What are the solutions to the equation (3/5)^2+y^2=1 ?

The solutions to the equation (3/5)^2+y^2=1 are y= 4/5 ,y=-4/5

Find the zeros of (3/5)^2+y^2=1

The zeros of (3/5)^2+y^2=1 are y= 4/5 ,y=-4/5

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(\frac{3}{5})^{2} + y^{2} = 1

y = \frac{4}{5}, y = - \frac{4}{5}

Decimal

y = 0.8, y = - 0.8

Solution steps

(\frac{3}{5})^{2} + y^{2} = 1

Expand

(\frac{3}{5})^{2} + y^{2} : \frac{9}{25} + y^{2}

\frac{9}{25} + y^{2} = 1

Move

\frac{9}{25}

to the right side

y^{2} = 1 - \frac{9}{25}

Simplify

1 - \frac{9}{25} : \frac{16}{25}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

y = \frac{16}{25}, y = - \frac{16}{25}

\frac{16}{25} = \frac{4}{5}

- \frac{16}{25} = - \frac{4}{5}

y = \frac{4}{5}, y = - \frac{4}{5}

- 5 y + 2 (y - 3) < - 9 y + 2 (3 y + 1)

(\frac{4}{3})^{4}

\frac{x}{- 2} > \frac{1}{3}

\frac{r ^{2}}{2} = r + \frac{1}{2}

3125

What are the solutions to the equation (3/5)^2+y^2=1 ?
The solutions to the equation (3/5)^2+y^2=1 are y= 4/5 ,y=-4/5
Find the zeros of (3/5)^2+y^2=1
The zeros of (3/5)^2+y^2=1 are y= 4/5 ,y=-4/5