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If-f-x-is-a-Continous-and-odd-function-Then-is-f-0-0-

Question Number 4873 by prakash jain last updated on 18/Mar/16 $$\mathrm{If}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{Continous}\:\mathrm{and}\:\mathrm{odd}\:\mathrm{function}. \\ $$$$\mathrm{Then}\:\mathrm{is}\:{f}\left(\mathrm{0}\right)=\mathrm{0}? \\ $$ Commented by prakash jain last updated on 18/Mar/16 $$\mathrm{Thanks}.\:\mathrm{That}\:\mathrm{is}\:\mathrm{what}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{but}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{think}\:\mathrm{of} \\…

Locate-the-critical-points-of-the-following-functions-and-state-the-nature-of-each-1-f-x-y-3x-4-2x-2-2xy-x-3y-2-6y-15-2-f-x-y-x-2-4xy-y-2-

Question Number 135933 by Engr_Jidda last updated on 17/Mar/21 $${Locate}\:{the}\:{critical}\:{points}\:{of}\:{the}\:{following} \\ $$$${functions}\:{and}\:{state}\:{the}\:{nature}\:{of}\:{each}. \\ $$$$\left(\mathrm{1}\right)\:{f}\left({x},{y}\right)=\mathrm{3}{x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{xy}−{x}+\mathrm{3}{y}^{\mathrm{2}} −\mathrm{6}{y}+\mathrm{15} \\ $$$$\left(\mathrm{2}\right)\:{f}\left({x},{y}\right)={x}^{\mathrm{2}} −\mathrm{4}{xy}+{y}^{\mathrm{2}} \\ $$ Answered by dhgt…

Evaluate-1-0-1-0-x-0-y-3x-2-2y-2-3z-2-dxdydz-2-2x-2-3-dx-3-x-5-x-2-10x-2-dx-

Question Number 135932 by Engr_Jidda last updated on 17/Mar/21 $${Evaluate}\:\left(\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{{x}} \int_{\mathrm{0}} ^{{y}} \left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{3}{z}^{\mathrm{2}} \right){dxdydz} \\ $$$$\left(\mathrm{2}\right)\:\int\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{3}} {dx} \\ $$$$\left(\mathrm{3}\right)\:\int\left(\frac{{x}−\mathrm{5}}{{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{2}}\right){dx}…

Solve-x-4-x-3-2ax-2-ax-a-2-0-a-R-

Question Number 70399 by Henri Boucatchou last updated on 04/Oct/19 $$\boldsymbol{{Solve}}\:\:\boldsymbol{{x}}^{\mathrm{4}} \:+\:\boldsymbol{{x}}^{\mathrm{3}} \:−\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} \:−\boldsymbol{{ax}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} =\:\mathrm{0},\:\:\boldsymbol{{a}}\:\in\:\mathbb{R} \\ $$ Commented by Prithwish sen last updated on 04/Oct/19…

Solve-for-x-x-1-1-1-1-

Question Number 4862 by prakash jain last updated on 18/Mar/16 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}=\sqrt{.\mathrm{1}+\sqrt{.\mathrm{1}+\sqrt{.\mathrm{1}+\sqrt{.\mathrm{1}+…}}}} \\ $$ Answered by Yozzii last updated on 18/Mar/16 $${Let}\:{us}\:{look}\:{at}\:{the}\:{general}\:{case}\:{of} \\ $$$$\:\:\:\:\:{x}=\sqrt{{a}+\sqrt{{a}+\sqrt{{a}+\sqrt{{a}+\sqrt{{a}+…}}}}}…

If-v-r-m-where-r-x-2-y-2-z-2-show-that-2-v-x-2-2-v-y-2-2-v-z-2-m-m-1-r-m-2-

Question Number 135935 by Engr_Jidda last updated on 17/Mar/21 $${If}\:\:{v}={r}^{{m}} \:{where}\:{r}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:}\:,\:{show}\:{that} \\ $$$$\frac{\partial^{\mathrm{2}} {v}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {v}}{\partial{y}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {v}}{\partial{z}^{\mathrm{2}} }={m}\left({m}−\mathrm{1}\right){r}^{{m}−\mathrm{2}} \\ $$ Answered…

Verify-that-the-following-functions-satisfies-the-mean-value-theorem-1-f-x-x-x-2-x-at-2-1-2-f-x-x-3-4x-2-2x-4-at-2-2-

Question Number 135929 by Engr_Jidda last updated on 17/Mar/21 $${Verify}\:{that}\:{the}\:{following}\:{functions}\:{satisfies} \\ $$$${the}\:{mean}\:{value}\:{theorem}. \\ $$$$\left(\mathrm{1}\right)\:\:\:{f}\left({x}\right)\:=\frac{{x}}{{x}^{\mathrm{2}} −{x}}\:\:\:{at}\:\left[−\mathrm{2},−\mathrm{1}\right] \\ $$$$\left(\mathrm{2}\right)\:\:{f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}\:\:\:{at}\left[−\mathrm{2},\mathrm{2}\right] \\ $$ Terms of Service Privacy…

montrer-que-sinA-sinB-sinC-4cos-A-2-cos-B-2-cos-C-2-

Question Number 70394 by Cmr 237 last updated on 04/Oct/19 $$\mathrm{montrer}\:\mathrm{que} \\ $$$$\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=\mathrm{4cos}\frac{\mathrm{A}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{C}}{\mathrm{2}} \\ $$ Answered by $@ty@m123 last updated on 05/Oct/19 $${LHS}=\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{sin}\:{C} \\ $$$$=\mathrm{2sin}\:\frac{\pi−{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{sin}\:{C}…