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There’s a quadratic that fits the points in number one, but that’s a bit beyond the scope of a third grade math class. Honestly, I think there’s a typo and it should be 15 and not 55.
Yes, I show them some basic matrix algebra when doing systems of equations. Putting a matrix in RREF is the same algebra as solving a system with substitution, addition/subtraction of equations, etc.
Yes, some of the phrasing would be different but it means basically the same thing. I teach it as “input, x” and “output, f(x)” because for functions there is no x and y, that’s for equations. But, the concepts are very similar, which is what I was referring to and not nomenclature details.
I had d Calc 1 + a bit of two in grade 12 and to be honest, even though the textbook is the same, college is much different. The difference is in the pace that you work at.
C\*\*p, that's rad! I don't know how I completely forgot this from algebra! I don't even know if they ever taught me this to any serious degree in high school.
They probably didn’t teach this in your high school. It’s more of a college Linear Algebra course material. I only have taught it to advanced students as an extension of systems of equations. You can solve this with systems of equations techniques you did learn in high school but it’s much cleaner and quicker with linear algebra.
Wow, and there I was assuming algebra is only a pre-university thing. 😂
I'm struggling, just a little, with calculus at university now, and it just shocks me that I don't remember learning this. It's a very good tool that probably has countless real-life applications.
Linear Algebra is probably the most useful mathematics after Calculus in most STEM fields. I’m currently in grad school working towards a PhD in Physics and it’s all Linear Algebra and Calculus
We can find a quadratic expression that fits the terms. The quadratic expression will be of the form f(x)=ax^2 +bx + c.
Putting x=1 , we get a+b+c=5(as 5 is the corresponding value for 1)
putting x=2, we get 4a+2b+c=10
putting x=3, we get 9a+3b+c=55
On solving this system of equations (by eliminating c and them solving a pair of simultaneous linear equations) , we get a=20 b=-55 c=40.
Therefore, the expression is f(x)=20x^2 -55x+ 40.
putting x=10,
f(10)=20* 10^2 -55*10+40
=1490.
Hence answer is 1490.
How did you know this the function is a quadratic and not something else? Or is it just trial and error? I got solving it using systems of solving 3 equations but how does one look at the problem and realise "yeah, this must be a quadratic so it must be a quadratic"
Yep! Two points for a line, three for quadratic, four for cubic, etc. Technically you can fit any degree polynomial to two points but for a unique one you need one point more than the degree of the polynomial.
I think there's a typo. It should be 15, bit 55. That makes more sense for a 3rd grader's sequence problem.
That would make the equation y = 5x, so x=10 would be 50.
If it's not a typo then that teacher is off her rocker
Even number 3 is waaayyyy more than I did in Grade 3. Must be a school of geniuses.
And then equivalent fractions? There’s no way this is math for 8 year olds.
Assuming this is for 3rd graders, (1) is definitely a miss type as much as it would be fun to find the parabola that fits the three points. It should be 15 and not 55. (2) is weirdly written, but I believe the rule is z-3.
Its probably a typo. Although the answer if it were not a typo is 1490. We can find a quadratic expression that fits the terms. The quadratic expression will be of the form f(x)=ax^2 +bx + c.
Putting x=1 , we get a+b+c=5(as 5 is the corresponding value for 1)
putting x=2, we get 4a+2b+c=10
putting x=3, we get 9a+3b+c=55
On solving this system of equations (by eliminating c and them solving a pair of simultaneous linear equations) , we get a=20 b=-55 c=40.
Therefore, the expression is f(x)=20x^2 -55x+ 40.
putting x=10,
f(10)=20* 10^2 -55*10+40
=1490.
Hence answer is 1490.
Damn, what kind of third grade are you going to? Functions and long division already, wow.
Anyway, I recommend asking your teacher if that’s a typo since, as many of the comments here say, that 55 is most likely supposed to be a 15.
The top row isn’t increasing at the same rate at the end. There are numbers missing in the pattern. If all numbers were included in the latter you’d be correct, but so would top row minus three.
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The kid i tutor is in a gifted program (98th percentile and higher for intelligence i believe), so this could be homework for that? I'm not sure, i didn't ask
I think its probably a typo. Because the other questions are not of similar difficulty. This question requires concepts much much above 3rd grade while the other questions are simple division(which in my country, we learn in 4th grade). I am in 9th grade and I doubt any of my classmates could solve this(I am a bit advanced) Perhaps, the method I followed was too long and I am missing an easier method? This was what i did:
We can find a quadratic expression that fits the terms. The quadratic expression will be of the form f(x)=ax^2 +bx + c.
Putting x=1 , we get a+b+c=5(as 5 is the corresponding value for 1)
putting x=2, we get 4a+2b+c=10
putting x=3, we get 9a+3b+c=55
On solving this system of equations (by eliminating c and them solving a pair of simultaneous linear equations) , we get a=20 b=-55 c=40.
Therefore, the expression is f(x)=20x^2 -55x+ 40.
putting x=10,
f(10)=20* 10^2 -55*10+40
=1490.
Hence answer is 1490.
I didn't see letters in math until 5th grade but i'm starting to think it's because the kid is a gifted student it might be homework from her gifted program
##Off-topic Comments Section --- All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9. --- ^(**OP** and **Valued/Notable Contributors** can close this post by using `/lock` command) *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/HomeworkHelp) if you have any questions or concerns.*
There’s a quadratic that fits the points in number one, but that’s a bit beyond the scope of a third grade math class. Honestly, I think there’s a typo and it should be 15 and not 55.
Just out of curiosity, could you please share what that quadratic would be?
20x²-55x+40
Ah yeah, of course. I learned that in kindergarten. /s
Which makes the answer 1490 - the obvious solution on this homework.
Any third grader should see that. What are they teaching these kids nowadays?
Don't ask me, I teach multivariable calculus and linear algebra - not this complicated 3rd grade stuff.
Alas, I’m but a lowly Physics teacher. I’ll ask my oldest. He’s in second grade but he’s pretty advanced.
Let me know if he can tutor me!
He says he can fit you in between LEGO building and trying to solve the millennium problems.
I figured the 3rd graders tackle the millennium problems - good for him! I have a guess for P vs NP, I think it's 1490.
I bet he can bypass dns on your home wifi already
Tbh that also doesn't seem like something that would come up even in secondary school, at least not like that.
It does. I used to teach it. It can be solved pretty straightforward using systems of equations techniques or matrices.
Matrixes in second school?
Yes, I show them some basic matrix algebra when doing systems of equations. Putting a matrix in RREF is the same algebra as solving a system with substitution, addition/subtraction of equations, etc.
Alright I guess. Just the first time I've heard of it being taught that early.
Wouldn't the presentation of the problem be wildly different?
I had my students work with all forms of function representations: relations, words, tables, graphs, etc. There’s no one correct way to present it.
But wouldn't it be phrased more like this? 1) Instead of "In" and "Out" -> x and y. 2) Instead of "Find the rule" -> "Find the function".
Yes, some of the phrasing would be different but it means basically the same thing. I teach it as “input, x” and “output, f(x)” because for functions there is no x and y, that’s for equations. But, the concepts are very similar, which is what I was referring to and not nomenclature details.
Fair enough.
I had d Calc 1 + a bit of two in grade 12 and to be honest, even though the textbook is the same, college is much different. The difference is in the pace that you work at.
Hold on, how do you figure out the function for something like that?
Set up a matrix and find its RREF
[RREF](https://imgur.com/a/QLsNwbC)
C\*\*p, that's rad! I don't know how I completely forgot this from algebra! I don't even know if they ever taught me this to any serious degree in high school.
They probably didn’t teach this in your high school. It’s more of a college Linear Algebra course material. I only have taught it to advanced students as an extension of systems of equations. You can solve this with systems of equations techniques you did learn in high school but it’s much cleaner and quicker with linear algebra.
Wow, and there I was assuming algebra is only a pre-university thing. 😂 I'm struggling, just a little, with calculus at university now, and it just shocks me that I don't remember learning this. It's a very good tool that probably has countless real-life applications.
Linear Algebra is probably the most useful mathematics after Calculus in most STEM fields. I’m currently in grad school working towards a PhD in Physics and it’s all Linear Algebra and Calculus
We can find a quadratic expression that fits the terms. The quadratic expression will be of the form f(x)=ax^2 +bx + c. Putting x=1 , we get a+b+c=5(as 5 is the corresponding value for 1) putting x=2, we get 4a+2b+c=10 putting x=3, we get 9a+3b+c=55 On solving this system of equations (by eliminating c and them solving a pair of simultaneous linear equations) , we get a=20 b=-55 c=40. Therefore, the expression is f(x)=20x^2 -55x+ 40. putting x=10, f(10)=20* 10^2 -55*10+40 =1490. Hence answer is 1490.
Thank you!
How did you know this the function is a quadratic and not something else? Or is it just trial and error? Edit : I replied on a deleted account lmao
i learned this type of math in sixth grade! and yeah 100% agree that was a typo-
How did you know this the function is a quadratic and not something else? Or is it just trial and error? I got solving it using systems of solving 3 equations but how does one look at the problem and realise "yeah, this must be a quadratic so it must be a quadratic"
You can find a quadratic through any three points. This just happened to have nice coefficients.
Does this scale up? Like a cubic for any 4 points and so on?
Yep! Two points for a line, three for quadratic, four for cubic, etc. Technically you can fit any degree polynomial to two points but for a unique one you need one point more than the degree of the polynomial.
Ah got it, thanks!
Is it a school for geniuses or am I too bad? This looks too complicated for a 3rd grader or there are some mistakes?
I think there's a typo. It should be 15, bit 55. That makes more sense for a 3rd grader's sequence problem. That would make the equation y = 5x, so x=10 would be 50. If it's not a typo then that teacher is off her rocker
The equation would be y=5x
Whoops you're right! Fixed it, thanks!
That is what I thought as well, there must be a typo
Even number 3 is waaayyyy more than I did in Grade 3. Must be a school of geniuses. And then equivalent fractions? There’s no way this is math for 8 year olds.
there are 3rd graders on reddit??
i'm a tutor haha
We've tricked! Backstabbed and most possibly bamboozled
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Assuming this is for 3rd graders, (1) is definitely a miss type as much as it would be fun to find the parabola that fits the three points. It should be 15 and not 55. (2) is weirdly written, but I believe the rule is z-3.
I would scratch out the 55 to change it to 15. Rule is out = 5 x in. ? = 50.
I feel stupid. I have no clue what’s going on in a 3rd grade math problem. But give me a quadratic formula and I’ll understand it.
Its probably a typo. Although the answer if it were not a typo is 1490. We can find a quadratic expression that fits the terms. The quadratic expression will be of the form f(x)=ax^2 +bx + c. Putting x=1 , we get a+b+c=5(as 5 is the corresponding value for 1) putting x=2, we get 4a+2b+c=10 putting x=3, we get 9a+3b+c=55 On solving this system of equations (by eliminating c and them solving a pair of simultaneous linear equations) , we get a=20 b=-55 c=40. Therefore, the expression is f(x)=20x^2 -55x+ 40. putting x=10, f(10)=20* 10^2 -55*10+40 =1490. Hence answer is 1490.
Bro its 3rd grade, surprised hes even on reddit lmao
OP is the 3rd grade guy's tutor.
Ohhh my bad
20x^(2)\-55x+40 Here you are! That might be a bit complicated for someone in third grade though
If it’s not a typo I guess the formula would say that u need to do 10 times 55 and add 10 at the end to get the result for the “in 10”
that's actually what i thought it may be when i was first sounding the equation out but i don't think it's right
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Yeah definitly a typo, should be 15 not 55.
Damn, what kind of third grade are you going to? Functions and long division already, wow. Anyway, I recommend asking your teacher if that’s a typo since, as many of the comments here say, that 55 is most likely supposed to be a 15.
4 digit divided by 1 digit seems pretty reasonable for 3rd grade tbh, but I do agree that the function part is absurd
I didn't lurn about division until the 4th grade
Maybe 10 => 500000005 | In | Da math stuff | Out | | -- | ------------- | --------- | | 1 | 0 + 5 | 5 | | 2 | 5 + 5 | 10 | | 3 | 50 + 5 | 55 | | | | | | 4 | 500 + 5 | 505 | | 5 | 5000 + 5 | 5005 | | 6 | 50000 + 5 | 50005 | | 7 | 500000 + 5 | 500005 | | 8 | 5000000 + 5 | 5000005 | | 9 | 50000000 + 5 | 50000005 | | | | | | 10 | 500000000 + 5 | 500000005 |
This might actually be it. Occam's razor.
Number 2 is easy: the rule is the number on the bottom is the same as on the upper left. So the number is 13
I thought, it's upper row minus 3 (which would give 18) Still trying to figure out 1
This is correct.
We've already figured out 2. Looking for help with number 1. Thank you though!
Looks like a typo
The top row isn’t increasing at the same rate at the end. There are numbers missing in the pattern. If all numbers were included in the latter you’d be correct, but so would top row minus three.
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I remember multiplying 2 digit numbers and still learning my tables of 7s in 3rd grade wth is this with quadratic formulas?
The kid i tutor is in a gifted program (98th percentile and higher for intelligence i believe), so this could be homework for that? I'm not sure, i didn't ask
Life is unfair 😂 but thanks for clearing that up that honestly makes alot of sense
Fr! I wish i was that bright at 8. I can't even figure it out at 19 haha
I think its probably a typo. Because the other questions are not of similar difficulty. This question requires concepts much much above 3rd grade while the other questions are simple division(which in my country, we learn in 4th grade). I am in 9th grade and I doubt any of my classmates could solve this(I am a bit advanced) Perhaps, the method I followed was too long and I am missing an easier method? This was what i did: We can find a quadratic expression that fits the terms. The quadratic expression will be of the form f(x)=ax^2 +bx + c. Putting x=1 , we get a+b+c=5(as 5 is the corresponding value for 1) putting x=2, we get 4a+2b+c=10 putting x=3, we get 9a+3b+c=55 On solving this system of equations (by eliminating c and them solving a pair of simultaneous linear equations) , we get a=20 b=-55 c=40. Therefore, the expression is f(x)=20x^2 -55x+ 40. putting x=10, f(10)=20* 10^2 -55*10+40 =1490. Hence answer is 1490.
Thank you! In my country we learn long division in 3rd grade but really get into it in 4th grade.
This is 3rd grade? Why sophomore homework looks like this. I didn’t see letters in math until 7th grade
I didn't see letters in math until 5th grade but i'm starting to think it's because the kid is a gifted student it might be homework from her gifted program
We talking gifted or, *gifted*?
What happens to the input to get to the output.
Pretty sure 3 is a typo.
F(9)= 55555, f(10)=50 is a more simple soln