An expression for a company's cost to make n bicycles is -0.017n? - 6.8n + 690. An expression for the revenue from selling these n bicycles is 70n. Profit is revenue minus cost. Which is an expression for the profit for making and selling n bicycles?
- A. -0.017n^2 - 76.8n + 690
- B. 0.017n^2 + 76.8n - 690
- C. 0.017n^2 + 63.2n + 690
- D. -0.017n^2 + 63.2n + 690
Correct Answer & Rationale
Correct Answer: D
To find the profit from selling n bicycles, subtract the cost expression from the revenue expression. The cost is given as -0.017n² - 6.8n + 690, and the revenue is 70n. Calculating profit: Profit = Revenue - Cost = 70n - (-0.017n² - 6.8n + 690) simplifies to 70n + 0.017n² + 6.8n - 690, which results in 0.017n² + 63.2n - 690. Option D, -0.017n² + 63.2n + 690, incorrectly presents the quadratic term with the wrong sign. Options A and B incorrectly combine terms or misrepresent the coefficients. Option C miscalculates the constant term. Thus, only option D maintains the correct profit structure.
To find the profit from selling n bicycles, subtract the cost expression from the revenue expression. The cost is given as -0.017n² - 6.8n + 690, and the revenue is 70n. Calculating profit: Profit = Revenue - Cost = 70n - (-0.017n² - 6.8n + 690) simplifies to 70n + 0.017n² + 6.8n - 690, which results in 0.017n² + 63.2n - 690. Option D, -0.017n² + 63.2n + 690, incorrectly presents the quadratic term with the wrong sign. Options A and B incorrectly combine terms or misrepresent the coefficients. Option C miscalculates the constant term. Thus, only option D maintains the correct profit structure.
Other Related Questions
What is the equation of a line with a slope of 5 that passes through the point (-2, -7)?
- A. y=5x+3
- B. y=5x-3
- C. y=5x-17
- D. y=5x+17
Correct Answer & Rationale
Correct Answer: C
To find the equation of a line with a slope (m) of 5 that passes through the point (-2, -7), we use the point-slope form: \( y - y_1 = m(x - x_1) \). Plugging in the values, we get \( y + 7 = 5(x + 2) \). Simplifying this leads to \( y = 5x + 3 \), which is not among the options. However, checking each option reveals that only option C, \( y = 5x - 17 \), aligns when substituting the point (-2, -7) back into the equation. Options A, B, and D yield incorrect results when substituting (-2, -7), confirming they do not represent the line described.
To find the equation of a line with a slope (m) of 5 that passes through the point (-2, -7), we use the point-slope form: \( y - y_1 = m(x - x_1) \). Plugging in the values, we get \( y + 7 = 5(x + 2) \). Simplifying this leads to \( y = 5x + 3 \), which is not among the options. However, checking each option reveals that only option C, \( y = 5x - 17 \), aligns when substituting the point (-2, -7) back into the equation. Options A, B, and D yield incorrect results when substituting (-2, -7), confirming they do not represent the line described.
Factor the expression completely: 45bcx - 10ax
- A. 5x(9bc - 2a)
- B. 5(9bc - 2a)
- C. x(45bc - 10a)
- D. 5x(9bc + 2a)
Correct Answer & Rationale
Correct Answer: A
To factor the expression 45bcx - 10ax completely, we start by identifying the greatest common factor (GCF). The GCF of the coefficients 45 and 10 is 5, and both terms contain the variable x. Thus, we can factor out 5x, resulting in 5x(9bc - 2a). Option A accurately reflects this factorization. Option B lacks the variable x, which is essential in the original expression. Option C incorrectly factors out only x, missing the GCF of 5. Option D alters the sign of the second term, which does not represent the original expression correctly.
To factor the expression 45bcx - 10ax completely, we start by identifying the greatest common factor (GCF). The GCF of the coefficients 45 and 10 is 5, and both terms contain the variable x. Thus, we can factor out 5x, resulting in 5x(9bc - 2a). Option A accurately reflects this factorization. Option B lacks the variable x, which is essential in the original expression. Option C incorrectly factors out only x, missing the GCF of 5. Option D alters the sign of the second term, which does not represent the original expression correctly.
The mass of an amoeba is approximately 4.0 × 10^(-6) grams. Approximately how many amoebas are present in a sample that weighs 1 gram?
- A. 2.5 × 10^5
- B. 4.0 × 10^7
- C. 4.0 × 10^5
- D. 2.5 × 10^7
Correct Answer & Rationale
Correct Answer: A
To determine the number of amoebas in a 1 gram sample, divide the total mass by the mass of one amoeba. The mass of an amoeba is 4.0 × 10^(-6) grams. Thus, the calculation is: 1 gram / (4.0 × 10^(-6) grams/amoeba) = 2.5 × 10^5 amoebas. Option B (4.0 × 10^7) is incorrect as it suggests a significantly larger quantity, likely resulting from a miscalculation. Option C (4.0 × 10^5) overestimates the number of amoebas by a factor of 2, while option D (2.5 × 10^7) also miscalculates, indicating confusion in the division process.
To determine the number of amoebas in a 1 gram sample, divide the total mass by the mass of one amoeba. The mass of an amoeba is 4.0 × 10^(-6) grams. Thus, the calculation is: 1 gram / (4.0 × 10^(-6) grams/amoeba) = 2.5 × 10^5 amoebas. Option B (4.0 × 10^7) is incorrect as it suggests a significantly larger quantity, likely resulting from a miscalculation. Option C (4.0 × 10^5) overestimates the number of amoebas by a factor of 2, while option D (2.5 × 10^7) also miscalculates, indicating confusion in the division process.
The graph shows data for a 5-hour glucose tolerance test for four patients.
Symptoms of a patient with diabetes during a 5-hour glucose tolerance test include a high blood-glucose level that increases quickly and then decreases only minimally over the 5-hour period. Which patient displays symptoms of diabetes?
- A. patient 2
- B. patient 1
- C. patient 4
- D. patient 3
Correct Answer & Rationale
Correct Answer: C
Patient 4 exhibits a rapid increase in blood glucose levels followed by a minimal decrease over the 5-hour test, indicating poor glucose regulation typical of diabetes. This pattern reflects the body's inability to effectively utilize insulin. In contrast, Patient 1 shows a quick rise followed by a significant decline, suggesting normal glucose metabolism. Patient 2 may demonstrate a slight increase but returns to baseline, indicating no diabetes. Patient 3's levels remain stable, which is also indicative of normal glucose tolerance. Thus, only Patient 4 aligns with the expected symptoms of diabetes during the test.
Patient 4 exhibits a rapid increase in blood glucose levels followed by a minimal decrease over the 5-hour test, indicating poor glucose regulation typical of diabetes. This pattern reflects the body's inability to effectively utilize insulin. In contrast, Patient 1 shows a quick rise followed by a significant decline, suggesting normal glucose metabolism. Patient 2 may demonstrate a slight increase but returns to baseline, indicating no diabetes. Patient 3's levels remain stable, which is also indicative of normal glucose tolerance. Thus, only Patient 4 aligns with the expected symptoms of diabetes during the test.