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a_sh-v [17]
3 years ago
15

Q#1: Suppose that you have two different algorithms for solving a problem. To solve a problem of size n, the first algorithm use

s exactly n*exp1-[n+2n+3n+4n]/nlne(10) operations and the second algorithm uses exactly n! operations. As n grows, which algorithm uses fewer operations?
Mathematics
1 answer:
svetlana [45]3 years ago
7 0

Answer:

The one of -[n+2n+3n+4n]/nlne(10) has fewer operations because the value of n is <u>static</u><u>.</u>

The one of n! ( factorial ) is factorised up to n∞ hence has infinity operations.

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Finding the average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11
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Answer:

The average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11) is 0.

Step-by-step explanation:

Given a function y, the average rate of change S of y=g(z) in an interval  (z_{s}, z_{f}) will be given by the following equation:

S = \frac{g(z_{f}) - g(z_{s})}{z_{f} - z_{s}}

In this problem, we have that:

g(z) = 4z^{2} - 5

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So z_{f} = 2, z_{s} = -2

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g(z_{s}) = g(-2) = 4*(-2)^{2} - 5 = 11

So

S = \frac{g(z_{f}) - g(z_{s})}{z_{f} - z_{s}} = \frac{11 - 11}{2 - (-2)} = \frac{0}{4} = 0

The average rate of change of g(z) =4x^2_5 between the points( -2,11) and (2,11) is 0.

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