Question Number 81944 by M±th+et£s last updated on 16/Feb/20 $${show}\:{that}\: \\ $$$${cot}\left(\mathrm{40}°\right)−{cot}\left(\mathrm{50}°\right)=\mathrm{2}{tan}\left(\mathrm{10}°\right) \\ $$$${cos}\left(\mathrm{70}°\right)\:{cos}\left(\mathrm{50}^{°} \right)\:{cos}\left(\mathrm{10}^{°} \right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$ Answered by TANMAY PANACEA last updated on…
Question Number 81942 by TawaTawa last updated on 16/Feb/20 Answered by TANMAY PANACEA last updated on 16/Feb/20 $${for}\:{bank}\:{p}\:{interest}=\frac{\mathrm{3000}×\mathrm{1}×{x}}{\mathrm{100}}=\mathrm{30}{x} \\ $$$${for}\:{bank}\:{Q}\:{interst}=\frac{\mathrm{2000}×\mathrm{1}×{y}}{\mathrm{100}}=\mathrm{20}{y} \\ $$$${as}\:{per}\:{second}\:{condition} \\ $$$${interest}\:{in}\:{bank}\:{p}=\frac{\mathrm{2000}×\mathrm{1}×{x}}{\mathrm{100}}=\mathrm{20}{x} \\…
Question Number 81943 by TawaTawa last updated on 16/Feb/20 $$\mathrm{Evaluate}:\:\:\:\:\left(\frac{\sqrt{\mathrm{30}\:+\:\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}}{\:\sqrt{\mathrm{8}}\:+\:\sqrt{\mathrm{5}}}\right)^{\mathrm{1}/\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147465 by gsk2684 last updated on 21/Jul/21 $${if}\:{n}>\mathrm{2},{n}\in{N}\:{then}\:{prove}\:{that}\: \\ $$$$\left\{\left(\mathrm{2}{n}−\mathrm{1}\right)^{{n}} +\left(\mathrm{2}{n}\right)^{{n}} \right\}<\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147453 by EDWIN88 last updated on 21/Jul/21 $$\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)…=? \\ $$ Commented by gsk2684 last updated on 21/Jul/21 $$\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{9}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{25}}\right)…=? \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)… \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\mathrm{5}}{\mathrm{4}}\:… \\…
Question Number 81910 by mr W last updated on 16/Feb/20 $${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right){a}_{{n}} −\mathrm{2}{a}_{{n}−\mathrm{1}} \\ $$$${find}\:{a}_{{n}} =? \\ $$ Commented…
Question Number 81913 by ahmadshahhimat775@gmail.com last updated on 16/Feb/20 Commented by john santu last updated on 16/Feb/20 $$\Rightarrow\sqrt{{x}}+\sqrt[{\mathrm{6}\:}]{{y}}\:+\frac{{y}\:\sqrt[{\mathrm{6}\:}]{{y}}+{x}\:\sqrt[{\mathrm{6}\:}]{{x}}}{\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt{{y}}}\:= \\ $$$$\frac{\sqrt[{\mathrm{6}\:}]{{x}^{\mathrm{5}} \:}\sqrt{{y}}\:+\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt[{\mathrm{6}\:}]{{y}^{\mathrm{4}} }+\sqrt[{\mathrm{6}\:}]{{y}^{\mathrm{7}} }+\sqrt[{\mathrm{6}\:}]{{x}^{\mathrm{7}} }}{\:\sqrt[{\mathrm{3}\:}]{{x}}\:\sqrt{{y}}} \\…
Question Number 16374 by gourav~ last updated on 21/Jun/17 $${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\…
Question Number 16373 by gourav~ last updated on 21/Jun/17 $${if}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}{i}^{{k}} +\underset{{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}{i}^{{p}} ={x}+{iy}\:{then}..\left({x},{y}\right){is}… \\ $$$${a}.\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${b}.\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${c}.\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${d}.\:\left(\mathrm{4},\mathrm{8}\right) \\…
Question Number 147432 by mathdanisur last updated on 20/Jul/21 $${if}\:\:\begin{cases}{\mathrm{2}{x}\:+\:{a}\:\:;\:\:{x}\:<\:−\mathrm{3}}\\{{x}^{\mathrm{2}} \:-\:\mathrm{4}\:\:;\:\:−\mathrm{3}\:\leqslant\:{x}\:<\:\mathrm{2}}\\{{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:\:;\:\:{x}\:\geqslant\:\mathrm{2}}\end{cases} \\ $$$${find}\:\:\:\mathrm{3}{a}\:-\:{b}\:=\:? \\ $$ Answered by liberty last updated on 21/Jul/21 $$\left(\bullet\right)\:\underset{{x}\rightarrow−\mathrm{3}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{x}\rightarrow−\mathrm{3}}…