Given that point [tex]\(P\)[/tex] is on the unit circle, has a [tex]\(y\)[/tex]-coordinate of [tex]\(-\frac{4}{8}\)[/tex], and lies in quadrant III, let's find the corresponding [tex]\(x\)[/tex]-coordinate.
1. Determine the given [tex]\(y\)[/tex]-coordinate:
[tex]\[
y = -\frac{4}{8} = -0.5
\][/tex]
2. Calculate [tex]\(y^2\)[/tex]:
[tex]\[
y^2 = (-0.5)^2 = 0.25
\][/tex]
3. Use the unit circle equation [tex]\(x^2 + y^2 = 1\)[/tex] to solve for [tex]\(x^2\)[/tex]:
[tex]\[
x^2 + 0.25 = 1
\][/tex]
[tex]\[
x^2 = 1 - 0.25
\][/tex]
[tex]\[
x^2 = 0.75
\][/tex]
4. Take the square root to determine [tex]\(x\)[/tex]. Since [tex]\(P\)[/tex] is in quadrant III, [tex]\(x\)[/tex] will be negative:
[tex]\[
x = -\sqrt{0.75}
\][/tex]
5. Simplify and round the result to one decimal place:
[tex]\[
x = -\sqrt{0.75} \approx -0.866 \approx -0.9
\][/tex]
So, the [tex]\(x\)[/tex]-coordinate of point [tex]\(P\)[/tex] is:
[tex]\[
x = -0.9
\][/tex]
