math equations

surface area prism

2(base area)+(length)(base perimeter)

binomial (success/failures)

distance from point and line

point (x₀,y₀) and line ax+by+c=0

|ax₀+by₀+c|/(sqrt(a²+b²))

distance from point and plane

point (x₀,y₀,z₀) and line ax+by+cz+d=0

|ax₀+by₀+cz₀+d|/(sqrt(a²+b²+c²))

lateral surface area of cylinder

2(pi)rh

surface area of cylinder

2(pi)r²+2(pi)rh

SA square pyramid

B+P(slant height)/2

cone surface area

(pi)r²+(pi)r(slant height)

sphere surface area

4(pir)r²

simple interest

I = Prt

amount with simple interest

A = P + Prt or P(1+rt)

compound interest (annual)

A = P(1+r)^t

solve for rate (compound)

r = (A/P)^(1/t)-1

linear scale factor

new length/og length

perimeter scale factor

same as linear

ex. scale factor is 4/3. small perimeter is 12. big perimeter is 12(4/3)=16

area scale factor

(linear scale factor)²

volume scale factor

(linear scale factor)³

side ratio to area ratio

1) square the side length ratio

2) set area ratio equal to area1/area2

pereimeter to linear factor

simply numerator and denominator

area using adjacent side vectors

|u_xv_y-u_yv_x|

rhombus area

d₁d₂/2

find second endpoint from midpoint and first endpoint

(2M_x-x, 2M_y-y)

rectangle has opposite vertices (a,b),(c,d) sides axis parallel. find the area

length: |a-c|

width: |b-d|

find perpendicularity with dot product

A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)

vector AB: <x₂-x₁, y₂-y₁> = <A,B>

vector BC: <x₃-x₂, y₃-y₂> = <C,D>

AC-BD=0 then perpendicular

find fourth vertice

x_D=A+C-B

equilateral triangle area

(s²sqrt(3))/4

diagonal length

sqrt(l²+w²+h²)

inscribed angle

(intercepted arc)/2

exterior angle from two secants

(far arc-near arc)/2

intersecting chords (length of each)

two secants (length of each line)

external(whole)=external(whole)

tangent-secant (length of each line)

c²=a(a+b)

in an inscribed square (square in circle), what is the diameter equal to

the square’s diagonal

angle of 2 intersecting chords

(arc1 + arc2)/2

at most k successes

add P(0)+P(1)+…+P(k)

at least k success

1-P(0)-P(1)…-P(k)

all successes

p^n

no successes

(1-p)^n

theroem for side lengths when circle is inscribed triangle

AF=AE

BF=BD

DC=CE

number of diagonals

n(n-3)/2

sum of interior angles

180(n-2)