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Transcrição

",{"_22":66,"_61":67,"_63":68},"be4289eb-71fd-46e1-85fb-9599700ce441","B","

Replicação

",{"_22":19,"_61":70,"_63":71},"C","

Crossing-over

",{"_22":73,"_61":74,"_63":75},"fe4fdd1e-aadf-4350-9de6-fb21345454c5","D","

Mutação

","solution_explanation","statement","

Qual dos seguintes processos é responsável pela troca de segmentos de DNA entre cromossomos homólogos durante a meiose, contribuindo para a variabilidade genética?

","status","pending_classification","hasRelatedVideo","relatedVideos",[84],{"_85":86,"_104":105,"_30":120,"_139":140,"_158":121},"content",{"_22":87,"_36":88,"_89":-5,"_90":91,"_92":93,"_94":-5,"_24":95,"_98":99,"_100":99,"_101":-5,"_102":103},"162087be-594f-460e-bbdd-a339ac2f10ad","Calor Latente - Parte 1","description","content_type_id","1051eee7-c62e-4b2c-9900-77a29fcec080","resource_id","f2b62fc2-7079-40e7-a5b5-c4a9769b205d","user_id",{"_96":97},"external_id","3659","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","calor-latente-parte-1","prometheusData",{"_106":107,"_113":114,"_115":116,"_117":118,"_6":119},"provider",{"_22":108,"_109":110,"_111":112},"06f120be-2194-4548-8336-7ba44ffbae14","key","grupo_q_proteus","kind","proteus","reference","bf8aacba-3de9-4b10-961c-47ada80dfe8c","state","completed","thumbnail_url","https://proteus.grupoq.io/videos/bf8aacba-3de9-4b10-961c-47ada80dfe8c/thumbnailsthumb.0000000.jpg","https://proteus.grupoq.io/videos/bf8aacba-3de9-4b10-961c-47ada80dfe8c/levels.m3u8",[121],{"_22":122,"_36":123,"_38":124,"_125":126,"_127":128,"_98":129,"_100":129,"_24":130,"_102":124,"_137":138},"96193fa0-9495-45b2-81cc-b547604354af","Física Térmica - Termologia","fisica-termica-termologia","tag_group_id","c1ce2cf2-b217-4cc7-8182-81c56bd3522d","parent_id","c8ab2a95-fd84-4d6f-ade0-835d34ba00f9","2025-04-29T03:44:27.560682+00:00",{"_131":132,"_133":134,"_135":136},"childrenCount",10,"content_count",0,"question_count",8575,"children",[],"video",{"_22":93,"_141":88,"_89":-5,"_142":143,"_144":145,"_6":-5,"_146":-5,"_106":145,"_147":148,"_117":-5,"_149":-5,"_150":-5,"_151":-5,"_152":-5,"_24":153,"_154":134,"_155":-5,"_79":156,"_98":157,"_100":157},"title","duration",17,"source_type","prometheus","iframe_url","provider_video_id","748a4adb-ad8f-4c5f-88a9-bc522d3b4246","format","quality","file_path","file_size",{"_96":97},"views_count","last_played_at","draft","2025-08-19T11:56:22.11908+00:00","tag","similarQuestions",[161,192,210],{"_18":162,"_20":21,"_22":163,"_57":164,"_76":180,"_77":181,"_30":182},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[165,168,171,174,177],{"_22":166,"_61":62,"_63":167},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_22":169,"_61":67,"_63":170},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_22":172,"_61":70,"_63":173},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_22":175,"_61":74,"_63":176},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_22":162,"_61":178,"_63":179},"E","y'' - 2y' + 6y = 0","Somente em y''−2y'+6y=0 o polinômio característico r²−2r+6=0 tem Δ<0, gerando raízes complexas conjugadas (1±i√5) e, portanto, solução em termos de exponencial modulando seno e cosseno. Nos demais itens (a–d) o discriminante é ≥0, produzindo raízes reais (distintas ou repetidas).","

Resolva as seguintes EDOs determinando a solução geral e, quando houver condições iniciais do PVI, a solução particular.

(a) y'' + 2y' - 3y = 0

(b) 6y'' - y' - y = 0

(d) 9y'' + 6y' + y = 0

(e) y'' - 2y' + 6y = 0

(f) 4y'' + 9y = 0

(g) 6y'' - 5y' + y = 0; \\, y(0) = 4; \\, y'(0) = 0

(h) y'' + 4y' + 4y = 0; \\, y(-1) = 2; \\, y'(-1) = 1

(i) y'' + y = 0; \\, y\bigg(\\frac{\\pi}{3}\\bigg) = 2; \\, y'\\bigg(\\frac{\\pi}{3}\\bigg) = -4

(j) y'' - 2y' - 3y = 3e^{2t}

(k) y'' + 9y = t^2 e^{3t} + 6

(l) 2y'' + 3y' + y = t^2 + 3 \\sin(t)

(m) y'' + y' - 2y = 2t; \\, y(0) = 0; \\, y'(0) = 1

(n) y'' - 2y' + y = te^t + 4; \\, y(0) = 1; \\, y'(0) = 1

(o) y'' + 4y = 3 \\sin(2t); \\, y(0) = 2; \\, y'(0) = -1

",[183],{"_22":184,"_36":185,"_38":186,"_125":126,"_127":187,"_98":188,"_100":188,"_24":189,"_102":186},"6516bbbd-f2a3-4710-8382-702240b17836","Mecânica na Engenharia de Petróleo","mecanica-na-engenharia-de-petroleo","b787e438-0ab3-454c-ab43-5d3d14c2720d","2025-04-29T03:44:35.328997+00:00",{"_131":190,"_133":134,"_135":191},4,12302,{"_18":193,"_20":21,"_22":194,"_57":195,"_76":207,"_77":208,"_30":209},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[196,199,202,204],{"_22":197,"_61":62,"_63":198},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_22":200,"_61":67,"_63":201},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_22":193,"_61":70,"_63":203},"

Verdadeiro, falso, verdadeiro

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Falso, falso, falso

","85,5 % decomposto → 14,5 % restante.\nN/N0 = 0,145 ⇒ t = ln(1/0,145)/(ln2/5730) ≈ 1,6×10⁴ anos.\nItens prováveis: (1) resta 14,5 % do C-14 (V); (2) idade ≈ 1,2×10⁴ anos ou 3 meias-vidas (F); (3) t½ do C-14 = 5730 anos (V). Sequência: V-F-V, correspondendo à alternativa c.","

Qual é a idade da madeira na qual 85,5\\% do C-14 se decompos?

",[183],{"_18":211,"_20":21,"_22":212,"_57":213,"_76":228,"_77":229,"_30":230},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[214,217,220,222,225],{"_22":215,"_61":62,"_63":216},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_22":218,"_61":67,"_63":219},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_22":211,"_61":70,"_63":221},"Indeterminado",{"_22":223,"_61":74,"_63":224},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_22":226,"_61":178,"_63":227},"2e06d948-d37e-4350-afc7-ced04941bcaf","Nenhum","Os polinômios e números fornecidos estão corrompidos (símbolos “????”), impossibilitando aplicar Ruffini e concluir se são raízes. Sem dados completos, o problema é indeterminado.","Emplea la regla de Ruffini para dictaminar si los siguientes números son o no raíces de los polinomios citados:
a) ????3 de ????4????5
b) ????2 de ????2????????2
c) ????1 de 2????????1
d) ????1 de 2????2????",[183],[183,232,239,249,256,263,264,272,278,283,289,296],{"_22":233,"_36":37,"_38":39,"_125":126,"_127":234,"_98":129,"_100":129,"_24":235,"_102":238},"6b6faf95-0e3a-4f58-abc8-e1444c48041d","92339f4f-ed20-4f4c-ad85-0c3de7e04ae0",{"_131":236,"_133":134,"_135":237},2,11846,"termodinamica-3",{"_22":240,"_36":241,"_38":242,"_125":126,"_127":243,"_98":244,"_100":244,"_24":245,"_102":248},"861ad829-750d-4a44-9569-2f18b430b4df","Engenharia Mecatrônica","engenharia-mecatronica","eb9315c5-bd04-4116-aa46-02c4fde320c0","2025-04-29T02:19:12.828432+00:00",{"_131":246,"_133":134,"_135":247},5,30614,"engenharia-mecatronica-1",{"_22":250,"_36":251,"_38":252,"_125":126,"_127":240,"_98":188,"_100":188,"_24":253,"_102":255},"88fe96b3-2a17-4198-90d5-ddda3d6f5d45","Mecânica","mecanica",{"_131":246,"_133":134,"_135":254},25741,"mecanica-3",{"_22":234,"_36":257,"_38":258,"_125":126,"_127":243,"_98":244,"_100":244,"_24":259,"_102":262},"Engenharia Mecânica","engenharia-mecanica",{"_131":260,"_133":134,"_135":261},11,36318,"engenharia-mecanica-2",{"_22":122,"_36":123,"_38":124,"_125":126,"_127":128,"_98":129,"_100":129,"_24":130,"_102":124},{"_22":265,"_36":266,"_38":267,"_125":126,"_127":-5,"_98":268,"_100":268,"_24":269,"_102":267},"9772e958-72c6-4499-ad47-877cf94d670d","Exatas","exatas","2025-04-29T01:56:45.284172+00:00",{"_131":270,"_133":134,"_135":271},18,135489,{"_22":273,"_36":37,"_38":39,"_125":126,"_127":184,"_98":274,"_100":274,"_24":275,"_102":277},"b21374de-aae5-4a4a-9bf3-0420e731797e","2025-04-29T04:00:14.680867+00:00",{"_131":134,"_133":134,"_135":276},8597,"termodinamica-5",{"_22":187,"_36":279,"_38":280,"_125":126,"_127":243,"_98":244,"_100":244,"_24":281,"_102":282},"Engenharia de Petróleo","engenharia-de-petroleo",{"_131":246,"_133":134,"_135":191},"engenharia-de-petroleo-1",{"_22":128,"_36":284,"_38":285,"_125":126,"_127":265,"_98":244,"_100":244,"_24":286,"_102":288},"Física","fisica",{"_131":260,"_133":134,"_135":287},13777,"fisica-4",{"_22":243,"_36":290,"_38":291,"_125":126,"_127":-5,"_98":292,"_100":292,"_24":293,"_102":291},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_131":294,"_133":134,"_135":295},29,351586,{"_22":297,"_36":37,"_38":39,"_125":126,"_127":250,"_98":298,"_100":298,"_24":299,"_102":300},"f5967cbe-d751-4b40-81f2-04eb5dead2a9","2025-04-29T04:00:13.215003+00:00",{"_131":134,"_133":134,"_135":276},"termodinamica-4","tagsTree",[303,307],{"_22":265,"_36":266,"_38":267,"_125":126,"_127":-5,"_98":268,"_100":268,"_24":269,"_102":267,"_137":304},[305],{"_22":128,"_36":284,"_38":285,"_125":126,"_127":265,"_98":244,"_100":244,"_24":286,"_102":288,"_137":306},[121],{"_22":243,"_36":290,"_38":291,"_125":126,"_127":-5,"_98":292,"_100":292,"_24":293,"_102":291,"_137":308},[309,315,319],{"_22":240,"_36":241,"_38":242,"_125":126,"_127":243,"_98":244,"_100":244,"_24":245,"_102":248,"_137":310},[311],{"_22":250,"_36":251,"_38":252,"_125":126,"_127":240,"_98":188,"_100":188,"_24":253,"_102":255,"_137":312},[313],{"_22":297,"_36":37,"_38":39,"_125":126,"_127":250,"_98":298,"_100":298,"_24":299,"_102":300,"_137":314},[],{"_22":234,"_36":257,"_38":258,"_125":126,"_127":243,"_98":244,"_100":244,"_24":259,"_102":262,"_137":316},[317],{"_22":233,"_36":37,"_38":39,"_125":126,"_127":234,"_98":129,"_100":129,"_24":235,"_102":238,"_137":318},[],{"_22":187,"_36":279,"_38":280,"_125":126,"_127":243,"_98":244,"_100":244,"_24":281,"_102":282,"_137":320},[321],{"_22":184,"_36":185,"_38":186,"_125":126,"_127":187,"_98":188,"_100":188,"_24":189,"_102":186,"_137":322},[323],{"_22":273,"_36":37,"_38":39,"_125":126,"_127":184,"_98":274,"_100":274,"_24":275,"_102":277,"_137":324},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

Comentários