Assuming we need to find i such that
1 ≤ i ≤ n and t[i]=i.
If we need to find only the first occurrence, we can do:
for i:1 to n {
if t[i]=i then return(i)
}
If exhaustive search is required, then put the results (values of i) in an array or a linked list, return the number of values found, and the array (or linked list).
Answer:
a = 5.81 m
Step-by-step explanation:
We can use the Pythagorean theorem
a^2 + b^2 = c^2
where b= 1.5 and c= 6
a^2 +1.5^2 = 6^2
a^2 +2.25 = 36
Subtract 2.25 from each side
a^2 +2.25-2.25 = 36-2.25
a^2 = 33.75
Take the square root of each side
sqrt(a^2) = sqrt(33.75)
a = 5.809475019
We only take the positive root because length must be positive
2^m=2^n take the natural log of both sides
m*ln2=n*ln2 divide both sides by ln2
m=n
