100 Math Concepts ACT
100 Math Concepts ACT
100 Math Concepts ACT
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
The factors of integer n are the positive integers 15. GREATEST COMMON FACTOR (GCF)
that divide into n with no remainder. The multi- To find the greatest common factor, break down
ples of n are the integers that n divides into with no both numbers into their prime factorizations and
remainder. 6 is a factor of 12, and 24 is a multiple take all the prime factors they have in common.
of 12. 12 is both a factor and a multiple of itself. 36 = 2 2 3 3, and 48 = 2 2 2 2 3.
11. PRIME FACTORIZATION What they have in common is two 2s and one 3, so
the GCF is = 2 2 3 = 12.
A prime number is a positive integer that has
exactly two positive integer factors: 1 and the inte-
ger itself. The first eight prime numbers are 2, 3, 5,
7, 11, 13, 17, and 19.
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
Ratios, Proportions, and Rates To find the average speed for 120 miles at 40 mph
and 120 miles at 60 mph, don’t just average the
36. SETTING UP A RATIO
two speeds. First figure out the total distance and
To find a ratio, put the number associated with the
the total time. The total distance is 120 + 120 =
word of on top and the quantity associated with
the word to on the bottom and reduce. The ratio 240 miles. The times are 3 hours for the first leg
20 5 and 2 hours for the second leg, or 5 hours total.
of 20 oranges to 12 apples is 12 which reduces to 3.
240
The average speed, then, is 5 48 miles per hour.
37. PART-TO-PART AND PART-TO-WHOLE
RATIOS
If the parts add up to the whole, a part-to-part
Averages
ratio can be turned into two part-to-whole ratios 41. AVERAGE FORMULA
by putting each number in the original ratio over To find the average of a set of numbers, add them
the sum of the numbers. If the ratio of males to up and divide by the number of numbers.
females is 1 to 2, then the males-to-people ratio is
Sum of the terms
1
1
Average =
Number of terms
1+2 = 3 and the females-to-people ratio is
2 2 2 To find the average of the five numbers 12, 15, 23,
= 3. Or, 3 of all the people are female.
1+2 40, and 40, first add them: 12 + 15 + 23 + 40 + 40
38. SOLVING A PROPORTION = 130. Then divide the sum by 5: 130 5 = 26.
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
Example: The charge for a phone call is r cents solve the two equations 4x 3y 8 and x y
for the first 3 minutes and s cents for 3, multiply both sides of the second equation by –3
each minute thereafter. What is the to get: –3x – 3y –9. Now add the equations; the
cost, in cents, of a call lasting exactly t 3y and the –3y cancel out, leaving: x = –1. Plug that
minutes? (t > 3) back into either one of the original equations and
you’ll find that y = 4.
Setup: The charge begins with r, and then
something more is added, depending 68. SOLVING AN EQUATION THAT
on the length of the call. The amount INCLUDES ABSOLUTE VALUE SIGNS
added is s times the number of min-
To solve an equation that includes absolute value
utes past 3 minutes. If the total num-
signs, think about the two different cases. For
ber of minutes is t, then the number
example, to solve the equation |x – 12| 3, think
of minutes past 3 is t – 3. So the charge
of it as two equations:
is r + s(t – 3).
x – 12 = 3 or x – 12 = –3
Intermediate Algebra x = 15 or 9
66. SOLVING A QUADRATIC EQUATION 69. SOLVING AN INEQUALITY
To solve a quadratic equation, put it in the To solve an inequality, do whatever is necessary to
ax2 bx c 0 form, factor the left side (if you both sides to isolate the variable. Just remember
can), and set each factor equal to 0 separately to that when you multiply or divide both sides by a
get the two solutions. To solve x2 12 7x, first negative number, you must reverse the sign. To
rewrite it as x2 7x 12 0. Then factor the left solve –5x + 7 < –3, subtract 7 from both sides to
side: get: –5x < –10. Now divide both sides by –5,
(x – 3)(x – 4) = 0 remembering to reverse the sign: x > 2.
x – 3 = 0 or x – 4 = 0 70. GRAPHING INEQUALITIES
x = 3 or 4
To graph a range of values, use a thick, black line
Sometimes the left side might not be obviously
over the number line, and at the end(s) of the
factorable. You can always use the quadratic for-
range, use a solid circle if the point is included or
mula. Just plug in the coefficients a, b, and c from
an open circle if the point is not included. The fig-
ax2 bx c 0 into the formula:
ure here shows the graph of –3 < x ≤ 5.
–b ±
b2–4ac
2a
–3 –2 –1 0 1 2 3 4 5
To solve x 2 4x 2 0, plug a = 1, b = 4, and
c = 2 into the formula:
Coordinate Geometry
–4
42–4
1
2
x= 71. FINDING THE DISTANCE BETWEEN TWO
21
POINTS
–4 ± 8
= = –2 ± 2 To find the distance between points, use the
2
Pythagorean theorem or special right triangles.
67. SOLVING A SYSTEM OF EQUATIONS
The difference between the x’s is one leg and the
You can solve for two variables only if you have two difference between the y’s is the other leg.
distinct equations. Two forms of the same equation
will not be adequate. Combine the equations in
such a way that one of the variables cancels out. To
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
e f
h g line 2
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
Triangles—General
3
80. INTERIOR ANGLES OF A TRIANGLE b° c°
c°
The three angles of any triangle add up to 180°.
4
100° 6
x° b°
a°
a° s
50°
In the figure above, x + 50 + 100 = 180, so x = 30. The triangles above are similar because they have
81. EXTERIOR ANGLES OF A TRIANGLE the same angles. The 3 corresponds to the 4 and the
6 corresponds to the s.
An exterior angle of a triangle is equal to the sum 3 6
= s
of the remote interior angles. 4
3s = 24
50°
s=8
A C
4 2
In the triangle above, 4 is the height when the 7 is
a°
chosen as the base.
1 1
c° Area = 2 bh = 2 (7)(4) = 14
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
Right Triangles
84. PYTHAGOREAN THEOREM 39
a
For all right triangles:
(leg1)2 + (leg2)2 = (hypotenuse)2
36
3
b
45° 45°
In the right triangle above, one leg is 30 and the q
hypotenuse is 50. This is 10 times 3-4-5. The
other leg is 40. If one leg is 3, then the other leg is also 3, and the
• 5-12-13 hypotenuse is equal to a leg times 2 , or 32.
If a right triangle’s leg-to-leg ratio is 5:12, or if
the leg-to-hypotenuse ratio is 5:13 or 12:13, then
it’s a 5-12-13 triangle and you don’t need to use
the Pythagorean theorem to find the third side.
Just figure out what multiple of 5-12-13 it is.
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
Other Polygons If PQRS is a square, all sides are the same length
as QR. The perimeter of a square is equal to four
86. SPECIAL QUADRILATERALS times the length of one side.
• Rectangle • Trapezoid
A rectangle is a four-sided figure with four right A trapezoid is a quadrilateral with one pair of
angles. Opposite sides are equal. Diagonals are parallel sides and one pair of nonparallel sides.
equal.
A E F
B
H G
D C
In the quadrilateral above, sides EF and GH are
Quadrilateral ABCD above is shown to have three parallel, while sides EH and GF are not parallel.
right angles. The fourth angle therefore also EFGH is therefore a trapezoid.
measures 90°, and ABCD is a rectangle. The
perimeter of a rectangle is equal to the sum of the 87. AREAS OF SPECIAL QUADRILATERALS
lengths of the four sides, which is equivalent to Area of Rectangle = Length Width
2(length + width). The area of a 7-by-3 rectangle is 7 3 = 21.
• Parallelogram
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A parallelogram has two pairs of parallel sides. A B
Opposite sides are equal. Opposite angles are
equal. Consecutive angles add up to 180˚. 3
110° 70° D C
L 6
In the figure above, s is the length of the side M
opposite the 3, so s = 3.
5
4
• Square
A square is a rectangle with 4 equal sides. K N
P Q
S R
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
T U
H E
G F
5
base1 + base2
Area of Trapezoid = (
height ) Circles
2
89. CIRCUMFERENCE OF A CIRCLE
Think of it as the average of the bases (the two
parallel sides) times the height (the length of the Circumference of a circle = 2πr
perpendicular altitude).
2 3
A B
13
5
D C
14
In the trapezoid ABCD above, you can use side AD Here, the radius is 3, and so the circumference is
2 + 14 2π(3) = 6π.
for the height. The average of the bases is 2 = 8,
so the area is 5 8, or 40. 90. LENGTH OF AN ARC
88. INTERIOR ANGLES OF A POLYGON An arc is a piece of the circumference. If n is the
The sum of the measures of the interior angles measure of the arc’s central angle, then the formula
of a polygon is (n – 2) 180, where n is the is:
number of sides. n
Length of an Arc = (2πr)
360 ( )
Sum of the angles = (n – 2) 180 degrees A
5
The eight angles of an octagon, for example, add B
up to (8 – 2) 180 = 1,080. O 72°
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
4
4
3
7
6
30°
5
6
4
In the figure above, the radius is 6 and the meas-
ure of the sector’s central angle is 30˚. The sector The volume of a 4-by-5-by-6 box is 4 5 6 =
30 1
has
360 or 12 of the area of the circle:
120
A cube is a rectangular solid with length, width,
( ) (π) (6 ) = ( ) (36π) = 3π
30
360
2 1
12 and height all equal. If e is the length of an edge of
a cube, the volume formula is:
Solids
Volume of a Cube = e 3
93. SURFACE AREA OF A RECTANGULAR SOLID
The surface of a rectangular solid consists of 3
pairs of identical faces. To find the surface area,
find the area of each face and add them up. If the
length is l, the width is w, and the height is h, the
formula is: 2
2
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
Opposite
95. VOLUME OF OTHER SOLIDS Sine =
Hypotenuse
Adjacent
Volume of a Cylinder = πr2h Cosine =
Hypotenuse
Opposite
Tangent =
Adjacent
B
5
17
8
2
A C
15
1 Adjacent
The volume of a cone where r = 3, and h = 6 is: Cotangent =
Tangent = Opposite
1
Volume = 3π(32)(6) = 18
1 Hypotenuse
4 Secant =
Cosine = Adjacent
Volume of a Sphere = πr3
3
If the radius of a sphere is 3, then: 1 Hypotenuse
4
Cosecant =
Sine = Opposite
Volume = 3 π(33) = 36π
D
Trigonometry
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96. SINE, COSINE, AND TANGENT OF ACUTE 2
ANGLES
To find the sine, cosine, or tangent of an acute E F
3
angle, use SOHCAHTOA, which is an abbreviation
for the following definitions: In the figure above:
2
cot D = 3
13
sec D = 2
13
csc D = 3
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ACT STUDY AIDS
100 Key Math Concepts for the ACT
98. TRIGONOMETRIC FUNCTIONS OF OTHER Because the triangle shown in the figure above is
ANGLES a 30-60-90 right triangle, we can determine that
3 1
To find a trigonometric function of an angle the coordinates of point P are – 2, – 2. The sine
1
greater than 90°, sketch a circle of radius 1 and is therefore – 2.
centered at the origin of the coordinate grid. Start
from the point (1, 0) and rotate the appropriate 99. SIMPLIFYING TRIGONOMETRIC
number of degrees counterclockwise. EXPRESSIONS
To simplify trigonometric expressions, use the
inverse function definitions along with the funda-
y mental trigonometric identity:
–1
210°
O x
1 The figure above shows a portion of the graph of
P(a, b) y = sin x.
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