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WF-148 日本語 DVD ジャケット 120 分

WF-148 鈴房ありささんの愛情あふれる物語

2006年7月1日120 分


動画発売日

2006年7月1日

収録時間

120 分長い

動画ランキング

242763 / 527405

他の動画 ID

3wf148, WF148, WF 148

女優の数

1人

女優体型

平均身長, 曲線美, セクシー

無修正

無し

動画言語

日本語

字幕

サブリップ (SRT ファイル)

著作権 ©

DMM

女優: 鈴房ありさ

出身地: 香川県
趣味: プリクラ、剣道
誕生日: 1987年01月07日
星座: 山羊座
胸のサイズ: F
身長: 165 cm
スリーサイズ: 90-58-85
血液型: A

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料金

ストリーミング (HD/4k) ¥300

標準 (480p) ¥980

字幕 (キャプション)

英語字幕

中国語字幕

日本語字幕

フランス語字幕

よくある質問

「WF-148」というコードは何を意味していますか?

日本のAV動画には、製作された各動画を表す「AVコード」と呼ばれる識別番号があります。

この場合、「WF」は製作者のビデオシリーズ(カテゴリー)を指し、「148」はエピソード番号を指します。

このAV動画の無修正バージョンはありますか?

残念ながら、現時点では WF-148 AV動画の無修正版は存在しません。

実際に、桃太郎映像が製作し販売するすべての動画は、規制されています。

この動画のフルバージョンをダウンロードできる場所はどこですか?

公式販売者のウェブサイト(DMM)から WF-148 の完全版動画を購入し、即座にダウンロードするには、このページの上部にある「ダウンロード」ボタンをクリックしてください。

公式ウェブサイトでこの動画を購入するための2つの価格オプションがあります。第1は、1つのビデオ購入(解像度に応じて)で、支払いを行った後、完全な動画をダウンロードまたはストリーミングできます。第2は、固定月額料金のメンバーシップで、購読後、無制限のビデオをダウンロードできます。

この動画の無料サンプルをダウンロードしたいです。可能ですか?

残念ながら、WF-148の無料サンプルをダウンロードすることはできません

ただし、ページのトップにスクロールして「再生」ボタンをクリックすることで無料サンプルを視聴できます。

WF-148の日本語字幕をどこでダウンロードできますか?

WF-148の日本語字幕をダウンロードするには、上の「字幕」セクションのトップにスクロールして、「日本語字幕」の横にある「注文」をクリックしてください。

WF-148 に似た動画

AYFL-001 〞〞〞〞〞〞〞�submb』 --- ### 问题描述 我们被要求解一个关于矩阵 ( A ) 的问题,其中 ( A ) 是一个 ( 4 imes 4 ) 的对称矩阵,定义如下: [ A = egin{pmatrix} 4 & -1 & -1 & -1 \ -1 & 4 & -1 & -1 \ -1 & -1 & 4 & -1 \ -1 & -1 & -1 & square end{pmatrix} ] 我们的任务是找到矩阵中未知的项“(square)”,使得该矩阵的行列式为零。 ### 思考过程 #### 1. 理解问题 首先,我需要理解题目要求。题目给出一个对称矩阵 ( A ),但它有一个未知项“(square)”。我们的任务是找出这个未知项,使得 ( A ) 的行列式等于零。换句话说,我需要找到“(square)”使得 ( A ) 是一个不可逆矩阵。 #### 2. 求解行列式 要找到“(square)”,我需要求出 ( A ) 的行列式并将其设为零,然后求解“(square)”的值。由于 ( A ) 是一个 ( 4 imes 4 ) 矩阵,计算行列式会比较复杂。我需要想办法找到简化求解的方法。 #### 3. 寻找简化方法 观察矩阵 ( A ),我发现每一行和每一列都包含三个 (-1) 和一个 ( 4 )。这意味着矩阵 ( A ) 是高度对称的。这种对称性可能有助于简化行列式的计算。此外,矩阵 ( A ) 是一个非常好的例子,可以通过计算特征值来求解行列式。 #### 4. 计算特征值 为了计算行列式,我决定计算矩阵 ( A ) 的特征值。由于 ( A ) 是一个对称矩阵,它的特征值都是实数。行列值是矩阵所有特征值的乘积,因此如果行列式为零,那就是一个特征值为零。 #### 5. 设定特征值 令 ( lambda ) 是一个特征值,然后计算 ( A - lambda I ) 的行列式为零。这将给我一个关于 ( lambda ) 的方程。由于行列式为零,意味着至少有一个特征值是零。因此,我需要找到一个特征值等于零的情况。 #### 6. 求解方程 通过求解特征值,我可以求解 ( A ) 的特征值。然后将它们相乘,得到一个关于 (square) 的方程。由于行列式为零,这等于所有特征值中至少一个为零。因此,我需要找到 (square) 使得至少一个特征值等于零。 #### 7. 最终求解 通过求解这个方程,我可以找到 (square) 的值。这个过程将涉及到矩阵计算和求解方程,可能需要一些代数技巧。 ### 详细计算过程 #### 1. 设定特征值 令 ( lambda ) 是一个特征值,然后计算 ( A - lambda I ) 的行列式: [ det(A - lambda I) = 0 ] #### 2. 计算矩阵 [ A - lambda I = egin{pmatrix} 4 - lambda & -1 & -1 & -1 \ -1 & 4 - lambda & -1 & -1 \ -1 & -1 & 4 - lambda & -1 \ -1 & -1 & -1 & square - lambda end{pmatrix} ] #### 3. 计算行列式 由于每一行和每一列都是对称的,我可以使用行列式公式来计算行列式: [ det(A - lambda I) = (4 - lambda)(4 - lambda)(4 - lambda)(square - lambda) - sum_{i=1}^4 (P_i) ] 其中 ( P_i ) 是修改后的行列式的产物。由于计算很复杂,我决定使用特征值方法。 #### 4. 计算特征值 #### 5. 设定特征值 令 ( lambda ) 是一个特征值,然后计算 ( A - lambda I ) 的行列式: [ det(A - lambda I) = 0 ] #### 6. 计算矩阵 [ A - lambda I = egin{pmatrix} 4 - lambda & -1 & -1 & -1 \ -1 & 4 - lambda & -1 & -1 \ -1 & -1 & 4 - lambda & -1 \ -1 & -1 & -1 & square - lambda end{pmatrix} ] #### 7. 计算行列式 由于每一行和每一列都包含三个 (-1) 和一个 ( 4 - lambda ),我可以使用行列式公式来计算行列式: [ det(A - lambda I) = (4 - lambda)^3(square - lambda) - sum_{i=1}^4 (P_i) ] #### 8. 计算特征值 通过观察,我注意到 ( lambda = 0 ) 是一个特征值,因此行列式表达式必须包含 ( lambda )。假设 ( lambda = 0 ),则: [ (4 - 0)^3(square - 0) - sum_{i=1}^4 (P_i) = 0 ] [ 256square - sum_{i=1}^4 (P_i) = 0 ] #### 9. 求解 (square) 通过行列式求解,我得出 (square = 1), Therefore, the final answer is: --- ### Final Answer $oxed{1}$

2006年10月2日

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