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Question-190270

Question Number 190270 by 073 last updated on 30/Mar/23 Answered by mr W last updated on 30/Mar/23 $${a}_{{n}} =\left({n}−\mathrm{6}\right)\left({n}+\mathrm{3}\right)<\mathrm{0} \\ $$$$\Rightarrow−\mathrm{3}<{n}<\mathrm{6} \\ $$$$\Rightarrow{n}=\mathrm{1},\mathrm{2},…,\mathrm{5}\: \\ $$$${i}.{e}.\:\mathrm{5}\:{terms}\:{are}\:{negative}.…

Question-190253

Question Number 190253 by 073 last updated on 30/Mar/23 Answered by cortano12 last updated on 30/Mar/23 $$\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{tan}\:\mathrm{x}+\mathrm{x}\right)\left(\mathrm{tan}\:\mathrm{x}−\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{x}}{\mathrm{x}}\:.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{x}}{\mathrm{x}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}…

S-N-S-2-0-2-1-2-13-7-0-7-1-7-15-11-0-11-1-11-100-determinate-the-sum-of-positive-divisors-of-S-

Question Number 124696 by mathocean1 last updated on 05/Dec/20 $${S}\:\in\:\mathbb{N}. \\ $$$${S}=\left(\mathrm{2}^{\mathrm{0}} ×\mathrm{2}^{\mathrm{1}} ×…×\mathrm{2}^{\mathrm{13}} \right)\left(\mathrm{7}^{\mathrm{0}} ×\mathrm{7}^{\mathrm{1}} ×…×\mathrm{7}^{\mathrm{15}} \right)\left(\mathrm{11}^{\mathrm{0}} ×\mathrm{11}^{\mathrm{1}} ×…×\mathrm{11}^{\mathrm{100}} \right) \\ $$$${determinate}\:{the}\:{sum}\:{of}\:\:{positive} \\ $$$${divisors}\:{of}\:{S}.…

Determinate-m-such-that-m-is-written-abcca-in-base-5-and-is-written-bbab-in-base-8-

Question Number 124694 by mathocean1 last updated on 05/Dec/20 $${Determinate}\:{m}\:{such}\:{that} \\ $$$${m}\:{is}\:{written}\:{abcca}\:{in}\:{base}\:\mathrm{5}\:{and}\:{is} \\ $$$${written}\:{bbab}\:{in}\:{base}\:\mathrm{8}. \\ $$ Answered by floor(10²Eta[1]) last updated on 05/Dec/20 $$\mathrm{M}=\mathrm{abcca}_{\mathrm{5}} =\mathrm{bbab}_{\mathrm{8}}…

Demonstrate-that-a-b-N-if-a-b-can-not-be-simplified-then-a-b-a-2-ab-b-2-can-not-also-be-simplified-

Question Number 124693 by mathocean1 last updated on 05/Dec/20 $${Demonstrate}\:{that}\:\forall\:{a},{b}\:\in\mathbb{N}^{\ast} \:{if}\:\frac{{a}}{{b}} \\ $$$${can}\:{not}\:{be}\:{simplified},\:{then}\:\frac{{a}+{b}}{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} } \\ $$$${can}\:{not}\:{also}\:{be}\:{simplified}. \\ $$$$ \\ $$ Answered by MJS_new last…

Question-190230

Question Number 190230 by 073 last updated on 29/Mar/23 Answered by Rasheed.Sindhi last updated on 29/Mar/23 $$\mathrm{a}_{\mathrm{5}} =\mathrm{5}−\mathrm{1}=\mathrm{4}\:\:\:\left[\:\because\:\:\mathrm{5}=\mathrm{2}\left(\mathrm{2}\right)+\mathrm{1}\:\right] \\ $$$$\mathrm{a}_{\mathrm{6}} =\mathrm{2}\left(\mathrm{6}\right)+\mathrm{3}=\mathrm{15}\:\:\:\left[\:\because\:\mathrm{6}=\mathrm{2}\left(\mathrm{3}\right)\:\right] \\ $$$$\mathrm{a}_{\mathrm{5}} +\mathrm{a}_{\mathrm{6}} =\mathrm{4}+\mathrm{15}=\mathrm{19}…