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Question Number 199821 by a.lgnaoui last updated on 09/Nov/23

$$\mathrm{ABFE}\mathrm{Care}$$$$\mathrm{determiner}\mathbf{x}\mathrm{en}\mathrm{fonction}\mathrm{de}\mathrm{a}\mathrm{etb}$$$$\mathrm{BC}=\mathbf{a}\mathrm{DE}=\mathbf{b}$$

Commented by a.lgnaoui last updated on 09/Nov/23

Commented by mr W last updated on 10/Nov/23

$$itissolved.$$$$x=\sqrt{a^2+b^2}$$$$$$$$seeQ199766$$

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