (A = J/3 + 4) incorrectly suggests that Amy's age is derived from Joe's, which contradicts the relationship given. Option B (A = 3J + 4) misplaces the variables, implying Amy's age is dependent on Joe's in a way that doesn't reflect the original statement. Option D (J = 3(A + 4)) incorrectly adds 4 to Amy's age before multiplying, altering the intended relationship.
To find the equation representing Joe's age in relation to Amy's, we start with the statement: Joe's age (J) is 4 more than 3 times Amy's age (A). This can be expressed mathematically as J = 3A + 4, which aligns with option C. Option A (A = J/3 + 4) incorrectly suggests that Amy's age is derived from Joe's, which contradicts the relationship given. Option B (A = 3J + 4) misplaces the variables, implying Amy's age is dependent on Joe's in a way that doesn't reflect the original statement. Option D (J = 3(A + 4)) incorrectly adds 4 to Amy's age before multiplying, altering the intended relationship.
Associative operations? Select ALL.
- A. Addition
- B. Subtraction
- C. Multiplication
- D. Division
- E. Exponentiation
Correct Answer & Rationale
Correct Answer: A,C
Associative operations allow the grouping of numbers in different ways without changing the result. Addition (A) and multiplication (C) are associative; for example, (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c). Subtraction (B) and division (D) are not associative; changing the grouping alters the result, such as in (a - b) - c ≠ a - (b - c) and (a ÷ b) ÷ c ≠ a ÷ (b ÷ c). Exponentiation (E) is also not associative, as (a^b)^c ≠ a^(b^c). Thus, only addition and multiplication qualify as associative operations.
Associative operations allow the grouping of numbers in different ways without changing the result. Addition (A) and multiplication (C) are associative; for example, (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c). Subtraction (B) and division (D) are not associative; changing the grouping alters the result, such as in (a - b) - c ≠ a - (b - c) and (a ÷ b) ÷ c ≠ a ÷ (b ÷ c). Exponentiation (E) is also not associative, as (a^b)^c ≠ a^(b^c). Thus, only addition and multiplication qualify as associative operations.
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.
Leslie descended 714 ft in 34 s, took 1 min 25 s to ground. Total distance?
- A. 1,270 feet
- B. 1,515 feet
- C. 1,785 feet
- D. 2,615 feet
Correct Answer & Rationale
Correct Answer: C
To determine the total distance Leslie descended, first convert the time taken to ground into seconds: 1 minute and 25 seconds equals 85 seconds. The total descent includes both the initial 714 feet and the additional distance covered during the 85 seconds. Using the average speed from the initial descent (714 ft in 34 s), we find the speed: 714 ft / 34 s ≈ 21 ft/s. Over 85 seconds, Leslie would descend approximately 21 ft/s × 85 s = 1,785 feet total. Option A (1,270 ft) underestimates the descent. Option B (1,515 ft) is also too low. Option D (2,615 ft) overestimates the total distance. Thus, C (1,785 ft) accurately reflects the total descent.
To determine the total distance Leslie descended, first convert the time taken to ground into seconds: 1 minute and 25 seconds equals 85 seconds. The total descent includes both the initial 714 feet and the additional distance covered during the 85 seconds. Using the average speed from the initial descent (714 ft in 34 s), we find the speed: 714 ft / 34 s ≈ 21 ft/s. Over 85 seconds, Leslie would descend approximately 21 ft/s × 85 s = 1,785 feet total. Option A (1,270 ft) underestimates the descent. Option B (1,515 ft) is also too low. Option D (2,615 ft) overestimates the total distance. Thus, C (1,785 ft) accurately reflects the total descent.