SAT 8

194	PART II / MATH REVIEW

E X A M P L E :

The statement “if John lives in Boston, then he lives in Massachusetts” is logically equivalent to which of the following?

I.If John lives in Massachusetts, then he lives in Boston.

II. If John does not live in Boston, then he does not live in Massachusetts.

III.If John does not live in Massachusetts, then he does not live in Boston.

(A)I only

(B)II only

(C)III only

(D)I and II only

(E)I, II, and III

The given statement is a conditional in the form “if p, then q” and it is a true statement. If you recall the properties of converse, inverse, and contrapositive statements, you can immediately determine that the contrapositive will also be true. The contrapositive is in the form “if not q, then not p” and coincides with answer III above.

If you don’t recall the properties of converse, inverse, and contrapositive statements, take a look at each statement individually. I and II are false statements so they cannot be logically equivalent to the given statement. The only possible answer is III.

The correct answer is C.

MATRICES

A matrix is a rectangular array of numbers written in brackets. Each number in a matrix is called an element. The dimensions of a matrix equal the number of rows by the number of columns, where the number of rows is always given first.

Two matrices are equal if the have the same dimensions and all corresponding elements are equal.

E X A M P L E :

What is the value of x, y, and z?

x − z 2x + z			−2 8
	− y		=	−5y 0
4		0

Set corresponding elements equal to each other to solve for the three variables. Let’s start with y.

4 − y = −5y

4 = −4y y = −1

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195

There are two elements written in terms of x and z. Set up a system to solve for x using the linear combination method to eliminate z.

x − z = −2

+ 2x + z = 8

3x

= 6

x = 2

Since x − z = −2, 2 − z = −2, so z = 4.

x = 2, y = −1, and z = 4.

(Answer)

To add matrices of the same dimension, simply find the sum of corresponding elements. Addition of matrices of different dimension is undefined.

A scalar is a real number in matrix terms. The scalar product of a real number n and the matrix X is the matrix nX. Each element of nX is n times its corresponding element in the given matrix X.

E X A M P L E :

Solve for the matrix X.

		1	0		1
X + 3				=				3
					2
		−2
			4		−1
								0

First, find the scalar product:

	3	0		1
			=				3
X +	−6			2
		12		−1
							0

Add the inverse of the second matrix to both sides.

				1					−3	0
	X =						3	+
				2
				−1					6	−12
							0
			−	5
	X =						3
				2						(Answer)
							−12	.
				5

The product of two matrices Xa × b and Yb × c is an a c matrix. Notice that the two matrices X and Y share a dimension of b. Two matrices can only be multiplied if the number of columns of the first matrix equals the number of rows of the second matrix. Multiplying a 2 2 matrix by a 2 1 matrix results in a 2 1 matrix with the following elements:

a b x		ax + by
		=
c d y		cx + dy

Matrix multiplication is not commutative.

The identity matrix is a matrix whose diagonal from the upper left corner to the lower right has elements of 1 and all other elements are 0.

196	PART II / MATH REVIEW

E X A M P L E :

A small shop has three full-time employees. The table below shows the number of hours worked by each employee during a given weekend. If James earns $8 an hour, Joe earns $7.50, and Celia earns $7, find a matrix expression that results in the total amount of money spent on salaries for each of the two days.

	Saturday	Sunday
James	8	4
Joe	6	6
Celia	4	2

You want to sum the three salaries for each day, so set up a matrix of the three hourly salaries.

[8 7.50 7]

Since this is a 1 × 3 matrix and the hours given in the table form a 3 × 2 matrix, their product will be a 1 × 2 matrix.

8	4
[8 7.50 7] 6	6	(Answer)
4	2

Although it was not asked in the problem, the product of the two matrices is:

[8(8) + 7.5(6) + 7(4) 8(4) + 7.5(6) + 7(2)] = [137 91]

This means that $137 was spent on salaries on Saturday and $91 was spent on salaries on Sunday.

:	a	b
	X =
	c	d

The determinant is a real number associated with each matrix. The determinant of a 2 × 2 matrix X is defined as:

det X =

a

b

= ad − bc

c

d

E X A M P L E :

Solve for x given

4 7

= 0

−2 x

4x − (−14) = 0

4x = −14

14

7

(Answer)

x = − 4 = −

2 .

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197

SEQUENCES

A sequence is a set of numbers listed in a certain order. You can also think of a sequence as a function whose domain is the set of consecutive positive integers. an is a term in the sequence, and n is the number of the term. Finite sequences have a limited number of terms, while infinite sequences continue indefinitely. Some examples of infinite sequences are as follows:

1,2,3,4,5, . . . n,. . .

2,4,6,8,10,. . . 2n . . .

3,9,27,81,243, . . . 3n , . . .

1,4,9,16,25, . . . n2. . .

1,4,7,10,13, . . . 3n − 2, . . .

In general, infinite sequences are in the form: a1, a2, a3, a4, a5, . . . an, . . . .

Arithmetic Sequences

If the difference between consecutive terms in a sequence is a constant, the sequence is an arithmetic sequence. The constant difference between terms is called the common difference and is represented by d. In a given arithmetic sequence a1, a2, a3, a4, a5, . . . , the difference between any two consecutive terms must equal d: a2 − a1 = d, a3 − a2 = d, a4 − a3 = d, etc.

The nth term of an arithmetic sequence can be found using the equation an = dn + c, where d is the common difference, n is the number of the term, and c is a constant. Alternately, the equation an = a1 + (n − 1)d (where a1 is the first term of the sequence) also represents the nth term of an arithmetic sequence.

E X A M P L E :

Find the nth term of the arithmetic sequence 6, 10, 14, 18, 22, . . . .

Because the sequence is arithmetic, there is a common difference between consecutive terms. d = 10 − 6 = 4. The nth term must be in the form:

an = dn + c. an = 4n + c.

Because a1 = 6, you can write an expression for the first term and solve for c.

6	= 4(1) + c.
2	= c.
The formula for the nth term is an = 4n + 2.		(Answer)

Notice the last example can also be solved using the equation an = a1 + (n − 1)d. Substituting a1 = 6 and d = 4, results in:

an = 6 + (n − 1)4.

an = 6 + 4n − 4 = 4n + 2.

198	PART II / MATH REVIEW

E X A M P L E :

How many numbers between 50 and 500 are divisible by 11?

Determine an arithmetic sequence to represent the numbers between 50 and 500 divisible by 11. The common difference between terms is 11. The first term of the sequence is 55 and the last is 495 (45 × 11 = 495).

an = a1 + (n − 1)d. an = 55 + (n − 1)11. 495 = 55 + (n − 1)11. 440 = (n − 1)11.

40 = n − 1.

n = 41 terms. (Answer)

Geometric Sequences

If the ratio between consecutive terms in a sequence is a constant, the sequence is a geometric sequence. The constant ratio between terms is called the common ratio and is represented by r. In a given geometric sequence a1, a2, a3, a4, a5, . . . , the ratio between any two consecutive terms

	must equal r:	a2	= r,	a3	= r,	a4	= r, etc. and r cannot equal zero.
		a		a		a
		1	2			3

The nth term of a geometric sequence can be found using the equation an = a1r n−1, where r is the common ratio, n is the number of the term, and a1 is the first term of the sequence. Using this formula, every geometric sequence can be written as follows:

a1, a2, a3, a4, a5,...an,... or
a a r1		, a r2 , a r3				, a r4			,...a r n−1,...
1,	1	1		1		1			1
E X A M P L E :
	Find the nth term of the geometric sequence: 3,									3	,	3	,	3	,	3	, . . .
										2		4		8
															16
	This is a geometric sequence because there is a common ratio.
	r =	3/2		3		1
				=		=		.
		3			6		2

Substitute the values r = 12 and a1 = 3 into the formula for the nth terms to get:

	a	= a r n−1.
	n		1
				1	n−1
	an = 3					(Answer)
				2
					.

E X A M P L E :

The second term of a geometric sequence is 12 and the 5th term is 1281 . Find the common ratio.

Start by writing the two given terms in an = a1r n−1 form.

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a2 = a1r1 =									1							and		a5 = a1r 4 =		1	.
								2											128
You can then write a5 in terms of a2.
	a2 × r × r × r = a5.
	1		× r		× r × r =											1	.
2															128
	1		r3 =			1				.
2					128
	r3 =				2			=			1			.
											64
128
	r = 3				1			=				1	.
											4
64
	1		.		(Answer)
4

*SERIES

A series is the sum of the terms of a sequence.
The sum of a finite arithmetic sequence is:	Sn =	n				+ an).
			(a1
		2
		1 − r n
The sum of a finite geometric sequence is:	Sn = a1						where r ≠ 1.
		1			− r
The sum of an infinite geometric sequence is: S =		a1				where −1 < r < 1.
		1 − r

E X A M P L E :

Find the sum of the terms in the sequence 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.

Of course, you could simply add these 15 terms to get an answer. A better way is to recognize that this is, in fact, an arithmetic sequence. n = 15, a1 = 2, and an = 30. Substitute these values into the formula for the sum of a finite arithmetic sequence to get:

Sn = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30

Sn = 2n (a1 + an )

Sn = 152 (2 + 30)

Sn = 152 (32)

Sn = 15(16) = 240

240.(Answer)

200	PART II / MATH REVIEW

E X A M P L E :

Find the sum of the integers from 1 to 200.

The integers from 1 to 200 form an arithmetic sequence having 200 terms. n = 200, a1 = 1, and an = 200. Substitute these values into the formula for the sum of a finite arithmetic sequence to get:

	Sn	= 1 + 2 + 3 + 4 + 5 + ... + 100
	Sn	=	n	(a1 + an )
			2
	Sn	=	200		(1 + 200)
			2
	Sn	= 100(201) = 20,100
	20,100.				(Answer)
	E X A M P L E :					7			49
	Find the sum of the geometric sequence: 1,							,			, . . .
						10			100

Divide the second term by the first term to determine the common ratio, r = 107 .

Since this is an infinite geometric sequence, find its sum by using the formula:

S =			a1
		1 − r
S =	1
	1 −							7
							10
S =		1
	3
	10
S =		10		.							(Answer)
	3

(v12

CHAPTER 10 / NUMBERS AND OPERATIONS

201

E X A M P L E :

10

Evaluate ∑5(−2)k .

k=1

The sigma notation reads “the sum of 5(−2)k for values of k from 1 to 10.” Essentially, you are finding the sum of the first 10 terms of a geometric series with a common ratio of −2. The series is: −10 + 20 − 40 + 80 − . . . + 5120.

		1 − rn
Sn = a1
			1 − r
S10 =	−10[1 − (−2)10 ]
			1 − (−2)
= −10(1 − 1024)
		3
= 3,410.			(Answer)

*VECTORS

A vector quantity is a quantity that has both size and direction. A vector is represented by an arrow whose length is proportional to the size of the vector quantity and whose direction is the direction of the vector quantity. For example, a vector can represent the path of a boat or plane. Equivalent vectors have the same size and direction.

If vector v is given by (v1, v2) and vector u is given by (u1, u2), then their resultant is:

(v1 + u1, v2 + u2). The norm is the length of a vector. The norm of vector v is

represented by v where v = = v22 ) . A vector whose norm equals one is the unit vector.

Two vectors v and u are perpendicular if their dot product equals zero. Their dot product is given by: v u = v1u1 + v2u2 = 0. It results in a real number, not another vector.

E X A M P L E :

Let v = (4, 5) and u = (−1, 2). What is the resultant of v and u? The resultant is the sum of v + u.

(v1 + u1, v2 + u2 ) (4 − 1,5 + 2)

(3, 7).

(Answer)

202	PART II / MATH REVIEW

E X A M P L E :

If v = (8, 2), then what is the norm of v?

v = (v12 + v22 ) = (82 + 22 ) = 68

2 17. (Answer)

E X A M P L E :

If u = (5, −2), find a vector perpendicular to u.

Let v = (x, y). For vectors to be perpendicular, their dot product must equal zero.

v . u = v1u1 + v2u2 = 0 5x − 2y = 0

One possible solution is v = (2, 5), since 5(2) − 2(5) = 0.

v = (2, 5).

(Answer)

*LIMITS

The limit of a function is the value that the function approaches as x approaches a given value. Limits occur in rational functions when the domain approaches an undefined value.

	E X A M P L E :
	If f(x) =	x2 − 9	, what value does the function approach as x approaches 3?
		x − 3

Of course, the function is undefined when x = 3 because the denominator cannot equal zero. There are many ways to go about solving this problem, so here are three methods:

(1)Factor the numerator and denominator. Then, simplify the expression and evaluate it when x = 3.

x2 − 9	=	(x − 3)(x + 3)	= x + 3
x − 3		x − 3

Substituting x = 3 results in 3 + 3 = 6.

(2) Substitute numbers close to 3 for x and evaluate the function.

Let x = 2.99. 2.992 − 9 = 5.99 2.99 − 3

(3) Graph the function on your calculator and see how the graph behaves around x = 3.

In any case, the function approaches 6 as x approaches 3.

(Answer)

CHAPTER 10 / NUMBERS AND OPERATIONS					203
E X A M P L E :
limx → 5		x3 − 6x2		+ 5x	=
		x2 − 25
Factor the numerator and denominator to get:
	x3 − 6x2 + 5x		=	x(x2 − 6x + 5)
	x2 − 25			(x − 5)(x + 5)

=x(x − 5)(x − 1) (x − 5)(x + 5)

=x(x − 1)

(x + 5)

Now, substitute x = 5 to find the limit as x approaches 5.

=	5(5 − 1)	=	20	= 2
			10
(5 + 5)
The limit as x approaches 5 is 2.					(Answer)
E X A M P L E :

If f (x)=	3x+ 3	, what value does the function approach as x approaches
	4x− 8

infinity?

Imagine that x is getting infinitely larger. You could substitute x = 1,000 or x = 10,000 into the function and evaluate it.

f (1,000) =	3(1000) + 3	= 0.75225
	4(1000) − 8

Or you could factor the numerator and denominator:

f (x) =	3x + 3		=	3(x + 1)
	4x − 8			4(x − 2)
The value of			(x + 1)		approaches 1 as x gets infinitely larger.
		(x − 2)
The function, therefore, approaches							3	.	(Answer)
			4