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Intermediate Algebra

by: Zechariah Hilpert

Intermediate Algebra MATH 101

Zechariah Hilpert
GPA 3.8


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This 21 page Class Notes was uploaded by Zechariah Hilpert on Thursday September 17, 2015. The Class Notes belongs to MATH 101 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/205296/math-101-university-of-wisconsin-madison in Mathematics (M) at University of Wisconsin - Madison.

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Date Created: 09/17/15
Math 101 Review of SOME Topics Spring 2007 Mehmet Haluk Sengiin May 16 2007 1 BASICS 11 Fractions 1 know you all learned all this years ago7 but I will still go over it Take a fraction7 say You can interpret this as follows you cut a cake in seven equal7 needless to say pieces and take 5 of those pieces Then 170 means you have 10 of those pieces Obviously7 seven of those pieces will give you a whole cake and you will have 3 such pieces left thus g is equal to 1 Another way to see that is 10 7 3 7 3 3 7 T f E 1 E A very common practice that we employ when we want to add two fractions is to change the way they look so that they have the same denominator Let s talk about changing the way a fraction looksGoing back to exampleremember we divided the cake into 7 pieces7 and took 5 of them Now if we divide every piece into 3 equal pieces7 in total you will get 21 pieces and the ones that you took will give you 15 pieces Hence 5 5 3 15 7 W X What we have just done is called 7 changing the denominator of a fraction but be careful7 while we are changing the bottom7 we are also changing the top lnfact7 we are multiplying BOTH the top and the bottom by the SAME nonzero numberl 111 Studied Examples and Problems 5 3 1 Add t 7 7 1 ion 7 4 We have 5 pieces of size one seventh and we have 3 pieces of size one fourth Since sizes of the pieces are different7 we cannot go ahead and say we have 8 pieces We need to adjust the sizes of the pieces7 in other words7 we need to put the fractions in COMMON DENOMlNATOR 2 Math 101 Spring 2007 the denominators are 4 and 7 A common multiple try to use the least common multiple in general for these two denominators is 28 So 4 mm Ilcn Now that they have the same denominator thus each piece has the same size we can combine the two fractions to 5 3 M lt l t 77 u 1p 1ca ion 7 4 Multiplication is straightforward we multiply the tops and the bottoms A A i 4 3 2 4 3 5 ASSIgnment Do as indicated 5 g E 5 113 14 5 03 2 Be Careful 7 2 y 13 F What should happen is the following 2 2 u a2 12 Order of Operations So it goes as follows paranthesis powerroot multiplicationdivision additionsubtraction As you see multiplication and diVision are in the same rank For these guys we work in order from left to right Same for addition and subtraction 121 Studied Examples and Problems 1 24326 First multiplication and diVision from left to right then addition Thus the order is 2432622 2 Here s an example lots of people did incorrectly in the third exam All these people worked out the problem until the end and got 37 36 This was the answer but they went ahead and did the substitution rst and got 73353 This is WRONG because multiplication comes before subtraction we have to multiply 36 and 53 rst which gives Review 3 3653 and then subtract that from 3 which gives 3 7 36 Thus there s no simpli cation whatsoever 3 7 3653 7 73353 3 722 7 722 Another common mistake in the left hand side abbreviated as LHS in what follows we square rst and then multiply by 71 thus LHS is 74 In the RHS you know what this stands for we have paranthesis rst and then power thus we take the square of 72 which is 4 2 Roots and Powers Given a real number a and a positive integer n we de ne a aaaa n times Then we extend this de nition to negative integers as Now let s talk about roots Given a positive real number a and a positive even integer n there are two numbers one opposite of the other whose nth power is a These two numbers are called the nth roots of a the positive one is called the principal nth root of a The principal one is denoted with 5 and then the negative one becomes 7 5 For example there are two square roots of 9 one is positive other is negative namely 3 and 73 The positive one is the principal one and thus we denote them as 3 and 7 3 7 When n is odd there is only one nth root of a Let s get back to exponentiation Let s extend exponentiation to rational exponents A an 5 This basically enables us to pass between exponent and radicals m a quot am 4 Math 101 Spring 2007 21 Main Rules 1 Gina a 5 Wm am a Double Exponentiation Rule a WW 7 bizl 22 Be Careful The nth root of a negative number is not real only if n is even Example V78 is not real but 37 72 is real Taking nth root and taking nth power are opposite of each other7 thus one would 77cancel77 the other But be careful when the index n is even 15 70in n is even Example 10973100 73l 3 BUT lot73101 73 23 Studied Examples and Problems 1 Simplifying Expressions with Exponents b732 2 6753 We start with the outmost exponents7 distributing them over all the factors inside the b714671371 paranthesis using rules number 3 and 4 above b7322 6753 b71471Cil371 Now we use the double exponentiation rule rule number 5 above Review b762 bl4 613 bee2 b14 C13 67103 Now we use rule number 1 on the Us and rule number 2 on the 0 s 5762gtlt14gt Clt1sgtelt710sgt bin4 C113 4m53m723 7 4m753 a2b5714 max3 b ASSIgnment Simplify aigbZVG 2 Simplifying Radicals7 numbers inside a V80 Since we are taking square root7 we need to the nd biggest factor of 80 that is a perfect square If you make the tree diagram that we saw in class7 you will see that 80 24 57 thus the biggest perfect square factor of 80 is 16 xET16 NE b 3 80 This time we need to nd the biggest perfect cube factor of 80 because we are taking cubic root It is 8 W e1025 c 1 80 This time we need to nd the biggest perfect fourth power factor of 80 because we are taking fourth root It is 16 41652 d 5 80 This time we need to nd the biggest perfect fth power factor of 80 because we are taking fth root From the tree diagram factorization we know that 80 24 5 Thus7 there s no perfect fth power factor of 80This means we cannot simplify 5 80 e Assignment Simplify the followings m7m 7 3 587 7E3 Math 101 Spring 2007 3 Simplifying Radicals variables inside a Vzlo As in the previous case we look for the biggest perfect square factor of 10 It is 10 itself because 10 m52 So 10 m5 Actually we have an easy alternative approach in this case that I want you to learn and use Let me explain Remember our connection between exponents and radicals n am a If you apply this to v 10 you get z m5 See next example 3 m10 Following our last approach this time we get Since 3 doesn t go into 10 z is not really a 7 simpli cation So 10 didn t work with 3 we go down by one and try 9 instead of 10 Obviously 3 goes into 9 so we should proceed as follows Vaml V3m9mm Wm3 5 33510 This time we get Since 4 doesn t go into 10 z is not really a 7 simpli cation So 10 didn t work with 4 we go down by one and try 9 instead of 10 Since 4 doesn t go into 9 so we should go down one more and try 8 Clearly 4 goes into 8 so it goes as follows V4m1 V4z8z m V4x zz vim d Assignment Simplify the followings Vz9 3y21 4 3421 V5 100 V3 101 Review 4 Simplifying Radicals7 general case a 3 24 22 3414 Greatest perfect cube factor of 24 is 8 We should take the 21 part of 22 and 3412 part of 3414 So it goes like this xa 24m22y14 3 8 3 21 m 3412 yz 2xy1 32 5 3my2 2m7y4 5 3my2 b Assignment Simplify v 28m97554y21z29 97232129 5128m100t36 5 Simplifying Expressions with Radicals a 95 5g 7 qu 40q4 First observe that 3 40g 38 5q3q 2g 3 5g qu 5g Now putting the two terms together 9W72q540q g ewq WETeAIq sq Observe that both terms have the factor 55 let s pull it out 955 7 4612 3 561 5561g 4f b Assignment Simplify 364zy2 327m4y57 237719 6 7 3m2p 4mp2 6 Rationalizing the Denominator Sometimes we can get rid of those ugly radicals in the denominator Look at the examples lt gt 2 3 a 5 Remember that multiplying BOTH the top and the bottom of a fraction by the same nonzero number will give an equivalent fraction And also remember that 2i by So we should multiply both the top and the bottom of 243243 flt2e 5i2fe 3 572 57E v v vs vm 5 Math 101 Spring 2007 3755 b 55 This time we will use the fact that WW WXWV a It goes like this 37W 37W 95 327W23W73WE33E7W VS 35 WW was WW 5 mi 45 Here is an important technique when we have the sum or the difference of two SQUARE root radicals we use the CONJUGATE ofthe denominator to do the rationalizing Given the sum or the difference of two square root radicals to get its radical we simply switch the sign the conjugate of f 7 5 is Now as usual we multiply BOTH the top and bottom 242 7 24 57 lt27 3gtlt 5gt 45 J45 w femw Now if you FOIL the bottom you will see that it will become an integer Actually this was the whole purpose of using the conjugate Maybe you have recognized the bottom is the right side of the 77Dil llerence of the Two Squares77 identity A27B2 A7 BA B So FOlL ing is a loss of time at this point we use the above identity instead 2 7 V5 2 i V5 f 7 mm V5 7 W2 7 2 2f7 3f7 3 5 7 2 2fifi T 275 T 73 We are done 4 2 f hint rst simplify the d Assignment Rationalize the denominator radical M 3 Review 3 Polynomials and Factorization A polynomial is a term or a nite sum of terms in which all variables have whole number exponents We will concentrate mostly on polynomials with one variable sometimes two variables In particular a polynomial in a variable t is a sum of the form ant anath a2 t2 a1t ao To each such polynomial we assign a number called the degree of the polynomial to show how 77big77 the polynomial is The degree of a polynomial in one variable is the biggest power of the variable that s present in the polynomial So while 11 7 z is a polynomial of degree 11 in z 5 is a polynomial of degree 0 in what variable 7 We can addsubtract and multiplydivide polynomials We will give examples in the Problems section Another important aspect is to factorize polynomials One can think of factorization as the opposite of multiplication Given two polynomials we get a 77bigger77 polynomial by multiplying them Factorization writes a 77big77 polynomial as a product 77smaller77 polynomials These small polynomials are called factors of the big guy example multiply three degree one polynomials we get a degree three polynomial W90 71W 1 W952 71 3 7 95 Now factorize the degree three polynomial to get its smaller factors back m3 7 m xm2 7 l 7 lm 1 31 Some VERY Useful Identities 1 Difference of Two Squares A2 7 B2 A 7 BA B 2 Square of a Sum A B2 A2 2AB B2 3 Square of a Difference A 7 B2 A2 7 2AB B2 As you can see now y2 mz iz 32 Studied Examples and Problems 1 pulling out the Greatest Common Factor a 28m3y3 7 42m4y22 We look at the common factors that both terms have Let s start with the numbers 28 and 42 the greatest common factor of these guys is 14 Now look at the variables the 2 03 Math 101 Spring 2007 most x we can take out from both terms is 3 and the most y we can take out is yz We cannot take any 2 out since the rst term has no 2 Thus we have 28m3y3 7 42m4y22 14x3y22y 7 3amp2 b Assignment Factor 8x3 7 4m 15x312 7 10z2y52 Smyz grouping When there is no common factor among the terms of the expression we may want to group terms that have common factos together 6am12bxa2b We have 4 terms with no common factor But we may group terms with common factors together there maybe a few grouping options they should all work don t worry just go with one In the above example one can go with grouping the rst two and the last two together 6am 121m a 2b 6am 121m a 2b 6za 21 a 2b 6m 1a 2b We could have grouped differently like this for example 6am 121m a 2b 6am a 121m 2b a6m 1 2b6m 1 a 2b6m 1 Assignment Factor 19qu 7 10 7 2q2 5192 m3 4m2 7 6m 7 24 the 2 bx 0 case Say we are given a polynomial of the form 2 bx c What we want to do is to factor it into two linear polynomail On other words we are looking for two numbers A and B such that z2bmc zAmB So how will we nd these A and B Let s FOIL the RHS and see what we really have on RHS z2bmcz2ABzAB So we see that in order to have RHS equal to LHS we need to have the coef cient of z and the constant terms on both sides to be equal bAB 0143 and Review 11 We are very close now with a little inspection we can see if the desired A and B exist Start with factors of 0 FIRST then see if you can nd a pair whose sum is I Look at the following examples a 2 7 8x 15 We are looking for A and B such that AB 15 and A B 78 We start with examining factors of 15 5 and 3 75 and 73 15 and 1 715 and 71 From these pairs the one we want is 75 and 73 as we want their sum to be 78 So 278x15m73m75 b 3mg 7 3m2 7 36m Well this is not the type of polynomial we are studying in this section But once you take out that common factor 3x from all three terms we will be ne 33 7 32 7 36m 3zz2 7 m 712 So let s nd A and B such that AB 712 and A B 71 Since the product is negative one should be positive and the other should be negative Also looking at their sum we see that their absolute values should be close By inspection we see that what we need is the pair 74 and 3 So m27x712x74m3 Thus 32 7 3m 7 36 3zz2 7 m 712 3m 7 4m 3 c NOTE Sometimes we do not have the desired A and B for example try 2 z 1 you will not succeed d Assignment Factor 2 7 4x 7 12 k2 7 11k 30 4193 24192 7 64p 4 Solving Equations An equation is basically two mathematical expressions set equal to each other Given an equation A B there are two way we can manipulate it to put it into different looking but EQUIVALENT form 1 ADDSUBTRACT If we add or subtract a real number C to BOTH sides of the original equality we get an equivalent equation to Math 101 Spring 2007 is equivalent to A C B C So this means we can 77move77 terms from one side to the other with the cost of SWITCHING its sign example 4m 7 5 3m 7 7 Add 5 to both sides 4x7553x775 4z3x775 Now subtract 3x from both sides 4m73z3x77573z Hence we get 4m 7 3m 77 5 So you see7 we 77moved77 the 75 from left hand side to right hand side as 5 and we 77moved77 the 3x from right to left as 73 Ideally you should use this 77moving77 shortcutl MULTIPLYDIVIDE If we multiply or divide BOTH sides of the original equality by a real number C 7 0 7 we get an equivalent equation AB is equivalent to ACBC andto AiB OTC This enables us to the following Say you have 5x10 To leave z alone7 we go ahead and divide both sides by 5 Review 13 5mig 5 5 andget x2 41 Linear Equations A linear equation in one variable is an equation of the form AzB0 For example 3x 7 5 4 7 5x is a linear equation although at rst it looks different than the above form Well after we move everything to one side we get 8x 7 9 O which ts our de nition So basically all we want is that the equation should involve only one variable and the exponents should all be 1 To solve such an equation we basically isolate the variable Here let s work on an example 395757269574475z778z72 First we do the necessary multiplications 3x7576z8475x778z16 and then collect all the terms that have the variable on one side and all the numbers on the other side REMEMBER when you move a term to the other side you SWITCH its sign 3x76z5m8x4716578 Now tidy up both sides 10x 24 Make a division to leave z alone now 7 24 7 12 a 10 5 411 Studied Examples and Problems The above example is enough for this topic so i ll just go ahead and assign some problems Assignment Solve 7x 7 34 7 2x 5 6x 7 3 4t 7 8 53 7 42t 2 14 Math 101 Spring 2007 42 Solving Equations By Factorization We will recall a very important property ZERO FACTOR PROPERTY It basically says that product of nonzero numbers is always nonzero In other words7 if a product is zero then at least one factor must be zero 421 to AB0 then A0 or B0 Studied Examples and Problems 7mg 2 76m We move everything to one side rst 0x37m276x Next step is to factor the RHS we take out the common factor x 0 7m7 6 Now let s factor the second factor on RHS we look for A and B such that AB 76 and A B 71 The pair we need is 73 and 2 27m76x73m2 Thus we get 0m37m276mm27x76mz73x2 Use the Zero Factor Property now7 x0 or x730 or x20 Thus x0 or x3 or m72 522252 Again we move everything to one side and we factor 5227252 0 52z 7 5 0 Review 15 By Zero Factor Property 3 Assignment Solve 48 3x2 7 z 4x2 7m 6253 5252 6t 5 43 Quadratic Equations A quadratic equation in one variable is an equation of the following form am2bmc0 Looking at its discriminant b2 7 400 we can see what kind of solutions we have There are 3 cases H b2 7 4m lt 0 Then our equation has no real solutions it has 77imaginary77 solutions 2 b2 7 4m 0 Then our equation has only one solution and it is real it is actually a rational number 3 b2 7 4m gt 0 Then our equation has two real solutions To nd the solutions we use the QUADRATIC ROOT FORMULA 7 7b i VI2 7 4m 7 2a This is a very powerful formula sometimes there are more basic approaches that solves the equation where using the quadratic root formula would be an overkill For example the SQUARE ROOT PROPERTY Given m2A with A positive there are two solutions miZ 16 Math 101 Spring 2007 431 Studied Examples and Problems 1 Quadratic Root Formula a 2 z 1 0 We rst look at its discriminant 12 7 4 1 1 17 4 73 lt 0 This means we do NOT have real solutions b m2 7x 1 We rst move everything to one side and get 2 z 7 1 O This time the discriminant is 12 7 4 1 71 1 4 5 gt 0 So we have two real solutions given as follows 711 m f d 2 c Assignment Solve m 2m 3 1 3m 7 4m 2 2m 7 5m 5 hint for both foil then move everything to one side Thus the two solutions are 71 5 717 fan 7 2 Square Root Property a 200 7 32 7121 0 One can FOIL and then apply the quadratic root formula but that s a waste of time for this one Observe that once you move the 7121 to the right you will get 2m 7 32 121 Now using the Square Root Property we get 2x73j121 2x73i11 So either 2x 7 3 11 or 2x 7 3 711 In the rst option we have x 7 and in the latter we have x 74 b Assignment Solve 2m 7 32 718 0 2m 32 18 0 3 Substitution Some equations are not quadratic but they can be treated the way we treat quadratic equations a 1075m540 Observe that 10 is the square of 5 Thus if we SUBSTITUTE u for 5 in the original equation we get u275u40 Review C 17 You can apply the quadratic root formula but you can easily factor the left hand side Always try to factor rstinstead of applying the quadratic root formula u74u710 So by zero factor property we get u4 or u1 Wait we are not done yet What we found is u what we want to solve for is z Using our substitution backwards we get m5 4 or m5 1 So there are two solutions z 541 and z 51 1 274272x27410 If you FOIL then you will get a degree four equation and we do not know how to solve them in this class Again we will be smart and do the following substitution 1 for 2 7 4 We get u 7 2v 7 1 0 We can factor the left hand side actually if you can factor don t use the quadratic formula go with the factorization u 7 12 0 So either by zero factor property or square root property 1 7 1 0 U 1 Again we are not done Remember that 1 stands for 2 7 4 So we have Use square root property now xig Assignment Solve 2123 711a13 12 0 23k 71 53k 71 72 18 Math 101 Spring 2007 44 Equations with Radicals Take an equation i will refer to this one as original equation like 7 523 2 V 2 To free the 2 on the right hand side from the radical we take the square of BOTH sides of the equality We get the following equation which i will call derived equation 52 3 22 T Now some of you hopefully will say wait a minute taking square of both hand sides is not one of the legal manipulations we can do to an equation Yes that s completely true If you look at the beginning of this section you will see that we are ONLY allowed to addsubtract from both hand sides or multiplydivide both hand sidesNOT to square both hand sideswhy do we do it then Well our motivation is the following the solutions of the original equation are amongst the solutions of the deriued equation The tricky part is that the derived equation in general will have more solutions than the original one So we have to CHECK every solution of the derived one to see if they are solutions of the original one Let s continue our example Let s solve the derived equation 52 3 2 7 Z 2 multiply both sides by 2 222 52 3 Move everything to one side 222 7 52 7 3 0 Use quadratic root formula 7775i 7527427375imi5i7 22 4 4 Z i 5 7 5 7 7 l 0 So there two solutlon 2 7 3 and 2 7 712 of the der1ved equatlon Now let s CHECK both of them to see if they solve the original equation or not 533 77 7 3 2 3 true Review 19 So z 3 is a solution of the original equation Now let s check the other one in 712 571223 false We don t even need to compute the right hand side to see that it is false Do you see why Because the square root of a number cannot be negative7 in particular the right hand side cannot be equal to 712 So 712 is NOT a solution of the original equation The conclusion is that our original equation has only one solution7 namely z 3 Assignment Solve v5 7 z 7 z 7 1 07 xk 7 xk 7 3 1 hint apply the method twice 45 Equations With Rational Expressions When we have equations with polynomial denominators7 we call it a rational equation We have to be careful when we try to get the solutions of such an equation because we need to avoid numbers which make a denominator 0 451 Studied Examples and Problems Let s solve the equation 2 7 1 6m 3m 1 7 m 3m 1 We FIRST start with identifying what z cannot be 3m17 0 and x7 0 Thus 71 m 7 and m 7 0 Now we start solving for m I will put everything to common denominator and will combine the two terms n the right hand side 2W M395 1 695W 3m 1 3m 2m 7 3m 1 7 6m2 3m T 1 We have two fractions that are equal to each other with the same denominator this means their numerators should be the same as well 20 Math 101 Spring 2007 2x3m176z2 Move everything to the left now 6m273z712m0 6x2 7 7 1 0 Use the quadratic root formula now and you will nd that 71 1 Note that the rst solutions is one of those 77load guys77 that we should avoid7 so I will not you shouldn t eithercount z 71 as a solution Thus we have only one solution7 namely z 1271 2 1 7 4 275m47 k2k76 k27k72 k24k3 5 3 7 ASSIgnment Solve 7 H 7 Review 5 Answers 11 231 232 233 234 235 236 321 322 323 411 421 4311 4312 4313 44 451 252414156135042 bins127 4m73 7 16m43 2m 2 2 233 393 3 37 77 yaw77 20 sag2 234m 3y7xgm y522W 2x20t7xsE 4 3w W 27712 7 3771219 W V613 aE3 12 3 i x 5 i 2V6 4x22 7 1 SzyC wzy 7 2xy32 2 q2 5P2 i 2 m 4m2 i 6 96 i 6X96 2 k i 5W i 6 4PP 8P i 2 2 t 71744 7124 756171 75 xg2 75 7 no real solutions 3 3M22 3 7 I m2V2 no real solutions 437323716r13 1 7 7 2 7 13


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