Menu Close

Author: Tinku Tara

Question-205083

Question Number 205083 by BaliramKumar last updated on 07/Mar/24 Answered by A5T last updated on 07/Mar/24 $$\mathrm{1}×\mathrm{1}×\mathrm{30};\mathrm{1}×\mathrm{2}×\mathrm{15};\mathrm{1}×\mathrm{3}×\mathrm{10};\mathrm{1}×\mathrm{5}×\mathrm{6}; \\ $$$$\mathrm{2}×\mathrm{3}×\mathrm{5};\:{There}\:{are}\:\mathrm{5}\:{ways}\:{upto}\:{permutation} \\ $$$${but}\:{if}\:{order}\:{matters}\:{there}\:{are},\:\mathrm{4}×\mathrm{3}!+\frac{\mathrm{3}!}{\mathrm{2}!}=\mathrm{27} \\ $$ Commented by…

a-3-b-3-c-3-d-3-a-b-c-d-N-i-3-3-4-3-5-3-6-3-HCF-3-4-5-1-other-example-

Question Number 205063 by BaliramKumar last updated on 07/Mar/24 $$\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{c}^{\mathrm{3}} \:=\:\mathrm{d}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a},\:\mathrm{b},\:\mathrm{c},\:\mathrm{d}\:\in\:\boldsymbol{\mathrm{N}} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{3}^{\mathrm{3}} \:+\:\mathrm{4}^{\mathrm{3}} \:+\:\mathrm{5}^{\mathrm{3}} \:=\:\mathrm{6}^{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\mathrm{HCF}\left(\mathrm{3},\:\mathrm{4},\:\mathrm{5}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{other}\:\mathrm{example}\:……….. \\ $$ Commented…

if-a-b-c-are-the-roots-of-f-x-x-3-2024x-2-2024x-2024-find-1-1-a-2-1-1-b-2-1-1-c-2-

Question Number 205073 by mr W last updated on 07/Mar/24 $${if}\:{a},\:{b},\:{c}\:{are}\:{the}\:{roots}\:{of} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{2024}{x}^{\mathrm{2}} +\mathrm{2024}{x}+\mathrm{2024} \\ $$$${find}\:\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}−{c}^{\mathrm{2}} }=? \\ $$ Answered by cortano12…

Question-205053

Question Number 205053 by Abdullahrussell last updated on 06/Mar/24 Commented by Frix last updated on 08/Mar/24 $${x}=\mathrm{3} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i} \\ $$$${z}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{i} \\ $$$${x}+{y}+{z}=\mathrm{6} \\ $$…

x-2-5x-6-0-amp-x-2-kx-1-0-have-a-common-root-then-k-

Question Number 205021 by BaliramKumar last updated on 06/Mar/24 $${x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\mathrm{6}\:=\:\mathrm{0}\:\&\:{x}^{\mathrm{2}} \:+\:{kx}\:+\:\mathrm{1}\:=\:\mathrm{0}\:{have}\:{a}\: \\ $$$${common}\:{root}\:\mathrm{then}\:\:{k}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 06/Mar/24 $${x}^{\mathrm{2}} \:+\:\mathrm{5}{x}\:+\mathrm{6}\:=\:\mathrm{0}\:\&\:{x}^{\mathrm{2}}…

For-what-value-of-k-can-be-expression-x-3-kx-2-7x-6-be-resolved-into-three-linear-factors-a-0-b-1-c-2-d-3-

Question Number 205018 by BaliramKumar last updated on 06/Mar/24 $$\mathrm{For}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\:'\mathrm{k}'\:\mathrm{can}\:\mathrm{be}\:\mathrm{expression}\:{x}^{\mathrm{3}} \:+\:{kx}^{\mathrm{2}} \:−\mathrm{7}{x}\:+\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{be}\:\mathrm{resolved}\:\mathrm{into}\:\mathrm{three}\:\mathrm{linear}\:\mathrm{factors}? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{c}\right)\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{d}\right)\:\mathrm{3} \\ $$ Commented by Rasheed.Sindhi last updated on 06/Mar/24…