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Free_Kalibri [48]
2 years ago
13

You are hiking in an area with steep trails. It takes you twice as long to hike uphill as it takes you to hike downhill. When yo

u get to the bottom of the trail, you take a 30-minute break. On your way back up the trail, you take a 15-minute break. Write an expression for the total time of your hike in hours. Explain what the terms represent in the expression.
Mathematics
1 answer:
baherus [9]2 years ago
7 0

Answer:

The total time spent will be:

(t + 1/2)hrs + (2t + 1/4)hrs = (3t + 3/4)hrs.

Step-by-step explanation:

Let the time it takes to hike downhill be represented by t in hrs.

This implies that the time it takes to hike uphill will be 2t, as stated in the question.

Since, a 30-minute break, which is 1/2hrs, was taken to hike downhill, then the total time spent for hiking downhill will be (t + 1/2)hrs.

Also, since a 15-minute, which 1/4hrs, was taken to hike uphill, then the total time spent for hiking uphill will be (2t + 1/4)hrs.

Therefore, the total time spent will be:

(t + 1/2)hrs + (2t + 1/4)hrs = (3t + 3/4)hrs.

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A right triangle has one vertex on the graph of y = 9 - x^2 , x > 0, at ( x , y ), another at the origin, and the third on th
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Answer:

  a.  A(x) = (1/2)x(9 -x^2)

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Step-by-step explanation:

a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...

  A(x) = (1/2)(x)(y)

  A(x) = (1/2)(x)(9-x^2)

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b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).

On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).

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c. A(2) = (1/2)(2)(9 -2^2) = 5

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d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).

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