Answer:
f(2n)-f(n)=log2
b.lg(lg2+lgn)-lglgn
c. f(2n)/f(n)=2
d.2nlg2+nlgn
e.f(2n)/(n)=4
f.f(2n)/f(n)=8
g. f(2n)/f(n)=2
Step-by-step explanation:
What is the effect in the time required to solve a prob- lem when you double the size of the input from n to 2n, assuming that the number of milliseconds the algorithm uses to solve the problem with input size n is each of these function? [Express your answer in the simplest form pos- sible, either as a ratio or a difference. Your answer may be a function of n or a constant.]
from a
f(n)=logn
f(2n)=lg(2n)
f(2n)-f(n)=log2n-logn
lo(2*n)=lg2+lgn-lgn
f(2n)-f(n)=lg2+lgn-lgn
f(2n)-f(n)=log2
2.f(n)=lglgn
F(2n)=lglg2n
f(2n)-f(n)=lglg2n-lglgn
lg2n=lg2+lgn
lg(lg2+lgn)-lglgn
3.f(n)=100n
f(2n)=100(2n)
f(2n)/f(n)=200n/100n
f(2n)/f(n)=2
the time will double
4.f(n)=nlgn
f(2n)=2nlg2n
f(2n)-f(n)=2nlg2n-nlgn
f(2n)-f(n)=2n(lg2+lgn)-nlgn
2nLg2+2nlgn-nlgn
2nlg2+nlgn
5.we shall look for the ratio
f(n)=n^2
f(2n)=2n^2
f(2n)/(n)=2n^2/n^2
f(2n)/(n)=4n^2/n^2
f(2n)/(n)=4
the time will be times 4 the initial tiote tat ratio are used because it will be easier to calculate and compare
6.n^3
f(n)=n^3
f(2n)=(2n)^3
f(2n)/f(n)=(2n)^3/n^3
f(2n)/f(n)=8
the ratio will be times 8 the initial
7.2n
f(n)=2n
f(2n)=2(2n)
f(2n)/f(n)=2(2n)/2n
f(2n)/f(n)=2
Answer:
9
Step-by-step explanation:
10-2×3+5
10-6+5
4+5
9
The answers to the question is A.
Answer:
How do we calculate the area of this quadrilateral using the coordinates of the four vertices? The answer is simple. We divide the quadrilateral in two triangles (using either of the diagonals), calculate the (positive value of) the areas of each triangle, and add these values to obtain the total area.
Step-by-step explanation:
jayfeather friend me
Answer:
B
The answer is b im doing a type limit.
