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Vikentia [17]
3 years ago
6

If a truck averages 12 miles to the gallon, how many pounds of carbon dioxide are emitted into the atmosphere when the truck tra

vels 170 miles?
Mathematics
1 answer:
77julia77 [94]3 years ago
3 0

Answer: 278.30 pounds (if the gasoline does not contain ethanol)

317.12 pounds (if the gasoline does contain ethanol)

Step-by-step explanation:

If the truck averages 12 miles per gallon, then in 170 miles it consumes:

170mi/12mi = 14.17 gallons.

We know that for one gallon consumed, the carbon dioxide emitted into the atmosphere is 19.64 pounds. (assuming it does not contain ethanol)

Then for 14.17 gallons consumed, we have a emission of:

14.17*19.64 pounds = 278.30 pounds

I the gasoline contains ethanlol, for a gallon the emiision is around  22.38 pounds.

In this case the total emission is:

14.17*22.38 pounds = 317.12 pounds

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Part A

The notation \lim_{x \to 2^{+}}f(x) means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

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Plug in x = 2 to find that...

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Then for the left hand limit \lim_{x \to 2^{-}}f(x), we'll involve x < 2 and we go for the first piece. So,

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===============================================================

Part B

Because \lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x) this means that the limit \lim_{x \to 2}f(x) does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since \lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3, this means the limit \lim_{x \to 4}f(x) does exist and it's equal to 3.

As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.

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