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数学 高校生

三角比の二次関数 sinθ180°=0なのに、変域で0≦t≦1 と、1になる理由がわからないです。教えてくれると助かります🙇

①との共通範囲は 1 2 ゆえに, √2 <sin0< を解いて 2 30°<0<45°, 135°<0<150° 2 <t<√2 2 ④ 150 (1) 0°≧0≦180°のとき (20°<8<90° のとき (1) cos20=1-sin' 0 であるから 練習 次の関数の最大値 最小値, およびそのときの0の値を求めよ。 y=4cos20+4sin0+5 y=2 tan²0-4 tan 0+3 (1-Vale &V)( =-4sin²0+4sin0+9 sin0=tとおくと, 0°≧0≦180°のとき yをtの式で表すと y=4cos20+4sin0+5=4(1-sin²0) +4sin0+5 ①の範囲において,yは t=1/23 で最大値 10, t=0, 1で最小値 9 をとる。 0°≦0≦180°であるから y=−4ť²+4t+9=−4(t²− t) + 9 = − 4( t - 12 - ) ² - 1203 +10 t=1/12 となるのは, sin0- 0= 1/1/2 から t=0 となるのは, sin0 = 0 から t=1 となるのは, sin0=1から よって ...... 2 3 [8] [9] y=2t2-4t+3=2(t2-2t)+3 0≤t≤1 0=30° 150°のとき最大値10 6=0°90° 180° のとき最小値 9 (2) tan0=t とおくと, 0°<0<90°のとき t>0 ① yをtの式で表すと 0° 0 <90° であるから t=1 となるのは, tan0=1から0=45° よって 881>> 0=30° 150° 0=0°, 180° 0=90° =2(t-1)'+1 ① の範囲において,yはt=1で最小値1を とり, 最大値はない。 2 1 最小 0 0=45°のとき最小値1, 最大値はない 135° 150° -1 √2 10. 1 2 I ←COS を消去して、 sin 0 だけの式で表す。 ←tの変域に注意。 y 最小 ユ (1) 類 自治医大] 30° -1 1x 45° E |最大 9 1 0 11 [32 YA 150° 1 最小 0 130° ←tの変域に注意。 y↑ 0 Caro 2 732 v31x 4章 練習 45° [図形と計量] 1x

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英語 高校生

間違ってるとこあったら教えてください

英語 7 次の英文を読み、1から4の ちから一つずつ選びなさい。 解答番号は 内に入れるのに最も適当なものを,それぞれ①~④のう 27 O others. 24 Nagisa was a nurse who was working in Zimbabwe, a country in Africa. One day, she got an email from her old high school homeroom teacher, Mr. Tamai. He wanted to ask was hesitant at first because she always had a fear of public speaking, she felt this would be a Nagisa to give his students a talk about what she was doing in Zimbabwe. Although Nagisa good chance to tell students about the joy of working abroad and helping people in need. The next time Nagisa went back to Japan, she visited Mr. Tamai's high school to speak with his students. She was very nervous, but to her relief, the students seemed to be very interested in her story. She talked about her job, her reasons for working in Zimbabwe, and both some good and bad things about working there. She shared her passion for helping After the talk, one of the students came to talk to Nagisa. He said, "I would like to work abroad and help people in the future like you, but I don't know what kind of job I would be able to do. Do you have any advice for me?" Nagisa said, "I think, doing something you like is the key. Keep doing it, and doors will open for you." (Ten years later) One sunny day, a group of Japanese farmers visited the village where Nagisa was living. They came to teach local people how to grow plants and vegetables. People in the village were eager to learn from them. Then, the youngest member of the farmers' group came to talk to Nagisa and said, "Hi, do you remember me? You gave a talk at my school ten years. ago. At that time, I liked growing plants and vegetables, but I didn't know how to use that to help others. You told me to keep doing what I liked and that has really opened doors for me to do what I'm doing now. Thank you." Hearing his words, Nagisa recognized who the young man was. She was surprised and pleased that her talk from ten years before was able to make a difference in this young man's life. 1 Nagisa was 24 a high school teacher. 2 afraid of public speaking. 3 scared of living abroad. 4 a doctor in Zimbabwe. 4 2 One thing Nagisa told Mr. Tamai's students was why she chose to work in Zimbabwe. how she learned a new language. 3 when she went to a high school in Africa. 4 what she did to impress local people. 3 One of the students said he wanted G (2) (3 to be a kind nurse like Nagisa. to teach Japanese culture in Africa. to open doors for other people. to help people overseas. 26 3 25 4 Ten years after her talk, Nagisa 27 made an appointment to meet one of her old friends in Africa. 2 became a farmer and taught local people how to grow vegetables. met one of Mr. Tamai's students again. 4 4 gave a small talk in her high school again.

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数学 高校生

マーカーを引いた部分の図の意味が分かりません💦 教えてください🙏

X コ 5 確率と漸化式 (1) 日本 例題 37 00000 される回数が奇数である確率pn をnの式で表せ。 1,2,3,4,5,6,7,8の数字が書かれた8枚のカードの中から1枚取り出し てもとに戻すことをn回行う。 このn回の試行で、数字8のカードが取り出 [産業医大 ] 基本30 CHART & SOLUTION 確率と漸化式LUTIONE 回目と(n+1) 回目に着目 確率が であるから, 偶数である確率は 1-pn 回の試行で, 数字 8 のカードが取り出される回数が奇数である (n+1)回の試行でpn+1 を求めるには, 次の2つの場合を考える。 7回の試行で奇数回で,(n+1)回目に8以外のカードを取り出す n回の試行で偶数回で,(n+1)回目に8のカードを取り出す 変形すると また (n+1)回の試行で8のカードが奇数回取り出されるのは, [1] n回の試行で8のカードが奇数回取り出され, (n+1) 回目に8のカードが取り出されない [2] n回の試行で8のカードが偶数回取り出され, (n+1) 回目に 8 のカードが取り出される のいずれかであり,[1], [2] は互いに排反であるから Pn+1=pn/1+(1-pn)・・ = 7 3 8 4 Pnt Pn+17 したがって 3 -12--³-(pm-12) pn Pi 11/27 - 12/17 - 31/12/1 8 Pn 3 n-1 3/3 84 n 1 1/3 p=²2 - 1 (3³) - (¹-(²) pn 24 S 8² よって、数列{ba-1/2 は初項 - 123 公比 1/23の等比数列で あるから -4-4-4/124 MOITUIG 8 回目 Pn 1-pn × 7 (n+1)回目 8 P+1 x. 8 inf. ① 確率の加法定理 事象A, Bが互いに排反 (A∩B=Ø) のとき P(AUB)=P(A)+P(B) ② 独立な試行 STで, Sでは事象A, T では 事象Bが起こる事象をC とすると P(C)=P(A)P(B) 3 a=a+₁ を解くと a=²1/22 は, 1枚目のカード が8の確率であるから p=1/ 405 1章 化式

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