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Category: Algebra

Find-the-largest-number-of-positive-integers-that-can-be-found-in-such-a-way-that-any-two-of-them-a-and-b-a-b-satisfy-the-next-inequality-a-b-ab-100-

Question Number 120809 by liberty last updated on 02/Nov/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{number}\:\mathrm{of}\:\mathrm{positive} \\ $$$$\mathrm{integers}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{found}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way} \\ $$$$\mathrm{that}\:\mathrm{any}\:\mathrm{two}\:\mathrm{of}\:\mathrm{them}\:{a}\:\mathrm{and}\:{b}\:\left(\:{a}\neq{b}\right)\: \\ $$$$\mathrm{satisfy}\:\mathrm{the}\:\mathrm{next}\:\mathrm{inequality}\:\mid{a}−{b}\mid\geqslant\frac{{ab}}{\mathrm{100}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-186329

Question Number 186329 by ajfour last updated on 03/Feb/23 Answered by ajfour last updated on 03/Feb/23 $${Cubic}\:{curve}:\:{y}={x}^{\mathrm{3}} −{x} \\ $$$${Parabola}:\:\:{x}={h}+\left({y}−{k}\right)^{\mathrm{2}} \\ $$$${x}={h}+\left({p}−{k}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{3}} −{p}={c}…

log-3-2-3-1-find-Characteristic-

Question Number 186305 by mustafazaheen last updated on 03/Feb/23 $$\mathrm{log}\:\left(\frac{\mathrm{3}.\bar {\mathrm{2}}}{\mathrm{3}.\bar {\mathrm{1}}}\right)\:\:\:\:\:\:\:{find}\:\mathrm{Characteristic}? \\ $$ Commented by MJS_new last updated on 03/Feb/23 $$\frac{\mathrm{3}+\frac{\mathrm{2}}{\mathrm{9}}}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{9}}}=\frac{\mathrm{29}}{\mathrm{28}}=\frac{\mathrm{29}}{\mathrm{2}^{\mathrm{2}} \mathrm{7}} \\ $$$$\mathrm{log}\:\frac{\mathrm{29}}{\mathrm{2}^{\mathrm{2}}…

solve-in-R-2log-8-x-1-3-log-4-x-1-2-

Question Number 186300 by mnjuly1970 last updated on 03/Feb/23 $$\:\:\: \\ $$$$\:\:\:\:\:\mathrm{solve}\:\:\mathrm{in}\:\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\:\lfloor\:\:\mathrm{2log}_{\:\mathrm{8}} \left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\rfloor\:=\:\mathrm{log}_{\:\mathrm{4}} \left({x}\:\right)+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by MJS_new…

x-p-x-3-x-1-3-0-x-4-px-3-x-2-p-1-3-x-p-3-0-x-2-ax-h-x-2-bx-k-0-a-b-p-h-k-ab-1-bh-ak-p-1-3-hk-p-3-say-ab-t-ah-bk-p-1-3-p-1-t-bh-ak-p-1-3-0-a-b-h-k-pt-

Question Number 186292 by ajfour last updated on 03/Feb/23 $$\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{px}^{\mathrm{3}} −{x}^{\mathrm{2}} +\left({p}−\frac{\mathrm{1}}{\mathrm{3}}\right){x}+\frac{{p}}{\mathrm{3}}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +{ax}+{h}\right)\left({x}^{\mathrm{2}} +{bx}+{k}\right)=\mathrm{0} \\ $$$${a}+{b}=−{p} \\ $$$${h}+{k}+{ab}=−\mathrm{1} \\…

Prove-that-R-m-n-C-m-n-m-Here-R-states-the-Ramsey-theory-

Question Number 186285 by Shrinava last updated on 03/Feb/23 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathrm{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$\mathrm{Here}\:\:\mathrm{R}\:\:\mathrm{states}\:\mathrm{the}\:\:\mathrm{Ramsey}\:\:\mathrm{theory} \\ $$ Commented by mr W last updated on 03/Feb/23…