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3 years ago

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The distance from New York City to Los Angeles is 4090 kilometers. a. [3 pts]What is the distance in miles? (You must use unit f

ractions. Round to the nearest mile and be sure to include units.) b. [3 pts]If your car averages 31 miles per gallon, how many gallons of gas can you expect to use driving from New York to Los Angeles? (You must use unit fractions. Round to one decimal place and be sure to include units.)

1 answer:

sergejj [24]3 years ago

6 0

Answer: a) 2556.25 miles

b) 82.5 gallons

Step-by-step explanation:

Given : The distance from New York City to Los Angeles is 4090 kilometers.

$=4090\times\dfrac{5}{8}=2556.25\text{ miles}$ [∵ 1 km= $\dfrac{5}{8}$ miles]

Hence, the distance in miles= 2556.25

If your car averages 31 miles per gallon, then the number of gallons of gas required to use driving from New York to Los Angeles = $\dfrac{\text{Distance}}{31}$

$=\dfrac{2556.25}{31}=82.4596774194\approx82.5\text{gallons}$

Hence, the required number of gallons = 82.5 gallons

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dmitriy555 [2]

$f(x)$ will be continuous at $x=a=7$ if
(i) $\displaystyle\lim_{x\to7}f(x)$ exists,
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The second condition is immediate, since $f(7)=8918$ has a finite value. The other two conditions can be established by proving that the limit of the function as $x\to7$ is indeed the value of $f(7)$ . That is, we must prove that for any $\varepsilon>0$ , we can find $\delta>0$ such that

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$|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|$

This is the quantity that we do not want exceeding $\varepsilon$ . Suppose we focus our attention on small values $\delta$ . For instance, say we restrict $\delta$ to be no larger than 1, i.e. $\delta\le1$ . Under this condition, we have

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Now, by the triangle inequality,

$|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275$

If $|x|$ , then this quantity is moreover bounded such that

$|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955$

To recap, fixing $\delta\le1$ would force $|x|$ , which makes

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and we want this quantity to be smaller than $\varepsilon$ , so

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which suggests that we could set $\delta=\dfrac{\varepsilon}{6955}$ . But if $\varepsilon$ is given such that the above inequality fails for $\delta=\dfrac{\varepsilon}{6955}$ , then we can always fall back on $\delta=1$ , for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set $\delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}$ .

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that $f(x)$ is indeed continuous at $x=7$ .

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