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Naya [18.7K]
3 years ago
5

A driver intends to complete 2 laps on a track that is exactly 1 mile around at an average speed of 60 mph. He gets off to a slo

w start and completes the 1st lap at an average of 30 mph. How fast must he drive the second lap in order to achieve his goal?
Mathematics
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:

Goal is not achievable

Step-by-step explanation:

There are Two laps and the total distance here is 2miles. Now, it is said that he hopes to travel at a speed of 60mph, thus the total time that it would take him to complete the 2 miles distance race is 2/60 = 1/30 hours. Thus if he travels at a constant speed of 60mph he would finish the race at 1/30hours.

This means he will take a time of 1/60 hours per lap.

Now, he travels 30mph to complete 1 mile, the time taken for this is 1/30h

We now need to know the speed he travels on the second lap.

Average speed = Total distance/Total time

60mph = 2/( 1/30 + T2)

60 = 2/( (1 + 30T2)/30)

60 = 2 divided by (1 + 30T2)/30

60 = 2 * (30)/1 + 30T2

60( 1 + 30T2) = 60

1 + 30T2 = 1

30T2 = 0

T2 = 0

This means that he cannot achieve his goal again as he had taken the time meant for the whole race in a single lap.

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Given the function h(x) =1/3 |x-6| +4, evaluate the function when x = - 3, - 2, and 0
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|x| = x for x ≥ 0

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--------------------------------------------------------------------------------

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

--------------------------------------------------------------------------------

h(x)=\dfrac{1}{3}|x-6|+4

Put the values of x to the equation of the function h(x):

x=-3\to h(-3)=\dfrac{1}{3}|-3-6|+4=\dfrac{1}{3}|-9|+4=\dfrac{1}{3}(9)+4=3+4=7\\\\x=-2\to h(-2)=\dfrac{1}{3}|-2-6|+4=\dfrac{1}{3}|-8|+4=\dfrac{1}{3}(8)+4=\dfrac{8}{3}+\dfrac{12}{3}=\dfrac{20}{3}\\\\x=0\to h(0)=\dfrac{1}{3}|0-6|+4=\dfrac{1}{3}|-6|+4=\dfrac{1}{3}(6)+4=2+4=6


6 0
3 years ago
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