Question

1.12 homework

1. a number line, drawn to scale, is shown below. the numbers 0 and 1 are marked, as are two other numbers a and b.

1. a - 1
2. -b
3. |a - b|
4. |b|
5. a + b
6. b - a

step a: circle all the values that will have a positive answer
step b: which expression will be larger: iii or vi? explain how you know.

1. show your work:

find the sum of -5\frac{1}{2}+ -2\frac{1}{6}
a. -7\frac{2}{9}
b. 3\frac{1}{3}
c. -3\frac{1}{3}
d. -7\frac{2}{3}

Explanation:

Step1: Analyze expressions for Step A

From the number - line, \(a>1\), \(b < 0\).

• For \(a - 1\): Since \(a>1\), then \(a - 1>0\).
• For \(-b\): Since \(b < 0\), \(-b>0\).
• For \(|a - b|\): The absolute - value of a number is non - negative. Also, \(a>0\) and \(b < 0\), so \(a - b=a+( - b)>0\), then \(|a - b|>0\).
• For \(|b|\): The absolute value of any non - zero number is positive, and since \(b

eq0\), \(|b|>0\).

• For \(a + b\): \(a>1\) and \(b < 0\), but we don't know the magnitudes precisely. However, if \(|b|>a\), \(a + b<0\), if \(|b|0\).
• For \(b - a\): \(b<0\) and \(a>0\), so \(b - a=b+( - a)<0\).

The positive expressions are \(a - 1\), \(-b\), \(|a - b|\), \(|b|\).

Step2: Compare expressions for Step B

Expression iii is \(|a - b|\) and expression vi is \(b - a\). Since \(|a - b|=a - b\) (because \(a>b\)) and \(b - a=-(a - b)\), and \(a - b>0\), then \(|a - b|>b - a\).

Step3: Solve the sum for question 2

First, convert the mixed numbers to improper fractions. \(-5\frac{1}{2}=-\frac{5\times2 + 1}{2}=-\frac{11}{2}\) and \(-2\frac{1}{6}=-\frac{2\times6+1}{6}=-\frac{13}{6}\).
Find a common denominator, which is 6. Then \(-\frac{11}{2}=-\frac{11\times3}{2\times3}=-\frac{33}{6}\).
\(-\frac{33}{6}+(-\frac{13}{6})=-\frac{33 + 13}{6}=-\frac{46}{6}=-\frac{23}{3}=-7\frac{2}{3}\).

Answer:

1. Step A: Circle \(a - 1\), \(-b\), \(|a - b|\), \(|b|\)

Step B: Expression iii (\(|a - b|\)) is larger because \(|a - b|=a - b>0\) and \(b - a=-(a - b)<0\)

1. D. \(-7\frac{2}{3}\)