What is 15=(\frac{x/(5+R(x)))/(R(x))}{15} ?

The solution to 15=(\frac{x/(5+R(x)))/(R(x))}{15} is Parabola with (h,k)=(-5625/4 ,-5/2),p= 1/900

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15=(\frac{x/(5+R(x)))/(R(x))}{15}

15 = \frac{\frac{\frac{x}{5 + R ( x )}}{R ( x )}}{15}

(h, k) = (- \frac{5625}{4}, - \frac{5}{2}), p = \frac{1}{900}

Solution steps

15 = \frac{\frac{\frac{x}{5 + y}}{y}}{15}

Rewrite

15 - \frac{\frac{\frac{x}{5 + y}}{y}}{15} = 0

in the standard form

4 \cdot \frac{1}{900} (x - (- \frac{5625}{4})) = (y - (- \frac{5}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{5625}{4}, - \frac{5}{2}), p = \frac{1}{900}

(x - 2)^{2} + (y - 0)^{2} = (x + 3)^{2}

y^{2} + 16 x - 6 y - 7 = 0

\frac{f ( x )}{2 x} \leq \frac{x ^{2} + 1}{2 x}

(y + 4)^{2} = - 42 (x + 2)

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

What is 15=(\frac{x/(5+R(x)))/(R(x))}{15} ?
The solution to 15=(\frac{x/(5+R(x)))/(R(x))}{15} is Parabola with (h,k)=(-5625/4 ,-5/2),p= 1/900