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Question Number 171759 Answers: 1 Comments: 0
$${solve}: \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:+\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\:\right)^{\frac{{x}}{\mathrm{2}}} \:+\:\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:−\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\right)^{\frac{{x}}{\mathrm{2}}} \:=\:\mathrm{2}^{\frac{{x}+\mathrm{2}}{\mathrm{4}}} .\:{find}\:{x} \\ $$
Question Number 171757 Answers: 0 Comments: 0
$${find}\:\left(\mathrm{2}{n}\right)! \\ $$
Question Number 171756 Answers: 0 Comments: 4
$${solve}\:{for}\:{real}\:{numbers}: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{7} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\sqrt{\mathrm{2}}\:{z}\left({x}+{y}\right) \\ $$
Question Number 171755 Answers: 0 Comments: 0
$${make}\:{R}\:{the}\:{subject}\:{of}\:{the}\:{formula}\:{if} \\ $$$${P}=\frac{{M}}{\mathrm{5}}\left({X}+{R}^{\mathrm{2}} \right)+\mathrm{2}. \\ $$
Question Number 171749 Answers: 2 Comments: 0
$${if}\:{a}+{b}+{c}=\mathrm{3},{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{5},{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{27}.\:{find}\:{a}^{\mathrm{100}} +{b}^{\mathrm{100}} +{c}^{\mathrm{100}} . \\ $$
Question Number 171742 Answers: 0 Comments: 3
$${solve}:\:\left(\mathrm{1}+{n}\right)!+\mathrm{2}\left({n}\right)!=\left({n}+\mathrm{2}\right)!−\mathrm{4}{n}! \\ $$
Question Number 171739 Answers: 0 Comments: 2
$${the}\:{result}\:{of}\:{dividing}\:\left(\frac{{x}^{{a}} }{{x}^{{b}} }\right)^{{a}+{b}} {by}\:\left(\frac{{x}^{{a}+{b}} }{{x}^{{a}−{b}} }\right)^{\frac{{a}^{\mathrm{2}} }{{b}\:}} {is} \\ $$
Question Number 171730 Answers: 1 Comments: 0
$${solve}\frac{\mathrm{8}^{{x}} −\mathrm{2}^{{x}} }{\mathrm{6}^{{x}} −\mathrm{3}^{{x}\:} }=\mathrm{2}.\:{find}\:{x} \\ $$$$ \\ $$
Question Number 171728 Answers: 1 Comments: 0
$$\frac{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }{{x}+{y}}=\mathrm{7} \\ $$$$\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}−{y}}=\mathrm{19}.\:{find}\:{x}\:{and}\:{y} \\ $$
Question Number 171727 Answers: 3 Comments: 0
Question Number 176836 Answers: 0 Comments: 3
$$\mathrm{s}\left(\mathrm{B}'−\mathrm{A}'\right)=\mathrm{4} \\ $$$$\mathrm{s}\left(\mathrm{B}−\mathrm{A}\right)=\mathrm{6} \\ $$$$\mathrm{s}\left(\mathrm{A}\right)=\mathrm{9} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{subsets}\:\mathrm{of}\:\mathrm{B}\:\mathrm{with} \\ $$$$\mathrm{at}\:\mathrm{most}\:\mathrm{2}\:\mathrm{elements}\:? \\ $$
Question Number 176839 Answers: 2 Comments: 0
$$\mathrm{If}\:{a}=\mathrm{2}+\sqrt{\mathrm{15}}\:\:\mathrm{Express} \\ $$$$\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{15}}+\frac{\mathrm{196}}{\mathrm{54}}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a} \\ $$
Question Number 171725 Answers: 1 Comments: 0
Question Number 171719 Answers: 1 Comments: 2
$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {x}}\:=\:\frac{\mathrm{48}}{\mathrm{35}} \\ $$
Question Number 171718 Answers: 0 Comments: 0
Question Number 171717 Answers: 1 Comments: 0
Question Number 171716 Answers: 0 Comments: 0
$${whats}\:{the}\:{formulla}\:{of}\:{E}\left({x}\right)\:{and}\:{Find}\: \\ $$$$\int_{\mathrm{1}} ^{\:\frac{\mathrm{5}}{\mathrm{2}}} {E}\left({x}^{\mathrm{2}} \right){dx} \\ $$
Question Number 171712 Answers: 0 Comments: 0
$${Calculate}: \\ $$$${I}\left(\alpha\right)=\int\frac{\mathrm{1}}{{t}^{\alpha} \left(\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)}{dt} \\ $$
Question Number 171708 Answers: 1 Comments: 0
$$\int_{\mathrm{0}} ^{\infty} \:\mathrm{2}{x}−\mathrm{3}\:{dx}=... \\ $$
Question Number 171706 Answers: 2 Comments: 0
Question Number 171705 Answers: 1 Comments: 0
Question Number 171694 Answers: 2 Comments: 0
Question Number 171693 Answers: 2 Comments: 0
Question Number 171688 Answers: 0 Comments: 0
$$\mathrm{In}\:\:\bigtriangleup\mathrm{ABC} \\ $$$$\mathrm{AA}^{'} \:,\:\mathrm{BB}^{'} \:,\:\mathrm{CC}^{'} \:-\:\mathrm{cevians} \\ $$$$\mathrm{AA}^{'} \:\cap\:\mathrm{BB}^{'} \:\cap\:\mathrm{CC}^{'} \:=\:\left\{\mathrm{P}\right\} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{min}\left(\left[\mathrm{APC}^{'} \right],\left[\mathrm{BPA}^{'} \right],\left[\mathrm{CPB}^{'} \right]\right)+\mathrm{min}\left(\left[\mathrm{APB}^{'} \right],\left[\mathrm{BPC}^{'} \right],\left[\mathrm{CPA}^{'} \right]\right)\:\leqslant\:\frac{\mathrm{Rr}\:\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Question Number 171684 Answers: 1 Comments: 0
Question Number 171681 Answers: 1 Comments: 0
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