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Question Number 45546 Answers: 2 Comments: 1
$$\mathrm{Simplify}:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:\:+\:\:\mathrm{2}}\:\:\:+\:\:\:\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{5}}\:\:−\:\mathrm{2}}\:\: \\ $$
Question Number 45539 Answers: 1 Comments: 2
Question Number 45543 Answers: 0 Comments: 0
Question Number 45527 Answers: 0 Comments: 2
Question Number 45520 Answers: 0 Comments: 1
$${let}\:{a}>\mathrm{0}\:{and}\:{b}>\mathrm{0}\:{calculate}\:\int\:\sqrt{{acos}^{\mathrm{2}} \theta\:+{bsin}^{\mathrm{2}} \theta}{d}\pi \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{2}{cos}^{\mathrm{2}} \theta\:+\mathrm{3}\:{sin}^{\mathrm{2}} \theta}{d}\theta\:. \\ $$
Question Number 45519 Answers: 0 Comments: 2
$${find}\:\:\int\:\sqrt{\mathrm{2}+{tan}^{\mathrm{2}} \theta}{d}\theta \\ $$
Question Number 45518 Answers: 2 Comments: 1
$${calculate}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{cos}\left({n}\theta\right)}{{n}^{\mathrm{2}} }\:\:{and}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{sin}\left({n}\theta\right)}{{n}^{\mathrm{2}} } \\ $$
Question Number 45517 Answers: 0 Comments: 0
$${find}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\:\:{with}\:\mid{x}\mid<\mathrm{1}\:\:. \\ $$
Question Number 45514 Answers: 0 Comments: 1
Question Number 45512 Answers: 0 Comments: 11
$$\mathrm{Calculate}:\:\:\:\frac{\left(\mathrm{2}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{4}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{6}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{8}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{10}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{12}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)}{\left(\mathrm{1}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{3}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{5}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{7}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\:\left(\mathrm{9}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{11}^{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$
Question Number 45511 Answers: 0 Comments: 0
$$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{ABC}\:,\:\:\angle\:\mathrm{ABC}\:=\:\mathrm{30}^{\mathrm{0}} \:,\:\:\mathrm{and}\:\:\mathrm{AC}\:=\:\mathrm{10}.\:\mathrm{A}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{to} \\ $$$$\mathrm{circumscribe}\:\mathrm{the}\:\mathrm{triangle}\:.\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$
Question Number 45506 Answers: 1 Comments: 0
$${If}\:\:\:{ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +\mathrm{2}{hxy}+\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$${be}\:{the}\:{equation}\:{of}\:{an}\:{ellipse},\:{find} \\ $$$${coordinates}\:{of}\:{its}\:{centre}. \\ $$
Question Number 45500 Answers: 1 Comments: 0
Question Number 45498 Answers: 1 Comments: 3
Question Number 45495 Answers: 1 Comments: 0
Question Number 45494 Answers: 0 Comments: 0
Question Number 45477 Answers: 0 Comments: 4
Question Number 45482 Answers: 1 Comments: 0
Question Number 45464 Answers: 1 Comments: 3
$$\boldsymbol{\mathrm{D}}\mathrm{ifferentiate}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{arctan}}\left(\frac{\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right) \\ $$
Question Number 45456 Answers: 1 Comments: 5
Question Number 45451 Answers: 1 Comments: 0
Question Number 45450 Answers: 1 Comments: 1
Question Number 45448 Answers: 0 Comments: 5
Question Number 45443 Answers: 0 Comments: 0
Question Number 45440 Answers: 1 Comments: 0
$${show}\:{that}\: \\ $$$$\frac{\mathrm{1}+\mathrm{2}{sin}\mathrm{2}\theta−{cos}\mathrm{2}\theta}{\mathrm{1}+{sin}\mathrm{2}\theta+{cos}\mathrm{2}\theta}\:\equiv\:{tan}\theta \\ $$
Question Number 45439 Answers: 2 Comments: 0
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{every}\:\mathrm{odd}\:\mathrm{integer}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\:\:\mathrm{8m}\:+\:\mathrm{1} \\ $$
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