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Question Number 141104 by Opredador last updated on 15/May/21 Answered by hknkrc46 last updated on 15/May/21 $$\left.\begin{matrix}{\sqrt{\mathrm{7}^{\sqrt{\mathrm{63}}} }\:=\:\sqrt{\mathrm{7}^{\mathrm{3}\sqrt{\mathrm{7}}} }\:=\:\mathrm{7}^{\sqrt{\mathrm{7}}} \sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }}\\{\mathrm{7}^{\sqrt{\mathrm{7}}} \:=\:\boldsymbol{{u}}}\end{matrix}\right\}\:\frac{\boldsymbol{{u}}\sqrt{\boldsymbol{{u}}}\:−\:\sqrt{\boldsymbol{{u}}}}{\boldsymbol{{u}}\:−\:\mathrm{1}} \\ $$$$=\:\frac{\sqrt{\boldsymbol{{u}}}\left(\boldsymbol{{u}}\:−\:\mathrm{1}\right)}{\boldsymbol{{u}}\:−\:\mathrm{1}}\:=\:\sqrt{\boldsymbol{{u}}}\:=\:\sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }…
Question Number 75440 by aliesam last updated on 11/Dec/19 Commented by MJS last updated on 11/Dec/19 $$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{good}\:\mathrm{at}\:\mathrm{this}\:\mathrm{but}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{true} \\ $$$$\mathrm{for}\:\mathrm{any}\:{x}\:\mathrm{at}\:\mathrm{least}\:{x}\geqslant\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \mathrm{e}^{−\left(\mathrm{2}{k}−\mathrm{1}\right){x}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{e}^{−\mathrm{2}{x}}…
Question Number 75421 by peter frank last updated on 10/Dec/19 $${If}\:{p}\:{is}\:{a}\:{point}\:{in}\:{the}\:{base} \\ $$$${AB}\:{of}\:\:{a}\:\:{triangle}\:\:{ABC} \\ $$$${such}\:{that}\:{AP}\:\::{PB}={P}:{Q} \\ $$$${prove}\:{that} \\ $$$$\left({p}+{q}\right)\mathrm{cot}\:\theta={q}\mathrm{cot}\:{A}−{p}\mathrm{cot}\:{B} \\ $$ Commented by som(math1967) last…
Question Number 9847 by 0942679167 last updated on 07/Jan/17 Commented by prakash jain last updated on 08/Jan/17 $$\mathrm{Can}\:\mathrm{u}\:\mathrm{please}\:\mathrm{type}\:\mathrm{the}\:\mathrm{question} \\ $$$$\mathrm{image}\:\mathrm{is}\:\mathrm{not}\:\mathrm{readble}. \\ $$$$\mathrm{Or}\:\mathrm{use}\:\mathrm{app}\:\mathrm{like}\:\mathrm{camscanner}\:\mathrm{to} \\ $$$$\mathrm{take}\:\mathrm{the}\:\mathrm{image} \\…
Question Number 9804 by FilupSmith last updated on 05/Jan/17 $$\mathrm{Express}\:{f}\left({x}\right)={x}^{{t}} {e}^{{x}} \:\mathrm{as}\:\mathrm{a}\:\mathrm{series} \\ $$ Commented by FilupSmith last updated on 07/Jan/17 $${x}^{{t}} ={e}^{{t}\mathrm{ln}\left({x}\right)} \\ $$$$\:\:\:\:=\underset{{n}=\mathrm{0}}…
Question Number 75327 by TawaTawa last updated on 09/Dec/19 $$\mathrm{If}\:\:\:\:\:\mathrm{T}_{\mathrm{n}\:+\:\mathrm{1}} \:\:\:=\:\:\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{T}_{\mathrm{n}} \\ $$$$\mathrm{Find}\:\mathrm{a}\:\mathrm{formular}\:\mathrm{for}\:\:\mathrm{T}_{\mathrm{n}} \:\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\:\mathrm{n} \\ $$$$\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{terms} \\ $$ Commented by malwaan last updated on 11/Dec/19…
Question Number 9776 by tawakalitu last updated on 01/Jan/17 $$\mathrm{A}\:\mathrm{plant}\:\mathrm{group}\:\mathrm{of}\:\mathrm{company}\:\mathrm{is}\:\mathrm{to}\:\mathrm{convey}\:\mathrm{a}\: \\ $$$$\mathrm{machinery}\:\mathrm{from}\:\mathrm{city}\:\mathrm{X}\:\mathrm{with}\:\mathrm{co}\:\mathrm{ordinate} \\ $$$$\left(\mathrm{300},\mathrm{200}\right)\:\mathrm{to}\:\mathrm{city}\:\mathrm{Z}\:\mathrm{with}\:\mathrm{co}\:\mathrm{ordinate}\:\left(\mathrm{700},\mathrm{900}\right). \\ $$$$\mathrm{The}\:\mathrm{machinery}\:\mathrm{can}\:\mathrm{be}\:\mathrm{air}\:\mathrm{lifted}\:\mathrm{at}\:\mathrm{a}\:\mathrm{cost}\:\mathrm{of} \\ $$$$#\mathrm{9}\:\mathrm{per}\:\mathrm{kilometer}\:\mathrm{or}\:\mathrm{through}\:\mathrm{city}\:\mathrm{Y}\:\mathrm{with} \\ $$$$\mathrm{co}\:\mathrm{ordinate}\:\left(\mathrm{700},\mathrm{200}\right)\:\mathrm{by}\:\mathrm{an}\:\mathrm{articulated}\:\mathrm{truck} \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{cost}\:\mathrm{of}\:#\mathrm{7}\:\mathrm{per}\:\mathrm{kilometer}.\: \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{route}\:\mathrm{that}\:\mathrm{is}\:\mathrm{least}\:\mathrm{expensive}. \\…
Question Number 9752 by tawakalitu last updated on 30/Dec/16 Answered by mrW last updated on 31/Dec/16 $$\mathrm{let}\:\mathrm{a}_{\mathrm{1}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{first}\:\mathrm{term}\:\mathrm{and} \\ $$$$\mathrm{d}\:\mathrm{the}\:\mathrm{common}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{A}.\mathrm{P}. \\ $$$$ \\ $$$$\mathrm{N}=\frac{\left[\mathrm{2a}_{\mathrm{1}} +\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right]×\mathrm{n}}{\mathrm{2}}…