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Vilka [71]
3 years ago
10

a wrestler is weighing in for their match this evening. their normal weight is 168 pounds, but that weight can vary by as much a

s 3.2 pounds. write an equation that models this situation and find their minimum and maximum weight.
Mathematics
1 answer:
Natalija [7]3 years ago
5 0

<em>Introduction </em>

<em>Wrestling is a wonderful activity with many advantages for the student-athlete. It is a sport that is highly </em>

<em>competitive, exciting and satisfying. It is a sport that provides for individual and team competition. It is - </em>

<em>and should be - fun. Unfortunately, the practice of losing weight by not eating, restricting fluid intake and </em>

<em>over-exercising reduces the sport's fun. This information is presented to help clear up misconceptions </em>

<em>regarding wrestling and weight loss. I also hope to give some guidance to those who desire to manage </em>

<em>their weight properly in preparation for and during the wrestling season. </em>

<em>History and Stigma </em>

<em>For too long, the wrestling community has unthinkingly accepted the myth that to be a good wrestler, you </em>

<em>must cut weight. The generally accepted thinking is something like this: if your natural weight is 135 </em>

<em>pounds, you may be a good wrestler at 135 pounds. But if you wrestle at 130 pounds, you'll be a better </em>

<em>wrestler. And if you can make it down to 125, you'll be a state champion. No facts support that widely held </em>

<em>view, yet wrestlers and parents subscribe to that faulty reasoning. Looking further back, many remember </em>

<em>the days that losing excessive weight was a specific practice and expectation among wrestlers. It was </em>

<em>supposed to teach sacrifice, commitment, and the idea of “No Pain, No Gain.” </em>

<em>Regardless of the current attitude of the majority of the wrestling community, the stigma that an unhealthy </em>

<em>loss of weight is a requirement of wrestling among outside observers sticks. The Virginia High School </em>

<em>League has followed national guidelines to establish rules that encourage healthy weight management </em>

<em>among wrestlers. </em>

<em>Current Regulations </em>

<em>Preseason weight certification is accomplished with three steps. The first is determining each wrestler’s </em>

<em>body fat percentage using skin fold calipers. Next, the wrestler completes a hydration test, to insure </em>

<em>against a dehydrated weight measurement, and the wrestler weighs in. The third step is the calculation of </em>

<em>the wrestler’s minimum wrestling weight based on 7% body fat for males and 12% for females. The </em>

<em>wrestler may not wrestle at a weight class below his minimum weight during the season. The wrestler is </em>

<em>also restricted from losing more then 1.5% body weight loss per week (official weigh-ins at certification, </em>

<em>matches, and tournaments are used for this calculation). A one-pound per month growth allowance is </em>

<em>provided to allow for the natural growth of this age group. The VHSL advises, and we have instituted, </em>

<em>daily weigh-ins before and after practice to monitor weight-loss and dehydration. </em>

<em>Additionally, the National Federation of State High School Associations has adopted several other rules to </em>

<em>guide coaches and wrestlers while managing their weight. Wrestlers are discouraged from wrestling at a </em>

<em>weight class more then one weight class above their certified weight. </em>

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Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation:  Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across  the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},

and from the right-angled triangle WAB, we have'

\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

8 0
3 years ago
PLZ CAN SOMEONE ANSWER THIS !!
Cerrena [4.2K]

Answer:

(-4,2)

Step-by-step explanation:

First we have to find a common multiple to use to add and find the system of equations. We need to eliminate one of the variables in order to solve.

Since 5 * 3 = 15, and 3 * -5 = 15, -15x and 15x will cancel each other out. Therefore x will be eliminated.

Before:

5x + 2y = -16

3x + 7y = 2

After:

15x + 6y = -48   <----------Now we solve for y

-15x + -35y = -10

+----------------------

-29y = -58

-29      -29

y = 2

Now what we do is we choose and equation and solve for x, since it is still unknown. Any equation is fine but I will choose the second one since it has easy numbers.

3x + 7y = 2

3x + 7(2) = 2

3x + 14 = 2

   -14     -14

----------------------

3x = -12

3        3

x = -4

Therefore our final answer is (-4,2)

5 0
3 years ago
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S_A_V [24]

Answer:

The one with the fraction is the answer.

Step-by-step explanation:

Quotient means dividing, so the answer has to have a fraction sign in it. We can't see the answer options properly, so whatever one with the fraction is the one you should choose. I think it's the third option that you wrote.

Hope this helped!

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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Mars2501 [29]

Answer:

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80*x=100

100/80=1.25

1.25*100=125

x=125%

3 0
3 years ago
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