Top Formulas Every ACT Math Student Should Memorize

Exponents

Exponents are a fundamental part of algebra and are tested frequently on the ACT. Mastering exponent rules will help you simplify expressions and solve equations more efficiently.

Rules of Exponents

1. Product of Powers Rule

When multiplying two exponents with the same base:

a^m × a^n = a^(m+n)

Example:
2^3 × 2^4 = 2^(3+4) = 2^7 = 128

2. Quotient of Powers Rule

When dividing exponents with the same base:

a^m ÷ a^n = a^(m−n)

Example:
5^6 ÷ 5^2 = 5^(6−2) = 5^4 = 625

3. Power of a Power Rule

When raising an exponent to another exponent:

(a^m)^n = a^(m×n)

Example:
(3^2)^4 = 3^8 = 6561

4. Power of a Product Rule

When raising a product to an exponent:

(ab)^n = a^n × b^n

Example:
(2×5)^3 = 2^3 × 5^3 = 8 × 125 = 1000

5. Zero Exponent Rule

Any non-zero base raised to the zero power is 1:

a^0 = 1

Example:
7^0 = 1

6. Negative Exponent Rule

A negative exponent means the reciprocal of the base raised to the positive exponent:

a^(-n) = 1 / a^n

Example:
2^(-3) = 1 / 2^3 = 1/8

7. Fractional Exponents

A fractional exponent represents a root:

a^(1/n) = √[n]a
a^(m/n) = √n

Example:
8^(1/3) = ³√8 = 2
16^(3/4) = (⁴√16)^3 = 2^3 = 8

Application Tips

• Look for like bases before applying exponent rules.

• Simplify negative and fractional exponents carefully.

• Use your calculator to verify tricky computations.

Statistics

The ACT also includes several basic statistics problems that you can solve easily if you know these concepts.

Mean (Average)

Add up all the numbers and divide by how many numbers there are.

Mean = (sum of values) / (number of values)

Example:
For 5, 8, 10, 4
Mean = (5 + 8 + 10 + 4) / 4 = 27 / 4 = 6.75

Median

The middle number is when the list is sorted in order.

• If the count is odd, the middle value

• If the count is even, the average of the two middle values

Example 1 (odd count):
List = 3, 6, 9
Median = 6

Example 2 (even count):
List = 2, 4, 6, 8
Median = (4 + 6) / 2 = 5

Mode

The number that appears most often.

Example:
List = 2, 3, 3, 4, 5
Mode = 3

If two or more numbers appear most frequently, you have multiple modes.

Range

The difference between the largest and smallest values.

Range = Maximum − Minimum

Example:
List = 2, 4, 7, 8, 10
Range = 10 − 2 = 8

Standard Deviation (conceptual)

This measures how spread out the numbers are from the mean.

While you won’t typically calculate it manually on the ACT, you should understand that:

• Low standard deviation = values close to the mean.

• High standard deviation = values spread out from the mean.

Summary Tips for Statistics

• Use order and grouping to help find medians and modes.

• Always double-check data for multiple modes.

• Remember: standard deviation is about consistency or variability.

Linear Equations

Linear equations form straight lines and are used to model real-world relationships. You will be tested on different forms and interpretations.

Standard Form of a Linear Equation

Ax + By = C

• A, B, and C are constants

• Commonly used to rearrange equations.

• To graph: convert to slope-intercept form.

Example:
3x + 4y = 12

Solve for y:
4y = −3x + 12 → y = (−3/4)x + 3

Slope-Intercept Form

y = mx + b

• m is the slope (rise over run)

• b is the y-intercept (where the line crosses the y-axis)

Example:
y = 2x − 5
Slope: 2
Y-intercept: −5

Point-Slope Form

y − y₁ = m(x − x₁)

Used when you know a point on the line and the slope.

Example:
Point: (2, 3), Slope: 4
Equation: y − 3 = 4(x − 2)

Convert to slope-intercept form:
y = 4x − 5

Finding Slope Between Two Points

Given two points (x₁, y₁) and (x₂, y₂):

m = (y₂ − y₁) / (x₂ − x₁)

Example:
Points: (1, 2), (4, 5)
m = (5 − 2) / (4 − 1) = 3 / 3 = 1

Parallel and Perpendicular Lines

Parallel lines: same slope
Example: Line 1: y = 2x + 3, Line 2: y = 2x − 4 (parallel)

Perpendicular lines: negative reciprocal slopes
Example: Line 1: slope = 2, Line 2: slope = −1/2 (perpendicular)

Intercepts

• x-intercept: set y = 0 and solve for x

• Y-intercept: set x = 0 and solve for y

Example:
Equation: 2x + 3y = 6
x-intercept: y = 0 → 2x = 6 → x = 3
y-intercept: x = 0 → 3y = 6 → y = 2

Graphing Tips

• Identify the slope and intercepts clearly

• Use rise/run from the slope to draw additional points.

• Sketch straight lines through plotted points.

Real-World Example

Suppose a plumber charges a $50 service fee plus $40 per hour.
Equation: y = 40x + 50
Where x = number of hours, y = total cost

To find the cost for 3 hours:
y = 40(3) + 50 = 170

Final Notes for Part 1

• Exponent rules are essential for simplifying algebraic expressions.

• Basic statistics like mean, median, and mode are common and easy to master.

• Linear equations form a core part of algebra, and their graphing is frequently tested.

• Memorize slope formulas and practice converting between equation forms.

Quadratic Equations

Quadratic equations are polynomials of degree two. They often appear in problems involving area, projectile motion, or optimization. Understanding the various forms and methods to solve quadratics is essential for ACT success.

Standard Form of a Quadratic Equation

A quadratic equation is generally written as:

ax² + bx + c = 0

Where:

• A, b, and c are constants

• a ≠ 0

Example:
2x² + 3x − 5 = 0

Solving Quadratic Equations

There are three main methods:

1. Factoring

Used when the equation is easily factorable.

Example:
x² − 5x + 6 = 0
(x − 2)(x − 3) = 0 → x = 2 or x = 3

Not all quadratics can be factored neatly, so you may need another method.

2. Completing the Square

Turn the equation into a perfect square trinomial.

Example:
x² + 6x = −5
x² + 6x + 9 = 4 → (x + 3)² = 4 → x = −1 or −5

3. Quadratic Formula

Use when factoring is difficult or impossible:

x = [−b ± √(b² − 4ac)] / (2a)

Example:
x² − 4x − 5 = 0
a = 1, b = −4, c = −5
x = [4 ± √(16 + 20)] / 2 = [4 ± √36] / 2 = (4 ± 6)/2 → x = 5 or −1

Discriminant

The discriminant helps determine the nature of the roots:

D = b² − 4ac

• If D > 0 → two real solutions

• If D = 0 → one real solution

• If D < 0 → no real solution (complex numbers)

Square of a Sum or Difference

(a + b)² = a² + 2ab + b²
(a-b)² = a² − 2ab + b²

These are useful when expanding or simplifying expressions.

Example:
(2x + 3)² = 4x² + 12x + 9

Difference of Squares

a² − b² = (a + b)(a − b)

Example:
x² − 9 = (x + 3)(x − 3)

This is a fast way to factor special binomials.

Cubic Equations (Occasionally Tested)

While rare on the ACT, you might encounter cubic identities:

Sum of Cubes

a³ + b³ = (a + b)(a² − ab + b²)

Difference of Cubes

a³ − b³ = (a − b)(a² + ab + b²)

Example:
x³ + 8 = (x + 2)(x² − 2x + 4)

Sequences and Patterns

These problems test your ability to understand and predict patterns in number lists. ACT problems typically involve arithmetic or geometric sequences.

Arithmetic Sequences

Each term increases by the same fixed amount (common difference, d).

nth Term Formula

an = a1 + (n − 1)d

Where:

• An = value of the nth term

• a1 = first term

• n = term number

• d = common difference

Example:
First term = 2, d = 3
a5 = 2 + (5 − 1)×3 = 2 + 12 = 14

Sum of n Terms

sn = (n / 2)(a1 + an)

Example:
Find the sum of the first 5 terms: 2, 5, 8, 11, 14
s5 = (5 / 2)(2 + 14) = (5 / 2)(16) = 40

Geometric Sequences

Each term is multiplied by a fixed number (common ratio, r).

nth Term Formula

an = a1 × r^(n−1)

Example:
a1 = 3, r = 2
a4 = 3 × 2^3 = 3 × 8 = 24

Sum of n Terms (finite)

sn = a1 × [(1 − r^n) / (1 − r)] when r ≠ 1

Example:
a1 = 3, r = 2, n = 4
s4 = 3 × [(1 − 2^4) / (1 − 2)] = 3 × [(1 − 16) / (−1)] = 3 × 15 = 45

Identifying Sequences

On the ACT, patterns may not be stated directly. You may need to:

• Find the pattern using a few terms

• Identify whether it’s arithmetic (add/subtract) or geometric (multiply/divide)

• Calculate terms or a sum based on the formula.

Recursive Formulas

Some sequences are given in terms of the previous term.

Example:
a1 = 2, an = an−1 + 5

To find a3:
a2 = 2 + 5 = 7
a3 = 7 + 5 = 12

Understand the starting point and use previous terms iteratively.

Functions

Functions describe relationships between inputs and outputs, commonly represented by f(x).

Function Notation

f(x) = expression

Example:
f(x) = 3x + 2
Then f(2) = 3(2) + 2 = 8

Domain and Range

• Domain: all valid inputs (x-values)

• Range: all resulting outputs (y-values)

Watch for undefined operations like:

• Division by 0 (excluded from the domain)

• The square root of a negative number (excluded from the domain for real functions)

Basic Function Operations

Addition

(f + g)(x) = f(x) + g(x)

Example:
f(x) = x + 2, g(x) = 3x
(f + g)(x) = x + 2 + 3x = 4x + 2

Subtraction

(f-g)(x) = f(x) − g(x)

Example:
f(x) = 2x, g(x) = x²
(f-g)(x) = 2x − x²

Multiplication

(f × g)(x) = f(x) × g(x)

Example:
f(x) = x, g(x) = x + 1
(f × g)(x) = x(x + 1) = x² + x

Division

(f ÷ g)(x) = f(x) ÷ g(x)

Make sure g(x) ≠ 0 to avoid undefined values.

Composition of Functions

(f ∘ g)(x) = f(g(x))

This means plugging the output of g into f.

Example:
f(x) = 2x + 1, g(x) = x − 3
(f ∘ g)(x) = f(g(x)) = f(x − 3) = 2(x − 3) + 1 = 2x − 6 + 1 = 2x − 5

Inverse Functions

The inverse of f(x), written f¹ (x), reverses the role of x and y.

Steps to find an inverse:

1. Replace f(x) with y

2. Swap x and y

3. Solve for y

4. Replace y with f¹ (x)

Example:
f(x) = 2x + 1
y = 2x + 1 → x = 2y + 1 → x − 1 = 2y → y = (x − 1)/2
f⁻¹(x) = (x − 1)/2

Graphing Functions

• f(x) = x is a diagonal line through the origin

• f(x) = x² is a parabola

• f(x) = √x is a half curve starting from the origin

• f(x) = |x| is a V-shaped graph

Tips for Function Questions

• Always identify what the function notation is asking for

• Substitute carefully

• Review inverse and composite functions, as they often appear on the ACT..

• Quadratics can be solved by factoring, completing the square, or using the quadratic formula.

• Sequences follow patterns, either arithmetic (add/subtract) or geometric (multiply/divide).

• Functions define input-output relationships and can be combined or composed.

Geometry Equations

Geometry problems on the ACT may involve calculating area, perimeter, angles, or volume. You won’t be given a formula sheet during the test, so memorizing these is crucial.

Lines and Angles

Complementary and Supplementary Angles

• Complementary: Two angles that add up to 90°
  • Example: If one angle is 40°, the other is 50°

• Supplementary: Two angles that add up to 180°
  • Example: If one angle is 120°, the other is 60°

Vertical Angles

Vertical (opposite) angles are equal when two lines intersect.

• If angle A = 50°, the angle opposite it is also 50°

Adjacent Angles

Adjacent angles share a common side and vertex. If they form a straight line, they are supplementary.

Parallel Lines and Transversals

When a transversal crosses parallel lines, it forms several types of congruent angles:

• Corresponding angles are equal

• Alternate interior angles are equal.

• Alternate exterior angles are equal.

• Consecutive interior angles are supplementary.

Triangles

Triangle Angle Sum

The sum of the angles in any triangle is always 180°.

Types of Triangles

• Equilateral: all sides and angles are equal (each angle = 60°)

• Isosceles: two equal sides and two equal angles

• Scalene: all sides and angles are different

• Right triangle: one 90° angle

Pythagorean Theorem

In a right triangle:

a² + b² = c²

Where:

• A and B are the legs

• c is the hypotenuse (longest side)

Example:
a = 3, b = 4
c² = 9 + 16 = 25 → c = √25 = 5

Special Right Triangles

These are shortcut triangles commonly tested on the ACT:

• 45°−45°−90° triangle:
  • Legs are equal, hypotenuse = leg × √2

• 30°−60°−90° triangle:
  • Hypotenuse = 2 × short leg

  • Long leg = short leg × √3

Example:
Short leg = 5
Long leg = 5√3
Hypotenuse = 10

Triangle Inequality Theorem

The sum of any two sides must be greater than the third side.

If two sides are 5 and 7, the third side must be between 2 and 12 (not inclusive).

Polygons

Interior Angle Sum

The sum of the interior angles of an n-sided polygon:

Sum = (n − 2) × 180

Example:
Hexagon (6 sides): (6 − 2) × 180 = 720°

Measure of Each Interior Angle (Regular Polygon)

Each angle = [(n − 2) × 180] / n

Example:
Octagon (8 sides): [(8 − 2) × 180] / 8 = 135°

Exterior Angles

The sum of all exterior angles of any polygon = 360°

Each exterior angle of a regular polygon = 360 / n

Example:
Regular pentagon: 360 / 5 = 72°

Circles

Circles are frequently tested on the ACT in both area and coordinate form.

Parts of a Circle

• Radius (r): distance from the center to any point on the circle

• Diameter (d): 2 × radius

• Circumference: distance around the circle

• Area: space inside the circle

Circle Formulas

• Circumference = 2πr or πd

• Area = πr²

Example:
Radius = 4
Circumference = 8π
Area = 16π

Arc Length and Sector Area

If θ is the central angle in degrees:

• Arc length = (θ / 360) × 2πr

• Sector area = (θ / 360) × πr²

Example:
r = 6, θ = 90°
Arc length = (90/360) × 2π(6) = (1/4)(12π) = 3π
Sector area = (1/4)(π)(36) = 9π

Standard Form of a Circle (on Coordinate Plane)

(x − h)² + (y-k)² = r²

Where:

• (h, k) is the center

• R is the radius

Example:
(x − 3)² + (y + 2)² = 25
Center = (3, −2), Radius = √25 = 5

Quadrilaterals

Parallelogram

• Opposite sides and angles are equal

• Area = base × height

Rectangle

• All angles = 90°, opposite sides equal

• Area = length × width

• Diagonals are equal

Square

• All sides equal, all angles = 90°

• Area = side²

• Perimeter = 4 × side

Trapezoid

• One pair of parallel sides

• Area = ½ × (base1 + base2) × height

Example:
Base1 = 6, Base2 = 10, Height = 4
Area = ½ × (6 + 10) × 4 = 32

Rhombus

• All sides equal

• Diagonals bisect each other at 90°

• Area = ½ × diagonal1 × diagonal2

Perimeter and Area

Perimeter

Sum of the side lengths.

Example:
Square with side 4 → Perimeter = 4 × 4 = 16

Area (2D Shapes)

• Rectangle = l × w

• Triangle = ½ × base × height

• Circle = πr²

• Trapezoid = ½ × (base1 + base2) × height

Surface Area and Volume (3D Shapes)

Rectangular Prism

• Volume = l × w × h

• Surface area = 2(lw + lh + wh)

Cube

• Volume = s³

• Surface area = 6s²

Cylinder

• Volume = πr²h

• Surface area = 2πr² + 2πrh

Sphere

• Volume = (4/3)πr³

• Surface area = 4πr²

Cone

• Volume = (1/3)πr²h

• Surface area = πr² + πrl
  • l = slant height

Coordinate Geometry

Distance Formula

Distance between two points (x₁, y₁) and (x₂, y₂):

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Midpoint Formula

Midpoint between two points:

M = ((x₁ + x₂)/2, (y₁ + y₂)/2)

Slope Formula

m = (y₂ − y₁) / (x₂ − x₁)

Used to determine line steepness or check for parallel/perpendicular lines.

Equation of a Line

• Point-slope: y − y₁ = m(x − x₁)

• Slope-intercept: y = mx + b

Transformations and Symmetry

Transformations include:

• Translation: shifting a figure

• Reflection: flipping across a line

• Rotation: turning around a point

• Dilation: resizing

Symmetry:

• A figure has line symmetry if it can be folded and the two halves match.

• A circle has infinite lines of symmetry.

• Know angle relationships (complementary, supplementary, vertical, corresponding)

• Memorize area, perimeter, and volume formulas.

• Understand properties of special triangles and polygons.

• Practice coordinate geometry for circles, lines, and distan.ce

• Apply formulas to real ACT geometry problems efficiently.

Trigonometry Equations

Trigonometry on the ACT is generally limited to right triangles, trigonometric ratios, and simple identities. Mastering a few key formulas will help you solve a wide variety of problems quickly and accurately.

Basic Trigonometric Ratios

Trigonometry begins with understanding the relationships between angles and sides in a right triangle.

Given a right triangle with:

• Opposite side (opposite the angle θ)

• Adjacent side (next to the angle θ)

• Hypotenuse (longest side opposite the right angle)

The three basic trigonometric ratios are:

Sine (sin)
sin(θ) = opposite / hypotenuse

Cosine (cos)
cos(θ) = adjacent / hypotenuse

Tangent (tan)
tan(θ) = opposite / adjacent

To remember these:
SOHCAHTOA
(Sine = Opposite / Hypotenuse, etc.)

Example:

In a triangle with:

• Opposite = 3

• Adjacent = 4

• Hypotenuse = 5

Then:

• sin(θ) = 3/5

• cos(θ) = 4/5

• tan(θ) = 3/4

These ratios are often tested through direct calculation or by using them to find missing sides or angles.

Reciprocal Trig Functions

Less commonly tested, but good to know:

• Cosecant (csc) = 1/sin(θ) = hypotenuse / opposite

• Secant (sec) = 1/cos(θ) = hypotenuse / adjacent

• Cotangent (cot) = 1/tan(θ) = adjacent / opposite

Pythagorean Identity

This identity is derived from the Pythagorean Theorem and always holds:

sin²(θ) + cos²(θ) = 1

If you know one of the values, you can always find the other:

Example:
If sin(θ) = 0.6
Then cos²(θ) = 1 − (0.6)² = 1 − 0.36 = 0.64 → cos(θ) = 0.8

Trig Ratios of Special Angles

These values are common and should be memorized:

For 30°, 45°, and 60°:

Angle	sin	cos	tan
30°	1/2	√3/2	√3/3
45°	√2/2	√2/2	1
60°	√3/2	1/2	√3

These can often be found in triangle setups without a calculator.

Right Triangle Rules

In addition to SOHCAHTOA, there are a few geometric principles to remember:

1. 30°-60°-90° Triangle

• Short leg = x

• Long leg = x√3

• Hypotenuse = 2x

2. 45°-45°-90° Triangle

• Legs = x

• Hypotenuse = x√2

These triangles help simplify trigonometry problems without needing trigonometric tables or calculator approximations.

Law of Sines and Law of Cosines

While not frequently tested on the ACT, these laws may help with certain advanced triangle problems:

Law of Sines:
a / sin(A) = b / sin(B) = c / sin(C)

Law of Cosines:
c² = a² + b² − 2ab cos(C)

Only use these if the problem involves non-right triangles and you are given enough sides/angles.

Trigonometric Applications

ACT questions involving trigonometry may include:

• Finding missing side lengths in right triangles

• Solving for unknown angles using inverse trig functions

• Word problems involving angles of elevation or depression

• Using trig identities to simplify expressions

Example Problem:

A ladder leans against a wall, forming a 60° angle with the ground. If the ladder is 10 feet long, how high up the wall does it reach?

Use sin(60°) = opposite / hypotenuse
√3/2 = height / 10
height = 10 × √3/2 = 5√3 ≈ 8.66 feet

Formulas with Diagrams

Many ACT math questions include diagrams such as polygons, coordinate grids, graphs, or 3D shapes. It’s critical to extract relevant information from the visual and apply the correct formula.

Coordinate Geometry with Diagrams

Use the following tools when interpreting coordinate-based diagrams:

• Distance Formula:
  √[(x₂ − x₁)² + (y₂ − y₁)²]
  Helps find side lengths or distances between points.

• Midpoint Formula:
  ((x₁ + x₂)/2, (y₁ + y₂)/2)
  Used for bisecting segments or finding centers.

• Slope Formula:
  (y₂ − y₁) / (x₂ − x₁)
  Identifies line steepness, perpendicularity, or parallelism.

• Equation of a Circle:
  (x − h)² + (y − k)² = r²
  Recognize the center (h, k) and radius r from the diagrams.

Interpreting Graphs

ACT problems often include:

• Linear graphs: recognize y = mx + b

• Quadratic graphs: identify parabolas and their vertex

• Piecewise functions: understand how graphs change behavior across intervals

Graph reading tip:

If a graph is labeled, use those values rather than calculating. For example, the slope can be read as “rise over run” from two marked points.

Area and Volume from Diagrams

Sometimes diagrams don’t provide dimensions directly, but give enough to deduce them using:

• Triangles in squares

• Circle segments

• 3D views with depth or height marked

Always label:

• Known lengths

• Right angles

• Height vs. slant height

• Hidden sides or segments in 3D views

Shaded Region Problems

These often involve:

• Subtracting one area from another

• Using composite shapes (e.g., a rectangle minus a triangle)

• Identifying areas of semicircles or sectors

Example:

A circle is inscribed in a square. Find the area of the shaded region outside the circle.

• Square area = side²

• Circle area = πr² (r = half the side of the square)

• Shaded area = square area − circle area

If side = 6:
Square = 36
Circle = π(3)² = 9π
Shaded = 36 − 9π

Leave answers in terms of π if instructed, or approximate π = 3.14 if needed.

Diagrams with Angles

In some diagrams:

• Exterior angles of polygons are tested

• Vertical and alternate angles are given or implied.

• Interior triangle angles need to be calculated.

Make sure to:

• Extend lines if necessary

• Look for supplementary or complementary angles.

• Mark congruent or equal angles clearly

Real-World Applications

Some ACT problems use diagrams in practical settings:

• Maps (distance between points)

• Architecture (angles in support beams)

• Engineering (cross-sections of pipes or cones)

• Physics-style motion problems (angles of launch, height, etc.)

Focus on setting up the correct relationships and identifying what is being asked.

• Trigonometric ratios (sine, cosine, tangent) are essential for solving right triangle problems.

• Use known values for special angles and right triangle properties.

• Understand the Pythagorean identity and basic trig transformations.

• Be able to interpret and extract data from diagrams, coordinate grids, and graphs.

• Practice identifying relationships visually, such as congruent angles or corresponding side lengths.s

• Use spatial reasoning for shaded areas, composite shapes, and 3D interpretations.

Final Thoughts

Preparing for the ACT Math Test requires more than just memorizing formulas—it demands consistent practice, strategic thinking, and a solid understanding of mathematical concepts. By mastering key areas like algebra, geometry, trigonometry, and data analysis, you equip yourself to tackle the test with confidence. Focus on applying formulas, interpreting diagrams, and solving real-world problems under timed conditions. Use a calculator wisely, avoid common traps in questions, and always review your mistakes to identify patterns. With daily effort and a targeted study approach, significant improvement is not only possible—it’s expected. Stay consistent, practice with purpose, and trust in the process; your hard work will pay off on test day.