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zloy xaker [14]
2 years ago
13

When seeking information on the internet about a variety of subjects the most useful place to look would be?

Computers and Technology
2 answers:
BaLLatris [955]2 years ago
8 0
I would use google because google could help you so you should use it
Anna35 [415]2 years ago
4 0
Well brainly.com of course! Welcome.

For your answer, i would go with: Websites with primary source information that is verifiable and has a good reputation.
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Thomas has signed a deal with a production house that allows them to use his images on their website. What is required when imag
Phoenix [80]
D. Fair use. Because you need to give credit to the owner.
4 0
2 years ago
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int) You are the head of a division of a big Silicon Valley company and have assigned one of your engineers, Jim, the job of dev
sergeinik [125]

Answer:

Correct option is E

Explanation:

a) 2n^2+2^n operations are required for a text with n words

Thus, number of operations for a text with n=10 words is 2\cdot 10^2+2^{10}=1224 operation

Each operation takes one nanosecond, so we need 1224 nanoseconds for Jim's algorithm

b) If n=50, number of operations required is 2\cdot 50^2+2^{50}\approx 1.12589990681\times 10^{15}

To amount of times required is 1.12589990681\times 10^{15} nanoseconds which is

1125899.90685 seconds (we divided by 10^{9}

As 1$day$=24$hours$=24\times 60$minutes$=24\times 60\times 60$seconds$

The time in seconds, our algortihm runs is \frac{1125899.90685}{24\cdot 60\cdot 60}=13.0312 days

Number of days is {\color{Red} 13.0312}

c) In this case, computing order of number of years is more important than number of years itself

We note that n=100 so that 2(100)^2+2^{100}\approx 1.267650600210\times 10^{30} operation (=time in nanosecond)

Which is 1.267650600210\times 10^{21} seconds

So that the time required is 1.4671881947\times 10^{16} days

Each year comprises of 365 days so the number of years it takes is

\frac{1.4671881947\times 10^{16}}{365}=4.0197\times 10^{13} years

That is, 40.197\times 10^{12}=$Slightly more than $40$ trillion years$

4 0
2 years ago
Assign decoded_tweet with user_tweet, replacing any occurrence of 'TTYL' with 'talk to you later'. Sample output with input: 'Go
nekit [7.7K]

Answer:

I am going to use the Python programming language to answer this. The source code is given below:

print("Enter your tweet here")

user_tweet = input()

decoded_tweet = user_tweet.replace('TTYL', 'talk to you later')

print("This is the decoded tweet: ")

print(decoded_tweet)

Explanation:

In the program the replace() module was used to replace 'TTYL' with 'talk to you later.'

Attached is the screenshot of the output.

8 0
2 years ago
Read 2 more answers
The number of operations executed by algorithms A and B is 40n2 and 2n3, respectively. Determine n0 such that A is better than B
Tom [10]

Answer:

Given that:

A= 40n^2

B = 2n^3

By given scenario:

40n^2=2n^3

dividing both sides by 2

20n^2=n^3

dividing both sides by n^2 we get

20 = n

Now putting n=20 in algorithms A and B:

A=40n^2

= 40 (20)^2

= 40 * (400)

A= 16000

B= 2n^3

= 2 (20)^3

= 2(8000)

B= 16000

Now as A and B got same on n = 20, then as given:

n0 <20 for n =20

Let us take n0 = 19, it will prove A is better than B.

We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.

8 0
2 years ago
how do you make a circuit so 1 switch will turn on/off all the lights(3 lights) and a second switch will change the lights from
expeople1 [14]

Parallel circuit. On a wire you put the lights and one switch, on the other you put a resistor and another switch. (and the third wire contains the generator)
7 0
1 year ago
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