這是鋼筋混凝土的題目，想請問第二題的繪圖題每一個case的圖跟V應該要怎麼計算？ 急需感謝🙏

一、梁剪力分析與設計 d=600-65:535 有一簡支懸臂梁 bXh = 400X600mm,承受均佈載重與集中載重,該梁考慮以下六種情況之載 重分佈,其中 WD=WL=30kN/m,Pu = 100 kN,如下圖所示: -3mm Wi Wu = 84 kN/m Pu P. Wi Wo 3mm P Pu Pu Wi WD 3 m 3.m 3 m P₁ P. P.. WL WD WL WD 3. m 3.m Pu 3 m 3. m P P. W₁ WD 3 m 3 m 3 m 3 m P.. Wi WD 3m- 3. m 3. m (1) 若該梁使用 fc = 30 MPa、f=420 MPa、fr = 300 MPa, 請算出該梁的Vc(30%) (2) 請以此六情況繪出該梁之剪力包絡線(30%) (3) 請設計該梁的剪力,若需配置剪力筋,請依課堂例題標準設計之(40%) 1) Vc = ± √f'c bwd = 195KN #

數學 重積分 🙏

Question_4 130176 a If ₁₁₂ (f(x, y)) = 96x3 y3 1+8xy+ then f(x,y)= ay 1. x2ln(1 + 8xy) 2. 3ln(1+8xy4) 3. In(1 + 8xy4) 4. 3x2ln(1+8xy4) Question_3 a 130178) If (f(x, y)) = 60x²y6 √1+5x³y4, then f(x,y)= ǝx 1+8y2 1. -(5x³y4)3/2 3 8y2 2. By ² (1+5x³y4) 8 3 3. (1+5x³y4)³/2 3 8 4.- (1+5x³y4)

數學 重積分 謝謝🙏

Question_2 130109 If₁ (f(x, y)) = 72xy³ e¹²x²y³, then f(x,y)= ǝx 1.72xy³ 12x²y³ 2. 3e 12x²y3 3. e12x²y3 4. 3y³e 12y³

數學 偏微分/重積分 謝謝🙏

Question_7 If z = e 3x5y5, where x = cost and y = sint, then , 1. 15x5y4e3x5y5 (-sint) + 15x4y5e3x5y5 (cost) 2. 15x4y5e3x5y5 (cost) + 15x5y4e3x5y5 (sint) 3. 15x4y5e3x5y5 (-sint) + 15x5y4e3x5y5 (cost) 4. 15x5y4e3x5y5 (cost) + 15x4y5e3x5y5 (sint) Question_8 If z = ex6y5, where x = cost and y , 1.5x yexy 5 2. 6x5y5 dz = dt dz = sint, then dt (-sint) + 6x5y5 ex6y5 (cost) (cost) +5x6y4exбy5 (sint) 3.6x5y5 ex6y5 (-sint) + 5x6y4 ex6y5 (cost) 4.5x6y4exys (cost) + 6x5y5 ex6y5 (sint) 1201@

靜力學 求出E, F的N, V, M 拜託了🙏

7-23. Determine the normal force, shear force, and moment at points E and F of the compound beam. Point E is located just to the left of the 800-N force. A 1200 N 800 N 400 N/m 5 3 E B C D 7 F −1.5 m―1 m2 m ►◄1 m→←–1.5 m →←−1.5 m- Prob. 7-23

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這題偏微分幫幫忙了🙏

Question_10 If z is defined implicitly as a function of x and y by the equation 2x² - 3y² + z² + 6xyz = 1, then Əz ax 2z+6xy 1. 4x+6yz 2. 3. - 2z+6xy 4x+6yz 4x+6yz 2z+6xy 4x+6yz 4. - 2z+6xy

有人解得出這三題偏微分題嗎？

Question_2 1.3x4y² ex¹³ 3 3 2.4x³y³4x³y³ 3.3x4y²e 3x4y² 4.4x³y³ ex¹y³ a dy (ex²y³) =

不知道怎麼解？請求大神幫忙

一、化學反應如下 CH4(g)+CO2(g) → 2CO(g)+2H2(g) (此題21分) 附錄資料 CH4 (g) CO2 (g) A Hom -74.81 kJ/mole -393.509 kJ/mole Som AGºm 186.264 J/mole-K -50.72kJ/mole J/mole-K -394.359kJ/mole CO(g) -110.525kJ/mole J/mole-K -137.168kJ/mole H2(g) 0 kJ/mole J/mole-K 0 kJ/mole (1)利用附錄資料求上述反應在25℃時的△,Sm、ArGm。 (2)利用附錄資料的△G數據,求上述反應在25°C時的△Gm。 (3)在25℃時,若始態 CH4 (g)和CO2(g)的分壓均為150 kPa,末態 CO (g)和H2(g)的分壓均為50kPa,求反應的△rSm、A.Gm。

想問這題要怎麼證明 不知道要怎麼證明單調跟有界

No. Date nic For use every "I nεN, let an = 1 + 1/2 + 1/3 + 111 + 1/1/1 - \n (n) Monotonic Sequence Theorem" to prove the sequence {an{m² converges.