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英語 高校生

空欄にはbが入るのですが、その理由を教えていただけませんか?

次の英文を読み, 後の問いに答えよ。 oh ni ai tuned" goizer o d Beauty is in the eye of the beholder. 1 This proverb was first recorded in the English language in its current form in the 19th century. However, (1). the concept of people viewing beauty differently from their own points of view has been around in most cultures of the world since ancient times. But what exactly is beauty, and is it really subjective? The definition in the Merriam-Webster dictionary is "the qualities in a person or a thing that give pleasure to the senses or the mind." This definition, however, does not mention whether there is a universal standard for beauty, or whether each individual person views beauty based on a totally different set of standards. Some of the arts seem to suggest the (2) if we consider the fact that everybody has their own favorite piece of music or painting that they consider to be beautiful. Nature, on the other hand, consistently comes up with scenes that are universally considered to be beautiful. There is little doubt that physical beauty, or beauty based on physical appearance of people, is personal. The ideal "beautiful woman" differs between cultures, and in many cases is based on fashion. Some cultures appreciate fatness, while others believe that body mutilation 2 represents beau example, body art in the form of piercings and tattoos is recognized as a sign of beauty in many countries of the world today, although there are also many people in these same countries who continue to ( 4 ) with this assessment. (3). For hana including Pythagoras believed that beauty was based on 1:1 11

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数学 高校生

⑵ですが、⑴でORが出たのでと思って写真にあるように解いてしまって答えが合わないのですが、自分がやったやり方だとダメなんですよね?

Check 例題 350 交点の位置ベクトル(1) △OAB において, 辺OA を 1:2に内分する点をP, 辺OB を 3:2に内 分する点をQ, AQ と BP の交点をRとする. 次の問いに答えよ. (1) OR を OA = d, OB = を使って表せ. (2) 線分 OR の延長と辺ABの交点をDとするとき, AD: DB を求め よ. 考え方 (1) R は AQ, BP 上の点より, AR: RQ=s: (1-s) BR: RP=t: (1-t) とおいて, OR を2通りで表す. à±0, 6±0, àxi zh, ma+nb=m'a+n'bm=m', を利用する. (2) 3点O, R, D が一直線上の点より, ODOR (kは実数) と表せることと,点Dは辺AB上の点 OCLAであることから, AD: DB=u: (1-u) とおいて, OD を2通りで表す. OR=(1-s)OA+sOQ 20 =(1-s)a+sb OR=(1-t)OB+tOP = (1-t)b+-ta m ①② より, A 3 (1-s)a+s6=ta +(1-t)b a = 0, 0, a と 較して, 1-s=1/31t, 2/23s=1-tより ₂T, OR=a+16 (1) AR: RQ=s: (1-s), BR: RP=t: (1-t) とお くと, m n=n' -²0) P 1-t. 0 R S= s=16, a=3p ①に代入して, OR=3(1-s)+ s 3 (別解) (①までは同じ)OP=pとおく.j=1234 P R S-R B -S t: D ここではBP 上の点より, 3(1-s)+1/23s=1,s= よって、①に代入して, OR = 1/23a+1/26 01A より 10 5 6 1-s BA A OR *** 1-t -U- -3187+AT P 0 は平行ではないから,係数を比がすべての敵を FLEGE R 1Q t D B 1-u (1-s)OA+SOQ s+(1-s) =(1-s) OA+soQ 0Q=OB=36 OP=OA=a B R は BP 上 [=06+APA 1 &G SAA&TA (S)

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