各位大神 求解 拜託🙏🆘

CA 梁柱 一截面積是原來的兩倍 22. Two girders are made of the same material. Girder A is twice as long as girder B and Bo DE C B³ has a cross-sectional area that is twice as great. The ratio of the mass density of girder A to the mass density of girder B is: 質量密度 A) 4 B) 2 C) 1 D) 1/2 E) 1/4 29. During a short interval of time the speed v in m/s of an automobile is given by v=at² + bt³, where the time t is in seconds. The units of a and b are respectively: A) m-s²; m-s4 s³/m; s4/m dv=at+bt² B) C) m/s²; m/s³ D) m/s³; m/s4 E) m/s4; m/s5 34. The position y of a particle moving along the y axis depends on the time t according to the equation B) C) D) E) 質固定,体2倍 #%# 40. The coordinate of an object is given as a function of time by x = 7t-31², where x is in meters and t is in seconds. Its average velocity over the interval from t = 0 to t = 2 s is: A) 5 m/s X₂=0, X₁₂=14-12=2 B) -5 m/s 11 m/s C) D) -11 m/s (2-0)/2=1 E) -14.5 m/s 41. The coordinate of a particle in meters is given by x(t) = 16t-3.01³, where the time t is in seconds. The particle is momentarily at rest at t = A) 0.75 s A y = at -bt². The dimensions of the quantities a and b are respectively: A) L²/T, L³/T2 B) L/T², L²/T C) L/T, L/T² D) L³/T, T²/L E) none of these 1.3 s 5.3 s 7.3 s 9.3 s V(t) = 16-9+² =- [9t²-16) =-(3t+4) 13t-4) =+= 0.75 45. The velocity of an object is given as a function of time by v = 4t-312, where v is in m/s and t is in seconds. Its average velocity over the interval from t=0 to t=2 s: A) is 0 B) is -2 m/s C) is 2 m/s D) is -4 m/s E) cannot be calculated unless the initial position is given 49. A particle moves along the x axis according to the equation x = 61² where x is in meters and t is in seconds. Therefore: E B) C) D) E) A) the acceleration of the particle is 6 m/s² t cannot be negative the particle follows a parabolic path each second the velocity of the particle changes by 9.8 m/s none of the above

請問17 然後 小數的1、2補數跟9、10補數如何計算呢thx!

TUTTI (e) 11000011 1.15 Find the 9's and the 10's complement of the following decimal (a) 65,234,035 (c) 87,000,367 1.16 (a) Find the 8's complement of (2360). (b) Convert (2360), to binary. (b) 56,783,223 (d) 99,999,000 (c) Find the 2's complement of the result in (b). Minib mol (d) Convert the answer in (c) to octal and compare with the answer in (a). 1.17 Perform subtraction on the given unsigned numbers using the 10's complement of the subtrahend. Where the result should be negative, find its 10's complement and affix a minus sign. Verify your answers. (a) 7,523-4.567 (c) 224-712 raison (b) 230 - 1,204 (d) 2,390 - 945

想請問這題要怎麼解🙏

8. a. The graph in the accompanying figure is made of line segments joined end to end. At which points of the interval [-4, 6] is f' not defined? Give reasons for your answer. b. Graph the derivative of f. The graph should show a step function. (-4,0) (0, 2) 1 (1, -2) y = f(x) (4,-2) (6. 2) 6 [§3.2-31]

第二題的更正分錄為什麼是這樣 為什麼會有保留盈餘

【回家練習】台中公司採用定期盤存制,近三年來淨利如下:X7 年$400,000, X8 年$320,000,X9 年$300,000,淨利逐年下降,經查發現以下資料 (1) X7 年期末存貨包括了承銷品$36,000扣 (2) X7 年 12 月 31 日收到一批進貨商品$50,000,故包含於期末盤點中,由於發 入帳 票於X8年1月3日收到,故於1/3 記錄進貨 x7年扣 (3) X8 年 12 月31日客戶訂購商品一批$48,000,當天即開立發票,並認列銷貨 收入,但因商品在X9年1月2日運出,因此 X8 年 12 月31日盤點存貨時, FESORIU 將此商品列入期末存貨之中 扣 (4) X8 年 12 月 31 日起運點銷貨$32,000,成本$24,000,因商品於當日已運出,加 故未包含在期末存貨之中,而於X9 年1月3日開立發票時記錄銷貨收入。 (5) X9 年1 月5日收到一批商品$40,000,經查該商品係 X8 年 12 月 30 日向供 應商購買,約定起運點交貨,供應商於 X8 年 12 月30日將商品交付運送, 故台中公司已於當日記錄進貨,但因商品未到達故未包括在期末存貨之中。 試作: 1. 計算正確淨利 2. X9 年結帳前發現上述錯誤應有之更正分錄 解: 原損益表淨利 調整: (1) (2) (3) (4) (5) 正確淨利 2. 更正分錄 銷貨成本 銷貨收入 保留盈餘 X7 年 $400,000 400 24 481000 存 481000 (36,000) (50,000) 凹 $314,000 X8 年 $320,000 26 $430,000 40,000 现收 321000 36,000 50,000 (48,000) 32,000 40,000 PAR 32,000 16,000 24,000 X9 年 $300,000 48,000 (32,000) (40,000) $276,000 存40000 銷 40000

請問這題該選B嗎🙏

17. Juan Pablo and Zak are competitors in a local market. Each is trying to decide if it is better to advertise on TV, on radio, or not at all. If they both advertise on TV, each will earn a profit of $8,000. If they both advertise on radio, each will earn a profit of $14,000. If neither advertises at all, each will earn a profit of $20,000, If one advertises on TV and other advertises on radio, then the one advertising on TV will earn $12,000 and the other will earn $10,000. If one advertises on TV and the other does not advertise, then the one advertising on TV will earn $22,000 and the other will earn $4,000. If one advertises on radio and the other does not advertise, then the one advertising on radio will earn $24,000 and the other will earn $8,000. If both follow their dominant strategy, then Juan Pablo will a. advertise on TV and earn $8,000. b. advertise on radio and earn $14,000. c. advertise on TV and earn $22,000. d. not advertise and earn $20,000.

這寫題目不懂！求解！拜託拜託謝謝

十一、 (1) 請求出由溶解1克的甘胺醯胺氯化氢(pKa 值 8.2,分子量為110.54) 與1克的甘胺醯 胺(分子量為74.08)配製成 0.1L 溶液的 pH 值? (2) 1 克的甘胺醯胺氯化氢(pKa 值 8.2,分子量 為 110.54),需加入多少克的甘胺醯胺(分子 量為 74.08)才能配製成 100mL、pH 值為 8 的溶液? (3) 如果(1)的溶液與 5mL 之 0.1M HC1混合,pH 會變為多少? (4) 如果(3)的溶液與 10mL 之 0.1M NaOH 混 合,pH 會變為多少?

拜託幫解答十一提所有題目！拜託 謝謝

十一、 (1) 請求出由溶解1克的甘胺醯胺氯化氢(pKa 值 8.2,分子量為110.54) 與1克的甘胺醯 胺(分子量為74.08)配製成 0.1L 溶液的 pH 值? (2) 1 克的甘胺醯胺氯化氢(pKa 值 8.2,分子量 為 110.54),需加入多少克的甘胺醯胺(分子 量為 74.08)才能配製成 100mL、pH 值為 8 的溶液? (3) 如果(1)的溶液與 5mL 之 0.1M HC1混合,pH 會變為多少? (4) 如果(3)的溶液與 10mL 之 0.1M NaOH 混 合,pH 會變為多少?

想請問這個表格上寫的 "fold reversal "是什麼意思 要怎麼翻譯比較好，謝謝（有上網查過了，但不太了解）

Pro DC 箱(W) 說明(H) nib).pdf x ↑ X gs 638 (3/14) 125% ... Table 1. Effect of lazertinib on reversing the MDR mediated by ABCB1 or ABCG2 in drug selected resistant cancer cells IC 50 ± SD (µM) (fold reversal) Compounds KB (parental) KBv200 (ABCB1) Vincristine 0.0059 +0.0323 (1.00) 0.2263 ± 0.0323 (1.00) +0.0625 µM Lazertinib 0.0053 ± 0.0003 (1.12) 0.0041 ± 0.0002 (1.43) 0.0782 +0.0182** (2.89) 0.0195 ±0.0105** +0.125 µM Lazertinib (11.60) 0.0090 0.0011** (25.21) +0.25 µM Lazertinib 0.0050 +0.0007 (1.54) 0.0042 ± 0.0002 (1,47) 0.0100 ± 0.0038** (22.60) +10 µM Verapamil Doxorubicin 0.0218 ± 0.0038 (1.00) 1.2091±0.4592 (1.00) +0.0625 µM Lazertinib 0.0216 ± 0.0093 (1.01) 0.4347±0.0807** (2.76) +0.125 uM Lazertinib SD alo and *p exp tor res an apy ure wh ing E Confe 100% M C 從表1年 對受質車 ABCG2 然而·L= 劑的IC5C 藥物敏感

我想詢問第39題 謝謝

每個積分 是以逆時針方向繞路徑。說 明答案的原因。 A DA 23.14 節 應用安培定律 並 第 或 -8 到 頭 的 E23.36 23.37。同軸電纜。半徑a的 實心導線透過內徑b和外徑 c傳導管軸上的隔離圓盤支 撐(圖 E23.37 )。這個中心 導線和管承載大小相等、方 向相反的電流I。電流均勻 分布在每個導線上的橫截 圖 E23.37 面。推導下列各處的磁場大小表示式:(a) 在中央實心 導線外部但是在管內部的點(a<r<b);(b) 在管外部的 點(r>c)。 23.38。重複習題 23.37 的問題,此時中央實心導線中的 電流是I,管中的電流是 I,並且這些電流朝著相同的 方向,而不是在相反的方向。 23:39 一螺線管內徑 12.0 cm 和外徑 15.0 cm。它攜帶 ● 1.50 A 的電流。必須有多少圈同樣間距的線圈,才能使 得它在這個線圈距離中心 14.0 cm 以內的點產生 3.75 mT 的磁場? ● 23.40 麻省理工學院弗朗西斯必特國家磁性實驗室已經 做出 37.2 T 的磁場。求出磁場在下列各處所需要的電 流:(a) 離長直導線 2.20 cm;(b) 在 100 圈、半徑 44.0 cm 的圓形線圈中心;(c) 在半徑 2.60 cm、長度 34.0 cm 和 40,000 圈的螺線管中心。

佳美公司於110/1/1 以$154，000 現金購入並啟用新設備。該設備之耐用年限為5年，估計殘值為＄4,000。估計於耐用期間內可生產50,000 單位產品。110年該設備生產產品9,000 單位、111年12,000 單位。試依下列各種折舊方法計算 110、111 ... 繼續閱讀