Pre-Algebra Chapter 3 Review. Chapter 3 Review 1)Write each decimal. a) three thousand one and three thousandths b) four and three tenths.
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Pre-AlgebraChapter 3 Review
Chapter 3 Review
1) Write each decimal.
a) three thousand one and three thousandths
b) four and three tenths
Chapter 3 Review
1) Write each decimal.
a) three thousand one and three thousandths
3001.003
b) four and three tenths
4.3
Answer
Chapter 3 Review
2) Round to the indicated place.
a) -3.089, nearest hundredth
b) 238.079, nearest unit
c) 1587.2364, nearest hundred
Chapter 3 Review
2) Round to the indicated place.
a) -3.089, nearest hundredth -3.09
b) 238.079, nearest unit 238
c) 1587.2364, nearest hundred 1600
Answer
Chapter 3 Review
3) Compare. Use >, < or =
a) 0.12 < 0.2
b) -4.7 < -4.6
c) 5.080 = 5.08
Chapter 3 Review
3) Compare. Use >, < or =
a) 0.12 < 0.2
b) -4.7 < -4.6
c) 5.080 = 5.08
Answer
Chapter 3 Review
4) Evaluate each expression for
a = 3.02 and b = -12.3
a) a – b =
b) │a│ + │b│ =
Chapter 3 Review
4) Evaluate each expression for
a = 3.02 and b = -12.3
a) a – b = 15.32
b) │a│ + │b│ = 15.32
Answer
Chapter 3 Review
5) Solve.
a) a + 0.03 = 3.75
b) 0.5 = x / 33
Chapter 3 Review
5) Solve.
a) a + 0.03 = 3.75 a = 3.72
b) 0.5 = x / 33 x = 16.5
Answer
Chapter 3 Review
6) Solve.
a) 0.9853n = 98.53
b) -15 – 3(2x – 5) = 1.2
Chapter 3 Review
6) Solve.
a) 0.9853n = 98.53 n = 100
b) -15 – 3(2x – 5) = 1.2 x = - 0.2
Answer
Chapter 3 Review
7) Write an equation and solve.
A pound of peanuts costs $2.87.
How many pounds can be bought with $20?
Chapter 3 Review
7) Write an equation and solve.
A pound of peanuts costs $2.87.
How many pounds can be bought with $20?
2.87p = 20
p = 6.97 lbs.
Answer
Chapter 3 Review
8) Write an equation and solve.
Three hundredths times a number is equal to the opposite of three and one hundredth.
Chapter 3 Review
8) Write an equation and solve.
Three hundredths times a number is equal to the opposite of three and one hundredth.
0.03n = -3.01
n = -100.3
Answer
Chapter 3 Review
9) How much more mass does a 2 L bottle of water have than a 2 L bottle of gasoline?
(dgas = 0.68 g/cm3)
Chapter 3 Review
9) How much more mass does a 2 L bottle of water have than a 2 L bottle of gasoline?
(dgas = 0.68 g/cm3)
d = m/v m = dv
mwater = (1)(2000) = 2000 g
mgas = (0.68)(2000) = 1360 g
Difference = 2000 – 1360 = 640 g or 0.64 kg
Answer
Chapter 3 Review
10)An object falls for 10 sec.
Ignoring air resistance, what is its velocity?
(agravity = 9.8 m/sec2)
Chapter 3 Review
10)An object falls for 10 sec.
Ignoring air resistance, what is its velocity?
(agravity = 9.8 m/sec2)
v = at
v = (9.8)(10) = 98 m/sec
Answer
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