Answer:
2094.5 muffins
Step-by-step explanation:
The problem is a little bit messy, but i’m guessing you meant that the elf ate 35% of the muffins, which 35% is 710 muffins. so since 35% of muffins is 710 muffins, let’s multiply that by 2 so we can get 70%.
so 710 x 2 is 1420. so 70% of the muffins is 1420 muffins! sadly that’s not all of the muffins. we’re still missing another 30%. since we don’t know how many muffins are in 30%, let’s just take 710 muffins - 5%. when you do that you get 674.5. so now that we know how much is in 30%, let’s take 1420 + 674.5. when you do that you get 2094.5! so you started off with 2094 and a half muffins!
The domain is the set of allowed inputs, in this case t values. The smallest t value allowed is t = 0. The largest is t = 165. So that's why the domain is
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The range is
since H = 0 is the smallest output of the function and H = 40,000 is the largest output. Like the domain, the range is the set of possible outputs of a function.
He worked on the car for 4 hours. 35+7+7+7+7=63.
1) The average increase in the level of CO2 emissions per year from years 2 to 4 is:
Average=[f(4)-f(2)]/(4-2)=(29,172.15-26,460)/2=2,712.15/2=1,356.075 metric tons. The first is false.
2) The average increase in the level of CO2 emissions per year from years 6 to 8 is:
Average=[f(8)-f(6)]/(8-6)=(35,458.93-32,162.29)/2=3,296.64/2=1,648.32 metric tons. The second is false.
3) The average increase in the level of CO2 emissions per year from years 4 to 6 is:
Average=[f(6)-f(4)]/(6-4)=(32,162.29-29,172.15)/2=2,990.14/2=1,495.07 metric tons. The third is false.
4) The average increase in the level of CO2 emissions per year from years 8 to 10 is:
Average=[f(10)-f(8)]/(10-8)=(39,093.47-35,458.93)/2=3,634.54/2=1,817.27 metric tons. The fourth is true.
Answer: Fourth option: The average increase in the level of CO2 emissions per year from years 8 to 10 is 1,817.27 metric tons.
