Answer:
y=2x-A
Step-by-step explanation:
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: 2(2y+5)
Explanation:
One possible factorization is 2(2y+5)=4y+10, which would correspond to a set of dimensions 2 by (2y+5) for the rectangle.
Note that this is only one of many possibilities to come up with the dimensions. For example another possibility would be 1*(4y+10), or 4*(y+2.5). In fact, in Real numbers, there are infinitely many possible rectangles with a particular area.
Answer:
The answer is "None of the above".
Step-by-step explanation:
In this question, both the team given data is the same, and when the coach calculates its value it will give the same value, that's why we can say that all the choices were wrong.