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Question-188418

Question Number 188418 by 073 last updated on 01/Mar/23 Answered by manxsol last updated on 01/Mar/23 $${T}_{{n}} ={i}^{\mathrm{4}{n}−\mathrm{3}} \\ $$$${T}_{{n}} =\frac{{i}^{\mathrm{4}{n}} }{{i}^{\mathrm{3}} }=\frac{\left({i}^{\mathrm{2}} \right)^{\mathrm{2}{n}} }{−{i}}=\frac{\mathrm{1}}{−{i}}×\frac{{i}}{{i}}=…

1-if-sinx-tgx-1-then-sin-4-x-tg-4-x-2-if-sinx-tgx-2-then-sin4x-tg4x-3-if-sinx-tgx-3-then-sin4x-sin-4-x-tg4x-tg-4-x-

Question Number 57345 by behi83417@gmail.com last updated on 02/Apr/19 $$\left.\mathrm{1}\right)\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{1},\boldsymbol{\mathrm{then}}:\:\:\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{tg}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}=? \\ $$$$\left.\mathrm{2}\right)\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{2},\boldsymbol{\mathrm{then}}:\:\:\boldsymbol{\mathrm{sin}}\mathrm{4}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{tg}}\mathrm{4}\boldsymbol{\mathrm{x}}=? \\ $$$$\mathrm{3}.\boldsymbol{\mathrm{if}}:\:\:\:\:\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{tgx}}=\mathrm{3},\boldsymbol{\mathrm{then}}:\:\:\frac{\boldsymbol{\mathrm{sin}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{sin}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}}+\frac{\boldsymbol{\mathrm{tg}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{tg}}^{\mathrm{4}} \boldsymbol{\mathrm{x}}}=? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…

find-the-value-of-k-such-that-k-x-2-y-2-y-2x-1-y-2x-3-0-is-a-circle-hence-obtain-the-centre-and-radius-of-the-resulting-circle-

Question Number 57332 by problem solverd last updated on 02/Apr/19 $$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{such}\:\mathrm{that} \\ $$$${k}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\left({y}−\mathrm{2}{x}+\mathrm{1}\right)\left({y}+\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{hence}\:\mathrm{obtain}\: \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{resulting}\:\mathrm{circle}. \\ $$ Commented by…

1-2-x-x-dx-

Question Number 57329 by naka3546 last updated on 02/Apr/19 $$\underset{−\mathrm{1}} {\int}\overset{\mathrm{2}} {\:}\:\mid{x}\mid\:\lfloor{x}\rfloor\:{dx}\:\:=\:\:\:? \\ $$ Commented by turbo msup by abdo last updated on 02/Apr/19 $${I}=\int_{−\mathrm{1}}…

Question-188392

Question Number 188392 by yawa7373 last updated on 28/Feb/23 Commented by Rasheed.Sindhi last updated on 28/Feb/23 $$\boldsymbol{{sir}}\:{mr}\:{W}\:,\:{please}\:{see}\:{my}\:{third}\:{answer} \\ $$$${to}\:{Q}#\mathrm{188327}\:{in}\:{which}\:{I}'{ve}\:{used} \\ $$$${derivative}. \\ $$ Commented by…

lim-n-1-n-2-2-n-2-3-n-2-n-n-2-

Question Number 122848 by ZiYangLee last updated on 20/Nov/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }+\frac{\mathrm{3}}{{n}^{\mathrm{2}} }+……+\frac{{n}}{{n}^{\mathrm{2}} }\right)=? \\ $$ Answered by nico last updated on 20/Nov/20 $$=\underset{{x}\rightarrow\mathrm{0}}…

Find-the-range-of-y-cos-2-x-3cos-x-5-

Question Number 122843 by ZiYangLee last updated on 20/Nov/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{y}=\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{3cos}\:{x}+\mathrm{5} \\ $$ Commented by Dwaipayan Shikari last updated on 20/Nov/20 $$−\mathrm{1}\leqslant{cosx}\leqslant\mathrm{1} \\ $$$$−\mathrm{3}\leqslant−\mathrm{3}{cosx}\leqslant\mathrm{3}\:\:\:\:\:\:\:\:\:\: \\…

Question-188360

Question Number 188360 by 073 last updated on 28/Feb/23 Answered by Rasheed.Sindhi last updated on 28/Feb/23 $$\:\:\mathrm{S}_{\mathrm{N}} =\frac{\mathrm{N}}{\mathrm{2}}\left({a}+{l}\right) \\ $$$$\:\:\:\mathrm{A}=\frac{{n}−\mathrm{2}}{\mathrm{2}}\left(\mathrm{1}+{n}−\mathrm{2}\right)=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{B}=\frac{{n}−\mathrm{14}}{\mathrm{2}}\left(\mathrm{15}+{n}\right)=\frac{\left({n}−\mathrm{14}\right)\left({n}+\mathrm{15}\right)}{\mathrm{2}} \\ $$$$\mathrm{A}−\mathrm{B}=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{2}}−\frac{\left({n}−\mathrm{14}\right)\left({n}+\mathrm{15}\right)}{\mathrm{2}}=\mathrm{42} \\…