Step-by-step explanation:
to compare them, we should all bring to the same denominator.
the lcm (lowest common multiplier) is the that denominator.
let's use the approach of the prime factors :
3 : 3
5 : 5
32 : 2×2×2×2×2
24 : 2×2×2×3
the lcm is the combination of the longest "streaks" of the prime factors.
so,
2×2×2×2×2 × 3 × 5 = 32×15 = 480
2/-3 = -2/3 = -320/480 (as 3×160 = 480)
-4/5 = -4×96 / 5×96 = -384/480 (as 5×96 = 480)
21/-32 = -21/32 = -315/480 (as 32×15 = 480)
-15/24 = -15×20 / 24×20 = -300/480 (24×20 = 480)
as
-384/480 < -320/480 < -315/480 < -300/480
we can say
-4/5 < -2/3 < -21/32 < -15/24
Step-by-step explanation:

In this case we have:
Δx = 3/n
b − a = 3
a = 1
b = 4
So the integral is:
∫₁⁴ √x dx
To evaluate the integral, we write the radical as an exponent.
∫₁⁴ x^½ dx
= ⅔ x^³/₂ + C |₁⁴
= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)
= ⅔ (8) + C − ⅔ − C
= 14/3
If ∫₁⁴ f(x) dx = e⁴ − e, then:
∫₁⁴ (2f(x) − 1) dx
= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx
= 2 (e⁴ − e) − (x + C) |₁⁴
= 2e⁴ − 2e − 3
∫ sec²(x/k) dx
k ∫ 1/k sec²(x/k) dx
k tan(x/k) + C
Evaluating between x=0 and x=π/2:
k tan(π/(2k)) + C − (k tan(0) + C)
k tan(π/(2k))
Setting this equal to k:
k tan(π/(2k)) = k
tan(π/(2k)) = 1
π/(2k) = π/4
1/(2k) = 1/4
2k = 4
k = 2
Option 3 is the best choice
This is impossible. The only two factors of this polynomial are 0 (multiplicity of 2) and 5.