Menu Close

Category: None

an-object-of-12cm-long-is-placed-at-the-centre-of-curvature-of-concave-mirror-whose-focal-length-is-40cm-the-location-of-the-image-formed-is-

Question Number 132899 by aurpeyz last updated on 17/Feb/21 $${an}\:{object}\:{of}\:\mathrm{12}{cm}\:{long}\:{is}\:{placed}\:{at}\:{the}\: \\ $$$${centre}\:{of}\:{curvature}\:{of}\:{concave}\:{mirror} \\ $$$${whose}\:{focal}\:{length}\:{is}\:\mathrm{40}{cm}.\:{the}\:{location} \\ $$$${of}\:{the}\:{image}\:{formed}\:{is}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

cot-142-5-

Question Number 132880 by faysal last updated on 17/Feb/21 $${cot}\left(\mathrm{142}.\mathrm{5}°\right)=? \\ $$ Commented by bramlexs22 last updated on 17/Feb/21 $$\mathrm{285}°=\:\mathrm{2}×\mathrm{142}.\mathrm{5} \\ $$$$\mathrm{285}°=\mathrm{270}°+\mathrm{15}° \\ $$$$\mathrm{cot}\:\mathrm{285}°\:=\:\mathrm{cot}\:\mathrm{285}° \\…

linear-magnification-for-convex-mirror-image-height-object-height-image-distance-object-distance-is-this-formula-correct-what-is-the-meaning-of-the-negative-sign-

Question Number 132872 by aurpeyz last updated on 17/Feb/21 $${linear}\:{magnification}\:{for}\:{convex} \\ $$$${mirror} \\ $$$$=\frac{{image}\:{height}}{{object}\:{height}}=\frac{−{image}\:{distance}}{{object}\:{distance}} \\ $$$$ \\ $$$${is}\:{this}\:{formula}\:{correct}?\:{what}\:{is}\:{the} \\ $$$${meaning}\:{of}\:{the}\:{negative}\:{sign}? \\ $$ Terms of Service…

n-360-1-or-n-360-for-plane-mirrors-inclined-an-angle-to-eachother-which-is-correct-

Question Number 132874 by aurpeyz last updated on 17/Feb/21 $${n}=\frac{\mathrm{360}}{\theta}−\mathrm{1}\:{or}\:{n}=\frac{\mathrm{360}}{\theta}\:{for}\:{plane} \\ $$$${mirrors}\:{inclined}\:{an}\:{angle}\:\theta\:{to} \\ $$$${eachother}.\:{which}\:{is}\:{correct}? \\ $$ Commented by Dwaipayan Shikari last updated on 17/Feb/21 $$\mu=\frac{\mathrm{360}°}{\theta°}−\mathrm{1}\:\:{or}\:\:\mu=\frac{\mathrm{2}\pi}{\theta}−\mathrm{1}\:\:\:\left({Radian}\right)…

dx-x-2-e-x-3-3-

Question Number 1799 by navajyoti.tamuli.tamuli@gmail. last updated on 30/Sep/15 $$\int\frac{{dx}}{{x}^{\mathrm{2}} .{e}^{−{x}^{\mathrm{3}} /\mathrm{3}} }=? \\ $$ Commented by 123456 last updated on 30/Sep/15 $$\mathrm{not}\:\mathrm{done}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{elementar}\:\mathrm{function} \\ $$…

Question-67307

Question Number 67307 by aliesam last updated on 25/Aug/19 Commented by mathmax by abdo last updated on 25/Aug/19 $${S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{cos}\left({n}\theta\right)}{\mathrm{2}^{{n}} }\:\Rightarrow\:{S}\:={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{in}\theta} }{\mathrm{2}^{{n}}…

7-couples-are-there-in-a-society-The-selection-of-pairs-are-to-be-done-for-a-mixed-double-badminton-tournament-In-how-many-ways-we-can-choose-these-pairs-

Question Number 1775 by lakshaysethi039 last updated on 26/Sep/15 $$\mathrm{7}\:{couples}\:{are}\:{there}\:{in}\:{a}\:{society}.\:{The}\:{selection}\:{of} \\ $$$$\:{pairs}\:{are}\:{to}\:{be}\:{done}\:{for}\:{a}\:{mixed}\:{double}\:{badminton} \\ $$$${tournament}.\:{In}\:{how}\:{many}\:{ways}\:{we}\:{can}\:{choose} \\ $$$$\:{these}\:{pairs}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com