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2 years ago

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"Please help! Using the table of range of values for each weight class, how can you express the Light weight class as an absolut

e-value? (Hint: Use w for the average weight in each weight class.)
A. |w - 45| ≤ 5
B. |w - 50| ≤ 5
C. |45 - w| ≥ 5
D. |w - 45| ≥ 5"

1 answer:

mr_godi [17]2 years ago

8 0

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

The answer is A. Below is the solution:

w - 45 <= 5 and <span>w - 45 <= -5
</span>w <= 50 <span> w<= 40
</span><span>the the class is 40 - 50</span>

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The magnitude and the direction of the resultant are approximately 57.871 and 198.676°.

<h3>How to determine the magnitude and the direction of a resultant</h3>

Vectors are elements with two given characteristics: Magnitude and direction. A resultant is derived from the sum of vectors, the magnitude is the norm of a vector and the direction is the orientation of a vector. The resultant and its characteristics are described below:

<h3>Resultant</h3>

$\vec r = \left(\sum\limits_{i=1}^{n}x_{i}, \sum\limits_{i=1}^{n}y_{i}\right)$ (1)

<h3>Magnitude</h3>

$\|\vec r\| = \sqrt{\left(\sum\limits_{i=1}^{n} x_{i}\right)^{2}+\left(\sum\limits_{i=1}^{n} y_{i}\right)^{2}}$ (2)

<h3>Direction</h3>

$\theta = \tan^{-1}\frac{\sum\limits_{i=1}^{n}y_{i}}{\sum\limits_{i=1}^{n}x_{i}}$ (3)

Where $\theta$ is resultant angle, measured in degrees.

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$\vec r = (-90\cdot \cos 40^{\circ}-60\cdot \sin 20^{\circ}+40\cdot \cos 30^{\circ}, 40\cdot \sin 30^{\circ}+60\cdot \cos 20^{\circ}-90\cdot \sin 40^{\circ})$

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$\|\vec r\| = \sqrt{(-54.824)^{2}+18.531^{2}}$

$\|\vec r\| \approx 57.871$

<h3>Direction</h3>

$\theta = \tan^{-1} \left(\frac{18.531}{-54.824}\right)$

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