Question

20(d+1,4)=140

$\frac{1}{6}$(h-$\frac{4}{7}$)=$\frac{2}{3}$

$1\frac1{3}{4}$*(g-1,2)=5,5
$1\frac{2}{3}$*(5x-$\frac{1}{2}$)=$\frac{7}{9}$

Answer 1

1)

$20(d + 1.4) = 140$

$d + 1.4 = 7$

$d = 7 - 1.4$

$d = 5.6$

2)

$\frac{1}{6} (h - \frac{4}{7} ) = \frac{2}{3}$

$h = \frac{4}{7} = 4$

$h = 4 + \frac{4}{7}$

$h = \frac{32}{7}$

3)

$1 \frac{1}{3} 4 \times (g - 1.2) = 5.5$

$\frac{4}{3} \times (g - \frac{6}{5} ) = \frac{11}{2}$

$\frac{4}{3} g - \frac{8}{5} = \frac{11}{2}$

$40g - 48 = 165$

$40g = 165 + 48$

$40g = 213$

$g = \frac{213}{40}$

4)

$1 \frac{2}{3} \times (5x - \frac{1}{2} ) = \frac{7}{9}$

$\frac{5}{3} \times (5x - \frac{1}{2} ) = \frac{7}{9}$

$\frac{25}{3} x - \frac{5}{6} = \frac{7}{9}$

$150x - 15 = 14$

$150x = 14 + 15$

$150x = 29$

$x = \frac{29}{150}$



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