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Question Number 207597 by hardmath last updated on 19/May/24

Answered by A5T last updated on 20/May/24

Commented by A5T last updated on 20/May/24

$$\frac{a}{x}=\frac{b}{y}\Rightarrow \frac{a}{b}=\frac{x}{y}\Rightarrow \frac{x}{x+y}=\frac{a}{a+b}$$$$\frac{z}{b}=\frac{a}{a+b}\Rightarrow z=\frac{ab}{a+b}$$$$\frac{d}{a}=\frac{\frac{ab}{a+b}}{b}=\frac{a}{a+b}\Rightarrow a-d=\frac{-ab}{a+b}$$$$c^2={(a-d)}^2+z^2=\frac{2{\left(ab\right)}^2}{{(a+b)}^2}\Rightarrow c=\frac{ab\sqrt{2}}{a+b}$$

Commented by hardmath last updated on 20/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

Answered by mr W last updated on 20/May/24

$$\frac{\frac{c}{\sqrt{2}}}{b-\frac{c}{\sqrt{2}}}=\frac{a}{b}\Rightarrow \frac{c}{\sqrt{2}b-c}=\frac{a}{b}\Rightarrow c=\frac{\sqrt{2}ab}{a+b}✓$$

Commented by hardmath last updated on 20/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{professor}$$

Commented by mr W last updated on 20/May/24

Commented by A5T last updated on 20/May/24

$$Usingareas:\frac{bc}{2\sqrt{2}}+\frac{ac}{2\sqrt{2}}=\frac{ab}{2}\Rightarrow c=\frac{\sqrt{2}ab}{a+b}$$

Answered by efronzo1 last updated on 20/May/24

$$\cancel{\underset{―}{⋖}}△\mathrm{ABC}=\cancel{\underset{―}{⋖}}△\mathrm{ABD}+\underset{―}{⋮}\underbrace{\mathrm{♠}\mathrm{♠}\mathrm{♠}\mathcal{H}}$$

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