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{"version":3,"file":"2N42-2-EUONGaAe.js","sources":["../../src/exercices/2e/2N42-2.js"],"sourcesContent":["import { choice, combinaisonListes, shuffle } from '../../lib/outils/arrayOutils'\nimport Exercice from '../deprecatedExercice.js'\nimport { listeQuestionsToContenu } from '../../modules/outils.js'\nimport { ajouteChampTexteMathLive } from '../../lib/interactif/questionMathLive.js'\nimport { setReponse } from '../../lib/interactif/gestionInteractif.js'\nexport const interactifReady = true\nexport const interactifType = 'mathLive'\nexport const titre = 'Express one variable in terms of others using mathematical formulas'\nexport const dateDePublication = '02/10/2023'\n/**\n *\n * @author Gilles Mora\n * 2N42-2\n */\nexport const uuid = '96bac'\nexport const ref = '2N42-2'\nexport default function ExprimerEnFonctionDesAutresFormules () {\n Exercice.call(this) // Héritage de la classe Exercice()\n this.titre = titre\n this.nbCols = 1\n this.nbColsCorr = 1\n this.spacing = 1\n this.spacingCorr = 1\n this.nbQuestions = 1\n this.sup = 1\n this.nouvelleVersion = function () {\n this.sup = parseInt(this.sup)\n this.listeQuestions = [] // Liste de questions\n this.listeCorrections = [] // Liste de questions corrigées\n let typesDeQuestionsDisponibles = []\n if (this.sup === 1) {\n typesDeQuestionsDisponibles = shuffle([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14])\n } else if (this.sup === 2) {\n typesDeQuestionsDisponibles = shuffle([15, 16, 17, 18, 19, 20, 21, 22])\n } else if (this.sup === 3) {\n typesDeQuestionsDisponibles = shuffle([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22])\n } // Tous les cas possibles sauf fractions\n const listeTypeDeQuestions = combinaisonListes(typesDeQuestionsDisponibles, this.nbQuestions)\n for (let i = 0, texte, texteCorr, reponse, cpt = 0, typesDeQuestions, choix1, nomV, choix; i < this.nbQuestions && cpt < 50;) {\n typesDeQuestions = listeTypeDeQuestions[i]\n texteCorr = ''\n switch (typesDeQuestions) {\n case 1: // côté carré en fonction de son perimètre\n texte = 'Express the side $c$ of a square in terms of its perimeter $P$.<br>'\n reponse = ['c=\\\\dfrac{P}{4}', '\\\\dfrac{P}{4}=c']\n texteCorr = 'The perimeter $P$ of a square as a function of its side $c$ is given by $P=4\\\\times c$.<br>We isolate $c$ in a member of the equality:<br>$ \\\\begin{aligned}P&=4\\\\times c\\\\\\\\\\\\dfrac{P}{4}&=c\\\\end{aligned}$ <br>Thus, $c=\\\\dfrac{P} {4}$'\n break\n case 2: // côté triangle équi en fonction de son perimètre\n texte = 'Express the side $c$ of an equilateral triangle in terms of its perimeter $P$.<br>'\n reponse = ['c=\\\\dfrac{P}{3}', '\\\\dfrac{P}{3}=c']\n texteCorr = 'The perimeter $P$ of an equilateral triangle as a function of its side $c$ is given by $P=3\\\\times c$.<br>We isolate $c$ in a member of the equality:<br> $\\\\begin{aligned}P&=3\\\\times c\\\\\\\\\\\\dfrac{P}{3}&=c\\\\end{aligned}$ <br>Thus, $c=\\\\dfrac{P }{3}$'\n break\n\n case 3:// rayon cercle en fonction de son perimètre\n texte = 'Express the radius $r$ of a circle as a function of its perimeter $P$.<br>'\n reponse = ['r=\\\\dfrac{P}{2\\\\pi}', '\\\\dfrac{P}{2\\\\pi}=r', 'r=\\\\dfrac{1}{2\\\\pi}\\\\times P', '\\\\dfrac{1}{2\\\\pi}\\\\times P=r']\n texteCorr = 'The perimeter $P$ of a circle as a function of its radius $r$ is given by $P=2\\\\pi r$.<br>We isolate $r$ in a member of the equality:<br>$ \\\\begin{aligned}P&=2\\\\times \\\\pi \\\\times r\\\\\\\\\\\\dfrac{P}{2\\\\times \\\\pi}&=r\\\\end{aligned}$ <br>Thus, $r=\\\\dfrac{P}{2\\\\times \\\\pi}$ or $r=\\\\dfrac{P}{2\\\\pi}$. '\n break\n\n case 4:// diamètre cercle en fonction de son perimètre\n texte = 'Express the diameter $d$ of a circle in terms of its perimeter $P$.<br>'\n reponse = ['d=\\\\dfrac{P}{\\\\pi}', '\\\\dfrac{P}{\\\\pi}=d', 'd=\\\\dfrac{1}{\\\\pi}\\\\times P', '\\\\dfrac{1}{\\\\pi}\\\\times P=d']\n texteCorr = 'The perimeter $P$ of a circle as a function of its diameter $d$ is given by $P=\\\\pi \\\\times d$.<br>We isolate $d$ in a member of the equality:<br>$\\\\begin{aligned}P&=\\\\pi \\\\times d\\\\\\\\\\\\dfrac{P}{\\\\pi}&=d\\\\end{aligned}$ <br>So, $d= \\\\dfrac{P}{\\\\pi}$. '\n break\n case 5: // perimètre d'un cercle en fonction de son diamètre\n texte = 'Express the perimeter $P$ of a circle as a function of its diameter $d$.<br>'\n reponse = ['P=\\\\pi\\\\times d', '\\\\pi\\\\times d=P']\n texteCorr = 'The perimeter $P$ of a circle as a function of its diameter $d$ is given by $P=\\\\pi \\\\times d$ or $P=\\\\pi d$.'\n break\n\n case 6:// perimètre d'un cercle en fonction de son rayon\n texte = 'Express the perimeter $P$ of a circle as a function of its radius $r$.<br>'\n reponse = ['P=2\\\\pi\\\\times r', '2\\\\pi\\\\times r=P']\n texteCorr = 'The perimeter $P$ of a circle as a function of its radius $r$ is given by $P=2\\\\pi\\\\times r$ or more simply $P=2\\\\pi r$.'\n break\n\n case 7 :// perimètre d'un cercle en fonction de son diamètre\n texte = 'Express the side $c$ of a square in terms of its area $A$.<br>'\n reponse = ['c=\\\\sqrt{A}', '\\\\sqrt{A}=c']\n texteCorr = 'The area $A$ of a square as a function of its side $c$ is given by $A=c^2$.<br>We isolate $c$ in a member of the equality:<br>$\\\\begin{aligned}A&=c^2\\\\\\\\\\\\sqrt{A}&=c\\\\end{aligned}$ <br>Thus, $c=\\\\sqrt{A}$. '\n break\n\n case 8:// diamètre d'un cercle en fonction de son perimètre\n texte = 'Express the diameter $d$ of a circle in terms of its perimeter $P$.<br>'\n reponse = ['d=\\\\dfrac{P}{\\\\pi}', '\\\\dfrac{P}{\\\\pi}=d', 'd=\\\\dfrac{2P}{2\\\\pi}', '\\\\dfrac{2P}{2\\\\pi}=d']\n texteCorr = 'The perimeter of a circle as a function of its radius $r$ is given by $P=2\\\\times \\\\pi\\\\times r$.<br>Or $r=\\\\dfrac{d}{2}$ , so the perimeter of a circle as a function of its diameter $d$ is given by $P=2\\\\times \\\\pi\\\\times \\\\dfrac{d}{2}$.<br>We isolate $d $ in a member of equality:<br>$\\\\begin{aligned}P&=2\\\\times \\\\pi\\\\times \\\\dfrac{d}{2}\\\\\\\\P&= \\\\pi\\\\times d\\\\\\\\\\\\dfrac{P}{\\\\pi}&=d\\\\end{aligned}$ <br>Thus, $d=\\\\dfrac{P}{\\\\pi}$. '\n break\n\n case 9:// rayon disque en fonction de son aire\n texte = 'Express the radius $r$ of a disk in terms of its area $A$.<br>'\n reponse = ['r=\\\\sqrt{\\\\dfrac{A}{\\\\pi}}', '\\\\sqrt{\\\\dfrac{A}{\\\\pi}}=r', 'r=\\\\dfrac{\\\\sqrt{A}}{\\\\sqrt{\\\\pi}}', '\\\\dfrac{\\\\sqrt{A}}{\\\\sqrt{\\\\pi}}=r']\n texteCorr = 'The area $A$ of a disk as a function of its radius $r$ is given by $A=\\\\pi\\\\times r^2$.<br>We isolate $r$ in a member of the equality :<br>$\\\\begin{aligned}A&=\\\\pi\\\\times r^2\\\\\\\\\\\\dfrac{A}{\\\\pi}&=r^2\\\\\\\\\\\\sqrt{ \\\\dfrac{A}{\\\\pi}}&=r\\\\end{aligned}$ <br>Thus, $r=\\\\sqrt{\\\\dfrac{A}{\\\\pi}}$ or again $ r=\\\\dfrac{\\\\sqrt{A}}{\\\\sqrt{\\\\pi}}$. '\n break\n\n case 10:// diamètre disque en fonction de son aire\n texte = 'Express the diameter $d$ of a disk in terms of its area $A$.<br>'\n reponse = ['d=\\\\sqrt{\\\\dfrac{4\\\\times A}{\\\\pi}}', '\\\\sqrt{\\\\dfrac{4\\\\times A}{\\\\pi}}',\n 'd=\\\\dfrac{2\\\\sqrt{A}}{\\\\sqrt{\\\\pi}}', '\\\\dfrac{2\\\\sqrt{A}}{\\\\sqrt{\\\\pi}}=d',\n 'd=2\\\\times\\\\sqrt{\\\\dfrac{A}{\\\\pi}}', 'd=2\\\\times\\\\sqrt{\\\\dfrac{A}{\\\\pi}}=d'\n ]\n texteCorr = 'The area $A$ of a disk as a function of its radius $r$ is given by $A=\\\\pi\\\\times r^2$.<br>As $r=\\\\dfrac{d}{2 }$, then $A=\\\\pi\\\\times \\\\left(\\\\dfrac{d}{2}\\\\right)^2$.<br>We isolate $d$ in a member of the equality: <br>$\\\\begin{aligned}A&=\\\\pi\\\\times \\\\left(\\\\dfrac{d}{2}\\\\right)^2\\\\\\\\A&=\\\\pi \\\\times \\\\dfrac{d^2}{4}\\\\\\\\4\\\\times A &=\\\\pi \\\\times d^2\\\\\\\\\\\\dfrac{4\\\\times A}{\\\\pi} & = d^2\\\\\\\\\\\\sqrt{\\\\dfrac{4\\\\times A}{\\\\pi}} &= d\\\\end{aligned}$ <br>So, $d=\\\\sqrt{ \\\\dfrac{4\\\\times A}{\\\\pi}}$ or $d=2\\\\dfrac{\\\\sqrt{A}}{\\\\sqrt{\\\\pi}}$. '\n break\n\n case 11:// longueur/largeur rectangle en fonction de son périmètre\n choix1 = choice([['length', 'L', 'width', 'l'], ['width', 'l', 'length', 'L']])\n texte = `Express the ${choix1[0]} $${choix1[1]}$ of a rectangle as a function of its perimeter $P$ and its ${choix1[2]} $${choix1[3]}$.<br>`\n reponse = [`${choix1[1]}=\\\\dfrac{P-2${choix1[3]}}{2}`, `\\\\dfrac{P-2${choix1[3]}}{2}=${choix1[1]}`, `${choix1[1]}=\\\\dfrac{P}{2}-${choix1[3]}`, `\\\\ dfrac{P}{2}-${choix1[3]}=${choix1[1]}`]\n texteCorr = `The perimeter $P$ of a rectangle as a function of its length and width is given by $P=2L+2l$.<br>We isolate $${choix1[1]}$ in a member of the equality:<br>$\\\\ begin{aligned}P&=2L+2l\\\\\\\\P-2${choix1[3]}&=2${choix1[1]}\\\\\\\\\\\\dfrac{P-2${choix1[3]}}{2}&=${choix1[1]}\\\\end{aligned}$ <br>Thus, $${choix1[1]} =\\\\dfrac{P-2${choix1[3]}}{2}$ or, for example $${choix1[1]}=\\\\dfrac{P}{2}-${choix1[3]}$. `\n break\n\n case 12:// longueur/largeur rectangle en fonction de son Aire\n choix1 = choice([['length', 'L', 'width', 'l'], ['width', 'l', 'length', 'L']])\n texte = `Express the ${choix1[0]} $${choix1[1]}$ of a rectangle in terms of its area $A$ and its ${choix1[2]} $${choix1[3]}$.<br>`\n reponse = [`${choix1[1]}=\\\\dfrac{A}{${choix1[3]}}`, `\\\\dfrac{A}{${choix1[3]}}=${choix1[1]}`]\n texteCorr = `The area $A$ of a rectangle as a function of its length and width is given by $A=L\\\\times l$.<br>We isolate $${choix1[1]}$ in one side of the equality:<br> $\\\\begin{aligned}A&=L\\\\times l\\\\\\\\\\\\dfrac{A}{${choix1[3]}}&=${choix1[1]}\\\\\\\\\\\\end{aligned}$ <br>Thus, $${choix1[1]}=\\\\dfrac{A}{${choix1[3]}}$. `\n break\n\n case 13:// Diamètre en fonction du rayon\n texte = 'Express the diameter $d$ of a circle in terms of its radius $r$.<br>'\n reponse = ['d=2\\times r', '2\\\\times r=d']\n texteCorr = 'The diameter of a circle is twice its radius. <br>Thus, $d=2\\\\times r$ or even $d=2r$. '\n break\n\n case 14:// rayon en fonction du diamètre\n texte = 'Express the radius $r$ of a circle as a function of its diameter $d$.<br>'\n reponse = ['r=\\\\dfrac{d}{2}', '\\\\dfrac{d}{2}=r', 'r=\\\\dfrac{1}{2}\\\\times d', '\\\\dfrac{1}{2}\\\\times d=r', 'r=0.5\\times d', '0.5 \\\\times d=r']\n texteCorr = 'The diameter of a circle is twice its radius, so the radius is half the diameter: <br>Thus, $r=\\\\dfrac{d}{2}$. '\n\n break\n // with the formulas* ******************************************** ******************************************* */\n case 15 : // aire triangle : hauteur ou base en fonction des autres\n choix1 = choice([['height', 'h', 'base', 'B'], ['Base', 'B', 'height', 'h']])\n texte = `The area $A$ of a triangle is given by: $A=\\\\dfrac{B\\\\times h}{2}$, with $B$ the length of one side and $h$ the height relative to this side.<br>Express $${choix1[1]}$ in terms of $A$ and $${choix1[3]}$.<br>`\n reponse = [`${choix1[1]}=\\\\dfrac{2\\\\times A}{${choix1[3]}}`, `\\\\dfrac{2\\\\times A}{${choix1[3]}}=${choix1[1]}`,\n `${choix1[1]}=2\\\\times\\\\dfrac{A}{${choix1[3]}}`, `2\\\\times\\\\dfrac{A}{${choix1[3]}}=${choix1[1]}`]\n texteCorr = `We isolate $${choix1[1]}$ in a member of the equality:<br>$\\\\begin{aligned}A&=\\\\dfrac{B\\\\times h}{2}\\\\\\\\A\\\\times 2&=B\\\\times h\\\\\\\\\\\\dfrac{A\\\\times 2}{${choix1[3]}}&= ${choix1[1]}\\\\end{aligned}$<br>An expression of $${choix1[1]}$ in terms of $A$ and $${choix1[3]}$ is $${choix1[1]}=\\\\dfrac{A\\\\times 2}{${choix1[3]}}$ or more simply $${choix1[1]}=\\\\dfrac{2A}{${choix1[3]}}$. `\n break\n\n case 16: // aire trapèze : petite base ou grande base en fonction des autres\n choix1 = choice([['b', 'B'], ['B', 'b']])\n texte = `The area $A$ of a trapezoid is given by: $A=\\\\dfrac{(b+B)\\\\times h}{2}$, with $B$ the length of the long base, $b$ the length of the small base and $h$ the height of the trapezoid.<br>Express $${choix1[0]}$ in terms of $A$, $${choix1[1]}$ and $h$.<br>`\n reponse = [`${choix1[0]}=2\\\\times\\\\dfrac{A}{h}-${choix1[1]}`, `2\\\\times\\\\dfrac{A}{h}-${choix1[1]}=${choix1[0]}`,\n`${choix1[0]}=\\\\dfrac{2\\\\times A}{h}-${choix1[1]}`, `\\\\dfrac{2\\\\times A}{h}-${choix1[1]}=${choix1[0]}`,\n`${choix1[0]}=\\\\dfrac{2\\\\times A-${choix1[1]}\\\\times h}{h}`, `\\\\dfrac{2\\\\times A-${choix1[1]}\\\\times h}{h}=${choix1[0]}`]\n texteCorr = `We isolate $${choix1[0]}$ in a member of the equality:<br>$\\\\begin{aligned}A&=\\\\dfrac{(b+B)\\\\times h}{2}\\\\\\\\2\\\\times A&=(b+B)\\\\times h\\\\\\\\\\\\dfrac{2A}{h}&=b+B\\\\\\\\\\\\dfrac{2A}{h}-${choix1[1]} &=${choix1[0]}\\\\end {aligned}$<br>An expression of $${choix1[0]}$ in terms of $A$, $${choix1[1]}$ and $h$ is $${choix1[0]}=\\\\dfrac{2A}{h}-${choix1[1]}$. `\n break\n\n case 17:// aire trapèze : hauteur en fonction des autres\n texte = 'The area $A$ of a trapezoid is given by: $A=\\\\dfrac{(b+B)\\\\times h}{2}$, with $B$ the length of the long base, $b$ the length of the small base and $h$ the height of the trapezoid.<br>Express $h$ in terms of $A$, $B$ and $b$.<br>'\n reponse = ['h=\\\\dfrac{2\\\\times A}{B+b}', '\\\\dfrac{2\\\\times A}{B+b}=h',\n 'h=2\\times \\\\dfrac{A}{B+b}', '2\\\\times \\\\dfrac{A}{B+b}=h']\n texteCorr = 'We isolate $h$ in a member of the equality:<br>$\\\\begin{aligned}A&=\\\\dfrac{(b+B)\\\\times h}{2}\\\\\\\\2\\\\times A&=(b+B)\\\\times h\\\\\\\\\\\\dfrac{2A}{b+B}&=h\\\\end{aligned}$<br>An expression of $h$ as a function of $A $, of $B$ and of $b$ is $h= \\\\dfrac{2A}{b+B}$. '\n break\n\n case 18:// moyenne arithmétique\n choix1 = choice([['a', 'b'], ['b', 'a']])\n texte = `The arithmetic mean of two numbers $a$ and $b$ is the number $m$ defined by: $m=\\\\dfrac{a+b}{2}$.<br>Express $${choix1[0]}$ as a function of $m $ and $${choix1[1]}$.<br>`\n reponse = [`${choix1[0]}=2\\\\times m - ${choix1[1]}`, `2\\\\times m - ${choix1[1]}=${choix1[0]}`\n ]\n texteCorr = `We isolate $${choix1[0]}$ in a member of the equality:<br>$\\\\begin{aligned}m&=\\\\dfrac{a+b}{2}\\\\\\\\2\\\\times m&=a+b\\\\\\\\2\\\\times m-${choix1[1]}&=${choix1[0]}\\\\end{aligned}$<br>An expression of $${choix1[0]}$ in terms of $m$ and $${choix1[1]}$ is $${choix1[0]}=2m-${choix1[1]}$. `\n break\n\n case 19:// moyenne géométrique\n choix1 = choice([['a', 'b'], ['b', 'a']])\n texte = `The geometric mean of two non-zero numbers $a$ and $b$ (of the same sign) is the number $m$ defined by: $m=\\\\sqrt{a\\\\times b}$.<br>Express $${choix1[0]} $ as a function of $m$ and $${choix1[1]}$.<br>`\n reponse = [`${choix1[0]}=\\\\dfrac{m^2}{${choix1[1]}}`, `\\\\dfrac{m^2}{${choix1[1]}}=${choix1[0]}`,\n `${choix1[0]}=\\\\dfrac{1}{${choix1[1]}}\\\\times m^2`, `\\\\dfrac{1}{${choix1[1]}}\\\\times m^2=${choix1[0]}`\n ]\n texteCorr = `We isolate $${choix1[0]}$ in a member of the equality:<br>$\\\\begin{aligned}m&=\\\\sqrt{a\\\\times b}\\\\\\\\m^2&=a\\\\times b\\\\\\\\\\\\dfrac{m^2}{${choix1[1]}}&=${choix1[0]}\\\\end{aligned}$<br>An expression of $${choix1[0]}$ in terms of $m$ and $${choix1[1]}$ is $\\\\dfrac{m ^2}{${choix1[1]}}$. `\n break\n\n case 20:// moyenne harmonique\n choix1 = choice([['a', 'b'], ['b', 'a']])\n texte = `The harmonic mean of two non-zero numbers $a$ and $b$ is the number $h$ defined by: $\\\\dfrac{1}{h}=\\\\dfrac{1}{2}\\\\left(\\\\ dfrac{1}{a}+\\\\dfrac{1}{b}\\\\right)$.<br>Express $${choix1[0]}$ in terms of $h$ and $${choix1[1]}$.<br>`\n reponse = [`${choix1[0]}=\\\\dfrac{1}{\\\\dfrac{2}{h}-\\\\dfrac{1}{${choix1[1]}}}`, `\\\\dfrac{1}{\\\\dfrac{2}{h}-\\\\dfrac{1}{${choix1[1]}}}=${choix1[0]}`,\n `${choix1[0]}=\\\\dfrac{${choix1[1]}\\\\times h}{2\\\\times ${choix1[1]}-h}`, `\\\\dfrac{${choix1[1]}\\\\times h}{2\\\\times ${choix1[1]}-h}=${choix1[0]}`,\n `${choix1[0]}=\\\\dfrac{1}{\\\\dfrac{2\\\\times ${choix1[1]}-h}{${choix1[1]}\\\\times h}}`, `\\\\dfrac{1}{\\\\dfrac{2\\\\times ${choix1[1]}-h} }{${choix1[1]}\\\\times h}}=${choix1[0]}`\n ]\n texteCorr = `We isolate $${choix1[0]}$ in a member of the equality:<br>$\\\\begin{aligned}\\\\dfrac{1}{h}&=\\\\dfrac{1}{2}\\\\left(\\\\dfrac {1}{a}+\\\\dfrac{1}{b}\\\\right)\\\\\\\\\\\\dfrac{2}{h}&=\\\\dfrac{1}{a}+\\\\dfrac{1 }{b}\\\\\\\\\\\\dfrac{2}{h}-\\\\dfrac{1}{${choix1[1]}}&=\\\\dfrac{1}{${choix1[0]}}\\\\\\\\\\\\dfrac{1}{\\\\dfrac{2}{h}-\\\\dfrac{1}{${choix1[1]}}}&=${choix1[0]}\\\\\\\\\\\\dfrac{1}{\\\\dfrac{2${choix1[1]}}{${choix1[1]}h}-\\\\dfrac{h}{ ${choix1[1]}h}}&=${choix1[0]}\\\\\\\\\\\\dfrac{1}{\\\\dfrac{2${choix1[1]}-h}{${choix1[1]}h}}&=${choix1[0]}\\\\\\\\\\\\dfrac{${choix1[1]}h}{2${choix1[1]}-h}&=${choix1[0]} \\\\end{aligned}$<br>An expression of $${choix1[0]}$ in terms of $h$ and $${choix1[1]}$ is $${choix1[0]}=\\\\dfrac{${choix1[1]}h}{2${choix1[1]}-h}$. `\n break\n\n case 21:// taux d'évolution F\n texte = 'The rate of change between two values $I$ and $F$ is the number $T$ defined by: $T=\\\\dfrac{F-I}{I}$.<br>Express $F$ as a function of $I $ and $T$.<br>'\n reponse = ['F=T\\\\times I+I', 'T\\\\times I+I=F',\n 'F=I\\\\times(T+1)', 'I\\\\times(T+1)=F']\n\n texteCorr = 'We isolate $F$ in a member of the equality:<br>$\\\\begin{aligned}T&=\\\\dfrac{F-I}{I}\\\\\\\\T\\\\times I&=F-I\\\\\\\\T \\\\times I+I&=F\\\\end{aligned}$<br>An expression of $F$ as a function of $T$ and $I$ is $F=T\\\\times I +I$ or again $ F=I(T+1)$. '\n break\n\n case 22:// taux d'évolution I\n texte = 'The rate of change between two values $I$ and $F$ is the number $T$ defined by: $T=\\\\dfrac{F-I}{I}$.<br>Express $I$ as a function of $F $ and $T$.<br>'\n reponse = ['I=\\\\dfrac{F}{T+1}', '\\\\dfrac{F}{T+1}=I']\n\n texteCorr = 'We isolate $I$ in a member of the equality:<br>$\\\\begin{aligned}T&=\\\\dfrac{F-I}{I}\\\\\\\\T\\\\times I+I&=F\\\\\\\\I(T+1)&=F\\\\\\\\I&=\\\\dfrac{F}{T+1}\\\\end{aligned}$<br>An expression of $I$ as a function of $T$ and of $F$ is $I=\\\\dfrac{F}{T+1}$. '\n break\n }\n texte += ajouteChampTexteMathLive(this, i)\n setReponse(this, i, reponse)\n if (this.questionJamaisPosee(i, typesDeQuestions, choix, nomV)) {\n // If the question has never been asked, we create another one\n this.listeQuestions.push(texte)\n this.listeCorrections.push(texteCorr)\n i++\n }\n cpt++\n }\n listeQuestionsToContenu(this)\n }\n this.besoinFormulaireNumerique = ['Possible cases', 3, '1: Without formula reminder\\n 2: With formula reminder\\n 3: 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