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RSB [31]
3 years ago
12

a truck is dispatched for delivery at 8:30 am to a customer 50 miles away. assume that the average speed of the truck is 25 mile

s per hour, and it takes 30 minutes to unload the truck. at what time can you expect the truck back?
Mathematics
1 answer:
igomit [66]3 years ago
6 0
1ndidhdbdunenwkwkjw dbhsbejdndjdjdndbhd
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198.4 ounches is the real answer

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Which number is farthest from 0? 8, -3, -12, 5
Vaselesa [24]

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Dahasolnce [82]
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8 0
3 years ago
Use the properties of limits to help decide whether the limit exists. If the limit​ exists, find its value. ModifyingBelow lim W
Sergeu [11.5K]

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The value of given limit problem is 0.

Step-by-step explanation:

The given limit problem is

lim_{x\rightarrow \infty}\dfrac{6x^3+5x-7}{6x^4-4x^3-9}

We need to find the value of given limit problem.

Divide the numerator and denominator by the leading term of the denominator, i.e., x^4

lim_{x\rightarrow \infty}\dfrac{\frac{6x^3+5x-7}{x^4}}{\frac{6x^4-4x^3-9}{x^4}}

lim_{x\rightarrow \infty}\dfrac{\frac{6}{x}+\frac{5}{x^3}-\frac{7}{x^4}}{6-\frac{4}{x}-\frac{9}{x^4}}

Apply limit.

\dfrac{\frac{6}{ \infty}+\frac{5}{ \infty}-\frac{7}{ \infty}}{6-\frac{4}{ \infty}-\frac{9}{ \infty}}

We know that \frac{1}{\infty}=0.

\dfrac{0+0-0}{6-0-0}

\dfrac{0}{6}

0

Hence, the value of given limit is 0.

8 0
3 years ago
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