Question

1. pentagons have 5 diagonals, as illustrated below. how many diagonals does the heptagon (7 sides) below have?

a. 7
b. 12
c. 14
d. 21
e. 28

1. after polling a class of 24 students by a show of hands, you find that 9 students play soccer and 21 students play basketball. given that information, what is the number of students in the class who must play both soccer and basketball?

a. 0
b. 1
c. 3
d. 6
e. 9

Explanation:

Question 19

Step1: Recall the formula for diagonals

The formula for the number of diagonals \(d\) of an \(n -\)sided polygon is \(d=\frac{n(n - 3)}{2}\).

Step2: Substitute \(n = 7\) into the formula

When \(n=7\), we have \(d=\frac{7\times(7 - 3)}{2}=\frac{7\times4}{2}\).

Step3: Calculate the value

\(\frac{7\times4}{2}=14\).

Step1: Use the principle of inclusion - exclusion

The principle of inclusion - exclusion states that \(|A\cup B|=|A|+|B|-|A\cap B|\), where \(|A|\) is the number of elements in set \(A\), \(|B|\) is the number of elements in set \(B\), \(|A\cup B|\) is the number of elements in the union of \(A\) and \(B\), and \(|A\cap B|\) is the number of elements in the intersection of \(A\) and \(B\).
Let \(A\) be the set of students who play soccer (\(|A| = 9\)) and \(B\) be the set of students who play basketball (\(|B|=21\)), and \(|A\cup B|\leq24\).

Step2: Rearrange the formula to solve for \(|A\cap B|\)

We can rewrite the formula as \(|A\cap B|=|A|+|B|-|A\cup B|\). Since \(|A\cup B|\leq24\), the minimum value of \(|A\cap B|\) occurs when \(|A\cup B|\) is maximum. The maximum value of \(|A\cup B|\) is \(24\).

Step3: Calculate \(|A\cap B|\)

Substitute \(|A| = 9\), \(|B| = 21\) and \(|A\cup B| = 24\) into the formula \(|A\cap B|=|A|+|B|-|A\cup B|\). Then \(|A\cap B|=9 + 21-24=6\).

Answer:

C. 14

Question 20