SAT math formulas not on the reference sheet

On test day

Open the to check a volume or a special right triangle. It prints no algebra at all, so the and formulas have to come from memory.
Write the formula on scratch paper before you put numbers in. For the slope through $(2,5)$ and $(6,13)$ , write $\frac{y_2-y_1}{x_2-x_1}$ first, then $\frac{13-5}{6-2}=2$ .
For a percent change, divide by the starting amount. Going from 40 to 50 is $\frac{10}{40}$ , a 25% increase.
For a 's highest or lowest point, use $x=-\frac{b}{2a}$ and keep the minus sign. For $y=x^2-10x+3$ , that's $x=5$ .
If a question gives $\sin x^\circ$ and asks for $\cos(90-x)^\circ$ , the answer is the same number.
Can't think which formula fits? Reread the question and your last line. Lots of algebra questions need no formula at all.

What the reference sheet gives you

You can open the reference sheet in either Math . This is everything it prints.

Shape	What the sheet prints
Circle	$A=\pi r^2$ and $C=2\pi r$
Rectangle	$A=\ell w$
Triangle	$A=\frac12bh$
Right triangle	$c^2=a^2+b^2$
Special right triangles	$30^\circ$ - $60^\circ$ - $90^\circ$ : $x$ , $x\sqrt3$ , $2x$ . $45^\circ$ - $45^\circ$ - $90^\circ$ : $s$ , $s$ , $s\sqrt2$
Box, cylinder, sphere	$V=\ell wh$ , $V=\pi r^2h$ , $V=\frac43\pi r^3$
Cone, rectangular pyramid	$V=\frac13\pi r^2h$ , $V=\frac13\ell wh$

It also says a circle has 360 degrees of , or $2\pi$ , and that a triangle's angles add to 180 degrees.

A few more facts follow from those. A square is a rectangle with equal sides, so its area is $s^2$ , and a cube's volume is $s^3$ . Halving the circle facts gives $180^\circ=\pi$ radians. Since a is $2r$ , the sheet's $C=2\pi r$ is the same as $C=\pi d$ .

How to use this list

Nothing below is on the sheet, so you need all of it in your head. Each row's name links to the lesson that teaches it, so open that lesson if a row doesn't make sense.

To study, cover the formula column and say each formula from its name. Then go the other way and say what question each formula answers, like "two points, and I need the slope."

Learn these four first: $y=mx+b$ , the percent multiplier $1+\frac{p}{100}$ , the slope formula and $x=-\frac{b}{2a}$ . They come up on more questions than all the geometry rows put together.

If a question needs a formula that isn't here or on the sheet, the question prints it. Plenty of questions need no formula at all, like solving the $3(x-4)=2x+5$ . If you're stuck on one of those, reread the question and your last line.

Lines and points

What you need	Formula	Example
Reading $y=mx+b$	Slope $m$ , $(0,b)$	$y=4x-7$ : slope 4, y-intercept $(0,-7)$
A rate and a starting value	$m$ is the amount per unit, and $b$ is the amount at the start	$25 to join plus $12 a month: $12x+25$
Slope from two points	$m=\frac{y_2-y_1}{x_2-x_1}$	$(2,5)$ and $(6,13)$ : $\frac{8}{4}=2$
Point-slope form	$y-y_1=m(x-x_1)$	Slope 2 through $(2,5)$ : $y-5=2(x-2)$
Slope of $ax+by=c$	$-\frac{a}{b}$	$4x+2y=9$ : $-\frac42=-2$
Parallel and perpendicular	Parallel: same slope. Perpendicular: flip the slope and change its sign. A flat line and an upright one are perpendicular too	$\frac23$ becomes $-\frac32$
Solutions of one equation	Simplify first. Same $x$ terms, different numbers: none. All the same: infinitely many. Otherwise one	$5x+2=5x-1$ : none
Solutions of a system	Same slope, different y-intercepts: none. Same line: infinitely many. Otherwise one	$y=2x+1$ and $y=2x-4$ : none
Flipping an inequality	Multiplying or dividing by a negative turns the sign around	$-3x\lt12$ gives $x\gt-4$

For a slope, subtract in the same order on top and bottom. $\frac{13-5}{2-6}$ gives $-2$ , the right size with the wrong sign.

No lesson teaches distance and midpoint on their own. The distance between two points is the Pythagorean theorem on a grid, with the change in $x$ and the change in $y$ as the legs. From $(1,2)$ to $(4,6)$ that's $\sqrt{3^2+4^2}=5$ .

\text{distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

The midpoint averages the two $x$ -coordinates and the two $y$ -coordinates. Between $(1,2)$ and $(5,8)$ it's $(3,5)$ .

\text{midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)

Percents and rates

What you need	Formula	Example
A percent of a number	$\frac{p}{100}$ times the number	15% of 60: $0.15\times60=9$
An increase or a decrease	Up $p\%$ : times $1+\frac{p}{100}$ . Down $p\%$ : times $1-\frac{p}{100}$	30% off $70: $70\times0.70=49$
Percent change	$\frac{\text{new}-\text{old}}{\text{old}}\times100$	40 to 50: $\frac{10}{40}\times100=25\%$
Two changes in a row	Multiply the multipliers	Up 10%, then down 10%: $1.1\times0.9=0.99$
A rate	"A per B" is $A\div B$ , and distance is $d=rt$	150 miles in 3 hours: 50 an hour
Proportional	$y=kx$ , with $k=\frac{y}{x}$	$y=18$ when $x=6$ : $k=3$
Density	$\frac{\text{mass}}{\text{volume}}$	60 grams in 12 cubic centimeters: 5 grams per cubic centimeter
Part of a whole	In a of $a$ to $b$ , the first part is $\frac{a}{a+b}$ of the whole	2 to 3: $\frac25$
Square and cubic units	Square or cube the conversion number	$1\text{ ft}^2=12^2\text{ in}^2$

A percent change is measured against where you started, so $\frac{10}{50}=20\%$ is wrong for 40 to 50. The second of two changes comes off the new amount, which is why up 10% and then down 10% leaves you 1% lower. When a question prints a conversion fact, use the one on the page.

Quadratics

What you need	Formula	Example
Vertex from $ax^2+bx+c$	$x=-\frac{b}{2a}$ , then put that $x$ in to get $y$	$y=x^2-10x+3$ : $x=-\frac{-10}{2(1)}=5$ , vertex $(5,-22)$
Vertex from $a(x-h)^2+k$	The vertex is $(h,k)$	$2(x+1)^2-6$ : $(-1,-6)$
Zeros from $a(x-p)(x-q)$	The zeros are $p$ and $q$ , and the vertex is halfway between	$(x-1)(x+5)$ : zeros 1 and $-5$ , vertex at $x=-2$
Quadratic formula	For $ax^2+bx+c=0$ : $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$	$2x^2+3x-5=0$ : $\frac{-3\pm7}{4}$ , so 1 or $-\frac52$
Discriminant	$b^2-4ac$ . Positive: two real solutions. Zero: one. Negative: none	$x^2+6x+9=0$ : $36-36=0$ , one
A line touching a parabola	Set them equal, move everything to one side, and touching once means $b^2-4ac=0$
Sum and product of the solutions	Sum $-\frac{b}{a}$ , product $\frac{c}{a}$	$3x^2-12x+5=0$ : sum 4, product $\frac53$
Completing the square	Halve the number with $x$ , square it, and add that to both sides	$x^2+6x=7$ becomes $(x+3)^2=16$
Difference of squares	$a^2-b^2=(a-b)(a+b)$	$x^2-49=(x-7)(x+7)$
Perfect squares	$a^2\pm2ab+b^2=(a\pm b)^2$	$x^2-8x+16=(x-4)^2$

and both flip the sign you see. The form subtracts $h$ , so $x+1$ means $h=-1$ , and $x+5$ is 0 when $x=-5$ . Read $a$ , $b$ and $c$ only after every term is on one side.

A question about something thrown upward usually prints $h(t)=-16t^2+vt+h_0$ , with height in feet. The $-16$ comes from gravity, $v$ is the starting upward speed, and $h_0$ is the height at $t=0$ . The object hits the ground when $h(t)=0$ .

Exponentials and powers

What you need	Formula	Example
Exponential model	$f(x)=a\cdot b^x$ : starting value $a$ , multiplier $b$	$300(2)^x$ starts at 300 and doubles
Percent growth or decay	Up $r\%$ : $b=1+\frac{r}{100}$ . Down $r\%$ : $b=1-\frac{r}{100}$	Down 8% a year from 500: $500(0.92)^t$
A change every $n$ units	$a\cdot b^{x/n}$	Doubles every 4 hours: $P_0\cdot2^{t/4}$
Powers	Same base only: $a^ma^n=a^{m+n}$ , $\frac{a^m}{a^n}=a^{m-n}$ , $(a^m)^n=a^{mn}$	$x^3\cdot x^4=x^7$
Zero and negative powers	$a^0=1$ , $a^{-n}=\frac{1}{a^n}$	$2^{-3}=\frac18$
Fractional powers	$a^{m/n}=\sqrt[n]{a^m}$	$27^{2/3}=\sqrt[3]{729}=9$

To check an like $\frac{t}{4}$ , put in one period. At $t=4$ it's $2^1$ , one doubling. $2^{4t}$ would give 16 doublings.

Data and probability

What you need	Formula	Example
Mean	Total divided by how many values	4, 7, 10: $21\div3=7$
A missing value	Total needed is the mean times the count	Mean 8 over 5 values: total 40
Two groups combined	$\frac{n_1m_1+n_2m_2}{n_1+n_2}$	10 averaging 80 and 30 averaging 60: $\frac{800+1800}{40}=65$
Median	Sort, then take the middle. With an even count, average the two middle values	9, 2, 6, 4 sorts to 2, 4, 6, 9: 5
Range and IQR	Range is largest minus smallest. The is $Q_3-Q_1$ , the spread of the middle half	A box from 20 to 32: IQR 12
Standard deviation	No formula. Values bunched near the mean give a smaller one
Probability	What you want over the group you're picking from
Not A	$1-P(A)$
A or B	$P(A)+P(B)-P(A\text{ and }B)$
A count	times the total	0.2 of 150: 30
Margin of error	Estimate plus or minus the margin	62% plus or minus 4: 58% to 66%
From an interval	Estimate is the middle, margin is half the width	30 to 38: estimate 34, margin 4
Residual	Actual minus predicted	Dot at 18, line at 21: $-3$

In "Of the students who chose B, what fraction are seniors?", you're picking from the students who chose B, so they go on the bottom. "Given that" shrinks the group the same way.

A is how far a dot sits above or below the , so a dot above the line has a positive one.

Right triangles and trigonometry

What you need	Formula	Example
Sine, cosine and tangent	$\sin=\frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos=\frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan=\frac{\text{opposite}}{\text{adjacent}}$	Sides 3, 4, 5, at the angle across from 3: $\sin=\frac35$ , $\tan=\frac34$
Sides to know by sight	3-4-5, 5-12-13, 8-15-17, 7-24-25, and their multiples	Legs 9 and 12: 15
Complementary angles	$\sin x^\circ=\cos(90-x)^\circ$	$\sin20^\circ=\cos70^\circ$
Sine squared plus cosine squared	$\sin^2\theta+\cos^2\theta=1$

Opposite and adjacent depend on which angle you stand at. The hypotenuse is always the longest side, across from the right angle, so find it before you match a side pattern.

The special triangles on the sheet give these values.

Angle	Sine	Cosine	Tangent
$30^\circ$	$\frac12$	$\frac{\sqrt3}{2}$	$\frac{\sqrt3}{3}$
$45^\circ$	$\frac{\sqrt2}{2}$	$\frac{\sqrt2}{2}$	1
$60^\circ$	$\frac{\sqrt3}{2}$	$\frac12$	$\sqrt3$

For an acute angle, sine and cosine are between 0 and 1, since a leg is shorter than the hypotenuse. Desmos starts in radians, so switch it to Degrees before you type $\sin(30)$ , as Desmos basics shows.

Circles

What you need	Formula	Example
The circle's equation	$(x-h)^2+(y-k)^2=r^2$ : center $(h,k)$ , $r$	$(x+2)^2+(y-5)^2=36$ : center $(-2,5)$ , radius 6
An arc's degrees	The same as its
Arc length	$\frac{\text{arc length}}{2\pi r}=\frac{\text{central angle}}{360^\circ}$	Radius 4, $90^\circ$ : $\frac14$ of $8\pi$ , or $2\pi$
Two arcs	The and add to $360^\circ$ , and to the whole	$100^\circ$ leaves $260^\circ$
Radians and degrees	To degrees, times $\frac{180}{\pi}$ . To radians, times $\frac{\pi}{180}$	$\frac{\pi}{3}\cdot\frac{180}{\pi}=60^\circ$
Tangent lines	A meets the radius at a right angle. Two tangent segments from one outside point are the same length

The center's signs are the opposite of the ones you see, so $x+2$ means $-2$ . The right side is the radius squared, so take its square root before you call it the radius.

Shapes and angles

What you need	Formula	Example
Prism	$V=Bh$ , where $B$ is the area of the base	Base area 12, length 10: 120
Perimeter	Add the sides: $2\ell+2w$ , or $4s$ for a square	7 by 3: 20
Surface area	Box $2(\ell w+\ell h+wh)$ . Cube $6s^2$ . Closed cylinder $2\pi r^2+2\pi rh$	Cube with edge 3: 54
Scale factor	Lengths times $k$ , areas times $k^2$ , volumes times $k^3$	Lengths doubled: area times 4, volume times 8
Two lines crossing	Angles side by side on a line add to $180^\circ$ . are equal
Parallel lines	A line crossing them makes two angle sizes that add to $180^\circ$	$50^\circ$ and $130^\circ$
Right triangle	The two smaller angles add to $90^\circ$
Outside angle	The two inside angles at the other corners, added	$35^\circ$ and $75^\circ$ : $110^\circ$
Two equal sides	The base angles are equal, each $\frac{180-A}{2}$ , where $A$ is the angle between the equal sides	$A=40^\circ$ : base angles $70^\circ$
Three equal sides	Every angle is $60^\circ$

A height is measured straight from the base, so a slanted side isn't the height. Halve a diameter before you use a formula that needs the radius.

Practice

Work through the three examples, then try five more. Before you put in any numbers, write the formula you need on scratch paper, or say the rule that gets you to the next line.

Worked example

Aiko loads 4,600 caps into a bottling line’s feeder before production starts. The feeder uses caps at a constant rate of 725 caps per hour; after 2 hours, 3,150 caps remain. The function $q$ gives the number of caps remaining $h$ hours after production starts.

Which equation defines $q$ ?

A
$q(h) = 4600h - 725$
B
$q(h) = 725h + 4600$
C
$q(h) = -725h + 4600$
D
$q(h) = -725h + 3150$

Show the answer

A
$q(h) = 4600h - 725$
This swaps the and . It uses the initial $4{,}600$ caps as the hourly and $-725$ , the hourly change, as the term.
B
$q(h) = 725h + 4600$
This drops the negative sign from the rate. Because the feeder uses $725$ caps per hour, the number remaining decreases by $725$ each hour.
C
$q(h) = -725h + 4600$
Correct answer
At $h=0$ , the feeder contains $4{,}600$ caps, so the term is $4{,}600$ . The amount decreases by $725$ caps per hour, so the is $-725$ , giving $q(h)=-725h+4600$ .
D
$q(h) = -725h + 3150$
This treats the $3{,}150$ caps remaining after $2$ hours as the value at $h=0$ . The term must represent the initial amount, which is $4{,}600$ .

Solution

Use the two stated amounts to calculate the hourly change:

$\dfrac{3150-4600}{2-0}=\dfrac{-1450}{2}=-725$ . The is $-725$ because the feeder loses $725$ caps each hour. At $h=0$ , it contains $4{,}600$ caps, so the is $4{,}600$ . Therefore, $q(h)=-725h+4600$ . Checking the given value, $q(2)=-725(2)+4600=3150$ , which matches the number of caps remaining after $2$ hours.

Report a problem

Worked example

At Dubois Dairy, the function $E(x)=2x^2-56x+434$ models energy use, in watt-hours, for a batch completed with $x$ churn cycles.

Which value of $x$ minimizes the energy use predicted by the model?

A
$14$
B
$-14$
C
$42$
D
$-28$

Show the answer

A
$14$
Correct answer
For a $ax^2+bx+c$ , the occurs at $x=-b/(2a)$ . Here, $x=-(-56)/(2\cdot2)=14$ , and the positive leading means the vertex is a minimum.
B
$-14$
This results from dropping the negative sign in the formula and calculating $b/(2a)=-56/4=-14$ .
C
$42$
This evaluates the at the , $E(14)=42$ , and reports the minimum energy use instead of the number of churn cycles.
D
$-28$
This omits the factor of $2$ in the and calculates $b/a=-56/2=-28$ .

Solution

The graph of $E(x)=2x^2-56x+434$ opens upward because the of $x^2$ is positive, so its gives the minimum energy use.

For $ax^2+bx+c$ , the vertex is $x=-b/(2a)$ . Substituting $a=2$ and $b=-56$ gives $x=-(-56)/(2\cdot2)=56/4=14$ . The model therefore predicts minimum energy use at $14$ churn cycles.