What are the solutions to the equation (-3/7)^2+y^2=1 ?

The solutions to the equation (-3/7)^2+y^2=1 are y=(2sqrt(10))/7 ,y=-(2sqrt(10))/7

Find the zeros of (-3/7)^2+y^2=1

The zeros of (-3/7)^2+y^2=1 are y=(2sqrt(10))/7 ,y=-(2sqrt(10))/7

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(- \frac{3}{7})^{2} + y^{2} = 1

y = \frac{2 10}{7}, y = - \frac{2 10}{7}

Solution steps

(- \frac{3}{7})^{2} + y^{2} = 1

Expand

(- \frac{3}{7})^{2} + y^{2} : \frac{9}{49} + y^{2}

\frac{9}{49} + y^{2} = 1

Move

\frac{9}{49}

to the right side

y^{2} = 1 - \frac{9}{49}

Simplify

1 - \frac{9}{49} : \frac{40}{49}

For

x^{2} = f (a)

the solutions are

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

y = \frac{40}{49}, y = - \frac{40}{49}

\frac{40}{49} = \frac{2 10}{7}

- \frac{40}{49} = - \frac{2 10}{7}

y = \frac{2 10}{7}, y = - \frac{2 10}{7}

(0.1)^{2}

144 x^{2} y - 9 y

7 x^{2} + 28 = 28 x

∣ 4 x + 13 ∣ < 3

- 2 x \geq - 10

What are the solutions to the equation (-3/7)^2+y^2=1 ?
The solutions to the equation (-3/7)^2+y^2=1 are y=(2sqrt(10))/7 ,y=-(2sqrt(10))/7
Find the zeros of (-3/7)^2+y^2=1
The zeros of (-3/7)^2+y^2=1 are y=(2sqrt(10))/7 ,y=-(2sqrt(10))/7