Suppose A = x ∈ R : −2 ≤ x

Question

Suppose A = x ∈ R : −2 ≤ x < 5 and B = (−1, 6]. Draw A and B on two separate number lines. Find A ∪ B and give the answer in interval notation. Write A ∩ B as one set using set builder notation.

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Question 1

1.1 Draw A and B on two separate number lines.
Set A: $A = \{x \in \mathbb{R} : -2 \le x < 5\}$
This represents an interval starting at -2 (inclusive) and ending at 5 (exclusive). On a number line, you would draw a closed circle at -2, an open circle at 5, and a line connecting them.

Set B: $B = (-1, 6]$
This represents an interval starting at -1 (exclusive) and ending at 6 (inclusive). On a number line, you would draw an open circle at -1, a closed circle at 6, and a line connecting them.

1.2 Find $A \cup B$ and give the answer in interval notation.

Identify the intervals for A and B.
$A = [-2, 5)$
$B = (-1, 6]$
Determine the union, which includes all elements present in either A or B.
The union starts at the smallest value in either set, which is -2 (from A).
The union ends at the largest value in either set, which is 6 (from B).
Since -2 is included in A, and 6 is included in B, both endpoints are included in the union.
$A \cup B = \boxed{[-2, 6]}$
1.3 Write $A \cap B$ as one set using set builder notation.
Identify the intervals for A and B.
$A = [-2, 5)$
$B = (-1, 6]$
Determine the intersection, which includes only elements common to both A and B.
The intersection starts at the largest of the lower bounds, which is -1 (since -1 is not included in B, it's not included in the intersection).
The intersection ends at the smallest of the upper bounds, which is 5 (since 5 is not included in A, it's not included in the intersection).
$A \cap B = (-1, 5)$
Write the intersection in set builder notation.
$A \cap B = \boxed{\{x \in \mathbb{R} : -1 < x < 5\}}$
Question 2

Calculate $5 \div \frac{3}{2} + 2 - \left(\frac{-4+9}{4}\right)$ without using a calculator.
Simplify the division and the expression inside the parenthesis.
$5 \div \frac{3}{2} = 5 \times \frac{2}{3} = \frac{10}{3}$ $\frac{-4+9}{4} = \frac{5}{4}$
Substitute these simplified values back into the original expression.
$\frac{10}{3} + 2 - \frac{5}{4}$
Find a common denominator for the fractions, which is 12.
$\frac{10}{3} + \frac{2}{1} - \frac{5}{4} = \frac{10 \times 4}{3 \times 4} + \frac{2 \times 12}{1 \times 12} - \frac{5 \times 3}{4 \times 3}$ $= \frac{40}{12} + \frac{24}{12} - \frac{15}{12}$
Perform the addition and subtraction.
$= \frac{40 + 24 - 15}{12} = \frac{64 - 15}{12} = \frac{49}{12}$
a) as an improper fraction
$\boxed{\frac{49}{12}}$
b) as a mixed fraction
Divide 49 by 12: $49 \div 12 = 4$ with a remainder of 1.
$\boxed{4 \frac{1}{12}}$
c) as a decimal fraction correct to two decimal places.
Convert the improper fraction to a decimal.
$\frac{49}{12} \approx 4.0833...$
Rounding to two decimal places:
$\boxed{4.08}$
Question 3

*a) A certain bank offers an interest rate of $12.5\%$ on a one-year fixed deposit and the interest is compounded at the end of the year.

Suppose A = x ∈ R : -2 ≤ x < 5 and B = (-1, 6]. Draw A and B on two separate number lines. Find A ∪ B and give the answer in interval notation. Write A ∩ B as one set using set builder notation.

Step-by-step solution

Identify the intervals for A and B.

Determine the union, which includes all elements present in either A or B.

Identify the intervals for A and B.

Determine the intersection, which includes only elements common to both A and B.

Write the intersection in set builder notation.

Simplify the division and the expression inside the parenthesis.

Substitute these simplified values back into the original expression.

Find a common denominator for the fractions, which is 12.

Perform the addition and subtraction.

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