Sejam !$ A !$ e !$ B !$ matrizes quadradas de ordem ímpar. Suponha que !$ A !$ é simétrica e que !$ B !$ é antissimétrica. Considere as seguintes afirmações:

I. !$ (A + B)^2 = A^2 + 2AB + B^2. !$

II. !$ A !$ comuta com qualquer matriz simétrica.

III. !$ B !$ comuta com qualquer matriz antissimétrica.

IV. !$ \det (AB) = 0 !$

É(são) VERDADEIRA(S):

Pretende-se distribuir 48 balas em 4 tigelas designadas pelas letras !$ A, B, C \ e \ D. !$ De quantas maneiras pode-se fazer essa distribuição de forma que todas as tigelas contenham ao menos 3 balas e a tigela !$ B !$ contenha a mesma quantidade que a tigela !$ D !$.

Leia atentamente o trecho destacado de “Presença”: “Ele pousou a mala no chão e pediu um apartamento. Por quanto tempo? Não estava bem certo, talvez uns vinte dias. Ou mais. O porteiro examinou-o da cabeça aos pés. Forçou o sorriso paternal, disfarçando o espanto com uma cordialidade exagerada, Mas o jovem queria um apartamento? Ali, naquele hotel?!”

Assinale a alternativa correta relativamente ao grifo do pronome demonstrativo e o uso da pontuação.

Acerca de Memórias de um sargento de milícias , é correto afirmar que:

Considere um triângulo !$ ABC !$ tal que !$ m(\overline {AB}) = 14 !$, !$ cos (\widehat{BAC}) = { \large 3 \over 5} !$ e !$ cos (\widehat{ABC}) = { \large 5 \over 13} !$.Então, o raio da circunferência inscrita ao triângulo é igual a:

A questão refere-se ao texto destacado a seguir.

When my family first moved to North Carolina, we lived in a rented house three blocks from the school where I would begin the third grade. My mother made friends with one of the neighbors, but one seemed enough for her. Within a year we would move again and, as she explained, there wasn’t much point in getting too close to people we would have to say good-bye to. Our next house was less than a mile away, and the short journey would hardly merit tears or even goodbyes, for that matter. It was more of a “see you later” situation, but still I adopted my mother’s attitude, as it allowed me to pretend that not making friends was a conscious choice. I could if I wanted to. It just wasn’t the right time.

Back in New York State, we had lived in the country, with no sidewalks or streetlights; you could leave the house and still be alone. But here, when you looked out the window, you saw other houses, and people inside those houses. I hoped that in walking around after dark I might witness a murder, but for the most part our neighbors just sat in their living rooms, watching TV. The only place that seemed truly different was owned by a man named Mr. Tomkey, who did not believe in television […].

To say that you did not believe in television was different from saying that you did not care for it. Belief implied that television had a master plan and that you were against it. It also suggested that you thought too much. When my mother reported that Mr. Tomkey did not believe in television, my father said, “Well, good for him. I don't know that I believe in it, either”. “That's exactly how I feel,” my mother said, and then my parents watched the news, and whatever came on after the news.

SEDARIS, David. Dress Your Family in Corduroy and Denim. Recurso

eletrônico. Boston: Little, Brown and Company, 2004, p. 5.

O termo still, destacado no trecho do primeiro parágrafo, “It was more of a ‘see you later’ situation, but still I adopted my mother’s attitude […]”, transmite a ideia de:

Seja !$ S ⊂ \mathbb R\ !$ o conjunto solução da inequação !$ (x^2 + x + 1)^{2x^2- x-1} ≤ 1 !$. Podemos afirmar que:

No experimento de dupla fenda de Young, suponha que a separação entre as fendas seja de 16 !$ μm !$. Um feixe de luz de comprimento de onda 500 nm atinge as fendas e produz um padrão de interferência. Quantos máximos haverá na faixa angular dada por !$ - 30º ≤ θ ≤ 30º !$?

Seja !$ z ∈ \mathbb C !$. Se a representação dos números !$ 4,z + 2 !$ e !$ z^2 !$ no plano complexo são vértices de um triângulo equilátero, então o comprimento do seu lado é igual a:

Seja !$ A !$ uma matriz real quadrada de ordem 2 tal que

!$ A \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} 1 & x \\ y & 0 \end{pmatrix} !$ e !$ A \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix} = \begin{pmatrix} x & 3 \\ y + 1 & 1 \end{pmatrix}. !$

Então, o traço da matriz !$ A !$ é igual a: