<span>Uphill distance = 1 4/5 = 9/5 = 18/10 km
Downhill Distance = 2 7/10 = 27/10 km
Total Uphill and Downhill = 18/10 + 27/10 = 45/10 = 4.5 km
Hence total level ground = Total race km - (Uphill + Downhill)
=10 - 4.5 = 5.5 km
Hence the answer is 5.5 km.</span>
The weight in 1980 is 
kilograms
<em><u>Solution:</u></em>
From 1980 to 1990, Lior’s weight increased by 25%
His weight is "k" kilograms in 1990
<em><u>To find: weight in 1980</u></em>
This is a percentage increase problem
Let "x" be the weight in kilograms in 1980
<em><u>The percentage increase is given by formula:</u></em>
Here,
Initial value in 1980 = x
Final value in 1990 = k
Percentage increase = 25 %
<em><u>Substituting the values in formula,</u></em>
Thus the weight in kilograms in 1980 is
100 kg becouse 1000 grams is egual 1 kg
Assume 0 < <em>x</em>/2 < <em>π</em>/2. Then
tan²(<em>x</em>/2) + 1 = sec²(<em>x</em>/2) ===> sec(<em>x</em>/2) = √(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - tan²(<em>x</em>/2))
===> cos(<em>x</em>/2) = 1/√(1 - <em>t</em> ²)
We also know that
sin²(<em>x</em>/2) + cos²(<em>x</em>/2) = 1 ===> sin(<em>x</em>/2) = √(1 - cos²(<em>x</em>/2))
Recall the double angle identities:
cos(<em>x</em>) = 2 cos²(<em>x</em>/2) - 1
sin(<em>x</em>) = 2 sin(<em>x</em>/2) cos(<em>x</em>/2)
Then
cos(<em>x</em>) = 2/(1 - <em>t</em> ²) - 1 = (1 + <em>t</em> ²)/(1 - <em>t</em> ²)
sin(<em>x</em>) = 2 √(1 - 1/(1 - <em>t</em> ²)) / √(1 - <em>t</em> ²) = 2<em>t</em>/(1 - <em>t</em> ²)
