Transcript 1.5

When the base includes more than one symbol, it is written in
parentheses.
Reading Math
A power includes a base and an exponent. The expression 23 is
a power of 2. It is read “2 to the third power” or “2 cubed.”
Ex 1: Write the expression in expanded form.
A.
(5z)2
B.
–s4
(5z)2
The base is 5z, and the
exponent is 2.
(5z)(5z)
5z is a factor 2 times. –(s  s  s  s)
–s4
The base is s, and the
exponent is 4.
s is a factor 4 times.
= –s  s  s  s
Ex 1:
C.
Write the expression in expanded form.
3h3(k + 3)2
3h3(k + 3)2
3(h)(h)(h) (k + 3)(k + 3)
There are two bases: h and k + 3.
h is a factor 3 times, and k + 3 is a factor 2
times.
Caution!
Do not confuse a negative exponent with a negative
expression.
Ex 2: Simplify the expression.
A.
3–2
The reciprocal of
.
B.
The reciprocal of
.
Ex 3: Simplify the expression. Assume all variables
are nonzero.
A.
3z7(–4z2)
3  (–4)  z7  z2
–12z7 + 2 = –12z9
B.
 y 3z 3

 z 53

3
(yz3 – 5)3 = (yz–2)3
y3(z–2)3
y3z(–2)(3)




 y 3 z 33 
 53 
 z

 y 3z 9 
  15 
 z 
3
 y 3z 9 
y
  15 6  
z6
z 
Scientific notation is a method of writing numbers by using
powers of 10. In scientific notation, a number takes a form m
 10n, where 1 ≤ m <10 and n is an integer.
Ex 4: Simplify the expression. Write the answer in scientific
notation.
B.
A.
(2.6  104)(8.5  107)
(2.6)(8.5)  (104)(107)
22.1  1011
3.0  10–11
2.21  1012
Ex 5: Light travels through space at a speed of about 3 
105 kilometers per second. Pluto is approximately
5.9  1012 m from the Sun. How many minutes, on
average, does it take light to travel from the Sun to
Pluto?
The speed of light in space is about 3  105 kilometers per
second.
The distance from the Sun to Pluto is 5.9  1012 meters.
First, convert the speed of light from
Use the relationship between time, distance, and speed to
find the number of minutes it takes light to travel from the
Sun to Pluto.
It takes light approximately 328 minutes to travel from the
Sun to Pluto.
Light travels at 3  105 km/s for 328(60) ≈ 19,666
seconds travels a distance of 5,899,560,000 =
5.89  109 km or 5.89  1012 m. The answer is
reasonable.