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posledela
3 years ago
10

a rectanular field is 65 meters wide and 115 meters long. a coach asks players to run from one corner to the opposite corner dia

gonally across field, then run back to where they started. how far do they have to run altogether

Mathematics
1 answer:
Eddi Din [679]3 years ago
8 0

Answer:

Players ran 264.20 m altogether.

Step-by-step explanation:

Diagram is attached for reference

Given:

Length of rectangular field (BC) = 115 m

Width of rectangular field (AB) = 65 m

a coach asks players to run from one corner to the opposite corner diagonally across field.

i.e In diagram from point A to Point C

So we will first find the length of diagonal AC.

Now we know that all angles of rectangle is 90°

Hence by Pythagoras theorem we get;

AC^2=AB^2+BC^2

Substituting the value we get;

AC^2 = 115^2+65^2 = 17450

Now taking square root on both side we get;

\sqrt{AC^2} = \sqrt{17450} \\\\AC \approx 132.10\ m

Now Players run back from where they started.

So players ran = 132.10 m + 132.10 m = 264.20 m

Hence Players ran 264.20 m altogether.

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Sharon inherited ½ of her grandfather’s estate. Out of sympathy she gave away ½ of the property that she inherited. How much pro
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Step-by-step explanation:

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3 years ago
What is the approximate measure of angle T in the triangle below 79° 84° 96° 101°
Anestetic [448]

The approximate measure of angle T in the given diagram is 79°. The correct option is the first option 79°

<h3>Law of Cosines</h3>

From the question, we are to determine the approximate measure of angle T

From the law of cosines, we can write that

cosT = (s² + u² - t²)/2su

From the diagram,

s = 3.9 cm

u = 2.7 cm

t = 4.3 cm

Thus,

cosT = (3.9² + 2.7² - 4.3²)/2(3.9)(2.7)

cosT = (15.21 + 7.29 - 18.49)/21.06

cosT = 4.01/21.06

cosT = 0.1904

T = cos⁻¹(0.1904)

T = 79.02°

T ≈ 79°

Hence, the approximate measure of angle T in the given diagram is 79°. The correct option is the first option 79°

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4 0
1 year ago
What is the average value of latex: y=\tan\left(\frac{x^2}{9}\right) y = tan ⁡ ( x 2 9 ) on the closed interval [1.25, 2]?
german
For this case we have the following function:
 f(x) =  tan(\frac{x^2}{9})
 The average rate of change is given by:
 AVR =  \frac{f(x2) - f(x1)}{x2- x1}
 Evaluating the function for the given interval we have:
 For x = 1.25:
 f(1.25) = tan(\frac{1.25^2}{9})
 f(1.25) = 0.18
 For x = 2:
 f(2) = tan(\frac{2^2}{9})
 f(2) = 0.48
 Then, replacing values we have:
 AVR = \frac{0.48 - 0.18}{2 - 1.25}
 AVR = 0.4
 Answer:
 
the average value of on the closed interval [1.25, 2] is:
 
AVR = 0.4



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Step-by-step explanation:

could you make me brainliest if this helped? :)

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