- a.Applying the order of operations, first per-
form the exponentiation. Then, compute the
quantities enclosed with parentheses. Finally,
compute the resulting quotient:
(3^2 + 6)( – 248) = (9 + 6)( –3) =
(15)( –3) = –5
- d.This one is somewhat more complicated
since we have an expression consisting of
terms within parentheses which are, in turn,
enclosed within parentheses, and the whole
thing is raised to a power. Proceed as follows:
(–2[1 – 2(4 – 7)])^2 = (–2[1 – 2( – 3)])^2
= (–2[1 – (–6)])^2
= (–2[7])^2
= (–14)^2
= 196
- b.Applying the order of operations, first com-
pute quantities enclosed within parentheses and
exponentiated terms on the same level from left
to right. Repeat this until all such quantities
are simplified. Then, multiply from left to right.
Finally, compute the resulting difference:
3(5 – 3)^2 – 3(5^2 – 3^2 ) = 3(2)^2 – 3(25 – 9) =
3(4) – 3(16) = 12 – 48 = –36
- a.Here we have an expression consisting of
terms within parentheses which are, in turn,
enclosed within parentheses. Proceed as
follows:
- (–2 – ( –11 – (– 3^2 – 5) – 2))=
–(–2 – (–11 – (–9 – 5) – 2))
= –(–2 –( –11 – (–14) – 2))
= –(–2 –(–11 + 14 – 2))
= –(–2 –(1))
= –(– 3)
= 3
28. c.Since h < 0, it follows that –h > 0. Since we
are also given that g > 0, we see that g– h= g+
(–h) is the sum of two positive numbers and
hence, is itself positive.
29. c.Observe that –g– h= –(g+ h). Since g < 0
and h< 0, it follows that g+ h< 0, so –(g+ h)
is positive.
30. d.First, note that since g < 0 and h< 0, it fol-
lows that g+ h must be negative, so –g– h=
–(g+ h) is positive. As such, we know that
–(g+ h) is larger than g+ h. Next, each of the
sums –g+ hand –g– hconsists of one positive
integer and one negative integer. Thus, while it
is possible for one of them to be positive, its
values cannot exceed that of –g– h since this
sum consists of two positive integers. As such,
we conclude that – g– h is the largest of the
four expressions provided.
31. c.First, note that since g < 0 and h< 0, it fol-
lows that g+ h must be negative and so, –g– h
= –(g+ h) is positive. As such, we know that
–(g+ h) is larger than g+ h. Next, each of the
sums –(g+ h) and g– hconsists of one posi-
tive integer and one negative integer. Thus,
while it is possible for one of them to be nega-
tive, its values cannot be smaller than g+ h
since this sum consists of two negative inte-
gers. As such, we conclude that g+ h is the
smallest of the four expressions provided.
32. d.First, note that since we are given that g < –2,
it follows that both gand – g^2 are negative, while
both –g and (–g)^2 are positive. Moreover, –g is
an integer larger than 2 (which follows by mul-
tiplying both sides of the given inequality by
–1). Squaring an integer larger than 2 produces
an even larger integer. As such, we conclude
that (–g)^2 is the largest of the four expressions
provided.
ANSWERS & EXPLANATIONS–
