Question Number 209359 by peter frank last updated on 08/Jul/24 Commented by mr W last updated on 08/Jul/24 $$“{with}\:{a}\:{vertical}\:{velocity}\:{v}''\:{should}\:{be} \\ $$$$“{with}\:{a}\:{velocity}\:{v}''. \\ $$$${answer}\:\frac{{v}^{\mathrm{2}} −{g}^{\mathrm{2}} {x}^{\mathrm{2}}…
Question Number 209385 by efronzo1 last updated on 08/Jul/24 Answered by Rasheed.Sindhi last updated on 08/Jul/24 $${p}\left({x}\right)={ax}^{\mathrm{2}} +{bx}+\mathrm{1}\Rightarrow{p}\left(\mathrm{1}\right)={a}+{b}+\mathrm{1} \\ $$$${q}\left({x}\right)={bx}^{\mathrm{2}} +{ax}+\mathrm{1}\Rightarrow{q}\left(\mathrm{1}\right)={b}+{a}+\mathrm{1} \\ $$$${p}\left(\:{q}\left(\mathrm{1}\right)\:\right)={q}\left(\:{p}\left(\mathrm{1}\right)\:\right) \\ $$$${p}\left({a}+{b}+\mathrm{1}\right)={q}\left({a}+{b}+\mathrm{1}\right)…
Question Number 209353 by alcohol last updated on 08/Jul/24 Answered by Berbere last updated on 08/Jul/24 $${f}\left({x}+{y}\right)={f}\left({x}\right).{f}\left({y}\right) \\ $$$${f}\left({x}\right)={f}\left(\frac{{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}\right)=\left({f}\left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\forall{x}\in\mathbb{R}\:{f}\left({x}\right)\geqslant\mathrm{0} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)\:\forall{y}\in\mathbb{R}\:{fixe}\:{x}\rightarrow{f}\left({x}+{y}\right)\:{est}\:{derivable} \\…
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Question Number 209341 by Tawa11 last updated on 07/Jul/24 $$\mathrm{solve}\:\:\:\:\mathrm{x}^{\mathrm{log}\:\mathrm{27}} \:\:+\:\:\mathrm{9}^{\mathrm{log}\:\mathrm{x}} \:\:=\:\:\:\mathrm{36} \\ $$ Answered by Frix last updated on 07/Jul/24 $${x}^{\mathrm{log}_{{b}} \:\mathrm{27}} ={x}^{\frac{\mathrm{3ln}\:\mathrm{3}}{\mathrm{ln}\:{b}}} \\…
Question Number 209342 by essaad last updated on 07/Jul/24 Answered by Berbere last updated on 07/Jul/24 $${U}_{{n}+\mathrm{1}} ={f}\left({U}_{{n}} \right) \\ $$$${x}\overset{{f}} {\rightarrow}\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}};{f}\:{increase} \\ $$$${f}\left(\left[\mathrm{0},\mathrm{1}\right]\right)=\left[\mathrm{0},\frac{\mathrm{3}}{\mathrm{4}}\right]\subset\left[\mathrm{0},\mathrm{1}\right]…
Question Number 209336 by Tawa11 last updated on 07/Jul/24 Commented by Tawa11 last updated on 07/Jul/24 $$\mathrm{Find}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shadded}. \\ $$ Answered by mr W last updated…
Question Number 209332 by efronzo1 last updated on 07/Jul/24 Answered by Frix last updated on 07/Jul/24 $${x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\left(\mathrm{16}−{k}\right){x}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{0}\:\:{x}_{\mathrm{2}} =\mathrm{4}−\sqrt{{k}}\:\:{x}_{\mathrm{3}} =\mathrm{4}+\sqrt{{k}} \\…
Question Number 209329 by Huy250 last updated on 07/Jul/24 $${Given}\:{an}\:{acute}\:{triangle}\:{with}\:{AB}\:<{AC} \\ $$$${is}\:{inscribed}\:{in}\:{the}\:{circle}\:\left({O}\right).\:{Let}\:{D} \\ $$$${and}\:{E}\:{be}\:{the}\:{midpoints}\:{of}\:{the}\:{minor}\:{arc} \\ $$$${and}\:{major}\:{arc}\:{BC},\:{respectively}.\:{Let}\:{I}\:{and}\:{J} \\ $$$${be}\:{the}\:{incenters}\:{of}\:{trianges}\:{ABD}\:{and}\:{ACD}, \\ $$$${respectively}.\:{Prove}\:{that}\:{EI}={EJ}. \\ $$ Commented by Huy250…
Question Number 209352 by Spillover last updated on 07/Jul/24 Answered by mr W last updated on 08/Jul/24 $${L}={mr}^{\mathrm{2}} \omega={constant} \\ $$$$\omega_{{max}} =\frac{{L}}{{mr}_{{min}} ^{\mathrm{2}} }=\frac{{L}}{{m}\left({a}−{c}\right)^{\mathrm{2}} }…