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3 years ago

Order of operation with Integers

Mathematics

1 answer:

vovikov84 [41]3 years ago

5 0

Answer:

Step-by-step explanation:

we have to follow PEDMAS.

what we have to do in each operation:

Parenthesis ~ 100 + 8 to the second power

Exponents ~ 10 to the 2nd power, 8 to the 2nd power, 2 to the 3rd power

Multiplication ~ none

Division ~ 10 to the 2nd power - (100 + 8 to the second power) divided by 2 to the 3rd power

Addition ~ 100 + 8 to the 2nd power

Subtraction ~ 10 to the 2nd power - (100+8 to the second power)

doing the work:

we start with P: but in what we have to do in the Parenthesis, there is an exponent and addition. we do the exponent first. 8 to the second power is 64. now we can add. 100 + 64 = 164.

E: we did the 2nd exponent, but we need the other 2 for later. 10 to the 2nd power is 100, and 2 to the 3rd power is 8.

we can skip M because there isn't any multiplication to do.

D: we have to do 10 to the 2nd power - (100 + 8 to the second power) first. 100-164= 64. now we can devide. 64 devided by 8 = 8

A: we already did the addition in the parenthesis part.

S: we already did the addition in the parenthesis part.

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3 years ago

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

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

$N = 15$

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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:

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