uc davis ; math placement

Réussis tes devoirs et examens dès maintenant avec Quizwiz!

multiplying fractions

(a/b)(c/d) = ac/bd

trigonometric equations

- this is complete and utter ass and there are so many rules and shit so let's just do some examples instead haha... - solve (sin^2)x = 1 for all 0 ≤ x < 2𝜋 > first, put everything on one side: (sin^2)x - 1 = 0 > then, factor: (sinx + 1)(sinx - 1) = 0 > then, solve and convert to radians (if specified) sinx = 1, -1 = sinx = {𝜋/2, 3𝜋/2} (notice that 2𝜋 is not included because it is not within the interval 0 ≤ x < 2𝜋) - solve cos2x = 1, where 0 ≤ x < 2𝜋 > first, use the double-angle formula for cos: cos2x = 1; (cos^2)x - (sin^2)x = 1; (1-(sin^2)x) - (sin^2)x = 1; 1 - 2(sin^2)x = 1; -2(sin^2)x = 0; (sin^2)x = 0; sin x = 0 > since sinx = 0 at x = 𝜋, the solution set is {0, 𝜋} - solve (sin^2)3x = 1 for 0 ≤ x < 2𝜋 > first, take the square root of bots sides to obtain sin3x ± 1; sin3x + 1, or sin3x -1 > the sinθ = 1 = 𝜋/2 and the sinθ = -1 = 3𝜋/2; hence, we have 3x = 𝜋/2, and 3x = 3𝜋/2, or x = 𝜋/6, and x = 𝜋/2 > this is not the complete set of solutions. the period of sin3x is 2𝜋/3 and this needs to be added to each solution until the values exceed 2𝜋. hence, another solution is 𝜋/6 + 2𝜋/3 = 5𝜋/6. continuting this process, we obtain: x = {𝜋/6, 𝜋/2, 5𝜋/6, 7𝜋/6, 3𝜋/2, 11𝜋/6}

inverses

- two functions are inverses if the components of every ordered pair of f(x) are in interchanged positions in the function g(x), and vise-versa. - to find the inverse of f(x), substitute y for f(x) in the function and then interchange all variables; xs become ys, and ys become xs. then, solve for y. - common notation for inverse of f(x) is f^(-1)(x) > NEVER means "reciprocal" - to prove two functions are inverses, must show that f(g(x)) = x, AND that g(f(x)) = x

solving absolute value inequalities

- with absolute value inequalities, you will always have two problems to solve - solve everything outside the parenthesis before the inside after splitting the equation > e.g. 2(x - 5) > 10 → x - 5 > 5 → x > 10

factoring by grouping

acx^3 + adx^2 +bcx + bd = ax^2(cx + d) + b(cx + d) = (ax^2 + b)(cx + d)

distributive law

ax + ay = a(x + y)

solving linear inequalities

can be treated like a linear equation, however, when multiplying or dividing both sides of an inequality by negative numbers requires the inequality sign to be reversed

absolute value

equivalent to two equations without the absolute value sign > e.g. |x + 3| = 7 → +(x + 3) = 7 or -(x + 3) = 7

adding fractions

find a common denominator ; a/b + c/d = a/b(d/d) + c/d(b/b) = (ad + bc)/bd

subtracting fractions

find a common denominator ; a/b - c/d = a/b(d/d) - c/d(b/b) = (ad - bc)/bd

second degree equations

solved by factoring or the quadratic formula

first degree equations

solved using addition, subtraction, multiplication, and division

quadratic formula

x = (-b ± √(b² - 4ac))/2a

simple trinomial

x^2 + (a + b)x + (a • b) = (x + a)(a + b)

cosine graph

y = cos(x)

sine graph

y = sin(x)

circles

- (x - h)^2 + (y - k)^2 = r^2 - (h, k) is the center - if equation is not in this form, must complete the square to put it in the correct form

measuring angles

- 360° and 2𝜋 radians in a circle - 180° = 1𝜋 radians - use dimensional analysis to convert btwn degrees and radians

canceling fractions

- ab/ad = b/d - (ab + ac)ad = (a(b + c))/ad = (b + c)/d

lines

- ax + by + c = 0 - x is x-intercept, and y is y-intercept

solving higher order inequalities

- cannot be solved in the same way as higher order equalities; must be able to account for all possible combinations of multiplying positive and negative terms - use a number line to solve for these > e.g. x^2 - 3x - 10 ≥ 0; (x - 5)(x + 2) ≥ 0; x = -2, 5; place the roots on a number line, using open circles for >/< and closed circles for ≥/≤; test each region to determine the sign of each term - never use a "closed circle" on a root when drawing the number line because it causes you to divide by 0

rationalizing fractions

- if the numerator or denominator is √a , multiply by √a/√a - if the numerator or denominator is √a - √b, multiply by (√a + √b)/(√a + √b) - if the numerator or denominator is √a + √b, multiply by (√a - √b)/(√a - √b)

dividing fractions

- invert and multiply ; (a/b)/(c/d) = a/b • d/c = ad/bc

sin and cos on the unit circle

- let (x, y) on the unit circle be (cosθ, sinθ) - cos0° = 1 - cos90° = 0 - cos180° = -1 - cos270° = 0 - sin0° = 0 - sin90° = 1 - sin180° = 0 - sin270° = -1 - tanθ = sinθ/cosθ > tan0° = 0/1 = 0 > tan90° = 1/0 = undefined > tan180° = 0/-1 = 0 > tan270° = -1/0 = undefined - csc, sec, tan, are just found by flipping the x and y coordinates.

log basics

- logb(1) = 0 - logb(b) = 1

inverse properties of logs

- logb(b^x) = x - b^(logb (x)) = x

laws of logarithms

- logb(x) + logb(y) = logb ( x • y) - logb(x) - logb(y) = logb(x/y) - n • logb(x) = logb (x^n)

operations with exponents

- multiplying like bases: b^n • b^m = b^(n + m) (add exponents) - dividing like bases: (b^n)/(b^m) = n^(n-m) (subtract exponents) - exponent of exponent: (b^n)^m = b^(n • m) (multiply exponents) - removing parenthesis: > (ab)^n = a^n • b^n > (a/b)^n = (a^n)/(b^n) - special conventions: > -b^n = -(b^n); -b^n ≠ (-b)^n > kb^n = k(b^n); kb^n ≠ (kb)^n b^n^m = b^(n^m) ≠ ((b^n)^m)

composition of a function

- one (or more) functions substituted into another function. - two types of notation: (f • g)(x) = f(g(x)) > first, find g(x). then, plug it into f(x). - order does matter! - f(g(x)) is generally not equal to g(f(x))

trigonometric identities

- pythagorean identities ; > (sin^2)θ + (cos^2)θ = 1 > 1 + (cot^2)θ = (csc^2)θ > (tan^2)θ + 1 = (sec^2)θ - addition formulas ; > sin(a ± b) = sina • cosb ± cosa • sinb > cos(a ± b) = cosa • cosb ∓ sina • sinb - double angle identities ; > sin(2a) = 2sina • cosa > cos(2a) = (cos^2)a - (sin^2)a - law of cosines: a^2 = b^2 + c^2 - 2bc • cosa - law of sines: sina/a = sinb/b = sinc/c

special triangles

- sin45° = 1/√2 = √2/2 - cos45° = 1/√2 = √2/2 - tan45° = 1/1 = 1 - sin30° = 1/2 - cos30° = √3/2 - tan30° = 1/√3 = √3/3 - sin60° = √3/2 - cos60° = 1/2 - tan60° = √3/1

trigonometry

- soh cah toa ; > sinθ = o/h > cosθ = a/h > tanθ = o/a = sinθ/cosθ - coh sha cao ; > cscθ = 1/sinθ = h/o > secθ = 1/cosθ = h/a > cotθ = 1/tanθ = a/o

range

- the "y-values" which are the result of all the domain values - to determine the range of a function ; > graph: you can visually determine the y-values > analysis: must find the domain first, and then answer: can y = 0? can y < 0? (be negative?) how far (to - ∞)? can y > 0? (be positive?) how far (to + ∞)?

reference angles

- the angle formed by the terminal side of the angle and the x-axis - determining the signs of trig functions ; > q i: all are + > q ii: sin/csc is + > q iii: tan/cot is + > q iv: cos/sec is + > can be remembered by the pneumonic device "cast," with c starting in q iv.

domain

- the set of inputs that can be in the function. excludes all restrictions. - when solving for the domain ; > cannot divide by 0 > cannot take the even root of a negative number > cannot take the logarithm of a negative or 0

properties of exponents

- whole number exponents: b^n = b • b • b... (n times) - zero exponent: b^0 = 1; b ≠ 0 - negative exponents: b^-n = 1/(b^n); b ≠ 0 - rational exponents (nth root): ^n√(b) = 1/(b^n); n ≠ 0, and if n is even, then b ≥ 0 - rational exponents: ^n√(b^m) = ^n√(b)^m = (b^(1/n))^m = b^(m/n); n ≠ 0, and if n is even, then b ≥ 0

difference of squares

- x^2 - a^2 = (x - a)(x + a) - x^4 - a^4 = (x^2 - a^2)(x^2 + a^2) = (x - a)(x + a)(x^2 + a^2)

sum or difference of cubes

- x^3 + a^3 = (x + a)(x^2 - ax + a^2) - x^3 - a^3 = (x - a)(x^2 + ax + a^2)

absolute values

- y - k = |x - h| - (h, k) is the vertex point

parabolas

- y = (ax)^2 + bx + c - must also find the vertex by ; > vsubx = (xsub1 + xsub2)/2 (midpoint of x-intercepts: xsub1 and xsub2) > vsubx = -b/2a (a and b are constants from the std. form) > with these methods, the y-coord of the vertex (vsuby) must be found by plugging vsubx back into the original formula; y - vsubv = (x - vsubx)^2

sin/cos graph translations/transformations

- y = a(sin/cos(bx ± n)) ± m > amplitude of y = a(sin/cos(x)) is a > in general, the period of y = sin/cos(bx) is 2𝜋/b > vertical shifts: y = sin/cos(x) ± m, in which m is the shift > horizontal shifts: y = sin/cos(x ± n), in which n is the shift.

translations

- y = f(x) + c is moved up c units from y = f(x) - y = f(x) - c is moved down c units from y = f(x) - y = f(x-c) is moved to the right c units from y = f(x) - y = f(x + c) is moved left c units from y = f(x)


Ensembles d'études connexes

Lesson 2 - Accounting for Bonds Payable

View Set

The Diversity of Life Chapter 1 Lesson 3-Domains and Kingdoms

View Set

Chapter 4: Understanding the interest rate

View Set

Introduction to Nutrition Chapter 7

View Set