Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

- StudyBlue
- Massachusetts
- Harvard University
- GMAT Quantitative

Aline G.

x ^{n - m}

(x^{n})/(x^{m})

x^{n + m}

x^{n}*x^{m}

Advertisement

(x^{n})/(y^{n})

(x/y)^{n}

(xy)^{n}

x^{n}*y^{n}

x^{yz}

(x^{y})^{z}

1/(x^{n})

x^{-n}

a√c + b√c

a√c + b√c = (a+b)√c

√x*√y

√xy

Example: √12*√3 = √36 = 6

a^{x/y}

a^{x/y }= ^{y}√a^{x}

Example: a^{2/3} = 3√a^{2}

√x/y

√x / √y

√a + √b

√a + √b ≠ √a+b

Advertisement

(√x)^{n}

(√x)^{n }= √x^{n}

1^{n}

1

x^{0}

1

x^{1 }=

x^{1 }= x

0^{n}

0, except 0^{0} = 1 or undefined

Time value of money (formula)

F_{V} = P_{V}(1 + i)^{n}

where F_{V is }future value, P_{V} is present value, i is the interest rate per period and n is the number of compounding periods

Distance (formula)

Distance = Rate*Time

Wage (formula)

Wage = Rate*Time

Combination (formula)

where k = number of objects selected from a pile of *n* objects

n = total number of objects from which
Permutation (formula)

where k = number of objects selected from a pile of *n* objects

n = total number of objects from which
(a/b)/(c/d)

(a/b) * (d/c)

Percent Change, as a decimal

(New - Old)/Old

Percent growth rate

N = (1 + G)O; N = New Value, 0 = Old Value, G = Growth Rate as decimal

Multiplying Odds and Evens

(Odd)(Even) = Even

(Odd)(Odd) = Odd

(Even)(Even) = Even

(Odd)(Odd) = Odd

(Even)(Even) = Even

+/- Odds and Evens

(Odd) ± (Even) = Odd

(Odd) ± (Odd) = Even

(Even) ± (Even) = Even

(Odd) ± (Odd) = Even

(Even) ± (Even) = Even

Multiplying Positives and Negatives

(Positive)(Positive) = Positive

(Positive)(Negative) = Negative

(Negative)(Negative) = Positive

(Positive)(Negative) = Negative

(Negative)(Negative) = Positive

Dividing Positives and Negatives

(Positive)/(Negative) = Negative

(Positive)/(Positive) = Positive

(Negative)/(Negative) = Positive

(Positive)/(Positive) = Positive

(Negative)/(Negative) = Positive

Sum of Interior Angles of Polygon

(n-2)(180)

where n = number of sides of a polygon

Central Angle of a Polygon

2(Inscribed Angle)

Area of a Square

A = a^{2 }where a is one side

Area of a Square

A = a^{2 }where a is one side

Area of a Rectangle

A = l*w where l is length and w is width

Area of a Parallelogram

A = b*h where b is base and h is height

Area of a Square

A = a^{2 }where a is one side

Area of a Rectangle

A = l*w where l is length and w is width

Area of a Parallelogram

A = b*h where b is base and h is height

Area of a Square

A = a^{2 }where a is one side

Area of a Rectangle

A = l*w where l is length and w is width

A = b*h where b is base and h is height

Area =

where base_{1} + base_{2} are the lengths of the parallel sides

Area = .5*(Diagonal_{1}*Diagonal_{2})

Area of a Circle

A = πr^{2}

If radius CA is 9cm and angle ACB is 60°, what is the area of the region shaded in black?

- Find the area

A = π(9)^{2}= 81π - Plug into the area of a sector formula

(^{60}/_{360})81π

(^{1}/_{6})(81π)

(^{81}/_{6})π

Circumference of a Circle

C = 2πr

= (^{degree of central angle}/_{360})Circumference

= (^{x}/_{360})2πr

If radius CB is 9cm and angle ACB is 60°, what is the length of arc AB?

- Find the circumference

C = 2π(9) = 18π - Plug into the length of an arc formula

(^{60}/_{360})(18π)

(^{1}/_{6})(18π)

3π

Central Angle of a Circle

2(Inscribed Angle)

- All inscribed angles with the same endpoints are equal
ADB = AEB

- Inscribed Angle = (

^{1}

- /

_{2}

- )(Central Angle)
ADB = (

^{1}

- /

_{2}

- )ACB

ADB is a right angle is because central angle ACB is 180° and ADB is an inscribed angle whose endpoints are the same as ACB.

ADB = ^{1}/_{2}(ACB)

ADB = ^{1}/_{2}(180) = 90

For the above to hold true: (1) C must be the center of the circle (2) AB must be a diameter of the center

Pythagorean Triples

3-4-5 triangles: 3^{2} + 4^{2} = 5^{2}

(Variations: 6-8-10, 9-12-15, 12-16-20)

5-12-13 triangles: 5 ^{2} + 12 ^{2} = 13 ^{2}^{}

(Variation: 10-24-26)^{}

8-15-17 triangles:8^{2} + 15 ^{2} = 17^{2}^{}

A triangle with angles 45°-45°-90° will have sides in the ratio of 1-1-2^{1/2}

A triangle with angles 30°-60°-90° will have sides in the ratio of 1-3^{1/2}-2

Area of Sector of a Circle

(x/360)πr^{2}

Perimeter of a Square

P = 4l (4x the length of one side)

Perimeter of a Rectangle

P = 2w + 2l

Perimeter of a Triangle

The perimeter of a triangle is the sum of its three sides

*The sum of any two sides of a triangle must be larger than the length of the third side

Perimeter of a Parallelogram

P = 2b + 2a, where a and b are the lengths of the non-parallel sides

A = .5bh or 1/2(b*h)

Pythagorean Theorem

A^{2} + B^{2} = C^{2} where A = one leg, B = the other leg, C = hypotenuse

Volume of a Cube

V = a^{3} where *a* is the length of a side

Volume of a Rectangular Solid

V = hwl

Volume of a Cylinder

V = πr^{2}h

Surface Area

Surface Area = 2(length)(width) + 2(height)(width) + 2(length)(height)

SA = 2lw + 2hw + 2lh

SA = 2lw + 2hw + 2lh

How to Solve: Absolute Values

1. Isolate the absolute value

X is Positive: x + 5 = 40

x = 35

X is Negative: x + 5 = - 40 [Make the expression opposite the absolute value negative]

x = - 45

2. Solve for x = positive AND x = negative

Example: |x + 5| + 20 = 60

|x + 5| = 40X is Positive: x + 5 = 40

x = 35

X is Negative: x + 5 = - 40 [Make the expression opposite the absolute value negative]

x = - 45

How to Solve: Absolute Value with Inequalities

1. Isolate the absolute value

X is Negative: 10x + 8 < (-1)(9 + 2x)

x < -17/12

2. Solve for x (same sign) positive AND x (switch sign) negative

|10x + 8| > 9 + 2x

X is Positive: 10x + 8 > 9 + 2x

x > 1/8

x > 1/8

X is Negative: 10x + 8 < (-1)(9 + 2x)

x < -17/12

Powers of 2

2^{1} = 2

2^{2} = 4

2^{3} = 8

2^{4} = 16

2^{5} = 32

2^{6} = 64

2^{7} = 128

2^{8} = 256

2^{9} = 512

2

2

2

2

2

2

2

2

Powers of 3

3^{1} = 3

3^{2} = 9

3^{3} = 27

3^{4} = 81

3^{5} = 243

3^{6} = 729

3

3

3

3

3

Powers of 4

4^{1} = 4

4^{2} = 16

4^{3} = 64

4^{4} = 256

4^{5} = 1,024

4

4

4

4

Powers of 5

5^{1} = 5

5^{2} = 25

5^{3} = 125

5^{4} = 625

5^{5} = 3,125

5

5

5

5

Order of Operations

P - Parentheses

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

Useful Radicals:

√121 =

√169 =

√225 =

√400 =

√625 =

√121 = 11

√169 = 13

√225 = 15

√400 = 20

√625 = 25

Positive Radicand and Even Root

^{Even}√Positive

there is only one solution and this solution is positive

(Technically there are 2 solutions, but the negative root is usually not applicable)

Positive Radicand and Odd Root

^{Odd}√Positive

there is only one solution and this solution is positive

Negative Radicand and Odd Root

^{Odd}√Negative

there is only one possible answer. Since the only way to have a negative product is by multiplying a negative number an odd number of times, the answer in Case 3 is always negative. In order to more clearly see this, notice that multiplying together a negative number an even number of times produces a positive number. Similarly, multiplying a positive number by itself an odd number of times produces a positive number.

Negative Radicand and Even Root

^{Even}√Negative

there is no real solution

quadratic formula

x=[−b±√(b2−4ac)]**/**2a

for quadratic of form ax^{2}+bx+c

Factoring Quadratic Equation ax^{2}+bx+c

x^{2} + 2x - 8 = 0

Find two numbers whose sum is b (or 2 in this example) and whose product is c (or -8 in this example).

Two such numbers are -2 and + 4, which add to +2 and multiply to -8.

x^{2} + 2x - 8 = 0

(x - 2)(x + 4) = 0

x = +2, -4 since these two numbers make each factor equal zero.

(x - 2)(x + 4) = 0

x = +2, -4 since these two numbers make each factor equal zero.

Reverse Factoring

From ax^{2} + bx + c = 0 to (x - a)(x - b) = 0

FOIL, which stands for **f**irst, **o**uter, **i**nner, **l**ast.

To acquire the quadratic form (ax^{2} + bx + c = 0) from the factored form [(x - a)(x - b) = 0]: (1) multiply the first terms, then the outer terms, then the inner terms, and finally the last terms (2) add each of the terms together and simplify.

Difference of Squares

a^{2} - b^{2}

(a + b)(a - b)

Example:

x^{2} - 4 = (x + 2)(x - 2)

a = x, b = +2

a = x, b = +2

A Plus B Squared

(a + b)^{2}

a^{2} + 2ab + b^{2}

Example:

x2 + 4x + 4 = (x + 2)2

a = x, b = +2^{}

a = x, b = +2

A Minus B Squared

(a - b)^{2}

a^{2} - 2ab + b^{2}

Example:

x^{2} - 4x + 4 = (x - 2)^{2}

a = x, b = +2

a = x, b = +2

Simplifying Expressions (remove common factor, example 1)

(210+560)/490 = ?

(70*3+70*8)/(70*7)= ?

70(3+8)/70(7) = ?

= 11/7

Simplifying Expressions (remove common factor, example 2)

(3x+27y+15z)/(x+9y+5z = ?

3(x+9y+5z)/(x+9y+5z) =

= 3

Simplifying Expressions (remove common factor, example 3)

(x^{2}-9)/(x+3) = 0, x = ?

(x+3)(x-3)/(x+3) = 0

(x-3)=0

x=3

simplifying radicals

when number inside a radical is a multiple of a perfect square, the expression can be simplified by factoring out the perfect square

√72 = √36x2 = 6√2

simplifying radicals (example)

(√125√28)/√35 = ?

(√25*5 √7*4)/√5√7

(√25√5√7√4/√5√7

√25√4

5*2 = 10

Complex Fraction Equality

(a/b)/(c/d)

(a/b)/(c/d) = (ad)/(bc)

Example: (1/2)/(3/4) =

4/6 = 1/3

Complementary Angles

Two angles whose sum is 90°

Supplementary Angles

Two angles whose sum is 180°

Angle Bisector

A ray that divides an angle into two equal parts

- Alternate Interior Angles Are Equal: K = L; O = J
- Alternate Exterior Angles Are Equal: H = M; N = I
- Corresponding Angles Are Equal: K = N; J = M; H = O; I = L
- Non-Alternate Interior Angles Are Supplementary: L + J = 180; K + O = 180

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly! I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU
StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2015 StudyBlue Inc. All rights reserved.

© 2015 StudyBlue Inc. All rights reserved.