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items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"数Ⅱ　図形と方程式　線分の内分点の軌跡"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"　実数qを用いて点Qの座標を表すと、点Q2が直線y=x+2上にあることから、Q(q, q+2)となります。これと点A(1, 6)を結ぶ直線を2:1に内分する点がPなので、内分点の公式により、その座標はP((2q+1)/3, (2q+10)/3)となります。このとき、点Pのx座標とy座標の関係を式で表すと、x=(2q+1)/3、y=(2q+10)/3=(2q+1)/3+9/3であることから、y=x+3となります。よって、求める点Pの軌跡は直線であり、その方程式はy=x+3であると求められます。記述の際は、このあとに「逆に、点Pが直線y=x+3にある時　とき…」といった感じで、求めた答えがちゃんと必要十分となるように逆からの検証を補充する必要があった気がしますが、そういった細かい記述の要素については流石に覚えていないので、ご自身で教科書等を参照してください。数式が見づらかったら申し訳ありません。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 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items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"微分の応用"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"X(t)に関して\n速度dx/dt=vとする。…①\nすると、加速度d^2x/dt^2=d/dt•(dx/dt)=dv/dt …②\nとなる。\n次にt(x)に関して\ndt/dx=1/(dx/dt)=(①を用いて)=1/v…③であり、\nd^2t/dx^2=d/dx•(dt/dx)=(③を用いて)=d/dx•(1/v)\n(これは合成関数の微分に相当するので)\n=-1/v^2•dv/dx=(vの変数としてのxはかなり扱いづらいので、tに変数変換して)=-1/v^2•dv/dt•dt/dx\nとなる。②、③を用いて変形すると、\nd^2x/dt^2=-v^3•d^2t/dx^2\nとなる。あとは①を代入して、答えは\n{}=-(dx/dt)^3となります。\n\nあってるかな、、？なんにせよこうゆうのにチャレンジしてみる姿勢は素晴らしいと思います。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary mr-1","children":["$undefined",[["$","path","0",{"fill":"none","d":"M0 0h24v24H0V0z","children":[]}],["$","path","1",{"d":"M12 6c1.1 0 2 .9 2 2s-.9 2-2 2-2-.9-2-2 .9-2 2-2m0 10c2.7 0 5.8 1.29 6 2H6c.23-.72 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items-center py-4","children":[["$","div",null,{"className":"flex-1 mr-3","children":[["$","div",null,{"className":"mb-1","children":"この数学の問題を教えて下さい🙇"}],["$","div",null,{"className":"text-xs text-caption line-clamp-2 mb-1","children":"自然数を8で割った余りは0〜7になるのは理解できると思います。\nそこで、nを自然数とすると、\n8で割った余りが\n0→8n\n1→8n 1\n2→8n 2\n3→8n 3\n4→8n 4\n5→8n 5\n6→8n 6\n7→8n 7\nとすることですべての自然数を表すことができます。問題で聞いているのは平方数ということなので、それぞれを2乗すると、\n\n0→64n^2=8×8n^2\n1→64n^2 16n 1=8(8n^2 2n) 1\n2→64n^2 32n 4=8(8n^2 4n) 4\n3→64n^2 48n 9=8(8n^2 6n 1) 1\n4→64n^2 64n 16=8(8n^2 8n 2)\n5→64n^2 80n 25=8(8n^2 10n 3) 1\n6→64n^2 96n 36=8(8n^2 12n 4) 4\n7→64n^2 112n 49=8(8n^2 14n 6) 1\n\nとなります。\nすべて(8n ○)^2という式になる以上、n^2とnの係数は8の倍数になるので、自然数部分である余りの2乗部分を8で割った時の余りが平方数の余りになります。\n\n長くなってすみません。わからなかったらまた質問してください。"}],["$","div",null,{"className":"flex mb-1","children":[["$","svg",null,{"stroke":"currentColor","fill":"currentColor","strokeWidth":"0","viewBox":"0 0 24 24","className":"text-subPrimary 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