the first thing we’re going to do is order numbers from least to greatest on a number line we’re ordering negative 2 5 6 0.9 1.2 and negative 1 and 1/5 so first negative 2 well that’s already on our number line so we’re going to put a dot there and just to remind ourselves we’ll say negative 2 5 6 well that’s going to be a little bit more than 1/2 and a little bit less than 1 so 5 6 would be somewhere around here 0.9 would also be a little bit less than 1 okay so we need to figure out between these 2 what comes first and we’ll do that in a little bit 1 point to be right about here and negative 1 and 1/5 here is negative 1 and 1/5 would be a little bit further so we have negative 1 and 1/5 so now when you need to determine between 5/6 and 0.9 which one comes first to do that it might be best to change 5/6 into a decimal so we say 5/6 can’t do that so we put decimal 0 put our decimal right at the top 6 goes into 50 8 times which is 48 subtract and get to bring down another 0 6 goes into 23 times should be 18 to bring down another 0 and that’s just going to repeat ok so comparing 0.9 and 0.8 3 repeating this would come first which means 5/6 is smaller than 0.9 so if we put them in order from least to greatest and be negative 2 negative 1 and 1/5 5 6 0.9 and 1.2 now we’re going to work with multiples and factors factors are numbers that divide evenly into a given number so if we’re looking at factors of 12 we want to find all the numbers that go into 12 with no remainder you can always start with 1 and whatever number you’re talking about so 1 in 12 2 goes into 12 and that’s paired up with 6 and then 3 is paired up with 4 so the factors of 12 are 1 2 3 4 6 and 12 multiples are the product of a given number and any nonzero whole number so it’s like we’re going to skip count by 12 so 12 would be the first multiple of 12 and then 12 times 2 is 24 12 times 3 is 36 12 times 4 is 48 and then 60 and then it keeps going on and on forever so notice there’s a set number of factors but there’s an infinite number of multiples so some things we can do with factors are multiples or find the greatest common factor which is called GCF or the least common multiple LCM of two numbers let’s try to find the GCF and the LCM of twenty and thirty six we can find these in two different ways the first method is to list so we are going to write down two and 36 we’re going to start by listing the factors for both of them per 20 it’d be 1 + 22 + 10 3 doesn’t go in but there’s 4 and 5 for 36 would be 1 + 36 2 + 18 3 and 12 4 9 and then 6 times 6 so we only write it once so if you’re

finding common factors well they both have one they both have 2 they both have 4 so the greatest common factor greatest means biggest common factor that would be 4 so you would say the GCF of 20 and 30 6 is 4 so let’s do multiples now so the multiples of 20 will be 20 40 60 80 100 and so on we’ll stop there for now 36 36 times 2 is 72 times 3 is 108 times 4 would be 144 and times 5 would be 180 okay that could go on and on also we see here that one of the multiples of 36 is 180 all the multiples of 20 are either going to end in 2040 60 80 or 0 0 so the next multiple here that would match up with something on this list would be 180 so our least common multiple of 20 and 36 is 180 now you can also find GCF and LCM by prime factoring only prime factor you want all of our numbers to be prime numbers people often use factor trees to prime factor but we can do it without one so we’re going to say 20 is 2 because that’s our smallest prime factor we’ll start with 2 2 times 10 10 can be broken up into 2 times 5 4 36 it would be 2 times 18 18 would be 2 times 9 and 9 is 3 times 3 for GCF what we’re going to do is find our common factors and circle them so we have a pair of 2’s and another pair of twos so we’re going to build our GCF we’re going to take one number from each pair and write it down and then we multiply them together so our GCF is 4 which we found before for LCM you start the same way you prime factor you circle your common factors you write your common factors down one from each pair but then you want to take all the extra numbers so 3 times 3 times 5 and now you’re going to multiply all of those together 2 times 5 gives you 10 and then 2 times 3 is 6 6 times 3 is 18 10 times 18 is 180 let’s use the distributive property to multiply 4 times 107 you can certainly sit there and do long multiplication but we’re going to split the number up so that’s easier to do mentally so we’re going to say 4 times 107 is the same thing as 4 inside the parentheses we’re going to split 107 up a nice number that’s close to 107 is 100 but to make sure we don’t change the value of our parentheses we’re going to say 100 plus 7 so this still equals 107 and we’re still multiplying it by 4 now the distributive property says if you have a number on the outside of parentheses and on the inside we have plus or minus we can take the number on the outside and multiply it by both things on the inside so that would mean this is 4 times 100 we bring down our plus and then 4 times 7 so we get 400 plus 28 and our answer is 428 here we’re going to apply rules of exponents in numerical expressions with rational exponents to write equivalent

expressions with rational exponents the first thing is when you’re multiplying the same base you can add the exponents together so we have 4 to the 5th times 4 to the 7th our base is 4 and our exponents are 5 and 7 so we’re going to say 4 and our exponent we’re going to add them together and 5 plus 7 is 12 so it’s 4 to the 12 power when we’re dividing the same base we can subtract the exponents so we have 7 to the 7th divided by 7 to the 3rd our base stays the same but now we subtract our exponents so our answer is 7 to the 4th anything besides zero raised to the power of zero equals one so negative six raised to the zero power is one when you raise an existing power to another power you multiply the exponents together so here we have three to the fifth but then we’re raising that to the sixth power so we’re going to say three and multiply our exponents together which equals three to the thirtieth now a number raised to a negative exponent means you send it to the denominator of a fraction so three to the negative second power means we’re going to create a fraction and then it’s going to go into the denominator but when you put it in the denominator your negative exponent becomes positive and we can’t just leave a blank in the numerator so we fill it in with a 1 so it equal one over three squared or if we want to simplify that a little bit further we’d say 1 over 9 ok a fractional exponent like 1 over N needs to take the nth root so for 16 to the 1/2 power it would be the square root of 16 which equals 4 finally a fractional exponent like M over N means to take the nth root first then raise it to the nth power so here this would mean to take the square root of 4 but then raise that to the third power the square root of 4 would be 2 so 2 to the third power will be 2 times 2 times 2 which equals Heat the last topic we’re going to talk about is absolute value the absolute value of a number is the numbers distance from zero on a number line a quick way to remember how to do absolute value is that it is always positive the absolute value of a number a is we’re in like so okay we kind of have two little sticks on the outside of your number so for looking at the absolute value of four here’s four here’s zero that would be four spots away so the absolute value of 4 is 4 4 looking at negative 4 once again here is 0 W 1 2 3 4 spots away so the absolute value of negative 4 is also 4 now the distance between two numbers on a number line is always positive take the absolute value of the difference so if we want to find the difference between negative 5 & 6 we’d say first let’s subtract negative 5 minus 6 that’s the same thing as adding the opposite so negative 5 plus negative 6 would be negative 11 but we want the distance so it’s the absolute value of that which means the distance between negative 5 & 6 is 11 now let’s move on to fractions when

adding and subtracting fractions you need to find a common denominator usually you try and find the least common denominator or the LCD that means it’s the least common multiple of the denominators so looking at 2/3 and 4/5 we need to find the least common multiple of the denominators which would be 3 and 5 and that would be 15 but we need to change our numerator so we can’t just leave them as 2 and 4 so we’re going to say is our original denominator was 3 3 times 5 gave us the 15 so 2 times 5 gives us a new numerator of 10 for 4/5 5 times 3 gave us 15 so 4 times 3 gives us 12 and now since we have the same denominator we can add our numerators our denominator stays the same we want to make sure we change that into a mixed number if we have an improper fraction so our answer would be 1 because 15 goes into 22 once and there’d be seven fifteenth’s leftover when multiplying fractions you want to cross reduce and divide out any common factors then you’re going to multiply your numerators and multiply your denominators so cross reducing means you’re looking on the diagonal here 3 goes into both 3 & 9 so I’m going to divide that out so 3 goes into 3 once and 3 goes into 9 3 times on this diagonal we can divide out of 5 5 goes into 5 once 5 goes into 10 twice so now we’re going to multiply 1 times 1 which gives us 1 and 2 times 3 which gives us 6 so our answer is 1/6 for division you multiply the first fraction by the reciprocal of the second fraction that really means you’re going to keep change flip so we’re going to keep 7/8 the same our division changes to multiplication we take the reciprocal of the second fraction which means we flip our numerator and our denominator so we’re going to say 20 over 13 so now that we’re multiplying we can follow these rules and we can cross reduce I see that 4 goes into both 8 and 20 so 4 goes into 8 two times and 4 goes into 20 five times nothing divides out between 7 and 13 so we multiply our numerators 7 times 5 which gives us 35 2 times 13 is 26 and if we change that back to a mixed number be 1 + 9 26 if you have mixed numbers you should change them to improper fractions before you start trying to solve the problem ok we’re not going to totally finish this problem but if we’re going to start it off we’d say 2 1 1/5 well 2 times 5 is 10 plus the 1 is 11 so 11 fifths 5 + 3 8 5 times 8 is 40 plus 3 is 43 so 43 ace and then you would continue on from there for decimals when you’re adding and subtracting you need to line up the decimal point and then you’re just going to add or subtract as usual so here we would say three point three four minus one point notice the decimal points are lined up one point seven one five but there’s a missing space here but if we have a terminating decimal which is the case we can always add a zero to the end because that doesn’t change the decimal and now we can subtract we can’t do zero minus five so we need to borrow so the four becomes a three and our zero becomes a 10 10 minus 5 is 5 3 minus 1 is 2 you need to borrow again so it becomes a 2 and that becomes 13 13 minus 7 is 6 we just bring down our decimal point and 2 minus 1 is 1 for multiplication we don’t want to line up the decimal points we’re going to set up the numbers as if we’re doing a long multiplication problem but line up the digits of your numbers along

the right hand side so looking at three point eight and two point one nine I see that this one has more digits so I’m going to put that one on top we don’t want to put three point eight here because that’s going to make life a little bit harder for us okay so we’re going to line up the digits along the right okay the digits line up now we can multiply 9 times 8 is 72 8 times 1 is 8 plus 7 is 15 8 times 2 is 16 plus 1 is 17 leave a space 3 times 9 is 27 3 times 1 is 3 plus 2 is 5 and 3 times 2 is 6 we add them together get two we get 12 we get 13 let me get eight and now we’re going to count how many decimal places were in our question we had two decimal places here and one here so our answer is going to have three total decimal places so 1 2 3 so 8 point 3 2 2 finally for division if we’re dividing by a decimal we want to slide the decimal over in our divisor until we get a whole number and then however many times we sit it over we’re going to do the same thing in the dividend so we’re going to do one point 4 3 divided by 1.1 however we can’t divide by a decimal so we slide it over once here so we have 11 which means we’re going to slide it over here okay so 11 goes into 14 once with a remainder of 3 and we bring down our 3 and 11 goes into 33 3 times with no remainder so our answer is 1.3 work with squares and square roots squaring the number means you’re multiplying a number by itself it means you have the power with an exponent of 2 so 5 squared would be 5 times 5 which equals 25 negative 8 quantity squared would be negative 8 times negative 8 we learned that a negative times a negative is a positive so be a positive 60 for this question looks like our last one however there’s no parentheses so what this means is we’re doing 8 squared but then on front of it we’re going to stick a little negative sign so this means negative 8 times 8 which would mean that this is 64 with a negative out front and then 2/3 squared would be 2/3 times 2/3 multiply our numerators multiply our denominators and we get 4/9 finding the square root of a number means finding a number that one multiplied times itself equals the given number so the square root of 81 means we need to find a number that when you multiply it by itself gives you 81 and the answer to that is 9 the square root of 121 means what number times itself equals 121 the answer to that is 11 now a lot of times the number under the square root is not a perfect square 81 and 121 are examples of perfect squares because when you square root it you get a nice whole number if we start from the beginning our perfect squares would be 1 squared which is 1 2 squared which is 4 3 squared which is 9 4 squared which is 16 5 squared which is 25 and so on but most of the time like I said these numbers are not perfect squares so we can leave these answers in radical form so what we’re going to do is see if we can find a perfect square that goes into our numbers well I know for 32 that that equals 4 times 8 so we can say that route 32 is really root 4 times root 8 but really root 8 could be broken up more so that would be root 4

and then root 8 would be root 4 times root 2 and now we have the square root of 4 that’s 2 again the square root of 4 is 2 and then we can’t really do anything with that root 2 so we leave it the same so our final answer would be 4 as 2 times 2 is 4 4 root 2 root 80 let’s see well root 80 would be root 4 times root 20 but root 20 would be root 4 times root 5 so this will be 2 this will be 2 leave that as root 5 and our answer is 4 root 5 now let’s look with cubes and cube roots cubing a number means you’ll have a power with an exponent of 3 so 5 Q to a be 5 times 5 times 5 which is 125 negative 2 quantity cubed would be negative 2 times a negative 2 times negative 2 these here give you a positive 4 so a positive 4 times a negative 2 is a negative 8 negative 2 cubed would be negative 2 times 2 times 2 this here gives you 8 so B negative 8 and then 2/3 cubed would be 2/3 times two-thirds times two-thirds 2 times 2 times 2 gives us 8 3 times 3 times 3 gives us 27 so it’s 8 27 so taking the cube root of a number means finding a number that when you multiply it by itself two other times you get whatever numbers underneath it so here we want to find a number that when you multiply it by itself again and then again one more time gives you eight well 2 times 2 would be 4 and times 2 again would be 8 so the cube root of 8 is 2 the cube root of a thousand if we took 10 and multiplied that by 10 it would be 100 and 100 times 10 again would be a thousand so the answer for the cube root of a thousand is 10 now there’s times when numbers are undefined undefined means when you have M over 0 and M doesn’t equal 0 that gives you a something else so if I have 8 over 0 its undefined this is because the fraction bar really means division and we can’t take 8 things and divide it up into 0 groups that just doesn’t make sense so whenever you have a fraction that has a denominator of 0 its undefined now certain numbers are undefined in certain sets so square roots of negative numbers are undefined in the set of real numbers okay you can’t have a square root of a negative number because a negative times a negative will give you a positive not a negative number okay there are imaginary numbers okay that will make this problem possible however in the set of real numbers square root of negative 81 is undefined now there’s times when numbers are undefined undefined means when you have M over zero and M doesn’t equal zero that gives you a something else so if I have 8 over 0 its undefined this is

because the fraction bar really means division and we can’t take 8 things and divide it up into zero groups that just doesn’t make sense so whenever you have a fraction that has a denominator of 0 its undefined now certain numbers are undefined in certain sets so square roots of negative numbers are undefined in the set of real numbers okay you can’t have a square root of a negative number because a negative times a negative will give you a positive not a negative number okay there are imaginary numbers okay that will make this problem possible however in the set of real numbers square root of negative 81 is undefined finally let’s work a little bit with scientific notation scientific notation is a way of writing really big numbers or really tiny decimals in a shorthand way so you don’t have to write out all the zeros inside the numbers so if we want to change from standard form to scientific notation we want to place a decimal somewhere in our number so that we create a number between one and 10 so where we would put this decimal here to create a number between 1 and 10 will be right there in between the nine in the two because nine is in between one and ten so we’re going to rewrite this we’re going to say nine point two eight can we chop off all those extra zeroes you’re always going to have times 10 to some power now it’s important to remember this here a positive exponent is going to give you a big number if you put a negative exponent means you’re talking about a decimal so here it was a big number so we’re going to have a positive exponent and to determine what that exponent is going to be we’re going to start with our original decimal and see how many spots we slid to get to our new decimal well you have to remember that on a whole number there’s an invisible decimal at the end of it okay so we slid it once twice three times four five six so would be nine point two eight times 10 to the sixth power for this one since it’s a decimal we know it’s going to have a negative exponent so we need to find a number between 1 and 10 that would happen by placing the decimal between the 6 and the 2 and then we’re going to count how many we moved so 1 2 3 4 5 so it’s 6 point 2 times 10 to the negative fifth to go back from scientific notation to standard form well we’re going to say let’s take that 2 point 7 8 and write it over here to give ourselves a little bit more room and since it’s 10 to the seventh power we want to move it 7 places now since it’s a positive 7 it means we’re creating a big number so we want to slide that decimal point to the right so we’re going to start here we move it seven times so 1 2 3 4 5 6 7 and with all those blank spots we’re going to fill in zeros so our final answer would be 27 million eight hundred thousand let’s rewrite 4.15 we need to move it five decimal places and since it’s a negative number we’re creating a decimal which means we move it to the left so removing it once twice three four five we fill in those decimals okay notice it’s not how many zeros we add to our number it’s how many times we slide that decimal place so our answers zero point zero zero zero zero four one five finally let’s multiply two numbers in scientific notation we have two point seven times ten to the seventh and three point one times 10 to the fifth now multiplication is commutative and associative means we can change up the groups and we can do whatever order we want so let’s group to point seven and three point one together and group ten to the seventh times 10 to the fifth together two point seven times three

point one be seven times two and then 3 times 7 is 21 three times two is six plus two is eight one decimal place here one decimal place here which means two decimal places in our answer so this gives us eight point three seven and we learned that when we multiply the same base we add our x1 Stu gether so be 10 to the twelfth power last thing we’re going to do is calculate weighted average according to a teacher syllabus 60% of Anna’s marking period average is based on her tests and quizzes 30% is based on homework and 10% is based on classwork she has a test in quiz average of 85 a homework average of 95 and a classwork average of 90 what is Anna’s marking period average rounded to the nearest whole number okay so her tests and quizzes were 60% and that was a grade of 85 30% was homework and her homework grade is a 95 and then 10% is classwork and our classwork was a 90 so what we’re going to do is multiply the percent by the respected grade however when we multiplied by percents we have to remember to change to the percent to a decimal so what we’re really doing here is 0.6 times 85 0.3 times 95 and 0.1 times 90 and then we’re going to add those up together so 85 times 0.6 5 times 6 is 30 8 times 6 is 48 plus 3 is 51 one decimal place so the first part is 51 95 times 0.3 that gives you 15 that gives you 28 with one decimal so twenty-eight point five and then point one times ninety is just going to be nine you can fill in those missing decimals so we add we get five we get 18 we get an 8 and rounded to the nearest whole number Anna’s final marking period grade is going to be in 89 so the four main measures of central tendency are mean median mode and range when you hear the word average that’s talking about the mean you add up the numbers and divide by however many numbers you had the median is the middle number you need to order your numbers and then find the one in the middle now if there’s two numbers in the middle you’re just going to find the mean of the two of them so you add those two numbers up and divide it by two and that’s your median the mode is the number that occurs the most often and the range is the biggest number minus the smallest number so we’re going to find the mean median mode and range of this set of data so for me we’re going to add up the numbers 4 plus 11 gives you 15 plus 2 is 17 plus 5 is 22 plus 6 is 28 plus 4 is 32 plus 3 is 35 so the sum was 35 and we’re going to divide by however many numbers we had which was 7 which means the mean is 5 for median we’re going to put our numbers in order so our smallest number is two then three and two fours then a 5 a 6 and 11 a lot of people like to cross numbers off so you start on one side you cross off a number on the other one and keep going crossing off the same amount on each side and then you get your number in the middle which is your median the mode is the number that occurs the most often the number four occurs twice well everything else only appears once which

means that four is our mode and the range is the biggest number minus the smallest number so that would be eleven minus two which gives you nine now often you’ll be given a set of data you’ll be told what the average is and then you’ll have to find a missing part of the data so for example it says Dan has received test grades of 90 80 65 and 95 what grade does he have to get on his fifth test to ensure his test average is an 85 so you need to figure out the global amount of points that Dan needs to get well if the average is 85 then we can say let’s take the 85 and multiply it by the five tests five times five is 25 eight times five is forty plus 2 is 42 so altogether his test grades need to add up to 425 well let’s figure out what he has so far we’re going to add up the ones that we were given it gives us 10 that’s 18 plus 6 is 24 plus 9 is 33 so so far he has 330 points he needs 425 he has 330 we’re going to subtract to figure out how many he needs on that last test we get 5 we borrow so Dan needs to get a 95 on his fifth test to make sure his test average is an 85 now we’re going to find the average number of siblings based on the frequency counts provided in the table now it’s often good to take the data and write it out so we can see it a little bit better so zero occurs once one occurs four times two occurs twice three occurs twice and for occurs once the average means we’re going to add them up and divide by 10 because there’s 10 numbers right here is 4 plus another 4 is 8 plus 6 is 14 plus 4 is 18 so the sum is 18 we’re going to divide that by 10 and the average is 1.8 okay let’s deal with a little bit of probability now probability is the number of favorable results over the total possible results and probability is always going to range between 0 and 1 0 means it’s definitely not going to happen a probability of 1 means it’s definitely going to happen but most of the time our probabilities are going to fall somewhere in between the two you can write probability as a fraction a decimal or as a percent but you have to make sure it’s between 0 and 1 so if we roll a 6-sided die once what is the probability of rolling an even number well if we have a six-sided die that means the total possible results are 6 because you can get a 1 a 2 a 3 a 4 or 5 or 6 well even numbers would be 2 4 & 6 so those would be the favorable results so we can have 3 possible good things over 6 total results which means our probability is 1/2 now a lot of times you’ll have to find probabilities for several things happening at once that’s what is called a compound event it’s a combination of two or more simple events and can be found by multiplying the two probabilities together now two things can happen you can have independent events which is events where one outcome does not affect the other outcome on the other hand dependent events means the result of the first outcome affects the probability of the second outcome okay so we have both cases demonstrated here if you roll a 6-sided die and flip a coin what is the probability of rolling a number bigger than 4 and having the coin land on heads now if we’re rolling

a six-sided die and flipping a coin those two things are totally separate okay flipping a coin is not going to affect rolling a die and rolling a die is not going to affect flipping a coin so these are independent events so we can just figure out each of the probabilities and then multiply them together to get our answer so for the die we want a number bigger than 4 once again for our die we have 6 total possible results and if we want a number bigger than 4 or options would be 5 or 6 which means there’s two favorable results for our coin if we want it to land on heads well there’s only two possible options heads or tails and heads would only be one favourable outcome so we have 2 6 times 1/2 we’re going to multiply those together we can either simplify now or simplify later but let’s multiply and we get 2 over 12 and then we can simplify that to 1 over 6 and that is the probability of rolling a number bigger than 4 and having the coin land on heads okay well that was an independent event let’s look at some dependent events there are four yellow marbles and six blue marbles in a bag what is the probability of choosing two blue marbles without replacement when you see without replacement that usually Clues you in to the fact that it’s dependent and you need to be careful because if you take out that first blue marble okay there’s not going to be the same number of marbles in that bag so the probability of taking that second marble is going to be affected by taking out the first marble okay so we’re going to say let’s assume that we reach our hand into this bag and pull out a blue marble okay well that probability 4 plus 6 there are 10 total marbles at first and there’s 6 blue ones okay so our first probability of grabbing a blue marble is 6 over 10 so now we need to assume that we have taken that marble out we have a blue marble okay which means that the number of marbles in our bag has gone down from 10 to 9 and if we took out a blue marble we don’t have 6 anymore we have 5 instead okay so let’s cross reduce in this case well simplify before we multiply 5 goes into 5 once 5 goes into 10 twice 3 goes into 6 twice and 3 goes into 9 3 times and actually I can simplify here more they reduce so really just 1 times 1 is 1 and 1 times 3 is 3 and that is our dependent probability the fundamental counting principle says if there are M Way is for one activity to occur and n ways for a second activity to her then multiplying M times n will give you the total possible outcomes of both occurring so for example a restaurant offers a three-course meal where you can choose one appetizer one entree and one dessert there are five appetizers to choose from eight entrees and three desserts how many possible three-course meals can you choose from in this case all we have to do is take the number that we have in each category and multiply everything together so we have five appetizers eight entrees and three desserts so we multiply 5 times 8 times 3 5 times 8 is 40 and 40 times 3 is 120 so there is 120 possible meals to choose from some other counting techniques are permutations and combinations these are both ways of ordering sets and the important question you need to ask yourself is does the

order of the set matter for permutation and the order matters some examples are races finishing 1st 2nd and 3rd choosing officers such as the president vice-president secretary and treasurer trying to calculate pin numbers and arranging books on a shelf two things can happen with permutation we can either order everything in the set or we can just choose to take a couple things from the set and order those in our first example it says in a raise of five runners how many ways can the runners finish so in this case we’re ordering all five runners and to do that we’re going to take the number what we have which is five and we say five factorial now what a factorial means is you’re going to take the number that you’re starting with and you’re going to keep multiplying by every number that’s smaller than it until you get to one so five factorial means 5 times 4 times 3 times 2 times 1 5 times 4 is 20 20 times 3 is 60 and 60 times 2 is 120 so they can place in 120 different ways now we’re going to have a similar question but in this case we only care about first second and third place and it says in a race with five runners how many ways can the runners finish first second or third in this case we’re not ordering all five of the runners just a couple of them in your calculator there’s often a button that will do this for you in that case we would say we’re taking our five runners there should be some function for permutation it’s usually NPR and if we’re choosing three people we would say three and the calculator will do it for us however right now we don’t have a calculator so we’re going to do it the old-fashioned way and that’s using this formula here we’re going to take n factorial which is how many runners we have so five runners and then this n minus K is the number of runners minus the number of people were actually taking in our set so it would look like five factorial over five minus three factorial okay which is really five factorial over 2 factorial now we’re not actually going to multiply anything it I’m just going to write it out so 5 times 4 times 3 times 2 times 1 and then over here on the bottom we have 2 times 1 and in these cases of permutations and you’ll find the same thing happens in combination we can often cancel out some common factors in our numerator and our denominator so over here we have twos num ones that we can cross off which means we are really just left with 5 times 4 times 3 5 times 4 is 20 and 20 times 3 is 60 so the runners can finish first second or third in 60 different ways now for combination the order does not matter for example if you’re choosing a committee which is just a group to represent people where no person has a different role than everyone else that’s a combination because it doesn’t matter in what order you choose those people you might also see it picking pairs or groups that’s a combination so let’s look at this example how many ways can two names be chosen from 10 in a raffle if only one entry per person is allowed so in this case we’re starting with 10 but we’re choosing two names so once again if we had a calculator we would say we’re taking 10 people the button you often press this n CR C for combination this time and two people at a time okay but let’s use our formula it’s very similar to our permutation formula except we’re adding an extra K factorial at the end so we’re going to say 10 factorial over 10 minus 2 factorial and our K is 2 factorial that would be 10 factorial 10 minus 2 is 8 and now let’s write it out we have 10 times 9 times 8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 and then

8 times 7 times 6 times 5 times 4 times 3 times 2 times 1 and then another 2 times 1 and while it looks like a lot we see we have an eight and an eight seven and a seven okay and a lot of our numbers will divide out so really we’re left with ninety 10 times 9 over 2 times 1 which is 2 and 90 over 2 is 45 this is a GED review video for expressions and polynomials first we’re going to deal with linear expressions and these are expressions that have one variable and that variable needs to be raised to the power of 1 now first we need to deal with some definitions terms are the parts of an expression that are added together like terms are terms that have the same variable raised to the same exponent for example if I have 2x and 3x those are like terms because they both have a variable of X and both X’s are raised to the first power finally a constant is a number with no variable attached for example 4 in an expression 2 constants are also considered like terms so first we’re going to add two expressions together so I have 4x plus 7 plus 8x minus 4 now the only thing we can do here is combine our like terms okay and once again like terms are terms that have the same variable raised to the same power so here I have 4x + 8 X I also have a positive 7 and while this says minus 4 we should know that minus and negative are really the same thing because we could change this to plus a negative 4 if we chose to do so so we have a positive 7 and a negative 4 so 4x plus 8x we add the coefficients together for X + 8 X gives us 12x + 7 plus a negative 4 is a positive 3 so our answer is 12x + 3 now if we’re subtracting we have 4x + 7 – 8 X – 4 now when you’re subtracting expressions you have to remember that this minus sign actually means we’re subtracting both things in our parenthesis so it might be helpful to rewrite the problem we have 4x plus 7 but now we’re subtracting 8x okay and here we’re subtracting a negative 4 which really means we are adding 4 so once again my like terms are 4x and in this case negative 8x and then 7 and positive 4 4x plus negative 8x gives me a negative 4x and 7 plus 4 is a positive 11 okay here we have an expression on the inside multiplied by a whole number on the outside this is what’s called the distributive property we need to take this number on the outside and multiply it by both things on the inside so we’re going to say 3 times 4x we always bring down whatever sign is there and then 3 times 7 well 3 times 4x is 12x and 3 times 7 is 21 once again not like terms so that is our final answer now I just want to back up one second because there’s actually another way of writing each of these answers it’s usually common to rate your answers with your variable first and then your constant after okay or in certain cases where we have multiple

terms going on we usually start with the biggest exponent and work our way down however it’s also possible to switch the order of our terms so for our first one instead of saying 12x plus 3 we could say 3 plus 12x okay for our second one we need to make sure that the sign in front of it stays with it when it moves okay so if we want to move this 4x we need to keep that minus sign in front of it and this positive 11 is going to go out front and since it’s out front we don’t need to put the positive sign in front of it and finally for this one there’s no negatives to worry about so we could also say if we chose 21 plus 12x okay we now have an example of factoring and this example that we have is kind of the most basic type of factoring we could do it’s pretty much the reverse distributive property here we had a number on the outside and we multiplied it to both things on the inside to get rid of our parentheses now we want to see if there’s a number that goes into both of our terms and we’re going to pull it out and kind of create some parentheses here so I have 8x minus 4 and I can say that full our can go into both 8x and for so I want to pull out the 4 which means in my parentheses I have to say well 4 times what gives me 8x and the answer that would be 2x okay bring down our minus 4 times what is 4 and that answer is 1 so 8x minus 4 factor would be 4 times in parenthesis 2 2x minus 1 this is a GED review video for expressions and polynomials ok you’ll often need to evaluate expressions by substituting an integer in for an unknown quantity so for example we’re going to evaluate 4 X plus 7 for x equals 3 I would recommend writing in your numbers first especially when your expressions are a lot bigger so you don’t make any careless mistakes so we’re going to say for this means multiplication when they’re squished together so 4 times 3 plus 7 following order of operations means your multiplication comes first so 12 plus 7 and that gives you 19 this one here is a little bit trickier we have a negative x so we write down our negative from here but our X happens to be negative 8 and then minus 7 now that negative negative 8 the two negatives really make it a positive so positive 8 minus 7 and our answer is 1 this is a GED review video for expressions and polynomials okay now let’s evaluate a polynomial same ideas before we’re first going to rewrite our expression but stick in the number that they give us so negative three times two squared plus seven times two minus three PEMDAS says exponents first so we’re doing that so negative three times four bring everything else down now we can do our multiplication negative 3 times 4 is negative 12 7 times 2 is 14 and then bring down your minus 3 negative 12 plus 14 is 2 and 2 minus 3 is negative 1 this is a GED review video for expressions

and polynomials okay here we’re moving on to polynomials which just means that we keep getting some more terms in our expression okay if there’s two terms it’s what’s called a binomial three terms would be a trinomial and anything more than that it’s just generally a polynomial okay so here we’re going to add two trinomials together okay our like terms are 4x cubed and negative 6x cubed we have 2x squared but no two x-squared over here we have a negative seven and a positive three and then a negative four X okay so putting them together for X cubed plus a negative 6x cubed is a negative 2x cubed there’s nothing to pair up with our 2x squared so that state is the same there’s nothing to pair up with our negative 4x so that also stays the same and then negative 7 plus 3 is a negative 4 okay here we are subtracting so let’s rewrite it we have negative 2x cubed plus 5x squared minus 7 but now we’re reversing the sign of everything in here minus a negative would be plus an 8 X cubed minus a negative 2x will be plus a 4 X and then minus a positive 3 is the same thing as just subtracting 3 okay we have our negative 2x cubed and our 8x cubed we have a 5x squared and nothing to match up with it we have a negative 7 and a negative 3 and then a 4 X so a negative 2x cubed plus 8x cubed is 6x cubed once again nothing goes with our x squared nothing to match up with our plain X and then negative 7 minus 3 is negative 10 okay now we’re going to multiply two polynomials and we need to follow what’s usually called foil and that stands for first outer inner last so we need to make sure we hit every pair of numbers that we have and that’s where this comes into play so the F stands for first and we’re going to multiply the first numbers in each of our parentheses so we have 4x times 8x that’s going to give us 32 x squared now we do the outer so that this number and that number there on the outside edges so 4x times a negative 4 is negative 16x now we do the inners start here and move on to there seven times a text is a 56 X and then we do this times the last number seven times negative four is negative 28 and really what we have to do now is combine our two middle terms so we just bring down our 32 x squared negative 16x plus 56 X is is positive 40 X and then minus 28 this is the GED review video for expressions and polynomials okay factoring okay we started one before example one is a little bit similar to that so we’re going to try and do the reverse distributive property so we’re going to see what goes into both of our terms we can pull out a 2

and we could also pull out an X ok so what times 2x would give us 6x squared well 2 times 3 gives us a 6 and x times X would give us the x squared okay minus 2 times 2 would give us the 4 and we have X and X so we don’t need any other X’s so this is our factored answer an example 2 it’s a trinomial so it’s a little bit different there’s nothing that goes into each of our terms but what we can do is kind of like a reverse foil we’re going to create two sets of parentheses okay well this one it’s not bad because in order to create x squared it’s just going to be X and x because x times X gives us the x squared but now we need to find two numbers that are going to multiply to negative 8 but then also add to negative 2 okay so our options for negative 8 would be 1 and 8 but none of those will work adding 2 negative 2 2 we could also have 2 & 4 and let’s say if we make that 4 negative 2 times negative 4 would be negative 8 & 2 plus negative 4 would give us negative 2 so we’re going to take these numbers and fill them in over here so we’re going to have an X plus 2 because that’s a positive 2 and then an X minus 4 and that is our factored answer the good thing about factoring is you could always go back and check to see if you did it correctly if we now file this we should get back our original question so x times X is x squared x times negative 4 is negative 4x 2 times X is 2x and 2 times negative 4 is negative 8 so we get x squared minus 2x minus 8 okay example three is very similar to example two but the problem is we don’t just have x squared out front so you have to play around with it a little bit it’s kind of like a guest in check okay it’s quite possible that we might have x + 6 X in our parentheses okay and then we need to figure out two numbers that multiply to negative 15 and then add to negative 1 okay but in this case we need to keep in mind that the numbers that we choose might also be multiplied by other numbers here so it’s not exactly like this we have to plug the numbers in and play with them okay so maybe we could try negative 3 and positive 5 because those multiply to negative 15 but what happens is we’ll get a positive 5x here and a negative 18x here which would be a negative 13 X so that doesn’t work okay maybe we could do one and 15 we’ll say this one’s negative this one’s positive in this case we’d have a negative 15 X on the outside and a positive 6x on the inside so that would add to negative 9x which isn’t negative 1x so like I said you have to play around with it okay we could also do 2x and 3x okay let’s try three and negative five and see if that works 2x times 3x is 6x squared 2x times negative 5 is negative 10 X 3 times 3x

is 9x and 3 times 5 is negative 15 and then these two together will give you your negative x so this is the correct factored answer ok in this case we have a polynomial with four terms nothing goes into each of them this method won’t work because this only works for trinomials okay but in this case we can kind of look at our pairs of terms and see if we can pull anything out so I’m going to treat these as a set and I’m going to treat these does a set looking at 3x cubed minus 2x squared I can actually pull out an x squared that would mean that I’d have an extra 3 and an extra X and then over here I’d have an extra – well now looking at my blue set I have 12 X minus 8 I could pull out a 4 which means I’d have 3x left I’d have a 2 and now from each of those sets we actually pulled out the same binomial so what we’re allowed to do is we can say well one of our pairs of factored terms is 3x minus 2 and then we’re going to combine our extra stuff together so we’re going to multiply that by x squared plus 4 and finally there are certain special cases this is one of them okay when you have a binomial but in the side binomial we have perfect squares what we can do is we can say let’s take the square root of each of them so the square root of x squared will be X the square root of 16 will be 4 and one of them is going to be plus and one of them is going to be minus because what happens there is we have a positive 4x and negative 4x and our middle plane X terms disappear okay now we have a polynomial in our numerator and our denominator so what we can do is we can factor both of our polynomials and see if we can simplify the expression at all okay so this one here on top we have X and X we need to find two numbers that multiply to negative 8 and add to negative 2 so that would be negative 4 and 2 on the bottom we can pull out a 2 which would be X minus 4 left and then since we have an X minus 4 in the numerator and X minus 4 in the denominator we can cross them off and we are left with X plus 2 over 2 this is a GED review video for expressions and polynomials okay now we’re going to add and multiply rational expressions okay so here we have 3x plus 7 over X plus 5 plus 2x over X minus 3 it’s just like if we were adding fractions with just numbers we need to find a common denominator and in this case our common denominator is going to be X plus 5 times X minus 3 all you need to do is multiply your two denominators together however whatever we do to the denominator we need to do the numerator so that means in this case for our first term okay if we had X plus 5 we need to multiply it by X minus 3 which means we

multiply our numerator by X minus 3 also so we foil we get 3x squared minus 9x plus 7x minus 21 okay and that would be 3x squared these two together give you minus 2x and minus 21 so that’s our first fraction our second fraction we will need to multiply it by X plus 5 which means we need to multiply the numerator by X plus 5 we get 2x squared and 2x times 5 is 10x so our second fraction gives us 2x squared plus 10x over X plus 5 times X minus 3 so we’re going to keep going and we’re going to combine our like terms together so 3x squared plus 2x squared is 5x squared negative 2x plus 10x is 8 X and then minus 21 and then we have x times X we’re just going to now foil this out so we know what it is x times X is x squared we’d have a negative 3x and a positive 5x so that’s a positive 2x and then a 5 times 3 is a negative 15 and that is our answer okay for multiplying we can first factor our polynomials and see if we can divide out any common factors okay so we’re going to have 4x nothing to factor here it’s just 2x plus 3 however on the top over here we can divide out a 2 and that will be 2x plus 3 over X so our 2x plus 3s will cancel our X’s will actually also cancel and all we’re left with is 4 times 2 which is 8 this is a GED review video for expressions and polynomials okay last example okay all we’re going to do here is once again substitute our Y in and evaluate the expression so we have 4 squared minus 4 over 3 times 4 this gives you 16 on the bottom we have 12 16 minus 4 is 12 and 12 divided by 12 is 1 this is the GED review video for linear equations there are certain steps we should follow when solving equations now not all equations will need to have all of these steps but it’s good to know what order you should do things in the first thing is if you see parenthesis you need to distribute you should then combine any like terms that are on the same side of the equal sign and notice it says no inverse operations inverse operations is just the opposite operation okay when we’re trying to move things from side to side that’s when we’re going to be doing the opposite but if we’re just dealing on the same side there’s no inverse operations needed the next thing is going to be to get all your variables to one side of the equation and all constants to the other side and that’s when your inverse operations will be needed and finally we’re going to multiply or divide to get the variable by itself okay now the first example is very basic X minus 8 equals 10 to get rid of minus 8 we must add 8 and whatever we do to

one side we need to do the other side so we get x equals 18 over here negative 7 minus x equals negative 5 we need to get rid of the negative 7 to get rid of negative 7 when we must add 7 and the reason we’re doing the inverse because negative 7 plus 7 equals 0 and if you have 0 with your X and it’s just going to be your X in this case though we actually have negative x so we need to bring negative x down and we have negative 5 plus 7 which is 2 and the problem is we don’t want negative x we want positive x and we’re allowed to switch that sign as long as on the other side we also switch the sign our next example is a two-step equation now we have variables on this side which means we should get all our constants to this side to get rid of minus 8 we’re going to add 8 so we get 3x ok negative 26 plus 8 is a negative 18 this right here is multiplication so the inverse would be to divide and by dividing we get 1x which is just X and x equals negative 6 here for looking at the left-hand side we have a lot going on so we need to combine our 4x and our negative 10x and notice since they’re on the same side of the equal sign we are just doing what the signs tell us to do there is no inverse operations yet so 4x minus 10x is negative 6x plus 5 and that equals negative 20 three two step equation let’s get our constants to the right side to get rid of plus five we subtract five so we have negative 6x equals negative 28 this actually isn’t going to vide into negative 28 evenly however that’s not always going to be the case so let’s practice we’re still going to get rid of our multiplication by dividing okay and we can either divide this out leave it as a decimal or we can leave it as a mixed number our answer is going to be positive because a negative divided by a negative is a positive so we don’t have to worry about those six goes into 28 4 times with 4 left over so x equals 4 and 2/3 and in our last example on the page we need to simplify both sides of our equation over here we need to distribute and on the right hand side we need to combine like terms so 3 times X is 3x 3 times 7 is 21 on the other side 9x minus 2x is 7x and we have our minus 3 okay now we need to get all our variables to one side and all our constants to the other side using inverse operations we actually have two options here we can either get these three X’s over here are these 7 X’s over there we could say let’s get rid of 7x by subtracting it which isn’t bad but in this case is going to make my coefficient negative so what I try to do is I try and rearrange it so that my X stays positive so instead I’m going to say let’s get this 3x over there by subtracting 3x which means we have 21 7 X minus 3 X is 4 X so going to my all my X’s are on the right side we still have minus 3 since my X’s are on the right side I need my constants on the left so to get rid of minus 3 we’re going to add 3 so of 24 equal to 4x and a final step is to divide by 4 and x equals 6 this is the

GED review video for linear equations okay we can sometimes use equations to help us solve word problems the total cost of repairing car is the sum of the amount paid for parts and the amount paid for labor you paid eighty two dollars for parts and forty dollars for each hour of labor the total cost to repair the car was 242 dollars how many hours did it take to repair the car so first we should identify the variable and the variable is going to be what the question is asking you to find in this case it’s asking us for hours so we can say that H equals the number of hours now the numbers that need to somehow go into our problem are 80 240 and 242 it’s often good to find a total if there is one in this case the total is 242 so we know we’re going to be adding something together that’s going to equal 242 now in our equations there’s usually going to be a number multiplied by the variable so you can look at the numbers you have and see which one does it make sense to multiply by our variable but it makes sense to multiply the cost for part it’s times the number of hours No so that 82 can just stay by itself what it makes sense to multiply the cost for each hour by the number of hours that it would so we can say for B each and that’s our equation and now to solve we can say minus 82 from both sides for th equals 160 if we divide by 40 we get four hours okay you spend $40 on clothes and buy four DVD movies your friend spends nothing on clothes and buys eight DVD movies you both spend the same amount of money that’s important all the DVDs cost the same amount how much does each DVD cost okay so our question asks for the cost of the DVD so we can say C equals cost of DVD remember you can use whatever variable you’d like so in this scenario we have one friend buying something and the other friend buying something and those two friends spent an equal amount of money okay well I have to deal with 44 and 8 does it make sense to multiply the cost of clothes times the cost of a DVD no they have no relationship what it makes sense to multiply the costs of a DVD times the number of DVDs you bought that would so plus 4 see your friend spends nothing on clothes and buys 8 DVD movies so that would be 8c and that’s our equation okay since we only have a number over here we should probably get our variables to this side so we can subtract 4 see we get 40 equals 4c the final step is to divide by four and each DVD costs $10 this is the GED review video for linear equations the last thing we’re going to talk about is systems of linear equations that’s when you have two equations with two variables and you need to find the x and the y that works in both of them and there are several ways that we can do this the first way is linear combination so here we have a negative 3x plus y equals negative 3 and X plus 2y equals E what we’re going to do is we’re going to take one of the equations and multiply it by some number so that if we add the two equations together either our X’s will disappear or are wise will disappear ok here I have a negative 3x so if I wanted the X’s to disappear I would need this to be a positive 3x

because negative 3 plus a positive 3 is zero so I’m going to take my second equation I’m going to multiply everything by three now my other option would have been to say well if I have two Y here I can make this so that I have a negative 2y on top that would also work but we’re going to try and get rid of our X’s so our first equation is going to stay the same 3 times X is 3x 3 times 2y is 6 y + 3 times 8 is 24 so now if I add these two together negative 3x plus 3x is 0 y plus 6 y is 7 y and negative 3 plus 24 is 21 and if I divide by 7 get a y equal to 3 and now all you need to do is choose one of your equations it doesn’t matter which ok we can say X plus 2y equals 8 we’ll use the second one we’re going to sub the Y that we just got into that equation so X plus 2 times 3 equals 8 which is really X plus 6 equals 8 one step equation we subtract 6 from both sides and x equals 2 so x equals 2 and y equals 3 is the solution to this system of equations you might also see it written as an ordered pair where your X is the x coordinate and the Y is the y coordinate here we have the same system of linear equations but we’re going to figure out our solution by using substitution now substitution means you’re going to take one of your equations and get one of the variables by itself on its own and then you’re going to take that and plug it into the other equation okay it doesn’t matter which one you do let’s use the bottom one for now so if we want X by itself that means this 2y has to get over here so we would subtract 2y we get X by itself and now we can’t combine these because they aren’t like terms so it’s just negative 2y plus 8 so now we’re going to take our other equation and we’re going to take what we just got and stick it in for the X so negative 3 times negative 2y plus 8 plus y equals negative 3 this gives you 6y minus 24 plus y equals negative 3 combine our Y’s together so 7y minus 24 equals negative 3 we can add 24 to both sides 7y equals 21 we can divide by 7 my equals 3 we then need to plug this back in so we can find our x so 2 times 3x plus 6 equals 8 subtract 6 and x equals 2 ok filing we can do this by graphing now I’m not going to go into a whole lot of detail about graphing here so if you need to please watch the separate video on graphing linear equations so to graph these I first need to put them in slope-intercept form so my first equation okay to get Y by itself I add 3x to the other side so y equals 3x minus 3 that’s in slope intercept form so this tells me the y-intercept so the y-intercept is at negative 3 the number for the X tells me the slope so I’m going to start at that y-intercept and use a slope of 3 which means 3 up 1 to the right 3 up 1 to the right ok and I’m going to connect them like so for my other one I need to subtract X from both

sides 2y equals negative x plus 8 and then dividing by two means that we would have negative 1/2 X plus 4 intercept at positive 4 our slope is negative 1/2 so starting here that means I’m going to go 1 and down 2 to the right 1 down to the right and connect my points and if we look at where they intersect the point they intersect at is 2/3 which means that is the solution to the system of equations