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zmey [24]
3 years ago
7

use the approximate operation count 2n^3 /3 for guassian elimination to estimate how much longer it takes to solve n equations i

n n unknowns if n is tripled
Mathematics
1 answer:
11111nata11111 [884]3 years ago
3 0

Answer:

the appropriate operation count is 27 times longer than the original one.

Step-by-step explanation:

Given the data in the question;

let us consider the approximate operation count for the system of n₁ equation in n₁ unknowns as; \frac{2}{3}n₁³ Or  

O₁ =  \frac{2}{3}n₁³

so if the value of the unknowns is tripled i.e, n₂ = 3n₁,

the appropriate operation count will be;

O₂  =  \frac{2}{3}n₂³

= \frac{2}{3}(3n_{1})³

= \frac{2}{3} × 27 × n₁³

= 27(  \frac{2}{3}n₁³  )

O₂ = 27(O₁)

Therefore, the appropriate operation count is 27 times longer than the original one.

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Answer:

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Step-by-step explanation:

We need to find the area of the frame itself. the radius of the inner circle is 12, but the frame has a radius of 4. You add 12 and 4 and you get 16. The formula for area of a circle is A=πr^2. So A= 3.14 x16^2. Then you have A= 3.14x 256 and you get 803.84. But we only want the area of the frame, so you have to get the area of the inner circle and subtract. You get A= 3.14 x 12^2. 3.14x144 is 452.16. 803.84-452.16 is 351.68. Hope this helped. :)

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