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Author: Tinku Tara

2C-4-n-35C-3-n-2-n-

Question Number 74129 by malwaan last updated on 19/Nov/19 $$\mathrm{2}\boldsymbol{{C}}_{\mathrm{4}} ^{\boldsymbol{{n}}} \:=\:\mathrm{35}\boldsymbol{{C}}_{\mathrm{3}} ^{\frac{\boldsymbol{{n}}}{\mathrm{2}}} \: \\ $$$$\Rightarrow\:\boldsymbol{{n}}\:=\:? \\ $$ Answered by MJS last updated on 20/Nov/19…

Factor-the-polynomial-c-2-x-2-b-3c-2-x-c-b-a-

Question Number 74121 by Raxreedoroid last updated on 19/Nov/19 $$\mathrm{Factor}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$$\left(\frac{{c}}{\mathrm{2}}\right){x}^{\mathrm{2}} +\left({b}−\frac{\mathrm{3}{c}}{\mathrm{2}}\right){x}+\left({c}−{b}+{a}\right) \\ $$ Commented by MJS last updated on 19/Nov/19 $$\left({x}+\frac{{b}}{{c}}−\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\left(\mathrm{2}{b}−{c}\right)^{\mathrm{2}} −\mathrm{8}{ac}}}{\mathrm{2}{c}}\right)\left({x}+\frac{{b}}{{c}}−\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\left(\mathrm{2}{b}−{c}\right)^{\mathrm{2}} −\mathrm{8}{ac}}}{\mathrm{2}{c}}\right)…

9-9-

Question Number 8582 by thomas last updated on 16/Oct/16 $$\mathrm{9}×\mathrm{9} \\ $$ Answered by Rasheed Soomro last updated on 17/Oct/16 $$\mathrm{9}×\mathrm{9} \\ $$$$=\left(\mathrm{10}−\mathrm{1}\right)\left(\mathrm{10}−\mathrm{1}\right) \\ $$$$=\left(\mathrm{10}−\mathrm{1}\right)^{\mathrm{2}}…

Question-139655

Question Number 139655 by mohammad17 last updated on 30/Apr/21 Answered by qaz last updated on 30/Apr/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} {x}\mathrm{tan}^{−\mathrm{1}} \left({x}\right){ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}}…

Find-the-volume-of-the-solid-that-lies-within-the-sphere-x-2-y-2-z-2-16-above-the-x-y-plane-and-below-the-cone-z-x-2-y-2-

Question Number 74117 by necxxx last updated on 19/Nov/19 $${Find}\:{the}\:{volume}\:{of}\:{the}\:{solid}\:{that}\:{lies} \\ $$$${within}\:{the}\:{sphere}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{16},\:{above} \\ $$$${the}\:{x}-{y}\:{plane}\:{and}\:{below}\:{the}\:{cone} \\ $$$${z}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$ Commented by…