Question Number 202726 by ajfour last updated on 02/Jan/24 $${pk}={b} \\ $$$${pq}+{sk}={c} \\ $$$${sq}={d} \\ $$$$\sqrt{{t}}=\frac{{t}−{q}}{{k}}=\frac{{t}−{s}}{{p}} \\ $$$${Given}\:{are}\:{b},\:{c},\:{d}.\:{Find}\:{t}. \\ $$ Commented by MATHEMATICSAM last updated…
Question Number 202651 by hardmath last updated on 31/Dec/23 Commented by Frix last updated on 31/Dec/23 $${y}=\frac{\mathrm{3cos}\:{x}\:+\mathrm{sin}\:{x}}{\mathrm{10}}+{c}_{\mathrm{1}} \mathrm{e}^{\mathrm{2}{x}} +{c}_{\mathrm{2}} \mathrm{e}^{{x}} \\ $$ Answered by Mathspace…
Question Number 202701 by cherokeesay last updated on 31/Dec/23 Commented by nikif99 last updated on 01/Jan/24 $${Be}\:{prepared}\:{for}\:{the}\:{next}\:{year},\:{too}. \\ $$$${Happy}\:{and}\:{prosperous}\:{new}\:{year}. \\ $$ Terms of Service Privacy…
Question Number 202638 by depressiveshrek last updated on 31/Dec/23 $$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$ Commented by AST last updated on 31/Dec/23 $${x}=\mathrm{1}.{Suppose}\:{x}\neq\mathrm{1},{let}\:{a}=\sqrt{{x}−\frac{\mathrm{1}}{{x}}};{b}=\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}} \\ $$$${a}−{b}=\mathrm{1}−\frac{\mathrm{1}}{{x}};\left({a}+{b}\right)^{\mathrm{2}} =\mathrm{1}+{x} \\ $$$${a}^{\mathrm{2}}…
Question Number 202616 by hardmath last updated on 30/Dec/23 $$\mathrm{If}\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{a}\:+\:\mathrm{b}}\:−\:\frac{\mathrm{1}\:−\:\mathrm{a}}{\mathrm{a}\:−\:\mathrm{b}}\:=\:\mathrm{x} \\ $$$$\mathrm{Find}\:\:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{b}}{\mathrm{a}\:−\:\mathrm{b}}\:+\:\frac{\mathrm{b}}{\mathrm{a}\:+\:\mathrm{b}}\:=\:? \\ $$ Answered by MATHEMATICSAM last updated on 30/Dec/23 $$\frac{{a}}{{a}\:+\:{b}}\:−\:\frac{\mathrm{1}\:−\:{a}}{{a}\:−\:{b}}\:=\:{x} \\ $$$$\Rightarrow\:\frac{{a}}{{a}\:+\:{b}}\:−\:\mathrm{1}\:−\:\frac{\mathrm{1}\:−\:{a}}{{a}\:−\:\:{b}}\:−\:\mathrm{1}\:=\:{x}\:−\:\mathrm{2} \\…
Question Number 202604 by MATHEMATICSAM last updated on 30/Dec/23 $$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{and}\: \\ $$$${y}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{2}} \:−\:{xy}\:+\:{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{1}}\:. \\ $$ Answered by Rasheed.Sindhi…
Question Number 202605 by hardmath last updated on 30/Dec/23 Commented by mr W last updated on 30/Dec/23 $${what}\:{do}\:{you}\:{mean}\:{when}\:{you}\:{write} \\ $$$$\underset{={b}} {{max}}\:\underset{={a}} {{min}}\:\left(…\right)\:? \\ $$ Commented…
Question Number 202633 by hardmath last updated on 30/Dec/23 Answered by cortano12 last updated on 31/Dec/23 $$\:\mathrm{5}\clubsuit\mathrm{6}\:=\:\mathrm{5}−\mathrm{6}=−\mathrm{1} \\ $$$$\:\left(\mathrm{5}\clubsuit\mathrm{6}\right)\clubsuit\mathrm{9}\:=\:−\mathrm{1}\clubsuit\mathrm{9}\:=\:−\mathrm{1}−\mathrm{9}=−\mathrm{10} \\ $$ Commented by hardmath last…
Question Number 202580 by hardmath last updated on 29/Dec/23 $$\mathrm{If} \\ $$$$\underset{\:=\:\boldsymbol{\mathrm{b}}} {\mathrm{max}}\:\underset{\:=\:\boldsymbol{\mathrm{a}}} {\mathrm{min}}\:\left(\mathrm{a}^{\mathrm{2}} \:-\:\mathrm{2ab}\:-\:\mathrm{b}^{\mathrm{2}} \:-\:\mathrm{2ax}\:+\:\mathrm{10bx}\right)\:=\:\mathrm{7} \\ $$$$ \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Terms of Service…
Question Number 202551 by aba last updated on 29/Dec/23 $$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{i}} =\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{1}} +\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{2}} +…\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}_{\mathrm{1}} +\mathrm{y}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} +…+\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \: \\ $$$$\mathrm{please}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:?…