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ELEN [110]
3 years ago
6

A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. W

hat are the smallest and largest areas that the field could be?
Physics
2 answers:
tatyana61 [14]3 years ago
5 0

Answer:

The smallest and largest areas could be 6400 m and 7500 m, respectively.

Explanation:

The area of a rectangle is given by:

A = l*w

Where:

l: is the length = 100 m

w: is the width

We can calculate the smallest area with the lower value of the width.

A_{s} = 100 m*64 m = 6400 m^{2}                            

And the largest area is:

A_{l} = 100 m*75 m = 7500 m^{2}  

Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.            

I hope it helps you!                        

Bezzdna [24]3 years ago
4 0

Answer:

the largest areas that the field could be is A_l=7587.75 m

the smallest areas that the field could be is A_s=6318.25 m

Explanation:

to the find the largest and the smallest area of the field measurement error is to be considered.

we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.

therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:

A_l= (L+0.5)(W+0.5)

(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m

To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:

A_s= (L-0.5)(W-0.5)

(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m

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<span>f(x) = 5.05*sin(x*pi/12) + 5.15

   First, you need to determine the period of the function. The period will be the time interval between identical points on the sinusoidal function. For this problem, the tide is rising and at 5.15 at midnight for two consecutive days. So the period is 24 hours. Over that 24 hour period, we want the parameter passed to sine to range from 0 to 2*pi. So the scale factor for x will be 2*pi/24 = pi/12 which is approximately 0.261799388. The next thing to note is the magnitude of the wave. That will simply be the difference between the maximum and minimum values. So 10.2 ft - 0.1 ft = 10.1 ft. And since the value of sine ranges from -1 to 1, we need to divide that magnitude by 2, so 10.1 ft / 2 = 5.05 ft.

   So our function at this point looks like f(x) = 5.05*sin(x*pi/12) But the above function ranges in value from -5.05 to 5.05. So we need to add a bias to it in order to make the low value equal to 0.1. So 0.1 = X - 5.05, 0.1 + 5.05 = X, 5.15 = X. So our function now looks like:
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   The final thing that might have been needed would have been a phase correction. With this problem, we don't need a phase correction since at X = 0 (midnight), the value of X*pi/12 = 0, and the sine of 0 is 0, so the value of the equation is 5.15 which matches the given value of 5.15. But if the problem had been slightly different and the height of the tide at midnight has been something like 7 feet, then we would have had to calculate a phase shift value for the function and add that constant to the parameter being passed into sine, making the function look like:
 f(x) = 5.05*sin(x*pi/12 + C) + 5.15
  where
 C = Phase correction offset.

   But we don't need it for this problem, so the answer is:
 f(x) = 5.05*sin(x*pi/12) + 5.15

   Note: The above solution assumes that angles are being measured in radians. If you're using degrees, then instead of multiplying x by 2*pi/24 = pi/12, you need to multiply by 360/24 = 15 instead, giving f(x) = 5.05*sin(x*15) + 5.15</span>
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igomit [66]

Answer:

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Marrrta [24]

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On the contrary, if the relationship between the magnitudes is inverse (when one magnitude increases the other decreases) the rule of three simple inverse applies.

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