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EastWind [94]
3 years ago
6

(2, - 9),(- 1, 4)= Plssss helppp me rnnn

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
7 0
We need a picture of the problem... then we can help you.
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What is the degree of vertex B and G?
Marat540 [252]

Answer:

degree of vertex B = 2

degree of vertex g = 4

Step-by-step explanation:

Using given picture we need to find about what is the degree of vertex B and G.

In graph theory, we know that the degree (or valency) of a vertex of a graph is the number of edges incident to the vertex.

So we just need to count how many edges are incindent on vertex B and G.

From picture we see that number of edges incident on vertex B = 2

Hence degree of vertex B = 2

From picture we see that number of edges incident on vertex G = 4

Hence degree of vertex g = 4

6 0
3 years ago
A country's population in 1990 was 132 million. In 1998 it was 137 million. Estimate the population in 2003 using the exponentia
Likurg_2 [28]
140 million.I guess.
8 0
4 years ago
Which expression is equivalent to 3^12 • 7^9
TEA [102]
The answers are...

B. (3 ^3) ^9 * (7 ^3) ^6
D. (3 ^3 + 3 ^9) * (7 ^6 + 7 ^3)
5 0
3 years ago
Help please quickkkkkk
allsm [11]

Answer:

5 \frac25

Step-by-step explanation:

2 4/5 + 2 3/5 = 4 7/5 = 4+1+2/5 = 5 2/5

6 0
3 years ago
Read 2 more answers
For question 25 please pick 1,2,3 or 4
Juli2301 [7.4K]

Answer:

4.

$x=\frac{2}{3} \pm  \frac{1}{6}i \sqrt{158}$

Step-by-step explanation:

18x^2-24x+87=0

Using Quadratic Formula

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:18\cdot \:87}}{2\cdot \:18}$

Solving the discriminant:

\Delta = \left(-24\right)^2-4\cdot \:18\cdot \:87\\\Delta = 576-6264\\ \Delta=-5688

$x= \frac{24 \pm \sqrt{-5688} }{36} $

$x= \frac{24 \pm \sqrt{5688i} }{36} $

Once \sqrt{5688} =\sqrt{2^3\cdot \:3^2\cdot \:79}=6\sqrt{158}

$x=\frac{24\pm6\sqrt{158}i}{36}$

Dividing the denominator and numerator by 6

$x=\frac{4\pm \sqrt{158}i}{6}$

Now rewrite it:

$x=\frac{4}{6} \pm  i\frac{\sqrt{158}}{6}$

$x=\frac{2}{3} \pm i \frac{\sqrt{158}}{6}$

or

$x=\frac{2}{3} \pm  \frac{1}{6}i \sqrt{158}$

5 0
3 years ago
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