howmany
7
5
kilos
aretherein2
7
1
2 \frac{1}{7} \: \div \: \frac{5}{7}2
7
1
÷
7
5
Convert 2 1/7 to improper fraction
(7 x 2) + 1 = 15 <--numerator
Copy the denominator, 7
2 1/7 = 15/7
To divide fractions, get the reciprocal of the dividend. Then multiply.
\frac{15}{7} \: \times \: \frac{7}{5} = \frac{105}{35}
7
15
×
5
7
=
35
105
Get the lowest term
105 and 35 is both divisible by 35
\frac{105}{35} \div \frac{35}{35} = \frac{3}{1} \: or \: 3
35
105
÷
35
35
=
1
3
or3
STEP 2
Since there are three 5/7 kilos, multiply it by the cost per 5/7 kilos which is 73.50
3 \times 73.50 = 220.503×73.50=220.50
Answers
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The angle between equi-potential surface and lines of force is always 90 degree.
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