File: /home/mmtprep/public_html/mathzen.mmtprep.com/assets/3G13-Wrdt2MTo.js
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$k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\dfrac{${c}}{${r}} = ${o}$.
`,this.correctionDetaillee&&(n=String.raw`$[${t}${a}]$ is the image of $[${t}${$}]$ and $${t} ${a} ${W} ${t} ${$}$ so it is ${N} and we have ${I}.<br> ${Q.solution}`,n+=String.raw`<br>The ratio of this homothety is ${le} quotient of the length of a segment "at arrival" by its length "at departure".<br>Let $k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\ dfrac{${c}}{${r}} = ${o}$.`);break;case"picture":x=String.raw`$${a}$ is the image of $${$}$ by a homothety of center $${t}$ and ratio $k=${o}$such that $ {${t}${$} = ${r}\text{ cm}}$.<br>${M}Calculate $${t}${a}$ ${V}. <br>${te.enonce}`,n=String.raw`
$${t}${a}= ${m} \times ${r} = ${c}~\text{cm}$.
`,this.correctionDetaillee&&(n=String.raw`
${I} donc $[${t}${a}]$ est ${N} de $[${t}${$}]$.
<br>${te.solution}
`,n+=String.raw`<br>A ${E} ratio scaling is a transformation which multiplies all the lengths by ${he} its ratio.<br>Let $${t}${a} = ${s}k \times ${t}${$}$.<br>So $${t}${a}= ${m} \times ${r} = ${c}~\text{cm }$.`);break;case"antecedent":x=String.raw`$${a}$ is the image of $${$}$ by a homothety of center $${t}$ and ratio $k=${o}$ such that $ {${t}${a} = ${c}\text{ cm}}$.<br>${M}Calculate $${t}${$}$ ${V}. <br>${$e.enonce}`,n=String.raw`$${t}${$}=\dfrac{${t}${a}}{${m}}=\dfrac{${c}}{${m}} = ${r}~\text{cm}$.`,this.correctionDetaillee&&(n=String.raw`
${I} donc $[${t}${a}]$ est ${N} de $[${t}${$}]$.
<br>${$e.solution}
`,n+=String.raw`<br>A ${E} ratio scaling is a transformation which multiplies all the lengths by ${he} its ratio.<br>Let $${t}${a} = ${s}k \times ${t}${$}$.<br>So $${t}${$}=\dfrac{${t}${a}}{${s}k}=\dfrac {${c}}{${m}} ${Te} = ${r}~\text{cm}$.`);break;case"image2steps":g=[String.raw`${t}${v} = ${D}\text{ cm}`,String.raw`${t}${a} = ${c}\text{ cm}`,String.raw`${t}${$} = ${r}\text{ cm}`],l=X([0,1,2]),S=g[l[0]],B=g[l[1]],ce=g[l[2]],x=String.raw`$${a}$ and $${y}$ are the respective images of $${$}$ and $${v}$ by a homothety${U} with center $${t}$ such that $ {${S}}$, $ {${B}}$ and $ {${ce}}$.<br>${M}Calculate $${t}${y}$ ${xe}.<br>${ae.enonce}`,n=String.raw`
$k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\dfrac{${c}}{${r}} = ${o}$ et $${t}${y}= ${m} \times ${D} = ${j}~\text{cm}$.
`,this.correctionDetaillee&&(n=String.raw`$[${t}${a}]$ is the image of $[${t}${$}]$ and $${t} ${a} ${W} ${t} ${$}$ so it is ${N} and we have ${I}.<br>${ae.solution}`,n+=String.raw`<br>The ratio of this homothety is${le} quotient of the length of a segment "at arrival" by its length "at departure".<br>Let $k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\ dfrac{${c}}{${r}} = ${o}$.<br>$[${t}${y}]$ is the image of $[${t}${v}]$.<br>But a scaling of ratio ${E} is a transformation which multiplies all the lengths by ${he} its ratio. <br>Let $${t}${y}= ${s}k \times ${t}${v}$.<br>So $${t}${y}= ${m} \times ${D} = ${j}~\text{cm}$.`);break;case"background2steps":g=[String.raw`${t}${y} = ${j}\text{ cm}`,String.raw`${t}${a} = ${c}\text{ cm}`,String.raw`${t}${$} = ${r}\text{ cm}`],l=X([0,1,2]),S=g[l[0]],B=g[l[1]],ce=g[l[2]],x=String.raw`$${a}$ and $${y}$ are the respective images of $${$}$ and $${v}$ by a ${U} scale with center $${t}$ such that $ {${S}}$, $ {${B}}$ and $ {${ce}}$.<br>${M}Calculate $${t}${v}$ ${xe}.<br>${ie.enonce}`,n=String.raw`$k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\dfrac{${c}}{${r}} = ${o}$and $${t}${v}=\dfrac{${t}${y}}{${m}}=\dfrac{${j}}{${m}} = ${D}~\ text{cm}$.`,this.correctionDetaillee&&(n=String.raw`$[${t}${a}]$ is the image of $[${t}${$}]$ and $${t} ${a} ${W} ${t} ${$}$ so it is ${N} and we have ${I}.<br>${ie.solution}`,n+=String.raw`<br>The ratio of this homothety is ${le} quotient of the length of a segment "at arrival" by its length "at departure".<br>Let $k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\ dfrac{${c}}{${r}} = ${o}$.<br>$[${t}${y}]$ is the image of $[${t}${v}]$.<br>But a scaling of ratio ${E} is a transformation which multiplies all the lengths by ${he} its ratio. <br>Let $${t}${y} = ${s}k \times ${t}${v}$.<br>So $${t}${v}=\dfrac{${t}${y}}{${s}k}=\dfrac{${j}}{${m}} ${Ee} = ${D}~\text{cm}$.`);break;case"areaImage":pe=A===_?"":"approximately",de=pe==="approximately"?"\\approx":"=",x=String.raw`A figure has the area $ {${d}\text{ cm}^2}$.<br>Calculate the area of its image by a scale of ratio $${s}${T}$ (round to the nearest $ {\text{mm}^2}$ close if necessary).`,n=String.raw`
$ {${ee}^2 \times ${d} ${de} ${_}~\text{cm}^2}$
`,this.correctionDetaillee&&(n=String.raw`A ${E} ratio scaling is a transformation that multiplies all areas by the square of its ratio.<br>$${ee}^2 \times ${d} = ${A}$<br>So the area of the image in this figure is ${pe} $ {${_}~\text{cm}^2}$.`);break;case"areaAntecedent":x=String.raw`The image of a figure by a homothety of ratio $${s}${T}$ has the area $ {${A}\text{ cm}^2}$.<br>Calculate the area of the starting figure.`,n=String.raw`$ {\dfrac{${A}}{${ee}^2} = ${d}~\text{cm}^2}$`,this.correctionDetaillee&&(n=String.raw`A ${E} ratio scaling is a transformation which multiplies all the areas by the square of its ratio.<br>Let $\mathscr{A}$ denote the area of the starting figure.<br>Hence $${ee}^2 \times \mathscr{A} = ${A}$.<br>Then $\mathscr{A}=\dfrac{${A}}{${ee}^2} = ${d}$.<br>So the area of the starting figure is $ {${d}~\text{cm}^2}$.`);break;case"areaReport":x=String.raw`A figure and its image by a ${E} ratio scale have respectively areas $ {${d}\text{ cm}^2}$ and $ {${A}\text{ cm}^2}$.<br>Calculate the ratio of the homothety.`,n=String.raw`$ {k=${s}\sqrt{\dfrac{${A}}{${d}}} = ${s}${T}}$`,this.correctionDetaillee&&(n=String.raw`A ${E} ratio scale is a transformation which multiplies all the areas by the square of its ratio.<br>Let $k$ be the ratio of this scale. We therefore have $k^2 \times ${d} = ${A}$, or again $k ^2=\dfrac{${A}}{${d}}$.<br>Hence $ {k=${s}\sqrt{\dfrac{${A}}{${d}}} = ${s}${T}}$.`);break;case"report2":g=[String.raw`${$}${a} = ${F}\text{ cm}`,String.raw`${t}${$} = ${r}\text{ cm}`],l=X([0,1]),S=g[l[0]],B=g[l[1]],x=String.raw`$${a}$ is the image of $${$}$ by a scaling ${U} with center $${t}$ such that $ {${S}}$ and $ {${B}}$.<br>${M} Calculate the ratio $k$ of this scaling ${V}.${re.enonce}`,n=String.raw`
$k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\dfrac{${c}}{${r}} = ${o}$.
`,this.correctionDetaillee&&(n=String.raw`$${t}${a} = ${je} = ${c}\text{ cm}$<br>$[${t}${a}]$ is the image of $[${t}${$}]$ and $${t} ${a} ${W} ${t} ${$}$ so it is ${N} and we have ${I}.<br> ${Q.solution}`,n+=String.raw`<br>The ratio of this homothety is ${le} quotient of the length of a segment "at arrival" by its length "at departure".<br>Let $k=${s}\dfrac{${t}${a}}{${t}${$}} = ${s}\ dfrac{${c}}{${r}} = ${o}$.`);break;case"frame":g=[String.raw`${t}${a} = ${c}\text{ cm}`,String.raw`${t}${$} = ${r}\text{ cm}`],l=X([0,1]),S=g[l[0]],B=g[l[1]],x=String.raw`$${a}$ is the image of $${$}$ by a homothety ${U} with center $${t}$ such that $ {${S}}$ and $ {${B}}$.<br>${M}Without performing calculations, what can we say about the ratio $k $ of this homothety?(choose the correct answer)<br>$\square\hphantom{a} k<-1 \hspace{1cm} \square\hphantom{a} -1 < k < 0 \hspace{1cm} \ square\hphantom{a} 0 < k < 1 \hspace{1cm} \square\hphantom{a} k > 1$.<br>${V}${Q.enonce}`,n=String.raw`
$${I}$.
`,this.correctionDetaillee&&(n=String.raw`$[${t}${a}]$ is the image of $[${t}${$}]$ and $${t} ${a} ${W} ${t} ${$}$ so it is ${N}.<br>Moreover $${a}${we}[${t};${$})$ therefore ${I}.<br> ${Q.solution}`);break;case"framerk2":g=[String.raw`${$}${a} = ${F}\text{ cm}`,String.raw`${t}${$} = ${r}\text{ cm}`],l=X([0,1]),S=g[l[0]],B=g[l[1]],x=String.raw`$${a}$ is the image of $${$}$ by a homothety ${U} with center $${t}$ such that $ {${S}}$ and $ {${B}}$.<br>${M}Without performing calculations, what can we say about the ratio $k $ of this homothety?(choose the correct answer)<br>$\square\hphantom{a} k<-1 \hspace{1cm} \square\hphantom{a} -1 < k < 0 \hspace{1cm} \ square\hphantom{a} 0 < k < 1 \hspace{1cm} \square\hphantom{a} k > 1$.<br>${V}${re.enonce}`,n=String.raw`$${I}$.`,this.correctionDetaillee&&(n=String.raw`$${t}${a} = ${je} = ${c}\text{ cm}$<br>$[${t}${a}]$ is the image of $[${t}${$}]$ and $${t} ${a} ${W} ${t} ${$}$ so it is ${N}.<br>Plus $ ${a}${we}[${t};${$})$ therefore ${I}.<br> ${Q.solution}`);break}this.questionJamaisPosee(Y,o)&&(this.listeQuestions.push(x),this.listeCorrections.push(n),Y++),Ae++}ze(this)}}export{it as dateDeModifImportante,at as dateDePublication,nt as default,st as ref,w as texNum,$t as titre,rt as uuid};
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