S=4P+4Q
S=4(P+Q)
S/4=P+Q
S/4-P=Q
Q=S/4-P
Answer:
Tyler is correct. The temperature dropped at a rate of about 4° per hour between 4 and 6, while the temperature dropped at about 2.25° per hour between 6 and 10.
Edit: Explanation
The question is asking about which window of time had a <em>faster</em> decline in temperature, not a larger total change in temperature.
In a 2 hour timeframe, the temperature dropped 8°. (4-6 PM)
In a separate 4 hour timeframe, the temperature dropped 9°. (6-10 PM)
To find which window had a faster change in temp, I took the total temperature drop for each timeframe, then divided it by the number of hours each drop took.
8° / 2 = 4° per hour for 4-6 PM
9° / 4 = 2.25° per hour from 6-10 PM
Since the speed at which the temperature dropped per hour was greater from 4-6 PM than 6-10 PM, Tyler was correct.
Answer:
The answer is A....im pretty sure
Step-by-step explanation:
Answer:
38 + 14 = 52
52 x 16 = 832
832 divided by 2 = 416
Formula of a trapezoid:
B1(base 1) + B2(base 2) x H(height) x 1/2(basically just dividing by 2)
