Question Number 186437 by aba last updated on 04/Feb/23 $$ \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{x}\sqrt{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} +\mathrm{1}}}}}\:}\:{dx} \\ $$$$\: \\ $$ Commented by…
Question Number 55368 by Tawa1 last updated on 22/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19 $$\left(\mathrm{800}−\mathrm{160}\right)^{\mathrm{2}} =\left(\mathrm{800}\right)^{\mathrm{2}} −\mathrm{2}{as}_{\mathrm{1}} \\ $$$$\mathrm{2}{as}_{\mathrm{1}} =\left(\mathrm{800}\right)^{\mathrm{2}} −\left(\mathrm{640}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{as}_{\mathrm{1}}…
Question Number 55335 by Tawa1 last updated on 21/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Feb/19 $$\frac{{a}}{{p}}=\frac{{b}}{{q}}\rightarrow{aq}={bp} \\ $$$${tw}\mathrm{0}\:{st}\:{line}\:{have}\:{no}\:{solution}\:{if}\:{they}\:{are}\:{parallel} \\ $$$${to}\:{each}\:{other}… \\ $$$${ax}+{by}={c} \\ $$$${y}=\frac{−{a}}{{b}}{x}+\frac{{c}}{{b}}\:\:{so}\:{slope}\:{m}_{\mathrm{1}}…
Question Number 55308 by Tawa1 last updated on 21/Feb/19 Answered by Rio Mike last updated on 21/Feb/19 $${PCl}_{\mathrm{5}\:} \underset{\bigtriangleup} {\rightarrow}\:{PCl}_{\mathrm{3}} \:+\:{Cl}_{\mathrm{2}} \\ $$$${from}\:{equation}\:\mathrm{1}{mol}\:\rightarrow\:\mathrm{24}{dm}^{\mathrm{3}} \\ $$$${at}\:{room}\:{temperature}\:{and}\:{pressure}…
Question Number 55301 by Otchere Abdullai last updated on 20/Feb/19 $${find}\:{the}\:{fourth}\:{term}\:{in}\:{the}\:{expansion} \\ $$$${of}\:\left(\frac{\sqrt{{x}}}{{y}^{\mathrm{2}} }−\frac{{y}}{\:\sqrt{{x}}}\right)^{\mathrm{6}} \\ $$ Commented by mr W last updated on 21/Feb/19 $$\left(\frac{\sqrt{{x}}}{{y}^{\mathrm{2}}…
Question Number 55245 by Tawa1 last updated on 20/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 20/Feb/19 $${F}_{\mathrm{1}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{20}×\mathrm{10}}{\mathrm{3}^{\mathrm{2}} } \\ $$$${F}_{\mathrm{2}} =\mathrm{9}×\mathrm{10}^{\mathrm{9}} ×\frac{\mathrm{16}×\mathrm{10}}{\mathrm{2}^{\mathrm{2}} }…
Question Number 55191 by Tawa1 last updated on 19/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19 $$\left.\mathrm{18}\right){at}\:{greatest}\:{heught}\:{vertical}\:{component}\:{of} \\ $$$${velocity}\:{is}\:{zero} \\ $$$$\mathrm{0}^{\mathrm{2}} =\left({usin}\theta\right)^{\mathrm{2}} −\mathrm{2}{gh}_{\mathrm{1}} \\ $$$$\mathrm{0}^{\mathrm{2}}…
Question Number 55190 by Tawa1 last updated on 19/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 19/Feb/19 $$\mathrm{0}={usin}\alpha−{gT}_{\mathrm{1}} \\ $$$$\mathrm{0}={usin}\beta−{gT}_{\mathrm{2}} \\ $$$$\frac{{usin}\alpha}{{usin}\beta}=\frac{{gT}_{\mathrm{1}} }{{gT}_{\mathrm{2}} } \\ $$$$\frac{{T}_{\mathrm{1}}…
Question Number 120727 by 77731 last updated on 02/Nov/20 Answered by TANMAY PANACEA last updated on 02/Nov/20 $${N}_{{r}} =\mathrm{2}+\mathrm{2}{cosx}−\mathrm{2}{cosnx}−{cos}\left({n}−\mathrm{1}\right){x}−{cos}\left({n}+\mathrm{1}\right){x} \\ $$$$=\mathrm{2}+\mathrm{2}{cosx}−\mathrm{2}{cosnx}−\mathrm{2}{cosnxcosx} \\ $$$$=\mathrm{2}\left(\mathrm{1}+{cosx}\right)−\mathrm{2}{cosnx}\left(\mathrm{1}+{cosx}\right) \\ $$$$=\mathrm{2}\left(\mathrm{1}+{cosx}\right)\left(\mathrm{1}−{cosnx}\right)…
Question Number 55174 by gunawan last updated on 18/Feb/19 $$\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{convergence} \\ $$$$\:\mathrm{of}\:\mathrm{series}\:\underset{{n}=\mathrm{0}} {\overset{\propto} {\Sigma}}\:\left(\frac{\mathrm{4}−\mathrm{2}{i}}{\mathrm{1}+\mathrm{5}{i}}\right)^{{n}} {z}^{{n}} \:\mathrm{is}… \\ $$$$ \\ $$ Commented by Abdo msup. last…