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Question-61048

Question Number 61048 by Gulay last updated on 28/May/19 Commented by Gulay last updated on 28/May/19 $$\mathrm{AB}=\mathrm{30}\:\:\:\:\mathrm{BD}=\mathrm{24}\:\:\:\mathrm{Find}\:\mathrm{DC}\:\:\:<\mathrm{B}=\mathrm{90} \\ $$$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$ Answered by mr W…

Question-61042

Question Number 61042 by aliesam last updated on 28/May/19 Answered by ajfour last updated on 28/May/19 $${let}\:\:{N}\bigtriangleup{x}=\mathrm{5}−\mathrm{2}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:{x}=\mathrm{2}+{n}\bigtriangleup{x}\:=\:\mathrm{2}+\frac{\mathrm{3}{n}}{{N}} \\ $$$$\int_{\mathrm{2}} ^{\:\mathrm{5}} \left({x}^{\mathrm{3}} −\mathrm{1}\right){dx}=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}}{{N}}\underset{{n}=\mathrm{1}}…

Solve-sin-cos-1-when-0-90-

Question Number 126575 by amns last updated on 22/Dec/20 $$\boldsymbol{{Solve}}:\:{sin}\theta\:+\:{cos}\theta\:=\:\mathrm{1},\:\boldsymbol{{when}}\:\mathrm{0}°\:\leqslant\:\theta\:\leqslant\:\mathrm{90}° \\ $$ Commented by amns last updated on 22/Dec/20 $${I}\:{need}\:{a}\:{help}\:{to}\:{solve}\:{it},\:\boldsymbol{{Help}}\:\boldsymbol{{me}}. \\ $$ Commented by benjo_mathlover…

Question-61030

Question Number 61030 by aanur last updated on 28/May/19 Commented by prakash jain last updated on 28/May/19 $${x}^{\mathrm{2}} =\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{4}} =\mathrm{9}×\mathrm{3}=\mathrm{27}=\mathrm{3}^{\mathrm{3}} \Rightarrow{x}=\mathrm{3}^{\mathrm{3}/\mathrm{4}} \\ $$$${y}^{\mathrm{2}}…

show-thatfor-z-C-z-1-2-2-z-2-z-1-2-2-This-formula-can-be-used-z-2-z-z-

Question Number 126553 by mathocean1 last updated on 21/Dec/20 $${show}\:{thatfor}\:{z}\in\mathbb{C}\: \\ $$$$\mid{z}+\mathrm{1}\mid^{\mathrm{2}} =\mathrm{2}\mid{z}\mid^{\mathrm{2}\:} \Leftrightarrow\mid{z}−\mathrm{1}\mid^{\mathrm{2}} =\mathrm{2}. \\ $$$${This}\:{formula}\:{can}\:{be}\:{used}: \\ $$$$\mid{z}\mid^{\mathrm{2}} ={z}×\overset{−} {{z}} \\ $$ Answered by…

f-x-x-f-x-x-2-1-f-2-

Question Number 192084 by sciencestudentW last updated on 07/May/23 $${f}\left({x}\right)+{x}\centerdot{f}\left(−{x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\sqrt{\mathrm{2}}\right)=? \\ $$ Answered by AST last updated on 07/May/23 $${x}^{\mathrm{2}} +\mathrm{1}=−\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)+\mathrm{2}=−\left(\mathrm{1}+{x}−{x}\left(\mathrm{1}+{x}\right)\right)+\mathrm{2} \\…