Question Number 145336 by physicstutes last updated on 04/Jul/21 $$\mathrm{Which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{linear}\:\mathrm{diophantine}\:\mathrm{equations}\:\mathrm{has} \\ $$$$\mathrm{positive}\:\mathrm{solutions}, \\ $$$$\mathrm{A}.\:{x}+\:\mathrm{5}{y}\:=\:\mathrm{7} \\ $$$$\mathrm{B}.\:{x}+\mathrm{5}{y}\:=\:\mathrm{3} \\ $$$$\mathrm{B}.\:{x}\:+\:\mathrm{5}{y}\:=\:\mathrm{2} \\ $$$$\mathrm{C}.\:{x}+\:\mathrm{5}{y}\:=\:\mathrm{1} \\ $$ Commented by Rasheed.Sindhi…
Question Number 145324 by mathdanisur last updated on 04/Jul/21 $$\int\:\frac{{dx}}{\mathrm{1}\:+\:{x}^{\mathrm{6}} }\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 04/Jul/21 $$\frac{\mathrm{1}}{\mathrm{6}}\mathrm{atan}\left(\frac{{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{atan}{x}+\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}\mathrm{ln}\mid\frac{{x}^{\mathrm{2}} +\sqrt{\mathrm{3}}{x}+\mathrm{1}}{{x}^{\mathrm{2}} −\sqrt{\mathrm{3}}{x}+\mathrm{1}}\mid+\mathrm{C} \\…
Question Number 79792 by mr W last updated on 28/Jan/20 $${JUST}\:{FOR}\:{FUN} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{2}}{\mathrm{3}},\:\mathrm{1},\:\frac{\mathrm{8}}{\mathrm{5}},\:\frac{\mathrm{8}}{\mathrm{3}},\:? \\ $$$${what}\:{do}\:{you}\:{think}\:{is}\:{the}\:{next}\:{number}\:? \\ $$$${why}? \\ $$ Commented by jagoll last…
Question Number 14245 by Ruth1 last updated on 30/May/17 $$\mathrm{pls}\:\mathrm{no};\:\mathrm{6},\mathrm{7},\mathrm{10},\mathrm{11},\mathrm{and}\:\mathrm{13} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 145314 by loveineq last updated on 04/Jul/21 $$\mathrm{Let}\:{a},{b}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\:=\:\left({a}+{b}\right)^{\mathrm{2}} \:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\left({a}+{b}\right)\sqrt{\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\left({b}+\mathrm{1}\right)^{\mathrm{3}} }\:\leqslant\:\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\left({b}+\mathrm{1}\right)^{\mathrm{2}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}}\left[\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\left({b}+\mathrm{1}\right)^{\mathrm{3}} \right] \\ $$ Commented by justtry last…
Question Number 14244 by Ruth1 last updated on 30/May/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17 $$\circledcirc\mathrm{10} \\ $$$$\frac{{jx}}{\mathrm{1}+{jy}}=\frac{\mathrm{3}{x}+{j}\mathrm{4}}{{x}+\mathrm{3}{y}}\Rightarrow{x}^{\mathrm{2}} {j}+\mathrm{3}{xyj}=\mathrm{3}{x}+{j}\mathrm{4}+\mathrm{3}{xyj}+\mathrm{4}{yj}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{4}\right)−\left(\mathrm{3}{x}−\mathrm{4}{y}\right)=\mathrm{0}\Rightarrow \\ $$$$\begin{cases}{{x}^{\mathrm{2}}…
Question Number 145304 by imjagoll last updated on 04/Jul/21 $$\:\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{1}!}\:+\frac{\mathrm{5x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{7x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{9x}^{\mathrm{4}} }{\mathrm{4}!}+…+\infty=? \\ $$ Answered by qaz last updated on 04/Jul/21 $$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}^{\mathrm{n}}…
Question Number 79757 by mathocean1 last updated on 27/Jan/20 $$\mathrm{And}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{is} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}{y}=\mathrm{0}. \\ $$$$\left({T}\right)\:\mathrm{is}\:\mathrm{his}\:\mathrm{his}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{at}\:\mathrm{M}\left({x}_{\mathrm{0}} ;{y}_{\mathrm{0}} \right) \\ $$$${passing}\:{by}\:{D}\left(\mathrm{2};\mathrm{1}\right). \\ $$$$ \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{y}\:\mathrm{verify}\:\mathrm{y}_{\mathrm{0}} ^{\mathrm{2}}…
Question Number 14211 by prakash jain last updated on 29/May/17 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{4} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{9} \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{zx}=\mathrm{16} \\ $$ Commented by…
Question Number 145280 by mathdanisur last updated on 03/Jul/21 $${if}\:\:\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{x}}+\mathrm{2}\:\:{find}\:\:\frac{\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{1}}{\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by puissant last updated on 03/Jul/21 $$\mathrm{x}^{\mathrm{3}} =\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\:\Rightarrow\:\mathrm{x}^{\mathrm{3}}…