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Question Number 193469 Answers: 0 Comments: 0
Question Number 193467 Answers: 3 Comments: 3
$$\mathrm{Proof}\:: \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\mathrm{sint}}+\sqrt{\mathrm{1}−\mathrm{sint}}}{\:\sqrt{\mathrm{1}+\mathrm{sint}}−\sqrt{\mathrm{1}−\mathrm{sint}}}\right)=\frac{\mathrm{t}}{\mathrm{2}}\: \\ $$
Question Number 193411 Answers: 2 Comments: 1
$$\:\:\:\:\: \\ $$$$ \\ $$
Question Number 193410 Answers: 2 Comments: 0
$$\:\:\begin{cases}{\mathrm{x}=\sqrt{\mathrm{3}−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\\{\mathrm{y}=\sqrt{\mathrm{3}+\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\end{cases}\: \\ $$$$\:\:\:\:\underbrace{\boldsymbol{{x}}} \\ $$
Question Number 193371 Answers: 3 Comments: 0
$$\mathrm{Reduce}\:\mathrm{to}\:\mathrm{first}\:\mathrm{order}\:\mathrm{and}\:\mathrm{solve}\:, \\ $$$$\mathrm{showing}\:\mathrm{each}\:\mathrm{step}\:\mathrm{in}\:\mathrm{detail}. \\ $$$$\mathrm{1}.\:\mathrm{y}''\:+\left(\mathrm{y}'\right)^{\mathrm{3}} \mathrm{siny}=\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{y}''=\mathrm{1}+\left(\mathrm{y}'\right)^{\mathrm{2}} \\ $$$$ \\ $$
Question Number 193363 Answers: 1 Comments: 0
$${y}\:=\:\mathrm{4}\:×\:\mathrm{10}^{\mathrm{2}{x}} \\ $$$$\mathrm{Express}\:{x}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{y},\:\mathrm{giving}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{simplified}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{log}\:\mathrm{base}\:\mathrm{10}. \\ $$
Question Number 193356 Answers: 2 Comments: 0
Question Number 193346 Answers: 2 Comments: 0
$$\sqrt{\mathrm{4x}−\mathrm{3}}−\sqrt{\mathrm{2x}−\mathrm{5}}=\mathrm{2} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$
Question Number 193339 Answers: 1 Comments: 0
$${Prove}\:{that}\:{a}\:{group}\:{G}\:{of}\:{prime}\:{order}\:{is}\:{cyclic}. \\ $$$$ \\ $$
Question Number 193314 Answers: 1 Comments: 0
$$ \\ $$$$\boldsymbol{{a}}\:,\boldsymbol{{b}},\:\boldsymbol{{c}}\:\:>\:\mathrm{0}\:\&\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} =\mathrm{3}\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\: \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}}\:\right)^{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} } \:}\leqslant\left(\mathrm{1}+\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}\right) \\ $$
Question Number 193296 Answers: 2 Comments: 1
$$\mathrm{If}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:=\:\mathrm{16},\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$$\mathrm{and}\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{20}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:? \\ $$
Question Number 193284 Answers: 1 Comments: 0
$$\:\:\:\mathrm{6}{y}\:−\mathrm{2}{xy}\:=\:\mathrm{4} \\ $$$$\:\:\:\:\mathrm{8}{z}\:−\:{yz}\:=\:\mathrm{9} \\ $$$$\:\:\:\mathrm{10}{x}\:−\:\mathrm{4}{xz}\:=\:\mathrm{8}\: \\ $$$${find}\:{x}+{y}\:+{z}\:=\:? \\ $$
Question Number 193238 Answers: 1 Comments: 0
$${s}={a}+{b}+{c}+{d}+..... \\ $$$${number}\:{terms}\::{n} \\ $$$$\left\{{a};{b};{c};{d}.....\right\}>\mathrm{0} \\ $$$$\left.{then}\:{E}={s}/\left({s}−{a}\right)+{s}/\left({s}−{b}\right)+{s}/{s}−{c}\right)+.... \\ $$$$\left.{a}\left.\right)\:{E}>={n}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:{b}\right){E}>={n}^{\mathrm{2}} /\left({n}−\mathrm{1}\right) \\ $$$$\left.{c}\left.\right)\:{E}>={n}/\left({n}+\mathrm{1}\right)\:\:\:\:\:\:{d}\right)\:{E}>={n}^{\mathrm{2}} /\left({n}+\mathrm{1}\right) \\ $$$$\left.{e}\right)\:{E}>={n}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Question Number 193221 Answers: 1 Comments: 0
$$ \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{cube}}\:\boldsymbol{{root}}\:\boldsymbol{{of}} \\ $$$$\mathrm{9}\boldsymbol{{ab}}^{\mathrm{2}} \:+\:\left(\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{24}\boldsymbol{{a}}^{\mathrm{2}} \right)\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{a}}^{\mathrm{2}} } \\ $$
Question Number 193213 Answers: 3 Comments: 0
$$\left(\frac{{a}^{{m}} }{{a}^{−{n}} }\right)^{{m}−{n}} \\ $$
Question Number 193197 Answers: 0 Comments: 0
$$\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{666}} {mod}\mathrm{1000}=? \\ $$
Question Number 193195 Answers: 0 Comments: 0
$$\left(−\mathrm{3}\right)^{\mathrm{666}} {mod}\mathrm{1000}=? \\ $$
Question Number 193182 Answers: 1 Comments: 0
$$ \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1}\: \\ $$$$ \\ $$$${find}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{w}^{\mathrm{2}} \right). \\ $$
Question Number 193087 Answers: 1 Comments: 0
Question Number 193130 Answers: 3 Comments: 1
$$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\mathrm{2}} −{c}=\sqrt{{c}−{x}} \\ $$
Question Number 193049 Answers: 0 Comments: 0
Question Number 193045 Answers: 1 Comments: 0
$${what}\:{is}\:{the}\:{HCF}\:{of}\: \\ $$$$\mathrm{8}{k}+\mathrm{1}\:{and}\:\mathrm{9}{k}\:?\:{where}\:{k}\:\in\:\mathbb{Z}^{+} \\ $$
Question Number 193026 Answers: 2 Comments: 0
Question Number 192958 Answers: 4 Comments: 0
Question Number 192957 Answers: 4 Comments: 2
$$ \\ $$$${bx}^{\mathrm{3}} =\mathrm{10}{a}^{\mathrm{2}} {bx}\:+\:\mathrm{3}{a}^{\mathrm{3}} {y}\:,\:{ay}^{\mathrm{3}} =\:\mathrm{10}{ab}^{\mathrm{2}} {y}\:+\:\mathrm{3}{b}^{\mathrm{3}} {x} \\ $$$${solve}\:{for}\:{x}\:{and}\:{y}\:{in}\:{terms}\:{of}\:\left({a}\:,\:{b}\right) \\ $$$${and}\:{solve}\:{for}\:{a}\:{and}\:{b}\:{in}\:{terms}\:{of}\:\:\left({x}\:,\:{y}\:\right) \\ $$
Question Number 192942 Answers: 2 Comments: 1
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