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an-object-is-placed-5cm-away-from-a-plane-mirror-the-mirror-is-rotated-tbrough-angle-20-0-about-the-point-of-incidence-calculate-the-shortest-distance-between-the-old-and-the-new-position-of-the-im

Question Number 133115 by aurpeyz last updated on 19/Feb/21 $${an}\:{object}\:{is}\:{placed}\:\mathrm{5}{cm}\:{away}\:{from}\:{a} \\ $$$${plane}\:{mirror}.\:{the}\:{mirror}\:{is}\:{rotated} \\ $$$${tbrough}\:{angle}\:\mathrm{20}^{\mathrm{0}} \:{about}\:{the}\:{point}\:{of} \\ $$$${incidence}.\:{calculate}\:{the}\:{shortest} \\ $$$${distance}\:{between}\:{the}\:{old}\:{and}\:{the}\:{new} \\ $$$${position}\:{of}\:{the}\:{image} \\ $$ Answered by…

Assuming-that-word-population-is-6-billion-and-total-world-trips-is-4-8-billion-a-what-is-the-country-potential-generation-index-CPG-I-of-the-country-b-interprete-your-results-in-a-above-

Question Number 133097 by otchereabdullai@gmail.com last updated on 18/Feb/21 $$\mathrm{Assuming}\:\mathrm{that}\:\mathrm{word}\:\mathrm{population}\:\mathrm{is}\: \\ $$$$\mathrm{6}\:\mathrm{billion}\:\mathrm{and}\:\mathrm{total}\:\mathrm{world}\:\mathrm{trips}\:\mathrm{is}\: \\ $$$$\mathrm{4}.\mathrm{8}\:\mathrm{billion} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{country}\:\mathrm{potential}\: \\ $$$$\mathrm{generation}\:\mathrm{index}\:\left(\mathrm{CPG}\:\mathrm{I}\right)\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{country} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{interprete}\:\mathrm{your}\:\mathrm{results}\:\mathrm{in}\:\left(\mathrm{a}\right)\:\mathrm{above}\: \\ $$$$ \\…

Question-67564

Question Number 67564 by azizullah last updated on 28/Aug/19 Commented by mathmax by abdo last updated on 28/Aug/19 $${F}\left(\mathrm{0}\right)=\mathrm{0}\:=\mathrm{1}+\mathrm{1}\:+\mathrm{2}\:+{f}\:\Rightarrow{f}\:=−\mathrm{4}\:\Rightarrow \\ $$$${F}\left({x}\right)\:=−\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}+\mathrm{1}}\:+\frac{\mathrm{2}}{{x}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{4}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:. \\…

Question-67561

Question Number 67561 by azizullah last updated on 28/Aug/19 Answered by Rasheed.Sindhi last updated on 28/Aug/19 $${Let}\:\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}−\mathrm{1}\right)}=\frac{{Ax}+{B}}{{x}^{\mathrm{2}} +\mathrm{1}}+\frac{{C}}{{x}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\left({Ax}+{B}\right)\left({x}−\mathrm{1}\right)+{C}\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${x}=\mathrm{1}\Rightarrow\mathrm{2}{C}=\mathrm{3}\Rightarrow{C}=\mathrm{3}/\mathrm{2} \\…

n-1-1-n-2-k-2-k-

Question Number 133075 by LUFFY last updated on 18/Feb/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{k}^{\mathrm{2}} }\right)\:{k}\in?????? \\ $$ Answered by Dwaipayan Shikari last updated on 18/Feb/21 $$\frac{\mathrm{1}}{\mathrm{2}{ik}}\underset{{n}=\mathrm{1}}…

Question-67516

Question Number 67516 by LPM last updated on 28/Aug/19 Commented by Prithwish sen last updated on 28/Aug/19 $$\mathrm{cos}^{\mathrm{2}} \mathrm{x}\:+\left(\mathrm{2cos}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{3x} \\ $$$$\mathrm{Simplyfying}\:\mathrm{we}\:\mathrm{get}, \\…

Question-133028

Question Number 133028 by mohammad17 last updated on 18/Feb/21 Answered by Dwaipayan Shikari last updated on 18/Feb/21 $$\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{1},\mathrm{1}\right)} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{x}−{y}}={x}+{y}=\mathrm{2} \\ $$ Terms of…

for-the-extremum-values-of-x-2-y-2-z-2-subject-to-the-constraints-ax-2-by-2-cz-2-1-and-lx-my-nz-0-show-that-the-stationary-points-setisfy-the-realation-l-2-1-a-1-m-2-1-b-1-n-2-

Question Number 133010 by rs4089 last updated on 18/Feb/21 $${for}\:{the}\:{extremum}\:{values}\:{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:{subject}\:{to}\:{the}\:{constraints} \\ $$$${ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +{cz}^{\mathrm{2}} =\mathrm{1}\:{and}\:{lx}+{my}+{nz}=\mathrm{0}\:. \\ $$$${show}\:{that}\:{the}\:{stationary}\:{points}\:{setisfy}\:{the}\:{realation} \\ $$$$\frac{{l}^{\mathrm{2}} }{\mathrm{1}+{a}\lambda_{\mathrm{1}} }+\frac{{m}^{\mathrm{2}} }{\mathrm{1}+{b}\lambda_{\mathrm{1}}…