這題的解答給的是C, 該怎麼求出，感謝！

6. Find the volume of the solid whose region is the intersection of the cube {(x, y, z) : |x| ≤ 1, |y| ≤ 1, || ≤ 1} and the ball {(x, y, z) = x² + y²+ z² < 2}. 16 (A) 32(√2-1). (B) (8-√2). 3 16 (C) (10- √2)TT. (D) √√2π. 3 32 3

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想問一下第七題

page 049 细胞質分裂前 間期结束~間期前→ 無法看到染色体 7. 細胞在兩次有絲分裂期之間,不會發生下列哪一個現象? (A)DNA 複製 (B)增加細胞質 的胞器數量 (C)染色質緊密纏繞成染色體 (D)增加細胞質的物質含量。 (E)核糖體製造 蛋白質 section 1-4 包含细陀黄開期開始、结束 AB8.有關染色體的敘述,下列何者正確?(A)在兩次細胞分裂之間,可看見絲狀的染色體 (B)染色體可用光學顯微鏡觀察 (C)複製完的姊妹染色分體,以著絲點連在一起 (D)在細 胞分裂時,紡錘絲直接附著於染色體的中節上而不需要其他蛋白質的協助 (E)真核生物 的同源染色體來自同一個親代。 算 B 9. 下列有關人類「同源染色體」的敘述中,何者錯誤?(A)是細胞內成對的染色體 (B)成 年人體內任何細胞中都含有成對的同源染色體 (C)一個來自父方一個來自母方 (D)遺傳 學上成對的等位基因分別位於同源染色體相對位置上 (E)正常的生殖細胞中不存有成對 的同源染色體。 2000精

求這題～ 答案為為 a .31cm b.866J c.1.15J

ping. What was the initial speed of the bullet? 8.43 A Ballistic Pendulum. A 12.0 g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 9.00 kg, suspended from a cord 70.0 cm long (see Example 8.8 in Section 8.3). Compute (a) the vertical height through which the pendulum rises, (b) the initial kinetic energy of the bullet, and (c) the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the wood. 150 kg block is attached A

為什麼用on?

• In the Notes Section (A) in (B) for on the right hand side, you can record important notes from class. (() on (D) above

綠色螢光筆的不是很了解 1. 為什麼會有5呢？ 2. 為何要加52000

dand 2. (LO 3, 4) On January 1, 2020, Hayslett AG had the following equity accounts. Share Capital-Ordinary (€10 par value, 260,000 shares issued and outstanding) Share Premium-Ordinary Retained Earnings During the year, the following transactions occurred. April 1 May 1 June 1 Aug. 1 31 Dec. 1 Declared a €1.50 cash dividend per share to shareholders of record on April 15, payable May 1. Paid the dividend declared in April. Announced a 2-for-1 share split. Prior to the split, the market price per share was €24. Declared a 10% share dividend to shareholders of record on August 15, distributable August 31. On August 1, the market price of the shares was €10 per share. Issued the shares for the share dividend. Declared a €1.50 per share dividend to shareholders of record on December 15, payable January 5, 2021. Determined that net income for the year was €600,000. 31 Instructions Solution 2. a. Apr. 1 May 1 June 1 Aug. 1 31 Dec. 1 31 1 Cash Dividends (260,000 × €1.50) Dividends Payable Dividends Payable Cash No journal entry needed for share split Share Dividends (52,000* × €10) Ordinary Share Dividends Distributable (52,000 × €5) Share Premium-Ordinary (52,000 × €5) *520,000 x.10 Ordinary Share Dividends Distributable Share Capital-Ordinary Cash Dividends (572,000** × €1.50) Dividends Payable **(260,000 × 2) + 52,000 Income Summary €2,600,000 1,500,000 3,200,000 Retained Earnings Retained Earnings Share Dividends Cash Dividends (€390,000+ €858,000) 390,000 390,000 520,000 | 260,000 858,000 600,000 1,768,000 390,000 390,000 260,000 260,000 Prepare dividend ent section. 260,000 858,000 600,000 520,000 1.248,000 14. Luz (a) r equ 15. pre 16- 17 for 18

求解第3、4題🙏 證明題總是卡住...

3. Let Bo and ₁ denote the least squares estimators of Bo and 3₁. Define SSE = Σ(Yi-Bo- i=1 B₁X₁)². Show that (a) E (SSE) = 0², (b) cov(Y, ₁X) = 0. 4. In Section 10.4, we defined a test statistic where s² = T = VSEZ SSE. In addition, it is known that T~: n-2' (a) Show that (1) can be re-written as 1 tn-2. T = r√n-2 (1) (2) (b) Use Part (a) to test Ho: p = 0 vs H₁ p 0. [Hint: (2) can be regarded as a test statistic under Ho.]

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

正合方程式的問題（工數） 有點毫無頭緒😵‍💫 求解

Section 1.3 Problems 測試是否為正合 In each of Problems 1-5, test the differential equation for exactness. If it is exact in some region R of the plane, find a potential function and a general solution, perhaps implicitly defined, that is valid for that region. If it is not exact on any region, do not attempt a solution. 1. 2y² + yexy + (4xy + xexy + 2y)y' = 0 2. 4xy + 2x + (2x² + 3y²)y' = 0 3. 4xy + 2x²y + (2x² + 3y²)y' = 0 4.2 cos(x+y) - 2x sin(x + y) - 2x sin(x + y)y' = 0 tlsx9 ² +y+ (3y² + x)y' = 0 5.

這個(c)的 inverse reciprocal trigonometric functions 是指什麼？(hahaha)

In a similar manner, one can define the 'arccosine' and 'arctangent' functions by 'restricting the domains'. The following table summarizes the domains and ranges of these inverse functions. Function Domain Range Function f(x) = sin¯¹(x)| (-1,1)|(−½, 3) || f(x) = sec¯¹(x) f(x) = cos ¹(x) [−1,1] f(x) = tan-¹(x) [0, π] Domain Range R (-)|| f(x) = cot-¹(x) x2! [-7,7- XZ1 OY f(x) = csc¯¹(x) 85-1 xs-1 but fuxito Exercise 4. (a) Sketch the graph y = tan-¹(x) (b) From (a), write down lim tan-¹(r) and lim tan-¹(x). x48 8118 (c) Look up the relevant section in the textbook (Stewart), write down the domains and ranges of the 'inverse reciprocal trigonometric functions' on the right hand side of the above table.