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Question-29953

Question Number 29953 by puneet1789 last updated on 14/Feb/18 Commented by math solver last updated on 14/Feb/18 $$\mathrm{i}\:\mathrm{am}\:\mathrm{getting}\:\mathrm{option}\left(\mathrm{3}\right)\:\:\mathrm{5}\sqrt{\mathrm{5}}−\mathrm{3}. \\ $$$$\mathrm{yes}\:\mathrm{sir},\:\mathrm{ur}\:\mathrm{right}\:\mathrm{i}\:\mathrm{also}\:\mathrm{did}\:\mathrm{with}\:\mathrm{same} \\ $$$$\mathrm{approach}. \\ $$ Commented…

A-1-1-2-1-3-4-1-5-6-1-37-38-1-39-40-B-1-21-40-1-22-39-1-23-38-1-39-22-1-40-21-A-B-

Question Number 161005 by vvvv last updated on 10/Dec/21 $$\boldsymbol{{A}}=\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{6}}+….+\frac{\mathrm{1}}{\mathrm{37}×\mathrm{38}}+\frac{\mathrm{1}}{\mathrm{39}×\mathrm{40}} \\ $$$$\boldsymbol{{B}}=\frac{\mathrm{1}}{\mathrm{21}×\mathrm{40}}+\frac{\mathrm{1}}{\mathrm{22}×\mathrm{39}}+\frac{\mathrm{1}}{\mathrm{23}×\mathrm{38}}+….+\frac{\mathrm{1}}{\mathrm{39}×\mathrm{22}}+\frac{\mathrm{1}}{\mathrm{40}×\mathrm{21}} \\ $$$$\frac{\boldsymbol{{A}}}{\boldsymbol{{B}}}=? \\ $$ Answered by Raxreedoroid last updated on 11/Dec/21 $${A}=\underset{{k}=\mathrm{1}} {\overset{\mathrm{39}}…

2-2-x-3-cos-x-2-4-x-2-dx-

Question Number 160969 by mkam last updated on 10/Dec/21 $$\int_{−\mathrm{2}} ^{\:\mathrm{2}} {x}^{\mathrm{3}} {cos}\left(\frac{{x}}{\mathrm{2}}\right)\:\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }\:{dx} \\ $$ Answered by MJS_new last updated on 10/Dec/21 $${g}\left({x}\right)={f}_{\mathrm{1}} \left({x}\right)×{f}_{\mathrm{2}}…

there-are-even-number-divided-all-odd-number-what-is-the-number-

Question Number 160970 by mkam last updated on 10/Dec/21 $${there}\:{are}\:{even}\:{number}\:{divided}\:{all}\:{odd} \\ $$$${number}\:{what}\:{is}\:{the}\:{number}\:? \\ $$ Answered by MJS_new last updated on 10/Dec/21 $$\mathrm{the}\:\mathrm{only}\:\mathrm{even}\:\mathrm{number}\:\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{divided} \\ $$$$\mathrm{by}\:{all}\:\mathrm{odd}\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{0} \\…