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Nana76 [90]
3 years ago
10

Draw the sequence of BSTs that results when you insert the keys E, A, S, Y, Q, U, E, S, T, I, O, N, in that order into an initia

lly empty tree. And draw the sequence of BSTs that results when you delete the keys from the tree one by one in the order they were inserted. (Note: here your BST should allow duplicate keys).

Engineering
1 answer:
jek_recluse [69]3 years ago
8 0

Answer:

answer is attached

Explanation:

An important special kind of binary tree is the binary search tree (BST). In a BST, each node stores some information including a unique key value, and perhaps some associated data. A binary tree is a BST iff, for every node n in the tree:

All keys in n's left subtree are less than the key in n, and

all keys in n's right subtree are greater than the key in n.

Note: if duplicate keys are allowed, then nodes with values that are equal to the key in node n can be either in n's left subtree or in its right subtree (but not both). In these notes, we will assume that duplicates are not allowed.

Here are some BSTs in which each node just stores an integer key:

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Simplify the expression below:<br><br> 313 + 12 =
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ANSWER :

325

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313 + 12 = 325

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A block of copper (SL model) is subjected to a change of state. The process is isentropic (that is, entropy does not change. Whi
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To develop this problem it is necessary to apply the definitions of entropy change within the bodies

The change of entropy in copper would be defined as

\Delta S= \frac{\delta Q}{T}

Where,

Q= Heat exchange

T = Temperature

For an incompressible substance, the change in the heat exchange is defined as

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Replacing in our equation we have that

\Delta S = \frac{mcdT}{T}

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Since  \Delta S = 0, then

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In this way for the change of enthalpy and internal energy you have to

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3 years ago
Consider a Carnot cycle executed in a closed system with 0.0058 kg of air. The temperature limits of the cycle are300 K and 940
cricket20 [7]

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\sum Q=\sum W

\sum Q=Q_{1-2}+Q_{3-4}

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d(Q)=d(W)

Q_{1-2}=mRT_1\ln {\frac{P_1}{P_2}}

Given P_1=2000KPa

P_3=20KPa

\left (\frac{T_2}{T_3}\right )^{\frac{\gamma }{\gamma -1}=\left (\frac{P_2}{P_3}\right )}

P_2=1089.06K

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\left (\frac{T_1}{T_4}\right )^{\frac{\gamma }{\gamma -1}=\left (\frac{P_1}{P_4}\right )}

Now we have to find P_4=36.72KPa

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Q_{net}=0.95-0.303=0.646KJ

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True or false? two countermeasures to reducing risks are to reduce speed and apply time management.
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Answer:

True

Explanation:

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