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2 years ago

Hi can someone solve the inequalities?

Mathematics

1 answer:

vodomira [7]2 years ago

8 0

8y - 5 < 3

8y < 5+3

8y<8

y<1

3p + 2 > -10

3p>-10-2

3p> -12

p>-4

-2 > m/6 -7

- m/6>+2-7

Multiply each member by 6

-m >-42+12

-m>-30

m<30

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3 0

3 years ago

Read 2 more answers

A) Find the value of N if, Kn has 105 edges.

blagie [28]

The number of edges can be calculated from the number of vertices.

There are 14 vertices for 105 edges
There are 200 vertices for 19900 edges

The variable N is used to always represent the number of vertices.

So, we represent the edges as:

$E \to Edges$

(a) The value of N for 105 edges

The relationship between N and E is:

$E = \frac{N \times (N - 1)}{2}$

Substitute 105 for E

$105 = \frac{N \times (N - 1)}{2}$

Multiply through by 2

$210 = N \times (N - 1)$

$210 = N^2 - N$

Rewrite as:

$N^2 - N - 210 = 0$

Expand

$N^2 +14N - 15N - 210 = 0$

Factorize

$N(N +14) - 15(N + 14) = 0$

Factor out N + 14

$(N - 15) (N + 14) = 0$

Solve for N

$N = 15$ or $N = -14$

The number of vertices (N) cannot be negative. So:

$N = 15$

(b) The value of N for 19900 edges

We have:

$E = \frac{N \times (N - 1)}{2}$

Substitute 19900 for E

$19900 = \frac{N \times (N - 1)}{2}$

Multiply through by 2

$39800 = N \times (N - 1)$

$39800= N^2 - N$

Rewrite as:

$N^2 - N - 39800= 0$

Expand

$N^2 +199N - 200N - 39800= 0$

Factorize

$N(N +199) - 200(N + 199) = 0$

Factor out N + 199

$(N + 199) (N - 200) = 0$

Solve for N

$N = 200$ or $N = -199$

The number of vertices (N) cannot be negative. So:

$N = 200$

Hence, there are 200 vertices for 19900 edges

Read more about vertices and edges at:

brainly.com/question/22118318

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2 years ago

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andreyandreev [35.5K]

The answer is to 50years

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2 years ago