Question

$4 \frac{1}{6 } + \frac{1}{7} \times (1 \frac{3}{4} ) {}^{2}$
помогите пожалуйста ​

Answer 1

$4\frac{1}{6}+\frac{1}{7}*(1\frac{3}{4})^{2} = 4\frac{1}{6}+\frac{1}{7}*(\frac{7}{4})^{2} = 4\frac{1}{6}+\frac{1}{7}*\frac{49}{16} = 4\frac{1}{6}+\frac{7}{16} = 4\frac{8}{48}+\frac{21}{48} = 4\frac{29}{48}$