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Question-109497

Question Number 109497 by rinasitorus_ last updated on 24/Aug/20 Answered by nurmaya_silaban last updated on 24/Aug/20 $$\left.\mathrm{1}\right){a}.\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{3}}{\mathrm{5}}=\:\frac{\mathrm{5}}{\mathrm{15}}+\frac{\mathrm{9}}{\mathrm{15}}=\frac{\mathrm{14}}{\mathrm{15}} \\ $$$$\:\:\:\:\:\mathrm{b}.\:\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{3}}{\mathrm{15}} \\ $$$$\:\:\:\:\:\mathrm{c}.\frac{\mathrm{2}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{7}}=\:\frac{\mathrm{14}}{\mathrm{21}}−\frac{\mathrm{9}}{\mathrm{21}}=\:\frac{\mathrm{5}}{\mathrm{21}} \\ $$$$\:\:\:\:\:\mathrm{d}.\frac{\mathrm{2}}{\mathrm{3}}:\frac{\mathrm{3}}{\mathrm{7}}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{7}}{\mathrm{3}}=\frac{\mathrm{14}}{\mathrm{9}} \\ $$$$…

Question-175028

Question Number 175028 by sciencestudent last updated on 16/Aug/22 Commented by mr W last updated on 17/Aug/22 $$\left({a}\right) \\ $$$${R}=\mathrm{990}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{750}}+\frac{\mathrm{1}}{\mathrm{680}}}=\mathrm{1346}.\mathrm{6}\Omega \\ $$ Terms of Service…

Question-109494

Question Number 109494 by mathdave last updated on 24/Aug/20 Answered by mathmax by abdo last updated on 24/Aug/20 $$\mathrm{for}\:\mid\mathrm{u}\mid<\mathrm{1}\:\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{du}}\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{n}} \:\Rightarrow \\ $$$$\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)\:=\sum_{\mathrm{n}=\mathrm{0}}…

1-sin-2x-cos-2x-sin-x-cos-x-0-2-lim-x-0-e-x-e-x-sin-x-3-lim-x-1-x-1-sin-x-1-

Question Number 109489 by bobhans last updated on 24/Aug/20 $$\left(\mathrm{1}\right)\:\mathrm{sin}\:\left(\mathrm{2}{x}\right)−\mathrm{cos}\:\left(\mathrm{2}{x}\right)−\mathrm{sin}\:\left({x}\right)+\mathrm{cos}\:\left({x}\right)=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{sin}\:{x}} \\ $$$$\left(\mathrm{3}\right)\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\mid{x}+\mathrm{1}\mid\:\mathrm{sin}\:\left({x}+\mathrm{1}\right) \\ $$$$ \\ $$ Answered by john…

n-0-1-n-7-6n-0-9-6n-2-0-7-6n-2-

Question Number 175023 by bubu last updated on 16/Aug/22 $$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\boldsymbol{\sum}}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{7}+\mathrm{6}\boldsymbol{\mathrm{n}}}\left[\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\frac{\mathrm{9}+\mathrm{6}\boldsymbol{\mathrm{n}}}{\mathrm{2}}\right)−\boldsymbol{\psi}^{\left(\mathrm{0}\right)} \left(\frac{\mathrm{7}+\mathrm{6}\boldsymbol{\mathrm{n}}}{\mathrm{2}}\right)\right] \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

A-large-lot-of-tires-contain-5-defectives-4-tires-are-to-be-chosen-for-a-car-Find-the-probability-that-you-find-a-2-defective-tires-before-4-good-ones-b-at-most-2-defective-tires-before-4-goo

Question Number 175011 by MathsFan last updated on 16/Aug/22 $$\mathrm{A}\:\mathrm{large}\:\mathrm{lot}\:\mathrm{of}\:\mathrm{tires}\:\mathrm{contain}\:\mathrm{5\%}\:\mathrm{defectives}. \\ $$$$\mathrm{4}\:\mathrm{tires}\:\mathrm{are}\:\mathrm{to}\:\mathrm{be}\:\mathrm{chosen}\:\mathrm{for}\:\mathrm{a}\:\mathrm{car}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{probability}\:\mathrm{that}\:\mathrm{you}\:\mathrm{find} \\ $$$$\:\left(\mathrm{a}\right)\:\mathrm{2}\:\mathrm{defective}\:\mathrm{tires}\:\mathrm{before}\:\mathrm{4}\:\mathrm{good}\:\mathrm{ones} \\ $$$$\:\left(\mathrm{b}\right)\:\mathrm{at}\:\mathrm{most}\:\mathrm{2}\:\mathrm{defective}\:\mathrm{tires}\:\mathrm{before}\:\mathrm{4}\:\mathrm{good}\:\mathrm{ones}. \\ $$ Commented by mr W last…

Question-109469

Question Number 109469 by mathdave last updated on 24/Aug/20 Commented by Her_Majesty last updated on 24/Aug/20 $${why}\:{do}\:{you}\:{first}\:{post}\:{a}\:{question}\:{with}\:{its} \\ $$$${solution}\:{and}\:{next}\:{post}\:{the}\:{same}\:{question} \\ $$$${again}??? \\ $$ Commented by…