Question Number 222662 by MrGaster last updated on 04/Jul/25 Commented by MrGaster last updated on 04/Jul/25 $$\exists\bigtriangleup{ABC},\angle{B}=\mathrm{90}°,{BA}={AC},{BD}=\mathrm{3}\wedge\angle{CED}=\mathrm{45}°,{CE}=\sqrt{\mathrm{2}}{AE},{CE}=? \\ $$ Commented by mr W last updated…
Question Number 222679 by fantastic last updated on 04/Jul/25 $${If}\:{x}=\underset{{x}=\mathrm{1}} {\overset{\mathrm{10}} {\prod}}{x}\:{then}\:\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} {x}}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{3}} {x}}+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{4}} {x}}…+\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} {x}}=?? \\ $$ Answered by Tawa11 last updated on 04/Jul/25…
Question Number 222673 by ajfour last updated on 04/Jul/25 Commented by ajfour last updated on 04/Jul/25 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Find}\:{small}\:{circle}\:{radius}\:\boldsymbol{{a}}. \\ $$$${Oh}!\:{i}\:{forgot}\:{to}\:{say}\:{almost}-\:{ABCD}\:{is}\:{sq}. \\ $$ Answered by mr W…
Question Number 222652 by Mingma last updated on 03/Jul/25 Commented by Ghisom last updated on 03/Jul/25 $$\left[\left({A}−{B}\right)\cup\left({B}−{C}\right)\cup\left({C}−{A}\right)\right]'= \\ $$$$=\left[\left({A}\cup{B}\cup{C}\right)−\left({A}\cap{B}\cap{C}\right)\right]'= \\ $$$$={A}\cap{B}\cap{C} \\ $$ Terms of…
Question Number 222638 by Mingma last updated on 03/Jul/25 Answered by gabthemathguy25 last updated on 03/Jul/25 $$\mathrm{cot}\:{A}\:+\:\mathrm{cot}\:{B}\:+\:\mathrm{cot}\:{C}\:=\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}{S}} \\ $$$${where}\:{a},{b},{c}\:=\:{side}\:{lengths},\:{S}={area}\:{of}\:{triangle} \\ $$$${S}={r}\centerdot{s}\:\Rightarrow\:\frac{\mathrm{1}}{{S}}=\frac{\mathrm{1}}{{rs}} \\…
Question Number 222639 by Mingma last updated on 03/Jul/25 Answered by gabthemathguy25 last updated on 03/Jul/25 $$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{8}\:{cm}=\mathrm{4}\:{cm} \\ $$$$\mathrm{Slant}\:\mathrm{height}\:=\:\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\mathrm{100}+\mathrm{16}}=\sqrt{\mathrm{116}}\approx\mathrm{10}.\mathrm{77}\:\mathrm{cm} \\ $$$$\mathrm{cos}\left(\theta\right)=\frac{\mathrm{10}.\mathrm{77}^{\mathrm{2}} +\mathrm{10}.\mathrm{77}^{\mathrm{2}} −\mathrm{8}^{\mathrm{2}}…
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Question Number 222646 by Tawa11 last updated on 03/Jul/25 Answered by som(math1967) last updated on 03/Jul/25 $$\:{AB}={BC}−{d},\:{AC}={BC}+{d} \\ $$$$\:\:{AC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{BC}^{\mathrm{2}} \\ $$$$\left({BC}+{d}\right)^{\mathrm{2}} =\left({BC}−{d}\right)^{\mathrm{2}} +{BC}^{\mathrm{2}}…
Question Number 222659 by Mingma last updated on 03/Jul/25 Answered by mr W last updated on 04/Jul/25 Commented by mr W last updated on 04/Jul/25…
Question Number 222636 by Mingma last updated on 02/Jul/25 Answered by mr W last updated on 04/Jul/25 $${d}=\sqrt{{a}^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }+{x}+\sqrt{{b}^{\mathrm{2}} −\left({b}−{x}\right)^{\mathrm{2}} } \\ $$$$\:\:=\sqrt{\mathrm{2}{ax}−{x}^{\mathrm{2}} }+{x}+\sqrt{\mathrm{2}{bx}−{x}^{\mathrm{2}}…