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Question Number 192590 by York12 last updated on 22/May/23 $$ \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{{m}^{\mathrm{2}} {n}}{\mathrm{3}^{{m}} \left(\mathrm{3}^{{n}} .{m}+\mathrm{3}^{{m}} .{n}\right)}\right]=\lambda\: \\ $$$${find}\:\lambda. \\ $$$$ \\…
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Question Number 61511 by Kunal12588 last updated on 03/Jun/19 $${cos}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\:{cos}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{2}} }=\mathrm{120}°=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{Find}\:\:{x} \\ $$ Answered by MJS last updated…
Question Number 192580 by Frix last updated on 21/May/23 $$\mathrm{This}\:\mathrm{might}\:\mathrm{be}\:\mathrm{helpful}: \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{calculate}\:{z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } \:\mathrm{with}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\in\mathbb{C}: \\ $$$$\mathrm{Transform}\:{z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{and}\:{z}_{\mathrm{2}} ={a}+{b}\mathrm{i}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}}…
Question Number 127047 by mohammad17 last updated on 26/Dec/20 $${how}\:{can}\:{graph}\:{this}\: \\ $$$$\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\:\right]\:{pleas}\:{sir}\:{help}\:{me}\:{with}\:{details}\:? \\ $$ Answered by benjo_mathlover last updated on 26/Dec/20 $${if}\:{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 61510 by Tony Lin last updated on 03/Jun/19 $${S}_{\mathrm{1}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{n}−\mathrm{16}{k}\right)\left(\mathrm{16}{n}+\mathrm{16}{k}\right)} \\ $$$${S}_{\mathrm{2}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\sqrt{\left(\mathrm{16}{k}−\mathrm{16}\right)\left(\mathrm{16}{k}+\mathrm{16}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{S}_{\mathrm{1}} +{S}_{\mathrm{2}} }{{n}^{\mathrm{2}} }=?…
Question Number 192582 by mathdave last updated on 21/May/23 Commented by mathdave last updated on 21/May/23 $${aAnyone}\:{to}\:{help}\:{me}\:{with}\:{this} \\ $$ Commented by Frix last updated on…
Question Number 127045 by benjo_mathlover last updated on 26/Dec/20 $$\:\:\frac{\mathrm{cos}\:\mathrm{1}°+\mathrm{cos}\:\mathrm{2}°+\mathrm{cos}\:\mathrm{3}°+…+\mathrm{cos}\:\mathrm{44}°}{\mathrm{sin}\:\mathrm{1}°+\mathrm{sin}\:\mathrm{2}°+\mathrm{sin}\:\mathrm{3}°+…+\mathrm{sin}\:\mathrm{44}°}\:=? \\ $$ Answered by liberty last updated on 26/Dec/20 $$\:\frac{\mathrm{cos}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{21}.\mathrm{5}°\right)+\mathrm{cos}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{20}.\mathrm{5}°\right)+…+\mathrm{cos}\left(\mathrm{22}.\mathrm{5}°+\mathrm{20}.\mathrm{5}°\right)+\mathrm{cos}\:\left(\mathrm{22}.\mathrm{5}°+\mathrm{21}.\mathrm{5}°\right)\:}{\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{21}.\mathrm{5}°\right)+\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}°−\mathrm{20}.\mathrm{5}°\right)+…+\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}°+\mathrm{20}.\mathrm{5}°\right)+\mathrm{sin}\:\left(\mathrm{22}.\mathrm{5}+\mathrm{21}.\mathrm{5}°\right)}\:= \\ $$$$\frac{\mathrm{2cos}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{2cos}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{2cos}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{0}.\mathrm{5}°}{\mathrm{2sin}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{2sin}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{2sin}\:\mathrm{22}.\mathrm{5}°\mathrm{cos}\:\mathrm{0}.\mathrm{5}°}\:= \\ $$$$\frac{\mathrm{cos}\:\mathrm{22}.\mathrm{5}°\left(\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{cos}\:\mathrm{0}.\mathrm{5}°\right)}{\mathrm{sin}\:\mathrm{22}.\mathrm{5}°\left(\mathrm{cos}\:\mathrm{21}.\mathrm{5}°+\mathrm{cos}\:\mathrm{20}.\mathrm{5}°+…+\mathrm{cos}\:\mathrm{0}.\mathrm{5}°\right)}\:= \\…