SAT 8
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48. D The largest angle of a triangle is opposite its longest side. Let C = the triangle’s largest angle. The side opposite C is side AB, and it measures 16 inches.
Recall that the Law of Cosines states: c2 = a2 + b2 − 2ab cos C,
where a, b, and c are the lengths of the sides of the triangle.
162 = 42 + 132 − 2(4)(13)cos C. 162 = 16 + 169 − 104 cos C.
71 = −104 cos C. cos C = − 10471 .
C= cos−1 − 10714 ≈ 133.1º.
49.C Recall that the absolute value of a complex
number is given by: a + bi = (a2 + b2 ). 4 + 2i = (42 + 22 ) = 20 = 2 5.
PART I / ABOUT THE SAT MATH LEVEL 2 TEST
50. E Using right triangle trigonometric ratios:
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Multiply the numerator and denominator by the LCD, c2b, to get:
a2b . c3
CHAPTER 3 / DIAGNOSTIC TEST |
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DIAGNOSE YOUR STRENGTHS AND WEAKNESSES
Check the number of each question answered correctly and “X” the number of each question answered incorrectly.
Algebra |
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Total Number Correct |
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10 questions |
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Solid Geometry |
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Total Number Correct |
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2 questions |
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Coordinate |
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Geometry |
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6 questions |
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Trigonometry |
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Total Number Correct |
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7 questions |
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Functions |
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Total Number Correct |
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15 questions |
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Data Analysis, |
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Statistics, |
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and Probability |
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Total Number Correct |
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4 questions |
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and Operations |
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Total Number Correct |
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6 questions |
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Number of correct answers − 14 (Number of incorrect answers) = Your raw score
___________________________ − 14 (_____________________________) =
36 |
PART I / ABOUT THE SAT MATH LEVEL 2 TEST |
Compare your raw score with the approximate SAT Math Test score below:
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SAT Math |
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Raw Score |
Approximate Score |
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Excellent |
43–50 |
770–800 |
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Very Good |
33–43 |
670–770 |
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Good |
27–33 |
620–670 |
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Above Average |
21–27 |
570–620 |
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Average |
11–21 |
500–570 |
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Below Average |
< 11 |
<500 |
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PART II
MATH REVIEW
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CHAPTER 4
ALGEBRA
This chapter provides a review of basic algebraic principles. On the Level 2 test, 48–52% of the questions relate to algebra and functions. That translates to about 20% of the test questions relating specifically to algebra and about 30% to functions. In reality, however, algebra is needed to answer nearly all of the questions on the test including coordinate geometry, solid geometry, and, especially, functions. The pie chart shows approximately how much of the Level 2 test is related directly to algebra.
Numbers and
Operations
12% Algebra
20%
Data Analysis,
Statistics, and
Probability
8%
Solid
Geometry
4%
Coordinate
Geometry
12%
Functions
30%
Trigonometry
14%
A wide variety of algebraic topics are covered in this chapter as an overall review of basic algebraic concepts. The topics are:
1.Evaluating Expressions
2.Fractions
a.Simplifying Fractions
b.Least Common Denominator
c.Multiplying Fractions
d.Using Mixed Numbers and Improper Fractions
e.Variables in the Denominator
39
40 |
PART II / MATH REVIEW |
3.Percentages
a.Converting Percentages to Decimals
b.Converting Fractions to Percentages
c.Percentage Problems
4.Exponents
a.Properties of Exponents
i.Rules of Exponents
b.Common Mistakes with Exponents
c.Rational Exponents
d.Negative Exponents
e.Variables in an Exponent
5.Real Numbers
a.Vocabulary
b.Properties of Real Numbers
i.Properties of Addition
ii.Properties of Multiplication
iii.Distributive Property
iv.Properties of Positive and Negative Numbers
6.Absolute Value
7.Radical Expressions
a.Roots of Real Numbers
b.Simplest Radical Form
c.Rationalizing the Denominator
d.Conjugates
8.Polynomials
a.Vocabulary
b.Adding and Subtracting Polynomials
c.Multiplying Polynomials
d.Factoring
i.Trinomials
ii.Difference of Perfect Squares
iii.Sum and Difference of Cubes
9.Quadratic Equations
a.Factoring
b.Quadratic Formula
c.Solving by Substitution
d.The Discriminant
e.Equations with Radicals
10.Inequalities
a.Transitive Property of Inequality
b.Addition and Multiplication Properties
c.“And” vs. “Or”
d.Inequalities with Absolute Value
11.Rational Expressions
a.Simplifying Rational Expressions
b.Multiplying and Dividing Rational Expressions
c.Adding and Subtracting Rational Expressions
d.Solving Equations with Rational Expressions
CHAPTER 4 / ALGEBRA |
41 |
12.Systems
a.Solving by Substitution
b.Solving by Linear Combination
c.No Solution vs. Infinite Solutions
d.Word Problems with Systems
13.*Binomial Theorem
EVALUATING EXPRESSIONS
Problems asking you to evaluate an expression represent the easiest of the algebra questions on the Math Level 2 test. To answer this type of question, simply substitute the value given for the variable.
E X A M P L E :
If x = n3 and n = 1 y, then find the value of x when y = 4.
2
Substitute y = 4 into the second equation to get n = 1 (4) = 2.
2
Now substitute n = 2 into the first equation to get x = 23 = 8.
The correct answer is 8.
FRACTIONS
Simplifying Fractions
Fractions are in simplest form when the numerator and denominator have no common factor other than 1. To simplify a fraction, factor both the numerator and denominator.
Don’t cancel terms that are not common factors. Below is a common mistake:
x2 − 4 ≠ x2 .
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x2 − 4 and 8 do not have a common factor, so this expression is already in simplest form.
E X A M P L E :
Simplify (3x + 12) .
(3x + 3y)
=3(x + 4) 3(x + y)
=(x + 4) x ≠ − y (x + y)
Factor the numerator and denominator. 3 is a common factor.
(This restriction is important because you cannot divide by zero!) (Answer)
*Denotes concepts that are on the Level 2 test only.
42 |
PART II / MATH REVIEW |
Least Common Denominator
The least common denominator (LCD) of two or more fractions is the least common multiple (LCM) of their denominators. To find the LCD:
1.Factor each denominator completely and write as the product of prime factors. (Factor trees are usually used for this.)
2.Take the greatest power of each prime factor.
3.Find the product of these factors.
E X A M P L E :
Find the LCD of 1 and 7 .
430
4 = 2 × 2 = 22. 30 = 2 × 3 × 5.
The greatest power of 2 is 22. The greatest power of 3 is 3, and the greatest power of 5 is 5.
22 × 3 × 5 = 60. |
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60 is the LCD. |
(Answer) |
The least common denominator is helpful when adding and subtracting
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The least common denominator is also used when simplifying complex fractions. A complex fraction is a fraction whose numerator or denominator contains one or more fractions. Find the LCD of the simple fractions and multiply the numerator and denominator of the complex fraction by it.
E X A M P L E : |
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We first start by finding a common denominator for the expression in the
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CHAPTER 4 / ALGEBRA |
43 |
Multiplying Fractions
a × c = ac
To multiply fractions, simply multiply straight across. b d bd .
E X A M P L E :
Simplify 2 × 3 × 4 .
5 6 7
=2 × 3 × 4 .
5 × 6 × 7
=. Divide by a common 210 factor to simplify.
=4 . (Answer)24
35
You can also simplify the fractions before multiplying to save time.
E X A M P L E : |
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=1 × 1 × 4 .
5 × 1 × 7
=4 .
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To divide by a fraction, multiply by its reciprocal. a ÷ c = a × d . This is b d b c
known as the Division Rule for Fractions. Of course, b, c, and d cannot equal zero because you cannot divide by zero.
E X A M P L E :
Simplify 18 ÷ 6 .
11
=18 × 11.
6
=3 × 11. Divide through by a common factor of 6. 1
= 3 × 11 = 33. |
(Answer) |
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Using Mixed Numbers and Improper Fractions
A mixed number represents the sum of an integer and a fraction.
3 1 = 3 + 1
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