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Arisa [49]
3 years ago
12

Check my first two question, what am I missing?

Mathematics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

  1. Node G has the highest degree (6), hence is the most central. Node B has the second highest degree (5), so is one of the most central. (You have also added nodes with degrees 4 and 3. I might not.)
  2. Node F has the same structure as Node D. (Your answer agrees.)
  3. See the right-most column of the attachment.

Step-by-step explanation:

1. Only one "actor" (G) has the highest degree, 6. That corresponds to selection E. Whether you include other actors in your list of "most central" depends on your interpretation of that term. My interpretation is that only the node with highest degree is of interest here.

2. Node that are structurally equivalent are ones with the same set of neighbors. Nodes D and F both have E and G as neighbors, so are structurally equivalent. Selection A is appropriate.

3. The problem seems to be to list the neighbors and add their degrees for any given actor. The attachment does that. The measure C(x) is listed in the rightmost column for each actor x listed in the leftmost column.

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x =  -  \frac{15}{7}




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1. Calculate the density of gold using the information given.
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For the given parallelogram, AP is an angle bisector of ∠A, AP ⊥ BP ,AB = 15 cm, and BP = 9 cm. Find the perimeter and the area
jeyben [28]

Answer:

Step-by-step Explanation:

Extend the sides AD and BP. This creates 2 congruent triangles using ASA. Triangles CBP and adjacent triangle are also congruent bc of ASA. This makes 3 congruent triangles (transitivity). That means just finding the area of the constructed large triangle, you find the area of the parallelogram. Since point P is the midpoint of B_, P_ also equals 9. Using the formula for the area of a triangle, and Pythagorean theorem (9-12-15), 12*18/2=108, which is the area of the parallelogram.

3 0
3 years ago
Consider the graph of the quadratic function y = 3x2 – 3x – 6. What are the solutions of the quadratic equation 0 = 3x2 − 3x − 6
son4ous [18]
Y = 3x^2 - 3x - 6 {the x^2 (x squared) makes it a quadratic formula, and I'm assuming this is what you meant...}

This is derived from:
y = ax^2 + bx + c

So, by using the 'sum and product' rule:

a × c = 3 × (-6) = -18

b = -3

Now, we find the 'sum' and the 'product' of these two numbers, where b is the 'sum' and a × c is the 'product':

The two numbers are: -6 and 3

Proof:

-6 × 3 = -18 {product}

-6 + 3 = -3 {sum}

Now, since a > 1, we divide a from the results

-6/a = -6/3 = -2

3/a = 3/3 = 1

We then implement these numbers into our equation:

(x - 2) × (x + 1) = 0 {derived from 3x^2 - 3x - 6 = 0}

To find x, we make x the subject of 0:

x - 2 = 0

OR

x + 1 = 0

Therefore:

x = 2

OR

x = -1

So the x-intercepts of the quadratic formula (or solutions to equation 3x^2 - 3x -6 = 0, to put it into your words) are 2 and -1.


We can check this by substituting the values for x:

Let's start with x = 2:

y = 3(2)^2 - 3(2) - 6
= 3(4) - 6 - 6
= 12 - 6 - 6
= 0 {so when x = 2, y = 0, which is correct}

For when x = -1:

y = 3(-1)^2 - 3(-1) - 6
= 3(1) + 3 - 6
= 3 + 3 - 6
= 0 {so when x = -1, y = 0, which is correct}
7 0
3 years ago
Read 2 more answers
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