Question Number 9977 by konen last updated on 20/Jan/17 Answered by sandy_suhendra last updated on 20/Jan/17 $$\mathrm{log}_{\mathrm{3}^{\mathrm{2}} } \:\left(\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} \:=\:\mathrm{log}_{\mathrm{9}} \:\left(\mathrm{x}−\mathrm{8}\right) \\ $$$$\left(\mathrm{x}−\mathrm{6}\right)^{\mathrm{2}} =\mathrm{x}−\mathrm{8} \\…
Question Number 9976 by konen last updated on 20/Jan/17 $$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{4a}+\mathrm{6b}+\mathrm{13}=\mathrm{0}\:\: \\ $$$$\Rightarrow\:\mathrm{a}+\mathrm{b}=? \\ $$ Answered by prakash jain last updated on 20/Jan/17 $${a}^{\mathrm{2}}…
Question Number 141050 by mathsuji last updated on 15/May/21 Answered by MJS_new last updated on 15/May/21 $$\sqrt{{p}}+\sqrt{{q}}=\sqrt{{r}} \\ $$$${p}+{q}+\mathrm{2}\sqrt{{pq}}={r} \\ $$$$\mathrm{2}\sqrt{{pq}}={r}−\left({p}+{q}\right) \\ $$$$\mathrm{4}{pq}=\left({r}−\left({p}+{q}\right)\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}}…
Question Number 9962 by konen last updated on 19/Jan/17 $$\mathrm{a},\mathrm{b}\:\in\mathrm{Z}^{+} \\ $$$$\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} =\mathrm{2b}+\mathrm{7a}+\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{a}+\mathrm{b}=? \\ $$ Commented by prakash jain last updated on…
Question Number 9959 by konen last updated on 19/Jan/17 $$\mid\mathrm{x}^{\mathrm{2}} −\mathrm{25}\mid=\mathrm{4}\mid\mathrm{x}−\mathrm{5}\mid\:\:\:\Rightarrow\Sigma\mathrm{x}=? \\ $$ Answered by mrW1 last updated on 19/Jan/17 $$\mid\left({x}−\mathrm{5}\right)\left({x}+\mathrm{5}\right)\mid=\mathrm{4}\mid{x}−\mathrm{5}\mid \\ $$$${x}=\mathrm{5}\:{or} \\ $$$$\mid{x}+\mathrm{5}\mid=\mathrm{4}\:{or}…
Question Number 141030 by iloveisrael last updated on 15/May/21 $$\:\:\:\:\:\:\:\:\:\:\rightarrow\langle\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{3}{n}}\:=?\:\rangle\leftarrow \\ $$ Commented by EDWIN88 last updated on 15/May/21 $$\:=\:\mathrm{1}−\mathrm{ln}\:\sqrt{\mathrm{3}}\:−\frac{\pi}{\mathrm{6}\sqrt{\mathrm{3}}} \\ $$$$\:\mathrm{S}=\frac{\mathrm{1}}{\mathrm{3}}\:\underset{\mathrm{n}=\mathrm{1}}…
Question Number 9953 by konen last updated on 18/Jan/17 $$\mathrm{x}−\mathrm{y}\:=\mathrm{8} \\ $$$$\mathrm{xy}=\mathrm{7}\:\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \:=? \\ $$ Answered by mrW1 last updated on 19/Jan/17 $$\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}}…
Question Number 141012 by ajfour last updated on 14/May/21 $$\:{s}^{\mathrm{2}} \left({s}+\mathrm{1}\right)^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\left({s}+\mathrm{1}\right) \\ $$$$\:\:+\frac{{c}^{\mathrm{3}} }{\mathrm{2}}\:=\:\mathrm{0}\:\:\:\:\:{solve}\:{for}\:{s}\:{in}\:{terms} \\ $$$${of}\:\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$ Terms of Service Privacy…
Question Number 141004 by loveineq last updated on 14/May/21 $$\mathrm{Let}\:\mathrm{0}\:\leqslant\:{a},{b}\:<\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}−{a}\right)\left(\mathrm{2}−{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\:\geqslant\:\frac{\mathrm{4}+{a}+{b}}{\mathrm{4}−{a}−{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141002 by EnterUsername last updated on 14/May/21 $$\mathrm{If}\:{a}>\mathrm{0}\:\mathrm{and}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{a}\mathrm{x}^{\mathrm{2}} +{b}\mathrm{x}+\mathrm{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:−\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{2},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}<\mathrm{0} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}>\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{4}{a}+\mathrm{2}\mid{b}\mid+{c}=\mathrm{0} \\ $$$$\left(\mathrm{D}\right)\:{a}+{b}={c} \\ $$ Answered by…