ALTERNATIVA 1<\/strong><\/p>\nComo \\(4,7 < x < 5,3\\)\u00a0e \\( – 2,1 < y < – 1,9\\), come\u00e7ando por utilizar a propriedade da p\u00e1g. 13 do manual vem:<\/p>\n
\\[\\begin{array}{*{20}{l}}{4,7 + \\left( { – 2,1} \\right) < x + y < 5,3 + \\left( { – 1,9} \\right)}& \\Leftrightarrow &{2,6 < x + y < 3,4}\\\\{}& \\Leftrightarrow &{\\left( {x + y} \\right) \\in \\left] {2,6;\\;3,4} \\right[}\\\\{}& \\Leftrightarrow &{\\left( {x + y} \\right) \\in \\left] {3 – 0,4;\\;3 + 0,4} \\right[}\\\\{}& \\Leftrightarrow &{\\begin{array}{*{20}{c}}{3 \\in \\left] {\\left( {x + y} \\right) – 0,4;\\;\\left( {x + y} \\right) + 0,4} \\right[}&{{\\rm{(Porqu\u00ea ?)}}}&{{\\rm{(Def}}{\\rm{. 2)}}}\\end{array}}\\end{array}\\]<\/p>\n
Deste modo, conclui-se que \\(3 = 5 + \\left( { – 2} \\right)\\)\u00a0\u00e9 uma aproxima\u00e7\u00e3o de \\(x + y\\)\u00a0com erro inferior a \\(0,4 = 0,3 + 0,1\\).<\/p>\n
ALTERNATIVA 2<\/strong><\/p>\nComo \\(5 – 0,3 < x < 5 + 0,3\\)\u00a0e \\( – 2 – 0,1 < y < – 2 + 0,1\\), come\u00e7ando por utilizar a propriedade da p\u00e1g. 13 do manual vem:<\/p>\n
\\[\\begin{array}{*{20}{l}}{5 – 0,3 + \\left( { – 2 – 0,1} \\right) < x + y < 5 + 0,3 + \\left( { – 2 + 0,1} \\right)}& \\Leftrightarrow &{3 – \\left( {0,3 + 0,1} \\right) < x + y < 3 + \\left( {0,3 + 0,1} \\right)}\\\\{}& \\Leftrightarrow &{\\left( {x + y} \\right) \\in \\left] {3 – 0,4;\\;3 + 0,4} \\right[}\\\\{}& \\Leftrightarrow &{\\begin{array}{*{20}{c}}{3 \\in \\left] {\\left( {x + y} \\right) – 0,4;\\;\\left( {x + y} \\right) + 0,4} \\right[}&{{\\rm{(Porqu\u00ea ?)}}}&{{\\rm{(Def}}{\\rm{. 2)}}}\\end{array}}\\end{array}\\]<\/p>\n
Deste modo, conclui-se que \\(3 = 5 + \\left( { – 2} \\right)\\)\u00a0\u00e9 uma aproxima\u00e7\u00e3o de \\(x + y\\)\u00a0com erro inferior a \\(0,4 = 0,3 + 0,1\\).<\/p>\n
ALTERNATIVA 3<\/strong><\/p>\nAnalogamente, \\(\\begin{array}{*{20}{l}}{5 – 0,3 + \\left( { – 2 – 0,1} \\right) < x + y < 5 + 0,3 + \\left( { – 2 + 0,1} \\right)}& \\Leftrightarrow &{3 – \\left( {0,3 + 0,1} \\right) < x + y < 3 + \\left( {0,3 + 0,1} \\right)}\\end{array}\\).<\/p>\n
Logo, \\(3 = 5 + \\left( { – 2} \\right)\\)\u00a0\u00e9 uma aproxima\u00e7\u00e3o de \\(x + y\\)\u00a0com erro inferior a \\(0,4 = 0,3 + 0,1\\). (Porqu\u00ea?) (Def. 2)
\n\u00ad<\/p>\n<\/li>\n
\nALTERNATIVA 1<\/strong><\/p>\nJ\u00e1 verificamos anteriormente que \\(4,7 < x < 5,3\\)\u00a0e \\( – 2,1 < y < – 1,9\\).<\/p>\n
Tendo em conta que \\(\\begin{array}{*{20}{c}}{ – 2,1 < y < – 1,9}& \\Leftrightarrow &{1,9 < – y < 2,1}\\end{array}\\)\u00a0e utilizando a propriedade da p\u00e1g. 13 do manual vem:
\n\\[\\begin{array}{*{20}{l}}{4,7 \\times 1,9 < – x \\times y < 5,3 \\times 2,1}& \\Leftrightarrow &{8,93 < – x \\times y < 11,13}\\\\{}& \\Leftrightarrow &{ – 11,13 < x \\times y < – 8,93}\\\\{}& \\Leftrightarrow &{\\left( {x \\times y} \\right) \\in \\left] { – 11,13;\\; – 8,93} \\right[}\\\\{}& \\Leftrightarrow &{\\left( {x \\times y} \\right) \\in \\left] { – 10 – 1,13;\\; – 10 + 1,07} \\right[}\\\\{}& \\Leftrightarrow &{\\begin{array}{*{20}{c}}{ – 10 \\in \\left] {\\left( {x \\times y} \\right) – 1,07;\\;\\left( {x \\times y} \\right) + 1,13} \\right[}&{{\\rm{(Nota**)}}}\\end{array}}\\end{array}\\]<\/p>\n
Deste modo, conclui-se que \\( – 10 = 5 \\times \\left( { – 2} \\right)\\)\u00a0\u00e9 uma aproxima\u00e7\u00e3o de \\(x \\times y\\)\u00a0com erro m\u00e1ximo inferior a \\(1,13\\).
\nC\u00e1lculo: \\(1,13 = 5 \\times 0,1 + 2 \\times 0,3 + 0,3 \\times 0,1\\). (Ver resolu\u00e7\u00e3o da Alternativa 2)<\/p>\n
ALTERNATIVA 2<\/strong><\/p>\nJ\u00e1 verificamos anteriormente que \\(5 – 0,3 < x < 5 + 0,3\\)\u00a0e \\( – 2, – 0,1 < y < – 2 + 0,1\\).<\/p>\n
Tendo em conta que \\(\\begin{array}{*{20}{c}}{ – 2 – 0,1 < y < – 2 + 0,1}& \\Leftrightarrow &{2 – 0,1 < – y < 2 + 0,1}\\end{array}\\)\u00a0e utilizando a propriedade da p\u00e1g. 13 do manual vem:<\/p>\n
\\[\\begin{array}{*{20}{l}}{\\left( {5 – 0,3} \\right) \\times \\left( {2 – 0,1} \\right) < – x \\times y < \\left( {5 + 0,3} \\right) \\times \\left( {2 + 0,1} \\right)}& \\Leftrightarrow &{10 – 5 \\times 0,1 – 2 \\times 0,3 + 0,3 \\times 0,1 < – x \\times y < 10 + 5 \\times 0,1 + 2 \\times 0,3 + 0,3 \\times 0,1}\\\\{}& \\Leftrightarrow &{ – 10 – 5 \\times 0,1 – 2 \\times 0,3 – 0,3 \\times 0,1 < x \\times y < – 10 + 5 \\times 0,1 + 2 \\times 0,3 – 0,3 \\times 0,1}\\\\{}& \\Leftrightarrow &{\\left( {x \\times y} \\right) \\in \\left] { – 10 – 5 \\times 0,1 – 2 \\times 0,3 – 0,3 \\times 0,1;\\; – 10 + 5 \\times 0,1 + 2 \\times 0,3 – 0,3 \\times 0,1} \\right[}\\\\{}& \\Leftrightarrow &{\\begin{array}{*{20}{c}}{ – 10 \\in \\left] {\\left( {x \\times y} \\right) – 5 \\times 0,1 – 2 \\times 0,3 + 0,3 \\times 0,1;\\;\\left( {x \\times y} \\right) + 5 \\times 0,1 + 2 \\times 0,3 + 0,3 \\times 0,1} \\right[}\\end{array}}\\\\{}& \\Leftrightarrow &{ – 10 \\in \\left] {\\left( {x \\times y} \\right) – 1,07;\\;\\left( {x \\times y} \\right) + 1,13} \\right[}\\end{array}\\]<\/p>\n
Deste modo, conclui-se que \\( – 10 = 5 \\times \\left( { – 2} \\right)\\)\u00a0\u00e9 uma aproxima\u00e7\u00e3o de \\(x \\times y\\)\u00a0com erro m\u00e1ximo inferior a \\(1,13\\).
\nC\u00e1lculo: \\(1,13 = 5 \\times 0,1 + 2 \\times 0,3 + 0,3 \\times 0,1\\).<\/p>\n
Nota**<\/strong>:
\n\\[\\begin{array}{*{20}{l}}{x’ \\in \\left] {x – {r_1},\\;x + {r_2}} \\right[}& \\Leftrightarrow &{x – {r_1} < x’ < x + {r_2}}\\\\{}& \\Leftrightarrow &{\\begin{array}{*{20}{c}}{x – {r_1} < x’}& \\wedge &{x’ < x + {r_2}}\\end{array}}\\\\{}& \\Leftrightarrow &{\\begin{array}{*{20}{c}}{x – {r_1} + {r_1} < x’ + {r_1}}& \\wedge &{x’ – {r_2} < x + {r_2} – {r_2}}&{{\\rm{(Porqu\u00ea ?)}}}\\end{array}}\\\\{}& \\Leftrightarrow &{\\begin{array}{*{20}{c}}{x < x’ + {r_1}}& \\wedge &{x’ – {r_2} < x}\\end{array}}\\\\{}& \\Leftrightarrow &{x’ – {r_2} < x < x’ + {r_1}}\\\\{}& \\Leftrightarrow &{x \\in \\left] {x’ – {r_2},\\;x’ + {r_1}} \\right[}\\end{array}\\]<\/p>\n<\/li>\n<\/ol>\n<< A defini\u00e7\u00e3o<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"Enunciado A defini\u00e7\u00e3o Enunciado Escreve uma aproxima\u00e7\u00e3o de\u00a0\\(\\pi \\) com erro inferior a\u00a0\\(0,01\\). Justifica a tua resposta. Considera que\u00a0\\(5\\) \u00e9 uma aproxima\u00e7\u00e3o de um n\u00famero real\u00a0\\(x\\) com erro inferior a\u00a0\\(0,3\\) e que\u00a0\\( – 2\\)...<\/p>\n","protected":false},"author":1,"featured_media":12736,"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-12722","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":3144,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Pi.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/12722","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=12722"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/12722\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/12736"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=12722"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=12722"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=12722"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=12722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2017\/10\/Pi.png\")
<\/a>Escreve uma aproxima\u00e7\u00e3o de\u00a0\\(\\pi \\) com erro inferior a\u00a0\\(0,01\\).