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English Senior High

英検一級のライティングの添削をお願いしたいです。 Should the Japanese government encourage more immigration? というトピックで書きました。よろしくお願いします。

The Japanese government should encourage more immigration Problems such as lack of diversity, Japanese companies retreating from the world declining birth rate 39 there is a because market, and aging population with a can be solved by increasing immigration. Japan, although it is one of the advanced countries, lack of diversity. Except of big cities like Tokyo, there are only few foreigners. I think this is a set back to Japan while other countries are stepping foward by changing its own system, Japan is refusing to make a change to its old style, and I believe this is because Japan doesn't have abundant. perspectives. foreign countries to. By encouraging more immigration, they can provide many different views and stimulate Jahanese people. as of what does other countries. a image Japanese companies are retreating from the world market. Many Companies in Japan are producing overeng ineered expensive products that only targets Japanese customers and it is losing its power in other countries : If Japan accepts more immigrants Japanese companies. will be able to get detailed regire to their product and stay in the world market. Aging population and a declining birth rate is a huge problem in Japan right now. Because of it, companies are losing workers and young people will have to saffer with a lot of taxes in the future. Encouraging immigration can increase population and solve. these problems more By encouraging more immigrations, many problems that Japan is facing can be solved.

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Mathematics Senior High

紫で囲んだところのように因数分解するのはどのようにしているんですか?

DATE fied from flask) fond F HAR 200 接線に垂直な直線 (法線) 点Pでない方を点Qとする、ただし、a≠0 とする。 曲線 y=x 上の点P(a, α²) における法線と、この曲線の交点のうち, (1) 法線の方程式を求めよ. Focus *[+2² halos $195. 接点で接線と垂直に交わる直線を法線と呼ぶ. (詳しくは数学Ⅲで学習) 点P(a, f(a)) における法線の傾きをmとすると, 接線の傾きが f'(a) のとき、 m.f'(a)=-1 つまり、m=f'(a) 1 frase (2-0)² + $99 ← fram thar (A) (-x)(o=o)G (1) f(x)=x2 とおくと,f'(x)=2x TEL より, 点Pにおける接線の傾きは, f'(a)=2a したがって, 点Pにおける法線の傾きをとすると 1 m・2a=-1より, m = __ (a+0) したがって, (2) 点Qの座標を求めよ. 1 微分係数と導関数 Px-a- CHERE (2) 曲線 y=x2 と直線y=- 2つの曲線① 2式からyを消去して、x=-x+α'+- BROOTRAN (x-2)(x+a+ 2a となる. 1 2a 接線の傾き f'(a)(0) ini よって, 点Pにおける法線の方程式は, y-a²=-2 / (x=a) £ y₁=y=-2/x+ a² + ²/²/2 2a x+a+1/12 の交点は連立方程式を解いて 交点のx座標を求め り、 る。 左辺に移項して因数 分解 点Pも交点の1つで 2a>=あるから,x=αる第6章 解になっている. 点Qのx座標は =0 (D)(8-DS) 1_22_1 "2a' --- a²+- *** V 4a² 1-2のとき、y=(-a-2 2a ·+1 することから よって、点Qの座標は, (-a- 4a² 法線の傾き [接線] まず, 接線の傾きを 考える. ( 接線の傾き) (法線の傾き) =-1 361 ジュー 2a 6)- 02 1 1030 f']]

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