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oee [108]
3 years ago
8

John paid $100 for renting a canoe for 5 hours. Which graph shows the relationship between the cost of renting a canoe for diffe

rent hours?
A graph is shown. The title of the graph is Canoe Renting. The horizontal axis label is Number of Hours. The horizontal axis values are 0, 1, 2, 3, 4, 5, 6. The vertical axis label is Cost in dollars. The vertical axis values are 0, 20, 40, 60, 80, 100, 120. Points are plotted on the ordered pairs 1,100 and 2,100 and 3,100.
A graph is shown. The title of the graph is Canoe Renting. The horizontal axis label is Number of Hours. The horizontal axis values are 0, 1, 2, 3, 4, 5, 6. The vertical axis label is Cost in dollars. The vertical axis values are 0, 20, 40, 60, 80, 100, 120. Points are plotted on the ordered pairs 1,20 and 2,40 and 3,60
A graph is shown. The title of the graph is Canoe Renting. The horizontal axis label is Number of Hours. The horizontal axis values are 0, 1, 2, 3, 4, 5, 6. The vertical axis label is Cost in dollars. The vertical axis values are 0, 20, 40, 60, 80, 100, 120. Points are plotted on the ordered pairs 1,10 and 2,20 and 3,30.
A graph is shown. The title of the graph is Canoe Renting. The horizontal axis label is Number of Hours. The horizontal axis values are 0, 1, 2, 3, 4, 5, 6. The vertical axis label is Cost in dollars. The vertical axis values are 0, 20, 40, 60, 80, 100, 120. Points are plotted on the ordered pairs 1,40 and 2,80 and 3,120.
Mathematics
1 answer:
Anarel [89]3 years ago
5 0

Answer:

The answer is B.

Step-by-step explanation:

You would pay about $20 each hour

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If we let x = distance batter has run at time t and  D = distance from second base to the batter at  time t, then we know \frac{dx}{dt}=31 ft/s and we want \frac{dD}{dt} when he is halfway (at x = 45).

Using Pythagoras theorem

D^2=90^2+(90-x)^2\\\text{Differentiating with respect to t}\\2D\frac{dD}{dt}=0+2(90-x)(-1)\frac{dx}{dt}\\2D\frac{dD}{dt}=-2(90-x)\frac{dx}{dt}

When x=45

D^2=90^2+(90-x)^2\\D^2=90^2+(90-45)^2\\D^2=10125\\D=\sqrt{10125}

2D\frac{dD}{dt}=-2(90-x)\frac{dx}{dt}\\2\sqrt{10125}\frac{dD}{dt}=-2(90-45)(31)\\\frac{dD}{dt}=\frac{-2(45)(31)}{2\sqrt{10125}} =-13.86ft/s

Thus, when the batter is halfway to first base, the distance between second base and the batter is  decreasing at the rate of about 13.86 ft/sec.

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