{"id":134705,"date":"2021-11-29T18:00:57","date_gmt":"2021-11-29T20:00:57","guid":{"rendered":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/?post_type=eaja&#038;p=134705"},"modified":"2021-12-22T10:28:17","modified_gmt":"2021-12-22T12:28:17","slug":"matematica-situacoes-problema-envolvendo-multiplos-e-divisores","status":"publish","type":"eaja","link":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/eaja\/matematica-situacoes-problema-envolvendo-multiplos-e-divisores\/","title":{"rendered":"Matem\u00e1tica &#8211; Situa\u00e7\u00f5es-problema envolvendo m\u00faltiplos e divisores"},"content":{"rendered":"\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Ol\u00e1! Esta aula de&nbsp;<strong>Matem\u00e1tica&nbsp;<\/strong>\u00e9 destinada a estudantes da<strong>&nbsp;5\u00aa S\u00e9rie<\/strong>&nbsp;da Eaja.<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"aligncenter size-full\"><img fetchpriority=\"high\" decoding=\"async\" width=\"720\" height=\"720\" src=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/ea-e1638186424369.jpg\" alt=\"\" class=\"wp-image-134706\" srcset=\"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/ea-e1638186424369.jpg 720w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/ea-e1638186424369-300x300.jpg 300w, https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-content\/uploads\/2021\/11\/ea-e1638186424369-150x150.jpg 150w\" sizes=\"(max-width: 720px) 100vw, 720px\" \/><figcaption>Matem\u00e1tica Pagar D\u00edgitos &#8211; Imagens gr\u00e1tis no Pixabay<\/figcaption><\/figure><\/div>\n\n\n\n<p class=\"has-black-color has-cyan-bluish-gray-background-color has-text-color has-background\" style=\"font-size:25px\">Nesta aula, voc\u00ea ir\u00e1 resolver situa\u00e7\u00f5es-problema do nosso cotidiano, que envolvam as ideias de m\u00faltiplo e de divisor de um n\u00famero natural. <\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-pale-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\">Assista \u00e0 videoaula do professor H\u00e9lio sobre essa tem\u00e1tica.<\/p>\n\n\n\n<figure class=\"wp-block-embed aligncenter is-type-video is-provider-youtube wp-block-embed-youtube wp-embed-aspect-16-9 wp-has-aspect-ratio\"><div class=\"wp-block-embed__wrapper\">\n<p class=\"responsive-video-wrap clr\"><iframe title=\"Situa\u00e7\u00f5es-problema Envolvendo M\u00faltiplos e Divisores | Matem\u00e1tica - aula 13 | 5\u00aa s\u00e9rie - Eaja\" width=\"1200\" height=\"675\" src=\"https:\/\/www.youtube.com\/embed\/DB6Q2cu9zhA?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<\/div><figcaption>Situa\u00e7\u00f5es-problema Envolvendo M\u00faltiplos e Divisores | Matem\u00e1tica &#8211; aula 13 | 5\u00aa s\u00e9rie &#8211; Eaja<br><\/figcaption><\/figure>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Antes de come\u00e7ar as situa\u00e7\u00f5es-problema, vamos a algumas defini\u00e7\u00f5es.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>M\u00faltiplo de um n\u00famero natural (Defini\u00e7\u00e3o): <\/strong>dizemos que um n\u00famero natural \u00e9 m\u00faltiplo de outro, caso o primeiro seja resultado da multiplica\u00e7\u00e3o do segundo por um n\u00famero natural qualquer.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Exemplos<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Exemplo 01<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">Os n\u00fameros 0, 5, 10, 15, 20, 25, 30, s\u00e3o todos m\u00faltiplos de 5, pois:<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">0 = 5 x 0 &nbsp; &nbsp; 5 = 5 x 1 &nbsp; &nbsp; 10 = 5 x 2 &nbsp; &nbsp; 15 = 5 x 3 &nbsp; &nbsp; 20 = 5 x 4 &nbsp; &nbsp; 25 = 5 x 5 &nbsp; &nbsp; 30 = 5 x 6<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Exemplo 02<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">Os n\u00fameros 0, 4, 8, 12, 16, 20 e 24, s\u00e3o todos m\u00faltiplos de 6, pois:<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">0 = 4 x 0 &nbsp; &nbsp; 4 = 4 x 1 &nbsp; &nbsp; 8 = 4 x 2 &nbsp; &nbsp; 12 = 4 x 3 &nbsp; &nbsp; 16 = 4 x 4 &nbsp; &nbsp; 20 = 4 x 5 &nbsp; &nbsp; 24 = 4 x 6<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Como obter m\u00faltiplos de um n\u00famero natural?<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Para obter os m\u00faltiplos de um n\u00famero natural, por exemplo os m\u00faltiplos de 8, basta multiplicar por 8 os n\u00fameros da sequ\u00eancia dos n\u00fameros naturais, ou seja, 0, 1, 2, 3, 4, 5, 6, etc. De uma maneira mais r\u00e1pida, basta acrescentar 8 unidades, a partir do zero, sucessivas vezes.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">M (6) = {0, 6, 12, 18, 24, 30, 36, &#8230;} L\u00ea-se: M\u00faltiplos de 6<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">M (9) = {0, 9, 18, 27, 36, 45, 54, &#8230;} L\u00ea-se: M\u00faltiplos de 9<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Situa\u00e7\u00f5es-problema com resolu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-light-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Situa\u00e7\u00e3o-problema 01<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Numa esta\u00e7\u00e3o de \u00f4nibus h\u00e1 dois trajetos, o \u00f4nibus que faz trajeto A&nbsp;passa a cada&nbsp;30&nbsp;minutos e o do trajeto B&nbsp;a cada 50. Considerando que os trajetos coincidem em determinado momento, em quantos minutos voltar\u00e3o a se encontrar na esta\u00e7\u00e3o?<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o&nbsp;<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Nesse caso basta encontrar o menor m\u00faltiplo comum, diferente de zero, entre 30 e 50.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">M (30) = {30, 60, 90, 120, <strong>150<\/strong>, 180, 210, &#8230;}<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">M (50) = {50, 100, 1<strong>50<\/strong>, 200, &#8230;}<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resposta:<\/strong> Depois de sa\u00edrem da esta\u00e7\u00e3o, os dois \u00f4nibus voltar\u00e3o a se encontrar depois de 150 minutos.<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-light-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Situa\u00e7\u00e3o-problema 02<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Paulo e H\u00e9lio treinam correndo numa pista, Paulo gasta 14 minutos para dar uma volta e H\u00e9lio 18 minutos. Se eles partem ao mesmo tempo da linha de sa\u00edda, a cada quanto tempo passam juntos num mesmo ponto?<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Mais uma vez, basta determinar o menor m\u00faltiplo comum, diferente de zero, entre 14 e 18.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">M (14) = {0, 14, 28, 42, 56, 79, 84, 98, 112, <strong>126<\/strong>, 140, &#8230;}<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">M (18) = {0, 18, 36, 54, 72, 90, 108, <strong>126<\/strong>, 144, &#8230;}<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resposta:<\/strong> Paulo e H\u00e9lio ir\u00e3o passar juntos, novamente, ap\u00f3s 126 minutos.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Divisores de um n\u00famero natural (Defini\u00e7\u00e3o): <\/strong>dizemos que um n\u00famero natural \u00e9 divisor ou fator outro caso a divis\u00e3o do segundo pelo primeiro seja exata. Uma divis\u00e3o \u00e9 exata quanto deixa resto zero.<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Exemplos<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">O n\u00famero 4 \u00e9 divisor de 12, pois 12:4 = 3 e deixa resto zero<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">O n\u00famero 16 n\u00e3o \u00e9 divisor de 5, pois 16:5=3 e deixa resto 1<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">O n\u00famero 28 \u00e9 divisor de 7, pois 28;7 = 4 e deixa resto zero<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Como obter os divisores de um n\u00famero natural?<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Para determinar, de uma maneira mais f\u00e1cil, os divisores de um n\u00famero natural, por exemplo 18, basta multiplicar os n\u00fameros naturais menores do que 18 e identificar aqueles cujo resultado (produto) \u00e9 18.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Exemplos<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Exemplo 01<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">Divisores de 15<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">1 x 18 = 18 &nbsp; &nbsp; 2 x 9 = 18 &nbsp; &nbsp; 3 x 6 = 18 &nbsp; &nbsp; 6 x 3 = 18 &nbsp; &nbsp; 9 x 2 = 18 &nbsp; &nbsp; 18 x 1 = 18<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">Logo os divisores de 18 s\u00e3o: 1, 2, 3, 6, 9 e 18<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Exemplo 02<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">Divisores de 20<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">1 x 20 = 20 &nbsp; &nbsp; 2 x 10 = 20 &nbsp; &nbsp; 4 x 5 = 20 &nbsp; &nbsp; 5 x 4 = 20 &nbsp; &nbsp; 10 x 2 = 20 &nbsp; &nbsp; 20 x 1 = 20<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-text-color\" style=\"font-size:25px\">Logo, os divisores de 20 s\u00e3o: 1, 2, 4, 5, 10 e 20<\/p>\n\n\n\n<p class=\"has-black-color has-cyan-bluish-gray-background-color has-text-color has-background\" style=\"font-size:25px\">Antes de come\u00e7ar a resolu\u00e7\u00e3o das situa\u00e7\u00f5es-problema, assista ao v\u00eddeo do canal do Prof. H\u00e9lio para compreender mais sobre divisores. Link: <a href=\"https:\/\/youtu.be\/Ln3o-qgVCPA\">https:\/\/youtu.be\/Ln3o-qgVCPA<\/a>&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-vivid-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Situa\u00e7\u00f5es-problema com resolu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-light-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Situa\u00e7\u00e3o-problema 01<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">(PMSC 1201\/001-Assistente Administrativo \u2013 2012) \u2013 Um escrit\u00f3rio comprou os seguintes itens: 40 marcadores de texto, 20 corretivos e 48 blocos de rascunho e dividiu esse material em pacotinhos, cada um deles contendo um s\u00f3 tipo de material, por\u00e9m todos com o mesmo n\u00famero de itens e na maior quantidade poss\u00edvel. Sabendo-se que todos os itens foram utilizados, determine o n\u00famero total de pacotinhos feitos?<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o&nbsp;<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Nesse caso, iremos determinar os divisores de 40, 20 e 48, e a resposta ser\u00e1 o maior desses divisores comuns.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">D (20) = {1, 2, <strong>4<\/strong>, 5, 10, 20} L\u00ea-se: divisores de 20<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">1 x 20 = 20 &nbsp; &nbsp; 2 x 10 = 20 &nbsp; &nbsp; 4 x 5 = 20 &nbsp; &nbsp; 5 x 4 = 20 &nbsp; &nbsp; 10 x 2 = 20 &nbsp; &nbsp; 20 x 1 = 20<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">D (40) = {1, 2, <strong>4<\/strong>, 5, 8, 10, 20, 40}<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">1 x 40 = 40 &nbsp; &nbsp; 2 x 20 = 40 &nbsp; &nbsp; 4 x 10 = 40 &nbsp; &nbsp; 5 x 8 = 40 &nbsp; &nbsp; 8 x 5 = 40 &nbsp; &nbsp; 10 x 4 = 40 &nbsp; &nbsp; 20 x 2 = 40 &nbsp; &nbsp; 40 x 1 = 40<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">D (48) = {1, 2, 3, <strong>4<\/strong>, 6, 8, 12, 16, 24 e 48}<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">1 x 48 = 48 &nbsp; 2 x 24 = 48 &nbsp; 3 x 16 = 48 &nbsp; 4 x 12 = 48&nbsp; &nbsp; 6 x 8 = 48&nbsp; &nbsp; 8 x 6 = 48&nbsp; &nbsp; 12 x 4 = 48&nbsp; &nbsp; 16 x 3 = 48&nbsp; &nbsp; 24 x 2 = 48&nbsp; &nbsp; 48 x 1 = 48<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resposta:<\/strong> O n\u00famero total de pacotinhos ser\u00e1 4. Para saber quantos marcadores, corretores e blocos em cada pacotinho, basta dividir a quantidades de cada item, por 4.<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-light-green-cyan-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Situa\u00e7\u00e3o-problema 02<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Um carpinteiro tem dois peda\u00e7os de madeira de 90&nbsp;e 120&nbsp;cent\u00edmetros cada um. Ele quer&nbsp;dividi-los&nbsp;em&nbsp;partes iguais com o maior tamanho poss\u00edvel<strong>&nbsp;<\/strong>sem que sobre nada<strong>. <\/strong>De quantos cent\u00edmetros deve ser cada uma das partes?<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resolu\u00e7\u00e3o<\/strong><\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Basta determinar o maior divisor comum entre 90 e 120.<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">Existe um m\u00e9todo para determinar o m\u00e1ximo divisor comum que irei apresentar em outra atividade. No momento vamos utilizar a multiplica\u00e7\u00e3o da sequ\u00eancia dos n\u00fameros naturais. Ap\u00f3s fazer essas multiplica\u00e7\u00f5es, encontramos:<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">D (90) = {1, 2, 3, 5, 6, 9, 10, 15, 18, <strong>30<\/strong>, 45, 90}<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\">D (120) = {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 24, <strong>30<\/strong>, 40, 60, 120}&nbsp;<\/p>\n\n\n\n<p class=\"has-black-color has-text-color\" style=\"font-size:25px\"><strong>Resposta:<\/strong> Ele deve cortar a madeira de 30 em 30 cent\u00edmetros.<\/p>\n\n\n\n<p class=\"has-text-align-center has-black-color has-pale-cyan-blue-background-color has-text-color has-background\" style=\"font-size:25px\"><strong>Problemas propostos<\/strong><\/p>\n\n\n\n<ol class=\"has-black-color has-text-color wp-block-list\" style=\"font-size:25px\"><li>Escreva os 10 primeiros m\u00faltiplos de 3, 4, 8 e 12.<\/li><li>Escreva todos os divisores naturais de: 8, 11, 18 e 36.<\/li><li>Qual \u00e9 o maior n\u00famero de 3 algarismos que \u00e9 m\u00faltiplo de 28?<\/li><li>Entre os elementos do conjunto {2, 3, 5, 8, 9, 10}, identifique os que s\u00e3o divisores de: 14, 28, 25, 45 e 54.<\/li><li>Quais s\u00e3o os divisores de 15 que tamb\u00e9m s\u00e3o divisores de 25?<\/li><li>Qual \u00e9 o menor m\u00faltiplo de 13 menor do que 100?<\/li><\/ol>\n\n\n\n<p class=\"has-black-color has-cyan-bluish-gray-background-color has-text-color has-background\" style=\"font-size:25px\">Assista aos v\u00eddeos no canal do Prof. H\u00e9lio para aprender um pouco mais. Link: <a href=\"https:\/\/www.youtube.com\/channel\/UCKmf3-WVHXY61WgWrz-OO4Q\">Professor Helio Roberto da Rocha &#8211; YouTube<\/a><\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<figure class=\"wp-block-table\"><table><tbody><tr><td><strong>Objetivos de Aprendizagem e Desenvolvimento:<\/strong><\/td><td>(EAJAMA0512) Resolver e elaborar situa\u00e7\u00f5es-problema que envolvam as ideias de m\u00faltiplo e de divisor de um n\u00famero natural.<\/td><\/tr><tr><td><strong>Refer\u00eancias:<\/strong><\/td><td>GIOVANNI J\u00daNIOR, Jos\u00e9 Ruy &#8211; A conquista da matem\u00e1tica: 6o ano: ensino fundamental: anos finais \/ Jos\u00e9 Ruy Giovanni J\u00fanior, Benedicto Castrucci. \u2014 4. ed. \u2014 S\u00e3o Paulo: FTD, 2018.SOUZA, Joamir Roberto de: Matem\u00e1tica realidade &amp; tecnologia: 6\u00ba ano: ensino fundamental: anos finais \/ Joamir Roberto de Souza. \u2013 1. ed. \u2013 S\u00e3o Paulo: FTD, 2018.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"has-cyan-bluish-gray-background-color has-background\" style=\"font-size:25px\"> Professor, essa aula segue a Matriz Estruturante para a Eaja 2021. Foi elaborada no ano de 2020, com a suspens\u00e3o das aulas presenciais devido a pandemia da Covid-19 e segue as orienta\u00e7\u00f5es de flexibiliza\u00e7\u00e3o curricular para o bi\u00eanio 2020\/2021 (Of\u00edcio Circular 149\/2020 Dirped). <\/p>\n","protected":false},"author":42,"featured_media":134706,"template":"","meta":{"_acf_changed":false,"ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"","ocean_second_sidebar":"","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"","ocean_custom_header_template":"","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"","ocean_menu_typo_font_family":"","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"default","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"default","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":""},"eaja_categoria":[104],"serie":[74],"eaja_componente":[78],"class_list":["post-134705","eaja","type-eaja","status-publish","has-post-thumbnail","hentry","eaja_categoria-2o-segmento-5a-e-6a-serie","serie-5a-serie","eaja_componente-matematica","entry","has-media"],"acf":[],"_links":{"self":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja\/134705","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja"}],"about":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/types\/eaja"}],"author":[{"embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/users\/42"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/media\/134706"}],"wp:attachment":[{"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/media?parent=134705"}],"wp:term":[{"taxonomy":"eaja_categoria","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja_categoria?post=134705"},{"taxonomy":"serie","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/serie?post=134705"},{"taxonomy":"eaja_componente","embeddable":true,"href":"https:\/\/sme.goiania.go.gov.br\/conexaoescola\/wp-json\/wp\/v2\/eaja_componente?post=134705"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}