Answer :
6.
To find the breadth of the rectangle, we will use the formula for the area of a rectangle, which is:
[tex]\[ \text{Area} = \text{Length} \times \text{Breadth} \][/tex]
Given:
- The area of the rectangle is [tex]\( 20 \frac{5}{7} \, \text{m}^2 \)[/tex].
- The length of the rectangle is [tex]\( 6 \frac{2}{3} \, \text{m} \)[/tex].
First, convert the mixed fraction for the area to an improper fraction:
[tex]\[ 20 \frac{5}{7} = 20 + \frac{5}{7} = \frac{140}{7} + \frac{5}{7} = \frac{145}{7} \][/tex]
Next, convert the mixed fraction for the length to an improper fraction:
[tex]\[ 6 \frac{2}{3} = 6 + \frac{2}{3} = \frac{18}{3} + \frac{2}{3} = \frac{20}{3} \][/tex]
With these fractions, we can now use the area formula to find the breadth. Rearrange the area formula to solve for the breadth:
[tex]\[ \text{Breadth} = \frac{\text{Area}}{\text{Length}} \][/tex]
Substitute the given values:
[tex]\[ \text{Breadth} = \frac{\frac{145}{7}}{\frac{20}{3}} = \frac{145}{7} \times \frac{3}{20} = \frac{145 \times 3}{7 \times 20} = \frac{435}{140} = \frac{87}{28} \approx 3.107142857142857 \, \text{m} \][/tex]
Thus, the breadth of the rectangle is approximately:
[tex]\[ 3.107142857142857 \, \text{m} \][/tex]
7.
To address the question involving the large can of milk, we should first determine the total volume of milk given.
Given:
- The large can contains [tex]\( 10 \frac{1}{2} \)[/tex] litres of milk.
Convert the mixed fraction to an improper fraction:
[tex]\[ 10 \frac{1}{2} = 10 + \frac{1}{2} = \frac{20}{2} + \frac{1}{2} = \frac{21}{2} = 10.5 \, \text{litres} \][/tex]
Now we can determine the necessary actions based on the requirement, which might further involve distributing the milk or any other desired operation. However, without the complete question, the exact action can't be accurately defined.
To find the breadth of the rectangle, we will use the formula for the area of a rectangle, which is:
[tex]\[ \text{Area} = \text{Length} \times \text{Breadth} \][/tex]
Given:
- The area of the rectangle is [tex]\( 20 \frac{5}{7} \, \text{m}^2 \)[/tex].
- The length of the rectangle is [tex]\( 6 \frac{2}{3} \, \text{m} \)[/tex].
First, convert the mixed fraction for the area to an improper fraction:
[tex]\[ 20 \frac{5}{7} = 20 + \frac{5}{7} = \frac{140}{7} + \frac{5}{7} = \frac{145}{7} \][/tex]
Next, convert the mixed fraction for the length to an improper fraction:
[tex]\[ 6 \frac{2}{3} = 6 + \frac{2}{3} = \frac{18}{3} + \frac{2}{3} = \frac{20}{3} \][/tex]
With these fractions, we can now use the area formula to find the breadth. Rearrange the area formula to solve for the breadth:
[tex]\[ \text{Breadth} = \frac{\text{Area}}{\text{Length}} \][/tex]
Substitute the given values:
[tex]\[ \text{Breadth} = \frac{\frac{145}{7}}{\frac{20}{3}} = \frac{145}{7} \times \frac{3}{20} = \frac{145 \times 3}{7 \times 20} = \frac{435}{140} = \frac{87}{28} \approx 3.107142857142857 \, \text{m} \][/tex]
Thus, the breadth of the rectangle is approximately:
[tex]\[ 3.107142857142857 \, \text{m} \][/tex]
7.
To address the question involving the large can of milk, we should first determine the total volume of milk given.
Given:
- The large can contains [tex]\( 10 \frac{1}{2} \)[/tex] litres of milk.
Convert the mixed fraction to an improper fraction:
[tex]\[ 10 \frac{1}{2} = 10 + \frac{1}{2} = \frac{20}{2} + \frac{1}{2} = \frac{21}{2} = 10.5 \, \text{litres} \][/tex]
Now we can determine the necessary actions based on the requirement, which might further involve distributing the milk or any other desired operation. However, without the complete question, the exact action can't be accurately defined.
