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Category: Mensuration

In-a-square-ABCD-a-triangle-APQ-inscribed-in-it-AP-4-cm-PQ-3-cm-and-AQ-5-cm-Point-P-is-on-the-side-BC-and-point-Q-is-on-side-CD-Find-the-area-of-the-square-ABCD-

Question Number 134097 by bobhans last updated on 27/Feb/21 $$\mathrm{In}\:\mathrm{a}\:\mathrm{square}\:\mathrm{ABCD}\:,\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{APQ}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{it}.\:\mathrm{AP}=\mathrm{4}\:\mathrm{cm}, \\ $$$$\mathrm{PQ}=\mathrm{3}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{AQ}=\mathrm{5}\:\mathrm{cm}.\:\mathrm{Point} \\ $$$$\mathrm{P}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:\mathrm{BC}\:\mathrm{and}\:\mathrm{point}\:\mathrm{Q} \\ $$$$\mathrm{is}\:\mathrm{on}\:\mathrm{side}\:\mathrm{CD}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{square}\:\mathrm{ABCD}. \\ $$ Answered by mr…

S-k-1-1-k-2-1-1-k-2-1-S-l-or-S-

Question Number 132230 by pticantor last updated on 12/Feb/21 $$ \\ $$$$ \\ $$$$\boldsymbol{{S}}=\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{k}}^{\mathrm{2}} −\mathrm{1}}}−\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{{k}}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$$$\boldsymbol{{S}}\:={l}\:{or}\:\boldsymbol{{S}}=\infty\:\:??? \\ $$ Answered by JDamian…

Let-P-be-a-probalistics-space-show-that-A-B-if-P-A-B-P-A-and-P-B-A-P-B-then-P-A-B-P-A-

Question Number 131847 by pticantor last updated on 09/Feb/21 $$\:\boldsymbol{{Let}}\:\left(\boldsymbol{\Omega},\digamma,\boldsymbol{{P}}\right)\:\boldsymbol{{be}}\:\boldsymbol{{a}}\:\boldsymbol{{probalistics}}\:\boldsymbol{{space}}\: \\ $$$$\boldsymbol{{show}}\:\boldsymbol{{that}},\: \\ $$$$\boldsymbol{{A}},\boldsymbol{{B}}\in\digamma \\ $$$$\boldsymbol{{if}}\:\left(\boldsymbol{{P}}\left(\boldsymbol{{A}}\mid\boldsymbol{{B}}\right)\leqslant{P}\left(\boldsymbol{{A}}\right)\:\boldsymbol{{and}}\:\boldsymbol{{P}}\left(\boldsymbol{{B}}\mid\boldsymbol{{A}}\right)\ll\boldsymbol{{P}}\left({B}\right)\right) \\ $$$${t}\boldsymbol{{h}}{en}\:\boldsymbol{{P}}\left({A}\mid\overset{\_} {\boldsymbol{{B}}}\right)\geqslant\boldsymbol{{P}}\left(\boldsymbol{{A}}\right) \\ $$$$ \\ $$ Answered by…