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Mnenie [13.5K]
2 years ago
9

Suppose the function H(t) gives the heart rate of a runner at various points in time (in minutes) during a 5 mile run, which tak

es the runner 57 minutes and 30 seconds to complete. The runner's heart rate never goes above 185 beats per minute. Describe an appropriate domain this function
Mathematics
1 answer:
Marrrta [24]2 years ago
5 0

Answer:

[0, 57.5]

<em>OR</em>

0 - 57.5

Step-by-step explanation:

Given the function H(t) which gives the heart rate as per time.

That means Heart rate of the runner changes according to time.

H(t) is a function which takes Time as input as observed by the function's representation. In bracket, there is t.

That also means time t is the input to the function H(t).

It is also given that during the time as the runner runs, the heart beat will change.

The output of the given function will be Heart beat of the runner.

Time taken to complete the run is given as 57 minutes and 30 seconds.

Domain of a function is the valid inputs that can be given to the function for which the output of the function is defined.

Here, time in minutes is the input to the function.

So, maximum time can be 57.5 minutes.

And as the runner will start from the time = 0 minutes

So, the valid inputs can be from <em>0 to 57.5 minutes.</em>

Therefore, the domain is [0, 57.5]

<em>OR</em>

0 - 57.5

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For a woman suspect whose shoe size was found at the crime scene, the equation which can be used to predict the estimated height is given as:

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(b) The estimated height of a woman who is of shoe size 7 is 64,33 inches.

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Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes
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14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

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At 5 pm:

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Now, the situation of distance traveled can be seen by the attached right angled \triangle AOB.

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = \dfrac{Total\ distance}{Total\ time}

\Rightarrow \dfrac{74.28}{5} = 14.86\ knots

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