QuestionFind the missing coordinate $y$ for the point $\left(-\frac{1}{4}, y\right)$ on the unit circle in quadrant III.

Studdy Solution

STEP 1

Assumptions1. The point lies on a unit circle. A unit circle is a circle with a radius of1. . The point is in the third quadrant. In the Cartesian coordinate system, the third quadrant is where both x and y are negative.
3. The x-coordinate of the point is $-\frac{1}{4}$ .
4. We are looking for the y-coordinate of the point.

STEP 2

The equation of a unit circle centered at the origin (0,0) is given by $x^2 + y^2 =1$

STEP 3

We know that the x-coordinate of our point is $-\frac{1}{}$ . We can substitute this value into the equation of the unit circle to solve for y.
$\left(-\frac{1}{}\right)^2 + y^2 =1$

STEP 4

implify the equation.
$\frac{1}{16} + y^2 =1$

STEP 5

Rearrange the equation to solve for $y^2$ .
$y^2 =1 - \frac{1}{16}$

STEP 6

implify the right side of the equation.
$y^2 = \frac{15}{16}$

STEP 7

To solve for y, we take the square root of both sides of the equation. Since the point is in the third quadrant where y is negative, we take the negative square root.
$y = -\sqrt{\frac{15}{16}}$

STEP 8

implify the square root. The square root of16 in the denominator is4.
$y = -\frac{\sqrt{15}}{4}$ So, the missing coordinate of the point $\left(-\frac{1}{4}, y\right)$ that lies on the graph of the unit circle in quadrant III is $y = -\frac{\sqrt{15}}{4}$ .

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QuestionFind the missing coordinate $y$ for the point $\left(-\frac{1}{4}, y\right)$ on the unit circle in quadrant III.

Studdy Solution

STEP 1

STEP 2

The equation of a unit circle centered at the origin (0,0) is given by $x^2 + y^2 =1$

STEP 3

We know that the x-coordinate of our point is $-\frac{1}{}$ . We can substitute this value into the equation of the unit circle to solve for y.
$\left(-\frac{1}{}\right)^2 + y^2 =1$

STEP 4

implify the equation.
$\frac{1}{16} + y^2 =1$

STEP 5

Rearrange the equation to solve for $y^2$ .
$y^2 =1 - \frac{1}{16}$

STEP 6

implify the right side of the equation.
$y^2 = \frac{15}{16}$

STEP 7

To solve for y, we take the square root of both sides of the equation. Since the point is in the third quadrant where y is negative, we take the negative square root.
$y = -\sqrt{\frac{15}{16}}$

STEP 8

implify the square root. The square root of16 in the denominator is4.
$y = -\frac{\sqrt{15}}{4}$ So, the missing coordinate of the point $\left(-\frac{1}{4}, y\right)$ that lies on the graph of the unit circle in quadrant III is $y = -\frac{\sqrt{15}}{4}$ .

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QuestionFind the missing coordinate yyy for the point (−14,y)\left(-\frac{1}{4}, y\right)(−41​,y) on the unit circle in quadrant III.

Studdy Solution

STEP 1

STEP 2

The equation of a unit circle centered at the origin (0,0) is given byx2+y2=1x^2 + y^2 =1x2+y2=1

STEP 3

We know that the x-coordinate of our point is −1-\frac{1}{}−1​. We can substitute this value into the equation of the unit circle to solve for y.(−1)2+y2=1\left(-\frac{1}{}\right)^2 + y^2 =1(−1​)2+y2=1

STEP 4

implify the equation.116+y2=1\frac{1}{16} + y^2 =1161​+y2=1

STEP 5

Rearrange the equation to solve for y2y^2y2.y2=1−116y^2 =1 - \frac{1}{16}y2=1−161​

STEP 6

implify the right side of the equation.y2=1516y^2 = \frac{15}{16}y2=1615​

STEP 7

To solve for y, we take the square root of both sides of the equation. Since the point is in the third quadrant where y is negative, we take the negative square root.y=−1516y = -\sqrt{\frac{15}{16}}y=−1615​​

STEP 8

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QuestionFind the missing coordinate $y$ for the point $\left(-\frac{1}{4}, y\right)$ on the unit circle in quadrant III.

The equation of a unit circle centered at the origin (0,0) is given by $x^2 + y^2 =1$

We know that the x-coordinate of our point is $-\frac{1}{}$ . We can substitute this value into the equation of the unit circle to solve for y.
$\left(-\frac{1}{}\right)^2 + y^2 =1$

implify the equation.
$\frac{1}{16} + y^2 =1$

Rearrange the equation to solve for $y^2$ .
$y^2 =1 - \frac{1}{16}$

implify the right side of the equation.
$y^2 = \frac{15}{16}$

To solve for y, we take the square root of both sides of the equation. Since the point is in the third quadrant where y is negative, we take the negative square root.
$y = -\sqrt{\frac{15}{16}}$