F(x)=x⁴-1
f'(x)=4x³
Newton’s Method: x[n+1]=x[n]-f(x[n])/f'(x[n]); x[n+1]=x[n]-(x[n]⁴-1)/4x[n]³
x₁=3.00390625
x₂=2.26215...
x₃=1.7182...
X'=X-(X⁴-1)/4X³=X-X/4+1/4X³ is a symbolic way of writing the recursive formula, where X' represents the next iteration.
When X'≈X, -X/4+1/4X³≈0; so X/4≈1/4X³; X≈1/X³, so X⁴≈1 and X⁴-1≈0. But this is f(x)≈0. Hence Newton’s Method converges to a solution.
The rate of change is x[n+1]-x[n]=-(x[n]⁴-1)/4x[n]³=x[n]/4-1/4x[n]³ or symbolically -X/4+1/4X³.
Note that the method converges to one solution. A different x₀ will possibly converge to the solution x=-1.
Answer:
th part of the field is covered in the second sprint.
Step-by-step explanation:
A football player is running the length of a 100 yard long football field.
Let player sprints x yards with the speed = 2 yards per second.
So time taken to cover x yards player will take time =
seconds
Now rest distance (100 - x) yards when covered with the speed of 4 yards per second, so time taken to cover this distance = 
=
seconds
Now total time taken by the player can be represented by the equation

Now we can solve this equation for the value of x.

x + 100 = 40×4
x + 100 = 160
x = 160 - 100 = 60 yards
And length of the second part will be = 100 - 60 = 40 yards
Now the fraction of the field covered by the player in second sprint will be
= 
=
or 40%
Therefore,
th part of the field was covered in second sprint.
5x - y = -4 is the standard form
Slope of k=(3-2)/5-6
=-1
parallel lines have equal slopes
so slope of line parallel to k=-1