Question Number 194209 by mnjuly1970 last updated on 01/Jul/23 $$ \\ $$$$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{\mathrm{1}+{kx}}\:+{x}\:=\:\mathrm{1}\:\:\:{has} \\ $$$$\:\:\:\:\:\:{two}\:{real}\:{roots}\:.\:\:\Rightarrow\:{k}=? \\ $$$$\:\:\:{kx}+\mathrm{1}=\:\mathrm{1}−\mathrm{3}{x}+\mathrm{3}{x}^{\mathrm{2}} −{x}^{\:\mathrm{3}} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:{x}^{\:\mathrm{3}} \:−\mathrm{3}{x}^{\:\mathrm{2}} +\:\left({k}+\mathrm{3}\right){x}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{0}…
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Question Number 194211 by horsebrand11 last updated on 30/Jun/23 $$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=? \\ $$ Answered by cortano12 last updated on 30/Jun/23…
Question Number 194204 by Rupesh123 last updated on 30/Jun/23 Answered by MM42 last updated on 30/Jun/23 $${if}\:\:{n}={m}=\mathrm{0}\Rightarrow\mathrm{2}{a}_{\mathrm{0}} −\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{a}_{\mathrm{0}} \right)\Rightarrow{a}_{\mathrm{0}} =\mathrm{1} \\ $$$${if}\:\:{m}=\mathrm{1}\:,\:{n}=\mathrm{0}\Rightarrow\mathrm{2}{a}_{\mathrm{1}} −\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{2}} +{a}_{\mathrm{0}} \right)…
Question Number 194207 by Aymoyali last updated on 30/Jun/23 $$\:\:\:\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\: \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{x}−\mathrm{4}} \:\:=\sqrt{\mathrm{3}\:} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194206 by Aymoyali last updated on 30/Jun/23 Answered by Skabetix last updated on 30/Jun/23 $$\left(\mathrm{3}.\mathrm{1}\right){e}^{\mathrm{2}} >\left(\mathrm{2}.\mathrm{3}\right)\prod^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 194201 by Shlock last updated on 30/Jun/23 Commented by BaliramKumar last updated on 30/Jun/23 $$\frac{\mathrm{3}\left(\mathrm{3}\sqrt{\mathrm{2}}\:−\:\mathrm{2}\right)}{\mathrm{7}} \\ $$ Commented by Rupesh123 last updated on…
Question Number 194197 by Mingma last updated on 30/Jun/23 Answered by mr W last updated on 01/Jul/23 Commented by mr W last updated on 01/Jul/23…
Question Number 194193 by Spillover last updated on 29/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 194194 by Spillover last updated on 29/Jun/23 Terms of Service Privacy Policy Contact: info@tinkutara.com