If one length is 6in, the other would be 4, since 6×4 is 24, which is the area of the rectangle. The length is 2 longer than the width, 6-4=2.
Answer:
f(x+2h)=−(f(x)+2hf′(x)+4h22f′′(x)+o(h3)==−f(x)−2hf′(x)−4h22f′′(x)−o(h3)
f(x−2h)=f(x)−2hf′(x)+4h22f′′(x)−o(h3)
8f(x+h)=8(f(x)+hf′(x)+h22f′′(x)+o(h3)=8f(x)+8hf′(x)+8h22f′′(x)+8o(h3)
−8f(x−h)=−8(f(x)−hf′(x)+h22f′′(x)−o(h3)=−8f(x)+8hf′(x)−8h22f′′(x)+8o(h3)
So
|f′(x)−112h[−f(x+2h)+8f(x+h)−8f(x−h)+f(x−2h)]|==|f′(x)−112h[−4hf′(x)−2o(h3)+16hf′(x)+16o(h3)]|==|f′(x)−f′(x)−14o(h3)12|=o(h3)?
2.
We just plug x+2h,x+h,x−h,x−2h to the approximation and get:
f′′(x)≈[−f(x+4h)−16f(x+3h)+16f(x+h)−130f(x)+64f(x+2h)+64f(x−h)+64f(x−2h)−64f(x−3h)]144h2
Step-by-step explanation:
Answer: 3.
The volume for all the boxes is 50in³ because 1 * 1 * 1 * 50 = 50.
Next, let's prime factorize 50 so that (using multiplication) we can find all the different bases.
50 ÷ 2 = 25
25 ÷ 5 = 5
5 ÷ 5 = 1
2 * 5² * 1
height x length x width
1 x 5 x 10
1 x 2 x 25
1 x 1 x 50
Linear, decrease by 2 for every decrease of x.