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英語 中学生

それぞれに埋まる単語教えて頂きたいです😭

◆次の英文を読んで,あとの問いに答えなさい。 <埼玉改〉 Ayako is a junior high school student and her homeroom teacher is Mr. Tanaka. He said to the students in class, "You will have work experience next month. So think about the future and the work experience you want. Then talk about these with your family. This is your homework. Next Monday you are going to talk about it with your classmates in class." That evening Ayako talked to her family about the homework. She said, "I have no idea about the future. What should I do?" Her mother said, "You don't need to worry about it now." Her brother said, "When I was a junior high school student, I was like you. But I was interested in environmental science when I was a third year student in high school. Now I'm studying it at college." Her father asked, "What are you interested in?" Ayako couldn't answer. He continued, "You should think about that first." Next Monday when class started, Mr. Tanaka said to the students, "Talk to the person next to you about the future and work experience. Then give some advice to each other." The person next to Ayako was Robert. He was from Australia. They enjoyed playing basketball at school and were good friends. "Well, Robert, do you have any ideas about the future and work experience?" Ayako asked. "Yes. I want to go to a hospital for work experience. When I was sick in the hospital, a nurse tried hard to help me in English. And she took good care of me. I was very glad. So I want to be a nurse in the future and help people," he said. "How about you, Ayako?" "I have no idea about the future," she said, and then talked to him about her father's advice. "Your father's advice sounds good," he said. "OK. I'll ask you some questions. What do you like 20 to do?" "I like to play basketball on the school's basketball team," she answered. "I like basketball too. Why are you interested in it?" he asked. "Because it's fun to play. And I'm the captain of my team, so I teach basketball to the younger members. I'm interested in that," she answered. "I think you are good at teaching basketball, and all the members like you," he said. "Oh, thank you," she said. "Well, Ayako, how about a school teacher as a future job?" he said. "I think you will take good care of students." 1 35 1. 25 "Thank you. I think it is an important job. OK. I'll think about it," she said. At the end of class 30 she was happy because she had an idea for her future job. She thought listening to her father's advice and talking with Robert were useful. The next week Ayako and Robert were talking again. He said, "Well, Ayako, did you think about work experience? Where do you want to go for it?" Ayako smiled and said, "I'm thinking about teaching at school as a future job, so I want to go to an elementary school for work experience. If I 35 work as a teacher, I need to learn a lot of things. I'm studying harder now."

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数学 高校生

222. なぜ解答3行目のような恒等式ができるのですか? また係数比較での計算の詳細は解答のように書くべきなんでしょうか??(大した計算ではないので「計算するとm=◯,n=◻︎とかでいいのかなと思いました。) 最後にu^2-2u-2=0、これはs,t以外の文字を用いて表... 続きを読む

340 0000 演習 例題2224次関数のグラフと2点で接する直線 [類 埼玉大 関数y=x(x-4) のグラフと異なる2点で接する直線の方程式を求めよ。 指針 次の①~③3 の考え方がある [ただしf(x)=x(x-4), s≠t]。 ③3 の考え方で解いてみよう ①点(t, f(t)) における接線が, y=f(x)のグラフと点 (s, f(s)) で接する。 (s, f(s)), (t, f(t)) におけるそれぞれの接線が一致する。 ③ y=f(x)のグラフと直線y=mx+nがx=s, x=tの点で接するとして、 f(x)=mx+n が 重解 s, tをもつ。 → f(x)-(mx+n)=(x-s)(x-t)^ 解答 y=x(x-4) のグラフと直線y=mx+nがx=s,x=t (st) の点で接するとすると、次のxの恒等式が成り立つ。 x³(x-4)-(mx+n)=(x-s)²(x-t)² (左辺)=x^-4x-mx-n (右辺)={(x-s)(x-t)}={x²-(s+t)x+st}2 =x^+(s+t)2x2+s2t2-2(s+t)x3-2(s+t)stx+2stx2 =x²-2(s+t)x+{(s+t)^+2st}x²-2(s+t)stx+s2t2 両辺の係数を比較して -4=-2(s+t) -m=-2(s+t)st ①から ③から ①, 0=(s+t)^2+2st ③, -n=s2t2 これ② から ④ から s+t=2 m=-8 s,tはμ²-2u-2=0の解で,これを解くと u=1±√3 よって, y=x(x-4)のグラフとx=1-√3,x=1+√3の点 で接する直線があり, その方程式は y=-8x-4 ... 2. 4 これを変形して よって, x2+2(t-2)x+3t2-8t=0 Aの判別式をDとすると st=-2 n=-4 4 x-4x²=(4t3-12t2)x-3t+8t tと異なる重解をもつことである。 (x-t)^{x2+2(1-2)x+3t2-8t}=0 別解y'=4x-12x2であるから,点(t, f(t-4)) における接線の方程式は y-t³(t-4)=(4t³-12t²)(x-t) Jħ5_y=(4t³-12t²)x-3t++8t³. この直線がx=s (s≠t) の点でy=x(x-4)のグラフと接するための条件は、 方程式 下の別解は、指針の①の え方によるものである。 YA <s≠t を確認する。 D=(t−2)²-1· (3t²-8t) = −2(t²—2t—2) これを解くと D=0 とすると t2-2t-2=0 このとき, Aの重解はs=-(t-2)=1+√3 (複号同順) t=1±√3は2-2t-2 = 0 を満たし -31¹+8t³ = -(t²-2t-2)(3t²-2t+2)-4--4 10 Aが,tと異なる重解 s をもてばよい。 t=1±√3 4t³-12t²=4(t²-2t-2)(t-1)-8=-8 ゆえに(*) から よって, s≠t である。 y=-&x-l E し 指 C y': おすこ こ f( f' 3 t

解決済み 回答数: 1