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Juli2301 [7.4K]
2 years ago
14

Suppose the numbers 7, 5, 1, 8, 3, 6, 0, 9, 4, 2 are inserted in that order into an initially empty binary search tree. The bina

ry search tree uses the usual ordering on natural numbers. What is the in-order traversal sequence of the resultant tree?.
Mathematics
1 answer:
tekilochka [14]2 years ago
8 0

In order traversal sequence of the resultant tree is 0 1 2 3 4 5 6 7 8 9.

<h3>What is tree transversal?</h3>

In this traversal method, the left subtree is visited first, then the root and later the right sub-tree. We should always remember that every node may represent a subtree itself.

If a binary tree is traversed in-order, the output will produce sorted key values in an ascending order.

Given:

7, 5, 1, 8, 3, 6, 0, 9, 4, 2 are inserted in that order into an initially empty binary search tree.

In order traversal:

In this traversal method, the left subtree is visited first, then the root and later the right sub-tree. We should always remember that every node may represent a subtree itself.

If a binary tree is traversed in-order, the output will produce sorted key values in an ascending order.

So, the sequence, 7, 5, 1, 8, 3, 6, 0, 9, 4, 2  are inserted in the order at the initially empty binary tree. And the binary search tree uses the usual ordering on natural numbers.

As, In-order traversal of a Binary search tree gives elements in increasing order.

Hence, the order of the elements in in order traversal is 0,1,2,3,4,5,6,7,8,9.

Learn more about in order traversal here:

brainly.com/question/12947940

#SPJ4

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What is the solution for x in the equation -4+5x-7=10+3x-2x in a fraction
Alex

Answer:

x = 21/4

Step-by-step explanation:

re-arrange the equation

-4+5x-7-3x = 10-2x (move 3x to the other side)

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5 0
4 years ago
A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A
julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

D^{2} = (90 - x)^{2} + 90^{2}   (1)

\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}

2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}  

D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}   (2)

Since:

D = \sqrt{(90 -x)^{2} + 90^{2}}

When x = 45 (the batter is halfway to first base), D is:

D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

100.62 \frac{d(D)}{dt} = -(90 - 45)*31          

\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

D^{2} = 90^{2} + x^{2}  

\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}

D\frac{dD}{dt} = x\frac{dx}{dt}   (3)

We have that D is:

D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63

By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:

\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s

Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.

I hope it helps you!

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