Question Number 141972 by iloveisrael last updated on 25/May/21 Commented by Dwaipayan Shikari last updated on 25/May/21 $${x}°+\mathrm{20}°+\left(\mathrm{360}°−\mathrm{140}°\right)+\mathrm{90}°=\mathrm{360}° \\ $$$$\Rightarrow{x}°=\mathrm{30}° \\ $$ Commented by I…
Question Number 76439 by mathocean1 last updated on 27/Dec/19 $$\mathrm{solve}\:\mathrm{in}\:\left[\mathrm{0};\mathrm{2}\pi\left[\:\right.\right. \\ $$$$\mathrm{1}+\mathrm{2sin3}{x}\leqslant\mathrm{0} \\ $$ Answered by john santu last updated on 27/Dec/19 $${sin}\mathrm{3}{x}=−\frac{\mathrm{1}}{\mathrm{2}}.\rightarrow{x}=\left\{\mathrm{10}^{{o}} ,\mathrm{50}^{{o}} ,\mathrm{130}^{{o}}…
Question Number 10902 by Nayon last updated on 01/Mar/17 $$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{dx}=? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$…
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Question Number 76436 by Master last updated on 27/Dec/19 Answered by john santu last updated on 27/Dec/19 $$\frac{\mathrm{1}}{{abc}}+\frac{{a}+{c}}{{ac}}=\frac{\mathrm{1}}{{b}} \\ $$$$\frac{{ab}+{bc}+\mathrm{1}}{{abc}}=\frac{\mathrm{1}}{{b}}\rightarrow{ab}+{bc}+\mathrm{1}={ac} \\ $$$${ab}+{bc}={ac}−\mathrm{1}\:\rightarrow{b}=\frac{{ac}−\mathrm{1}}{{a}+{c}} \\ $$$${by}\:{substitusi} \\…
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Question Number 10900 by Saham last updated on 01/Mar/17 Commented by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{the}\:\mathrm{picture}\:\mathrm{isn}'\mathrm{t}\:\mathrm{clear}\:\mathrm{and}\:\mathrm{too}\:\mathrm{small} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 76434 by benjo last updated on 27/Dec/19 $$\mathrm{if}\:\mathrm{sec}\:\mathrm{x}\:−\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{4},\:\mathrm{then}\:\mathrm{find}\: \\ $$$$\mathrm{sin}\:\mathrm{x}. \\ $$ Answered by john santu last updated on 27/Dec/19 $${using}\::\:\mathrm{sec}\:^{\mathrm{2}} {x}−\mathrm{tan}\:^{\mathrm{2}} {x}\:=\mathrm{1}…
Question Number 10899 by Saham last updated on 01/Mar/17 Answered by sandy_suhendra last updated on 02/Mar/17 Commented by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{q}_{\mathrm{1}} =\mathrm{q}_{\mathrm{2}}…
Question Number 10898 by Saham last updated on 01/Mar/17 Answered by sandy_suhendra last updated on 02/Mar/17 $$\mathrm{E}^{\mathrm{2}} =\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} +\left(\mathrm{pc}\right)^{\mathrm{n}} \\ $$$$\mathrm{m}^{\mathrm{2}} \mathrm{c}^{\mathrm{b}} =\mathrm{E}^{\mathrm{2}} −\left(\mathrm{pc}\right)^{\mathrm{n}}…