Step 3: Calculating time Assume that the cheetah travels an average of 40 mph to go from its resting place to a rock near a river. On the return trip to its resting place, the cheetah travels an average of 70 mph. If the cheetah traveled for 15 minutes, how many minutes did the return trip take to the nearest minute and second?

a. Set up the table as follows. Label the rows "To the River" and "From the River". Label the columns Distance, "Rate", and "Time (in Hours)." Let t represent the unknown quantity problem. Fill in the table.

b. From the table, set up an equation relating the distances.

c. Solve the problem. Write the answer in a complete sentence, stating it in terms of minutes and seconds.

Part A

See the diagram below for the table.

Here's how to set up the table.

Let t be the time it takes to make the return trip.

15 min = (15 min)*(1 hr/60min)

15 min = (15/60) hr

15 min = 1/4 hr

Since the cheetah traveled 15 minutes total, this means it traveled 1/4 of an hour total.

If it took t hours to go back home then it took 1/4 - t hours to get to the river. The values in the time column must add up to 1/4 hour.

The speeds given to us are placed in the table as is.

Multiply those speeds with the time expressions to get the distance expressions.

For the first distance we get

d = r*t

d = 40*(1/4-t)

d = 40*(1/4)+40*(-t)

d = 10-40t

Representing the distance traveled going from home to the river.

The second distance equation is

d = r*t

d = 70*t

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Part B

The equation is 10-40t  = 70t

Simply set each distance equal to one another. This works because the cheetah travels the same distance from home to the river, and then back again.

In other words,

distance from home to river = distance from river to home

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Part C

Let's solve the equation formed in part B

10-40t = 70t

10-40t+40t = 70t+40t .... adding 40t to both sides

10 = 110t

110t = 10

110t/110 = 10/110 .... dividing both sides by 110

t = 1/11 hours

t = 60/11 minutes .... multiply by 60 to convert from hrs to min

t = 3600/11 seconds ... multiply by 60 to convert from min to sec

So,

3600/11 sec = 327.2727 sec

3600/11 sec = (300 sec) + (27.2727 sec)

3600/11 sec = (300/60) min + (27.2727 sec)

3600/11 sec = 5 min + 27.2727 sec

3600/11 sec = 5 min,  27.2727 sec

Approximately 5 min,  27.2727 sec

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.