Question Number 140248 by EnterUsername last updated on 05/May/21 $$\mathrm{Let}\:{p}\:\mathrm{and}\:{q}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{having}\:\mathrm{no}\:\mathrm{positive} \\ $$$$\mathrm{common}\:\mathrm{divisors}\:\mathrm{except}\:\mathrm{unity}.\:\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,…,\:{z}_{{q}} \:\mathrm{be}\:\mathrm{the} \\ $$$${q}\:\mathrm{values}\:\mathrm{of}\:{z}^{{p}/{q}} ,\:\mathrm{where}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{complex}\:\mathrm{number}.\:\mathrm{Then} \\ $$$$\mathrm{the}\:\mathrm{product}\:{z}_{\mathrm{1}} {z}_{\mathrm{2}} …{z}_{{q}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{A}\right)\:{z}^{{p}}…
Question Number 140241 by mathdanisur last updated on 05/May/21 $$\int\left(\mathrm{4}{ctg}^{\mathrm{3}} {x}−\mathrm{5}{ctg}^{\mathrm{2}} {x}+\mathrm{7}{ctgx}\right){e}^{{x}} {dx}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 140225 by mondli last updated on 05/May/21 $${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}=\mathrm{0}\:\: \\ $$ Answered by MJS_new last updated on 05/May/21 $${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\:\Rightarrow\:{x}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}} \\ $$…
Question Number 9141 by tawakalitu last updated on 20/Nov/16 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{if} \\ $$$$\left(\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} \:+\:\left(\sqrt{\mathrm{2}\:−\:\sqrt{\mathrm{3}}}\right)^{\mathrm{x}} \:=\:\mathrm{4} \\ $$ Commented by tawakalitu last updated on 20/Nov/16 $$\mathrm{please}\:\mathrm{help}. \\…
Question Number 140205 by EnterUsername last updated on 05/May/21 $${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{rhombus}.\:\mathrm{Its}\:\mathrm{diagonals}\:{AC}\:\mathrm{and}\:{BD}\:\mathrm{inter}- \\ $$$$\mathrm{sect}\:\mathrm{at}\:{M}\:\mathrm{and}\:\mathrm{satisfy}\:{BD}=\mathrm{2}{AC}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{points}\:{D}\:\mathrm{and} \\ $$$${M}\:\mathrm{are}\:\mathrm{represented}\:\mathrm{by}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{1}+{i}\:\mathrm{and} \\ $$$$\mathrm{2}−{i},\:\mathrm{respectively},\:\mathrm{then}\:{A}\:\mathrm{is}\:\mathrm{represented}\:\mathrm{by} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{3}−{i}/\mathrm{2}\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{3}+{i}/\mathrm{2}\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{1}+\mathrm{3}{i}/\mathrm{2}\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{1}−\mathrm{3}{i}/\mathrm{2} \\ $$ Answered by mr W last…
Question Number 140196 by mathdanisur last updated on 05/May/21 $${Calculate}:\:\sqrt{\mathrm{3}}\:{cosec}\:\mathrm{20}°−{sec}\:\mathrm{20}° \\ $$ Answered by liberty last updated on 05/May/21 $$\:\frac{\sqrt{\mathrm{3}}}{\mathrm{sin}\:\mathrm{20}°}−\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:=\: \\ $$$$\frac{\mathrm{2}\left(\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{20}°−\mathrm{sin}\:\mathrm{20}°\right)}{\mathrm{sin}\:\mathrm{40}°}\:= \\ $$$$\frac{\mathrm{4}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{20}°−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°\right)}{\mathrm{sin}\:\mathrm{40}°}\:= \\…
Question Number 9128 by tawakalitu last updated on 20/Nov/16 Commented by tawakalitu last updated on 20/Nov/16 $$\mathrm{Note}:\:\:#\:\mathrm{means}\:\:\mathrm{Naira}\:\mathrm{in}\:\mathrm{Nigeria}. \\ $$$$\mathrm{while}:\:\:\mathrm{K}\:\mathrm{means}\:\:\mathrm{Kobo}\:\mathrm{in}\:\mathrm{Nigeria} \\ $$$$\mathrm{and} \\ $$$$\mathrm{100}\:\mathrm{kobo}\:=\:\mathrm{1}\:\mathrm{Naira}\:\:\:\:\left(\mathrm{100k}\:=\:#\mathrm{1}\right) \\ $$…
Question Number 140198 by EnterUsername last updated on 05/May/21 $$\mathrm{Let}\:{a}>\mathrm{0}\:\mathrm{and}\:\mid{z}+\left(\mathrm{1}/{z}\right)\mid={a}\:\left({z}\neq\mathrm{0}\:\mathrm{is}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}\right). \\ $$$$\mathrm{Then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{values}\:\mathrm{of}\:\mid{z}\mid\:\mathrm{are} \\ $$$$\left(\mathrm{A}\right)\:\frac{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{2}{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}} \\ $$$$\left(\mathrm{C}\right)\:\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}−{a}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{4}}−\mathrm{2}{a}}{\mathrm{2}} \\ $$ Answered by Dwaipayan…
Question Number 74663 by TawaTawa last updated on 28/Nov/19 $$\mathrm{If}\:\:\:\:\mathrm{x}^{\mathrm{x}} \:\mathrm{y}^{\mathrm{y}} \:\mathrm{z}^{\mathrm{z}} \:\:\:=\:\:\:\mathrm{c}\:\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{at}\:\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{y}\:\:=\:\:\mathrm{z} \\ $$$$\:\:\:\:\:\:\frac{\partial^{\mathrm{2}} \mathrm{z}}{\partial\mathrm{x}\partial\mathrm{y}}\:\:\:=\:\:\:−\:\left(\mathrm{x}\:\mathrm{log}\:\mathrm{ex}\right)^{−\mathrm{1}} \\ $$ Answered by mind is power last updated…