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4 years ago

What does the Sort feature do with a database

1 answer:

6 0

Nothing. If you SELECT data from a database, and add an ORDER BY clause, it just affects the ordering of the data that is returned to the user, but the data in the database remains unaltered.

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Discuss your views on multiple backgrounds. What are the advantages and disadvantages of having multiple backgrounds? Your submi

PSYCHO15rus [73]

Answer:

It is certainly possible to apply different backgrounds to elements. And these are being layered one over another. The first background is on the top and the last one at the back. And we can get a clue with a little designing knowledge that the last background can only have the background color.

And by the end of the process, we are up with the dynamic background design which is good enough to fit each of the web pages.

And it's quite simple to specify the multiple backgrounds. You can set multiple backgrounds to an element, and each of them. And these backgrounds are being layered atop over one and other, the first one being at the top. and the last one is listed at the back. However, as discussed, we can only have the last background to include the background color.

This can be done with the help of a shorthand background property, as well as various sub-properties of background like background-color. This is definitely possible.

There are some disadvantages, however, as well. The website loading speed can decrease, and hence SEO strategies can fail. Also, the visual effect that these multiple backgrounds creates can become boring after a certain period of time.

Explanation:

Please check the answer section.

7 0

3 years ago

Declared inside a function and is only available within the function in which it is declared.

vitfil [10]

Answer:

2b2t

Explanation:

2b2t

3 1

3 years ago

What type of malware actually evolves, changing its size and other external file characteristics to elude detection by antivirus

Free_Kalibri [48]

Answer:

This type of malware are called Polymorphic Malware.

6 0

3 years ago

100 POINTS PLEASE HELP

Oksana_A [137]

Answer:

hope this helped ,do consider giving brainliest

Explanation:

import numpy as np

#PrintArray Function

def printArray(array):

for i in range(len(array)):

for j in range(len(array[i])):

print(array[i][j], end= " ")

print()

#Flip horizontal function

def flipHorizontal(array):

#reversing the order of arrays

array2 = np.fliplr(array).copy() printArray(array2)

#Flip Vertical function

def flipVertical(array):

#Preserving the order of array and reversing each array.

array3 = np.flipud(array).copy() printArray(array3)

#Main() function def main():

array = [[0,2,0,0,0],[0,2,0,0,0],[0,2,2,0,0],[0,2,0,2,0],[0,2,0,0,2]]

print("The array: \n")

printArray(array)

print("\nFlipped horizontally: \n") flipHorizontal(array)

print("\nFlipped vertically: \n") flipVertical(array)

if __name__=="__main__":

main()Explanation:

6 0

2 years ago

Read 2 more answers

find_cow(name, cows) Given a name and a Python list of Cow objects, return the Cow object with the specified name. If no such Co

nadya68 [22]

Using the computational knowledge in python it is possible to write a code that Given a name and a Python list of Cow objects

<h3>Writting the code in python:</h3>

def load_cows(filename):

cow_dict = dict()

f = open(filename, 'r')

for line in f:

line_data = line.split(',')

cow_dict[line_data[0]] = int(line_data[1])

return cow_dict

def greedy_cow_transport(cows, limit=10):

trips = []

cowsCopy = cows.copy()

sortedCows = sorted(cowsCopy.items(), key=lambda x: x[1], reverse = True)

while sum(cowsCopy.values()) > 0:

ship = []

total = 0

for cow, value in sortedCows:

if cowsCopy[cow] != 0 and value + total <= limit:

ship.append(cow)

total += value

cowsCopy[cow] = 0

trips.append(ship)

return trips

def brute_force_cow_transport(cows,limit=10):

trips = []

possibilities = []

for i in power_list:

ship = []

for j in i:

ship_weights = []

for k in j:

ship_weights.append(cows[k])

#print(ship_weights)

ship.append(sum(ship_weights))

#print(ship)

if all(d <= limit for d in ship):

possibilities.append(i)

pruned_possibilities = []

for k in possibilities:

if k not in pruned_possibilities:

pruned_possibilities.append(k)

# now find the minimum list length:

min_list_len = min(map(len, pruned_possibilities))

for l in pruned_possibilities:

if len(l) == min_list_len:

return l

def compare_cow_transport_algorithms():

greedy_start = time.time()

greedy_results = greedy_cow_transport(cows, limit = 10)

greedy_end = time.time()

print('Greedy Algorithm time:', greedy_end -greedy_start)

brute_force_start = time.time()

brute_force_results = brute_force_cow_transport(cows, limit = 10)

brute_force_end = time.time()

print('Brute force time:', brute_force_end - brute_force_start)

print('Greedy Algorithm results:', greedy_results)

print('Number of trips returned by Greedy Algorithm:', len(greedy_results))

print('Brute Force Algorithm results:', brute_force_results)

print('Number of trips returned by Brute Force Algorithm:', len(brute_force_results))

See more about python at brainly.com/question/13437928

#SPJ1

4 0

2 years ago