Question Number 74726 by ajfour last updated on 29/Nov/19 Commented by ajfour last updated on 29/Nov/19 $${Find}\:{the}\:{sides}\:{of}\:{the}\:{squares}, \\ $$$${p},\:{q},\:{r}\:\:{in}\:{terms}\:{of}\:{a},\:{b},\:{c}\:. \\ $$ Commented by mr W…
Question Number 140260 by EnterUsername last updated on 05/May/21 $$\mathrm{Passage}:\:\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{three}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{representing}\:\mathrm{the}\:\mathrm{points}\:{A},\:{B}\:\mathrm{and}\:{C},\:\mathrm{respectively},\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{Argands}\:\mathrm{plane}\:\mathrm{and}\:\angle{BAC}=\alpha,\:\mathrm{then} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{z}_{\mathrm{3}} −{z}_{\mathrm{1}} }{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }=\left(\frac{{AC}}{{AB}}\right)\left(\mathrm{cos}\alpha+{i}\mathrm{sin}\alpha\right) \\ $$$$\left({i}\right)\:\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}…
Question Number 9189 by raj bunsha last updated on 22/Nov/16 $${x}+{y}+{z}=−\mathrm{6} \\ $$$${xy}+{xz}+{yz}=\mathrm{11} \\ $$$${xyz}=−\mathrm{6} \\ $$$${then}\:{wbat}\:{are}\:{thd}\:{value}\:{of}\:{x}\:{yand}\:{z} \\ $$ Answered by mrW last updated on…
Question Number 9188 by geovane10math last updated on 22/Nov/16 $${Thanks}! \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9187 by geovane10math last updated on 22/Nov/16 $${Thanks}! \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 74723 by necxxx last updated on 29/Nov/19 $$\mathrm{3}{xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5} \\ $$$${find}\:{the}\:{second}\:{derivative} \\ $$ Commented by mathmax by abdo last updated on 29/Nov/19…
Question Number 74720 by chess1 last updated on 29/Nov/19 Commented by mathmax by abdo last updated on 29/Nov/19 $${changement}\:\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}\:={t}\:{give}\:{x}−\mathrm{1}\:={t}^{\mathrm{2}} \left({x}+\mathrm{1}\right)\:\Rightarrow\left(\mathrm{1}−{t}^{\mathrm{2}} \right){x}=\mathrm{1}+{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}}…
Question Number 74716 by chess1 last updated on 29/Nov/19 Answered by Tanmay chaudhury last updated on 29/Nov/19 $${x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)×{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)×…{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{25}} }\right)=\mathrm{1} \\ $$$${x}^{\mathrm{25}} ×{A}=\mathrm{1} \\ $$$${x}=\left(\frac{\mathrm{1}}{{A}}\right)^{\frac{\mathrm{1}}{\mathrm{25}}}…
Question Number 140249 by ghoku last updated on 05/May/21 $$\underset{\mathrm{0}\:} {\overset{\mathrm{1}} {\int}}\underset{\mathrm{0}\:} {\overset{\mathrm{2}} {\int}}\left({k}\overset{\mathrm{2}} {{z}}+{y}\right){dxdy}=\mathrm{9}\:{k}=?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 140248 by EnterUsername last updated on 05/May/21 $$\mathrm{Let}\:{p}\:\mathrm{and}\:{q}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{having}\:\mathrm{no}\:\mathrm{positive} \\ $$$$\mathrm{common}\:\mathrm{divisors}\:\mathrm{except}\:\mathrm{unity}.\:\mathrm{Let}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,…,\:{z}_{{q}} \:\mathrm{be}\:\mathrm{the} \\ $$$${q}\:\mathrm{values}\:\mathrm{of}\:{z}^{{p}/{q}} ,\:\mathrm{where}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{complex}\:\mathrm{number}.\:\mathrm{Then} \\ $$$$\mathrm{the}\:\mathrm{product}\:{z}_{\mathrm{1}} {z}_{\mathrm{2}} …{z}_{{q}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{A}\right)\:{z}^{{p}}…