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# ABLE MATH REVIEW

Write the value of the underlined digit.

ten thousands
thousands
hundreds
tens
ones
tenths
hundredths
thousandths
ten thousandths
hundred thousandths

1a. 326 __________ b. 16.56 __________ c. 34,687 __________ d. 3.097 __________

Round to the nearest hundred.

2a. 51 __________ b. 789 __________ c. 922 __________ c. 3,651 __________

Find each answer. (Watch the operation signs )

Simplify. (Reduce the fractions)

Add or subtract. Simplify. (Watch the operation signs)

Multiply or divide. Use cancellation when possible. Simplify.

Write each decimal as a fraction or mixed number.

8a. .5 = _______ b. 5.25 = _______ c. 7.02 = _______ d. .625 = _______

Write each fraction as a decimal.

Find each answer. Write zeros as needed.

Change each percent to a decimal and then to a fraction.

Change each fraction to a decimal and then to a percent.

13a. 65% of what number is 26

b. What percent of 50 is 10?

c. 41% of 320 is

d. 75% of 36 is

e. 20 is 80% of what number?

f. 18% of what number is 9

g. What percent of 100 is 33?

h. 240 is 80% of what number?

Simplify.

Evaluate each expression if m = -3, n = 6, and p = 4.

Solve.

1a) three hundreds
b) six hundredths
c) four thousands
d) zero tenths

# Answers to MATH 130 Notebook Problems and Remarks

Answers to Some of the Notebook Problems

The problems graded this time were
4.2 # 11, 15, 21
5.1 # 15, 19
5.2 # 11, 27, 45
5.3 # 11, 13
6.1 # 23, 25

4.2 #11 (a) GCF(280, 168) = GCF( 23 · 5 · 7, 23 · 3 · 7 ) = 23 · 7 = 56.
(b) GCF(12, 15, 125) = GCF( 22 · 3, 3 · 5, 53 ) = 1.

4.2 #15 (a) LCM(22, 56) = LCM( 2 · 11, 23 · 7 ) = 23 · 7 · 11 = 616.
(b) LCM(6, 38, 16) = LCM( 2 · 3, 2 · 19, 24 ) = 24 · 3 · 19 = 912.
(c) LCM(30, 42) = LCM( 2 · 3 · 5, 2 · 3 · 7 ) = 2 · 3 · 5 · 7 = 210.

4.2 #21 (a) LCM(7, 15, 81) = LCM( 7, 3 · 5, 34 ) = 34 · 5 · 7 = 2835 seconds ≈ 47.25
minutes, i.e., less than one hour.

(b) # of births = 2835/7 = +405, # of deaths = 2835/15 = −189, and # of immigrants
= 2835/81 = +35. So the net change of population was an increase of 251 over the period
of 2835 seconds.

(c) LCM(7, 15, 81, 900) = LCM( 7, 3 · 5, 34, 32 · 22 · 52 ) = 22 · 34 · 52 · 7 = 56, 700 seconds
= 945 minutes = 15 hours and 45 minutes.

5.1 #15 (a) 30° F + (−2 hours) · (−6° F/hour) = 30 + 12 = +42° F .
(b) 12° F + (−6 hours) · (+4° F/hour) = 12 − 24 = −12° F .

5.2 #23(a)
5.2 #23(b)

5.3 #11(f)

5.3 #11(g)

5.3 #11(h)

5.3 #13 The opposite of −4 is 4, and the reciprocal of −4 is −1/4.

6.1 #23 (a) A fraction a/b in lowest terms can be expressed as a terminating decimal
exactly when b has only 2’s and 5’s in its prime factorization. 7/17 can only be expressed,
therefore, as a repeating decimal.
· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·
Some Remarks on Pythagoras.

Last time we saw that certain triples of whole numbers (x, y, z) satisfy the relation
x2 + y2 = z2, namely the relation of the lengths of the sides of a right triangle given by
the Pythagorean Theorem. These were (3,4,5) and (5,12,13) and (8,15,17), and they are
called Pythagorean triples. How do we find other examples with whole numbers? One way
uses the following algorithm: if u and v are any whole numbers with no common factor (
GCF( u, v ) = 1) and not both are odd, then we can generate Pythagorean triples (x, y, z)
as follows:

x = u2 − v2,
y = 2uv,
y = u2 + v2 .

Thus, for example, (7,24,25) and (21,20,29) are two more Pythagorean triples.

Who was Pythagoras? Pythagoras (570-500 BC) was born in Samos, a Greek island
off the coast of what is now Turkey. It is believed that he traveled and studied in the
Persian empire which extended at that time from Greece to the Indus Valley and included
ancient Mesopotamia. In about 525 BC, Pythagoras founded a society in Crotone, Italy,
known as the ‘Pythagorean Brotherhood’. The latter was a mixture of political party and
religious cult. They believed in reincarnation and practiced vegetarianism. Pythagoras
taught that ‘all is number’, in the sense that everything could be understood in terms
of whole numbers and their ratios. The so-called Pythagorean Theorem had been known
much earlier, since it is known that it was understood by the Babylonians. It is probable
that Pythagoras or some member of the brotherhood gave a rigorous proof of it for the first
time. Ironically, someone discovered which probably was very disturbing and a tough
challenge to the philosophy of the Pythagoreans. The followers of Pythagoras did a fair
amount of work in arithmetic (perfect numbers, amicable numbers, figurative numbers),
geometry (regular solids), astronomy, and music – four subjects later forming the medieval
quadrivium. (In the university curriculum of the Middle Ages, these subjects were meant to
follow the trivium: the ‘trivial’ subjects of grammar, rhetoric, and logic.) It was no doubt
this school which established for the first time a systematic study of geometry, creating a
general theory of geometry where statements were backed up by proofs.

All information here is obtained from the book “The Heritage of Thales”, by W.S. Anglin
and J. Lambek (Springer-Verlag, 1995, chapter 7).
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# The Factor Tree

## 2.3 Order of Operations & Evaluating Algebraic Expressions ORDER OF OPERATIONS

We have discussed the mathematical operations of addition, subtraction, multiplication, division,
evaluating exponents, and finding roots. The majority of the problems to this point involved
only one of these operations. What happens when you have a problem that contains more than
one of these operations? Which operation do you perform first? One might have the tendency to
perform the operations in order from left to right, because that is how we read. This is not
necessarily the case. Not all languages are read from left to right, some are read top to bottom.
Mathematics is a universal language. Because of this, the Order of Operations were
established. The Order of Operations guarantee that the steps will be done in the same order
regardless the language you speak or the country you are in. The following is a table that
outlines the steps to the Order of Operations.

 ORDER OF OPERATIONS 1. Simplify within parentheses ( ) and other grouping symbols, such as brackets [ ], braces { }, or the fraction bar ––. (When more than one pair of grouping symbols occur within a problem, work the innermost set of grouping symbols first.) 2. Evaluate an expression involving exponents or roots . 3. Multiply or divide in order from left to right. 4. Add or subtract in order from left to right.

Example 1 Simplify 5 + 2•8

There are no parentheses, exponents or roots, so we
will begin with multiplication.

Example 2 Simplify 24 − (5 − 2)•6

Start by simplifying within the parentheses.
Multiply.
Subtract.

Example 3 Simplify

Start by simplifying within the parentheses.
Evaluate the exponents.
Multiply.

Example 4 Simplify 75 ÷5•3− 30 + 3

There are no parentheses, exponents, or roots. Begin
with multiplication and division. Perform these
operations in order from left to right. Because the
operation of division is on the left, do this first.
Multiply.
Add or subtract in order from left to right.

Example 5 Simplify 5•9 − (4 + 8) ÷ 2 − (7× 4)

Start by simplifying within the parentheses.
Multiply.
Divide.
Subtract. Work left to right.
Subtract.

Example 6 Simplify

The fraction bar serves as a grouping symbol.
Completely simplify the top and completely simplify
the bottom before you divide. Start by simplify within
the parentheses on the bottom.
Evaluate the exponent.
Multiply.
Add or subtract in order from left to right.
Divide.

Example 7 Simplify

Start by simplifying within the innermost parentheses.
Find the value of the root.
Subtract within the innermost parentheses.
Multiply
Subtract.
Multiply.

Example 8 Simplify

Start by simplifying within the innermost
Simplify within the brackets. Evaluate the
exponent.
Multiply within the brackets.
Subtract within the brackets.
Multiply within the braces.
Subtract within the braces.
Multiply.

AVERAGE
A common application problem for the Order of Operation is finding average. To find the
average, find the sum of your values, then divide that sum by the total number of values. For
example, Beth’s test scores are 97%, 81%, 91%, and 71%. Find her average test score.

Example 9 Find the average of 1, 9, –20, and –6.

Find the sum of all the values and divide that sum by
the number of values.
Divide.

Example 10 Find the average of 7, 9, 0, 15, –3, –8, and –6.

Find the sum of all the values and divide that
sum by the number of values.
Divide.

Example 11 Brian, the center for the Badgers basketball team, scored 25 points, 15
points, 10 points, 23 points, and 45 points in the last five basketball
games. Find his average points per game.

Find the sum of all the values and divide that
sum by the number of values.
Divide.

Brian averaged 23.6 points per game.

EVALUATING ALGEBRAIC EXPRESSIONS

We will now apply the Order of Operations when simplifying algebraic expressions.
An algebraic expression is any single variable or number or any grouping of variables and
numbers without an equal sign. Examples of algebraic expressions include:

To simplify an algebraic expression, replace the variable(s) with their given numerical value(s).
Follow the Order of Operations to simplify the expression. Finding the value of the expression is
also called evaluating the expression for the variable.

# SKETCH OF SOLUTIONS TO MATH HOMEWORK

Part I

2.

4. s = -3, t = 2. s and t are not unique: take s = 8 t = -5

16.

Using these equations we get:

2 = 26(34) - 7(126)

7. We may assume (why?) that and with
then and where
and . Since we get

Another proof that does not use the fundamental theorem of arithmetics.
Set d = gcd(a, b) and let a = a'd and b = b'd. Then gcd(a', b') = 1 (why?)
and lcm(a', b') = a'b' = ab/d2 (why?). Then lcm(a, b) = lcm(a'd, b'd) =
dlcm(a', b') = ab/d.

8. Since a | c, c = am for some m ∈ Z, also, since b | c and (since
(a, b) = 1) we must have by Euclid's lemma that b | m i.e. m = bn
therefore c = abn so ab | c. Example when gcd(a, b) ≠ 1: a = b = 2 and
c = 6.

14.

18. Suppose for some m ∈ Z, then

but this is impossible.

19. Suppose there are finitely many prime numbers and is their complete
list. Then, by exercise 18, is not divisible by any of these
primes. However, by the fundamental theorem of arithmetic it should be a
represented
as a product of these prime numbers — a contradiction.

26. By induction on n (using the second PMI)
Base:
Inductive step: Suppose then

Part II

13.- Suppose d = 1 then there exist s, t ∈ Z such that as+nt = 1 i.e. nt = 1-as
i.e 1 ≡ as mod n thus, s is a solution to ax ≡ 1 mod n.
Suppose now ax ≡ 1 mod n has a solution s, then as-1 = nk for some
k ∈ Z i.e. as - nk = 1 therefore gcd(a, n) = 1

28. n3- n = n(n2 - 1) = n(n - 1)(n + 1) = (n - 1)n(n + 1). Since n - 1, n
and n + 1 represent different cosets (mod 3) one of them must be the zero
coset (mod 3). Hence 3|(n3 -n). Also, one of these numbers must be even,
hence 2|(n3 - n). By exercise 8, since gcd(2, 3) = 1, 6|(n3 - n).

39. = 1 it works since (0, 6, 1, 8, 1, 2, 2, 1, 4, 1) · (10, 9, 8, 7, 6, 5, 4, 3, 2, 1) =
154 and 154 ≡ 0 mod 11

40. x ≡ 7 mod 11

46. (1) a ~ a: True since a - a = 0 ∈ Z
(2) a ~ b => b ~a: If a - b ∈ Z then b- a = -(a - b)∈ Z
(3) a ~b, b ~c => a ~c: If a-b ∈Z and b-c ∈ Z then (a-b)+ (b-c) ∈ Z
i.e. a - c ∈ Z
The equivalence classes are given by sets of the form r + Z, r ∈ R

47. No, -2 ~ 0 and 2 ~ 0 but

48. (1) a ~a: True since a + a = 2a which is even
(2) a ~ b => b a: If a + b = 2m then b + a = 2m
(3) a ~b, b ~c => a ~c: If a + b = 2m and b + c = 2k then a + c =
2m - b + 2k - b = 2(m + k - b)
There are two equivalence classes:
[0] = The set of all even integers.
[1] = The set of all odd integers.

22.1 Introduction

In grade 10, the basics of solving linear equations, quadratic equations, exponential equations
and linear inequalities were studied . This chapter extends on that work. We look at different

22.2 Solution by Factorisation

The solving of quadratic equations by factorisation was discussed in Grade 10. Here is an example
to remind you of what is involved.

 Worked Example 103: Solution of Quadratic Equations Question: Solve the equation 2x2 − 5x − 12 = 0. Answer Step 1 : Determine whether the equation has common factors This equation has no common factors. Step 2 : Determine if the equation is in the form ax2 + bx + c with a > 0 The equation is in the required form, with a = 2, b = −5 and c = −12. Step 3 : Factorise the quadratic 2x2 − 5x − 12 has factors of the form: (2x + s)(x + v) with s and v constants to be determined. This multiplies out to 2x2 + (s + 2v)x + sv We see that sv = −12 and s + 2v = −5. This is a set of simultaneous equations in s and v, but it is easy to solve numerically. All the options for s and v are considered below. We see that the combination s = 3 and v = −4 gives s + 2v = −5. Step 4 : Write the equation with factors (2x + 3)(x − 4) = 0 Step 5 : Solve the equation If two brackets are multiplied together and give 0, then one of the brackets must be 0, therefore 2x + 3 = 0 or x − 4 = 0 Therefore, or x = 4 Step 6 : Write the final answer The solutions to 2x2 − 5x − 12 = 0 are or x = 4.

It is important to remember that a quadratic equation has to be in the form ax2 + bx + c = 0
before one can solve it using these methods.

 Worked Example 104: Solving quadratic equation by factorisation Question: Solve for a: a(a - 3) = 10 Answer Step 1 : Rewrite the equation in the form ax2 + bx + c = 0 Remove the brackets and move all terms to one side. a2 - 3a - 10 = 0 Step 2 : Factorise the trinomial (a + 2)(a - 5) = 0 Step 3 : Solve the equation a + 2 = 0 or a - 5 = 0 Solve the two linear equations and check the solutions in the original equation. Step 4 : Write the final answer Therefore, a = -2 or a = 5 Worked Example 105: Solving fractions that lead to a quadratic equation Question: Solve for b: Answer Step 1 : Put both sides over the LCM Step 2 : Determine the restrictions The denominators are the same, therefore the numerators must be the same. However, b ≠ −2 and b ≠ −1 Step 3 : Simplify equation to the standard form Step 4 : Factorise the trinomial and solve the equation Step 5 : Check solutions in original equation Both solutions are valid Therefore, or b = 1

 Exercise: Solution by Factorisation Solve the following quadratic equations by factorisation. Some answers may be left in surd form.

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