Suppose there are n chairs in a row. We want to compute the number of ways to put 2 students into seats so that they are not nex
t to each other. Assume that students are interchangeable: e.g., if n = 3, the solution should be 1, because the only way to accomplish this is to use the first and last chairs. For both parts to this problem, you must explain why your answer is correct: it is not sufficient to compute a few values directly and look for a pattern.
a. Write a recurrence relation describing the number of ways to put the 2 students into seats so that they are not next to each other.
b. Repeat part a, but now suppose the chairs are in a circle (hint: it might help if you label the chairs 1, ... n, even though they are in a circle without a 'start' or 'end').
a. Write a recurrence relation describing the number of ways to put the 2 students into seats so that they are not next to each other.
b. Repeat part a, but now suppose the chairs are in a circle (hint: it might help if you label the chairs 1, ... n, even though they are in a circle without a 'start' or 'end').
1 answer:
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Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness = 92 Mpa√m
yield strength σ = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length = 1/π(
/ Yσ )²
we substitute
= 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
= 1/π( 92 Mpa√m / (517.5 Mpa )²
= 1/π( 0.177777 )²
= 1/π( 0.03160466 )
= 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection
Answer:21.3%
Explanation:
Given
80 % reduction in tool life
According to Taylor's tool life
=c
where V is cutting velocity
T=tool life of tool
80 % tool life reduction i.e. New tool Life is 0.2T
Thus
=1.213V
Thus a change of 21.3 %(increment) is required to reduce tool life by 80%