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Question Number 171846 Answers: 0 Comments: 2
$${solve}: \\ $$$${x}^{\mathrm{9}} −\mathrm{2}{x}^{\mathrm{5}} −\mathrm{3}{x}=\mathrm{0} \\ $$
Question Number 171845 Answers: 0 Comments: 0
$${n}_{{c}_{{r}+\mathrm{1}} } +\:{n}_{{c}_{{r}} } ={n}+\mathrm{1}_{{c}_{{r}\:\:.{solve}\:{for}\:\:{n}\:{and}\:{r}.} } \\ $$
Question Number 171844 Answers: 1 Comments: 0
$${solve}: \\ $$$${x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}=\mathrm{0} \\ $$
Question Number 171843 Answers: 0 Comments: 0
$${find}\:{a}: \\ $$$$\left(\mathrm{2}:\mathrm{3}:\mathrm{4}:\mathrm{5}\right){a}=\mathrm{5}:\mathrm{6}:\mathrm{7}:\mathrm{8} \\ $$
Question Number 171841 Answers: 1 Comments: 0
$${solve}: \\ $$$$\mathrm{2}^{\left(\mathrm{5}{x}−\mathrm{20}\right)} =\mathrm{1}! \\ $$
Question Number 171840 Answers: 0 Comments: 0
$${solve}: \\ $$$${a}^{\mathrm{3}} +\mathrm{9}{ab}^{\mathrm{2}} =\mathrm{26} \\ $$$${a}^{\mathrm{2}} {b}+{ab}^{\mathrm{3}} =\mathrm{15} \\ $$
Question Number 171839 Answers: 0 Comments: 0
$${solve}\:{the}\:{simultaneous}\:{eqn} \\ $$$${y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$$${y}={b}_{\mathrm{1}} {x}+{c}_{\mathrm{1}} \\ $$
Question Number 171838 Answers: 0 Comments: 0
$${solve}: \\ $$$$\mathrm{4}^{{x}^{\mathrm{2}} } +\mathrm{3}{x}=\mathrm{6} \\ $$
Question Number 171837 Answers: 1 Comments: 0
$${solve}: \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{2} \\ $$$${x}+{y}=\mathrm{5}{xy} \\ $$
Question Number 171836 Answers: 1 Comments: 0
$${solve}: \\ $$$${x}+\mathrm{3}{y}=\mathrm{1} \\ $$$${xy}={y}^{\mathrm{2}} −\mathrm{3} \\ $$
Question Number 171835 Answers: 1 Comments: 0
$${solve}\:{for}\:{x}: \\ $$$${x}^{\mathrm{5}} +{x}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$
Question Number 171834 Answers: 0 Comments: 0
$${solve}\:{for}\:{x}: \\ $$$${x}^{{x}^{{x}^{{x}} } } =\mathrm{2} \\ $$
Question Number 171833 Answers: 0 Comments: 0
$${solve}: \\ $$$${ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$
Question Number 171830 Answers: 0 Comments: 0
$${solve}\:{for}\:{x},{y},{z} \\ $$$${xy}+\mathrm{3}{x}+\mathrm{2}{y}=−\mathrm{6} \\ $$$${yz}+{y}+\mathrm{3}{z}=−\mathrm{3} \\ $$$${zx}+\mathrm{2}{z}+{x}=\mathrm{2} \\ $$
Question Number 171829 Answers: 1 Comments: 0
$${please}\:{i}\:{need}\:{cubic}\:{formula} \\ $$$$ \\ $$
Question Number 171828 Answers: 0 Comments: 2
$${solve}\:{for}\:{x}: \\ $$$$\mathrm{2}{log}_{{x}} \mathrm{4}={log}_{\mathrm{2}} {x} \\ $$$$ \\ $$
Question Number 171827 Answers: 0 Comments: 0
$${solve}\:{for}\:{x}\:{and}\:{y}: \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xy}=\mathrm{5} \\ $$$${xy}+\mathrm{3}{y}^{\mathrm{2}} =\mathrm{10} \\ $$
Question Number 171826 Answers: 1 Comments: 0
$${if}\:{a}+{b}+{c}=\mathrm{0},{find}\:{the}\:{value}\:{of}\: \\ $$$$\frac{{a}^{\mathrm{2}} }{{bc}}+\frac{{b}^{\mathrm{2}} }{{ca}}+\frac{{c}^{\mathrm{2}} }{{ab}} \\ $$
Question Number 171825 Answers: 1 Comments: 0
$${solve}\:{for}\:{x}\:{and}\:{y}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{10} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$
Question Number 171824 Answers: 0 Comments: 0
$${solve}\:{for}\:{a},{b},{c},{d} \\ $$$${a}+{b}=\mathrm{9} \\ $$$${ab}+{c}+{d}=\mathrm{29} \\ $$$${ad}+{bc}=\mathrm{3} \\ $$$${cd}=\mathrm{18} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Question Number 171823 Answers: 1 Comments: 0
$${solve}\:{for}\:{y}: \\ $$$${y}^{\mathrm{4}} −\mathrm{34}{y}^{\mathrm{2}} +\mathrm{225}=\mathrm{0} \\ $$
Question Number 171822 Answers: 0 Comments: 1
$${solve}\:{for}\:{x}: \\ $$$${x}^{{x}^{{x}^{\mathrm{2017}} } } =\mathrm{2017} \\ $$
Question Number 171821 Answers: 0 Comments: 1
$${solve}\:{for}\:{x}: \\ $$$${x}\sqrt{{x}}\:−\mathrm{11}\sqrt{{x}}\:=\mathrm{10},\:{find}\:{x}−\sqrt{{x}} \\ $$
Question Number 171820 Answers: 0 Comments: 1
$${solve}\:{for}\:{x}: \\ $$$${x}^{{x}} =\mathrm{2}^{\mathrm{2048}} \\ $$
Question Number 171816 Answers: 0 Comments: 0
$${solve}\:{for}\:{x}: \\ $$$$\frac{\mathrm{3}\left({x}−\mathrm{2}\right)+\mathrm{4}\sqrt{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}}}{\mathrm{2}\left({x}^{\mathrm{3}} −\mathrm{1}\right)}=\mathrm{1} \\ $$
Question Number 171815 Answers: 1 Comments: 0
$${solve}\:{for}\:{x}: \\ $$$$\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{{x}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}\:}\:{x}+\mathrm{4}} =\mathrm{1} \\ $$
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