第五題是因為沒除2被扣分嗎

X = 109. (Y+ JY²41) =>2²=Y+SY²³1 5²(x) = 2^²=√²+1 4. (10%) 證明 f(x)=log2 (2+VⅠ? + 1) 為奇函數,並求其反函數 5. (10%)寫出三角函數積化和差(三條) 公式。 (1) sin (α(+B) + sin(2-³)=2 sinx cUSB (05(x-B)-(056d+ ° (2) (05 (α-B) + CO₂(X+B) =Zcosx.cosB³ = 2 sind sing (3) 6. (10%) 將 sin(cos a + tan-y) 化簡為 工, y 的(不含三角函數之)表示式,其中 -1 ≤x≤1, y ∈ P (b) (5%) sin(tan-¹x) (c) (5%) sin (2 tan-¹ x) (d) (5%) sin(cos-¹2) 1 7. (10%) 證明 2 tan' X 6. sin (d+B) = sind cosB+ cosa sin B ()() + (x)( 2x X41 = COS x²-1 tt x2+10 面 Jity AUF-X² Thoxe Storipy THIN x -10

求過程 算到最後卡住了

To 1-3 -1 301 | R₂ + ZR₁ Brok: 02 6 1-5 0 - R3-R1 03-924 |-|| R4 +R₁ N 2₁ R₁ - R₂ R3+5R₂ T 0-1 3 1 3 2 01-302 ött 00 32 | 1 。 O 0622 -R2- 0 CONCE 1-3 -1 30 17 000 5-51-4 0000622 200 1-10-1 22 11

請問c怎麼寫？

e 2. Simplify each expression. Write your answer without negative exponents. (a) √200 √32 = 10√2 - 4√₂ = 6,W² (b) (3a³b³) (4ab²)² = 48 (a³ b') (ab ²)² = 48 9²³ 6² 95 2632 216 2443 L (c) ( 3x³/2y X² ,-1/2 -8 943 -1 3 (XXX)" (X^X^)" (²/²)^ x4 y

深綠色箭頭為什麼那邊要用Re1不能用Rπ 淺綠色螢光筆框框處為什麼有負號

如圖所示的三級串接負回授结 構,已知Emiz=100mA/V, ign = 20 mA/V, h% = 100 台 「如,試求其迴路增益。 [4] =r=h g = 100 100=1kQr₁ = h 電子學第四冊 IV V=0 A₁B₁ = 例題 14-8 =-Bz2 關於迴路增益 AA 測試,其相關電路 關閉信號源 其中,因Q 共射放大器為單向,該處的輸入電阻與其他無關,所以 最佳切開點選在「Q的輸入端」 R = R | = R || 1+h -V =100× R9kQ X T R1002 x x x 關聯V跳 vi I I x X 分析步驟: (a)切開閉迴路,切口看人等效電阻R=r=k2 10002 R R₂ + R₁₁ 5 5+13.75 R. 392 {R x101x -Rex(1+h)x R (b) 在響應端並聯一電阻R,並在測試端加上電壓源V ey (C)為了便於分析,於是先定義出各處等效電阻,加以整理,如圖所示 者,分別是: 1₁ I V V bi x 0.1 0.649 + 0.1 R T V co Re5kn 第十四章 負回授放大器 1- R₁ = ₁; +(1+h) (R₁ + R₂)||R₁|=5+ 101×00.649||0.1) rex 21 =13.75k2 (d)迴路增益分析:算法如下所列 Q方面:採用轉導放大,將信號轉成電流信號,再計算進化的 分流。不過,對於Q 而言,此分流為倒流者 Q方面:為射極追隨器,採用電流放大論點,再以分流觀念,計算 流入Q輸入端的電流 V R, www 6402 =h/g=100 20=5k$2. 1000 AI -=100|| 392 Q方面:為共基放大器,因輸入端有分支,遂用轉導放大觀念牙計 算出響應端電壓V RE Rask R₂ www 6402 (R+R) + R₂ Rea0.6k 1002 Re × 0.009×100X R 0.6k25R2 9×1 9+1 V 1000 Re ×R ×g (RRY = 291.27 說明: (1)此迴路增益A3與R 無關,這是因為不考慮r 之故。 (2)若考慮時,則將會干擾R 與此刻的人,而干擾的程度與尺, 門 1. 21 有關。因此,這時的迴路增益A 值將有關於R,所幸,此關聯 性甚弱,遂可忽略之。

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

求解

Chrome 檔案 Sylves Scores M Gmail ▸YouTube | 地圖 Pearson eText Purchase Options Study Area User Settings Course Home Ghe binding energy of a two-par X + openvellum.ecollege.com/course.html?courseld=17667848&OpenVellumHMAC=9c8d5dd7baa0c70ec3a75db4e52f2d39#10001 Document Sharing Course Tools 檢視 歷史記錄 書籤 設定檔 分頁 視窗 說明 單 > 翻譯 The position of a 280 g object is given (in meters) by x = 5.1 t³ - 8.0 t²-44 t, where t is in seconds. https://session.masteringphysics.com/myct/itemView?assignment ProblemID=176946106 10月 13 Part A tv Determine the net rate of work done on this object at t = 2.0s. Express your answer using two significant figures. Pt 2.0s= = Submit Part B VAΣO Pt=4.18= = Request Answer Determine the net rate of work done on this object at t = 4.1s. Express your answer using two significant figures. VAE P Pearson ? 1999+ ? W Caright © 2022 Pearson Education Inc. All rights reserved. Terms of Use | Privacy Policy. Permissions Contact Us | W G3 ✩ 10月13日 週四 上午 12:38 I Constants 更新

求解

Chrome 檔案 M Gmail 3 My Courses Course Home Syllabus Scores YouTube Pearson eText Study Area Purchase Options User Settings Course Home X Ghe binding energy of a two-par x + openvellum.ecollege.com/course.html?courseld=17667848&OpenVellumHMAC-9c8d5dd7baa0c70ec3a75db4e52f2d39#10001 Document Sharing Course Tools 單 檢視 歷史記錄 書籤 設定檔 分頁 視窗 說明 地圖 > 翻譯 <L.8 Problem 8.51 Part A W = Submit VAE Provide Feedback How much work would be required to move a satellite of mass m from a circular orbit of radius r₁= 2re about the Earth to another circular orbit of radius r2 = 3rE? (rE is the radius of the Earth.) Express your answer in terms of the variable m and constants rE, ME, and G. Request Answer https://session.masteringphysics.com/myct/itemView?assignment ProblemID=176946298 10月 13 tv ? 10月13日 週四 上午12:37 - 999+ ■ 8 of 10 更新: Constants Next >

求解

Chrome 檔案 M Gmail 3 My Courses Course Home Syllabus Scores YouTube Pearson eText Study Area Purchase Options Course Home User Settings Document Sharing Course Tools 單 openvellum.ecollege.com/course.html?courseld=17667848&OpenVellumHMAC=9c8d5dd7baa0c70ec3a75db4e52f2d39#10001 檢視 歷史記錄 書籤 設定檔 分頁 視窗 說明 地圖 > 翻譯 X Ghe binding energy of a two-par x + <L.8 Problem 8.99 A satellite is in an elliptic orbit around the Earth. Its speed at the perigee A is 8526 m/s (Figure 1). Figure 16,460 km B 8230 km 8230 km 10月 13 A 1 of 1 14.255 km tv UB = Use conservation of energy to determine its speed at B. The radius of the Earth is 6380 km. Express your answer in meters per second. Submit Part B VC = VAΣ Submit Request Answer Use conservation of energy to determine the speed at the apogee C. Express your answer in meters per second. VAE Request Answer ? 1999+ m/s ? G m/s 10月13日 週四 上午12:37 ■ 10 of 10 Constants 更新

求救

Chrome 檔案 M Gmail Course Home Syllabus Scores YouTube Pearson eText Study Area Purchase Options Course Home User Settings Document Sharing Course Tools 單 openvellum.ecollege.com/course.html?courseld=17667848&OpenVellumHMAC=9c8d5dd7baa0c70ec3a75db4e52f2d39# 10001 檢視 歷史記錄 書籤 設定檔 分頁 視窗 說明 地圖 > 翻譯 X Ghe binding energy of a two-par x + <L.8 Problem 8.75 The binding energy of a two-particle system is defined as the energy required to separate the two particles from their state of lowest energy to r= ∞o. The potential energy of the two atoms in a diatomic (two-atom) molecule can be written b U(r) = − 6 +12, where r is the distance between the two atoms and a and b are positive constants. 10月 13 Part A Determine the binding energy for this molecule. Express your answer in terms of the variables a and b. Ebind= Submit Provide Feedback VAEO tv JA xa Xb b √x x x X Incorrect; Try Again; 3 attempts remaining Previous Answers Request Answer 1999+ C |X| ? X.10n 10月13日 週四 上午12:33 I 9 of 10 Constants 更新 Next >

請問5跟8

7. /5. 4/7/2$5=1g 在一個體積可調整的反應器中,於 27 ℃、1 大氣壓,注入 10 毫升的 A2 氣體與 30 毫升 Bz氣體(A與B為兩種原子)。假設恰好完全反應,產生甲氣體。 已知甲的分子式與其實驗式相同, 則下列哪一個是甲的分子式? (A)AB (B)AB2 (C) AB3 (D) A₂B3 (E) A₂B. 6. 碳與氧可形成兩種不同的化合物,這兩種化合物中碳和氧的質量比不同。若將碳的質量 固定時,兩化合物中氧的質量之間成一簡單整數比,此稱為倍比定律。 下列各組物質,何者符合倍比定律? (A)C60、C30只有 有個 (B)Pb304、PbQ 来到 (E)Al(OH)3、Al2O3 (D) GaCl3 AIC13 准 (C)SiO2、CO2 15. 反應 N2(g)+3 H2(g) +2 NH3(g)的係數比 1:3:2,其中所代表的意義為下列何項? (多選) (A)質量比 (B)分子數比 (C)原子數比 (D)莫耳數比(E)同狀況下之體積比 有 X、Y 兩元素相結合成甲、乙兩化合物,已知化合物甲 3.0 克中含有 X 為 1.4 克,化 合物乙 27 克中含有 Y 為 20 克。若知甲化學式為 XY,則乙化學式應為下列何者? (A)X,Y, (B)X2Ys (C) XY2 (D) X,Y