What are the solutions to the equation (4/7)^2+y^2=1^2 ?

The solutions to the equation (4/7)^2+y^2=1^2 are y=(sqrt(33))/7 ,y=-(sqrt(33))/7

Find the zeros of (4/7)^2+y^2=1^2

The zeros of (4/7)^2+y^2=1^2 are y=(sqrt(33))/7 ,y=-(sqrt(33))/7

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(\frac{4}{7})^{2} + y^{2} = 1^{2}

y = \frac{33}{7}, y = - \frac{33}{7}

Solution steps

(\frac{4}{7})^{2} + y^{2} = 1^{2}

Expand

(\frac{4}{7})^{2} + y^{2} : \frac{16}{49} + y^{2}

Expand

1^{2} : 1

\frac{16}{49} + y^{2} = 1

Move

\frac{16}{49}

to the right side

y^{2} = 1 - \frac{16}{49}

Simplify

1 - \frac{16}{49} : \frac{33}{49}

For

x^{2} = f (a)

the solutions are

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

y = \frac{33}{49}, y = - \frac{33}{49}

\frac{33}{49} = \frac{33}{7}

- \frac{33}{49} = - \frac{33}{7}

y = \frac{33}{7}, y = - \frac{33}{7}

0 \leq x + 3 < 5

6 x^{2} + 2 x - 2

(- \frac{1}{10} - x) (- \frac{1}{5} - x) - \frac{3}{400} = 0

13 y - 8 y + 15 \leq - 7 + 32

(- 4 b^{2} + 8 b) + (- 4 b^{3} + 5 b^{2} - 8 b)

What are the solutions to the equation (4/7)^2+y^2=1^2 ?
The solutions to the equation (4/7)^2+y^2=1^2 are y=(sqrt(33))/7 ,y=-(sqrt(33))/7
Find the zeros of (4/7)^2+y^2=1^2
The zeros of (4/7)^2+y^2=1^2 are y=(sqrt(33))/7 ,y=-(sqrt(33))/7