Answer:
Given,
P = (22, 1, 42, 10)
Q = (20, 0, 36, 8)
a. Formula for Euclidean Distance :
distance = ((p1-q1)^2 + (p2-q2)^2 + ... + (pn-qn)^2)^(1/2)
Now,
distance = ( (22-20)^2 + (1-0)^2 + (42 - 36)^2 + (10-8)^2) ) ^(1/2)
=( (2)^2 + (1)^2 + (6)^2 + (2)^2 ) ) ^(1/2)
=(4+1+36+4)^(1/2)
=45^(1/2)
Distance = 6.7082
b.Manhattan distance :
d = |x1 - x2| + |y1 - y2|
d = |22- 20| + |1 - 0|
d = |2| + |1|
Explanation:
Is advertising influencing her?
What are her motivations?
Has she compared prices?
Is she buying at the right time?
Answer:
Question is incomplete.
Assuming the below info to complete the question
You have a collection of n lockboxes and m gold keys. Each key unlocks at most one box. Without a matching key, the only way to open a box is to smash it with a hammer. Your baby brother has locked all your keys inside the boxes! Luckily, you know which keys (if any) are inside each box.
Detailed answer is written in explanation field.
Explanation:
We have to find the reachability using the directed graph G = (V, E)
In this V are boxes are considered to be non empty and it may contain key.
Edges E will have keys .
G will have directed edge b1b2 if in-case box b1 will have key to box b2 and box b1 contains one key in it.
Suppose if a key opens empty box or doesn’t contain useful key means can’t open anything , then it doesn’t belongs to any edge.
Now, If baby brother has chosen box B, then we have to estimate for other boxes reachability from B in Graph G.
If and only if all other boxes have directed path from box B then just by smashing box B we can get the key to box b1 till last box and we can unlock those.
After first search from B we can start marking all other vertex of graph G.
So algorithm will be O ( V +E ) = O (n+m) time.
My school requires us to use Arial and double space it but we have to use font size 30 for title and 24 for the paragraphs.
Answer:
<div id="header">
<h1>Waketech</h1>
</div>
<header><h1>Waketech</h1></header>
Explanation:
I think thats the answer your welcome