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Leona [35]
4 years ago
8

The longest_word function is used to compare 3 words. It should return the word with the most number of characters (and the firs

t in the list when they have the same length). Fill in the blank to make this happen.
Engineering
1 answer:
NARA [144]4 years ago
4 0

Answer:

len(word2) >= len(word1) and len(word2) >= len(word3):

Question with blank is below

def longest_word(word1,word2,word3):

   if len(word1) >= len(word2) and len(word1) >= len(word3):

       word = word1

   elif _________________________________________

       word = word2

   else:

       word = word3

   return word

print(longest_word("chair","couch","table"))

print(longest_word("bed","bath","beyond"))

print(longest_word("laptop","notebook","desktop"))

print(longest_word("hi","cat","Cow"))

Explanation

In line 1 of the code word1, word2, and word3 are the parameters used to for the defining the longest_word function. They will be replaced by 3 words to be compared. The code that is filled in the blank is len(word2) >= len(word1) and len(word2) >= len(word3): It is a conditional statement that is true only if the number of characters in the string of word2 is greater than or equal to word1 and word2 is greater than that of word3 .

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Answer:

Below is the required code.

Explanation:

%% Newton Raphson Method

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ep=10;

i=0;

cc=input('Condition of convergence:\n');

while ep>=cc

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temp=x;

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fprintf('x = %6f and error = %6f at iteration = %2f \n',x,ep,i);

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Command Window:

(a) First Root:

Initial guess:

1.5

Condition of convergence:

0.01

x = -1.815662 and error = 3.315662 at iteration = 1.000000

x = -0.644115 and error = 1.171547 at iteration = 2.000000

x = 0.208270 and error = 0.852385 at iteration = 3.000000

x = 0.434602 and error = 0.226332 at iteration = 4.000000

x = 0.451631 and error = 0.017029 at iteration = 5.000000

x = 0.451732 and error = 0.000101 at iteration = 6.000000

The solution x = 0.451732

>>

Second Root:

Initial guess:

3.5

Condition of convergence:

0.01

x = 3.300299 and error = 0.199701 at iteration = 1.000000

x = 3.305650 and error = 0.005351 at iteration = 2.000000

The solution x = 3.305650

>>

(b) Guess x=0.5:

Initial guess:

0.5

Condition of convergence:

0.01

x = 0.450883 and error = 0.049117 at iteration = 1.000000

x = 0.451732 and error = 0.000849 at iteration = 2.000000

The solution x = 0.451732

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Guess x=1.75:

Initial guess:

1.75

Condition of convergence:

0.01

x = 227.641471 and error = 225.891471 at iteration = 1.000000

x = 218.000998 and error = 9.640473 at iteration = 2.000000

x = 215.771507 and error = 2.229491 at iteration = 3.000000

x = 217.692636 and error = 1.921130 at iteration = 4.000000

x = 216.703197 and error = 0.989439 at iteration = 5.000000

x = 216.970438 and error = 0.267241 at iteration = 6.000000

x = 216.971251 and error = 0.000813 at iteration = 7.000000

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Guess x=3.0:

Initial guess:

3

Condition of convergence:

0.01

x = 3.309861 and error = 0.309861 at iteration = 1.000000

x = 3.305651 and error = 0.004210 at iteration = 2.000000

The solution x = 3.305651

>>

Guess x=4.7:

Initial guess:

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Condition of convergence:

0.01

x = -1.916100 and error = 1.051861 at iteration = 240.000000

x = -0.748896 and error = 1.167204 at iteration = 241.000000

x = 0.162730 and error = 0.911626 at iteration = 242.000000

x = 0.428332 and error = 0.265602 at iteration = 243.000000

x = 0.451545 and error = 0.023212 at iteration = 244.000000

x = 0.451732 and error = 0.000187 at iteration = 245.000000

The solution x = 0.451732

>>

Explanation:

The two solutions are x =0.451732 and 3.305651 within the range 0 < x< 5.

The initial guess x = 1.75 fails to determine the solution as it's not in the range. So the solution turns to unstable with initial guess x = 1.75.

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Answer:

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Complete question

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w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of  x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

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