Question Number 207065 by manxsol last updated on 05/May/24 $$ \\ $$$$\:\:\:{f}\left({x}\right)=\left[{cos}\mathrm{2}{x}+{cos}\mathrm{3}{x}\right]\left[{cos}\mathrm{4}{x}+{cos}\mathrm{6}{x}\right]\left[\left[{cosx}+{cos}\mathrm{5}{x}\right]\right. \\ $$$${evaluar}\:\:\:{f}\left(\frac{\mathrm{2}\pi}{\mathrm{13}}\right)\:\: \\ $$ Answered by Berbere last updated on 06/May/24 $$\left.\mathrm{4}{cos}\left({x}\right){cos}\left(\mathrm{5}{x}\right){cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{3}{x}\right).\left[{cos}\left(\mathrm{2}{x}\right)+{cos}\left(\:\mathrm{3}{x}\right)\right]\right]{a}\mathrm{2}\left(\right. \\…
Question Number 207060 by efronzo1 last updated on 05/May/24 $$\:\:\downharpoonleft\underline{\:} \\ $$ Commented by mr W last updated on 05/May/24 $${i}\:{think}\:{you}\:{mean}\:{a}_{{n}+\mathrm{2}} . \\ $$ Answered…
Question Number 207035 by mr W last updated on 04/May/24 Answered by A5T last updated on 04/May/24 $${BE}=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{10}}\Rightarrow{EF}={FB}=\sqrt{\mathrm{10}} \\ $$$${Let}\:{the}\:{line}\:{through}\:{F}\:{parallel}\:{to}\:{BC}\:{meet}\:{AB},{DC} \\ $$$${at}\:{H},{I}\:{resp}.,{then}\:{BH}=\mathrm{3}={DI};\:{FH}=\mathrm{1} \\…
Question Number 207042 by necx122 last updated on 04/May/24 $${Prove}\:{that}\:{the}\:{sum}\:{of}\:{a}\:{square}\:{function} \\ $$$${is}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}} \\ $$ Answered by Rasheed.Sindhi last updated on 04/May/24 $${p}\left({n}\right): \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}}…
Question Number 207043 by necx122 last updated on 04/May/24 $${Solve}\:{for}\:{the}\:{domain}\:{and}\:{range}\:{of}\:{f}\left({x}\right)\:{if} \\ $$$${f}\left({x}\right)\:=\:\sqrt{\frac{\mathrm{3}{x}−\mathrm{5}}{{x}+\mathrm{3}}} \\ $$ Answered by Frix last updated on 04/May/24 $$\mathrm{1}.\:{x}+\mathrm{3}\neq\mathrm{0} \\ $$$${x}\neq−\mathrm{3} \\…
Question Number 207021 by efronzo1 last updated on 03/May/24 $$\:\:{If}\:\:{a}_{{n}+\mathrm{1}} =\mathrm{2}−\mathrm{5}{a}_{{n}} \:{and}\:{a}_{\mathrm{4}} =\:−\mathrm{8} \\ $$$$\:\:{prove}\:{that}\:{a}_{\mathrm{43}} −{a}_{\mathrm{30}} \:{divisible}\:{by}\:\mathrm{5} \\ $$ Commented by mr W last updated…
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Question Number 207018 by cherokeesay last updated on 03/May/24 Answered by mr W last updated on 03/May/24 Commented by cherokeesay last updated on 03/May/24 $${R}=?…
Question Number 206997 by efronzo1 last updated on 03/May/24 $$\mathrm{Given}\:\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\:\mathrm{3}−\mathrm{6a}_{\mathrm{n}} \:\mathrm{and}\:\mathrm{a}_{\mathrm{4}} =−\mathrm{9} \\ $$$$\:\mathrm{Find}\:\mathrm{a}_{\mathrm{n}} =?\: \\ $$ Answered by mr W last updated on…
Question Number 206999 by efronzo1 last updated on 03/May/24 $$\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{6}^{\mathrm{20}} −\mathrm{1}\:=\:\mathrm{0}\:\left(\mathrm{mod}\:\mathrm{7}\right) \\ $$ Commented by Frix last updated on 03/May/24 $$\mathrm{Prove}\:\mathrm{that}\:{n}^{\mathrm{2}{k}} −\mathrm{1}=\mathrm{0}\left(\mathrm{mod}\left({n}+\mathrm{1}\right)\right): \\ $$$${n}^{\mathrm{2}{k}} −\mathrm{1}=\left({n}+\mathrm{1}\right)\underset{{j}=\mathrm{1}}…