Question Number 74142 by MASANJAJ last updated on 19/Nov/19 Answered by Rio Michael last updated on 19/Nov/19 $$\:{T}_{\mathrm{7}} \:=\:{a}\:+\:\mathrm{6}{d}\:=\:\mathrm{6}\:−−−\left(\mathrm{1}\right) \\ $$$${T}_{\mathrm{18}} \:=\:{a}\:+\:\mathrm{17}{d}\:=\:\mathrm{22}−−−\left(\mathrm{2}\right) \\ $$$$\:{eqn}\left(\mathrm{2}\right)\:−\:{eqn}\left(\mathrm{1}\right)\:\Rightarrow\:\mathrm{11}{d}\:=\:\mathrm{16} \\…
Question Number 8607 by tawakalitu last updated on 17/Oct/16 Answered by ridwan balatif last updated on 18/Oct/16 $$\mathrm{23}.\:\:\mathrm{C}=\mathrm{0}.\mathrm{50}\mu\mathrm{F}=\mathrm{5}×\mathrm{10}^{−\mathrm{7}} \mathrm{F} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{L}=\mathrm{0}.\mathrm{050H} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{f}=\frac{\mathrm{100}}{\pi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\omega=\mathrm{2}\pi\mathrm{f}…
Question Number 74143 by MASANJAJ last updated on 19/Nov/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74140 by MASANJAJ last updated on 19/Nov/19 Answered by Rio Michael last updated on 19/Nov/19 $${let}\:{he}\:{element}\:{be}\:{m} \\ $$$$\:\:{f}\left({m}\right)\:=\:\mathrm{2} \\ $$$$\Rightarrow\:\:\mathrm{4}{m}−\mathrm{2}=\mathrm{2} \\ $$$${m}\:=\:\mathrm{1}\: \\…
Question Number 74141 by MASANJAJ last updated on 19/Nov/19 Commented by TawaTawa last updated on 19/Nov/19 $$\left.\mathrm{d}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:\:=\:\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{Forth}\:\mathrm{term}\:\:=\:\:\:\mathrm{ar}^{\mathrm{3}} \:\:=\:\:\mathrm{40} \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\mathrm{5r}^{\mathrm{3}} \:\:=\:\:\mathrm{40}…
Question Number 74138 by MASANJAJ last updated on 19/Nov/19 Commented by mathmax by abdo last updated on 19/Nov/19 $${f}\left({x}\right)={x}^{\mathrm{2}} −{x}−\mathrm{2}\:={x}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{x}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}\:=\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow{minf}\left({x}\right)=−\frac{\mathrm{9}}{\mathrm{4}}\:\:\:\:{and}\:{no}\:{maximum} \\…
Question Number 139672 by mathlove last updated on 30/Apr/21 Answered by MJS_new last updated on 30/Apr/21 $$\left({x}^{{a}} \right)^{{b}} ={x}^{{ab}} \\ $$$$\left({x}^{{a}} \right)^{{a}} ={x}^{{a}^{\mathrm{2}} } \\…
Question Number 74139 by MASANJAJ last updated on 19/Nov/19 Commented by TawaTawa last updated on 19/Nov/19 $$\mathrm{a}\::\:\mathrm{b}\:\:=\:\:\frac{\mathrm{21}}{\mathrm{4}}\:,\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\:\:\:\:\mathrm{b}\::\:\mathrm{c}\:\:=\:\:\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\therefore\:\:\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:\:=\:\:\frac{\mathrm{21}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\:\:\:\frac{\mathrm{b}}{\mathrm{c}}\:\:=\:\:\frac{\mathrm{7}}{\mathrm{3}} \\ $$$$\therefore\:\:\:\:\:\:\mathrm{a}\:\:=\:\:\frac{\mathrm{21b}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}\:\:\:=\:\:\frac{\mathrm{3b}}{\mathrm{7}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\::\:\mathrm{b}\::\:\mathrm{c}\:\:\:=\:\:\:\frac{\mathrm{21b}}{\mathrm{4}}\::\:\mathrm{b}\::\:\frac{\mathrm{3b}}{\mathrm{7}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\::\:\mathrm{b}\::\:\mathrm{c}\:\:\:=\:\:\:\frac{\mathrm{21b}}{\mathrm{4}}\:×\:\mathrm{28}\::\:\mathrm{b}\:×\:\mathrm{28}\::\:\frac{\mathrm{3b}}{\mathrm{7}}\:×\:\mathrm{28}…
Question Number 8602 by tawakalitu last updated on 17/Oct/16 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{that}\:\mathrm{will}\:\mathrm{make} \\ $$$$\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{13x}^{\mathrm{3}} \:+\:\mathrm{6x}^{\mathrm{2}} \:+\:\mathrm{px}\:+\:\mathrm{q}\:\: \\ $$$$\mathrm{a}\:\mathrm{perfect}\:\mathrm{square} \\ $$ Answered by sandy_suhendra last updated on…
Question Number 74137 by MASANJAJ last updated on 19/Nov/19 Commented by TawaTawa last updated on 19/Nov/19 $$\left(\mathrm{16a}\right) \\ $$$$\:\:\:\:\:\:\mathrm{5}\:\mathrm{painters}\:\mathrm{will}\:\mathrm{spend}\:\mathrm{less}\:\mathrm{hours} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\mathrm{1}\:\mathrm{painter}\:\:\:=\:\:\mathrm{2}\:×\:\mathrm{6}\:\:=\:\:\mathrm{12}\:\mathrm{hrs} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}\:\mathrm{painters}\:\:\:\:=\:\:\:\frac{\mathrm{12}}{\mathrm{5}}\:\mathrm{hrs} \\ $$…