### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# Note for MATH 113 at UMass

### View Full Document

## 14

## 0

## Popular in Course

## Popular in Department

This 7 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Massachusetts taught by a professor in Fall. Since its upload, it has received 14 views.

## Similar to Course at UMass

## Popular in Subject

## Reviews for Note for MATH 113 at UMass

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 02/06/15

1 Order of Operations a PEMDAS Parenthesis Exponents Multiply Divide Add Subtract 2 Mental math techniques a count onbackbegin with the subtrahend and count up to the minuend i nd the missing addend by counting up from the subtrahend Record the count for each power of ten The difference is the sum of the values from the count b choose compatible numbers look for numbers that add equally to a number oftens 51 1970 c break apart numbers when you break apart numbers to add the 1039s and 139s separately 1 4836 1 First add the tens 4030 70 2 Next add the ones 8614 3 48 3684 d use compensation changing one number to make it easier28 30 11 10 etc e equal additions For each place value an equal amount is added to both the minuend and subtrahend to form a basic fact subtraction The actual subtraction may be completed by either the take away or missing addend approach 1 6 7 49 1 67 1 49 1 a 68 50 18 3 Estimation strategies a Rounding round each number to the 100s b substitute compatible numbers c frontend with and without adjustment use front part of numbers 1 350 220 1 Add thefront digits in each number a 300 200 500 2 Adjust to account for the remaining digits a 50 20 70 3 500 70570 d Clustering When all numbers are near each other i 39384340 1 40 x 40 160 e range estimation estimate a range that addition falls under 4 different algorithms for whole number addition in any base a standard algorithm i begin with the righthand column and proceed to the left one column at a time The exchange is recorded at the top of the next column 437 l 11 521 368 693 958 425 458 793 1151 5 Lattice Addition Algorithm a This is basically the whole group algorithm6rec90rd 2d in a different format 4 3 7 3 6 8 s 71 42 5 458 0 0 0 0 9 508 7 813 101411 9 5 8 7 93 1151 b expanded algorithm i Write the number in expanded notation and then perform the operations 437400307 36830060 8 693 600 90 3 521500201425400205 458400 50 8 900508 7008013 100014011 958 70080103 100010040101 700903 1000100501 793 1151 6 different algorithms for whole number subtraction in any base a standard algorithm Exchanges completed only on the minuend The actual subtraction may be completed by either the takeaway or missingaddend approach 785 1113 214813 353 1317 3493 432 1247 1527 758 1966 489 1 b expanded algorithm c adding the complement algorithm i In base ten the complement of a whole number is found by subtracting each digit from 9 Numbercomplement 72 27 99 4 72 7 27 358 642 999 4 358 7 642 10294 89705 99999 4 10294 7 89705 ii Change the problem to an addition problem by adding to the subtrahend and subtracting that amount at the end of the problem The procedure is to replace the subtrahend with its complement then adding the two values Finally subtract the amount that was added by dropping the lead one and adding one to the result 7 8 5 Add 999 to the total value of the problem 7 8 5 6 4 6 That is 999 7 353 646 The 646 is the complement of353 l 4 3 1 Since this value is 999 larger than the actual difference subtract 999 We do this l subtraction by noting that 999 1000 7 1 So we subtract 1000 and add 1 4 3 2 l 2 4 7 3 4 9 3 7 7 5 8 7 l 5 2 7 Add 9999 to the total value 1 2 4 7 3 4 9 3 9 2 4 l 8 4 7 2 That is 9999 71527 8472 l 0 4 8 8 l l 9 6 5 l 1 Since 10000 7 l 9999 we remove the 9999 4 8 9 l 9 6 6 we added earlier by subtracting 10000 and adding 1 7 different algorithms for whole number multiplication in any base a standard algorithm i This is basically an abbreviation of the partial products algorithm 4 7 XI3 1 4 147x3141 4 7 047x10470 6 1 1 b expanded algorithmpartial products i This algorithm is similar to the partial sums algorithm for addition The procedure is to multiply one pair of digits at a time 21 7x321 70 7x1070 40x10400 6 11 ii Note that with this algorithm it does not matter the order in which digits are multiplied commutative property c lattice algorithm i This is basically the partial products algorithm recorded in a different format 47Xl36ll 4 7 0 O4 O71 612213 8 different algorithms for whole number division in any base a standard algorithm long division i The procedure is repeated subtraction where the largest possible power of ten multiple is subtracted each timeand the quotient is written above the dividend b expanded algorithm scaffolding i This is a more efficient version of repeated subtraction The procedure is to subtract multiples of the divisor ii Note that the multiple chosen maybe any number that is less than the dividend 89l7 17 89 734 2 55 734 2 21 717 l 4 5 89l75R4 9 determine whether or not a relation rule of assignment is a function a Definitions i A relation is a set of ordered pairs where the first components of the ordered pairs are the input values and the second components are the output values ii A function is a relation that assigns to each input number EXACTLY ONE output number 1 Be careful Not every relation is a function A function has to t the above definition to a tee b Vertical line test no 2 points of y can correspond to one point of X 10 evaluate a function for a particular value of a variable a To evaluate a function we insert a given X value a number in the domain and see what number we get which is a number in a range i To evaluate f4 f4 24 8 1 We just evaluatedfX for the value X 4 ll understand the concept of a function machine a allows you to eXperiment with various functions 12 solve an equation for one variable in equations that contain two variables a Use substitution 13 set of natural numbers a counting numbers starts with one 14 factors and multiples a Factor a number that is multiplied to get a product i l 3 9 are factors of9 b Multiple the product of a quantity by an integer i 36 45 63 are some multiples of9 15 divisibility rules for the numbers 2 3 4 5 6 8 9 10 and 11 2 even numbers 3 if the sum of all digits is divisible by 3 4 last two digits in your number divisible by 4 5 ends in 5 or0 6 divisible by 2 and 3 8 the last 3 digits are divisible by 8 10 ends in a 0 11 If you sum every second digit and then subtract all other digits and the answer is 0 or divisible by ll 16 the Factor Test Theorem use to nd and list all distinct factors of a number a If fa 0 then Xa is a factor of fX l7 prime number vs composite number vs the number one a prime natural number that has exactly two distinct natural number divisors l and itself b Composite positive integer which has a positive divisor other than one or itself c Number 1 neither prime or composite 18 the Factor Test Theorem use to tell whether or not a number is prime a 19 the Fundamental Theorem of Arithmetic aka Unique Factorization Theorem a anyinteger greater than 1 can be written as a unique product up to ordering of the factors of prime numbers L 6936 23 x31gtlt172 1200 24 x 31 x 52 20 Different ways to find GCF and LCM a list out factors and multiples b use prime factorizations To Find the LCM of two natural numbers rst nd the prime factorization of each number Then take each or the prunes that are Factnrs of either of the given numbers The LCM IS the product of these artmes each raised to the greatest power of the prime that occurs in either of the prime c factorizations d Euclidean Algorithm to find GCF i Divide larger number by smaller ii If remainder is not 0 divide first divisor by remainder 1 16573 2478 6 Rl705 a 24781705 1 R 773 e use the GCFLCM Theorem P qorhrDPpp e 71m u m h 01 mm q m puumu W3 m 133 M1119 anlqul nu mp Wu mu um 7 U YUUJ 7 2UliU 7 m Knowm Hun 1N1 m 7m MW smug pumqmnn le uluanmpJnnnprud1qzu5l pur um su mm mg n mnr lpJn mnde w 55 us mm m m mm mt z 1 mm ml 4 mm m mi 5 uuuuumm amud m4 inmoqlt unu mp v 57 7n 39ur mun 3W w suunuumsvj mm m L m pm mu mf put mg mp mlusuua Wm put 13 mp u nm uuuaauuu 3 no UL

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

#### "Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.