Question Number 190076 by Spillover last updated on 26/Mar/23 Answered by som(math1967) last updated on 27/Mar/23 $${a}.\:\:\:{f}^{\boldsymbol{{l}}} \left({x}\right)=\:\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\:\frac{\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)}−\frac{\mathrm{1}}{\mathrm{1}+\boldsymbol{{sinx}}}}{\boldsymbol{{h}}} \\ $$$$\:=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{sinx}}−\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)}{\boldsymbol{{h}}\left\{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)\right\}\left(\mathrm{1}+\boldsymbol{{sinx}}\right)} \\ $$$$=\underset{\boldsymbol{{h}}\rightarrow\mathrm{0}} {\boldsymbol{{lim}}}\:\frac{−\mathrm{2}{sin}\left(\frac{{h}}{\mathrm{2}}\right){cos}\left({x}+\frac{{h}}{\mathrm{2}}\right)}{{h}\left(\mathrm{1}+\boldsymbol{{sinx}}\right)\left\{\mathrm{1}+\boldsymbol{{sin}}\left(\boldsymbol{{x}}+\boldsymbol{{h}}\right)\right\}}…
Question Number 190073 by Spillover last updated on 26/Mar/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190075 by Spillover last updated on 26/Mar/23 Answered by PowerMaths last updated on 27/Mar/23 $$\left({a}\right)\:{p}_{\mathrm{1}} =\:\bot^{{r}} \:{from}\:{A}\left(\mathrm{1},\mathrm{3}\right)\:{to}\:{line}\:=\frac{\mid\mathrm{3}×\mathrm{1}+\mathrm{4}×\mathrm{3}−\mathrm{9}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}\:=\:\frac{\mathrm{6}}{\mathrm{5}}\:=\:\mathrm{1}.\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{p}_{\mathrm{2}} \:=\:\bot^{{r}} \:{from}\:{B}\left(\mathrm{2},\mathrm{7}\right)\:{to}\:{line}\:=\frac{\mid\mathrm{3}×\mathrm{2}+\mathrm{4}×\mathrm{7}−\mathrm{9}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}}…
Question Number 124522 by bemath last updated on 03/Dec/20 Answered by mr W last updated on 04/Dec/20 $${CD}=\mathrm{25}−\mathrm{16}=\mathrm{9} \\ $$$${AC}^{\mathrm{2}} ={BC}×{CD}=\mathrm{16}×\mathrm{9} \\ $$$${AC}=\mathrm{4}×\mathrm{3}=\mathrm{12} \\ $$$${AB}^{\mathrm{2}}…
Question Number 58972 by hovea cw last updated on 02/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 190039 by Shrinava last updated on 26/Mar/23 $$\mathrm{b}+\sqrt{\mathrm{b}}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{b}>\mathrm{0} \\ $$$$\mathrm{find}\:\:\:\mathrm{2b}^{\mathrm{2}} −\mathrm{14b}\:+\:\mathrm{28}\:=\:? \\ $$ Answered by cortano12 last updated on 26/Mar/23 $$\:\sqrt{\mathrm{b}}\:=\:\mathrm{3}−\mathrm{b}\Rightarrow\mathrm{b}=\mathrm{9}−\mathrm{6b}+\mathrm{b}^{\mathrm{2}}…
Question Number 190038 by Shrinava last updated on 26/Mar/23 $$\mathrm{Find}:\:\:\:\frac{\mathrm{21}\:+\:\sqrt{\mathrm{4a}\:−\:\mathrm{3}}}{\mathrm{4a}\:+\:\sqrt{\mathrm{3}\:−\:\mathrm{4a}}} \\ $$ Answered by Rasheed.Sindhi last updated on 26/Mar/23 $$\mathrm{Find}:\:\:\:\frac{\mathrm{21}\:+\:\sqrt{\mathrm{4a}\:−\:\mathrm{3}}}{\mathrm{4a}\:+\:\sqrt{\mathrm{3}\:−\:\mathrm{4a}}} \\ $$$${Assuming}\:{the}\:{value}\:{is}\:{real} \\ $$$$\Rightarrow \\…
Question Number 58963 by Tony Lin last updated on 02/May/19 $${solve}\:{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+{i}\right){x}−\mathrm{5}+\mathrm{14}{i}=\mathrm{0} \\ $$ Answered by tanmay last updated on 02/May/19 $${x}^{\mathrm{2}} −\mathrm{2}{x}\left(\mathrm{1}+{i}\right)+\left(\mathrm{1}+{i}\right)^{\mathrm{2}} −\mathrm{5}+\mathrm{14}{i}−\left(\mathrm{1}+{i}\right)^{\mathrm{2}} =\mathrm{0}…
Question Number 190032 by Shrinava last updated on 26/Mar/23 $$\mathrm{x}\:>\:\mathrm{0} \\ $$$$\mathrm{xy}\:−\:\mathrm{18}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2x}\:+\:\mathrm{y}\right)_{\boldsymbol{\mathrm{min}}} \:=\:? \\ $$ Answered by cortano12 last updated on 26/Mar/23 $$\:\mathrm{z}\:=\mathrm{2x}+\frac{\mathrm{18}}{\mathrm{x}}\:=\:\mathrm{2x}+\mathrm{18x}^{−\mathrm{1}}…
Question Number 190035 by Shrinava last updated on 26/Mar/23 $$\mathrm{10}^{\mathrm{93}} \:−\:\mathrm{93} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers} \\ $$ Answered by BaliramKumar last updated on 26/Mar/23 $$\mathrm{999999}…….\mathrm{999999907}\:\:\:\:\:\:\left(\mathrm{93}\:{digits}\right) \\ $$…