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Question-196015

Question Number 196015 by Ahmed777hamouda last updated on 15/Aug/23 Commented by Ahmed777hamouda last updated on 15/Aug/23 $$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{{i}}_{\boldsymbol{{D}}{p}} ,{i}_{{Dn}} \\ $$ Terms of Service Privacy Policy…

Question-196007

Question Number 196007 by sonukgindia last updated on 15/Aug/23 Answered by mr W last updated on 15/Aug/23 $${AC}=\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\overset{\frown} {{BC}}=\left(\pi−\alpha\right){R} \\ $$$$\left(\pi−\alpha\right){R}=\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\left(\pi−\alpha\right)\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\mathrm{1}…

the-family-A-has-5-members-and-the-family-B-has-4-members-there-are-6-personsfrom-other-families-in-how-many-ways-can-you-arrange-these-15-persons-around-a-round-table-such-that-no-member-from-fami

Question Number 195964 by mr W last updated on 15/Aug/23 $${the}\:{family}\:{A}\:{has}\:\mathrm{5}\:{members}\:{and}\:{the} \\ $$$${family}\:{B}\:{has}\:\mathrm{4}\:{members}.\:{there}\:{are}\: \\ $$$$\mathrm{6}\:{personsfrom}\:{other}\:{families}. \\ $$$${in}\:{how}\:{many}\:{ways}\:{can}\:{you}\:{arrange} \\ $$$${these}\:\mathrm{15}\:{persons}\:{around}\:{a}\:{round}\:{table} \\ $$$${such}\:{that}\:{no}\:{member}\:{from}\:{family}\:{A} \\ $$$${and}\:{no}\:{member}\:{from}\:{family}\:{B}\:{are} \\ $$$${next}\:{to}\:{each}\:{other}?…

lim-x-1-1-2-1-x-1-3-1-x-1-3-with-out-l-pital-rule-

Question Number 195982 by mathlove last updated on 14/Aug/23 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\left[\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}−\sqrt{{x}}\right)}−\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{1}−\sqrt[{\mathrm{3}}]{{x}}\right)}\right]=? \\ $$$${with}\:{out}\:{l}'{pital}\:{rule} \\ $$ Answered by MM42 last updated on 14/Aug/23 $${lim}_{{x}\rightarrow\mathrm{1}} \:\left(\frac{\mathrm{1}+\sqrt{{x}}}{\mathrm{2}\left(\mathrm{1}−{x}\right)}−\frac{\mathrm{1}+\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}{\mathrm{3}\left(\mathrm{1}−{x}\right)}\right)…