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velikii [3]
2 years ago
6

Select the order in which the vertices are visited in a pre-order traversal of the tree shown below.Question 5 options:e, i, b,

h, f, a, c, d, ge, i, b, h, f, a, d, c, gb, i, f, h, a, d, g, c, eb, i, f, h, d, g, c, a, ePrevious PageNext Page
SAT
1 answer:
Aleksandr [31]2 years ago
3 0

Based on the vertices and the pre-order traversal order the tree would be visited is e, i, b, h, f, a, c, d, g.

<h3>What order will the tree be visited?</h3>

When using the pre-order traversal, the order would be from the root child node to the left child node to right child node.

The root node is the highest node which is e. From here you'll then proceed to the left roots first which means you'll go to i and then b.

From there you go to h and then f. The next nodes would be a and c.

Continuing from left to right, you'll then go to d and then g.

You'll therefore start from "e" and move to "i" and then "b" to "h" and then "f" and "a" to "c" and then to "" and finally to "g"

Find out more on the pre-order traversal at brainly.com/question/14559964.

#SPJ1

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A line passes through point (-8, -5) and has a slope of
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Answer:

-5x+4y=20

Explanation:

A line passes through point (-8, -5) and has a slope of  5/4

Write an equation in Ax+By=C form for this line.

Use integers for A, B, and C.​

So we have a point and a line we can use the point slope formula

y-y1=m(x-x1)

y-(-5)=(5/4)(x- -8)

y+5=(5/4)(x+8)

y=5/4x+10-5

y=5/4x+5

Then we multiply 4 both sides

4y=5x+20

-5x+4y=20

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What mass of glucose (C6H12O6) should be dissolved in 12. 0 kg of water to obtain a solution with a freezing point of -5. 8 ∘C?.
bogdanovich [222]

The mass of glucose solute dissolved in the solution is 6.739 Kg.

Recall that;

ΔT = K m i

ΔT = Freezing point depression

K =Freezing point depression constant = 1.86°C/mol

m = molality of the solution

i = Van't Hoff factor = 1 (molecular solution)

We have to find the freezing point depression from;

Freezing point depression = Freezing point of pure solvent - Freezing point of solution

Freezing point of pure water = 0°C

Freezing point of solution = -5. 8 ∘C

Freezing point depression = 0°C - (-5. 8 ∘C) = 5. 8 ∘C

Now;

m = ΔT/K i

m = 5. 8 ∘C/ 1.86°C/mol × 1

m = 3.12 m

But molality = number of moles of solute/mass of solvent in Kg

Molar mass of solute = 180 g/mol

Let the mass of solute be m

3.12 = m/180/12

3.12 = m/180 × 12

m = 3.12 × 180 × 12

m = 6739 g or 6.739 Kg

Learn more: brainly.com/question/6249935

8 0
2 years ago
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