Answer:
The World Wide Web Consortium provide a simple online tool (https://validator.w3.org/) that automatically check your HTML code and point out any problems/errors your code might have, such as missing closing tags or missing quotes around attributes.
Explanation:
your question is not clear
hope it helps
Answer:
def sum_1k(M):
s = 0
for k in range(1, M+1):
s = s + 1.0/k
return s
def test_sum_1k():
expected_value = 1.0+1.0/2+1.0/3
computed_value = sum_1k(3)
if expected_value == computed_value:
print("Test is successful")
else:
print("Test is NOT successful")
test_sum_1k()
Explanation:
It seems the hidden part is a summation (sigma) notation that goes from 1 to M with 1/k.
- Inside the <em>sum_1k(M)</em>, iterate from 1 to M and calculate-return the sum of the expression.
- Inside the <em>test_sum_1k(),</em> calculate the <em>expected_value,</em> refers to the value that is calculated by hand and <em>computed_value,</em> refers to the value that is the result of the <em>sum_1k(3). </em>Then, compare the values and print the appropriate message
- Call the <em>test_sum_1k()</em> to see the result
A terminology which best describe Kevin's goal in terms of expansion and contraction as needs change is: A. Scalability.
<h3>What is scalability?</h3>
Scalability can be defined as a measure of the ability of a system to change (expansion or contraction) in performance and cost as a result of changes in demands of application and system processing, especially in a network architecture.
In this context, we can logically deduce that a terminology which best describe Kevin's goal in terms of expansion and contraction as needs change is scalability.
Read more on scalability here: brainly.com/question/14301721
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Complete Question:
Kevin would like to ensure that his software runs on a platform that is able to expand and contract as needs change. Which one of the following terms best describes his goal? A. Scalability B. Elasticity C. Cost effectiveness D. Agility
Answer:
String word = "George slew the dragon";
int pos = word.indexOf("dr");
String drWord = word.substring(pos, pos+4);
System.out.println(drWord);
Explanation:
Assuming dr is always there, we don't have to check the validity of 'pos'. Normally, you would!
