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devlian [24]
3 years ago
10

A number c decreased by 9 is less than or equal to 20. Find c?

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
8 0
C - 9 < = 20
Add 9 to both sides
c -9 + 9 <= 20 + 9
c <= 29
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A person leaves her camp at 7:00 a.m. to hike back to her car. The distance from the car in kilometers y after x hours of hiking
FromTheMoon [43]

Answer:

The x axis in the function represents, the number of hours after 7:00 A.M. , the person reaches her car. The person reaches the car at 1:00 P.M.

Step-by-step explanation:

The x axis denotes the no. of hours and and the y axis denotes the distance from the car.

X Intercept is a point where the line intersects the X axis, we can easily notice the fact that at that point, y=0 ie. The person has reached his/her respective car.

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Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

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Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

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so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
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