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Question-9947

Question Number 9947 by ridwan balatif last updated on 18/Jan/17 Commented by sandy_suhendra last updated on 18/Jan/17 $$\mathrm{x}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{x}}}\:\:\Rightarrow\:\mathrm{x}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\:\mathrm{x}^{\mathrm{2}} =\mathrm{3x}+\mathrm{1}\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{y}=\mathrm{3}\:+\:\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{y}}}}\:\Rightarrow\:\mathrm{y}=\mathrm{3}+\frac{\mathrm{1}}{\mathrm{y}}\:\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{3y}+\mathrm{1}\Rightarrow\mathrm{y}^{\mathrm{2}}…

s-2-s-1-2-c-2-s-s-1-2-c-2-2-s-1-c-3-2-0-solve-for-s-in-terms-of-0-lt-c-lt-2-3-3-

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…

Two-parallel-plate-capacitor-seperated-by-a-di-electric-of-thickness-5mm-They-acquire-a-charge-of-55-milli-coloumb-when-a-voltage-180-volt-is-connected-accross-them-calculate-electric-field-streng

Question Number 9936 by Tawakalitu ayo mi last updated on 16/Jan/17 $$\mathrm{Two}\:\mathrm{parallel}\:\mathrm{plate}\:\mathrm{capacitor}\:\mathrm{seperated}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{di}\:\mathrm{electric}\:\mathrm{of}\:\mathrm{thickness}\:\mathrm{5mm}.\:\mathrm{They}\:\mathrm{acquire} \\ $$$$\mathrm{a}\:\mathrm{charge}\:\mathrm{of}\:\mathrm{55}\:\mathrm{milli}\:\mathrm{coloumb},\:\mathrm{when}\:\mathrm{a}\:\mathrm{voltage} \\ $$$$\mathrm{180}\:\mathrm{volt}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{accross}\:\mathrm{them}.\: \\ $$$$\mathrm{calculate}\:\mathrm{electric}\:\mathrm{field}\:\mathrm{strength}. \\ $$ Terms of Service…

Let-0-a-b-lt-1-Prove-that-1-4-2-a-2-b-1-a-1-b-4-a-b-4-a-b-

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

lim-n-n-1-n-

Question Number 75469 by aliesam last updated on 11/Dec/19 $$\underset{{n}\rightarrow\infty} {{lim}}\left(\:\sqrt{{n}+\mathrm{1}}−\sqrt{{n}}\:\right) \\ $$ Commented by MJS last updated on 11/Dec/19 $$\mathrm{0} \\ $$$$\mathrm{because}\:{n}+\mathrm{1}\sim{n}\:\mathrm{for}\:\mathrm{large}\:{n} \\ $$…

Question-9929

Question Number 9929 by ridwan balatif last updated on 16/Jan/17 Answered by mrW1 last updated on 18/Jan/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{4}{x}\centerdot\mathrm{tan}^{\mathrm{2}} \:\mathrm{3}{x}+\mathrm{6}{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{sin}\:\mathrm{3}{x}\centerdot\mathrm{cos}\:\mathrm{2}{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{4}\left(\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{4}{x}}\right)\mathrm{tan}^{\mathrm{2}}…

If-a-gt-0-and-one-root-of-ax-2-bx-c-0-is-less-than-2-and-the-other-is-greater-than-2-then-A-4a-2-b-c-lt-0-B-4a-2-b-c-gt-0-C-4a-2-b-c-0-D-a-b-c-

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…