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

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.

A答案，求救

A laser beam of intensity S and cross-sectional area A is completely absorbed by a particle of mass m for a period At. What is the change in speed of the particle? (A) SAAt/mc; (B) SAAt/m; (C) SA/mc; (D) SAAt/c; (E) None.

［高一生物］求解第七題。C 有絲分裂和減數分裂初期染色質不都會濃縮成染色體嗎？（可能答案有誤或此題無解）

page 049 section 1 - 4 Susan 7. 細胞在兩次有絲分裂期 之間,不會發生下列哪一個現象? (ADNA 複製(B) 增加細胞質 有絲 的胞器數量 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 1 8. 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 真核生物 的同源染色體來自同一個親代。

完全不懂

In Exercises 53–56, find the volume of the solid with the given base R and the indicated shape of every cross section taken per- pendicular to the x-axis. 53. Cross section: a square y -X = y2 2 R x 4 - -2

21題 教教我QAQ

Section 11.3 Angular Momentum J-s behind rmv a he ur 265 = ? a ter 30. Example 11.1: A 58.1 Iw COMP string and whirled at 18. Express the units of angular momentum (a) using only the funda- momentum of magr mental units kilogram, meter, and second; (b) in a form involving the circular path. Fir newtons; (c) in a form involving joules. marw horizontal and (b) the 19. Use data from Appendix E to make an order-of-magnitude esti- 31. Example 11.2. A stai mate for the angular momentum of our Solar System about the Max 27 the end of its lifetime galactic center. mu = 17.293 of radius 4.96 X 10' 20. A gymnast of rotational inertia 63 kg•m? is tumbling head over heels dwarf of radius 4.21 with angular momentum 460 kg•m?/s. What's her angular speed? 17.99 acted on the core, fine A 660-g hoop 95 cm in diameter is rotating at 170 rpm about its 32. Example 11.2: Astro 22. A 1.3-th-diameter golf ball has mass 45 g and is spinning at central axis. What's its angular momentum? L: 1W=0-66*(0-95) x 140x277.10 km and determin core that collapsed to 3000 rpm. Treating the golf ball as a uniform solid sphere, what's ing with a period of 4 its angular momentum? 18-3 33. Example 11.2. The s 10-598 rpm With her arms outstre Section 11.4 Conservation of Angular Momentum tational inertia is 3.5 23. A potter's wheel with rotational inertia 6.20 kg•m? is spinning (Fig. 11.6b), her rot freely at 20.0 rpm. The potter drops a 2.50-kg lump of clay onto her final spin rate? L = 6 2 x 2-1 = 12-99~13 - 2x2 22 13= 60.48)" x 2.5 W W = 2 . X z W = 20x2T 60 ²2.1 ~ 1 12.574