Question Number 72871 by TawaTawa last updated on 04/Nov/19 Answered by ajfour last updated on 04/Nov/19 Commented by ajfour last updated on 04/Nov/19 $${Top}\:{vertex}\:{of}\:{triangle}\: \\…
Question Number 72838 by Rio Michael last updated on 03/Nov/19 $${given}\:{that}\:\:\:{f}\left({x}\right)\:=\:\frac{\mid{x}\:−\mathrm{2}\mid}{\mathrm{1}−\mid{x}\mid} \\ $$$${check}\:{if}\:{f}\:{is}\:{continuous}\:{a}\:{x}\:=\:\mathrm{2} \\ $$$${hence}\:\:{write}\:{f}\left({x}\right)\:{as}\:{a}\:{pairwise}\:{function}\: \\ $$ Commented by mathmax by abdo last updated on…
Question Number 72841 by indalecioneves last updated on 03/Nov/19 $${Could}\:{someone}\:{help}\:{me}\:{on}\:{this}\:{question}? \\ $$$${Knowing}\:{that}\:{the}\:{area}\:{of}\:{a}\:{circle}\:{segment}\:{is}\:{given}\:{by}\:{A}={R}^{\mathrm{2}} \left(\theta−{sin}\theta\right)/\mathrm{2}.\:{Where}\:{A}=\mathrm{7}{m}^{\mathrm{2}} ;\:{R}^{\mathrm{2}} =\frac{\mathrm{28}}{\pi}. \\ $$$${What}\:{is}\:{the}\:{best}\:{answer}\:{for}\:{the}\:{angle}\:{value}\:\left({degree}\right) \\ $$$$\left.{a}\right)\:\mathrm{85}°<\theta<\mathrm{90}° \\ $$$$\left.{b}\right)\:\mathrm{95}°<\theta<\mathrm{100}° \\ $$$$\left.{c}\right)\:\mathrm{105}°<\theta<\mathrm{110}° \\ $$$$\left.{d}\right)\:\mathrm{115}°<\theta<\mathrm{120}°…
Question Number 72837 by Rio Michael last updated on 03/Nov/19 $${find}\: \\ $$$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\:+\:\left[{x}\right] \\ $$ Commented by mathmax by abdo last updated on 03/Nov/19…
Question Number 138365 by cherokeesay last updated on 12/Apr/21 Answered by mr W last updated on 12/Apr/21 $${a}={b}\:\mathrm{sin}\:\alpha \\ $$$$\frac{{a}}{{b}}=\mathrm{sin}\:\alpha \\ $$$$\underset{\alpha\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}}{{b}}=\mathrm{0} \\ $$…
Question Number 7289 by FilupSmith last updated on 21/Aug/16 $$\mathrm{For}\:\mathrm{any}\:\mathrm{number}\:{x}>\mathrm{1}:{x}\in\mathbb{Z} \\ $$$${x}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\mathrm{a}\:\mathrm{combination} \\ $$$$\mathrm{of}\:\mathrm{numbers}\:\mathrm{multiplied}\:\mathrm{together}. \\ $$$$\mathrm{e}.\mathrm{g}. \\ $$$$\mathrm{10}=\mathrm{5}×\mathrm{2} \\ $$$$\mathrm{20}=\mathrm{5}×\mathrm{4}=\mathrm{5}×\mathrm{2}×\mathrm{2} \\ $$$$\mathrm{100}=\mathrm{10}×\mathrm{10}=\mathrm{5}×\mathrm{2}×\mathrm{5}×\mathrm{2} \\ $$$$\: \\…
Question Number 72824 by Rio Michael last updated on 03/Nov/19 $${fnd}\:{all}\:{integers}\:{n}\:{for}\:{which}\: \\ $$$$\:\mathrm{13}\mid\:\mathrm{4}\left({n}^{\mathrm{2}} \:+\:\mathrm{1}\right) \\ $$ Answered by mind is power last updated on 03/Nov/19…
Question Number 72805 by oluwaponmilesulaimon last updated on 03/Nov/19 $${PARTIAL}\:{VARIATION} \\ $$$${The}\:{success}\:{rate}\:{of}\:{government}\:{variws}\: \\ $$$${inversly}\:{as}\:{the}\:{number}\:{of}\:{corrupt}\:{mi} \\ $$$${nded}\:{individual}\:{and}\:{varies}\:{directly} \\ $$$${as}\:{the}\:{number}\:{of}\:{clean}\:{minded}\:{individal} \\ $$$$.{if}\:\:{the}\:{goverment}\:{attain}\:\mathrm{95\%}\:{success} \\ $$$${rate}\:{when}\:{there}\:{are}\:{two}\:{corrupt}\:{minded} \\ $$$${and}\:\mathrm{75\%}\:{success}\:{rate}\:{when}\:{there}\:{are} \\…
Question Number 72806 by Rio Michael last updated on 03/Nov/19 $$\underset{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\rightarrow\mathrm{0}} {\:{show}\:{that}\:\:\mathrm{lim}}\:\left[\:{x}\right]\:\:{does}\:{not}\:{exist}. \\ $$$${Hence}\:{define}\:\:\left[{x}\right]\:\:{and}\:{sketch}\:{a}\:{graph}\:{for}\: \\ $$$$\:{y}\:=\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\left[{x}\right] \\ $$ Commented by mathmax by abdo last…
Question Number 138299 by cherokeesay last updated on 12/Apr/21 Answered by bemath last updated on 12/Apr/21 $${BD}\:=\:{r}_{\mathrm{1}} +\mathrm{2}\:\Leftrightarrow\:{r}_{\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2} \\ $$$${r}_{\mathrm{1}} =\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\:=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$${shaded}\:{area}=\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}}…