To determine the value of [tex]\( y \)[/tex] for the point [tex]\( P\left(-\frac{3}{5}, y\right) \)[/tex] that lies on the unit circle in the second quadrant, we start with the equation of a unit circle:
[tex]\[
x^2 + y^2 = 1
\][/tex]
Given [tex]\( x = -\frac{3}{5} \)[/tex], substitute this value into the equation:
[tex]\[
\left(-\frac{3}{5}\right)^2 + y^2 = 1
\][/tex]
Now, calculate [tex]\( \left(-\frac{3}{5}\right)^2 \)[/tex]:
[tex]\[
\left(-\frac{3}{5}\right)^2 = \frac{9}{25}
\][/tex]
Next, substitute [tex]\( \frac{9}{25} \)[/tex] in place of [tex]\( \left(-\frac{3}{5}\right)^2 \)[/tex]:
[tex]\[
\frac{9}{25} + y^2 = 1
\][/tex]
To solve for [tex]\( y^2 \)[/tex], subtract [tex]\( \frac{9}{25} \)[/tex] from both sides of the equation:
[tex]\[
y^2 = 1 - \frac{9}{25}
\][/tex]
Convert 1 into a fraction with the same denominator:
[tex]\[
1 = \frac{25}{25}
\][/tex]
Now perform the subtraction:
[tex]\[
y^2 = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}
\][/tex]
Thus,
[tex]\[
y^2 = \frac{16}{25}
\][/tex]
To find [tex]\( y \)[/tex], take the square root of both sides:
[tex]\[
y = \sqrt{\frac{16}{25}}
\][/tex]
Since [tex]\( y \)[/tex] is in the second quadrant, where [tex]\( y \)[/tex] values are positive:
[tex]\[
y = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5}
\][/tex]
Therefore, the value of [tex]\( y \)[/tex] is:
[tex]\[
y = \frac{4}{5}
\][/tex]
Thus, [tex]\( y = 0.8 \)[/tex]
