- b.The point A is exactly halfway between –2
and –3 on the number line; therefore, its value
is –2.5. - c.Since the digit in the thousandths place is 8,
we round the digit in the hundredths place up
by 1, resulting in 117.33. - b.We must determine the value ofn for which
1 n 00 300 = 400. The value ofn that satisfies
this equation is 133^13 . So, we conclude that
133 ^13 % of 300 results in 400. - d.Starting with 0.052, moving the decimal
place to the left one unit to obtain 0.0052 is
equivalent to dividing 0.052 by 10. Therefore,
0.0052 is smaller than 0.052. - c.The phrase “400% of 30” is equivalent to the
mathematical expression ^4100 ^00 30. Simplifying
this expression yields 120. - c.Note that x = ^38 = 0.375, which satisfies the
condition 0.34 < x< 0.40. It also satisfies the
condition 156 < x < 290 , which is seen by per-
forming the following two comparisons using
cross multiplication:
Comparison 1: 156 ^38 Cross multiplying yields
the true statement 40 < 48, so the original
inequality is true.
Comparison 2:^38 290 Cross multiplying yields
the true statement 60 < 72, so the original
inequality is true.- b.22.5% is equivalent to ^2120 . 05 , which is equal
to 0.225. - d.Note that ^25 = 0.40 and ^37 ≈0.42857. So,^37 is
not less than ^25 . - b.To see that – 0.01 < – 0.005, first convert
both to their equivalent fractional form:
–0.01 = – 1100 – 0.005 = – 10500 = – 2100 Next, we compare these two fractions. To this
end, note that – 1100 – 2100 is equivalent to 1100
2100 . Cross multiplying in the latter inequal-
ity yields the true statement 200 > 100, so the
inequality is true. Since –0.005 is clearly less
than 1.01, we conclude that –0.005 is between
–0.01 and 1.01.- b.Observe that ^58 – ^25 = ^2450 – ^1460 = 490 = 0.225.
- c.Observe that (3.09 1012 )3 = 3. 309 1012
= 1.03 1012. Alternatively, you could first
rewrite 3.09 1012 as 3,090,000,000,000 and
divide by 3 to obtain 1,030,000,000,000, which
is equivalent to 1.03 1012. - b.Move the decimal place to the right until
just after the first nonzero digit; each place
moved contributes an additional –1 power of
10. Doing so in 0.0000321 requires that we
move the decimal place 5 units to the right, so
that 0.0000321 is equivalent to 3.21 10 –5. - c.We must determine the value ofnfor which
1 n 00 ^89 = ^13 . Solve for n, as follows:
1 n 00 ^89 = ^13
1 n 00 = ^98 ^13 = ^38
n= 100^38 = ^308 ^0 = 37.5Thus, we conclude that 37.5% of^89 is ^13 .Set 6 (Page 10)- a.Apply the order of operations as follows:
–2(–3)^2 + 3(–3) – 7 = –2(9) –9 – 7 = –18 – 9 –7
= –34- b.Apply the order of operations as follows:
(–2)7(–2) (^2) + (–2) = 4 – 2–14 = –^142 = –7
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ANSWERS & EXPLANATIONS–