Finite Math Examples

$\frac{\sqrt{y} + L \cdot e^{- L (\sqrt{y} \cdot z + \sqrt{y - y - z^{2}})}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 1

Subtract $y$ from $y$ .

$\frac{\sqrt{y} + L \cdot e^{- L (\sqrt{y} \cdot z + \sqrt{0 - z^{2}})}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 2

Subtract $z^{2}$ from $0$ .

$\frac{\sqrt{y} + L \cdot e^{- L (\sqrt{y} z + \sqrt{- z^{2}})}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 3

Simplify each term.

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Step 3.1

Reorder $- 1$ and $z^{2}$ .

$\frac{\sqrt{y} + L \cdot e^{- L (\sqrt{y} z + \sqrt{z^{2} \cdot - 1})}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 3.2

Pull terms out from under the radical.

$\frac{\sqrt{y} + L \cdot e^{- L (\sqrt{y} z + z \sqrt{- 1})}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$\frac{\sqrt{y} + L \cdot e^{- L (\sqrt{y} z + z i)}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 4

Apply the distributive property.

$\frac{\sqrt{y} + L \cdot e^{- L (\sqrt{y} z) - L (z i)}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 5

Remove parentheses.

$\frac{\sqrt{y} + L \cdot e^{- L \sqrt{y} z - L (z i)}}{2 \sqrt{y - y \cdot z^{2}}}$

Step 6

Remove parentheses.

$\frac{\sqrt{y} + L e^{- L \sqrt{y} z - L z i}}{2 \sqrt{y - y \cdot z^{2}}}$