😎Big thanks to JW Math Tutoring who went through the work to find these changes + provide screenshots! Please make sure to check out his channel! JW Math Tutoring: ruclips.net/channel/UCWZe0ktD7bP-b4clsq2CLjg Also, in case you haven't noticed - this video was made at 1 AM at night.. lol. If you catch any errors/mistakes, please report them here in the comments! Note: Question 2 in this video is NEW, but not unique (has been tested before)
I think there’s an easier and faster way to do the last question without using desmos. You can find the sum of the solutions for (2x^2-16x+6p) using -b/a (the sum of the solutions for a quadratic) and get 8. Then, for the (3x+p) part, you can solve for x and get x= -1/3p. Then finally, you can solve for p by getting the equation -1/3p + 8 = 20/3, and you would get p = 4
I also did quite same but approach was different coz I did not know that formula. I used the ultimate root eqn for quadratic x = [-b±√(b2-4ac)]/2a , this was same because when you add both roots, you are left with (-b/2a)-(-b/2a) which equals -b/a 🤣
Here's an easier approach for Q. 15, i) 3x+p= gives us x=-p/3 ii) 5x^2-45=0 gives us x= 3 and -3 (We can ignore this as sum =0) iii) 2x^2-16x+6p, For iii, we can apply the sum property which is -b/a leaving us with -(-16)/2= 8 (this isn't a solution but the sum of solutions) Now we can add our solutions : -p/3+8=20/3 which gives us p=4
@PrepworksEducation or @safal808 Could you please explain to me what -b/a is. Because when I looked up what sum property was I didn't find that formula.
@@PrepworksEducation Thanks for responding, I still just had one question though. Can you only use this if there's multiply quadratics equating to 0 and we already know the sum of the solutions?
@@blueberrymateen7200 the altitude theorem is basically just understanding how the two triangles are similar to each other. Triangle ABD is similar to triangle BCD, so AD/BD = BD/DC, because the two triangles are similar and their sides are in proportion. the altitude theorem is just a formula version of this.
Thank you so much after watching to your videos, now I'm pretty sure that I will perform good in free live full length test on 10th and 11th August held by Mentomind.
There's an error at 0:02 when you say, "exciting updates". hahahhahahahahahhaa just kidding i love your videos so much hahahahahahhahahahah thank you, youare the raisin i got a 5 on physics 1 and world and comp sci a and chemistry last year
BRO the moment you see an altitude you should at least of EUCLIDEAN THEOREM. which is basically in our case here. (BD)^2= DC x DA We have the value of BD by using Pythagorean as you used and side DA already given its all now left for DC to be calculated which is (3) ALSO.
thank you man I have taken the dsat twice now and tmmrw will be my third, and this video really helped breakdown some of the harder mod 2 questions, I have watched this video twice now lmao
Hey, for both of these questions it does not matter what you set the slider for d or n to. For 17, You would just have to adjust your other answer choices with those variables. For 6, we were solving for n, which we got as 1. I set “p” equal to 1 because it was simpler to work with. Since “p” was present in both the radius and coordinate portion, it serves as sort of a scale factor. The same “ratio” is maintained throughout, allowing us to set the slider to anything!
great video! quick question, where can we learn these concepts + the fundamentals of how to work these questions out?? i did quite a bit of practice on khan academy, but there weren’t a lot of questions like THIS. thank you!
Hey! Admittedly, there's no easy work around. The concepts behind the problems are not the issue, but some of the "hard" questions are meant to be a puzzle. They can be a challenge to everyone regardless of their mathematical background. Here is the SAT Question bank: satsuitequestionbank.collegeboard.org/digital/search Use the link above to search for more "hard" questions to practice (some are not on the same level as the ones in the video, but still good)!
Yes, that’s part of it! Also, it may be helpful to touch up on formulas/core concepts. Some examples include Vieta’s formula, understanding parts of a quadratic, the discriminant, congruence/similarity theorems etc.
For the first question, for the sake of time on the SAT can we assume that the common base is always going to be the starting # and then multiply by the given factor get our total and then solve each answer choice with x=c and see if our total matches?
For the first question you reviewed in the video, isn't option C equivalent to option B? When X is 6 because of the 1/c relationship, you would get 9 times 48 again. Could you explain why option B is correct but option C isn't? Thanks.
Hereyhow I solved 1: 48*(3)^(x+3) is 27 * (48*3^x), so if the function was just 48*(3)^x, c would be 3, but it's actually 48*(3)^.5x, so c needs to be 6, and therefore B is the correct answer since it has 1/6 as a coefficient for x
What if the slider in question 4 was not set for d being equal to 1? I changed it and my x-intercept was different. The question says that d is a constant, but that could mean any constant right?
can you do a more detailed explanation of the last question I don't understand why you had to subtract to get -4/3 or why we couldn't just use the old negative solution for 3x+p on the graph
Hey! So, for the first part we want a sum of 20/3 for our solution. We found that for the quadratic part of the expression, the sum of the solution is always positive 8, regardless of p (instead of using the desmos slider, you can also use the equation -b/a) So, for the “3x + p”, that determines the rest of the sum. 20/3 = 6.6667 Therefore, since we are currently at a sum of 8, we need a negative sum of 1.3333 to get us to a total sum of 20/3. Could you clarify your second question/put a timestamp?
Hey! That value you are referring to was actually given info from the problem. “C”, the common side length (side BD) had a value of 11, which we found using pythagorean theorem.
Sure thing! So, it says for each increase in value of “x” by “c” it will increase the function value by a factor of 27. This means if we start at x = 0, all the exponents will be = 0 - because anything multiplied by 0 is 0! Knowing this, it wasn’t 1/c to be displayed as a coefficient of “x”. All the equations here have a value multiplied by “x” to get the exponent. For example, in option A, it is (1/2) * x ^ In this case “2” is our “c” value. The “c” value and “x” value has to be the same. However if x = 2, then our exponent = 1, and our entire thing is not scaled by a factor of 27. “as a coefficient of” = this part may be confusing, but think of something like 2x. The “2” is the coefficient of “x”
“displays 1/c as a coefficient of x” This part indicates that 1/c is going to be in front of x. “each increase in value of x by c increases function value by a factor of 27” We are given all the “c” values. This cannot change. The parameter says that every time we increase by x by “c”, the whole function increases by a scale of 27 - so we just replace all the x values in the function with the c values. Only one of them when replaced will yield a factor of 27.
For Q1, could I also just look at the “growth rate” inside the parenthesis and immediately choose B since it is the only one showing its growth factor as 27, or 3^3, and the other ones aren’t? Or do I still have to go through the whole process that you showed us?
In the last question when you cancelled out 5x-45, you would have to have solved it beforehand in depth to know its irrelevant, what if a kid doesn't know that during the exam and still plugs it into demos and gets the wrong ans because of it. How can this be avoided?
Hey, you can still just plug the entire thing into desmos. Then, you would find the roots to that part as well -3 and +3: It would be displayed as: (0, 3) (0, -3) So, the sum here still cancels out as 3 + (-3) = 0
There’s quite a few! The SAT went from long-form essays that attached several questions to short passages containing no more than 150 words per question. It also no longer directly assesses vocabulary and includes more cross-text connections.
Ah, I see. These math questions aren’t too different from the ones we’ve seen before and don’t represent official changes. The DSAT hasn’t changed since it was officially released around a year ago.
Hey, we can set "d" equal to anything because it is present in both the equation and answer choices. So, it acts like a scale factor. I just chose "1" because it was easy to work with!
I can safely say none of this came on the the test this August it felt too easy of a test. So either I studied everything(which I didnt) or the test was no made challenging those time😭👍
Hey, what do you mean by picking any point of the line? In reality, we could just graph the entire thing into desmos granted we set p = to a numerical value (any), then just add all the solutions that appear. My method avoids some mishaps that occur with rounding/inputting decimals just to be safe!
Hey! You could use -b/a (as some viewers have pointed out) to find the sum of the solutions for quadratic part. Then, you would not need to graph any part of the expression!
@@PrepworksEducation also -b/a wouldn't work because it would be confusing to do and if you expand it it wouldnt be a quadratic function it'd be a quad function (x^4). you could just find the values of x and on the third factor you can use sum of solution and add all the x's values and set it up equal to 20/3 and you get p.
Hey! Here’s an alternative. Instead of using desmos, you can just use the formula -b/a to find the solutions for the quadratic part. Then, complete the rest of the steps as shown in the video!
@@PrepworksEducation Mate you do realize you could've solved the last problem so much friggin easier. just intorduce the formula -b/a and bam u solved it lmao I hate when teachers overcomplicated easy things. We get that desmos exists now and is some life-changing hack but that doesn't mean its the easiest way to go around solving everything. believe it or not, old school is the best way for many of these problems.
@@idkwhy77 You can solve most SAT problems in multiple ways - the issue here is, we can't illustrate every method for each problem or the video would be far too long. Whatever works for you (or is most intuitive) might not be the same for another person!
What if the slider for question 4 was not set to d being equal to 1? I changed it and my x-intercept was different. The question said d was a constant, but that could mean any constant right?
😎Big thanks to JW Math Tutoring who went through the work to find these changes + provide screenshots! Please make sure to check out his channel!
JW Math Tutoring: ruclips.net/channel/UCWZe0ktD7bP-b4clsq2CLjg
Also, in case you haven't noticed - this video was made at 1 AM at night.. lol. If you catch any errors/mistakes, please report them here in the comments!
Note: Question 2 in this video is NEW, but not unique (has been tested before)
the link doesnt work
It must’ve broke or something… :/
I think there’s an easier and faster way to do the last question without using desmos. You can find the sum of the solutions for (2x^2-16x+6p) using -b/a (the sum of the solutions for a quadratic) and get 8. Then, for the (3x+p) part, you can solve for x and get x= -1/3p. Then finally, you can solve for p by getting the equation -1/3p + 8 = 20/3, and you would get p = 4
That is the more effective method! The desmos method is still there in case you forget -b/a, or other formulas like Vieta’s formula! 👍
Exactly how I did it
@@PrepworksEducation vieta formula is the sum and product formula
I was not actually aware of that! Thank you for correcting my false attribution :)
I also did quite same but approach was different coz I did not know that formula. I used the ultimate root eqn for quadratic x = [-b±√(b2-4ac)]/2a , this was same because when you add both roots, you are left with (-b/2a)-(-b/2a) which equals -b/a 🤣
thank you for being a good samaritan and using desmos to “cheat” saves me so many stupid formulas, goat for that will be subbing
Appreciate it, but it is worth knowing some of them!
Here's an easier approach for Q. 15,
i) 3x+p= gives us x=-p/3
ii) 5x^2-45=0 gives us x= 3 and -3 (We can ignore this as sum =0)
iii) 2x^2-16x+6p,
For iii, we can apply the sum property which is -b/a leaving us with -(-16)/2= 8 (this isn't a solution but the sum of solutions)
Now we can add our solutions :
-p/3+8=20/3
which gives us p=4
Great method that saves a ton of time! If students forget the -b/a formula, the desmos method in the video is still a valid approach!
@PrepworksEducation or @safal808
Could you please explain to me what -b/a is. Because when I looked up what sum property was I didn't find that formula.
-b/a finds the sum of the solutions for a quadratic equation!
-b/a finds the sum of the solutions for a quadratic equation!
@@PrepworksEducation Thanks for responding, I still just had one question though. Can you only use this if there's multiply quadratics equating to 0 and we already know the sum of the solutions?
thank you for the reminder of the similar triangles theorem!!!!
Of course!
Please make more august sat hard questions like these please 🙏. This video really helped me learn the altitude theorem
We'll try our best! Happy to help.
@@fedal5956 same bro I’ve never heard of the Altitude Theorem until I saw this vid. College Board tryna screw us over lol
You don't need to know it! There are always multiple ways to solve problems.
@@blueberrymateen7200 the altitude theorem is basically just understanding how the two triangles are similar to each other. Triangle ABD is similar to triangle BCD, so AD/BD = BD/DC, because the two triangles are similar and their sides are in proportion. the altitude theorem is just a formula version of this.
Yup! It's just an application of common knowledge.
Good luck to all you SAT test takers.
🙌
Thank you so much after watching to your videos, now I'm pretty sure that I will perform good in free live full length test on 10th and 11th August held by Mentomind.
Best of luck!
question no.8 you can just use the Euclidean theory
11^2=(121/3)x and done x=3
here is the rule:
BD^2=(AD)(DC)
AB^2=(AD)(AD+DC)
BC^2=((DC)(AD+DC)
Great insight!
Thank you for this
Of course!
I wondered why number 6 of module 2 was so hard. I thought it got harder as you advance. Thanks for this video
No problem!!
Last one can be solved with vieta’s generalized formulas
Indeed! It’s a nice short cut. In case you forget it desmos will always be there!!
There's an error at 0:02 when you say, "exciting updates". hahahhahahahahahhaa just kidding i love your videos so much hahahahahahhahahahah thank you, youare the raisin i got a 5 on physics 1 and world and comp sci a and chemistry last year
I appreciate your kind words! Super glad you were able to get 5s on your AP Exams as well! 🔥
BRO the moment you see an altitude you should at least of EUCLIDEAN THEOREM. which is basically in our case here.
(BD)^2= DC x DA
We have the value of BD by using Pythagorean as you used and side DA already given its all now left for DC to be calculated which is (3) ALSO.
There’s many ways to solve this problem! 😄
thank you man I have taken the dsat twice now and tmmrw will be my third, and this video really helped breakdown some of the harder mod 2 questions, I have watched this video twice now lmao
No problem - you will do great! Perhaps these last minute tips will also help!
ruclips.net/video/fqvQDqdhAfI/видео.html
For question 6 and 17, could you explain why you set d and n equal to 1? Thank you!
Hey, for both of these questions it does not matter what you set the slider for d or n to.
For 17, You would just have to adjust your other answer choices with those variables.
For 6, we were solving for n, which we got as 1. I set “p” equal to 1 because it was simpler to work with. Since “p” was present in both the radius and coordinate portion, it serves as sort of a scale factor.
The same “ratio” is maintained throughout, allowing us to set the slider to anything!
great video! quick question, where can we learn these concepts + the fundamentals of how to work these questions out?? i did quite a bit of practice on khan academy, but there weren’t a lot of questions like THIS. thank you!
Hey! Admittedly, there's no easy work around. The concepts behind the problems are not the issue, but some of the "hard" questions are meant to be a puzzle. They can be a challenge to everyone regardless of their mathematical background.
Here is the SAT Question bank: satsuitequestionbank.collegeboard.org/digital/search
Use the link above to search for more "hard" questions to practice (some are not on the same level as the ones in the video, but still good)!
@@PrepworksEducation i see, i’m assuming that just practicing a lot will help with understanding these questions..? thank you so much for your help!
Yes, that’s part of it! Also, it may be helpful to touch up on formulas/core concepts.
Some examples include Vieta’s formula, understanding parts of a quadratic, the discriminant, congruence/similarity theorems etc.
This was awesome man thanks so much
No problem!
For the first question, for the sake of time on the SAT can we assume that the common base is always going to be the starting # and then multiply by the given factor get our total and then solve each answer choice with x=c and see if our total matches?
For the first question you reviewed in the video, isn't option C equivalent to option B? When X is 6 because of the 1/c relationship, you would get 9 times 48 again. Could you explain why option B is correct but option C isn't? Thanks.
Hey! We want a scale of 27 - it was actually my fault for misreading 3^3 as equivalent to 9 💀 - so 48 x 27 would be what we are looking for!
@PrepworksEducation that's a very weird question, but thanks for the answer
A lot of SAT questions are designed to trick you
Hereyhow I solved 1: 48*(3)^(x+3) is 27 * (48*3^x), so if the function was just 48*(3)^x, c would be 3, but it's actually 48*(3)^.5x, so c needs to be 6, and therefore B is the correct answer since it has 1/6 as a coefficient for x
in question 2 Why did you choose to substitute with number 5?
It made the fractions easier to work with as it simplified to whole numbers. You can use any value for “r” though 👍
@@PrepworksEducation oh kkkkkk thank you
No problem! Make sure to check out our new video that goes through the five hardest SAT problems ever made!
What if the slider in question 4 was not set for d being equal to 1? I changed it and my x-intercept was different. The question says that d is a constant, but that could mean any constant right?
Yes, any constant would work.
my desmos calculator doesn"t show the points inside circles it only show the points on circle so i cant get the center coordinates due to the reason
I don’t believe it automatically does that. I eyeballed it and just labeled that coordinate for the video 👍
can you do a more detailed explanation of the last question I don't understand why you had to subtract to get -4/3 or why we couldn't just use the old negative solution for 3x+p on the graph
Hey! So, for the first part we want a sum of 20/3 for our solution.
We found that for the quadratic part of the expression, the sum of the solution is always positive 8, regardless of p (instead of using the desmos slider, you can also use the equation -b/a)
So, for the “3x + p”, that determines the rest of the sum.
20/3 = 6.6667
Therefore, since we are currently at a sum of 8, we need a negative sum of 1.3333 to get us to a total sum of 20/3.
Could you clarify your second question/put a timestamp?
u can just graph or solve -x/3 + 8 -20/3 . i think doing the subtraction is an unnecessary step
That’s a fair point. I just wanted to break the problem down into more steps 👍
@@PrepworksEducation thank you so much
@@gabeowser9881 thanks
For any question says some random variable is a constant you can use desmo and set them to 1???
for question 6 circle question i got n=7 when I set p=2 instead of 1. Would there be multiple answers to this question?
Hey, I just checked using desmos and I got n = 3.
If p = 2, the radius of the circle is 6 units.
6 = 2n
n = 3
i think u didnt calculate the radius correctly
That may be the issue. Make sure to check the units on the axes - each box for me counted for 2 units!
bro thank you so much
No problem
It would be helpful if you make each problem bigger... I can barely see each answer choice and the question.
Apologies for the formatting! Filmed this on whim 🫤
@@PrepworksEducation all good!! just a suggestion/tip
Thanks!
thanks a lot man
You're welcome!
Why is D wrong?
For number 8 on the practice test 3, how did you know that 11(130)/3 is C? (I know its not “(130)” but I don’t know how to type square root, LOL)
Hey! That value you are referring to was actually given info from the problem.
“C”, the common side length (side BD) had a value of 11, which we found using pythagorean theorem.
Is there another way to do the triangle question? Or is it a must you know the theorem
There are multiple ways! Pythagorean theorem + trig, or you can combine pythagorean theorem + geometric mean
question no.8 you can just use the Euclidean theory
11^2=(121/3)x and done x=3
here is the rule:
BD^2=(AD)(DC)
AB^2=(AD)(AD+DC)
BC^2=((DC)(AD+DC)
Always lots of ways to solve - use whatever you feel most comfortable with.
I don’t understand q1 in this video pls explain explain the exponent part like why should c & x be equal
Sure thing!
So, it says for each increase in value of “x” by “c” it will increase the function value by a factor of 27.
This means if we start at x = 0, all the exponents will be = 0 - because anything multiplied by 0 is 0!
Knowing this, it wasn’t 1/c to be displayed as a coefficient of “x”.
All the equations here have a value multiplied by “x” to get the exponent. For example, in option A, it is (1/2) * x
^ In this case “2” is our “c” value. The “c” value and “x” value has to be the same. However if x = 2, then our exponent = 1, and our entire thing is not scaled by a factor of 27.
“as a coefficient of”
= this part may be confusing, but think of something like 2x. The “2” is the coefficient of “x”
@@PrepworksEducation ohhh thank you
Of course!
@@PrepworksEducation i still don't get why x and c should be equal, what part of the question suggests that they should be?
“displays 1/c as a coefficient of x”
This part indicates that 1/c is going to be in front of x.
“each increase in value of x by c increases function value by a factor of 27”
We are given all the “c” values. This cannot change.
The parameter says that every time we increase by x by “c”, the whole function increases by a scale of 27 - so we just replace all the x values in the function with the c values.
Only one of them when replaced will yield a factor of 27.
For Q1, could I also just look at the “growth rate” inside the parenthesis and immediately choose B since it is the only one showing its growth factor as 27, or 3^3, and the other ones aren’t? Or do I still have to go through the whole process that you showed us?
That works! I just wanted to explain it further because the question is worded weirdly!
I don't get the 8 (first) question, aren't 3 and 9 factors of 27 as well? lol please help an international student out
No worries! By factor, they don't mean the factors of 27. They mean a SCALE factor of 27.
@@PrepworksEducation oh that makes sense, thanks!!!
No problem!
In the last question when you cancelled out 5x-45, you would have to have solved it beforehand in depth to know its irrelevant, what if a kid doesn't know that during the exam and still plugs it into demos and gets the wrong ans because of it. How can this be avoided?
Hey, you can still just plug the entire thing into desmos. Then, you would find the roots to that part as well -3 and +3:
It would be displayed as:
(0, 3)
(0, -3)
So, the sum here still cancels out as 3 + (-3) = 0
What are the changes to the RW Section?
There’s quite a few! The SAT went from long-form essays that attached several questions to short passages containing no more than 150 words per question. It also no longer directly assesses vocabulary and includes more cross-text connections.
@@PrepworksEducation Thanks but I meant, any changes in the DSAT RW, like the intro of these new math questions
Ah, I see. These math questions aren’t too different from the ones we’ve seen before and don’t represent official changes. The DSAT hasn’t changed since it was officially released around a year ago.
Same question for Q17, why/how u set that to 1????
Hey, we can set "d" equal to anything because it is present in both the equation and answer choices. So, it acts like a scale factor. I just chose "1" because it was easy to work with!
Can i use desmos on the SAT paper? Because that would be easy. what about the dSAT form, can i still use desmos?
Yes. The digital SAT is tested in bluebook and DESMOS is built DIRECTLY into it.
I can safely say none of this came on the the test this August it felt too easy of a test. So either I studied everything(which I didnt) or the test was no made challenging those time😭👍
🙌
How about if in the question 15 you graph the equation in desmos and then just pick any point of the line ?
Hey, what do you mean by picking any point of the line? In reality, we could just graph the entire thing into desmos granted we set p = to a numerical value (any), then just add all the solutions that appear.
My method avoids some mishaps that occur with rounding/inputting decimals just to be safe!
@@PrepworksEducationokay okay
Yup!
Thanks a lot! I don’t understand why they word the questions so weirldy especially in the first one
No problem! And I agree, the wording for some questions are really weird!
is there any way to do number 2 algebrically or without plugging in answers for R?
Yes, you would plug in the coordinates into corresponding x, y values. You would work with the “r” values as replacements for x and y in the system.
Next time, can you crop the video so it just shows one question at a time, ocd is going crazy, overall good vid and good explanations though
Appreciate it! Formatting is always a struggle but we’ll take that into consideration!
for question 3 could you have used any number for p or just 1?
Yes, you could’ve used any number!
@@PrepworksEducation dude this question had me so cooked thanks
how could you do question 6 algebarically?
Hey! You could use -b/a (as some viewers have pointed out) to find the sum of the solutions for quadratic part. Then, you would not need to graph any part of the expression!
@@PrepworksEducation no i meant practice test #2 question 6
@@PrepworksEducation also -b/a wouldn't work because it would be confusing to do and if you expand it it wouldnt be a quadratic function it'd be a quad function (x^4). you could just find the values of x and on the third factor you can use sum of solution and add all the x's values and set it up equal to 20/3 and you get p.
Divide both sides by 8, complete the square, convert to circle formula.
“to find the sum of the solutions for the quadratic part”
-(-16)/2 = 8
8 + x = 20/3
x = -4/3
How would this not work?
Sorry but I don`t really understand the explanation for the last question. Can someone please re-explain or another easier explanation?
Hey! Here’s an alternative. Instead of using desmos, you can just use the formula -b/a to find the solutions for the quadratic part. Then, complete the rest of the steps as shown in the video!
@@PrepworksEducation ohh okay thank youu
No problem!
these were really difficult
There is a mix of easy, medium, and hard questions on the SAT!
Q6 - why the f the p is 6?
Do you mean Q6 of PT 2? I set "p" to 1 for the same reason as Q17. It acts like a scale factor & "1" is just easier to work with :)
holy shit plz never cook again at 1 am.. These explanations are terribly difficult to understand for some questions, some are understandable.
ANYTHING FOR THE VIEWERS!
@@PrepworksEducation Mate you do realize you could've solved the last problem so much friggin easier. just intorduce the formula -b/a and bam u solved it lmao
I hate when teachers overcomplicated easy things. We get that desmos exists now and is some life-changing hack but that doesn't mean its the easiest way to go around solving everything. believe it or not, old school is the best way for many of these problems.
@@idkwhy77 You can solve most SAT problems in multiple ways - the issue here is, we can't illustrate every method for each problem or the video would be far too long. Whatever works for you (or is most intuitive) might not be the same for another person!
@@idkwhy77 are you retarded? You can't even use -b/a for that sum of the solution problem.
@@idkwhy77Bro can you explain whats -b/a. I also didn’t understood his explanations
Q6 - why the fk the p is 6?
Do you mean Q6 of PT 2? I set "p" to 1 for the same reason as Q17. It acts like a scale factor & "1" is just easier to work with :)
Your explaining is terrible I'm sorry
Hey, we don't like hearing that. How should we improve?
What if the slider for question 4 was not set to d being equal to 1? I changed it and my x-intercept was different. The question said d was a constant, but that could mean any constant right?
Yes, any constant would work.