Question Number 23537 by math solver last updated on 01/Nov/17 Commented by math solver last updated on 01/Nov/17 $$\mathrm{q}.\mathrm{7}\:? \\ $$ Commented by ajfour last…
Question Number 89070 by ajfour last updated on 15/Apr/20 $$\frac{\sqrt{{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} }}\:=\:\left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){x} \\…
Question Number 23535 by gopikrishnan005@gmail.com last updated on 01/Nov/17 $${if}\:{a}\:{compound}\:{statement}\:{is}\:{made}\:{up}\:{of}\:{three}\:{simple}\:{statements}\:{then}\:{the}\:{number}\:{of}\:{rows}\:{in}\:{the}\:{truth}\:{table}\:{is} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 23531 by ajfour last updated on 01/Nov/17 Commented by ajfour last updated on 01/Nov/17 $${If}\:{the}\:{system}\:{consisting}\:{of}\:{a} \\ $$$${ring}\:{of}\:{radius}\:\boldsymbol{{R}},\:{mass}\:\boldsymbol{{M}},\:{mounted} \\ $$$${on}\:{a}\:{horizontal}\:{axis}\:{with}\:{a}\:{disc} \\ $$$${of}\:{radius}\:\boldsymbol{{r}},\:{mass}\:\boldsymbol{{m}}\:{connected}\: \\ $$$${as}\:{shown},\:{through}\:{a}\:{rod}\:{of}\:{mass}\:{m}_{\mathrm{0}}…
Question Number 89064 by john santu last updated on 15/Apr/20 $$\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{…}}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{…}}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{…}}}}}}\:=\:… \\ $$ Answered by M±th+et£s last updated on 16/Apr/20 $$\mathrm{3}=\sqrt{\mathrm{1}+\mathrm{8}}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{16}}}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{25}}}} \\ $$$$=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{36}}}}}=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{49}}}}}……} \\ $$$$=…………=\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+\mathrm{6}\sqrt{\mathrm{1}+…..}}}}}}…
Question Number 154596 by mathdanisur last updated on 19/Sep/21 $$\mathrm{let}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{z}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }}\:+\:\sqrt{\mathrm{2}}\:\geqslant\:\mathrm{2}\:\sqrt{\frac{\mathrm{z}}{\mathrm{x}+\mathrm{y}}}\:+\:\frac{\sqrt{\mathrm{2xy}}}{\mathrm{x}+\mathrm{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 89063 by talminator2856791 last updated on 15/Apr/20 $$\:\mathrm{let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{2}{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{4} \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{region}\:\mathrm{enclosed}\:\mathrm{by}\:{f}\:',\:{f}\:''\:\mathrm{and}\:{f}\:''' \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 154599 by mathdanisur last updated on 19/Sep/21 Commented by mr W last updated on 21/Oct/21 $${see}\:{Q}\mathrm{157292} \\ $$ Answered by amumath last updated…
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Question Number 154593 by Niiicooooo last updated on 19/Sep/21 Commented by Niiicooooo last updated on 19/Sep/21 Answered by ARUNG_Brandon_MBU last updated on 19/Sep/21 $${S}=\underset{{n}=\mathrm{1}} {\overset{\infty}…