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英語 高校生

下線部を訳す問題で、赤い所の単語が分かりませんでした。 そんな時は、どのように訳せば良いですか?

91 5 次の狩猟に関する英文を読み、以下の設問に答えよ。 (配点 60点) It's November, opening morning of deer hunting season in Wisconsin I'm in my treestand just inside the woods, /very close to open land which does not allow hunting. White-tailed deer live on the open land all year, and my treestand is just above a route they often use to escape when feeling threatened. and/A As they move I see, six white-tailed "does with a 10-point buck in the open land/ farther away, am curious:/What would they do if I shot into the ground? My gunshot echoes in the narrow valley making it difficult to pinpoint the source of the noise. After the sound settles, does burst through a gap in the woods and disappear into the bushes below my stand/ I hold my breath as the buck quarters toward me I feel lucky but also regretful in a clearing only 25 yards away. I take the shot. that my anticipated long day in the woods is over, with plenty of processing work (2) ahead. Admittedly, along with luck, my understanding of resident deer habits helped me punch my buck tag. A modern hunter with knowledge of whitetail behavior and sophisticated modern weaponry can successfully ambush deer. /That raises questions about human hunting capabilities. Do modern humans have the のうりょくこ capabilities physical and sensory of ancient hunters? Or have we lost those skills because of our reliance on technology? My short answer to both questions is yes. (3) Recent analyses from archaeological sites in Olduvai Gorge, in East Africa's 考古学 Great Rift Valley, established the capability of humans living nearly 2 million years ago to ambush "wildebeest-size prey using simple wooden spears at close range. I believe that humans today still possess the capabilities of the ancients. Those skills

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英語 高校生

答えをおしえていただきたいです。

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数学 高校生

125番の問題について、上が解答、下が問題なのですが、これは解答の通りに場合分けしないとだめですか? 単純に、t²+(b-1)t-b+1=0が一つ以上解を持てばいいとして、D=b²+2b-3≧0で、b≦-3,1≦bとなり、b>-1より、b≧1と求めるのはやはり不十分ですか?

直線 AT の傾きは 125 同様に、直線BT の傾きは ∠ATB が直角であるための条件は t2+ (b-1)t-6 +1 0 ...... (A) すなわち 求める条件は、 2次方程式 (A) が-1<< b の範囲に少なくとも1つ の実数解をもつことである。 f(t)=12+(b-1-6+1とし、 2次方程式 (A) の判別式をDとする D=(b-1)-4(1-b)=(6+3)(b-1) と また,y=f(t) のグラフの軸は、直線=1 2 (F(b) = 26³-26+1=2(6-1)² + ==> 0 1 20 777 40 xx=eの座標では 共通範囲を求めて 軸≦-1 かつf(-1)<0 軸の条件から であるから、求める条件は,次の [1] または [2] が成り立つことである。 [1] D≧0かつ1<軸くり DZ0 から 軸の条件から t²-1 +-(-1) t+b (+3)(6-1)≧0 1-b -1<L f(-1) < 0 から 共通範囲を求めて [1] または [2] から 1-b 2 ≤-1 ·=t-1 b3 3-26 < 0 b≥3 b≧1 (t-1)(t+b)=-1 <b ゆえに b-3, 1≦b よって 1/32<b<3 ゆえに 63 3 ゆえに 6/12/0 二次関数の場合分け 126 (1) OA'=OB'=AB2=2 から s2+f2=2 ①-② から 2s+2t-2=0 ③を①に代入して s2+(-s+1)=2 整理すると 2s2-2s-1=0 (s-1)+(t-1)=2 (2) よってt=-s+1 よって S= 1+√3 2 key] 垂直 傾きの積が1 あれば、その解の 長の座標では、 AT,BTが垂直になる Chi 4 TON 21 12 T 0 th O B 直角を作るグラフ 牛(大)のグラフ (LATBが直角である (3) 定義域-1<大く) (04) ここに解があれば 直角である Clear 125 曲線 y=x2 上に2点A(-1,1),B(b, b2)をとる。 ただし 6> -1 とする。 このとき、次の条件を満たするの範囲を求めよ。 条件: y=x2 上の点T(t, t2) (−1 <t<b) で, ∠ATBが直角になるもの が存在する。 [ 16 名古屋大〕 x

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