Question Number 200353 by faysal last updated on 17/Nov/23 $$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} +\mathrm{3px}^{\mathrm{2}} +\mathrm{3qx}+\mathrm{r}=\mathrm{0}\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{harmonic}\:\mathrm{progression},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{2q}^{\mathrm{3}} =\mathrm{r}\left(\mathrm{3pq}−\mathrm{r}\right) \\ $$ Answered by jabarsing last updated on…
Question Number 200350 by faysal last updated on 17/Nov/23 Commented by ajfour last updated on 18/Nov/23 comments attached, how can he?! Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200351 by faysal last updated on 17/Nov/23 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{3}} −\mathrm{12x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{10}=\mathrm{0}\: \\ $$$$\mathrm{by}\:\mathrm{cardon}'\mathrm{s}\:\mathrm{method} \\ $$ Answered by Frix last updated on 17/Nov/23 $$\mathrm{The}\:\mathrm{name}\:\mathrm{is}\:{Cardano} \\…
Question Number 200336 by faysal last updated on 17/Nov/23 Answered by cortano12 last updated on 17/Nov/23 $$\:\:\begin{array}{|c|c|c|}{}&\hline{\mathrm{1}}&\hline{\mathrm{2}}&\hline{−\mathrm{3}}&\hline{\mathrm{5}}&\hline{−\mathrm{2}}&\hline{\mathrm{3}}\\{\mathrm{x}=\mathrm{2}}&\hline{\ast}&\hline{\mathrm{2}}&\hline{\mathrm{8}}&\hline{\mathrm{10}}&\hline{\mathrm{30}}&\hline{\mathrm{56}}\\{}&\hline{\mathrm{1}}&\hline{\mathrm{4}}&\hline{\mathrm{5}}&\hline{\mathrm{15}}&\hline{\mathrm{28}}&\hline{\mathrm{59}}\\\hline\end{array} \\ $$$$\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\mathrm{59} \\ $$$$\mathrm{the}\:\mathrm{quotient}\:\mathrm{is}\:\mathrm{h}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{3}} +\mathrm{5x}^{\mathrm{2}} +\mathrm{15x}+\mathrm{28} \\…
Question Number 200330 by hardmath last updated on 17/Nov/23 $$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:=\:\sqrt{\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} } \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 17/Nov/23…
Question Number 200240 by universe last updated on 16/Nov/23 $$\mathrm{the}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function} \\ $$$${f}\left({x},{y}\right)\:=\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{3}{y}+\mathrm{11} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200242 by cherokeesay last updated on 16/Nov/23 Answered by cortano12 last updated on 17/Nov/23 $$\:\frac{\left({ab}+{c}\right){x}}{{b}−\frac{{c}}{{a}}\:}\:−\frac{\left({ab}−{c}\right){x}}{{b}+\frac{{c}}{{a}}}\:=\:\frac{{ab}−{c}}{{b}+\frac{{c}}{{a}}}\:−\frac{{ab}+{c}}{{b}−\frac{{c}}{{a}}} \\ $$$$\:\frac{{ax}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{ax}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{a}\left({ab}−{c}\right)}{{ab}+{c}}\:−\frac{{a}\left({ab}+{c}\right)}{{ab}−{c}} \\ $$$$\:\:\frac{{x}\left({ab}+{c}\right)}{{ab}−{c}}\:−\frac{{x}\left({ab}−{c}\right)}{{ab}+{c}}\:=\:\frac{{ab}−{c}}{{ab}+{c}}−\frac{{ab}+{c}}{{ab}−{c}} \\ $$$$\:\frac{\left({ab}+{c}\right)\left({x}+\mathrm{1}\right)}{{ab}−{c}}\:=\:\frac{\left({ab}−{c}\right)\left({x}+\mathrm{1}\right)}{{ab}+{c}} \\ $$$$\:\left({ab}+{c}\right)^{\mathrm{2}}…
Question Number 200302 by Calculusboy last updated on 16/Nov/23 Answered by Rasheed.Sindhi last updated on 18/Nov/23 $$\sqrt{{a}+{bx}}\:+\sqrt{{b}+{cx}}\:+\sqrt{{c}+{ax}}\:\:=\sqrt{{b}−{ax}}\:+\sqrt{{c}−{bx}}\:+\sqrt{{a}−{cx}}\: \\ $$$${a}+{bx}\geqslant\mathrm{0}\:\wedge\:{b}+{cx}\geqslant\mathrm{0}\:\wedge\:{c}+{ax}\geqslant\mathrm{0} \\ $$$$\left({a}+{b}+{c}\right)+\left({a}+{b}+{c}\right){x}\geqslant\mathrm{0}\Rightarrow{x}\geqslant−\frac{{a}+{b}+{c}}{{a}+{b}+{c}}=−\mathrm{1} \\ $$$$\begin{array}{|c|}{{x}\geqslant−\mathrm{1}}\\\hline\end{array} \\ $$$${b}−{ax}\geqslant\mathrm{0}\:\wedge\:{c}−{bx}\geqslant\mathrm{0}\:\wedge\:{a}−{cx}\geqslant\mathrm{0}…
Question Number 200186 by Calculusboy last updated on 15/Nov/23 Answered by ajfour last updated on 15/Nov/23 $$\frac{{y}}{{x}}={p}\:,\:\frac{{z}}{{y}}={q}\:,\:\frac{{x}}{{z}}={r}\:=\frac{\mathrm{1}}{{pq}} \\ $$$$\mathrm{1}+{p}+{p}^{\mathrm{2}} =\left(\mathrm{2}+{p}\right)\left(\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{q}+{q}^{\mathrm{2}} =\left(\mathrm{2}+{q}\right)\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{r}+{r}^{\mathrm{2}}…
Question Number 200168 by hardmath last updated on 15/Nov/23 $$\mathrm{If}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{86}\:\:\mathrm{and}\:\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{x}\:−\:\mathrm{4} \\ $$$$\mathrm{Then}\:\mathrm{find}:\:\:\mathrm{g}\left[\mathrm{f}^{−\mathrm{1}} \left(\mathrm{g}\left(\mathrm{14}\right)\right)\right]\:=\:? \\ $$ Commented by jazeee last updated on 15/Nov/23 $$…