5 Academy's 52 ACT® Math Problem Types You'll See | 5 Academy's Strategies and Tips for ACT® Math

so this is a new video series that i wanted to do where i discuss all of the act math skills and how you can master them with real problem examples that we discussed the idea here is that we want to master these skills because they recur from exam to exam the more you get exposed to these skills and understand which ones you are struggling with the more likely you are to get them right and to keep yourself from making the same mistakes again and again because that's going to keep you from improving so in this video we're going to be going over the first 18 skills and then obviously we'll get to the rest of them in the next few videos at the end of the video i'll provide you with two methods for learning these skills the first one will take you a few months and the second one will take you just a few weeks so we'll get to that at the end of the video first let's get to the first 18 skill all right and just so you guys know all these problems that you're going to see in this video are updated for the 2023 act math exam so the content here is going to be just like what you'll see on the test this upcoming year all right so let's get into it all these companies profits are directly proportional this is related to the skill obviously to the number of motherboards that are sold he profited this much off of this many motherboards how many motherboards did he sell if he made this much profit so the key thing behind the skill of proportions is that you need to keep in mind that this ratio of one thing to the other is always going to be true so for every six thousand dollars there's going to be 25 motherboards that means that if you have 21 840 dollars in your numerator of this ratio and your unknown number of motherboards is labeled x then that means that you can just set up this equation like this and solve for x okay you can also think of it as okay how much left to multiply 6000 by to get to this number and i can do the same thing multiply this number by that same quantity that i multiplied 6000 by to 25. so for example i could do 21840 and divide it by 6000 this will give you 3.64 if you do it on a calculator and if i multiply 25 by 3.64 or i solve this equation for x i'll get the same answer of 91 i believe so that's how problems like this work and that's how you can do this very quickly um so essentially that's how proportions work you need to keep in mind this ratio that you're given is always going to be true and you use that as you solve the problem micah's backyard is shaped like the image below to the nearest whole number what's the perimeter of the backyard so perimeter problems uh it's a simple enough skill when they give you a visual like this you really just add up the side lengths in this case you have to also calculate a side length using pythagorean theorem as far as area if they ask you the area of this shape what you need to do for complex shapes like this is um most likely break the shape into multiple different things so you have a rectangle here you have a rectangle here here and you have a triangle here and then you kind of add up all those different areas as far as area goes for something like a square or a rectangle or a triangle those are probably the most common shapes that you're going to see on the test these are formulas that you should probably just memorize so a triangle is half base times height rectangle is length times width and square is just length squared for perimeter of a square you have 4l for perimeter of a rectangle you have 2l plus 2w and then for a perimeter of a triangle you just add up the sides okay and there isn't really a go-to formula for that biking team a charges its team members this much dollars for membership plus this much of membership fee this team charges two months for uh monthly membership and this much for a membership fee after how many months is the price of biking team z's membership twice that of team a so if i create an equation for each of these it would look like a linear equation and that's why these type of problems are called applications of linear equations you're using the y equals mx plus b form essentially you have a rate and an initial value and you're using that to describe a real-world scenario in this case it's a financial scenario of calculating the price of a monthly membership okay so problems like this you are developing this formula and applying it okay so let's make up a formula for each of these brands so it says vikings team a charges three dollars per month your rate is always going to be your slope so three dollars per month i'll just call it y1 is equal to 3x where x is the number of months plus your initial fee which is three as well and then for biking team z you have uh y two equals what two dollars per month plus ten dollar membership fee okay now it's asking uh after how many months after essentially how many values of x is the price of one membership double that of the other so we want the price of y two z to be double that of y one two times y one okay uh and i can just essentially substitute y1 and y2 into this equation essentially substitute 3x plus 3 and 2x plus 10. so i'll end up getting 2x plus 10 equals 2 times 3x plus 3. this turns into 2x plus 10 equals 6x plus 6. this terms if i subtract 2x on both sides and then subtract 6 on both sides i get 4 equals 4x and that means x is equal to 1. so after one month you will see that the value of this is double the value of this so your answer should be a you can just double check that this right here would equal 12 this right here would equal 6. and that makes perfect sense the other way to do this is just plug in all the answer options and see which one is actually the correct one this next skill is all about simple algebra and distributive property the whole goal here is to take a complex and complicated statement of terms like this where this is a term this is a term this is a term term term term and to shrink it down to uh essentially a much smaller more reduced version in this case you're actually solving for x so uh you're going to end up with just x equals a number okay and these are the numbers you see here so the way you can do this is keep in mind you can combine like terms okay so for example if i want to add x plus x squared can i combine this and get 2x squared or 2x no you can't you can't combine these two because they have a different exponent okay or what about sine of x plus x does this become sine of 2x does it become 2 sine of x no it does not because this and this they don't have the same degree this is a part of a sine function you can't add these two together this is the most simplified form of this equation but for example if i give you x squared plus 14x squared can you add these together what does that equal well since they both have the same exponent and they're just in the form of x to the something you can combine them and get 15 x squared you imagine that there's a one in front of this and you can combine okay so that's that's one of the main rules you need to keep in mind here combine like terms the other thing is distributive property so when you have a times b plus c this equals a times b which is essentially a to distribute to the b and then you add that to a distributed to the c a times c okay that's how you can do distributive property and that's what you're doing right here so you can use those rules to solve this problem i recommend you try it yourself i'll just quickly do it i guess plus 12 x minus 75 x squared equals 9x minus 15. um so these two cancel out we end up with uh 3x equals 5 uh if i'm not mistaken so minus 9x minus 9 oh sorry 15 not uh and then we get x equals five okay simple enough and this is supposed to be a minus fifteen actually so we're going to turn that into a minus five all right moving on let the function w be defined as this what is w at negative two so this skill is all about substituting values into polynomials you're given a value negative two essentially x equals negative two and you're given a polynomial which is w to the x w of x equals 2x plus 1 all to the third the way that you do this is you obviously are going to plug in a number here for x so w at negative 2 is equal to 2 times negative plus 1 all to the third right so how do you solve something like this it really problems like this end up a lot of the time being assigned game it's a game of putting in the right signs and not making a sign mistake as you can see by some of the answer options here they're just reverse signs of the other so as long as you get your signs correctly you're probably going to get the problem right at least 50 chance that you will okay so how do we do this so the way you're going to do this is first start with again keep in mind your order of operations so whatever's uh in the parentheses you always start with that first so i see these two within the parentheses i have 2 times negative 2 plus 1. um since these two are multiplication i'm going to do that first so this turns into 2 times negative 2 which is negative 4 and then add that to one and then all that to the third i can combine negative four and three and one and i get three negative all to the third and then i finish that off by doing the exponent so if i do negative three times negative three times negative three you can end up getting negative 27. you can do that on your calculator if you're not sure okay simple enough it's just a matter of plugging in your signs in the right place and not making a sign mistake the next skill is about analyzing frequency charts so i have two frequency chart problems here i'm going to show you i'll try not to spend too much time on this but really what a frequency chart is is we give you different categories of data it doesn't have to be numerical right obviously here we have just grades and then we give you the frequency or the number of things or the number of people in this case that are in each category okay so if i asked you for example how many students scored above an f well you look at the total number of students you look at how many got an f and then you look at everyone that's above so it would be 33 right simple enough but what if i gave you a more complex problem so here we have the frequency of gas mileage for 30 cars in a parking lot and then i ask you how many vehicles have gas mileage between 20 and 30 miles per gallon so would you just say oh it's 24 well not quite because this includes 20 plus it also includes 30 plus and it includes 40 plus because you see these numbers are just pluses they're not ranges they're just greater than or less than a certain value right so if i want to find the value that's between 20 and 30 what i would do is i would take the value of 20 plus which includes all of these right and then i'll take the value of 30 plus which includes all of these and then i would subtract the corresponding values so i would do 24 which is all of these cards it's 20 plus 30 plus 40 plus and then subtract 17 which just includes everything above 30. and this will leave me with what's the difference which is just seven cars that are going to be in this range right here between okay you can think of it as out of all the cars out of all 30 of the cars 29 of them are greater than 10 in miles per gallon 20 of them are greater than sorry 24 of them are greater than 20 miles per gallon 17 of them are greater than 30 and just 3 are greater than 40 okay so if you find the the the difference between each of these you find how many are actually in the range of 10 to 20 20 to 30 30 to 40. okay that's how you do a problem like this this type of example has showed up on past exams and it's very likely that you'll see it again seven points are shown below if they're connected to make this shape a b c d e f g what is the sum of the interior angle measures of the shape so let's just make a shape really quick so what do we notice this shape has how many sides it's not drawn perfectly but i'll just do it again really quick a b c d e f g so we have one two three four five six seven okay a seven sided shape what is going to be the interior angle measure of a seven sided shape we obviously cannot measure these but what we do know is that uh we have this equation 180 times n minus 2. this equation basically says for a polygon with n sides where n equals 7 in this case the interior angle measures is just going to be this value right here 180 times 5 in this case is going to give us i believe 900 so the interior angle measures add up to 900 okay and you can plug in n equals to oh sorry not two but three and four and you'll see that for a triangle you get 180 and for a square you get 360. so this formula makes sense i recommend you memorize this and keep it to heart there are more type of problems you'll get about calculating angle measures some having to do with like parallel lines alternate exterior angles i can go through all the examples here you can check out our website if you want to see that but this is a solid example of one very important rule that you should know about angle measures moving on so now the next skill is about finding mean median and mode with a list of numbers in this case we have a list of numbers here so how do we find let's just try this problem and we'll kind of demonstrate how not only will you be asked to find mean median in mode but you might be asked to modify it so or extract information about a data set given the mean median in mode so this person took four quizzes the average of those four is 90. she recalls that the first four three scores of the four were these uh so what's the fourth score okay so in order to do this problem correctly you need to keep in mind the formula for a mean so the average or the mean is going to be the sum of the values divided by the number of values in this case we know that the the average is what this right here is going to be 90. this right here is going to be something we'll figure that out in a second this has to be 4. so we know that the sum has to be equal to 90 times 4 which is 360. essentially i know and what is the sum right the sum is score 1 plus score 2 plus 3 plus 4 right so i know this score i know this i know this and i don't know this and i know the sum as well so 360 has to equal 78 plus 86 plus 99 plus x if i subtract these values and solve for x i get 360 minus 78 minus 86 minus 99 and end up getting x equals 97. okay that's her one of her higher scores so essentially what's the point of this problem it's asking you can you take this formula right here and extract values from the sum or the individual test score values given what you know about the average and the number of values okay that's what the scale is all about the next problem type is about solving problems in context of money this is a very common skill really any problem dealing with rates of money or quantities of money being added or subtracted or multiplied will fit into this category so let's go through a typical example this person has a four bedroom house with 12 windows each bedroom will be given a 23 ceiling fan and each window will be given this much dollars worth of curtains which of the following is the total cost of ceiling fans and curtains in this person's house so let's take it piece by piece so we have the fan cost let's call that f so the fan cost is going to be what it's going to be uh one fan per bedroom right each bedroom will be given a fan and there's four bedrooms so four bedrooms times the cost per fan in each bedroom that's gonna be 23. that turns into four times 23 which is um i believe 92 okay and then we have the curtain cost the curtain cost is going to be uh we have 12 windows and each window will be given 40 worth so 40 times 12 curtains is going to give you 480 worth of curtain that's a lot of curtains if we add these together we get 480 plus 92 which is 572 and that's your total cost okay so in this case you're dealing with rates 23 and 40 namely and you have to recognize you have to multiply those rates by a particular value corresponding to the quantity of things you're buying and get an output value of total cost and that's essentially what the skill here is dealing with money the next skill is deriving information about geometric figures so if i give you something like a cube and i tell you some property about it can you extract other properties about that cube like area surface area perimeter side length the total sum of the side lengths etc so let's do a typical example the sum of a certain cube's edge lengths is 96 inches what is the area of one side of the cube okay edge length so how do we find the how do we correlate edge lengths with the area of a cube so what's the what's the area of one side of a cube well if we know a cube is just a bunch of squares right in a three-dimensional shape so if i take one of these shapes like i said it's just a square with side length l so the area of one side length is just going to be l squared okay now what i need to do in order to calculate this is just find one thing l so how do i find l well i know that all of these side lengths added together is going to be 96 so the game is just if i know how many side lengths there are i could divide 96 by the number of side lengths and find the value of l and then just square l and get my area of one face okay so let's see the best way to do this at least if you don't have any clue how many edge links there are is to visualize this at least in your brain but try to visualize it on paper so here i see i have a cube right let me draw the parts of the cube that we can't really see right so at this point i've drawn all the edge lengths now it's just a matter of counting them so if i count the front face obviously that's four edges i can count the back face that's four more and then i have one two three four edge lengths connecting the front and back faces that's 4 times 3 so this number is going to be 12. that means this number here is going to be 96 divided by 12 which i believe should be 8. so l is equal to 8. that means my area is going to be 8 squared which is 64. so one face of this cube is going to be 64 inches squared moving on properties of exponents this is one of the most important skills because there's a lot of exponent rules and some of the problems will require you to play around with these exponent fractions or these exponent expressions that can get pretty complex so let's go over a typical example like this one so which of the following essentially are equivalent to what we see here okay so what do i notice actually all these are negative when your number up here is actually positive so these are all actually going to be incorrect this problem is really just testing you on a rule that says that a over b to the c is equal to b over a to the negative c okay this is a property of exponents that you need to essentially memorize and the only one only answer option that really agrees with this is this where you flip the numerator and denominator as we did from a to b to b over a and you change the sign of the exponent from negative x to positive x or vice versa and these two expressions should be equal you can test this out by doing 5 over 6 to the third and it should be equal to 6 over 5 to the negative 3. all right there are more properties of exponents i won't be able to go over them here but just look up on google images properties of exponents and commit those to heart we have specific problem sets on our site about 20 problems that you can go over specifically for problems uh related to properties of exponents and different types of properties of excellence so i would highly recommend checking that out if you have the time the next skill is about finding sine cosine and tangent with visuals okay so the game is all about using sohcahtoa rules and also i guess chochacao if you know what this is i'll define all of these and finding these ratios given this triangle given a right triangle like this okay so what is sohcahtoa this is sine theta equals opposite over hypotenuse cosine of theta equals adjacent over our hypotenuse tan of theta equals opposite over adjacent how do you define opposite adjacent and hypotenuse well so the opposite side relative to the angle is the easiest to identify it is the side that the angle opens up to okay so this angle is opening up in this direction and it opens up to the side that is on the other side so this is your opposite in this case okay the hypotenuse is the next easiest and this is going to be the longest side now in this case you don't know if this is the longest side right i know it's drawn really weirdly it says 20 and 101. forget about the scale but keep in mind the properties of the triangle so this right here is going to be your longest side how do you know that for a fact it's because the largest angle is 90 degrees and the largest angle always opens up to the largest side so whatever side the angle 90 degrees the right angle here opens up to that is going to be your hypotenuse okay so hypotenuse and then the last remaining side is going to be your adjacent side okay adjacent side is the side that's touching the angle so that's essentially the side that's adjacent now you might say okay the hypotenuse is also touching but that's not the adjacent side because the hypotenuse is obviously the hypotenuse so it's a different side it's categorized differently okay and then here you have some more rules this is cosecant this is secant and this is cotangent which are essentially if you look at each one it's just a flipped version of whatever's above it so o h turns to h o a h turns to h a o a turns to a o okay very straightforward now if this is asking about cotangent which is cal so that's going to be adjacent which is 20 over opposite opposite we don't know judging by the answer options is probably 99 it's the third value so it's going to just be adjacent 20 over 99 opposite so your answer should be a based on this the next skill is about similar and special right triangles so um first of all i should define what similar and special right triangles are so special right triangles are in these two right triangles that you see right here this one is a right isosceles it's right because one of the angles is 90 degrees it's isosceles because two of these side lengths have the same length okay it's also known as the 45 90 triangle you should commit this to heart um the other triangle is the 30 60 90. okay uh in the side lengths for this you should also commit to heart so the smallest angle opens up to the smallest side of half in length the largest angle second largest angle sorry opens up to a square root of three over two which is the middle side length in terms of length and then the largest angle of 90 degrees opens up to the longest side which is the hypotenuse okay similar triangles are any two triangles that have the same proportions of side length so essentially they are two triangles that are uh basically scale versions of one another so for example these two triangles here they have the same angle measures essentially okay corresponding angle measures this is the same as this this is the same as this this is obviously the same as this okay so these are two similar triangles and the notable thing here is that this side length here and this side length here have the same relation as this side length here and this side length here they are just scalar multiples of one another so times some number x times some number x okay so that's how you can solve similar right triangles similar triangles sorry now special right triangles let's just go through this problem it's a good demonstration of this so the following is a right isosceles triangle essentially the same thing as this the hypotenuse length is one so literally the same thing as this suppose angle a was expanded to 60 degrees with the hypotenuse remaining the same length okay so this triangle is really turning into this side b's length would increase by a factor of what so side b what side b it's going from this to this so to find the factor of increase what you could just do is divide this by this right here okay square root of 3 over 2 divided by square root of 2 over 2. that turns into when you divide two fractions you can really just multiply by the reciprocal square root of 3 times 2 over square root of 2. this turns into square root of 3 over square root of 2. that's the correct answer but we don't see it in any of the answer options so what you can do is turn this into uh square root of three um i believe what you have to do is turn this into multiply by square root of two over two so this becomes square root of six over two and the way you do that is just by you take this original version and you multiply by square root of two over square root of two from square root of six over two okay a bag has three red blocks five green blocks and seven pink blocks how many red blocks must be added to the bag for the probability of randomly drawing a red block to increase to 60 so probability problems like this test you on your knowledge of probabilities you need to keep in mind that the probability of any event x is the same thing as the fraction or the ratio of the number of desired outcomes divided by the number of total outcomes okay that's the basic probability formula and you need to use this for all probability problems that you're going to come across namely when you see something where you have to increase or change a probability so for example we're looking for the probability of drawing a red block so originally what is that the number of desired outcomes there's three of them the number of total outcomes it's three plus five plus seven so that should be what eight plus seven that's 15 okay and we're trying to get this probability to be 60 now obviously this is way less than 60 it's closer to 20 so how do we increase the probability well you need to add red blocks to the bag so if i add one red block what happens to my probability well it'll increase to 4 over 16 which from 20 that's an increase of 5 up to 25 right but that's still we're not we're not quite there yet we need to keep adding them uh keep adding blocks so let's try adding you know maybe four blocks let's see what happens there three plus four over fifteen plus four that's seven over nineteen what does that get you seven over nineteen is thirty 37 okay so we're getting there right now what we need to essentially do is keep adding blocks until we get the right number the way you can do this very quickly is just make an equation so you can just do um 3 plus x some unknown number x divided by 15 plus some unknown number x and then make that equal to 0.6 and you can solve this for x i believe your answer should be 12 or 15. i'm not exactly sure what the answer is but this will get you the right answer okay so this type of process taking into account this formula probability of x equals desired over total is really the key and then using this formula to add or subtract values from this and this is how you solve this the next problem type is conditional and joint probability so what you're doing in problems like this is you're taking probabilities or percents and combining them to get an answer it's pretty straightforward so let's go through this example at this high school this many students graduate and this many students who graduate go on to university so what percent of students at grant go on to university so if i have all students at grant i know that ninety percent of them graduate and ten percent don't and then out of the ninety percent sixty percent of them um university and then 40 don't go to university so i want to find what is the uh percent of people that are in this category well is it just going to be 60 or 90 not quite because this is 90 of students and this is 60 of the 90 so what i really need to do is i need to multiply the two percents together 0.6 times 0.9 that's going to give me 0.45 5 4 sorry and that's the answer okay so it's a matter of combining your percentages which of the following is the result of reflecting the point about this other point so the best way to do problems about locating and moving a point in the coordinate plane is to draw it out and visualize it because when you're moving stuff like literally in a in a two-dimensional plane it's a good idea to have an idea of what that two-dimensional plane looks like and where things might be moving so let's plot this point four comma negative five that's probably down here and then we have negative six comma five negative six comma five somewhere up here okay so they're a bit far out um so it might be kind of hard to conceptualize where these points are going but really what we're trying to do is uh we're reflecting about this point so we're trying to flag this point all the way across to somewhere here okay we need to find where exactly that new point is going to be so the way you can do that is um think of what do you have to do in order to get from this point to this point how much do you have to go left and then how much you have to go up so in terms of left when we think if we're going from um sorry if we're going from four to negative six what we did is we went four this way and then six this way again so that looks like a total of negative ten and then we go up how much we go from negative five up to five that looks like we're going plus ten so in order to get to the next point we're gonna have to go up 10 and left 10 again so from negative 6 comma 5 we're going to be going if i could add a point essentially you'd be adding on another negative 10 and then a positive 10 to go left and up respectively this gives you a result of negative 16 and 15. so that's how you can do this type of problem it's moving points reflecting which is just a matter of doing the same movement essentially again in the same direction okay which of the following is the least common multiple so this type of problem least common multiple greatest common factor might seem a little challenging but really it's the same thing every time and it is a very easy way to reserve reverse solve the problem by just doing it backwards using the answer options and checking which one is right okay so the least common multiple between these the way you can solve this very quickly is just take the smallest answer option that you're given divide it by all three of these numbers and see if it gives you a whole number for each so do you get a whole number so i can just check that on my calculator 192 divided by 32 that gives me six 192 divided by 24 that gives me 8 192 divided by 18. that does not give me a whole number so that means this is not a least common multiple or a multiple at all you do the same thing with all of the interruptions going from smallest to largest and you should find one of the smaller values that actually end up working so let's try it with 288 288 divided by 32. that works 280 divided by 24. that works 288 divided by 18. that works so your answer is b it's the smallest value that's given that is a multiple okay and our last skill for the day is polynomial arithmetic sounds kind of complicated really it's just combining like terms okay we already went over this before in the one of the skills early on i think it's about distributive property or something but here i mean it's a similar skill there is an element of distributive property so in this problem we are subtracting one polynomial from another so how do you do this type of thing um well i would recommend if you're given a subtraction problem and you have negative signs in the actual polynomial that you distribute this negative across to all the terms and then can convert this to an addition problem so it'll turn into 9x squared minus 5xy minus 3y squared and then minus negative x squared that turns into plus x squared minus 3xy so that turns into minus 3xy minus negative 12y squared that turns into plus 12y squared so if you add all these up you can combine like terms now combining like terms means like i said in the above example if they have the same exponent if they have the same base all right so i can't combine x y with y squared because the bases are not the same this has an x y in it this has a y um i cannot combine x squared and y squared even though they have the same exponent because they don't have the same base so i can only combine 9x squared with x squared that gives me 10x squared so i immediately know d and e are out i can do my negative 5xy with minus 3xy negative 8xy and immediately i know this has to be my answer because it's the only answer option with these two terms okay so that's how you do this particular problem and all 18 of the first 18 skills all right so those are the first 18 skills of the 52 on the ecd math section i hope the problem breakdowns and the examples were helpful in terms of getting an understanding of the kind of skills that you're going to see on these exams at least a part of those skills now let's get into the actual strategies the first one like i said will take you a few months most likely if you're using a typical study plan where you study maybe one or one or two practice tests a week so what are you doing in this in this method the first method you're going to take a practice test one or two a week and not only just take it with you know with the timing and everything and you're not just going to grade it but when you actually are finished with it you want to look over each and every individual mistake that you made and learn as much as possible from that mistake so you want to look over the mistake what mistake did you make uh what was the correct answer why was it correct what was the skill being tested in that problem and you want to do this complete analysis of each and every error that you make in a practice exam because the thing is there's certain skill weaknesses that you have your unique skill set is different from someone else's so you need to understand that skill set and where you are lacking in certain skills and if you do this exercise from practice exam to practice exam after maybe 10 or 11 practice exams you'll start to notice patterns in the type of things that you're getting wrong and this will give you a better understanding of how you can improve in those specific areas just by seeing those kinds of problems and taking note of the problems that you're struggling with from each exam that'll help you to just do better in terms of getting those problems right so use this method to just basically make the most of your practice tests as you're going through your act prep process now that's the first method and like i said if you do about one or two practice tests per week it'll take you a few months that's the typical study plan for most students the second method is where you use the problem sets that are on our website we have full length problem sets for each skill on the act math exam so there's 52 problem sets we provide you with diagnostic quizzes you take the diagnostics and based on how you perform the diagnostics will provide you with problem sets that are there to fulfill your weak areas because the diagnostic it finds what your weaknesses are and you don't have to do any of that work on your own in terms of identifying the weaknesses the software does all that for you for five dollars a month you can access this right now on our website it's linked below and i'm sure it can help you expedite that process of improving your math score if you're someone that's cramming or you have a little bit of time left before the exam then i recommend that you try this approach it's only five dollars right now and it can help you expedite that process of improving your math score really quickly so that's what i'll leave you guys with remember the two methods first is make the most of your practice exams i'll understand what your mistakes are and why you're making them second method is use our course it will help you as well and i will leave you guys with that again like i said there are six free practice exams and a free tutoring session linked below if you want to get some tutoring and some official exams and that's about it so good luck with the act and i'll see you in the next video where we discuss the next 14 act math skills

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Id: SHISfXpB9pM

Length: 35min 46sec (2146 seconds)

Published: Fri Jun 03 2022