The world's highest suspension bridge spans the Arkansas River at a height of 1,053 feet above the water. If a ball is dropped from the bridge. The height of the ball, In feet, after t seconds can be modeled by the equation f(t)= -16(t)^2 + 1053. How many feet above the water is the ball 7 seconds after being dropped?
Correct Answer & Rationale
Correct Answer: A
To determine the height of the ball 7 seconds after being dropped, substitute \( t = 7 \) into the equation \( f(t) = -16(t)^2 + 1053 \). Calculating this gives \( f(7) = -16(7)^2 + 1053 = -16(49) + 1053 = -784 + 1053 = 269 \) feet. Option A provides this correct height of 269 feet. Other options are incorrect because they result from miscalculations or incorrect substitutions into the equation. For example, using an incorrect value for \( t \) or failing to properly apply the formula leads to heights that do not reflect the physics of the scenario.
To determine the height of the ball 7 seconds after being dropped, substitute \( t = 7 \) into the equation \( f(t) = -16(t)^2 + 1053 \). Calculating this gives \( f(7) = -16(7)^2 + 1053 = -16(49) + 1053 = -784 + 1053 = 269 \) feet. Option A provides this correct height of 269 feet. Other options are incorrect because they result from miscalculations or incorrect substitutions into the equation. For example, using an incorrect value for \( t \) or failing to properly apply the formula leads to heights that do not reflect the physics of the scenario.
Other Related Questions
Acceleration, a, in meters per second squared (m/s^2), is found by the formula a = (V2 - V1)/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 using the formula \( a = \frac{(V2 - V1)}{t} \), substitute the values: \( V1 = 14 \, \text{m/s} \), \( V2 = 8 \, \text{m/s} \), and \( t = 4 \, \text{s} \). This results in \( a = \frac{(8 - 14)}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference and time. Option D (-12) incorrectly computes the acceleration by misapplying the formula or misinterpreting the values. Thus, the only accurate calculation reflects a deceleration, resulting in -1.5 m/s².
To find acceleration using the formula \( a = \frac{(V2 - V1)}{t} \), substitute the values: \( V1 = 14 \, \text{m/s} \), \( V2 = 8 \, \text{m/s} \), and \( t = 4 \, \text{s} \). This results in \( a = \frac{(8 - 14)}{4} = \frac{-6}{4} = -1.5 \, \text{m/s}^2 \). Option A (1.5) is incorrect as it does not account for the decrease in velocity. Option C (4.5) miscalculates the difference and time. Option D (-12) incorrectly computes the acceleration by misapplying the formula or misinterpreting the values. Thus, the only accurate calculation reflects a deceleration, resulting in -1.5 m/s².
A shipping box for a refrigerator is shaped like a rectangular prism. The box has a depth of 34.25 inches (in.), a height of 69.37 in., and a width of 32.62 in. To the nearest hundredth cubic inch, what is the volume of the shipping bax?
- A. 2,262.85
- B. 77,502.59
- C. 136.24
- D. 25,834.20
Correct Answer & Rationale
Correct Answer: B
To determine the volume of a rectangular prism, the formula \( V = \text{length} \times \text{width} \times \text{height} \) is applied. Given the dimensions—depth (length) of 34.25 in., width of 32.62 in., and height of 69.37 in.—the calculation yields a volume of approximately 77,502.59 cubic inches. Option A (2,262.85) is far too small, indicating a miscalculation. Option C (136.24) is implausibly low, likely resulting from using incorrect units or dimensions. Option D (25,834.20) is also incorrect, as it does not reflect the correct multiplication of the given dimensions. Thus, only option B accurately represents the calculated volume.
To determine the volume of a rectangular prism, the formula \( V = \text{length} \times \text{width} \times \text{height} \) is applied. Given the dimensions—depth (length) of 34.25 in., width of 32.62 in., and height of 69.37 in.—the calculation yields a volume of approximately 77,502.59 cubic inches. Option A (2,262.85) is far too small, indicating a miscalculation. Option C (136.24) is implausibly low, likely resulting from using incorrect units or dimensions. Option D (25,834.20) is also incorrect, as it does not reflect the correct multiplication of the given dimensions. Thus, only option B accurately represents the calculated volume.
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.
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.