{"id":2759,"date":"2023-03-03T09:32:10","date_gmt":"2023-03-03T08:32:10","guid":{"rendered":"http:\/\/matma.com.pl\/?p=2759"},"modified":"2023-05-06T10:04:52","modified_gmt":"2023-05-06T08:04:52","slug":"zadanie-29-0-6-zbior-rozszerzenie","status":"publish","type":"post","link":"http:\/\/matma.com.pl\/?p=2759","title":{"rendered":"zadanie 29 (0-6) zbi\u00f3r rozszerzenie"},"content":{"rendered":"<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%;\">Korzystaj\u0105c z twierdzenia Pitagorasa mamy:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?5^2+d^2=13^2\" alt=\"5^2+d^2=13^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?25+d^2=169\" alt=\"25+d^2=169\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?d^2=144\" alt=\"d^2=144\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?d=12\" alt=\"d=12\" align=\"absmiddle\" \/><\/td>\n<td style=\"width: 50%;\"><img fetchpriority=\"high\" decoding=\"async\" class=\"wp-image-2763 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-01-104655-300x157.jpg\" alt=\"\" width=\"395\" height=\"207\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-01-104655-300x157.jpg 300w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-01-104655.jpg 587w\" sizes=\"(max-width: 395px) 100vw, 395px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Za\u0142\u00f3\u017cmy, \u017ce Janusz cze\u015b\u0107 drogi p\u00f3jdzie g\u0142\u00f3wn\u0105 drog\u0105 po czym skr\u0119ci i p\u00f3jdzie &#8222;na skr\u00f3ty&#8221; do domu. Zobrazujmy tak\u0105 ewentualno\u015b\u0107<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%;\"><img decoding=\"async\" class=\"wp-image-2762 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-01-105910-300x136.jpg\" alt=\"\" width=\"415\" height=\"188\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-01-105910-300x136.jpg 300w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-01-105910.jpg 677w\" sizes=\"(max-width: 415px) 100vw, 415px\" \/><\/td>\n<td style=\"width: 50%;\"><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?5^2+\\left&amp;space;(&amp;space;12-x&amp;space;\\right&amp;space;)^2=y^2\" alt=\"5^2+\\left ( 12-x \\right )^2=y^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?25+\\left&amp;space;(&amp;space;12-x&amp;space;\\right&amp;space;)^2=y^2\" alt=\"25+\\left ( 12-x \\right )^2=y^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?y=\\sqrt{25+\\left&amp;space;(&amp;space;12-x&amp;space;\\right&amp;space;)^2}\" alt=\"y=\\sqrt{25+\\left ( 12-x \\right )^2}\" align=\"absmiddle\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Obliczmy teraz czas jaki Janusz potrzebuje na pokonanie kolejnych odcinkach drogi:<\/p>\n<ul>\n<li>odcinek <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x\" alt=\"x\" align=\"absmiddle\" \/> pokona z pr\u0119dko\u015bci\u0105 <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?5\\;&amp;space;\\mathrm{km\/h}\" alt=\"5\\; \\mathrm{km\/h}\" align=\"absmiddle\" \/>. Zatem potrzebuje czasu: <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t_{1}=\\frac{x}{5}\" alt=\"t_{1}=\\frac{x}{5}\" align=\"absmiddle\" \/><\/li>\n<li>odcinek <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\left&amp;space;(12-x&amp;space;\\right&amp;space;)\" alt=\"\\left (12-x \\right )\" align=\"absmiddle\" \/> pokona z pr\u0119dko\u015bci\u0105 <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?3\\;&amp;space;\\mathrm{km\/h}\" alt=\"3\\; \\mathrm{km\/h}\" align=\"absmiddle\" \/>. Zatem potrzebuje czasu: <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t_{2}=\\frac{\\sqrt{25+\\left&amp;space;(&amp;space;12-x&amp;space;\\right&amp;space;)^2}}{3}\" alt=\"t_{2}=\\frac{\\sqrt{25+\\left ( 12-x \\right )^2}}{3}\" align=\"absmiddle\" \/><\/li>\n<\/ul>\n<p>Opiszmy funkcj\u0105 czas potrzebny Januszowi na dotarcie do domu<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t(x)=\\frac{x}{5}+\\frac{\\sqrt{25+\\left&amp;space;(&amp;space;12-x&amp;space;\\right&amp;space;)^2}}{3}\" alt=\"t(x)=\\frac{x}{5}+\\frac{\\sqrt{25+\\left ( 12-x \\right )^2}}{3}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t(x)=\\frac{x}{5}+\\frac{\\sqrt{25+144-24x+x^2}}{3}\" alt=\"t(x)=\\frac{x}{5}+\\frac{\\sqrt{25+144-24x+x^2}}{3}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t(x)=\\frac{x}{5}+\\frac{\\sqrt{169-24x+x^2}}{3}\" alt=\"t(x)=\\frac{x}{5}+\\frac{\\sqrt{169-24x+x^2}}{3}\" align=\"absmiddle\" \/><\/p>\n<p>Okre\u015blmy dziedzin\u0119 naszej funkcji<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?12-x\\geq&amp;space;0\\;&amp;space;\\;&amp;space;\\wedge&amp;space;\\;&amp;space;\\;&amp;space;x\\geq&amp;space;0\\;&amp;space;\\;&amp;space;\\wedge&amp;space;\\;&amp;space;\\;&amp;space;25+\\left&amp;space;(&amp;space;12-x&amp;space;\\right&amp;space;)^2\\geq&amp;space;0\" alt=\"12-x\\geq 0\\; \\; \\wedge \\; \\; x\\geq 0\\; \\; \\wedge \\; \\; 25+\\left ( 12-x \\right )^2\\geq 0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x\\leq&amp;space;12\\;&amp;space;\\;&amp;space;\\wedge&amp;space;\\;&amp;space;\\;&amp;space;x\\geq&amp;space;0\\;&amp;space;\\;&amp;space;\\wedge&amp;space;\\;&amp;space;\\;&amp;space;x\\in&amp;space;\\mathbb{R}\" alt=\"x\\leq 12\\; \\; \\wedge \\; \\; x\\geq 0\\; \\; \\wedge \\; \\; x\\in \\mathbb{R}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?D_{t}:x\\in&amp;space;\\left&amp;space;[&amp;space;0,12&amp;space;\\right&amp;space;]\" alt=\"D_{t}:x\\in \\left [ 0,12 \\right ]\" align=\"absmiddle\" \/><\/p>\n<p>Obliczymy kolejno pochodna naszej funkcji <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t'(x)\" alt=\"t'(x)\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t'(x)=\\frac{1}{5}+\\frac{1}{3}\\frac{1}{2\\sqrt{169-24x+x^2}}\\cdot&amp;space;\\left&amp;space;(&amp;space;2x-24&amp;space;\\right&amp;space;)\" alt=\"t'(x)=\\frac{1}{5}+\\frac{1}{3}\\frac{1}{2\\sqrt{169-24x+x^2}}\\cdot \\left ( 2x-24 \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t'(x)=\\frac{1}{5}+\\frac{2\\left&amp;space;(&amp;space;x-12&amp;space;\\right&amp;space;)}{6\\sqrt{169-24x+x^2}}\" alt=\"t'(x)=\\frac{1}{5}+\\frac{2\\left ( x-12 \\right )}{6\\sqrt{169-24x+x^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t'(x)=\\frac{1}{5}+\\frac{\\left&amp;space;(&amp;space;x-12&amp;space;\\right&amp;space;)}{3\\sqrt{169-24x+x^2}}\" alt=\"t'(x)=\\frac{1}{5}+\\frac{\\left ( x-12 \\right )}{3\\sqrt{169-24x+x^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t'(x)=\\frac{3\\sqrt{169-24x+x^2}+5\\left&amp;space;(&amp;space;x-12&amp;space;\\right&amp;space;)}{3\\sqrt{169-24x+x^2}}\" alt=\"t'(x)=\\frac{3\\sqrt{169-24x+x^2}+5\\left ( x-12 \\right )}{3\\sqrt{169-24x+x^2}}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?D_{t'}:x\\in&amp;space;\\left&amp;space;[&amp;space;0,12&amp;space;\\right&amp;space;]\" alt=\"D_{t'}:x\\in \\left [ 0,12 \\right ]\" align=\"absmiddle\" \/><\/p>\n<p>Obliczymy miejsca zerowej pochodnej:<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{3\\sqrt{169-24x+x^2}+5\\left&amp;space;(&amp;space;x-12&amp;space;\\right&amp;space;)}{3\\sqrt{169-24x+x^2}}&amp;space;=0\" alt=\"\\frac{3\\sqrt{169-24x+x^2}+5\\left ( x-12 \\right )}{3\\sqrt{169-24x+x^2}} =0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?3\\sqrt{169-24x+x^2}+5x-60=0\" alt=\"3\\sqrt{169-24x+x^2}+5x-60=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?3\\sqrt{169-24x+x^2}=60-5x\" alt=\"3\\sqrt{169-24x+x^2}=60-5x\" align=\"absmiddle\" \/><\/p>\n<p>Podnie\u015bmy obustronnie r\u00f3wnanie do kwadratu<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?9\\left&amp;space;(&amp;space;169-24x+x^2&amp;space;\\right&amp;space;)=\\left&amp;space;(60-5x&amp;space;\\right&amp;space;)^2\" alt=\"9\\left ( 169-24x+x^2 \\right )=\\left (60-5x \\right )^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?1521-216x+9x^2=3600-600x+25x^2\" alt=\"1521-216x+9x^2=3600-600x+25x^2\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?16x^2-384x+2079=0\" alt=\"16x^2-384x+2079=0\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\Delta&amp;space;=\\left&amp;space;(&amp;space;-384&amp;space;\\right&amp;space;)^2-4\\cdot&amp;space;16\\cdot&amp;space;2079=147456-133056=14400\" alt=\"\\Delta =\\left ( -384 \\right )^2-4\\cdot 16\\cdot 2079=147456-133056=14400\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\sqrt{\\Delta&amp;space;}=120\" alt=\"\\sqrt{\\Delta }=120\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x_{1}=\\frac{384-120}{2\\cdot&amp;space;16}=\\frac{264}{32}=8\\frac{1}{4}\" alt=\"x_{1}=\\frac{384-120}{2\\cdot 16}=\\frac{264}{32}=8\\frac{1}{4}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x_{1}=\\frac{384+120}{2\\cdot&amp;space;16}=\\frac{504}{32}=15\\frac{3}{4}\" alt=\"x_{1}=\\frac{384+120}{2\\cdot 16}=\\frac{504}{32}=15\\frac{3}{4}\" align=\"absmiddle\" \/><\/p>\n<table style=\"border-collapse: collapse; width: 100%; height: 216px;\">\n<tbody>\n<tr style=\"height: 216px;\">\n<td style=\"width: 50%; height: 216px;\">Uwzgl\u0119dniaj\u0105c dziedzin\u0119 pochodnej funkcji <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?D_{t'}:x\\in&amp;space;\\left&amp;space;[&amp;space;0,12&amp;space;\\right&amp;space;]\" alt=\"D_{t'}:x\\in \\left [ 0,12 \\right ]\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t'(x)&gt;0\\;&amp;space;\\;&amp;space;\\Leftrightarrow&amp;space;\\;&amp;space;\\;&amp;space;x\\in&amp;space;\\left&amp;space;[&amp;space;0,8\\frac{1}{4}&amp;space;\\right&amp;space;)\" alt=\"t'(x)&gt;0\\; \\; \\Leftrightarrow \\; \\; x\\in \\left [ 0,8\\frac{1}{4} \\right )\" align=\"absmiddle\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t'(x)&lt;0\\;&amp;space;\\;&amp;space;x\\in&amp;space;\\left&amp;space;(8\\frac{1}{4},12\\right&amp;space;]\" alt=\"t'(x)&lt;0\\; \\; x\\in \\left (8\\frac{1}{4},12\\right ]\" align=\"absmiddle\" \/><\/td>\n<td style=\"width: 50%; height: 216px;\"><img decoding=\"async\" class=\"wp-image-2767 aligncenter\" src=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-02-151453-300x127.jpg\" alt=\"\" width=\"394\" height=\"167\" srcset=\"http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-02-151453-300x127.jpg 300w, http:\/\/matma.com.pl\/wp-content\/uploads\/2023\/03\/Zrzut-ekranu-2023-03-02-151453.jpg 617w\" sizes=\"(max-width: 394px) 100vw, 394px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Funkcja <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t(x)\" alt=\"t(x)\" align=\"absmiddle\" \/>\u00a0jest malej\u0105ca w przedziale <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x\\in&amp;space;\\left&amp;space;[&amp;space;0,8\\frac{1}{4}&amp;space;\\right&amp;space;]\" alt=\"x\\in \\left [ 0,8\\frac{1}{4} \\right ]\" align=\"absmiddle\" \/> oraz rosn\u0105ca w przedziale <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x\\in&amp;space;\\left&amp;space;[&amp;space;8\\frac{1}{4},12&amp;space;\\right&amp;space;]\" alt=\"x\\in \\left [ 8\\frac{1}{4},12 \\right ]\" align=\"absmiddle\" \/>. Zatem najmniejsz\u0105 warto\u015b\u0107 funkcja osi\u0105gnie dla argumentu <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?x=8\\frac{1}{4}\" alt=\"x=8\\frac{1}{4}\" align=\"absmiddle\" \/> (maksimum)<\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t\\left&amp;space;(8\\frac{1}{4}&amp;space;\\right&amp;space;)&amp;space;=t\\left&amp;space;(\\frac{33}{4}&amp;space;\\right&amp;space;)\" alt=\"t\\left (8\\frac{1}{4} \\right ) =t\\left (\\frac{33}{4} \\right )\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t\\left&amp;space;(&amp;space;\\frac{33}{4}&amp;space;\\right&amp;space;)=\\frac{\\frac{33}{4}}{5}+\\frac{\\sqrt{25+\\left&amp;space;(&amp;space;12-\\frac{33}{4}&amp;space;\\right&amp;space;)^2}}{3}=\\frac{33}{20}+\\frac{\\sqrt{25+\\left&amp;space;(&amp;space;\\frac{15}{4}&amp;space;\\right&amp;space;)^2}}{3}\" alt=\"t\\left ( \\frac{33}{4} \\right )=\\frac{\\frac{33}{4}}{5}+\\frac{\\sqrt{25+\\left ( 12-\\frac{33}{4} \\right )^2}}{3}=\\frac{33}{20}+\\frac{\\sqrt{25+\\left ( \\frac{15}{4} \\right )^2}}{3}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?t\\left&amp;space;(&amp;space;\\frac{33}{4}&amp;space;\\right&amp;space;)=\\frac{33}{20}+\\frac{\\sqrt{25+\\frac{225}{16}}}{3}=\\frac{33}{20}+\\frac{\\sqrt\\frac{{625}}{16}}{3}=\\frac{33}{20}+\\frac{25}{12}=\\frac{99+125}{60}=\\frac{224}{60}\" alt=\"t\\left ( \\frac{33}{4} \\right )=\\frac{33}{20}+\\frac{\\sqrt{25+\\frac{225}{16}}}{3}=\\frac{33}{20}+\\frac{\\sqrt\\frac{{625}}{16}}{3}=\\frac{33}{20}+\\frac{25}{12}=\\frac{99+125}{60}=\\frac{224}{60}\" align=\"absmiddle\" \/><\/p>\n<p><img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?\\frac{224}{60}=3\\:&amp;space;h\\:&amp;space;44\\:&amp;space;min\" alt=\"\\frac{224}{60}=3\\: h\\: 44\\: min\" align=\"absmiddle\" \/><\/p>\n<p>Najkr\u00f3tszy czas potrzebny Januszowi na dotarcie do domu to <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?3\" alt=\"3\" align=\"absmiddle\" \/> godziny i <img decoding=\"async\" src=\"http:\/\/latex.codecogs.com\/gif.latex?44\" alt=\"44\" align=\"absmiddle\" \/> minuty.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Korzystaj\u0105c z twierdzenia Pitagorasa mamy: Za\u0142\u00f3\u017cmy, \u017ce Janusz cze\u015b\u0107 drogi p\u00f3jdzie g\u0142\u00f3wn\u0105 drog\u0105 po czym skr\u0119ci i p\u00f3jdzie &#8222;na skr\u00f3ty&#8221; do domu. Zobrazujmy tak\u0105 ewentualno\u015b\u0107 Obliczmy teraz czas jaki Janusz potrzebuje na pokonanie kolejnych odcinkach drogi: odcinek pokona z pr\u0119dko\u015bci\u0105 . Zatem potrzebuje czasu: odcinek pokona z pr\u0119dko\u015bci\u0105 . Zatem potrzebuje czasu: Opiszmy funkcj\u0105 czas [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-2759","post","type-post","status-publish","format-standard","hentry","category-najnowsze-pliki"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.1 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>zadanie 29 (0-6) zbi\u00f3r rozszerzenie - Matma.com.pl<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/matma.com.pl\/?p=2759\" \/>\n<meta property=\"og:locale\" content=\"pl_PL\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"zadanie 29 (0-6) zbi\u00f3r rozszerzenie - Matma.com.pl\" \/>\n<meta property=\"og:description\" content=\"Korzystaj\u0105c z twierdzenia Pitagorasa mamy: Za\u0142\u00f3\u017cmy, \u017ce Janusz cze\u015b\u0107 drogi p\u00f3jdzie g\u0142\u00f3wn\u0105 drog\u0105 po czym skr\u0119ci i p\u00f3jdzie &#8222;na skr\u00f3ty&#8221; do domu. Zobrazujmy tak\u0105 ewentualno\u015b\u0107 Obliczmy teraz czas jaki Janusz potrzebuje na pokonanie kolejnych odcinkach drogi: odcinek pokona z pr\u0119dko\u015bci\u0105 . Zatem potrzebuje czasu: odcinek pokona z pr\u0119dko\u015bci\u0105 . 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