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AlgebraQuestion and Answers: Page 312
Question Number 44918 Answers: 2 Comments: 5
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}.\mathrm{2}.\mathrm{3}}\:+\:\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}.\mathrm{4}}\:+\:\frac{\mathrm{5}}{\mathrm{3}.\mathrm{4}.\mathrm{5}}\:+\:...\: \\ $$
Question Number 44898 Answers: 1 Comments: 1
$${If}\:\:\:\:{x}^{\mathrm{4}} +{px}^{\mathrm{3}} +{qx}^{\mathrm{2}} +{rx}+\mathrm{5}\:=\:\mathrm{0} \\ $$$${has}\:{four}\:{real}\:{roots},\:{then}\:{find} \\ $$$$\:{the}\:{minimum}\:{value}\:{of}\:\boldsymbol{{pr}}. \\ $$
Question Number 44892 Answers: 1 Comments: 2
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{formular}\:\mathrm{for}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{kth}\:\mathrm{power}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{number} \\ $$
Question Number 44876 Answers: 1 Comments: 0
Question Number 44819 Answers: 0 Comments: 7
Question Number 44783 Answers: 1 Comments: 0
$$\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{97}.......\mathrm{1} \\ $$$$\left({x}^{\mathrm{2}} \right)^{\mathrm{3}} +\left({y}^{\mathrm{2}} \right)^{\mathrm{3}} =\mathrm{793}.......\mathrm{2} \\ $$$${solve}\:{the}\:{simultaneous}\:{equation} \\ $$
Question Number 44773 Answers: 0 Comments: 0
$$ \\ $$
Question Number 44765 Answers: 0 Comments: 0
Question Number 44763 Answers: 1 Comments: 1
Question Number 44729 Answers: 4 Comments: 0
Question Number 44691 Answers: 1 Comments: 1
Question Number 44570 Answers: 1 Comments: 0
Question Number 44543 Answers: 1 Comments: 3
$${If}\:\:{y}\:={f}\left({x}\right)\:=\:{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${and}\:\:{at}\:{some}\:{x},\:{say}\:\:{x}=\:{p} \\ $$$$\int_{\mathrm{0}} ^{\:\:{p}} {ydx}\:=\:{y}\left({p}\right)=\:{y}\:'\left({p}\right)\:=\:{y}\:''\left({p}\right)=\:{p}\:, \\ $$$${then}\:{find}\:\boldsymbol{{p}}\:. \\ $$
Question Number 44502 Answers: 1 Comments: 0
$$\mathrm{If}\:\mathrm{a}>\mathrm{b},\mathrm{and}\:\mathrm{c}>\mathrm{d},\mathrm{prove}\:\mathrm{that}\:\mathrm{a}−\mathrm{c}\:\mathrm{may}\:\mathrm{be}\:\mathrm{greater}\:\mathrm{than}, \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{or}\:\mathrm{less}\:\mathrm{than}\:\mathrm{b}−\mathrm{d}. \\ $$$$ \\ $$
Question Number 44444 Answers: 1 Comments: 0
$${simplify}\:\:\:\:\sqrt{\left(\mathrm{4}{x}^{\mathrm{2}} {y}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\left(\mathrm{8}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$
Question Number 44397 Answers: 2 Comments: 4
$${If}\:{x}\:{is}\:{nearly}\:{equal}\:{to}\:\mathrm{1}\:{then} \\ $$$$\frac{{mx}^{{m}} −{nx}^{{n}} }{{m}−{n}}= \\ $$
Question Number 44384 Answers: 2 Comments: 6
$$\mathrm{Let}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$${a}\:>\:{b}\:>\:\mathrm{0} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\sqrt{\mathrm{2}}{a}^{\mathrm{3}} \:+\:\frac{\mathrm{3}}{{ab}\:−\:{b}^{\mathrm{2}} } \\ $$
Question Number 44355 Answers: 0 Comments: 0
Question Number 44200 Answers: 0 Comments: 4
Question Number 44095 Answers: 1 Comments: 0
Question Number 44062 Answers: 1 Comments: 0
Question Number 44038 Answers: 2 Comments: 0
$$\mathrm{How}\:\mathrm{many}\:\mathrm{times}\:\mathrm{does}\:\mathrm{the}\:\mathrm{digit}\:\mathrm{6}\:\mathrm{appear}\:\mathrm{when}\:\mathrm{writing}\:\mathrm{from}\:\:\mathrm{6}\:\mathrm{to}\:\mathrm{400}\:? \\ $$
Question Number 44029 Answers: 0 Comments: 0
Question Number 43971 Answers: 1 Comments: 0
Question Number 43902 Answers: 2 Comments: 4
$$\sqrt{{a}−{b}}\:+\:\sqrt{{a}+{b}}\:=\:{c} \\ $$$$\sqrt{{a}−{c}}\:+\:\sqrt{{a}+{c}}\:=\:{b} \\ $$$${Solve}\:{for}\:{real}\:{b},\:{and}\:{c}\:;\:{in}\:{terms}\: \\ $$$${of}\:{real}\:{a}. \\ $$
Question Number 43894 Answers: 0 Comments: 0
$$\mid{z}\mid=\mid{Arg}\:\left(\frac{{a}}{{b}}\pi\right)\mid=\mathrm{1}\wedge{k},\:{n}\in\mathbb{Z}\wedge{b}\neq\mathrm{0}\leqslant{k}<{n}: \\ $$$${x}^{{n}} ={z}\Rightarrow{x}={e}^{\frac{\mathrm{2}{k}+{a}}{{bm}}\pi{i}} \\ $$$$\mathrm{To}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{please}. \\ $$
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