A) An inequality to represent the temperatures the pilot can take off and land in is .
B) The graph is shown below, it shows the temperature range in which the aircraft can takeoff or land.
C) The pilot could not have taken off on June 1990, the temperature in Phoenix, Arizona as the temperature is above the operating limitations.
Let the operating temperature be .
The aircraft cannot operate if the temperature is at or below ° F, or at or above ° F.
So, the inequality will be .
In the graph, the shaded region represents the temperature in which the temperatures the pilot can take off and land in. Here, we can see open circles at the end points as they are not included in the interval.
The temperature cannot be at or above 118° F. So, the pilot could not have taken off on June 1990, the temperature in Phoenix, Arizona as the temperature is above the operating limitations.
Answer: 0.88 should be the answer
H(t) = -16t² + 60t + 95
g(t) = 20 + 38.7t
h(1) = -16(1²) + 60(1) + 95 = -16 + 60 + 95 = -16 + 155 = 139
h(2) = -16(2²) + 60(2) + 95 = -16(4) + 120 + 95 = -64 + 215 = 151
h(3) = -16(3²) + 60(3) + 95 = -16(9) + 180 + 95 = -144 + 275 = 131
h(4) = -16(4²) + 60(4) + 95 = -16(16) + 240 + 95 = -256 + 335 = 79
g(1) = 20 + 38.7(1) = 20 + 38.7 = 58.7
g(2) = 20 + 38.7(2) = 20 + 77.4 = 97.4
g(3) = 20 + 38.7(3) = 20 + 116.1 = 136.1
g(4) = 20 + 38.7(4) = 20 + 154.8 = 174.8
Between 2 and 3 seconds.
The range of the 1st object is 151 to 131.
The range of the 2nd object is 97.4 to 136.1
h(t) = g(t) ⇒ 131 = 131
It means that the point where the 2 objects are equal is the point where the 1st object is falling down while the 2nd object is still going up.
Answer:
1. True
2. False
3. False
4. False
Step-by-step explanation:
the length of CD+BC make sense because if u put those lines together it would equal BD, the others are wrong because the lengths don't add up. Please source me in your answer