Answer:
[tex]x=\frac{\sqrt{40}}{7}[/tex]
Explanation: Provided a point on a unit circle, that has a y-coordinate of y=(-3/7) and lies in the fourth quardrand, we have to find the x-coordinate of the point, the visulization of the problem is as follows:
Using the Pythagorean theorem, the following equation can be constructed:
[tex]\begin{gathered} x^2+(-\frac{3}{7})^2=1^2 \\ x^2+(-\frac{3}{7})^2=1^\text{ }\Rightarrow(1) \end{gathered}[/tex]
Simplifying the equation (1) gives the following answer:
[tex]\begin{gathered} x^2+(-\frac{3}{7})^2=1^ \\ x^2=1-(-\frac{3}{7})^2 \\ x=\frac{\sqrt{40}}{7} \end{gathered}[/tex]
In conclusion, the answer is:
[tex]x=\frac{\sqrt{40}}{7}[/tex]
