Question Number 2001 by Fitrah last updated on 29/Oct/15 $${Prove}\:{that}\:: \\ $$$$\frac{\mathrm{1}}{\mathrm{15}}\:<\:\frac{\mathrm{1}}{\mathrm{2}}\:\centerdot\:\frac{\mathrm{3}}{\mathrm{4}}\:\centerdot\:\frac{\mathrm{5}}{\mathrm{6}}\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:\centerdot\:\frac{\mathrm{99}}{\mathrm{100}}\:<\:\frac{\mathrm{1}}{\mathrm{10}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 1999 by Fitrah last updated on 29/Oct/15 $${if}\: \\ $$$$\left(\mathrm{1}\:+\:{tan}\:\mathrm{1}°\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{2}°\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{3}°\right)…… \\ $$$$\left…..\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{44}°\right)\left(\mathrm{1}\:+\:{tan}\:\mathrm{45}°\right)\overset{} {\:}=\:\:\left\{\:\:\:\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{50}}\:+\:\mathrm{7}\right)}\:−\:\sqrt[{\mathrm{3}}]{\left(\sqrt{\mathrm{50}}−\mathrm{7}\right)}\:\overset{\left({x}\:−\:\mathrm{7}\right)} {\right\}} \\ $$$$\: \\ $$$${find}\:{x}\:=\:…? \\ $$ Answered by Yozzi…
Question Number 67535 by mr W last updated on 28/Aug/19 Commented by mr W last updated on 28/Aug/19 $${find}\:{the}\:{area}\:{of}\:{the}\:{square}\:{A}=? \\ $$ Commented by Prithwish sen…
Question Number 67514 by mr W last updated on 28/Aug/19 Commented by mr W last updated on 28/Aug/19 $${side}\:{length}\:{of}\:{regular}\:{hexagon}\:{is}\:\mathrm{1}. \\ $$$${find}\:{the}\:{yellow}\:{area}=? \\ $$ Answered by…
Question Number 1928 by Yozzi last updated on 24/Oct/15 $${Prove}\:{that},\:{if}\:{p}>{q}>\mathrm{0}\:{and}\:{x}\geqslant\mathrm{0},\:{then} \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{{p}}\left(\frac{{x}^{{p}} }{{p}+\mathrm{1}}−\mathrm{1}\right)\geqslant\frac{\mathrm{1}}{{q}}\left(\frac{{x}^{{q}} }{{q}+\mathrm{1}}−\mathrm{1}\right).\: \\ $$ Commented by Rasheed Soomro last updated on 24/Oct/15 $${Prove}\:{that},\:{if}\:{p}>{q}>\mathrm{0}\:{and}\:{x}\geqslant\mathrm{0},\:{then}…
Question Number 67430 by TawaTawa last updated on 27/Aug/19 Commented by MJS last updated on 27/Aug/19 $$\mathrm{coordinate}\:\mathrm{method} \\ $$$$\mathrm{turn}\:\mathrm{the}\:\mathrm{triangle}\:\rightarrow\:{CA}\:\mathrm{is}\:\mathrm{the}\:\mathrm{base} \\ $$$$\mathrm{side}\:\mathrm{length}\:={s} \\ $$$${s}=\mathrm{8}+{x}\:\Rightarrow\:{x}={s}−\mathrm{8} \\ $$$${C}=\begin{pmatrix}{−\frac{{s}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:{A}=\begin{pmatrix}{\frac{{s}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\mathrm{0}}\\{\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}}\end{pmatrix}…
Question Number 67422 by mr W last updated on 27/Aug/19 Commented by Prithwish sen last updated on 27/Aug/19 $$\bigtriangleup\mathrm{AMN}\:\mathrm{is}\:\mathrm{an}\:\mathrm{isoceles}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{AM}=\mathrm{AN} \\ $$$$\mathrm{AM}+\mathrm{BM}=\:\mathrm{AN}+\mathrm{D}^{'} \mathrm{N}=\mathrm{16} \\ $$$$\mathrm{Area}\:\mathrm{ABMND}^{'} =\:\mathrm{Area}\:\mathrm{of}\:\mathrm{ABCD}\:−\mathrm{Area}\:\mathrm{of}\:\bigtriangleup\mathrm{AMN}.…
Question Number 67386 by mr W last updated on 26/Aug/19 Commented by mr W last updated on 26/Aug/19 $${Find}\:{AB}=? \\ $$$${Find}\:{shaded}\:{area}=? \\ $$ Commented by…
Question Number 67350 by TawaTawa last updated on 26/Aug/19 Commented by TawaTawa last updated on 26/Aug/19 $$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}.\: \\ $$ Answered by mr W last updated…
Question Number 132869 by syamilkamil1 last updated on 17/Feb/21 Commented by syamilkamil1 last updated on 17/Feb/21 $${what}\:{is}\:{the}\:{radius}\:{of}\:{the}\:{circle}\:,\:{if}\:{length}\:{the}\:{square}\:\mathrm{2}\:{centimetres}? \\ $$ Commented by bramlexs22 last updated on…