請問第29題怎麼算

In Exercises 21 through 36, solve the given equation for x. 21. 4* = 53 POS 22. log₂ x = 4 2 23. log3(2x - 1) = 2 24. 32x-1= 17 minsgol gai 25. 2 e = e0.06x 27. 3 = 2 + 5e 29. ln x = t 50 - 4x + C 1 2 Lo = Qoe 0 28. -2 ln x = b 26. 30. 5 = 3 ln x - - 1.2x 1 - ln x 2

請問這題怎麼算

9. A rectangle is inscribed in a right triangle, as 000,00 shown in the accompanying figure. If the triangle 0 has sides of length 5, 12, and 13, what are the

第二小題、第3小題求過程🥺 答案：-42、2

3₁ % f(1) = -2₁ f^(1)=3, 9(₁) = 2; g'(1) = -1 731) Jak # 求下列問題 + x ( faj 9(x)} | X= = dx (2) / (f² (4x-3) +(3ײ-5)) |x-1 32 +₂ ( f (57) + = (2-1)) |x-1= in d

怎麼化簡這個式子？

(c) 1 (x+h)² ht 1 2 X

如圖，為什麼將e^(-2y)積分後會變成-1/2（e^-2y)，求過程

dv=e²² dy v=-10²2²2y

只需要告訴我第二個怎麼微分就行-(2x)^1/2的部分 需要微分兩次～ 謝謝🙏

BOT12 45. y = 2 3x √2x + √2x - 1 6√x

請問這題要怎麼算～

1 89. Let f be a function such that f'(x): If 1 + x² 0₁5 2 LIFE AND Qurbliet 1 + x² dig(x) = f(2x + 1), what is g'(x)?sup inslissxs) =

不知道怎麼算(b) 謝謝！

4. 試決定k的值,使f(x)為連續函數: k(x+1) 當æ<3 (a) f(x) = ((b) g(x) = = x² 2 當>3 √2x +5 − √√√x + 6 æ-1 k 呆 k 當T>1 當 æ < 1

想問第4題畫線處 分母(tan²x)-1 =(sin²x/cos²x)-1 =(sin²x-cos²x)/cos²x 為什麼把1/cos²x移到分子以後 分母不是sin²x-cos²x ? 感謝！

x² Example 4 Find lim x 0 sec X 1 SOLUTION The evaluation of this limit requires a little imagination. Since both the numerator and denominator tend to zero as x tends to zero, it is not clear what happens to the fraction. However, we can rewrite the fraction in a more amenable form by multiplying both numerator and denominator by sec x + 1. x2 x² sec x 1 lim x-0 secx 1 - = = x2 sec x 1 - x² (secx + 1) sec² x 1 lim (sin x-0 sin x sec x + 1 sec x + 1) - = Since each of these factors has a limit as x tends to 0, the fraction we began with has a limit: = x² (secx + 1) tan² x - 1 x² cos²x(secx + 1) sin² X 2 (*) ² (cos²x)(secx + 1). 2 ·lim cos²x lim (secx + 1) = (1)(1)(2) = 2. x→0 x ●

求解第24題

t (a), explain wh 2+x-6 x-2 lim (x + 3) imit, if it exists. 12. lim x² - 4x x-1 x²-3x - 4 14. lim 2x² + 3x + 1 2 x-1 X² 2x - 3 16. lim 18. lim h→0 20. lim X→-1 22. lim u-2 24. lim t-0 (2 + h) - 8 h 3x - 4 28. lim h→0 x² + 2x + 1 .4. - 1 26. lim x--4 (1 t 4u + 1 u - 2 3 t² + t x²+9-5 x + 4 1 (x + h)² h 1 X Illustrate tion of th 33. If 4x 34. If 2x - 35. Prove 36. Prove 37-42= explain 37. lim 39. 41. 43