There is not enough information to solve the problem.
Answer:
n = 29 iterations would be enough to obtain a root of
that is at most
away from the correct solution.
Step-by-step explanation:
You can use this formula which relates the number of iterations, n, required by the bisection method to converge to within an absolute error tolerance of ε starting from the initial interval (a, b).

We know
a = -2, b = 1 and ε =
so

Thus, n = 29 iterations would be enough to obtain a root of
that is at most
away from the correct solution.
<u>You can prove this result by doing the computation as follows:</u>
From the information given we know:
This is the algorithm for the Bisection method:
- Find two numbers <em>a</em> and <em>b</em> at which <em>f</em> has different signs.
- Define

- If
then accept c as the root and stop - If
then set <em>c </em>as the new<em> b</em>. Otherwise, set <em>c </em>as the new <em>a</em>. Return to step 1.
We know that
and
so we take
and
then 
Because
we set
as the new <em>b.</em>
The bisection algorithm is detailed in the following table.
After the 29 steps we have that
hence the required root approximation is c = -0.50
Answer: Not p.
Reasoning: It is neither "and p" nor "or p" because it isn't a statement with more than one variable in it to combine p with. It is neither "p is false" nor "p is true" because negation of a variable has nothing to do with the truth value of the claim.
Hope this was helpful.
Answer:
Proved
Step-by-step explanation:
Given
![1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)](https://tex.z-dn.net/?f=1%2B5%2B9%2B...%2B%5B4%28n-1%29%2B1%5D%2B%5B4n%2B1%5D%20%3D%20%28n%2B1%29%282n%2B1%29)

Required
Prove by induction
![1+5+9+...+[4(n-1)+1]+[4n+1] = (n+1)(2n+1)](https://tex.z-dn.net/?f=1%2B5%2B9%2B...%2B%5B4%28n-1%29%2B1%5D%2B%5B4n%2B1%5D%20%3D%20%28n%2B1%29%282n%2B1%29)
Increment n by 1 on both sides
![1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+1+1)(2(n+1)+1)](https://tex.z-dn.net/?f=1%2B5%2B9%2B...%2B%5B4%28n-1%29%2B1%5D%2B%5B4n%2B1%5D%2B%5B4%28n%2B1%29%2B1%5D%20%3D%20%28n%2B1%2B1%29%282%28n%2B1%29%2B1%29)
Simplify the right hand side
![1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+2+1)](https://tex.z-dn.net/?f=1%2B5%2B9%2B...%2B%5B4%28n-1%29%2B1%5D%2B%5B4n%2B1%5D%2B%5B4%28n%2B1%29%2B1%5D%20%3D%20%28n%2B2%29%282n%2B2%2B1%29)
![1+5+9+...+[4(n-1)+1]+[4n+1]+[4(n+1)+1] = (n+2)(2n+3)](https://tex.z-dn.net/?f=1%2B5%2B9%2B...%2B%5B4%28n-1%29%2B1%5D%2B%5B4n%2B1%5D%2B%5B4%28n%2B1%29%2B1%5D%20%3D%20%28n%2B2%29%282n%2B3%29)
Group the left hand side
![(1+5+9+...+[4(n-1)+1]+[4n+1])+[4(n+1)+1] = (n+2)(2n+3)](https://tex.z-dn.net/?f=%281%2B5%2B9%2B...%2B%5B4%28n-1%29%2B1%5D%2B%5B4n%2B1%5D%29%2B%5B4%28n%2B1%29%2B1%5D%20%3D%20%28n%2B2%29%282n%2B3%29)
Recall that
----[Given]
So; Substitute
for
on the left hand side
![(n+1)(2n+1)+[4(n+1)+1] = (n+2)(2n+3)](https://tex.z-dn.net/?f=%28n%2B1%29%282n%2B1%29%2B%5B4%28n%2B1%29%2B1%5D%20%3D%20%28n%2B2%29%282n%2B3%29)
Open All Brackets

Collect Like Terms


<em>Notice that the expression on both sides are equal;</em>
<em>Hence, the given expression has been proven</em>
Answer:
6 and 4
Step-by-step explanation: