Question Number 118482 by peter frank last updated on 17/Oct/20 $$\mathrm{If}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{focal}\:\:\mathrm{chord}\:\mathrm{of} \\ $$$$\mathrm{parabola}\:\mathrm{meet}\:\mathrm{the} \\ $$$$\mathrm{tangent}\:\mathrm{at}\:\mathrm{the}\:\:\mathrm{vertex} \\ $$$$\mathrm{in}\:\mathrm{C},\mathrm{D}.\mathrm{prove}\:\mathrm{that}\:\mathrm{CD} \\ $$$$\mathrm{substends}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angle} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{focus} \\ $$…
Question Number 118452 by 2004 last updated on 17/Oct/20 $$\boldsymbol{{Question}}: \\ $$$$\mathrm{2}^{\boldsymbol{{x}}} +\mathrm{2}^{\mathrm{2}\boldsymbol{{x}}+\mathrm{1}} +\mathrm{1}=\boldsymbol{{y}}^{\mathrm{2}} \:\:\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{equation}}\:\boldsymbol{{if}} \\ $$$$\boldsymbol{{x}},\boldsymbol{{y}\epsilon}\mathbb{Z} \\ $$ Commented by prakash jain last updated…
Question Number 118435 by bramlexs22 last updated on 17/Oct/20 $${If}\:\mathrm{4}\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:−\mathrm{9}\:\sqrt[{\mathrm{8}\:}]{{x}^{\mathrm{9}} }\:+\:\mathrm{4}\:=\:\mathrm{0}\:,\:{then}\: \\ $$$$\:\sqrt[{\mathrm{4}\:}]{{x}^{\mathrm{9}} }\:+\:\sqrt[{\mathrm{4}\:}]{{x}^{−\mathrm{9}} }\:=? \\ $$ Answered by benjo_mathlover last updated on 18/Oct/20…
Question Number 118391 by mathocean1 last updated on 17/Oct/20 $${show}\:{that}\:{if}\:\:^{} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:{can}\:{be}\:{divised} \\ $$$${by}\:\mathrm{7},\:{a}+{b}\:{can}\:{also}\:{be}\:{divised}\:{by}\:\mathrm{7}. \\ $$ Answered by mindispower last updated on 17/Oct/20 $${a}^{\mathrm{2}}…
Question Number 118371 by MJS_new last updated on 17/Oct/20 $$\mathrm{old}\:\mathrm{problem}\:{question}\:\mathrm{118120} \\ $$$$\mathrm{tan}\:\mathrm{tan}\:{x}\:=\mathrm{tan}\:\mathrm{3}{x}\:−\mathrm{tan}\:\mathrm{2}{x} \\ $$$$\mathrm{let}\:{t}=\mathrm{tan}\:{x} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\mathrm{tan}\:{t}\:=\frac{{t}^{\mathrm{5}} +\mathrm{2}{t}^{\mathrm{3}} +{t}}{\mathrm{3}{t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\mathrm{for}\:{t}\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\left(\mathrm{approximating}\right) \\ $$$${t}_{\mathrm{0}} =\mathrm{0}…
Question Number 118346 by 1549442205PVT last updated on 17/Oct/20 $$\mathrm{Let}\:\mathrm{30}\:\mathrm{furniture}\:\mathrm{sets}\:\mathrm{arrive}\:\mathrm{at}\:\mathrm{two}\:\mathrm{city} \\ $$$$\mathrm{stations}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:,\mathrm{15}\:\mathrm{sets}\:\mathrm{for}\:\mathrm{each}\:\mathrm{station} \\ $$$$\mathrm{All}\:\mathrm{funiture}\:\mathrm{sets}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{delivered} \\ $$$$\mathrm{to}\:\mathrm{two}\:\mathrm{furniture}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D},\mathrm{and}\:\mathrm{10} \\ $$$$\mathrm{sets}\:\mathrm{must}\:\mathrm{be}\:\mathrm{delivered}\:\mathrm{to}\:\mathrm{store}\:\mathrm{C},\mathrm{and} \\ $$$$\mathrm{to}\:\mathrm{store}\:\mathrm{D}−\mathrm{20}.\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{delivery} \\ $$$$\mathrm{one}\:\mathrm{furniture}\:\mathrm{set}\:\mathrm{from}\:\mathrm{the}\:\mathrm{station}\:\mathrm{A}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D}\:\mathrm{costs}\:\mathrm{1}\:\mathrm{and}\:\mathrm{3}\:\mathrm{monetary} \\…
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Question Number 183864 by Michaelfaraday last updated on 31/Dec/22 Commented by MJS_new last updated on 31/Dec/22 $$\left.{b}\right) \\ $$ Commented by Michaelfaraday last updated on…
Question Number 118322 by mr W last updated on 16/Oct/20 $${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$ Commented by malwaan last updated on 17/Oct/20 $$\frac{{yz}+{xz}+{xy}}{{xyz}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\:\left({x}\neq\mathrm{0};{y}\neq\mathrm{0};{z}\neq\mathrm{0}\right)…
Question Number 183848 by Michaelfaraday last updated on 30/Dec/22 Commented by MJS_new last updated on 31/Dec/22 $${y}=−{a}\vee{y}=−{b} \\ $$$$\mathrm{look}\:\mathrm{at}\:\mathrm{it},\:\mathrm{it}'\mathrm{s}\:\mathrm{plain}\:\mathrm{to}\:\mathrm{see} \\ $$ Answered by a.lgnaoui last…