Menu Close

Category: None

Question-67918

Question Number 67918 by aliesam last updated on 02/Sep/19 Answered by $@ty@m123 last updated on 02/Sep/19 $${Let}\:\alpha+\beta={x} \\ $$$$\mathrm{tan}\:{x}=\frac{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{1}\Rightarrow{x}=\mathrm{45}^{\mathrm{o}} \:\:…\left(\mathrm{1}\right) \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}\:\&\:\mathrm{cos}\:\beta=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}\:..\left(\mathrm{2}\right) \\ $$$$\therefore\:\mathrm{cos}\:\left(\mathrm{12}\alpha+\mathrm{13}\beta\right)=\mathrm{cos}\:\left(\mathrm{12}{x}+\beta\right) \\…

an-object-3cm-high-is-placed-5cm-away-from-the-pole-of-a-concave-spherical-mirror-of-radius-of-curvature-25cm-The-position-and-orientaion-and-size-of-the-image-are-a-8-33cm-upright-and-5cm-b-5cm

Question Number 133446 by aurpeyz last updated on 22/Feb/21 $$ \\ $$$${an}\:{object}\:\mathrm{3}{cm}\:{high}\:{is}\:{placed}\:\mathrm{5}{cm}\:{away} \\ $$$${from}\:{the}\:{pole}\:{of}\:{a}\:{concave}\:{spherical} \\ $$$${mirror}\:{of}\:{radius}\:{of}\:{curvature}\:\mathrm{25}{cm}. \\ $$$${The}\:{position}\:{and}\:{orientaion}\:{and}\:{size}\: \\ $$$${of}\:{the}\:{image}\:{are} \\ $$$$\left({a}\right)−\mathrm{8}.\mathrm{33}{cm}\:{upright}\:{and}\:\mathrm{5}{cm} \\ $$$$\left({b}\right)\mathrm{5}{cm}\:{upright}\:{and}\:\mathrm{8}.\mathrm{33}{cm} \\…

What-exactly-does-f-C-C-mean-

Question Number 2367 by Filup last updated on 18/Nov/15 $$\mathrm{What}\:\mathrm{exactly}\:\mathrm{does}\:{f}:\mathbb{C}\rightarrow\mathbb{C}\:\mathrm{mean}? \\ $$ Answered by RasheedAhmad last updated on 18/Nov/15 $${A}\:{function}\:\:{f}\:\:{whose}\:{domain}\:{and} \\ $$$${range}\:{both}\:{are}\:\mathbb{C}\:\left({set}\:{of}\:{complex}\right. \\ $$$$\left.{numbers}\right) \\…

dx-x-2-3x-9-x-2-5x-7-

Question Number 2334 by Syaka last updated on 16/Nov/15 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{3}{x}\:+\:\mathrm{9}\right)\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\:\mathrm{7}}\:}\:\:=\:? \\ $$ Commented by prakash jain last updated on 16/Nov/15 $${x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{7}=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}…

1-2-1-2-2-1-2-2-2-

Question Number 2326 by Filup last updated on 16/Nov/15 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}}×…=? \\ $$ Commented by Rasheed Soomro last updated on 16/Nov/15 $$\mathcal{N}{ice}\:\mathcal{A}{pproach}! \\ $$ Commented by…

x-2-x-6-0-

Question Number 67844 by aliesam last updated on 01/Sep/19 $${x}^{\mathrm{2}} +\mid{x}\mid−\mathrm{6}=\mathrm{0} \\ $$ Commented by gunawan last updated on 01/Sep/19 $$\mathrm{case}\:\mathrm{1}\:{x}<\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{6}=\mathrm{0} \\…