All categories

3 years ago

Which point on the unit circle corresponds to -pi/6 ?

Mathematics

1 answer:

xeze [42]3 years ago

8 0

Answer:

(srt(3)/2, -1/2)

Step-by-step explanation:

x = rcos theta

r = 1 on unit circle

x = 1 cos (-pi/6)

x = sqrt(3)/2

y = r sin theta

y =1 sin (-pi/6)=

y = -1/2

You might be interested in

Which equation represents a proportional relationship between the x and y values?

Airida [17]

Answer:

D) $y=-2x$

Step-by-step explanation:

A linear relationship is of the form $y=mx+b$ , where, $m$ is the slope and $b$ is the y-intercept (constant).

For a proportional relationship, the value of $b=0$ and thus it is of the form $y=mx$

Let us check each option and express it in the form above.

Option A:

$y=2$

This can be written as $y=0x+2$ . So, $b\ne 0$ .

Since, b \ne 0, therefore, it is not a proportional relationship.

Option B:

$y=14x+1$

Here, $b=1$ . So, $b\ne 0$ .

Since, b \ne 0, therefore, it is not a proportional relationship.

Option C:

$y=x-4$

Here, $b=-4$ . So, $b\ne 0$ .

Since, b \ne 0, therefore, it is not a proportional relationship.

Option D:

$y=-2x$

This can be rewritten as $y=-2x+0$

There is no y-intercept on this. So, $b=0$

Since, $b = 0$ , therefore, it is a proportional relationship.

Therefore, the correct option is D.

8 0

3 years ago

Read 2 more answers

Pls helpppppppppppp

Deffense [45]

-0.25 don’t know but I try

5 0

3 years ago

Lost as usual yep I am

denpristay [2]

Answer:

y = -3x -4

Step-by-step explanation:

A perpendicular line has a slope that is the negative reciprocal of that of the given line. When the equation starts out in standard form, a line with negative reciprocal slope can be written by swapping the x- and y-coefficients and negating one of them.

The given x- and y-coefficients have the ratio 1:-3, so we can use the coefficients 3 and 1 for our purpose.

The usual process of making the line go through a given point can be used. That is, we can translate the line from the origin to the desired point by subtracting the point coordinates from x and y. Then we have ...

3(x+3) +(y-5) = 0

This is "an" equation. It is in no particularly recognizable form. It can be rearranged to the form y = mx + b:

3x +9 +y -5 = 0 . . . . . eliminate parentheses

y = -3x -4 . . . . . subtract terms that are not "y"