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Category: Arithmetic

Question-120560

Question Number 120560 by peter frank last updated on 01/Nov/20 Answered by mathmax by abdo last updated on 01/Nov/20 $$\mathrm{I}=\int\:\:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int\:\:\frac{\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}{\mathrm{1}+\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx}\:=\int\:\:\frac{\mathrm{2}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{3}+\mathrm{cos}\left(\mathrm{2x}\right)}\mathrm{dx} \\ $$$$=_{\mathrm{tanx}=\mathrm{t}} \:\:\:\:\int\:\:\frac{\mathrm{2}−\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}}…

It-takes-Musa-3-days-to-buid-a-room-while-it-takes-John-4-days-How-many-days-will-it-takes-both-of-them-

Question Number 120502 by Engr_Jidda last updated on 31/Oct/20 $$\mathrm{It}\:\mathrm{takes}\:\mathrm{Musa}\:\mathrm{3}\:\mathrm{days}\:\mathrm{to}\:\mathrm{buid}\:\mathrm{a}\:\mathrm{room} \\ $$$$\mathrm{while}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{John}\:\mathrm{4}\:\mathrm{days}. \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{days}\:\mathrm{will}\:\mathrm{it}\:\mathrm{takes}\:\mathrm{both}\:\mathrm{of}\:\mathrm{them}? \\ $$ Answered by Olaf last updated on 31/Oct/20 $$\mathrm{3}×\mathrm{4}\:=\:\mathrm{12}. \\…

Determine-the-sum-of-the-1st-nth-term-of-the-Sequence-1-4-10-22-46-Almighty-Formula-

Question Number 120363 by Lordose last updated on 30/Oct/20 $$ \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{1st}\:\mathrm{nth}\:\mathrm{term} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{Sequence} \\ $$$$ \\ $$$$\:\mathrm{1},\mathrm{4},\mathrm{10},\mathrm{22},\mathrm{46},…. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bigstar\mathrm{Almighty}\:\mathrm{Formula} \\ $$ Answered…

Question-54765

Question Number 54765 by peter frank last updated on 10/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19 $$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}×\mathrm{24}−\mathrm{1}} {xcos}^{\mathrm{2}×\mathrm{27}−\mathrm{1}} {xdx} \\ $$$${formula}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}}…

Question-185805

Question Number 185805 by Rupesh123 last updated on 27/Jan/23 Answered by MJS_new last updated on 28/Jan/23 $$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2023}} ={x} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}×\frac{{x}^{\mathrm{2}} +{x}^{−\mathrm{2}} +\mathrm{1}}{{x}+{x}^{+\mathrm{1}} }=\frac{\mathrm{1}+{n}^{\mathrm{2}} }{{n}}\:\Rightarrow\:{n}=\left(\frac{\sqrt{\mathrm{3}}{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\pm\mathrm{1}}…

Correction-to-the-last-assignment-1-1-81-x-2-27-1-x-1-3-4-x-2-3-3-1-x-1-3-4x-8-3-3-3x-3-4x-8-3-3-3x-4x-8-3-3x-C-L-T-4x-3x-3-8-x-5-x-5-2-9-x-1-729-9-x-

Question Number 120241 by olalekanoriyomi last updated on 30/Oct/20 $$\boldsymbol{\mathrm{C}}{orrection}\:\boldsymbol{{to}}\:\boldsymbol{{the}}\:\boldsymbol{{last}}\:\boldsymbol{{assignment}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{81}^{\boldsymbol{{x}}−\mathrm{2}} }\:=\:\mathrm{27}^{\mathrm{1}−\boldsymbol{{x}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}\left(\boldsymbol{{x}}−\mathrm{2}\right)} }=\mathrm{3}^{\mathrm{3}\left(\mathrm{1}−\boldsymbol{{x}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}\boldsymbol{{x}}−\mathrm{8}} }=\mathrm{3}^{\mathrm{3}−\mathrm{3}\boldsymbol{{x}}} \\ $$$$\mathrm{3}^{−\mathrm{4}\boldsymbol{{x}}+\mathrm{8}} =\mathrm{3}^{\mathrm{3}−\mathrm{3}\boldsymbol{{x}}} \\…

Find-the-sum-to-infinity-1-1-2-cos-1-4-cos-2-1-8-cos-3-

Question Number 54700 by peter frank last updated on 09/Feb/19 $$\mathrm{F}{i}\mathrm{nd}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\:}\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+………. \\ $$$$ \\ $$ Commented by maxmathsup by imad last updated on…