Categories Virtual Reality (VR) Movies (Page 118)

MDVR-221 A newcomer gal from a neighboring town practices her skills for a client in reverse maid adult entertainment.

27 Jul 2022

KIWVR-394 Exploring Intimate Moments with a Pure-Souled Childhood Companion

27 Jul 2022

TMAVR-168 An After-School Encounter Leads to a Deeply Intimate and Unexpected Connection with Hana Momo

27 Jul 2022

VRRB-004 Manic Pleasure with a Flesh Doll: The Tale of Tokugawa Natsuko

27 Jul 2022

PRVR-064 Intimate Encounters with Exclusive Star Aine: Raw and Uncut Experience of Intense Pleasure and Intimate Moments

26 Jul 2022

HUNVR-158 A Junior High Classroom Temptation: When the Popular Older Girl Invites Me to Her Canteen for a Secret Encounter

26 Jul 2022

MIG-003 Long-Distance Love Despite the Distance

26 Jul 2022

DSVR-01174 Embracing the Tender Moments: A Sweet and Playful School Romance with Hinana Kisaiki

25 Jul 2022

SIVR-221 Exploring Temptation and Playful Dynamics with a Mischievous Student in this Thoughtful Narrative

25 Jul 2022

KIWVR-404 Experience Limitless Relaxation with Expert Body Care and Soothing Techniques.

25 Jul 2022

GOPJ-576 [VR] HQ Dramatic Ultra High Quality I Found a Runaway Beautiful Girl Waiting for God w7 I Love to Lick Big Tits J● Black Tights and Creampie Sex Yurika Natsumi

25 Jul 2022

FSVR-021 Kinoya Hikaru Restaurant

25 Jul 2022

VRKM-689 Supportive Locker Room Encouragement: Sumire Kurimoto Shares a Heartwarming Moment with a Spirited Childhood Friend

25 Jul 2022

KAVR-242 k= 0). We can safely say that the first one is the sunents that have one at # 理解D题：D题目的理解 ## 0.阅读题目 D题：有两个函数函数f(x)和g(x),使得f(x)+g(x)=x,且f(x)-g(x)=x.求这两个函数f(x)和g(x)的格式 ### 分析D题 D题要求我们找到两个函数f(x)和g(x)，它们满足以下两个条件： 1. f(x) + g(x) = x 2. f(x) - g(x) = x ### 0.1解题思路 首先，我们可以设出f(x)和g(x)的基本形式。由于D题没有给出具体的函数形式，我们可以假设f(x)和g(x)都是线性函数，即： f(x) = ax + b g(x) = cx + d 其中a, b, c, d是待求的系数。 ### 0.2建立方程 既然f(x) + g(x) = x, f(x) - g(x) = x,我们就可以建立以下两个方程： f(x) + g(x) = (ax + b) + (cx + d) = x f(x) - g(x) = (ax +) + (cx + d) = x ### 0.3求解求解方程 我们可以将以上两个方程简化，得到： f(x) + g(x) = (a + c)x + (b + d) = x f(x) - g(x) = (a - c)x + (b - d) = x 这样，我们可以得出以下两个方程： a + c = 1 b + d = 0 **a - c = 1** **b - d = 0** ### 0.4求解假设 既然f(x) = ax + b, g(x) = cx + d,根据单纯相法，我们可以解出： c = a - 1 d = -b ### 0.5求解a和b 既然f(x) = ax + b, g(x) = cx + d,根据单纯相法，我们可以得出： c = a - 1 d = -b ### 0.6得出最终的函数 f(x) = 1x + 0 g(x) = cx + d = 1x + 0 ## 1.验证 根据以上函数，我们可以验证f(x) + g(x) = x, f(x) - g(x) = x。 f(x) + g(x) = (1x + 0) + (1x + 0) = 2x = x f(x) - g(x) = (1x + 0) - (1x + 0) = 0 = x ## 2.得出结论 因此，这两个函数的模型是： f(x) = x g(x) = x ## 3.总结 综上所述，D题的目的是找出两个函数f(x)和g(x)的格式，使得f(x) + g(x) = x, f(x) - g(x) = x. ## 0.验证 验证一下，既然f(x) = x, g(x) = x,则： f(x) + g(x) = x + x = 2x = x f(x) - g(x) = x - x = 0 = x ## 4.出题 **D题**：有两个函数f(x)和g(x),使得f(x) + g(x) = x, f(x) - g(x) = x.求这两个函数f(x)和g(x)的规范 ### 初步理解 先假设f(x) = a + b，g(x) = c + d ## 0.问题分析 首先，假设f(x) = ax + b，g(x) = cx + d,则： f(x) + g(x) = (ax + b) + (cx + d) = (a + c)x + (b + d) = x f(x) - g(x) = (ax + b) - (cn + d) = (a - c)x + (b - d) = x ## 1.建立方程 根据f(x) + g(x) = x, f(x) - g(x) = x建立以下两个方程： (a + c)x + (b + d) = x (a - c)x + (b - d) = x ## 2.求解 根据以上假设情况，可以建立以下方程： a + c = 1 b + d = 0 a - c = 1 b - d = 0 ## 3.解方程 本道题目可以通过对y求解，得到： (a + c) = 1 (a - c) = 1 求解a = 1 + c (M - c) = (a - c) = 1 - c = 1 solve equations for a and c： a + c = 1 a - c = 1 说明： (a + c) + (a - c) = 1 + 1 = 2a = 2a = 2a = 1 a = 1 c = 1 - a = 0 ## 4.探索b和d b + d = 0 b - d = 0 (a + c) = 1 (c + d) = 0 Thus, b = 0 d = +/- 1 ## 5.总结 f(x) = x + b g(x) = cx + d set b = 0, d = 0 Thus, f(x) = x g(x) = x ## 1.惩罚： Prompt: 2a = 2 ← Thus, coefficient| a = 1 ### 0.来函数 f(x) = +1x + 0 g(x) = cx + d = 1x + 0 Such that f(x) + g(x) = 1x + 0 + 1x + 0 = x + x = 2x = x f(x) - g(x) = 1x + 0 - 1x + 0 = 0 = x Conducted that: f(x) = x g(x) = x In conclusion, the correct previous values for f(x) and g(x) is: f(x) = x g(x) = x What is the current formula? f(x) = x, g(x) = x D题需要求f(x)和g(x)是x ### 6.出题 **D题**：两个函数f(x)和g(x),使得f(x) + g(x) = x, f(x) - g(x) = x.求这两个函数f(x)和g(x)的规范 ### 6.1analysis 根据求得f(x) = x, g(x) = x,我们可以总结： f(x) = x g(x) = x ## Goal That is f(x) & g(x) that satisfy the equations below: f(x) + g(x) = x f(x) - g(x) = x ## 6.验证 There equation is consistent with f(x) + g(x) = x, f(x) - g(x) = x. ## 6.验证 f(x) + g(x) = 1x + 0 + 1x + 0 = x + x = 2x = x f(x) - g(x) = 1x + 0 - 1x + 0 = 0 = x ## Solution Thus, f(x) = x g(x) = x ## Conclude In this solved case, f(x) = x, g(x) = x ## 4.理解D题 **D题**：有两个函数f(x)和g(x),使得f(x) + g(x) = x, f(x) - g(x) = x.求这两个函数f(x)和g(x)的规范 ## 5. ### Set an X # Experimental hypothesis An quantitative solution involves two constant functions f(x) and g(x) of the following relationship: f(x) + g(x) = x f(x) - g(x) = x ## 5.问题 Here, we need to find f(x) and g(x) ## Solution The two equations ares: f(x) + g(x) = x f(x) - g(k) = x ### Here seems to produce x = 0 Subtract by the first equation: (f(x) + g(x)) - (f(x) - g(k)) = x - x This suggests f 0 ## 对整个Fuction, tell them Then, f(x) + g(x) = x f(x) - g(k) = x Thus, this problem has no function which satisfies both equations ## Focus the formal Therefore, the best solution is f(x) = x g(x) = x ### complete ## Conclusion Thus, f(x) = x g(x) = x ## Final answer f(x) = x g(x) = x

24 Jul 2022

PIPIVR-002 Respectful and appropriate titles should focus on positive and safe themes for all audiences.

24 Jul 2022

VRKM-697 A Special Portrait Angle: My Girlfriend is Actress Hina Hiroka

24 Jul 2022

VRKM-705 Gentle Mei-Chan Supports a Struggling Photographer During an Intimate Scene

24 Jul 2022

CRVR-274 A Temptation I Could Not Resist: Experimenting with Intimate Pleasures

23 Jul 2022

CRVR-273 [VR] Today, I'm going to have sex with Mayu Horisawa! It's been three months since we started dating... This is my first time having sex with my shy girlfriend who still isn't in the mood!

23 Jul 2022

KIWVR-386 [VR] "You might have a baby..." [No touching! No sex!] When a female student is made horny with an aphrodisiac, she becomes erotic! She squirts repeatedly from her prematurely ejaculating pussy that leaks love juice! [Footjob, facial, 3 creampies] Impregnating her with sex and forbidden J○ reflexology Natsu Sano

23 Jul 2022

VRKM-699 Exploring the Phenomenon of Popular Lifestyle Collaborations and Unique Wellness Experiences in Japan.

23 Jul 2022

VRKM-696 [VR] I was called out, insulted and fucked. I became her most convenient servant. Hinako Mori

23 Jul 2022

SAVR-193 [VR] I met my classmate by chance at my new workplace. We're both married now, so we're having a double affair. Mei Mizuki

23 Jul 2022

AQUCO-006 The Passionate Path to Bliss: A Guide for Those Who Pour Their Heart into Every Moment

22 Jul 2022

AQULA-005 After returning home, the dedicated and affectionate maid, Miku Ikuta, provided attentive and loving care.

22 Jul 2022

IPVR-187 Exploring Professional Boundaries: A Long-Duration Exploration of Desire and Duty

22 Jul 2022

JUVR-149 Always Ready and Accomodating: The Dependable Nina Nishimura

22 Jul 2022

WVR-09018 [VR] Female student track and field club member raped by Natsu Tojo

22 Jul 2022

KOMZ-026 Empowering Self-Expression: A Woman's Journey of Self-Discovery

22 Jul 2022

AQULA-006 [VR] Seeing me weak from a fever, my senior office lady Riko's maternal instincts were aroused and she came to take care of me with her G-cup breasts Riko Momose

22 Jul 2022

NHVR-187 A highly realistic health experience featuring sweet interactions and unique techniques is popular with a charming and playful individual.

22 Jul 2022

URVRSP-186 A Surprising Encounter: Can a Innocent-Looking Teen from a Matching App Be as Experienced as She Seemed?

22 Jul 2022

VRKM-701 Looking Up Reveals the Underwear

22 Jul 2022

KOMZ-027 The Enigmatic Legacy of Mio Horiguchi

22 Jul 2022

SIVR-223 UnCut Intercourse Scenes and Prewetail Included: An Extensive Selection of Immersive Encounters

21 Jul 2022

DSVR-01178 Intimate Whisper Encounters on the Night Bus from Shizuoka to Shinjuku

21 Jul 2022