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zmey [24]
2 years ago
6

Suppose that you have two different algorithms for solving a problem of size n. The first algorithm uses exactly n(log n) operat

ions and the second algorithm uses exactly n 3/2 operations. As n grows, determine which algorithm uses fewer operations?
Mathematics
1 answer:
Mumz [18]2 years ago
5 0

Answer:

Algorithm 1 uses:

n*log(n) operations.

While algorithm 2 uses:

n^(3/2) operations.

We want to see, as n grows, which algorithm uses fewer operations.

So we would want to first solve:

n*log(n) = n^(3/2)

This will give us the exact value of n such that the number of operations is the same in both algorithms.

dividing both sides by n we get:

log(n) = n^(3/2)/n = n^(3/2 - 1) = n^(1/2)

where we can use:

log(n) = ln(n)/ln(10)

ln(n) = ln(10)*n^(1/2)

This equation actually has no solutions.

This happens because the right side is always larger than the left side.

Then, the same thing happens for our two initial equations:

n^(3/2) is always larger than n*log(n), as you can see in the graph below, where n^(3/2)  is represented with the orange graph:

So we can conclude that the fist algorithm uses less operations as n grows.

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we conclude that the 'bait and switch' technique is not a promotional tactic used by a seller.

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From the given options, the 'bait and switch' technique is not a promotional tactic used by a seller.

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6 0
2 years ago
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2 years ago
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Answer:

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Answer:

x

=

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Explanation:

(

sin

x

)

2

=

3

(

cos

x

)

2

(

sin

x

)

2

=

3

(

1

−

(

sin

x

)

2

)

(

sin

x

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2

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(

sin

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2

4

(

sin

x

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2

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(

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4

sin

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±

(

√

3

2

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x

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(

2

π

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−

π

3

x

=

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3

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2

π

3

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4

π

3

,

5

π

3

If this was in the region

0

≤

x

≤

2

π

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3 years ago
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