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:
a) x = -270
b) y = 304/9 or y = 33.78
c) x = -37.8
Step-by-step explanation:
y varies directly with x so y = kx
a)
y = kx
k = y/x
k = 10/-30 = -1/3
So
y = -1/3 x
If y = 90 then
90 = -1/3 x
x = 90 *(-3) = -270
b)
k = y/x
k = -19/-4.5
k = 38/9
So
y = 38/9 x
If x = - 8 then
y = 38/9 *(-8)
y = 304/9 or y = 33.78
c)
k = y/x
k = -4.3 / 12.9
k = - 1/3
So
y = -1/3 x
If y = 12.6 then
12.6 = -1/3 x
x = 12.6 *(-3)
x = -37.8
Average rate of change: r=[f(b)-f(a)]/(b-a)
r=-60→[f(b)-f(a)]/(b-a)=-60
b=5; f(b)=-213; a=1; f(a)=27
(-213-27)/(5-1)=(-240)/4=-60
Answer: The <span>two points in the table which create an interval with an average rate of change of -60 are:
x f(x)
1 27
5 -213</span>
