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Question Number 189757 by Shrinava last updated on 21/Mar/23 $$\mathrm{Find}:\:\:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\mid\:\mathrm{sin}\:\mathrm{4x}\:\mid}{\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{4x}}}\:\:\:=\:\:\:? \\ $$$$\mathrm{Find}:\:\:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\boldsymbol{\pi}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\mid\:\mathrm{sin}\:\mathrm{6x}\:\mid}{\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{6x}}}\:\:\:=\:\:\:? \\ $$ Answered by Frix last updated on 21/Mar/23 $$\mathrm{Use}\:{t}=\mathrm{tan}\:\mathrm{2}{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{and}\:{t}=\mathrm{tan}\:\mathrm{3}{x}\:\mathrm{in}\:\mathrm{the}…
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Question Number 58682 by hhghg last updated on 27/Apr/19 $$\mathrm{3}\frac{\mathrm{1}}{\mathrm{5}}+\mathrm{2}\frac{\mathrm{1}}{\mathrm{15}} \\ $$ Answered by MJS last updated on 27/Apr/19 $$\mathrm{3}+\frac{\mathrm{1}}{\mathrm{5}}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{1}^{×\mathrm{3}} }{\mathrm{5}^{×\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}}= \\ $$$$=\mathrm{5}+\frac{\mathrm{3}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{3}+\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{4}}{\mathrm{15}}=\mathrm{5}\frac{\mathrm{4}}{\mathrm{15}} \\…
Question Number 189748 by normans last updated on 21/Mar/23 $$ \\ $$$$\:\:\:\:\:\boldsymbol{{x}}−\sqrt{\boldsymbol{{x}}}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\boldsymbol{{x}}=?? \\ $$ Commented by mr W last updated on 21/Mar/23 $${x}>\mathrm{0}…
Question Number 124210 by bramlexs22 last updated on 01/Dec/20 $$\:{If}\:{x}\:+\:\sqrt{{x}}\:=\:\mathrm{2020}\: \\ $$$${then}\:{x}\:+\:\frac{\mathrm{2020}}{\:\sqrt{{x}}}\:=\:? \\ $$ Answered by malwan last updated on 01/Dec/20 $${x}\:+\:\sqrt{{x}}\:=\:\mathrm{2020}\:\left(\boldsymbol{\div}\sqrt{{x}}\right) \\ $$$$\Rightarrow\sqrt{{x}}\:+\:\mathrm{1}\:=\:\frac{\mathrm{2020}}{\:\sqrt{{x}}} \\…
Question Number 58669 by hhghg last updated on 27/Apr/19 $$\left\{\left[\mathrm{3}×\left(\mathrm{5}+\mathrm{5}\right)\right]+\mathrm{5}\right\}+\left\{\left[\mathrm{4}+\left(\mathrm{5}×\mathrm{4}\right)+\mathrm{5}\right]\right\} \\ $$ Answered by hovea cw last updated on 27/Apr/19 $$\left[\mathrm{3}×\mathrm{10}+\mathrm{5}+\mathrm{9}+\mathrm{20}\right. \\ $$$$=\mathrm{30}+\mathrm{34} \\ $$$$=\mathrm{64}…
Question Number 58663 by mr W last updated on 27/Apr/19 Answered by ajfour last updated on 27/Apr/19 $$\:\mathrm{100T}_{\mathrm{n}+\mathrm{1}} =\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{T}_{\mathrm{n}} +\mathrm{T}_{\mathrm{n}−\mathrm{1}} +…+\mathrm{T}_{\mathrm{1}} \right) \\ $$$$\:\mathrm{and}\:\:\mathrm{T}_{\mathrm{1}} +\mathrm{T}_{\mathrm{2}}…
Question Number 58644 by hhghg last updated on 27/Apr/19 $$\mathrm{6}+\mathrm{3}^{\mathrm{2}} ×\mathrm{4} \\ $$ Answered by hovea cw last updated on 27/Apr/19 $$\mathrm{use}\:\mathrm{bodmas} \\ $$$$\mathrm{6}+\mathrm{9}×\mathrm{4} \\…
Question Number 58641 by hhghg last updated on 27/Apr/19 $$\mathrm{What}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{4}}? \\ $$ Commented by Kunal12588 last updated on 27/Apr/19 $$\:{it}\:{is}\:{a}\:{fraction} \\ $$ Answered by hovea…