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olga2289 [7]
4 years ago
13

On an indoor circular track of circumference 50 feet, Joneal starts at point $S$, runs in a counterclockwise direction, and then

stops when he has run exactly one mile (5280 feet). On which quarter of the circle, $A$, $B$, $C$ or $D$, did Joneal stop
Mathematics
1 answer:
Archy [21]4 years ago
4 0

Answer:

He stoped on 3th quarter,i.e, $C$.

Step-by-step explanation:

He ran 105 full circles ( 5280/50=105 ( rst= 30ft) ). So in the last circles he started from point S to run 30ft more.

The quarter of the circle is long 50ft/4= 12,5ft. So for 30 feets he must run 2 quarters, its 25ft. The last 5ft he ran on the 3th quarter, so he stoped on C.

This answer $C$, if $C$ is the 3th quadrant, i don't see the picture of the track.

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A $16,000 robot depreciates linearly to zero in 10 years. (a) find a formula for its value as a function of time, t, in years.
SVETLANKA909090 [29]

We are given, cost of the robot for 0 number of year = $16,000.

0 represents initial time of the robot.

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The problem is about the number of the years and cost of the robot over different number of years.

So, we could take x coordinate by number of hours and y coordinate for y number of hours.

So, from the problem, we could make two coordinates for the given situation.

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In order to find the function of time, we need to find the rate at which robot rate depreciates each year.

Slope is the rate of change.

So, we need to find the slope of the two coordinates we wrote above.

We know, slope formula

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Plugging values in formula, we get

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Brecasue of depreciation we got a negative number for slope or rate of change.

Therefore, rate of depreciation is $1600 per year.

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So, we can setup an a function

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But the problem is asked to take the variable t for time.

Replacing x by t, we get

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3 years ago
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