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AlgebraQuestion and Answers: Page 2
Question Number 225239 Answers: 0 Comments: 0
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$$\:\:\:{tgx}+{tgy}+{tgz}={A} \\ $$$$\:\:\:{tg}^{\mathrm{3}} {x}+{tg}^{\mathrm{3}} {y}+{tg}^{\mathrm{3}} {z}=? \\ $$
Question Number 224989 Answers: 1 Comments: 0
$$\mathrm{Calculate} \\ $$$${D}_{{n}} =\begin{vmatrix}{{x}_{\mathrm{1}} +{a}_{\mathrm{1}} ^{\mathrm{2}} }&{{a}_{\mathrm{1}} {a}_{\mathrm{2}} }&{\ldots}&{{a}_{\mathrm{1}} {a}_{{n}} }\\{{a}_{\mathrm{2}} {a}_{\mathrm{1}} }&{{x}_{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} }&{\ldots}&{{a}_{\mathrm{2}} {a}_{{n}} }\\{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{{a}_{{n}} {a}_{\mathrm{1}} }&{{a}_{{n}} {a}_{\mathrm{2}} }&{\ldots}&{{x}_{{n}} +{a}_{{n}} ^{\mathrm{2}} }\end{vmatrix} \\ $$
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$${Find}\:{for}\:{acute}\:\theta,\:\mathrm{sin}\:\theta\:{and}\:\mathrm{cos}\:\theta \\ $$$${in}\:{terms}\:{of}\:\mathrm{0}<{k}<\mathrm{1}, \\ $$$${if}\:\:\:\:\frac{\mathrm{sin}\:\theta\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{sin}\:\theta\right)}={k}. \\ $$
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$$\:\:\:\left(\mathrm{2}{x}^{\mathrm{3}} +{x}−\mathrm{3}\right)^{\mathrm{3}} =\mathrm{3}−{x}^{\mathrm{2}} \\ $$
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$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{sin}\left(\mathrm{54}°\right)\:=\:\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{4}} \\ $$
Question Number 224629 Answers: 4 Comments: 0
Question Number 224614 Answers: 1 Comments: 0
$${A}\:{binary}\:{operation}\:\ast\:{is}\:{defined}\:{on}\:{a}\:{set} \\ $$$${real}\:{numbers},\:{R}\:{by} \\ $$$${x}\ast{y}\:=\:\mathrm{2}{x}\:+\:\mathrm{2}{y}\:−\frac{{xy}}{\mathrm{3}}\:. \\ $$$${find}: \\ $$$$\left({i}\right)\:{The}\:{inverse}\:{of}\:{x}\:{under}\:{the}\:{operation}\:\ast \\ $$$$\left({ii}\right)\:{Truth}\:{set}\:{when}\:{m}\ast\mathrm{7}=−\mathrm{2}\ast{m} \\ $$
Question Number 224591 Answers: 1 Comments: 0
$$\mathrm{p}\:{is}\:{a}\:{prime}\:{number} \\ $$$${prove}\:{that}\:{if}\:\mathrm{p}^{\mathrm{2}} +\mathrm{8}\:{is}\:{prime}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{p}^{\mathrm{3}} +\mathrm{4}\:{is}\:{also}\:{prime} \\ $$
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Question Number 224539 Answers: 3 Comments: 0
$${a}^{{x}} ={m},\:{a}^{{y}} ={n}\:,{a}^{\mathrm{2}} =\left({m}^{{y}} {n}^{{x}} \right)^{{z}} \\ $$$${prove}\:{xyz}=\mathrm{1} \\ $$
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