Rachel is setting up tables for a party. Four of the tables are covered with red tablecloths, and eight of the tables are covere
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Notice the graph, the domain is just the horizontal area "used up" over the x-axis, so, the graph goees from ![\bf -\cfrac{5x}{2}\quad to\quad \cfrac{5x}{2}\implies domain\implies \left[-\cfrac{5x}{2}\ ,\ \cfrac{5x}{2}\right]](https://tex.z-dn.net/?f=%5Cbf%20-%5Ccfrac%7B5x%7D%7B2%7D%5Cquad%20to%5Cquad%20%5Ccfrac%7B5x%7D%7B2%7D%5Cimplies%20domain%5Cimplies%20%5Cleft%5B-%5Ccfrac%7B5x%7D%7B2%7D%5C%20%2C%5C%20%20%20%5Ccfrac%7B5x%7D%7B2%7D%5Cright%5D)
the range is just, the vertical area "used up" over the y-axis, so, the graph goes to 2 and down to -2, thus![\bf range\implies [2\ ,\ -2]](https://tex.z-dn.net/?f=%5Cbf%20range%5Cimplies%20%5B2%5C%20%2C%5C%20-2%5D)
the range is just, the vertical area "used up" over the y-axis, so, the graph goes to 2 and down to -2, thus
Euler's method uses the recurrence relation
to approximate the value of the solution
to the ODE 
.
With a step size of
, there will only be two steps necessary to find the approximate value of 
based on the initial point 
. See the attached table below for the computation results.
To demonstrate how the table is generated: Since, you are using 
.
The next point then uses
to approximate the value of the solution
With a step size of
To demonstrate how the table is generated: Since
The next point then uses