Question

1. multiply. show your work for part b. you can use a generic rectangle or change to improper fractions. a. $\frac{1}{4}\cdot\frac{4}{9}$ b. $3\frac{2}{3}\cdot4\frac{5}{7}$

Explanation:

Step1: Multiply numerators and denominators for part a

For \(\frac{1}{4}\cdot\frac{4}{9}\), use the rule \(\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}\). So, \(a = 1\), \(b = 4\), \(c = 4\), \(d=9\). Then \(\frac{1\times4}{4\times9}\)

Step2: Simplify the fraction for part a

Cancel out the common factor of 4 in the numerator and denominator. \(\frac{1\times\cancel{4}}{\cancel{4}\times9}=\frac{1}{9}\)

Step3: Convert mixed numbers to improper fractions for part b

For \(3\frac{2}{3}\), use the formula \(a\frac{b}{c}=\frac{ac + b}{c}\). So \(3\frac{2}{3}=\frac{3\times3+ 2}{3}=\frac{9 + 2}{3}=\frac{11}{3}\). For \(4\frac{5}{7}\), \(4\frac{5}{7}=\frac{4\times7+5}{7}=\frac{28 + 5}{7}=\frac{33}{7}\)

Step4: Multiply the improper fractions for part b

\(\frac{11}{3}\cdot\frac{33}{7}=\frac{11\times33}{3\times7}\)

Step5: Simplify the fraction for part b

First, note that \(33\div3 = 11\). So \(\frac{11\times33}{3\times7}=\frac{11\times11}{1\times7}=\frac{121}{7}\)

Step6: Convert the improper fraction to a mixed number for part b (optional)

\(121\div7 = 17\) with a remainder of 2. So \(\frac{121}{7}=17\frac{2}{7}\)

Answer:

a. \(\frac{1}{9}\)
b. \(\frac{121}{7}\) (or \(17\frac{2}{7}\))