First, let x be the number of hours it will take for Hector to finish the same job alone. This means that every hour, Hector can do 1/x of the job. The amount of work done by Shawn and Hector in 6 hours should be equal to 1 complete work. This can be expressed as,
(1/9 + 1/x)(6) = 1
The value of x is 18. Thus, Hector can to 1/18 of the job in one hour.
Answer:
The answer is B
Step-by-step explanation:
The best way to solve is to plug in the x points into each equation to see if there is only one that works for the first point and if more than one works do the same for the second point
Answer:
18e⁶
⁵/₁₂₈
Step-by-step explanation:
Rₙ(x) = f⁽ⁿ⁺¹⁾(c) / (n+1)! (x − a)ⁿ⁺¹, and a < c < x.
f(x) = eˣ, a = 0, and n = 1. Thus R₁ is:
R₁(x) = f"(z)/2! x²
R₁(x) = eᶻ/2 x²
|R₁| is a maximum when |f"(z)| is a maximum. On the domain 0 < z < 6, that maximum is e⁶. At x = 6, the upper bound of |R₁| is:
|R₁| = 18e⁶
This time, f(x) = 1 / √(1 + x) = (1 + x)^-½. a = 0, and n = 2.
R₂(x) = f⁽³⁾(z)/3! x³
Find f⁽³⁾(x):
f'(x) = -½ (1 + x)^-³/₂
f"(x) = ¾ (1 + x)^-⁵/₂
f⁽³⁾(x) = -¹⁵/₈ (1 + x)^-⁷/₂
On the domain -½ < z < 0, |f⁽³⁾(z)| is a maximum at z = 0.
|f⁽³⁾(z)| = ¹⁵/₈
Therefore, at x = -½, the upper bound of R₂ is:
|R₂| = (¹⁵/₈)/6 |(-½)³|
|R₂| = ⁵/₁₂₈
This is what I got, but you might want to double check me because I'm not that good at math (this is what I got for the discontinuity);
1. f(x)= x^2+6+8/x+4
*x+4=0
-4 -4
x= -4
This means that x cannot equal -4
2. f(x)= (x+2)(x-4)/(x+2)(x+2)
*Note: When I did my calculations, I cancelled out all (x+2) binomials except for one in the denominator*
f(x)= x-4/x+2
*x+2=0
-2 -2
x= -2
This means x cannot equal -2
*You may want to check my work, but I believe that your answer is going to be either C or D. Personally, I assume it's C, but that's just me. Anyway, it's probably going to be C or D, or at least that's what I think.*
What is (B∩A)c where A={5,11,3,8,9} , B={99,5,11,2,4} and Universal Set={5,16,4,12,2,11,99,3,1,9,8}?
Anna71 [15]
Answer:
yi8yf9yf9yf9t86g8yg9uufhhhhhhohigydydxddddyfoyciyc
ouvouf9ufoyouyuou0u
