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:
all work is shown and pictured
Answer:
72 words per minute
Step-by-step explanation:
42 × y = 1 × 3024
42y = 3024
42y ÷ 42 = 3024 ÷ 42
y = 72
X^2/(x- 9 = 81/(x - 9)
This is the equation for which you want the solution.
Multiplying both sides of the equation by (x - 9) we get
x^2(x - 9)/(x - 9) = 81(x - 9)/(x - 9)
So the (x - 9) goes out from both the denominator and the numerator and then the simplified equation becomes
x^2 = 81
x ^2 = (9)^2
x = 9
So the value of the unknown variable x comes out to be 9.
