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4 years ago

10

Lola Jean Ramos just turned 21 on June 19th and she loves to go dancing. From the given scenarios, which one describes the best

way for Lola to create a 6 character password?
The answer is C.

2 answers:

GrogVix [38]4 years ago

4 0

The answer is C. Hope this helped.

Rashid [163]4 years ago

3 0

Answer:

the answer is c

Step-by-step explanation:

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4x_25+x+14+6x_51=180

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3 years ago

Find the six trigonometric function values of the angle θ in standard position, if the terminal side of θ is defined by x + 2y =

Black_prince [1.1K]

Answer:

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

Step-by-step explanation:

First, we need to draw the terminal position of the given angle. To do so, we need to find a point that lies on the straight line $x + 2y= 0, x\geq 0$

If we choose $x = 2$ (we can do so because of the condition $x \geq 0$ , which means that any positive value is suitable for $x$ ), then we have

$2 +2y = 0\implies 2 = -2y \implies y = -1$

Therefore, the terminal side of the angle $\theta$ is passing through the origin and the point $(2,-1)$ and now we can draw it.

The angle $\theta$ is presented below.

The distance of the point $(2,-1)$ from the origin equals

$r = \sqrt{2^2 + (-1)^2} = \sqrt{5}$

Now, we can determine the values of the six trigonometric function, by using their definitions.

$\sin \theta = \frac{y}r} = \frac{-1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{\sqrt{5}}{5}\\\\\cos \theta = \frac{x}{r} = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = -\frac{2\sqrt{5}}{5} \\\\\tan \theta = \frac{y}{x} = \frac{-1}{2} = -\frac{1}{2} \\\\\cot \theta = \frac{x}{y} = \frac{2}{-1} = -2\\\\\sec \theta = \frac{r}{x} = \frac{\sqrt{5}}{2} \\\\\csc \theta = \frac{r}{y} = \frac{\sqrt{5}}{-1} = -\sqrt{5}$

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3 years ago

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yKpoI14uk [10]

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3 years ago

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Fynjy0 [20]

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3 years ago

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