What are the solutions to the equation (6/7)^2+y^2=1 ?

The solutions to the equation (6/7)^2+y^2=1 are y=(sqrt(13))/7 ,y=-(sqrt(13))/7

Find the zeros of (6/7)^2+y^2=1

The zeros of (6/7)^2+y^2=1 are y=(sqrt(13))/7 ,y=-(sqrt(13))/7

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(\frac{6}{7})^{2} + y^{2} = 1

y = \frac{13}{7}, y = - \frac{13}{7}

Decimal Notation

y = 0.51507 \dots, y = - 0.51507 \dots

Solution steps

(\frac{6}{7})^{2} + y^{2} = 1

Expand

(\frac{6}{7})^{2} + y^{2} : \frac{36}{49} + y^{2}

\frac{36}{49} + y^{2} = 1

Move

\frac{36}{49}

to the right side

y^{2} = 1 - \frac{36}{49}

Simplify

1 - \frac{36}{49} : \frac{13}{49}

For

x^{2} = f (a)

the solutions are

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

y = \frac{13}{49}, y = - \frac{13}{49}

\frac{13}{49} = \frac{13}{7}

- \frac{13}{49} = - \frac{13}{7}

y = \frac{13}{7}, y = - \frac{13}{7}

30 x - 18

k^{7} + 18 \leq 20

2 x + 8 = x + 25 + 8

(- 10)^{2}

∣ 5 x - 23 ∣ = - 17

What are the solutions to the equation (6/7)^2+y^2=1 ?
The solutions to the equation (6/7)^2+y^2=1 are y=(sqrt(13))/7 ,y=-(sqrt(13))/7
Find the zeros of (6/7)^2+y^2=1
The zeros of (6/7)^2+y^2=1 are y=(sqrt(13))/7 ,y=-(sqrt(13))/7