Question Number 143008 by mohammad17 last updated on 08/Jun/21 $${Solve}\::\:\:{x}={p}^{\mathrm{3}} −{p}+\mathrm{2}\:\:,\:{y}^{'} ={p} \\ $$ Commented by mohammad17 last updated on 09/Jun/21 $${how}\:{sir}\:{can}\:{you}\:{give}\:{me}\:{steb}\:{by}\:{steb}\:{please}? \\ $$ Answered…
Question Number 142992 by otchereabdullai@gmail.com last updated on 08/Jun/21 Commented by otchereabdullai@gmail.com last updated on 08/Jun/21 $$\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{we}\: \\ $$$$\mathrm{have}\:\mathrm{S}=−\mathrm{20m}\:\mathrm{in}\:\mathrm{option}\:\left(\mathrm{b}\right) \\ $$ Commented by Olaf_Thorendsen last…
Question Number 77464 by naka3546 last updated on 06/Jan/20 $$\underset{\:\:\mathrm{0}} {\int}\:\overset{\frac{\pi}{\mathrm{2}}} {\:}\mathrm{ln}\:\left(\mathrm{2}\:\mathrm{cos}\:{x}\right)\:{dx}\:\:=\:\:? \\ $$ Commented by kaivan.ahmadi last updated on 06/Jan/20 $${u}={ln}\left(\mathrm{2}{cosx}\right)\Rightarrow{du}=−{tgxdx} \\ $$$${dv}={dx}\Rightarrow{v}={x} \\…
Question Number 142977 by mohammad17 last updated on 08/Jun/21 Commented by mohammad17 last updated on 08/Jun/21 $${can}\:{you}\:{solve}\:{by}\:{Gramar}\:{rulle} \\ $$ Commented by mohammad17 last updated on…
Question Number 11902 by FilupS last updated on 04/Apr/17 $$\mathrm{Assuming}\:\mathrm{it}\:\mathrm{rained}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{rate}, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{rain}\:\mathrm{fell}\:\mathrm{at}\:\mathrm{angle}\:\theta\:\mathrm{to}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\left(\mathrm{see}\:\mathrm{diagram}\right),\:\mathrm{determine}\:\mathrm{if}\:\mathrm{walking}\:\mathrm{or} \\ $$$$\mathrm{running}\:\mathrm{causes}\:\mathrm{you}\:\mathrm{to}\:\mathrm{get}\:\mathrm{more}/\mathrm{less}\:\mathrm{wet}, \\ $$$$\mathrm{or}\:\mathrm{of}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{difference}\:\mathrm{for}: \\ $$$$\: \\ $$$$\mathrm{1}.\:\:\:\theta=\mathrm{90}°\:\:\left(\mathrm{downwards}\right) \\ $$$$\mathrm{2}.\:\theta<\mathrm{90}°\:\:\left(\mathrm{the}\:\mathrm{rain}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{on}\:\mathrm{the}\right. \\…
Question Number 11899 by ahmet last updated on 04/Apr/17 $${f}'\left({x}\right)=\left\{_{\mathrm{3}\:\:\:\:\:;{x}>\mathrm{2}} ^{\mathrm{2}{x}\:\:;\:{x}\leqslant\mathrm{2}} \right. \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{1}\:{ise}\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{3}\right)=? \\ $$$${czm}\because\:{f}\left({x}\right)=\left\{_{\mathrm{3}{x}+{c}_{\mathrm{2}} \:;\:{x}>\mathrm{2}} ^{{x}^{\mathrm{2}} +{c}_{\mathrm{1}} \:;{x}\leqslant\mathrm{2}} \right. \\ $$$${f}\left(\mathrm{2}\right)={x}^{\mathrm{2}} +{c}_{\mathrm{1}} \:{dir}…
Question Number 142971 by 0731619 last updated on 08/Jun/21 Commented by wassel last updated on 11/Jun/21 $$\left(\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}}−\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}=−\frac{\mathrm{3}}{\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{2}}{i}^{\mathrm{2}} \\ $$$$\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}=\pm{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}\:\:}\: \\ $$ Answered by Rasheed.Sindhi…
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Question Number 142945 by Rankut last updated on 07/Jun/21 $$\int_{\mathrm{0}} ^{\mathrm{0}.\mathrm{5}} \sqrt[{\mathrm{3}}]{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{3}} }\boldsymbol{{dx}} \\ $$$$ \\ $$ Answered by mindispower last updated on 07/Jun/21 $$\left(\mathrm{1}+{y}\right)^{{a}}…
Question Number 11862 by Mr Chheang Chantria last updated on 03/Apr/17 $$\boldsymbol{{Lesson}}\mathrm{1}.\:\boldsymbol{\mathrm{AM}}−\boldsymbol{\mathrm{GM}}\:'\:\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{inequality}}\:\left(\boldsymbol{\mathrm{Cauchy}}\right) \\ $$$$\boldsymbol{\mathrm{form}}\::\:\frac{\boldsymbol{{a}}_{\mathrm{1}} +\boldsymbol{{a}}_{\mathrm{2}} +…+\boldsymbol{{a}}_{\boldsymbol{{n}}} }{\boldsymbol{{n}}}\:\geqslant\:\sqrt[{\boldsymbol{{n}}}]{\boldsymbol{{a}}_{\mathrm{1}} \boldsymbol{{a}}_{\mathrm{2}} …\boldsymbol{{a}}_{\boldsymbol{{n}}} } \\ $$$$\boldsymbol{{where}}\:\boldsymbol{{a}}_{\mathrm{1}} ,\boldsymbol{{a}}_{\mathrm{2}} ,….,\boldsymbol{{a}}_{\boldsymbol{{n}}} >\mathrm{0}…