Question Number 207930 by hardmath last updated on 30/May/24 $$\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:\:\:\mathrm{number}\:\mathrm{series} \\ $$$$\mathrm{a}_{\boldsymbol{\mathrm{k}}+\mathrm{3}} ^{\mathrm{2}} \:\:+\:\:\mathrm{a}_{\boldsymbol{\mathrm{k}}} \:\:=\:\:\mathrm{a}_{\boldsymbol{\mathrm{k}}+\mathrm{2}} \:\:+\:\:\mathrm{a}_{\boldsymbol{\mathrm{k}}+\mathrm{7}} \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{k}}\:=\:? \\ $$ Commented by mr W…
Question Number 207919 by efronzo1 last updated on 30/May/24 $$\:\:\:\:\downharpoonleft\underline{\:} \\ $$ Commented by Frix last updated on 30/May/24 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{0}\vee\mathrm{15} \\ $$…
Question Number 207920 by efronzo1 last updated on 30/May/24 $$\:\:\:\:\mathrm{Given}\:\mathrm{p},\mathrm{q}\:,\mathrm{r}\:\mathrm{and}\:\mathrm{s}\:\mathrm{real}\:\mathrm{positive}\: \\ $$$$\:\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\:\mathrm{r}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\\{\mathrm{p}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} −\mathrm{ps}\:=\:\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} +\mathrm{qr}.}\end{cases} \\ $$$$\:\:\mathrm{Find}\:\:\frac{\mathrm{pq}+\mathrm{rs}}{\mathrm{ps}+\mathrm{qr}}\:. \\…
Question Number 207885 by hardmath last updated on 29/May/24 $$\mathrm{Find}: \\ $$$$\boldsymbol{\mathrm{i}}^{\mathrm{4}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{8}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{12}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{16}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{20}} \:+\:\boldsymbol{\mathrm{i}}^{\mathrm{24}} \:+…+\:\boldsymbol{\mathrm{i}}^{\mathrm{100}} \:=\:? \\ $$ Commented by mr W…
Question Number 207897 by hardmath last updated on 29/May/24 $$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{12}\:\:\centerdot\:\:\mathrm{13}\:\:\centerdot\:\:\mathrm{14}\:\:\centerdot\:\:\mathrm{15}\:\:+\:\:\mathrm{1}}\:\:=\:\:? \\ $$ Answered by Frix last updated on 29/May/24 $$\sqrt{{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)+\mathrm{1}}= \\ $$$$=\sqrt{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}}…
Question Number 207876 by hardmath last updated on 29/May/24 $$\mathrm{Find}:\:\:\:\mathrm{ln}\:\left(\frac{\mathrm{2}\:\mathrm{tg}\:\mathrm{22},\mathrm{30}°}{\mathrm{1}\:−\:\mathrm{tg}^{\mathrm{2}} \:\mathrm{22},\mathrm{30}°}\right)\:=\:? \\ $$ Commented by mr W last updated on 29/May/24 $${use}\:{your}\:{calculator}! \\ $$ Answered…
Question Number 207866 by efronzo1 last updated on 29/May/24 $$\:\:\:\mathrm{If}\:\mathrm{the}\:\mathrm{system}\:\begin{cases}{\mathrm{y}=−\mathrm{mx}^{\mathrm{2}} −\mathrm{2}}\\{\mathrm{4x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\:\mathrm{4}}\end{cases} \\ $$$$\:\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\: \\ $$$$\:\:\mathrm{them}\:\mathrm{m}\:=\: \\ $$$$\:\:\left(\mathrm{A}\right)\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\left(\mathrm{B}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\:\left(\mathrm{C}\right)\:\mathrm{1}\:\:\:\:\left(\mathrm{D}\right)\:\sqrt{\mathrm{2}}\:\:\:\left(\mathrm{E}\right)\:\sqrt{\mathrm{3}} \\ $$$$\:\: \\ $$ Answered by…
Question Number 207852 by hardmath last updated on 28/May/24 Answered by MM42 last updated on 28/May/24 $${y}\left({y}+{x}+\mathrm{2}\right)={x}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} +{xy}+\mathrm{2}{y}−{x}^{\mathrm{2}} =\mathrm{0}={f}\left({x},{y}\right) \\ $$$${x}=\mathrm{4}\Rightarrow{y}^{\mathrm{2}} +\mathrm{6}{y}−\mathrm{16}=\mathrm{0}\Rightarrow{y}=\mathrm{2} \\…
Question Number 207845 by hardmath last updated on 28/May/24 $$\mathrm{Find}: \\ $$$$\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}\:+…+\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{40}} \\ $$ Answered by Frix last updated on 28/May/24 $$\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{k}=\mathrm{1}} {\overset{{j}}…
Question Number 207846 by kgmxdd last updated on 28/May/24 Answered by Frix last updated on 30/May/24 $${x}=\frac{\mathrm{2}}{\mathrm{3}−{x}} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$$${x}=\mathrm{1}\vee{x}=\mathrm{2} \\ $$ Commented…