Formula and notation reference

Use the guidance first, attempt each worked example, then open its solution. Practice sets and timed rehearsals keep their original boundaries.

6.1 Distinguish supplied information from knowledge to retrieve

The first reference page gathers the relationships shown on the standard SAT Math reference sheet in an original layout. It is the only reference page intended for use during the fixed rehearsals. The other pages are study references: relationships to know or derive, not extra material supplied by the test. [9]

Before substituting Ask

Meaning Which symbol represents the quantity I need? Is the input distinct from the output?

Units Are the quantities measured in compatible units? Is the requested result linear, squared, or cubed?

Conditions Are the denominator and leading coefficient nonzero? Is the angle in radians? Is similarity justified?

Form Would a factored form, vertex form, ratio, or equation combination reveal the target more directly?

Check Can I substitute, estimate, test an endpoint, or compare with a geometric or statistical bound?

A useful retrieval exercise

Cover the formula column and reconstruct a relationship from its meaning. Then state its restrictions and solve one of the reference examples. Reading a formula until it looks familiar is not the same as choosing it correctly on a fresh problem.

The reference is deliberately selective. It consolidates the core relationships used across this series, not every theorem in high-school mathematics. Explanations and challenging applications remain in the earlier content guides.

6.2 Provided-formula practice sheet

PROVIDED RELATIONSHIPS / available from Reference during rehearsals

An original arrangement of the mathematical relationships shown on the standard SAT Math reference sheet. This is not a screenshot of Bluebook. Additional study formulas follow in this reference section. [9]

Figure or relationship

Formula

Circle

A = πr2, C = 2πr

Rectangle

A = lw

Triangle

A = 12 bh; h is perpendicular to the chosen base.

Right triangle

c2 = a2 + b2; c is the hypotenuse.

45°–45°–90°

Side lengths s, s, s2; the last is the hypotenuse.

30°–60°–90°

Side lengths x, x3, 2x, opposite the respective angles.

Rectangular prism

V = lwh

Right circular cylinder

V = πr2 h

Sphere

V = 43πr3

Right circular cone

V = 13πr2 h

Right rectangular pyramid

V = 13 lwh

Full circle

360° = 2π radians

Triangle angle sum

180°

Keep the symbols grounded A is area, C is circumference, V is volume, r is radius, and l, w, h, b indicate the relevant lengths. Use consistent units. A diameter is twice a radius; it is not a substitute for r. In a solid, the perpendicular height need not be a slanted edge.

6.3 Algebra and foundations

KNOW OR DERIVE / not the supplied formula sheet

Relationship

Meaning and conditions

m = y2 − y1x2 − x1

Slope; requires x2 ≠ x1. Use consistent subtraction order.

y = mx + b

m is slope; b is the output at x = 0.

y − y1 = m(x − x1)

Point-slope form for a nonvertical line.

Ax + By = C

Standard form; not both A, B zero. If B ≠ 0, slope is −A/B.

Parallel / perpendicular lines

Distinct parallel nonvertical lines have equal slopes. Perpendicular finite slopes have product −1; horizontal and vertical lines form the separate case.

x = h; y = k

Vertical line; horizontal line. Vertical slope is undefined; horizontal slope is zero.

ax + b = c

If a ≠ 0, x = (c − b)/a. If a = 0, compare b and c to distinguish no solutions from every real input.

Multiply or divide an inequality

Reverse its direction when multiplying or dividing by a negative number. Do not divide by an unknown-sign expression without analyzing cases.

Linear system solution counts

One intersection, no intersection, or the same line. Check constants as well as variable coefficients for proportionality.

ab + cd = ad + bcbd

Requires b, d ≠ 0. A common denominator combines entire numerators.

ab ÷ cd = adbc

Requires b, c, d ≠ 0. Division by zero is undefined.

a(b + c) = ab + ac

Distribute to every term, including a negative sign outside the parentheses.

Scientific notation

a × 10n with 1 ≤ |a| < 10 for a nonzero number; n is an integer.

A target can be an expression If a problem supplies 2x − 3 = 11 and asks for 2x + 4, use the difference of 7 between the expressions. Solving every variable is not a requirement. A calculator graph or table also needs to be read for the requested coordinate or value.

6.4 Advanced Math

KNOW OR DERIVE / preserve every domain restriction

Relationship

Meaning and conditions

(a ± b)2 = a2 ± 2ab + b2

Do not omit the middle term.

a2 − b2 = (a − b)(a + b)

Difference of squares.

xm xn = xm+n; (xm)n = xmn

For rational-exponent use without extra cases, take x > 0. Integer-exponent cases allow broader domains.

xm/xn = xm−n; x−n = 1/xn

Require a nonzero base and defined real expressions.

xp/q = xpq

For a clean real-valued rule, take x > 0, integer p, and positive integer q. Odd-root negative-base cases require care.

x2 = |x|

The principal square root is nonnegative.

ax2 + bx + c = 0

x = −b ± b2 − 4ac2a

Requires a ≠ 0 to use quadratic rules.

The quadratic formula. A real root requires a nonnegative radicand.

D = b2 − 4ac

D > 0: two distinct real roots; D = 0: one; D < 0: none. Restrictions may reject algebraic roots.

h = −b/(2a); k = f (h)

Vertex (h, k); minimum when a > 0, maximum when a < 0.

f(x) = a(x − h)2 + k

Vertex form. Factored form a(x − r)(x − s) displays zeros when defined.

r + s = −b/a; rs = c/a

Root sum and product, including a repeated root counted twice.

|u| = d

If d > 0, solve u = d or u = −d. If d = 0, solve u = 0. If d < 0, no real solution.

Q(t) = Q0bt/T

b > 0 is the multiplier per interval of length T > 0. Growth p% gives b = 1 + p/100; decay gives b = 1 − p/100 > 0.

g(x) = a f (b(x − h)) + k

For a, b ≠ 0, a point (u, v) maps to (h + u/b, av + k). Check the new domain as well as the shape.

Restrictions to retrieve: denominator ≠ 0; even-root radicand ≥ 0; original sign conditions before squaring; nonzero leading coefficient before discriminants; stated contextual domains before choosing a root.

6.5 Additional geometry and trigonometry

KNOW OR DERIVE / not additional supplied formulas

Relationship

Meaning and conditions

Vertical / supplementary angles

Vertical angles are equal. A linear pair sums to 180°. Parallel-line angle rules require parallel lines.

Triangle relationships

Exterior angle equals the sum of the remote interior angles. Equal sides face equal angles. Similarity needs justified matching angles or side ratios.

sin θ = oppositehypotenuse

Right triangle, acute angle θ; label sides relative to that angle.

cos θ = adjacenthypotenuse

tan θ = oppositeadjacent.

sin θ = cos(90° − θ)

Complementary acute-angle relationship.

rθ; 12 r2θ

Arc length; sector area. These versions require θ in radians.

θ360°(2πr); θ360°(πr2)

Degree versions of arc length and sector area.

Inscribed angle / tangent

An inscribed angle is half its intercepted arc measure. A tangent is perpendicular to the radius at the tangency point.

(x − h)2 + (y − k)2 = r2

Circle center (h, k) and radius r > 0. Complete both squares if needed.

d = (x2 − x1)2 + (y2 − y1)2

Coordinate distance; derived from a right triangle.

Midpoint (x1 + x22, y1 + y22)

Average coordinates separately.

Similar figures: length factor s > 0

Perimeter factor s; area factor s2; volume factor s3. Only use the rule after similarity is established.

Parallelogram A = bh

Use perpendicular height, not a slanted side.

Trapezoid A = 12(b1 + b2)h

b1, b2 are the parallel bases.

Prism V = Bh; pyramid V = Bh/3

B is base area, not a base edge. The general forms extend the supplied rectangular-base cases.

Surface area

Add exposed faces only; remove shared interior faces. For a sphere, S = 4πr2.

6.6 Data analysis, rates, and percentages

KNOW OR DERIVE / interpretation is part of the calculation

Relationship

Meaning and conditions

rate = quantity/time

Average speed is total distance divided by total time, not generally the average of speeds.

Ratio a : b

Represent matching quantities as ak, bk for a common scale factor k.

Unit conversion

Multiply by conversion fractions equal to 1. Square or cube the factor for area or volume units.

percent change = new − oldold 100%

Uses the original nonzero base. A positive change is an increase; a negative one is a decrease.

New = old ×(1 ± p/100)

Multiply successive factors; divide by the factor to reverse a change.

Percentage points

The arithmetic difference of percentages. Not the same as relative percent change.

x¯ = ∑ xin

Mean; total = nx¯. Count frequencies when data are in a table.

Combined mean = ∑njxj¯∑nj

Pool group totals and counts. Equal weighting of group means requires equal group sizes.

Median / range / IQR

Median is the middle of sorted data; range is maximum minus minimum; IQR is Q3 − Q1.

yi = axi + b

Mean becomes ax¯ + b; standard deviation becomes |a| times the old one. Adding b alone does not change spread.

Residual = observed − predicted

For a fitted model, positive residual means the observation is above its prediction.

P(A | B) = P(A ∩ B)P(B)

Requires P(B) > 0. In a count table, restrict the denominator to B.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

“At least one”; subtract the overlap counted twice.

P(Ac) = 1 − P(A)

Complement: outcomes not in A.

Independent events

P(A ∩ B) = P(A)P(B). Sampling without replacement usually changes the second conditional probability.

6.7 Inference, notation, and final checks

KNOW OR DERIVE / keep the claim no stronger than the evidence

Idea or symbol

Meaning and caution

Sample to population

Estimate a population count with population size × sample proportion when the sampling method supports that inference. An estimate is not an exact census.

Estimate ± margin

Reported interval endpoints are estimate minus margin and estimate plus margin. Match fraction units or percentage-point units.

Larger random samples

With method, confidence level, and other relevant factors comparable, larger samples generally yield smaller margins of error. They do not automatically remove selection bias.

Random sampling

Helps justify generalizing to the population sampled, subject to study quality and uncertainty.

Random assignment

Helps justify a causal treatment comparison in a well-conducted experiment. It does not automatically make a volunteer sample representative.

Association

Two variables vary together. This alone does not identify a cause; confounding or reverse direction may be possible.

f(x)

Output of function f at input x, not the product f · x. f (a + h) changes the input; f (a) + h changes the output.

(x, y)

Ordered pair: input/horizontal coordinate first, output/vertical coordinate second.

= versus ≈

Exact equality versus approximation. Preserve exact expressions until final rounding is required.

≠, <, ≤

Not equal; strictly less; less than or equal. Equality is included only in the last relation.

[a, b], (a, b)

As intervals, square brackets include endpoints and parentheses exclude them. Context distinguishes an open interval from an ordered pair.

∩ and ∪

Intersection (both) and union (at least one).

√, |x|, π

Principal nonnegative square root; distance of x from zero; the circle constant.

yˆ, x¯

A model’s predicted value; the mean of the data values.

The final-answer audit

Target: Is this the quantity asked for? Conditions: Is it allowed in the original problem? Units: Are they compatible and of the correct dimension? Entry: Did you preserve the sign, fraction, and requested percent or coefficient? Reasonableness: Does the value fit a bound, estimate, or independent substitution?

15 worked examples

6.01. Derive slope rather than guessing a sign

Find the slope of the line through (−3, 8) and (5, −4).

Show worked solutionHide worked solution for example 6.01

Work. Use matching subtraction order in numerator and denominator: m = (−4 − 8)/(5 − (−3)) = −12/8 = −3/2. The line falls as x increases, consistent with a negative slope.

Answer: −32

Check. An input increase of 8 produces an output decrease of 12.

Key takeaway. The slope formula is a relationship to know or derive; it is not on the standard supplied geometry reference.

6.02. Rearrange a formula without changing its meaning

For nonzero h, solve A = 12(b1 + b2)h for b2.

Show worked solutionHide worked solution for example 6.02

Work. Multiply both sides by 2 and divide by h: 2A/h = b1 + b2. Subtract b1 to obtain b2 = 2A/h − b1. The assumption h ≠ 0 justifies the division.

Answer: b2 = 2Ah − b1

Check. Substituting back gives 12[b1 + (2A/h − b1)]h = A.

Key takeaway. A formula reference does not remove the need to know which variable the question asks for.

6.03. Use the supplied Pythagorean relationship correctly

A right triangle has hypotenuse 17 and one leg 8. Find the other leg.

Show worked solutionHide worked solution for example 6.03

Work. Let the missing leg be b > 0. The hypotenuse belongs alone on the squared side: 82 + b2 = 172. Thus b2 = 289 − 64 = 225 and b = 15.

Answer: 15

Check. 82 + 152 = 64 + 225 = 289 = 172.

Key takeaway. Identifying the hypotenuse is the reasoning step; the supplied equation cannot identify it for you.

6.04. Use the quadratic formula with a signed coefficient

Solve 2x2 + 3x − 2 = 0.

Show worked solutionHide worked solution for example 6.04

Work. Here a = 2, b = 3, and c = −2. The formula gives x = [−3 ± 9 − 4(2)(−2)]/4 = (−3 ± 5)/4. Thus x = 1/2 or x = −2.

Answer: 12 or −2

Check. The factorization (2x − 1)(x + 2) = 0 independently gives the same roots.

Key takeaway. Keep the entire numerator over 2a. Factoring can be a faster alternative when the factors are visible.

6.05. Connect discriminant and root count

How many distinct real solutions does 5x2 + 2x + 3 = 0 have?

Show worked solutionHide worked solution for example 6.05

Work. The leading coefficient is nonzero, so the quadratic discriminant applies. D = 22 − 4(5)(3) = 4 − 60 = −56 < 0. There are no real solutions.

Answer: 0

Check. Completing the square gives 5(x + 1/5)2 + 14/5 = 0, impossible over the reals.

Key takeaway. The word real matters. This guide counts real solutions unless stated otherwise.

6.06. Use a vertex formula with the correct target

At what input does f(x) = −2x2 + 12x + 1 reach its maximum?

Show worked solutionHide worked solution for example 6.06

Work. For ax2 + bx + c, the vertex input is −b/(2a). Here it is −12/(2(−2)) = 3. Since a < 0, this vertex is a maximum. The maximum output would be 19, but it is not requested.

Answer: x = 3

Check. f(x) = −2(x − 3)2 + 19.

Key takeaway. The same vertex contains an input and an output. Label both before deciding which one to enter.

6.07. Respect the domain in exponent rules

For x > 0, simplify x5/2x1/2.

Show worked solutionHide worked solution for example 6.07

Work. For the same nonzero base, subtract exponents: x5/2−1/2 = x2. The condition x > 0 also ensures the real square-root denominator is defined and nonzero.

Answer: x2

Check. At x = 4, the original value is 32/2 = 16, matching x2.

Key takeaway. An exponent identity is used under assumptions. A numerical check illustrates it but is not its proof.

6.08. Build a growth formula from an interval

A quantity starts at 500 and follows an exponential growth model, increasing by 8% every 2 years. Write a model for its value after t ≥ 0 years.

Show worked solutionHide worked solution for example 6.08

Work. The initial factor is 500, the multiplier for one growth interval is 1.08, and the number of intervals is t/2. Thus Q(t) = 500(1.08)t/2 in the stated model.

Answer: Q(t) = 500(1.08)t/2

Check. Q(0) = 500 and Q(2) = 540.

Key takeaway. A growth model combines three choices: initial value, interval multiplier, and number of intervals.

6.09. Derive an arc formula from a full circle

A radius-15 circle has a central angle of 72°. Find the corresponding minor arc length.

Show worked solutionHide worked solution for example 6.09

Work. The arc is 72/360 = 1/5 of the full circumference. Thus s = (1/5)(2π · 15) = 6π. This derives the degree-version arc relationship from the supplied circumference formula.

Answer: 6π

Check. The full circumference is 30π, and five such arcs complete it.

Key takeaway. When you forget an arc formula, use the fraction of a full turn.

6.10. Distinguish sector area from arc length

A circle has radius 6 and a sector with central angle π/3 radians. Find the sector area.

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Work. The angle is (π/3)/(2π) = 1/6 of a turn. The sector is one-sixth of the circle area: (1/6)(π62) = 6π. Equivalently, A = 12 r2θ.

Answer: 6π square units

Check. The full circle has area 36π, so the result is one-sixth of it.

Key takeaway. Arc length has linear units; sector area has squared units. Similar-looking angle formulas measure different things.

6.11. Apply a volume formula with a radius, not a diameter

A cylinder has diameter 10 and height 9. Find its volume.

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Work. The radius is 5. Using V = πr2 h, obtain V = π(5)2(9) = 225π. Using 10 as the radius would multiply the volume by 4.

Answer: 225π cubic units

Check. Base area is 25π; nine such area-units of height give 225π.

Key takeaway. Before substituting, label the geometric quantities in the formula on the figure or in the wording.

6.12. Use a trigonometric ratio with labeled sides

In a right triangle, an acute angle θ has opposite side 7 and adjacent side 24. Find sin θ.

Show worked solutionHide worked solution for example 6.12

Work. The hypotenuse is 72 + 242 = 25. Sine uses opposite over hypotenuse, so sin θ = 7/25. The ratio 7/24 is tangent, not sine.

Answer: 725

Check. The value is between 0 and 1, as required for the sine of an acute angle.

Key takeaway. Label sides relative to the specified angle; a side can be opposite one acute angle and adjacent to the other.

6.13. Convert a mean into a missing total

Five values have mean 18. Four of them sum to 71. Find the fifth value.

Show worked solutionHide worked solution for example 6.13

Work. The mean relationship gives total = 5(18) = 90. Subtract the known subtotal: 90 − 71 = 19.

Answer: 19

Check. (71 + 19)/5 = 18.

Key takeaway. The total-count-mean triangle is often more useful than repeatedly writing out an average formula.

6.14. Use a probability union with an overlap correction

In a group of 100 people, 42 speak language A, 35 speak language B, and 12 speak both. A person is selected at random. Find the probability that the person speaks at least one of the languages.

Show worked solutionHide worked solution for example 6.14

Work. Adding the two counts counts the 12 bilingual people twice. Subtract their overlap once: 42 + 35 − 12 = 65. The probability is 65/100 = 13/20.

Answer: 1320

Check. The union count 65 exceeds each separate count, 42 and 35, but does not exceed their sum 77 or the group size 100.

Key takeaway. At least one is a union; both is the overlap. A two-way table can supply the same count.

6.15. Attach the correct assumption to a reference statement

For nonzero roots r and s of 4x2 − 12x + 5 = 0, find 1/r + 1/s.

Show worked solutionHide worked solution for example 6.15

Work. The sum is r + s = 12/4 = 3 and the product is rs = 5/4. Therefore 1/r + 1/s = (r + s)/(rs) = 3/(5/4) = 12/5. The nonzero product justifies the division.

Answer: 125

Check. From 4r2 − 12r + 5 = 0 and r ≠ 0, a reciprocal substitution yields the reciprocal-root quadratic 5u2 − 12u + 4 = 0, whose root sum is 12/5.

Key takeaway. Build a requested expression from known relationships, carrying every nonzero condition with it.