{"id":12807,"date":"2017-10-29T02:33:50","date_gmt":"2017-10-29T02:33:50","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=12807"},"modified":"2022-01-15T23:24:46","modified_gmt":"2022-01-15T23:24:46","slug":"escreve-um-valor-aproximado","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=12807","title":{"rendered":"Escreve um valor aproximado"},"content":{"rendered":"<p><ul id='GTTabs_ul_12807' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_12807' class='GTTabs_curr'><a  id=\"12807_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_12807' ><a  id=\"12807_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_12807'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Escreve um valor aproximado, por excesso, a menos de uma cent\u00e9sima, do n\u00famero\u00a0\\(\\sqrt 5 + \\sqrt 7 \\).<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_12807' onClick='GTTabs_show(1,12807)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_12807'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><br \/>\n<!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Calculo3.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"12809\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=12809\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Calculo3.png\" data-orig-size=\"198,134\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"Calculo3\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Calculo3.png\" class=\"alignright size-full wp-image-12809\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Calculo3.png\" alt=\"\" width=\"198\" height=\"134\" \/><\/a>Recorrendo \u00e0 calculadora, podemos indicar \\(4,89\\) como\u00a0um valor aproximado, por excesso, a menos de uma cent\u00e9sima, do n\u00famero\u00a0\\(\\sqrt 5 + \\sqrt 7 \\).<\/p>\n<\/p>\n<p>Vamos agora resolver a quest\u00e3o recorrendo ao m\u00e9todo dos quadrados perfeitos.<\/p>\n<p>Consideremos para o erro das parcelas a metade do indicado para a soma, isto \u00e9, seja\u00a0\\(r = 0,005 = \\frac{5}{{1000}} = \\frac{1}{{200}}\\), donde\u00a0\\(n = 200\\).<\/p>\n<p>Assim, temos:<\/p>\n<p>\\[\\begin{array}{*{20}{c}}{\\begin{array}{*{20}{c}}{{{447}^2} &lt; {{200}^2} \\times 5 &lt; {{448}^2}}\\\\{{{\\left( {\\frac{{447}}{{200}}} \\right)}^2} &lt; 5 &lt; {{\\left( {\\frac{{448}}{{200}}} \\right)}^2}}\\\\{2,235 &lt; \\sqrt 5 &lt; 2,240}\\end{array}}&amp;{}&amp;{\\begin{array}{*{20}{c}}{{{529}^2} &lt; {{200}^2} \\times 7 &lt; {{530}^2}}\\\\{{{\\left( {\\frac{{529}}{{200}}} \\right)}^2} &lt; {{200}^2} \\times 7 &lt; {{\\left( {\\frac{{530}}{{200}}} \\right)}^2}}\\\\{2,645 &lt; \\sqrt 7 &lt; 2,650}\\end{array}}\\end{array}\\]<\/p>\n<p>Donde se conclui:<\/p>\n<p>\\[\\begin{array}{*{20}{c}}{2,235 &lt; \\sqrt 5 &lt; 2,240}\\\\{2,645 &lt; \\sqrt 7 &lt; 2,650}\\\\\\hline{4,880 &lt; \\sqrt 5 + \\sqrt 7 &lt; 4,890}\\end{array}\\]<\/p>\n<p>Logo, \\(\\sqrt 5 + \\sqrt 7 \\approx 4,89\\), por excesso, a menos de uma cent\u00e9sima.<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_12807' onClick='GTTabs_show(0,12807)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Escreve um valor aproximado, por excesso, a menos de uma cent\u00e9sima, do n\u00famero\u00a0\\(\\sqrt 5 + \\sqrt 7 \\). Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":20307,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,258],"tags":[443,442],"series":[],"class_list":["post-12807","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-os-numeros-reais","tag-enquadramento","tag-valor-aproximado"],"views":1113,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/9V1Pag024-3_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/12807","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=12807"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/12807\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20307"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=12807"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=12807"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=12807"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=12807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}