QuestionFind the missing coordinate $x$ for the point $\left(x,-\frac{1}{5}\right)$ on the unit circle in quadrant IV.

Studdy Solution

STEP 1

Assumptions1. The point $\left(x,-\frac{1}{5}\right)$ lies on the unit circle. . The point is in quadrant IV.
3. The equation of a unit circle is $x^ + y^ =1$ .

STEP 2

We know that the point $\left(x,-\frac{1}{5}\right)$ lies on the unit circle. So, it satisfies the equation of the unit circle. We can substitute the given y-coordinate into the equation of the unit circle to solve for x.
$x^2 + \left(-\frac{1}{5}\right)^2 =1$

STEP 3

implify the equation.
$x^2 + \frac{1}{25} =1$

STEP 4

Subtract $\frac{1}{25}$ from both sides of the equation to isolate $x^2$ .
$x^2 =1 - \frac{1}{25}$

STEP 5

implify the right side of the equation.
$x^2 = \frac{24}{25}$

STEP 6

Take the square root of both sides of the equation to solve for x. Remember that the square root of a number has two solutions, one positive and one negative. However, since we know that the point is in quadrant IV, where x is positive, we only consider the positive solution.
$x = \sqrt{\frac{24}{25}}$

STEP 7

implify the square root.
$x = \frac{4\sqrt{6}}{5}$ The missing coordinate is $x=\frac{4\sqrt{6}}{5}$ .

Was this helpful?

QuestionFind the missing coordinate $x$ for the point $\left(x,-\frac{1}{5}\right)$ on the unit circle in quadrant IV.

Studdy Solution

STEP 1

Assumptions1. The point $\left(x,-\frac{1}{5}\right)$ lies on the unit circle. . The point is in quadrant IV.
3. The equation of a unit circle is $x^ + y^ =1$ .

STEP 2

We know that the point $\left(x,-\frac{1}{5}\right)$ lies on the unit circle. So, it satisfies the equation of the unit circle. We can substitute the given y-coordinate into the equation of the unit circle to solve for x.
$x^2 + \left(-\frac{1}{5}\right)^2 =1$

STEP 3

implify the equation.
$x^2 + \frac{1}{25} =1$

STEP 4

Subtract $\frac{1}{25}$ from both sides of the equation to isolate $x^2$ .
$x^2 =1 - \frac{1}{25}$

STEP 5

implify the right side of the equation.
$x^2 = \frac{24}{25}$

STEP 6

STEP 7

implify the square root.
$x = \frac{4\sqrt{6}}{5}$ The missing coordinate is $x=\frac{4\sqrt{6}}{5}$ .

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the missing coordinate xxx for the point (x,−15)\left(x,-\frac{1}{5}\right)(x,−51​) on the unit circle in quadrant IV.

Studdy Solution

STEP 1

Assumptions1. The point (x,−15)\left(x,-\frac{1}{5}\right)(x,−51​) lies on the unit circle. . The point is in quadrant IV.3. The equation of a unit circle is x+y=1x^ + y^ =1x+y=1.

STEP 2

STEP 3

implify the equation.x2+125=1x^2 + \frac{1}{25} =1x2+251​=1

STEP 4

Subtract 125\frac{1}{25}251​ from both sides of the equation to isolate x2x^2x2.x2=1−125x^2 =1 - \frac{1}{25}x2=1−251​

STEP 5

implify the right side of the equation.x2=2425x^2 = \frac{24}{25}x2=2524​

STEP 6

STEP 7

implify the square root.x=465x = \frac{4\sqrt{6}}{5}x=546​​The missing coordinate is x=465x=\frac{4\sqrt{6}}{5}x=546​​.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionFind the missing coordinate $x$ for the point $\left(x,-\frac{1}{5}\right)$ on the unit circle in quadrant IV.

Assumptions1. The point $\left(x,-\frac{1}{5}\right)$ lies on the unit circle. . The point is in quadrant IV.
3. The equation of a unit circle is $x^ + y^ =1$ .

implify the equation.
$x^2 + \frac{1}{25} =1$

Subtract $\frac{1}{25}$ from both sides of the equation to isolate $x^2$ .
$x^2 =1 - \frac{1}{25}$

implify the right side of the equation.
$x^2 = \frac{24}{25}$

implify the square root.
$x = \frac{4\sqrt{6}}{5}$ The missing coordinate is $x=\frac{4\sqrt{6}}{5}$ .