ALL MATH Flashcards
natural numbers
positive not zero
whole numbers
zero and positive number
integer
positive and negative
rational
fraction, repeating/terminating deci
irrational
nonterminating deci, square root, pi
distance
rate x time
exponential function
y=nb^x
b=constant ratio: y2/y1
n= solve for by plugging in ordered pair. y intercept
-n: flect over x axis
y=ab^x + n
up n
y=a^(x+n)
shift left
even function
symmetric across y-axis
(-x,y)
make x negative, y should be the same as before
f(-x)=f(x)
all degree/powers is even
odd function
(-x,-y)
symmetric about the origin
make x negative, y will b opposite of before.
f(-x)= - f(x)
all degrees are odd
growth/decay
y= a(1+r)^t
y=a(1-r)^t
a= initial amt
finding final amount
compounded interest
A=p(1+r/n)^nt
P: initial amt
n: # compounded per year
arithmetic sequence
An= An-1 + d, where A1 =______
d= second # - first # A1 = first term in sequence
arithmetic sequence
calculate
An= A1 + d(n-1)
n: term #
geometric sequence
An=An-1( r ) where A1= ____
An= a1 ( r )^n-1
measure of center
mean x~
not resistant to outliers
median-resistant
measures of variation
- range
- mean absolute variation (= actual # - mean)
- IQR (Q3-Q1)
Boxplot
min,Q1, med,Q3,max
bigger side=skewd
correlation
r [-1,1]
strength of linear relationship
strong correlation= straightline, -1,1
weak: 0
scatterplot residuals
actual y-predicted y (from best line fit)
then graph, if it appears random then line best fit is appropriate
reflecting over a line
y=a
(x, 2a-y)
x=a
(2a-x,y)
y=x
(y,x)
rotate counterclockwise
90 (-y,x)
180 (-x,-y)
270 (y,-x)
proofs of triangle
- reflexive property
- sum of 2 sides of a triangle is bigger than the 3rd
- vertical angles
exterior angle theorem
exterior angle=sum of 2 nonadjacent angle
transitive propert
a=b, b=c, a=c
supplementary
180
perpendicular bisector theorem
a point on a perpendicular bisector is equidistant from the endpoints of the segment
based angle theorem
if two sides of an angle are congruent, then the angles opposite are congruent
congruent triangle
sss sas asa aas hl
cpctc
used to prove a side/angle of 2 triangles are congruent. first prove that the angles are congruent
similar triangles
angles are congruent sides are proportional aa~ sss~ sas~
incenter
angle bisector
incircle
equidistant from sides at a right angle
circumcenter
perpendicular bisector
equidistant from vertices
centroid
medians, vertex to midpoint
2x point x
orthocenter
altitude: perpendicular segment from vertex to opp side
tangent to a circle
perpendicular to radius
circumference of a circle
2pir
pi* d
area of a circle
pi*r^2
calculate arc length
arc length = arc degree
circumference = 360
area of a sector
area of sector= arcdegree
area of circle 360
distance formula
/(x2-x1)^2 + (y2-y1)^2
midpoint
x1+x2 /2
partitioning a line segment
finding a point that is 2/5 distance from A-B
x= fraction(x2-x1)+ x1
for y, replace x with y
circle
center: (h,k)
(x-h)+(y-k) = r*
ellipse
center: (h,k)
(x-h)* + (y-h)* = 1
a* b*
a*= how long it is leftnright b*= up and down (total) focus is on the longest axis c= /a*-b* c= distant from center to foci
hyperbola
center (h,k)
(x-h)* - (y-k)* =1
a* b*
left and right
(y-k)- (x-h) =1
b* a*
up, down
c=/a+b
slope of asymptote: b/a
parabolas
up/down
y-k=1/4p(x-h)*
directrix-p-vertex-p-focus
left/right
x-h=1/4p(y-k)*
directrix-focus=focus-point
arc length=S
angle degree= s/r
degrees to radians
pi/180
coterminal
add/minus 360
reference angle
acute/always positive
trig
sin: odd, y
cos: even, x
tan: odd, y/x
to find asymptote, state the first x value then see when the next one is=n.
function trig
asin(bx-c) +d
a: amplitude max+min / 2
b: period 2pi/b (how long one cycle is
c: phase shift
boundaries
left: bx-c=o
right: bx-c=2pi
cosectant
graph sine
asymptote where sign touches the midline
secant
graph sign
tangent/cotangent
period: pi/b
divide period/domain into 4 sections. 1 and 3 is amplitude
inverse of trigs
switch x and y
is it a function: pass horizontal line test
sin2A
2sinAcosA
cos2A
cosA-sinA
1-2sinA
2cosA-1
tan2A
2tanA
1-tan*A
sin(x+y)
sinXcosY + cosXsinY
same sign
cos(x+y)
cosXcosY - sinXsinY
opposite sign
tan (x+y)
tanX+tanY
1-tanXtanY
same
opp
pythagorean identity
sin0 + cos0 =1
tan0 + 1= sec0
1 + cot0 = csc0
finding missing side/angle of triangle
a = b
sinA sinB
ssa
A is acute
a<b></b>
a=bsinA one solution
a>bsinA two
a
A is cute
a>_ b
one solution
A is right/obtuse
ab one
law of cosine
a=b + c* - 2bc x cosA
to find area of triangle
.5absinC
heron’s formula to find area of a triangle when given only sides
/s(s-a)(s-b)(s-c)
s=.5(a+b+c)
complex numbers
imaginary
standard form
a+bi
+/-: add like terms, distribute -1
x: foil
divide: multiply by conjugate
complex number polar form
R: modulus: magniture
argument: direction/angle
rcis0
+/-: convert to standard form
x: multiply modulus
add argument
divide: divide modulus
subtract argument
exponent: raise exponent to modulus
multiply exponent to argument
vectors
directio and magnitude/length [v]
component form
adding two vertex=resultant
linear combination form of vectors
-2i+8j
direction magnitude form of vectors
[v]
unit vector, magnitude=1
component to direct/mag form: use tangent
magnitude of vector
/x+y
finding unit vector
1/[v] x V
or
vector with weight
weight is vector going straight down
force/tension=magnitude
velocity vector
speed(cos0i + sin0j)
find vector given initial and terminal point
.
permuatation
order matters
nPr= n!/(n-r)!
n: # of things you choose from
r: actual # of things you chose
combination
order does not matter
nCr= n! / r!(n-r)!
how many different ways/outcomes
multiply the number of possibilities
flip a coin 4 times, possible outcomes?
2x2x2x2
how many diff ways can the letters b arranged
=number of each letter
total# ! #! x #!
P(AuB)
P(A) + P(B) - P(AnB)
independent if
P(AnB)= P(A) x P(B)
P(A given B)
P(AnB)
P(B)
more C/P
#uhave C uwant x _C_ total C needed
prediction based on rate
nCx . p^x (1-p)^n-x
binompdf
at most /
binomcdf
at least
1-binomcdf
finding expected value from frequency table to probability distribution
x(Px) + x(Px)
graphing log
passes through (1,1)
log(x-n)
shift right
logx+n
up n
-logx
reflect over x
domain: (x-n)>0
asymptote (x-n)=0
x intercept: set it to zero
absolute value
piecewise
turn left into negative
replace lines with paranthesis
to find the turning point, set paranthesis to zero
number before parantesis is slope for right
i
/-1
i*
-1
i^3
-i
i^4
1
i^high number
divide exponent by 4 .25= i .5 = i* .75=i^3 no decimal= i^4
complex conjugate
(a+bi)(a-bi)
factoring
ac= x/a
b =+
quadratic formula
b*-4ac=0 one real
0 two real
parabola up/down
vertex form
y=a(x-h)*+k
to find vertex from standard form:-b/2a
a: positive up
a: strech/shrink
- a: reflect overx
inequalities with polynomials and shading a number line
find zeros
x shade away
midpoint formula
(x1+x2/2, y1+y2/2)
multiplying dividing poynomial fraction
cancel things out diagonal
adding/subtracting polynomial fraction (rational)
find common denominator than multiply each fraction by it, cancel things out
rational expression
horizontal asymptote
end behavior
based on degree
N>D none
N
rational expression
vertical asymptote
domain
set denominator to zero
rational expression
zeros
set numerator to zero
deviation
n=sample size
variation: (each data entry-mean)*
calcualting standard deviations
mean+/- mean deviation
two standard deviation
mean+/- 2(msd)
a-b
(a+b)(a-b)
a+b
(a+bi)(a-bi)
(a+b)*
a+2ab+b
(a-b)*
a-2ab+b
(a+b)^3
a^3+ 3ab+3ab+ b^3
(a-b)^3
a^3- 3ab+3ab - b^3
a^3 + b^3
(a+b)(a-ab+b)
s o ap
pascal triangle
apand (a+b)^
^ 0 (1) 1 (11) 2 (121) 3 (1331) 4 (14641) 5 (1,5,10,10,5,1)
sum of finite geometric series
a1(1-r^)
1-r
^= nth term
sum of infinite geometric series
a1
1-r
dividing polynomials
divide Xs= put on top
multiply answer by left of box
use upside down box for x+a
critical values
vertical asy,ptotes
(y interecpts)
roots
radical graph
domain: [0, infinity)
/x+n
shift left
starting point: x+n=0
/-x = reflect over y
-/x = reflect over x
3/x goes both ways
