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victus00 [196]
1 year ago
13

Help!

Computers and Technology
1 answer:
aleksandrvk [35]1 year ago
6 0

Answer:

it may be linear gradient.

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Do you think social media should affect presidential elections
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No, not really cuz I think that would be too personal for the president.

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2 years ago
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Because collecting the adjustment data requires time, the adjusting entries are often a.estimated and recorded earlier than the
Yanka [14]

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b. entered later but dated as of the last day of the period.

Explanation:

Because collecting the adjustment data requires time, the adjusting entries are often entered later but dated as of the last day of the period.

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Which of the following best describes the protocols used on the Internet?
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C: The protocols of the Internet are open and used by all devices connected to the network

Explanation:  Hope this helps.  

There are billions of devices connected to the Internet, and hundreds of different kinds of devices: laptops, tablets, phones, refrigerators, handheld credit card readers, and so on. Protocols (standards) ensure that the variety of devices interact with each other smoothly.  There are a lot of protocols! The Internet was designed with several layers of abstraction that sort the protocols according to what part of the process they support.

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2 years ago
What is the number system that uses only the numbers 0 and 1?
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It is c because its C
4 0
3 years ago
Suppose you wish to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm. Your algo
Kazeer [188]

Answer:

The number of subproblems are given as T(n)=a*T(n/8)+\theta(n^2) while the value of T(n) to be less than S(n) is for 342.

Explanation:

The number of subproblems are given as

T(n)=a*T(n/8)+\theta(n^2)

Asymptotic running time for Strassen’s algorithm is S(n)=\theta(n^{log(7)})

Now, when a increases, number of subproblems determines the asymptotic running time of the problem and case 1 of master theorem applies. So, in worst case, asymptotic running time of the algorithm will be

T(n)=\theta(n^{logb(a)})=\theta(n^{log8(a)})=\theta(n^{log_{2}(a^{1/3})})

Now, for T(n) to be smaller than S(n)

n^{loga^{1/3}}

So,

log(a^{(1/3)})

So,

a=342

5 0
3 years ago
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