Menu Close

Category: Others

Calculate-0-pi-4-tan-x-tan-2-x-tan-x-tan-2-x-cos-x-dx-

Question Number 132124 by Chhing last updated on 11/Feb/21 $$ \\ $$$$\:\:\:\mathrm{Calculate} \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \sqrt{\frac{\mathrm{tan}\left(\mathrm{x}\right)+\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{tan}\left(\mathrm{x}\right)−\mathrm{tan}^{\mathrm{2}} \left(\mathrm{x}\right)}}\:\mathrm{cos}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$\:\: \\ $$ Commented by liberty…

jumlah-akar-akar-persamaan-x-2-2-x-3-0-sama-dengan-a-10-b-3-e-4-c-1-d-0-

Question Number 1039 by tera last updated on 22/May/15 $${jumlah}\:{akar}−{akar}\:{persamaan}\:\left[{x}\right]^{\mathrm{2}} −\mathrm{2}\left[{x}\right]−\mathrm{3}=\mathrm{0}\:{sama}\:{dengan} \\ $$$${a}.−\mathrm{10}\:\:\:\:\:\:\:\:\:\:{b}.−\mathrm{3}\:\:\:\:\:\:{e}.\mathrm{4} \\ $$$${c}.−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:{d}.\mathrm{0} \\ $$ Commented by prakash jain last updated on 22/May/15…

arg-z-a-arg-z-z-1-arg-z-z-1-kpi-a-R-z-1-C-z-1-0-z-C-k-Z-the-locus-of-z-are-

Question Number 651 by 123456 last updated on 19/Feb/15 $$\mathrm{arg}\left(\mathrm{z}−\mathrm{a}\right)−\mathrm{arg}\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)−\mathrm{arg}\left(\mathrm{z}−\bar {\mathrm{z}}_{\mathrm{1}} \right)={k}\pi \\ $$$${a}\in\mathbb{R} \\ $$$${z}_{\mathrm{1}} \in\mathbb{C},\Im\left(\bar {{z}}_{\mathrm{1}} \right)\neq\mathrm{0} \\ $$$${z}\in\mathbb{C} \\ $$$${k}\in\mathbb{Z} \\…

Evaluate-x-x-1-dx-

Question Number 66548 by AnjanDey last updated on 17/Aug/19 $$\boldsymbol{{Evaluate}}:\int\sqrt{{x}\sqrt{{x}+\mathrm{1}}\:}{dx} \\ $$ Commented by AnjanDey last updated on 19/Aug/19 $${Please}\:{evaluate}\:{this}\:{integral}…{It}'{s}\:{very}\:{needful}\:{and}\:{urgent}… \\ $$ Terms of Service…

3-3x-4-3-x-2-4-

Question Number 991 by Madan pd gupta last updated on 13/May/15 $$\mathrm{3}\left(\mathrm{3}{x}−\mathrm{4}\right)+\mathrm{3}\left({x}−\mathrm{2}\right)=\mathrm{4} \\ $$ Answered by prakash jain last updated on 13/May/15 $$\mathrm{9}{x}−\mathrm{12}+\mathrm{3}{x}−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{12}{x}=\mathrm{18}\Rightarrow{x}=\frac{\mathrm{18}}{\mathrm{12}}=\frac{\mathrm{3}}{\mathrm{2}}…