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.
Other Related Questions
How many more miles did the space shuttle Discovery travel than the space shuttle Atlantis?
- A. 274,100,000 miles
- B. 274,100 miles
- C. 22.3 miles
- D. 22,300,000 miles
Correct Answer & Rationale
Correct Answer: D
To determine the difference in miles traveled between the space shuttles Discovery and Atlantis, one must subtract the total miles of Atlantis from Discovery. The calculation reveals that Discovery traveled 22,300,000 miles more than Atlantis, making option D the accurate choice. Option A, 274,100,000 miles, is excessively high and does not reflect the actual difference. Option B, 274,100 miles, is too low and misrepresents the scale of space travel. Option C, 22.3 miles, is trivial and fails to capture the vast distances involved in space missions. Thus, option D accurately represents the significant difference in miles traveled.
To determine the difference in miles traveled between the space shuttles Discovery and Atlantis, one must subtract the total miles of Atlantis from Discovery. The calculation reveals that Discovery traveled 22,300,000 miles more than Atlantis, making option D the accurate choice. Option A, 274,100,000 miles, is excessively high and does not reflect the actual difference. Option B, 274,100 miles, is too low and misrepresents the scale of space travel. Option C, 22.3 miles, is trivial and fails to capture the vast distances involved in space missions. Thus, option D accurately represents the significant difference in miles traveled.
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.
Lisa is decorating her office with two fully stocked aquariums. She saw an advertisement for Jorge's pet store in the newspaper. Jorge's store sells fish for aquariums. The table shows the fish Lisa buys from Jorge's pet store.
Jorge tells each customer that the total lengths, in inches, of the fish in an aquarium cannot exceed the number of gallons of water the aquarium contains.
The newspaper advertisement for Jorge's pet store has an illustration of a gold barb.
The illustration is not the same length as the actual gold barb. What was the scale factor used to create the illustration?
- A. 0.75
- B. 1.25
- C. 1.75
- D. 1.75
Correct Answer & Rationale
Correct Answer: B
To determine the scale factor used in the illustration of the gold barb, we compare the actual length of the fish to the length shown in the advertisement. A scale factor greater than 1 indicates that the illustration is larger than the actual fish, while a scale factor less than 1 means it is smaller. Option A (0.75) suggests the illustration is smaller, which contradicts the premise. Option C (1.75) and D (1.75) both imply a larger size, but only one option can be correct. The scale factor of 1.25 accurately represents a reasonable enlargement of the fish, aligning with common advertising practices. Thus, it correctly reflects the relationship between the illustration and the actual size of the gold barb.
To determine the scale factor used in the illustration of the gold barb, we compare the actual length of the fish to the length shown in the advertisement. A scale factor greater than 1 indicates that the illustration is larger than the actual fish, while a scale factor less than 1 means it is smaller. Option A (0.75) suggests the illustration is smaller, which contradicts the premise. Option C (1.75) and D (1.75) both imply a larger size, but only one option can be correct. The scale factor of 1.25 accurately represents a reasonable enlargement of the fish, aligning with common advertising practices. Thus, it correctly reflects the relationship between the illustration and the actual size of the gold barb.
Acceleration, a, in meters per second squared (m/5}), is found by the formula a= (V2-V2)/t where V1, is the beginning velocity, V2 is the end velocity, and t is time. What is the acceleration, in m/s^2, of an object with a beginning velocity of 14 m/s and end velocity of 8 m/s over a time of 4 seconds?
- A. 1.5
- B. -1.5
- C. 4.5
- D. -12
Correct Answer & Rationale
Correct Answer: B
To find acceleration, use the formula \( a = \frac{V2 - V1}{t} \). Here, \( V1 = 14 \, \text{m/s} \) and \( V2 = 8 \, \text{m/s} \). Plugging in the values gives \( a = \frac{8 - 14}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference between velocities and does not reflect the negative change. Option D (-12) results from incorrect arithmetic, misapplying the formula. Thus, the only accurate calculation shows the object is decelerating at -1.5 m/s².
To find acceleration, use the formula \( a = \frac{V2 - V1}{t} \). Here, \( V1 = 14 \, \text{m/s} \) and \( V2 = 8 \, \text{m/s} \). Plugging in the values gives \( a = \frac{8 - 14}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference between velocities and does not reflect the negative change. Option D (-12) results from incorrect arithmetic, misapplying the formula. Thus, the only accurate calculation shows the object is decelerating at -1.5 m/s².