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2 years ago

5

Find the domain and range of

title="y = \frac{2x}{1 + x^2}" alt="y = \frac{2x}{1 + x^2}" align="absmiddle" class="latex-formula">. PLEASE show all steps. Thank you.

1 answer:

Vinvika [58]2 years ago

8 0

Hi,

Work:

Equation;

$y = \frac{2x}{1 + {x}^{2} }$

Root = (0, 0)
Domain: x = R
Vertical Intercept = (0, 0)

Hope this helps.
r3t40

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Step-by-step explanation:

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Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

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Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .

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$\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2$ .

In this case, the equation would be:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(\sqrt{56}\right)^2$ .

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$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56$ .

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Answer:

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$0\leq \theta \leq \dfrac{\pi}{2}$

Theta lie in first quadrant.

Multiply both sides by 2

$0\leq 2\theta \leq \pi$

2theta lie in I and II quadrant.

$\tan2\theta$ is positive in I and negative in II quadrant.

$\sin\theta=\dfrac{1}{3}$

$\cos\theta=\dfrac{2\sqrt{2}}{3}$

$\sin2\theta=2\sin\theta\cos\theta$

$\sin2\theta=2\cdot \dfrac{1}{3}\cdot \dfrac{2\sqrt{2}}{3}=\dfrac{4\sqrt{2}}{9}$

$\cos2\theta=\pm \sqrt{1-\sin^22\theta}=\pm\dfrac{7}{9}$

$\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}$

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Answer:

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