Question

for exercises 8 - 10, draw a diagram, find the appropriate b - 8. if 40 houses in a community all need direct lines to one another telephone service, how many lines are necessary? is that practical describe two models: first, model the situation in which direct lin house to every other house and, second, model a more practical 9. if each team in a ten - team league plays each of the other teams fo season, how many league games are played during one season? w figures can you use to model teams and games played? 10. each person at a party shook hands with everyone else exactly on 66 handshakes. how many people were at the party? review for exercises 11 - 19, identify the statement as true or false. for each false explain why it is false or sketch a counterexample. 11. the largest chord of a circle is a diameter of the circle. 116 chapter 2 reasoning in geometry

Explanation:

8.

Step1: Use combination formula

The number of lines (or connections) between \(n\) objects where each object is connected to every other object is given by the combination formula \(C(n,2)=\frac{n(n - 1)}{2}\). Here \(n = 40\).
\[C(40,2)=\frac{40\times(40 - 1)}{2}\]

Step2: Calculate the result

\[C(40,2)=\frac{40\times39}{2}= 780\]
A more practical model could be a central - switching model where all 40 houses are connected to a central telephone exchange. So instead of 780 lines, we would have 40 lines (one from each house to the exchange).

Step1: Apply combination formula

The number of games in a league where each of \(n\) teams plays each other team once is given by \(C(n,2)=\frac{n(n - 1)}{2}\). Here \(n=10\).
\[C(10,2)=\frac{10\times(10 - 1)}{2}\]

Step2: Compute the value

\[C(10,2)=\frac{10\times9}{2}=45\]
We can model teams as points and games as line - segments connecting the points.

Step1: Set up equation

Let the number of people at the party be \(n\). The number of hand - shakes is given by \(C(n,2)=\frac{n(n - 1)}{2}\), and we know it is 66. So we have the equation \(\frac{n(n - 1)}{2}=66\).
\[n(n - 1)=132\]
\[n^{2}-n - 132=0\]

Step2: Solve quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b=-1\), \(c=-132\)), we use the quadratic formula \(n=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(-1)^{2}-4\times1\times(-132)=1 + 528 = 529\). Then \(n=\frac{1\pm\sqrt{529}}{2}=\frac{1\pm23}{2}\). We get two solutions \(n_1=\frac{1 + 23}{2}=12\) and \(n_2=\frac{1-23}{2}=-11\). Since the number of people cannot be negative, we discard \(n_2\).

Answer:

780 lines are needed for the first model. A more practical model is a central - switching model with 40 lines.

9.