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Question Number 41495 by Dawajan Nikmal last updated on 08/Aug/18 $$\mathrm{The}\:\mathrm{equation}\:\pi^{{x}} =−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{9}\:\mathrm{has} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18 $${no}\:{solution} \\ $$…
Question Number 41291 by Mayur last updated on 04/Aug/18 $$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{parallelogram}\:\mathrm{whose} \\ $$$$\mathrm{diagonals}\:\mathrm{intersect}\:\mathrm{at}\:{P}\:\mathrm{and}\:\mathrm{ley}\:{O}\:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{origin},\:\mathrm{then}\:\overset{\rightarrow} {{OA}}+\overset{\rightarrow} {{OB}}+\overset{\rightarrow} {{OC}}+\overset{\rightarrow} {{OD}}\: \\ $$$$\mathrm{equals} \\ $$ Answered by MJS…
Question Number 41141 by Kishan Daroga last updated on 02/Aug/18 $$\mathrm{If}\:{X}=\begin{bmatrix}{\mathrm{3}}&{−\mathrm{4}}\\{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{X}^{{n}} \mathrm{is} \\ $$ Commented by math khazana by abdo last updated on 04/Aug/18 $${let}\:{A}\:=\:\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:−\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\:\:{the}\:{caracteristic}\:{polynome}\:{of}…
Question Number 41138 by Kishan Daroga last updated on 02/Aug/18 $$\mathrm{A}\:\mathrm{parallelogram},\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{whose}\: \\ $$$$\mathrm{sides}\:\mathrm{are}\:\mathrm{12}\:\mathrm{cm}\:\mathrm{and}\:\mathrm{8}\:\mathrm{cm},\:\mathrm{has}\:\mathrm{one}\: \\ $$$$\mathrm{diagonal}\:\:\:\mathrm{10}\:\mathrm{cm}\:\mathrm{long}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{length} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{diagonal}. \\ $$ Commented by ajfour last updated on…
Question Number 106666 by deep last updated on 06/Aug/20 $$\mathrm{In}\:\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{8}\:\mathrm{and}\:\:\mathrm{5}{x}+{Ky}=\mathrm{3},\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:{K}\:\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{given}\:\mathrm{system} \\ $$$$\mathrm{of}\:\mathrm{equation}\:\mathrm{has}\:\mathrm{infinte}\:\mathrm{solution}. \\ $$ Commented by bemath last updated on 06/Aug/20 $$\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:\mathrm{3}}\\{\mathrm{5}\:\:\:\:\:\:\mathrm{k}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{3}}\end{pmatrix} \\…
Question Number 106664 by deep last updated on 06/Aug/20 $$\mathrm{Factorise}:\:\:\:{x}^{\mathrm{6}} \:+\:\mathrm{64}{y}^{\mathrm{6}} \\ $$ Answered by som(math1967) last updated on 06/Aug/20 $$\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} +\left(\mathrm{4y}^{\mathrm{2}} \right)^{\mathrm{3}} \\…
Question Number 106665 by deep last updated on 06/Aug/20 $$\:\mathrm{Find}\:\mathrm{a}\:\mathrm{fourth}\:\mathrm{proportional}\:\mathrm{to} \\ $$$$\:\mathrm{3},\:\mathrm{12}\:\mathrm{and}\:\mathrm{15} \\ $$ Answered by nimnim last updated on 06/Aug/20 $${Let}\:{x}\:{be}\:{the}\:{fourth}\:{proportional}. \\ $$$$\therefore\:\:\:\:\:\:\mathrm{3}:\mathrm{12}::\mathrm{15}:{x} \\…
Question Number 41079 by nadeem last updated on 01/Aug/18 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two} \\ $$$$\mathrm{larger}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{24}\:\mathrm{and}\:\mathrm{22},\:\mathrm{respectively}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{are}\:\mathrm{in}\:\mathrm{AP},\:\mathrm{then}\:\mathrm{the}\:\mathrm{third} \\ $$$$\mathrm{side}\:\mathrm{is} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 01/Aug/18…
Question Number 40321 by faaiz last updated on 19/Jul/18 $$\mathrm{If}\:{abc}=\mathrm{8}\:\mathrm{and}\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{then} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{ab}+{bc}+{ca}. \\ $$ Commented by math khazana by abdo last updated on 19/Jul/18 $$\Rightarrow{abc}\left(\frac{\mathrm{1}}{{a}}+\:\frac{\mathrm{1}}{{b}}\:+\frac{\mathrm{1}}{{c}}\right)=\frac{\mathrm{3}}{\mathrm{2}}{abc}\Rightarrow…