Question Number 209193 by Tawa11 last updated on 03/Jul/24 A pin 6cm high is placed in front of a diverging lens of focal length 15cm,…
Question Number 209129 by Adeyemi889 last updated on 02/Jul/24 Answered by A5T last updated on 02/Jul/24 $$\frac{{x}^{\mathrm{3}} +\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}={x}+\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{1}}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{{x}+\mathrm{1}}\Rightarrow{x}+\mathrm{3}=\left({A}+{B}\right){x}+{A}−{B}…
Question Number 209023 by Spillover last updated on 30/Jun/24 Commented by Spillover last updated on 01/Jul/24 $${let}\:{u}=\frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\:\:\:\:\:\:\frac{{du}}{{dx}}=\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{5}{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dx}}\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\right)=\frac{\frac{\mathrm{4}}{\left(\mathrm{1}+\mathrm{5}{x}\right)^{\mathrm{2}} }}{\mathrm{1}+\left(\frac{\mathrm{4}{x}}{\mathrm{1}+\mathrm{5}{x}}\right)^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{1}+\mathrm{10}{x}+\mathrm{25}{x}^{\mathrm{2}} } \\…
Question Number 209015 by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 02/Jul/24 Answered by Spillover last updated on 02/Jul/24 Answered by…
Question Number 209016 by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 02/Jul/24 Answered by Spillover last updated on 02/Jul/24 Terms of…
Question Number 209030 by Spillover last updated on 30/Jun/24 Answered by aleks041103 last updated on 30/Jun/24 $${moment}\:{generating}\:{function}\:{is} \\ $$$${m}\left({t}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{m}_{{k}} }{{k}!}{t}^{{k}} ={m}_{\mathrm{0}} +{m}_{\mathrm{1}} {t}+……
Question Number 209026 by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 07/Jul/24 $${f}\left({x}\right)=\left(\frac{\mathrm{4}}{{x}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{x}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−{x}} \:\:\: \\ $$$${x}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$${f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{4}}{\mathrm{1}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−\mathrm{1}} =\left(\frac{\mathrm{4}}{\mathrm{1}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}}…
Question Number 209027 by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 05/Jul/24 $${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}^{{k}+\mathrm{1}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},…..{n} \\ $$$${from}\:{Moment}\:{generating}\:{function}\left({MGF}\right) \\ $$$${M}_{{x}} \left({t}\right)={E}\left({e}^{{tx}} \right) \\…
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Question Number 208969 by Spillover last updated on 29/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by Spillover last updated on 30/Jun/24 Answered by…