Answer:
acorns
Step-by-step explanation:
20x=9y
9/20 to 5/9
5/9 of the acorns collected are from Hannah
9/20 of the conkers collected are from Hannah
5/9 is larger than 9/20
Hannah collected more acorns
Answer:
multiplication
Step-by-step explanation:
hope this helps you!
Answer:
w= 6
Step-by-step explanation:
I just started by making an educated guess using the values already given. Then I inserted that into the problem to see if it worked.
L= 2w - 5
I used 6 as a random, educated guess for the value of w.
L = 2(6) - 5
L = 12-5
L = 7
Then, multiply L by 2 to account for both side lengths of the rectangle.
7(2)= 14
Subtract that value from the total perimeter to find what the width must equal.
26 - 14 = 12
Divide that answer by 2 since there are two sides for width.
12/2 = 6
I know this was kind of long, but I hope it helps! :)
Answer:
Step-by-step explanation:
The details that complete the question are:
and
Required
Determine how far the ball travelled
Distance is calculated using:
Substitute values for x's and y's
<em>Hence, the ball travelled a distance of 63.717 units</em>
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
