Answer :

First, rewrite both numbers as improper fractions.

[tex]3\frac{3}{7}=3+\frac{3}{7}=\frac{3\cdot7}{7}+\frac{3}{7}=\frac{21}{7}+\frac{3}{7}=\frac{24}{7}[/tex][tex]1\frac{3}{4}=1+\frac{3}{4}=\frac{1\cdot4}{4}+\frac{3}{4}=\frac{4}{4}+\frac{3}{4}=\frac{7}{4}[/tex]

Next, multiply both numbers. To do so, multiply the numerators and denominators sparatedly:

[tex]3\frac{3}{7}\times1\frac{3}{4}=\frac{24}{7}\times\frac{7}{4}=\frac{24\cdot7}{7\cdot4}[/tex]

Since the factor 7 appears both in the numerator and the denominator, we can cancel it out:

[tex]3\frac{3}{7}\times1\frac{3}{4}=\frac{24\cdot7}{7\cdot4}=\frac{24}{4}[/tex]

Since 24 divided by 4 is 6, then:

[tex]3\frac{3}{7}\times1\frac{3}{4}=6[/tex]