Page 173 - ArithBook5thEd ~ BCC

Example 215. The formula for the distance s (in feet) that an object falls in a time t (in seconds) is

2
s =16t .
Use the formula to determine the distance an object falls in (a) 2seconds; (b) 4seconds.

Solution. We substitute t =2 and t = 4 into the formula. In 2 seconds the object falls

2
s =16(2) =64 feet,
while, in 4 seconds it falls
2
s =16(4) =256 feet.

Example 216. The formula which converts temperature in degrees Celsius ( C) to temperature in
◦
◦
degrees Fahrenheit ( F)is
9
F = C +32.
5
Find the temperature in F if a thermometer reads −10 C.
◦
◦
Solution. Substituting (−10) for C in the formula, we find
9
F = (−10) + 32
5
✚❃
9 ✚ 10 2
= − 1 · +32
✚ 5 ✚❃ 1
= −18 + 32 = 14.

The temperature is 14 F.
◦
Recall that the Pythagorean theorem is a relation between the side-lengths of a right triangle. If c
denotes the length of the hypotenuse (the long side opposite the right angle) and a and b the lengths
of the two shorter sides (legs), the relation is

2
2
2
a + b = c .
From this relation, we can extract three formulae:
!
2
2
a = c − b ;
!
2
2
b = c − a ;
!
c = a + b 2
2
The first two are formulae for the length of one leg in terms of the length of the other leg and the length
of the hypotenuse. The last is a formula for the length of the hypotenuse in terms of the lengths of the
two legs.

Example 217. The hypotenuse of a right triangle is 169 feet, and one its legs is119 feet. How long is
the other leg?

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