D. 192in^2
This would be a simple area problem with a triangle. REMEMBER THIS EQUATION: a=bh*1/2 (b for base and h for height, these are multiplied together then that answer is halved out.)
So we just need to plug in our values into the equation, so the equation would look like a= (16)*(24)*1/2. 16 times 24 would then give you 384, you could either divide by 2 or multiply by 0.5 to get the next answer, as long as your HALVING the answer.
So we have our bh value so now we can multiply by 1/2 which will give us 384*1/2 which leaves us with 192.
I have also attached a photo of doing a longer(ish) way than this, that also proves that this equation works. Either one will provide you an answer.
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:
what do you want to be answered
Step-by-step explanation: