What is the answer to solvefor y,x=sqrt(64-y^2) ?

The answer to solvefor y,x=sqrt(64-y^2) is y=sqrt(-x^2+64) {0<= x<= 8},y=-sqrt(-x^2+64) {0<= x<= 8}

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solvefor y,x=sqrt(64-y^2)

solve for

y, x = 64 - y^{2}

y = - x^{2} + 64 {0 \leq x \leq 8}, y = - - x^{2} + 64 {0 \leq x \leq 8}

Solution steps

x = 64 - y^{2}

Square both sides:

x^{2} = 64 - y^{2}

x^{2} = 64 - y^{2}

Solve

x^{2} = 64 - y^{2} : y = - x^{2} + 64, y = - - x^{2} + 64

y = - x^{2} + 64, y = - - x^{2} + 64

Verify Solutions:

y = - x^{2} + 64 {0 \leq x \leq 8}, y = - - x^{2} + 64 {0 \leq x \leq 8}

The solutions are

y = - x^{2} + 64 {0 \leq x \leq 8}, y = - - x^{2} + 64 {0 \leq x \leq 8}

25 x = \frac{x ^{2}}{25}

20 + x = x

2 x - 1 + x + 4 = 0

2 p^{2} + 17 = 3 - p

6 x + 7 = 7

What is the answer to solvefor y,x=sqrt(64-y^2) ?
The answer to solvefor y,x=sqrt(64-y^2) is y=sqrt(-x^2+64) {0<= x<= 8},y=-sqrt(-x^2+64) {0<= x<= 8}