Algebraic expressions? Select ALL.
- A. 2*(x+3)+4
- B. 4=x^2
- C. x=3y+7
- D. 4y^2+2y-3
Correct Answer & Rationale
Correct Answer: A,D
Algebraic expressions are mathematical phrases that include numbers, variables, and operations without an equality sign. Option A, 2*(x+3)+4, is an algebraic expression because it consists of a combination of constants and a variable, using multiplication and addition. Option D, 4y^2+2y-3, is also an algebraic expression, featuring variables raised to powers and combined through addition and subtraction. Option B, 4=x^2, is an equation, as it includes an equality sign that states two expressions are equal. Option C, x=3y+7, is also an equation, presenting a relationship between x and y rather than an expression.
Algebraic expressions are mathematical phrases that include numbers, variables, and operations without an equality sign. Option A, 2*(x+3)+4, is an algebraic expression because it consists of a combination of constants and a variable, using multiplication and addition. Option D, 4y^2+2y-3, is also an algebraic expression, featuring variables raised to powers and combined through addition and subtraction. Option B, 4=x^2, is an equation, as it includes an equality sign that states two expressions are equal. Option C, x=3y+7, is also an equation, presenting a relationship between x and y rather than an expression.
Other Related Questions
36 pencils in equal groups? Select THREE.
- A. 3
- B. 4
- C. 5
- D. 6
- E. 8
Correct Answer & Rationale
Correct Answer: A,B,D
To determine how many equal groups can be formed from 36 pencils, we need to identify the factors of 36. Option A (3) is valid because 36 ÷ 3 = 12, resulting in 12 pencils per group. Option B (4) is also correct since 36 ÷ 4 = 9, yielding 9 pencils per group. Option D (6) works as well, as 36 ÷ 6 = 6, giving 6 pencils per group. Options C (5) and E (8) are incorrect because 36 is not divisible by 5 (36 ÷ 5 = 7.2, which is not a whole number) and 8 (36 ÷ 8 = 4.5, also not a whole number). Thus, only 3, 4, and 6 are valid factors of 36.
To determine how many equal groups can be formed from 36 pencils, we need to identify the factors of 36. Option A (3) is valid because 36 ÷ 3 = 12, resulting in 12 pencils per group. Option B (4) is also correct since 36 ÷ 4 = 9, yielding 9 pencils per group. Option D (6) works as well, as 36 ÷ 6 = 6, giving 6 pencils per group. Options C (5) and E (8) are incorrect because 36 is not divisible by 5 (36 ÷ 5 = 7.2, which is not a whole number) and 8 (36 ÷ 8 = 4.5, also not a whole number). Thus, only 3, 4, and 6 are valid factors of 36.
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.
Eraser 20g in mg?
- A. 1.002
- B. 0.02
- C. 2,000
- D. 20
Correct Answer & Rationale
Correct Answer: D
To convert grams to milligrams, one must remember that 1 gram equals 1,000 milligrams. Therefore, 20 grams can be calculated as follows: 20 g x 1,000 mg/g = 20,000 mg. Option A (1.002 mg) is incorrect as it significantly underestimates the conversion. Option B (0.02 mg) is also wrong; it suggests a conversion error by not accounting for the unit scale correctly. Option C (2,000 mg) miscalculates the conversion by a factor of ten. Option D correctly represents 20 grams as 20,000 milligrams, aligning with the proper conversion calculation.
To convert grams to milligrams, one must remember that 1 gram equals 1,000 milligrams. Therefore, 20 grams can be calculated as follows: 20 g x 1,000 mg/g = 20,000 mg. Option A (1.002 mg) is incorrect as it significantly underestimates the conversion. Option B (0.02 mg) is also wrong; it suggests a conversion error by not accounting for the unit scale correctly. Option C (2,000 mg) miscalculates the conversion by a factor of ten. Option D correctly represents 20 grams as 20,000 milligrams, aligning with the proper conversion calculation.
3/4 as sum of unit fractions?
- A. 1/8 + 1/8 + 1/8 + 1/4 + 1/4
- B. 2/8 + 1/4 + 4/16
- C. 5/8 + 2/16
- D. 1/2 + 1/4
Correct Answer & Rationale
Correct Answer: D
To express \( \frac{3}{4} \) as a sum of unit fractions, each option must be evaluated for its total. Option A totals \( \frac{3}{8} + \frac{1}{2} = \frac{3}{8} + \frac{4}{8} = \frac{7}{8} \), which exceeds \( \frac{3}{4} \). Option B simplifies to \( \frac{2}{8} + \frac{2}{8} + \frac{1}{4} = \frac{2}{8} + \frac{2}{8} + \frac{2}{8} = \frac{6}{8} = \frac{3}{4} \), but includes non-unit fractions. Option C simplifies to \( \frac{5}{8} + \frac{1}{4} = \frac{5}{8} + \frac{2}{8} = \frac{7}{8} \), again exceeding \( \frac{3}{4} \). Option D correctly adds \( \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \) using unit fractions only.
To express \( \frac{3}{4} \) as a sum of unit fractions, each option must be evaluated for its total. Option A totals \( \frac{3}{8} + \frac{1}{2} = \frac{3}{8} + \frac{4}{8} = \frac{7}{8} \), which exceeds \( \frac{3}{4} \). Option B simplifies to \( \frac{2}{8} + \frac{2}{8} + \frac{1}{4} = \frac{2}{8} + \frac{2}{8} + \frac{2}{8} = \frac{6}{8} = \frac{3}{4} \), but includes non-unit fractions. Option C simplifies to \( \frac{5}{8} + \frac{1}{4} = \frac{5}{8} + \frac{2}{8} = \frac{7}{8} \), again exceeding \( \frac{3}{4} \). Option D correctly adds \( \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} \) using unit fractions only.