By inspection, the equation for the horizontal asymptote is y = -3.
Considering the described changes, the temperature at 6 pm was of -14.5ºF.
<h3>How to find the temperature at 6 pm?</h3>
We look at the temperature for each interval, as follows:
- At 9 am, the temperature was -8.4°F.
- Between 9am and noon, the temperature increased by 2.6°F, hence, at noon, the temperature was of -8.4ºF + 2.6ºF = -5.8ºF.
- Between noon and 3pm, the temperature rose 1.2°F, hence, at 3 pm, the temperature was of -5.8ºF + 1.2ºF = -4.6ºF.
Between 3pm and 6pm, the temperature went down 9.9°F, hence at 6 pm, the temperature was given as follows:
-4.6ºF - 9.9ºF = -14.5ºF.
The temperature at 6 pm was of -14.5ºF.
A similar problem, involving change of temperatures, is given at brainly.com/question/3066572
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Answer:
B: 6 units
Step-by-step explanation:
just took the test
I don't know what the first two blanks are asking for, but the last one should be Alternate Exterior Angles Property of Congruence (AEAPC).
