Answer:
Assuming this is 0 based array indexing, it would look like this...
[ 5, 6, 10, 7, 3, 2.5 ]
[ 0, 1, 2, 3, 4, 5, 6 ]
The index of 7 would be 3.
I believe the correct answer from the choices listed above is option D. <span>When planning the structure of a spreadsheet, columns are for group items and rows are for single items. Hope this answers the question. Have a nice day.</span>
Def firstChars( word ):
if( word ): # make sure word has characters
print( word[ 0 ] ) # print the first character (Python 3 syntax)
Answer:
A and C
Explanation:
Option A:
In IPv6 there is a rule to reduce an IPv6 address when there are two or more consecutive segments of zeros just one time. This rule says that you can change the consecutive zeros for “::”
Here is an example
How to reduce the following IPv6 address?
ff02:0000:0000:0000:0000:0000:0000:d500
Ans: ff02::d500
Example 2:
2001:ed02:0000:0000:cf14:0000:0000:de95
Incorrect Answer -> 2001:ed02::cf14::de95
Since the rule says that you can apply “::” just one time, you need to do it for a per of zero segments, so the correct answer is:
Correct Answer -> 2001:ed02::cf14:0:0:de95
Or
2001:ed02:0:0:cf14::de95
Option C:
Since in IPv6 there are 
available addresses which means 340.282.366.920.938.463.463.374.607.431.768.211.456 (too many addresses), there is no need of NAT solution, so each device can have its own IP address by the same interface to have access through the internet if needed. If not, you can block the access through internet by the firewall.
Answer:
See explaination
Explanation:
Keep two iterators, i (for nuts array) and j (for bolts array).
while(i < n and j < n) {
if nuts[i] == bolts[j] {
We have a case where sizes match, output/return
}
else if nuts[i] < bolts[j] {
this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1
}
else {
this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1
}
}
Since we go to each index in both the array only once, the algorithm take O(n) time.
