Sure, let's break down each part of your question step by step.
1. Multiplying [tex]\( \frac{5}{6} \)[/tex] and [tex]\( \frac{34}{5} \)[/tex]:
[tex]\[
\frac{5}{6} \times \frac{34}{5} = \frac{5 \times 34}{6 \times 5} = \frac{170}{30} \approx 5.6667
\][/tex]
So the result is approximately [tex]\( 5.6667 \)[/tex].
2. Finding the value of [tex]\( N \)[/tex] in the statement [tex]\( \frac{4}{7} \times \frac{63}{5} = N \)[/tex]:
[tex]\[
N = \frac{4}{7} \times \frac{63}{5} = \frac{4 \times 63}{7 \times 5} = \frac{252}{35} \approx 7.20
\][/tex]
So, [tex]\( N \approx 7.20 \)[/tex].
3. Multiplying [tex]\( \frac{2}{9} \)[/tex] and 4:
[tex]\[
\frac{2}{9} \times 4 = \frac{2 \times 4}{9} = \frac{8}{9} \approx 0.8889
\][/tex]
So the product is approximately [tex]\( 0.8889 \)[/tex].
4. Multiplying [tex]\( \frac{5}{8} \)[/tex] and [tex]\( \frac{3.4}{5} \)[/tex]:
[tex]\[
\frac{5}{8} \times \frac{3.4}{5} = \frac{5 \times 3.4}{8 \times 5} = \frac{17}{40} \approx 0.425
\][/tex]
So the product is approximately [tex]\( 0.425 \)[/tex].
5. Multiplying [tex]\( 4 \frac{5}{6} \)[/tex] and [tex]\( \frac{4}{5} \)[/tex]:
First, convert [tex]\( 4 \frac{5}{6} \)[/tex] to an improper fraction:
[tex]\[
4 \frac{5}{6} = \frac{4 \times 6 + 5}{6} = \frac{24 + 5}{6} = \frac{29}{6}
\][/tex]
Now multiply [tex]\( \frac{29}{6} \)[/tex] by [tex]\( \frac{4}{5} \)[/tex]:
[tex]\[
\frac{29}{6} \times \frac{4}{5} = \frac{29 \times 4}{6 \times 5} = \frac{116}{30} \approx 3.8667
\][/tex]
So the product is approximately [tex]\( 3.8667 \)[/tex].
