help-28

[woeful pasture]

did u get jacobian as initially -2(x^2 + y^2)?

[patent valve]

Yes, this is J(x,y) actually. Since finding J(x,y) was easier, I found J(x,y), and then found the J(u,v) as 1/J(x,y) which is the same thing.

[woeful pasture]

shouldnt the integrand be

[ -\frac{1}{2} \sqrt{4u^2 + v^2}]

[glossy valve]

k

[patent valve]

Yes, but we need the absolute value by plugin the J(u,v) in the integral. So my equation is right(i suppose)

[woeful pasture]

im not following?

shouldnt the square root thingy be in the numerator?

[patent valve]

yes, I think (99%) my findings is correct so far, I'm asking if is there better way(better change of variable) to do that.

[woeful pasture]

[\int_{v=1}^{9} \int_{u=2}^{4} \frac{1}{2}\sqrt{v^2 + 4u^2} du dv]

[glossy valve]

k

[woeful pasture]

then u can do trig sub, no?

i think this is good enough substitution

[patent valve]

This is wrong, the square is not in the numerator.

Notice J(u,v) = 1 / J(x,y), but you found J(u,v) = J(x,y) which is not right.

[woeful pasture]

do u mean

[patent valve]

No, sorry it must be in the denominator. Sorry, am a little tired 😦

[woeful pasture]

${J(u,v) = \frac{\partial(u,v)}{\partial(x,y)}}$?

[glossy valve]

k

[patent valve]

No, actually
${J(u,v) = \frac{\partial(x,y)}{\partial(u,v)} }$

[glossy valve]

idiot_max

[woeful pasture]

ye sorry

but it think its a notation not a fraction..

[patent valve]

No worries. Yes, it's a notation, but the relation between J(x,y), and J(u,v) is defined as the following:

${J(u,v) = \frac{1}{J(x,y)} }$

[glossy valve]

idiot_max

[woeful pasture]

so, ${J(u,v) = -2(x^2 + y^2)}$?

[glossy valve]

k

[patent valve]

No, it's actually J(x,y) you found.

[woeful pasture]

oh

i made a mistake

[patent valve]

No worries, this is fine. I find the notation a little confusing myself too.

[woeful pasture]

but i think that change of variable is the best tho

[patent valve]

Probably yes, though the integral looks not very friendly. So that's why am asking if is there any better change of system regarding my findings.

[full forum]

@patent valve Has your question been resolved?

[patent valve]

Thank you so much for following up, really appreciate it. I think the change of system I found is fair enough, and it was the obvious one, so I'm going to consider it done.
Math bless us all 🍻🤘💯

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[idle bridge]

(a) Find, with reasons, all ways to arrange the numbers 1, 3, 5, 7, 9, 11 into two groups of three so that the sum of each group is a different prime number
(10) about
Conditions
The sum of each group must be a prime number
2 groups of 3
The sum of everything is 1+3+5+7+9+11 = 36

The smallest possible sum for a group is 1+3+5 = 9
The largest possible sum for a group is 7+9+11=27
However, 9 and 27 are not prime numbers -> we find the prime numbers in between

The possible prime numbers between 9 and 27 to act as the sums of the groups of 3 are 11, 13, 17, 19, 23.

Find the combinations for each of the groups.
11: 1+3+7
13: 1+3+9, 1+5+7
17: 1+5+11, 1+7+9, 3+5+9
19: 3+5+11, 3+7+9

(b) Can the numbers 1, 2, 3, …, 9 be arranged into three groups of three so that the sum of each group is a different prime number? Explain.
The total sum of 1+2+3+...+9 = 45

Splitting into 3 groups → p+q+r=45

Of group
Smallest possible sum = 1 + 2 + 3=6
Largest possible sum = 7 + 8 + 9 =24

[unreal wren]

What are you struggling with?

[full forum]

@idle bridge Has your question been resolved?

[idle bridge]

I'm struggling with continuing my solution and finding the best method to do both problems

[unreal wren]

So I will help with part a first

The sum of the numbers are 36, right?

And you found 8 possible combination for groups of sum 11,13,17,and 19

We need to choose 2 groups such that their sum is 36, what must those 2 combination s be?

[idle bridge]

is there only 2 combinations?

perhaps 17, 19; i cannot find any other combinations

[unreal wren]

Just show me your answer for it first

[unreal wren]

Yup correct

But which exactly? (The 2 groups of 3 numbers)

What are the groupings of 1,3,5,7,9,11?

[idle bridge]

I do not understand

[unreal wren]

Is it, for example : 1,3,5 with sum 7 and 7,9,11 with sum 27?

[idle bridge]

yes but this wasnt the method i used

[unreal wren]

What are the groups of 3 numbers

[unreal wren]

I mean, you found that it must be sum 17 and 19 right?

And there are many ways to group the numbers to get sum 17 or 19, which exactly did you use?

[idle bridge]

so this was the main part of my approach. you got 2 groups of 3 numbers

the sums could be The possible prime numbers between 9 and 27 to act as the sums of the groups of 3 are 11, 13, 17, 19, 23.

not sure what to do after this mainly

[unreal wren]

Example, 17 can be 1+5+11, 1+7+9, or 3+5+9

So, you alr calculated the possible prime numbers and how to get those sums

[idle bridge]

are you sure its only 17, 19 for the two groups?

[unreal wren]

Like this one

[idle bridge]

how do i prove this?

[unreal wren]

Yes

[unreal wren]

It's obvious

[idle bridge]

will my method be correct?

[unreal wren]

Yes

First, just tell me the groupings

[unreal wren]

There are many ways to get sum of 17, same thing with 19. You must state which one you used

[full forum]

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[idle bridge]

.reopen

[full forum]

✅

[unreal wren]

Ok do you get what I mean by this?

[idle bridge]

Not exactly

Do you mean find all possible combinations because there's obviously more than 1

[unreal wren]

Yes

The sum of the 2 groups must be 17 and 19 tho,but the way you achieve the sum 17 or 19 can be different

[idle bridge]

I got these so far: 17: 1+5+11, 1+7+9, 3+5+9 19: 3+5+11, 3+7+9

[unreal wren]

Yup

So, how do they pair together? Ex : 1+5+11 and 3+7+9

[idle bridge]

so there're only two pairs?

[unreal wren]

Yes

[unreal wren]

I alr gave you one

[idle bridge]

but i have a feeling there's more to this question. or is 2 the answer

[unreal wren]

Only 2

I am certain

[idle bridge]

What about question (b), it seems to be a bit harder with 3*3 combinations

(b) Can the numbers 1, 2, 3, …, 9 be arranged into three groups of three so that the sum of each group is a different prime number? Explain.

[unreal wren]

Wait I just realised you haven't calculated the ways to get 23 as a sum

You calculated the ways to get 11,13,17,19 and noted that 23 was possible but you didn't count the possible combinations for 23

Calculate the 2 ways to get 23 first

[idle bridge]

Why is 23 possible?

[unreal wren]

There is a way to get that number by summing three numbers from 1,3,5,7,9,11

Look for them

A trio of odd numbers less than 13 which sum to 23

[idle bridge]

i dont understand

[unreal wren]

Look for 3 numbers in the set {1, 3,5,7,9,11} which sum to 23

[idle bridge]

Are you saying these two methods are possible to get 36
13, 23
17, 19

[unreal wren]

Yes

But, as with 17 and 19,there are many ways to pair them

[idle bridge]

i still dont understand

[unreal wren]

Okay, for 17 and 19 I already gave this example right?

Look for the other possible pair 17 and 19

[idle bridge]

Maybe for 17, 19
We've got this right?
17: 1+5+11, 1+7+9, 3+5+9
19: 3+5+11, 3+7+9

so 1, 5, 11, 3, 7, 9 or 1, 7, 9, 3, 5, 11

So 2

[unreal wren]

Yess

Now look for ways to get the sum of 23

Ex : 5+7+11

[idle bridge]

Combinations
11: 1+3+7
13: 1+3+9, 1+5+7
17: 1+5+11, 1+7+9, 3+5+9
19: 3+5+11, 3+7+9
23: 3+9+11

So i think 3, 9, 11; match with 13: 1, 5, 7

[idle bridge]

yes or no?

[unreal wren]

Yes, what does 5+7+11 match with?

[unreal wren]

Yup, but u forgot 5+7+11

[idle bridge]

oh yes true, 1, 3, 9

So my process is basically correct, right?

for (a)

[unreal wren]

Yup

It's basically the same method for part b

But I will help you out if you still need it

[idle bridge]

Yeah i need help

[unreal wren]

Note all the prime numbers between 6 and 24

[idle bridge]

I also did this:
Smallest possible sum = 1 + 2 + 3=6
Largest possible sum = 7 + 8 + 9 =24
7, 11, 13, 17, 19, 23
But theres no total sum

[unreal wren]

There is a total sum here? It's 45

[idle bridge]

oh yes true.

but its still harder for me to do it

[unreal wren]

Look for three prime numbers that sum to 45

(If you can find any)

[idle bridge]

its ve ry hard to systematically do it

[unreal wren]

Are you allowed to use calculator?

You could just brute force

[idle bridge]

no i cannot use calculator

i need a systematic way to do it proper;y

without erorr

[unreal wren]

So, no brute force huh.

Well the sum of the last digits of the prime numbers must be equal to 5

[idle bridge]

yeah?

so what is the list of numbers

[unreal wren]

What do you mean?

[idle bridge]

the prime numbers that add to 45?

[unreal wren]

I am just a helper not a calculator

[idle bridge]

cannot find any combination of 3 groups from this that add to 45

(b) Can the numbers 1, 2, 3, …, 9 be arranged into three groups of three so that the sum of each group is a different prime number? Explain.
The total sum of 1+2+3+...+9 = 45

Splitting into 3 groups → p+q+r=45

Of group

I also did this:
Smallest possible sum = 1 + 2 + 3=6
Largest possible sum = 7 + 8 + 9 =24
7, 11, 13, 17, 19, 23
But theres no total sum

[unreal wren]

The question never mentioned that there ARE solutions

Then explain (prove) that there aren't any solutions, if that is your answer

[idle bridge]

ah... i see. so one final check of my method, try to see any holes in it

i find the total sum

[unreal wren]

Notice that, in part a they said FIND and in part b they used CAN. That is a clue to whether the question even has solutions or not

[idle bridge]

i find the minimum and maximum sums of the groups then i find the prime number sums that work

[unreal wren]

Yea

But then prove that there aren't any trio of prime numbers that sum to 45

[idle bridge]

I try to make sure the prime numbers are different and add the total (the groups)
Then yeah

[idle bridge]

can you elaborate on this

[unreal wren]

You don't know how to prove that?

Doesn't the question in part b say "explain"? So just casually explain, show, that the sum of the prime numbers can't be 45

Just write out your thought process

[idle bridge]

does my method not already work?

what method of proof do i use to definitely provethat there aren't any trio of prime numbers that sum to 45

[unreal wren]

That is beyond me (I am bad at proofs)

All i can think of is brute forcing

Writing out every sum to show that none of them are 45

[idle bridge]

could i use proof by contradiction for this

by assuming that it does work?

i still plan to:
i find the total sum
i find the minimum and maximum sums of the groups then i find the prime number sums that work
I try to make sure the prime numbers for the sums are different and add the total (the groups)

all good this method?

[unreal wren]

I have no idea about proofs

Wait for another helper ig

[idle bridge]

i think this is good for now

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[idle bridge]

could anyone quickly check these 2 qs thanks

[hidden harbor]

that looks fine

[idle bridge]

im not sure if expand and simplify means my working for (a)

also for (b), not particularly sure if my working is right and if it meets the requirements

[hidden harbor]

well, on (a) you're technically using the notable products instead of expanding, you'd need to do the three steps of (x^5+1)(x^5+1), the product, and your result after adding the similar terms

[idle bridge]

i dont understand that method

[hidden harbor]

on (b) i'd say it does meet requirements, since you're using the difference of squares on step 2, and you're writing it as a product of two factors on step 3

okay, so you know that something squared means something multiplied by itself, right?

[idle bridge]

yes?

[hidden harbor]

$(x^5+1)^2=(x^5+1)(x^5+1)=x^5\cdot x^5+x^5\cdot 1+1\cdot x^5+1\cdot 1$

[glossy valve]

LordFelix

[hidden harbor]

which then you simplify to what you have

[idle bridge]

so what are the three steps here

is it ok to skip the middle tep

[hidden harbor]

the first step would be writing the square as a product of itself

the second step would be doing the actual multiplications (that i've left indicated)

the third step would be adding the similar terms, which is what you have as result

[idle bridge]

what about is (b) correct

[hidden harbor]

yes

[idle bridge]

is there any other answers. why would they give a as a subq before b

[hidden harbor]

thing is, your results are correct, the only question would be if (a) is allowed to be done in a single step (i'd allow it, since notable products are a thing that should be learned)

if you're not sure, you can just put every step in between

[idle bridge]

ok yep

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[tame cradle]

Hello

[idle bridge]

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[drifting shale]

lets do a proof by parity and around 13 cases.

each group is prime, and each prime except 2 is odd. so the split of [1,3,5,7,9] in groups is 1-1-3
the only triples that sum to a prime are:
1+3+7=11
1+3+9=13
1+5+7=13
1+7+9=17
and then the rest of the knapsack has to be be equal to 45-p = even number which is made of two primes. down here are the cases, since one group is <= the other.
1+3+7=11. 45-11=34.
34-17=17 X.
34-15=19 X.
34-13=21 X.
34-11=23 X.
34-9=25 X.
34-7=27 X.
1+3+9=13. 45-13=32.
32-15=17 X.
32-13=19 X.
32-11=21 X.
32-9=23 X.
32-7=25 X.
1+5+7=13
same as before
1+7+9=17
45-17=28
28-13=15 X.
28-11=17 X.
28-9=19 X.
28-7=21 X.

each case has either a repetition or not a prime

[drifting shale]

also, i do not how to prove something is prime, so these questions are usually disproving primes

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[manic bramble]

I'm trying to give a definition of the "Cartesian Product". I have 2 set builders for the definition. Based on the wikipedia page, the 2nd one is correct. However, my mind is having a hard time differentiating between the 2 definitions. In the first one, it reads to me as creating only 1 n-tuple in resulting set; but, the second one also reads the same to me. The difference between them is the "forall" in the second.

[manic bramble]

To me, the first version produces a set: {(a, b, c, ..., n)}. The second also reads to me as producing the same.

That is, the forall doesn't read to me as materially changing the outcome.

Where am I going wrong?

[full forum]

@manic bramble Has your question been resolved?

[sharp lagoon]

The part to the left of | doesn't use i. On the right of |, you have a_i is in A, but it's unclear which i values that applies to. For example, you could say that a_1 and a_2 come from A and the others have some other restriction. The forall i tells which subscripts a_i applies to.

[manic bramble]

But, how is that different between the 2 notations, and how does that imply all the permutations of the tuples?

[sharp lagoon]

Well, if you say i is in {1, ..., n}, that doesn't require that all i values from 1 to n must be used.

Hmm, wait.

So, the thing on the right is what's true of one value on the left.

Like { x | x in reals and x + 2 < 3 } has the "x in reals and x + 2 < 3" talking about one x value on the left.

If you say that a_i is in A, that doesn't tell which i values are used.

There's no i on the left side.

So, we need to specify which values can be filled in for i.

If you say that i is in {1, ..., n}, that means that there's one value i that's in {1, ..., n}.

So, maybe a_i refers to a_1.

There's only one i.

If you say for all i in {1, .., n}, that means that you're sort of iterating through all those i values.

[manic bramble]

So, you're saying that, in a set like {x^2 | x in R}, that means a single x square? Vs {x^2 | forall x in R} means all x-squared?

[sharp lagoon]

No.

On the right, yes.

On the left, no.

The part on the right are restrictions on each potential x value.

Emphasis on the fact that 'value' is singular.

[manic bramble]

I mean, how many elements of the set are produced.

[sharp lagoon]

So, if you want to say that x is mustard, no, mustard is not in the reals. If you want to say that x is 5 + 6i, no, x is not in the reals.

Any elements that pass the test on the right of |.

The left side is the form of the elements.

Like {(a, b) | ...} has (a, b) as the form of the elements.

The ... is the restrictions on each element.

And the resulting set is every element that passes the restrictions on the right.

[manic bramble]

Yes, I understand that.

My point is, | x in R means any x in R. So then, why would x in {1, ..., n} mean a single element, and not all the elements in {1,...n}.

[sharp lagoon]

No, it doesn't.

It means that a specific x value must be in R to be in the set.

Like if x = 3, that's the specific x value.

Does it pass the restrictions? Yes, it's a real number.

So, that's one of the elements in the resulting set.

[manic bramble]

Ok, then x in {1, ..., n} means any x in that set, no?

[sharp lagoon]

It means that you fill in a particular x and see if it's in that set.

If so, that particular x is in the resulting set.

So, n + 1 wouldn't pass, so it wouldn't be in the set.

[manic bramble]

right. So that means x could be {1, 2, ..., n}

So a_i could be any of {1, 2, ..., n}

So, how is that different than forall i in {1,...,n}

[sharp lagoon]

The idea is that you give exactly one value for each variable to the part on the right.

The part on the right tells you whether the value is in the set or not.

So, if you do i in {1, ..., n}, that's going to be exactly one i value.

Much like if you do {(a, b) | a in reals, b in naturals}.

You supply exactly one value for b on the right.

That tells you whether that b passes the test.

[manic bramble]

right

[sharp lagoon]

So, if you have {(a_1, a_2, a_3) | i in {1, 2, 3}}, you give exactly one value for i to the part on the right.

[manic bramble]

yes

[sharp lagoon]

So, i can only have one value, not all three.

While you're still over on the right side, figuring out whether all the conditions are met, i has exactly one value.

Just like {x | x in the reals} has exactly one value for x.

If it meets the conditions, then that x can be used on the left side.

[manic bramble]

yes

[sharp lagoon]

When you do forall, that changes.

i can have every value in the set it's an element of.

So, with {(a_1, ..., a_n) | a_i in A, for all i in {1, ..., n}}, i doesn't have just one value that we're testing to make sure it can be used on the left side.

[manic bramble]

So, do all the possible a_i in A (on the right) apply to each individual a_1, a_2 on the left?

[sharp lagoon]

Yes.

I would write the notation a bit differently.

[manic bramble]

as in a_1 = a_1 in A, a_1 = a_2 in A, etc

[sharp lagoon]

{(a_1, ..., a_n) | for all i in {1, ..., n}, a_i is in A}

That makes it a bit clearer, since it's just a standard quantified statement.

Have you done quantifiers outside of sets?

[manic bramble]

not much

[sharp lagoon]

x is in reals is a proposition that can be true or false.

for all i in {1, ..., n}, a_i is in A is also a proposition that can be true or false.

[manic bramble]

What I find confusing is that (a_1, a_2) on the left seems to implicitly bind to the corresponding a_i on the right. But, you imply that "forall" kind of breaks that binding.

[sharp lagoon]

No, the bindings don't change.

a_1 on the left is still a_1 on the right.

The trick is that for all is like a loop in programming.

It loops over all the i values and, for each one, checks whether the proposition after the period or whatever is true.

If all of them are true, the whole for all statement is true.

So, you can think of for all i in {1, ..., n}. a_i is in A as a_1 is in A and a_2 is in A and a_3 is in A ... and a_n is in A.

If all the anded parts are true, then the whole proposition with all those ands in it are true.

[manic bramble]

So, with L=(a_1) and R=(a_1 in A), those two a_1's match, but if 'i' on the right changes, say 2, then how does a_2 in A still binds to a_1 on the right?

[sharp lagoon]

I'm not sure what you mean.

If you have {a_1 | a_1 in A}?

[manic bramble]

yeah

[sharp lagoon]

In that case, your set is just equal to A.

a_1 is the form of the results and a_1 must be an element of A.

So, you throw everything and the kitchen sink (literally) as possibilities for a_1, and it throws out everything that's not in A.

The quantifier is basically a way of shortening long propositions.

a_1 is in A and a_2 is in A and a_3 is in A and ... and a_n is in A.

That can be shortened to for all i in {1, 2, 3, ..., n}. a_i is in A.

[manic bramble]

I'm heading now 2 things that sound contradicting to me:

1. {(a_1, a_2) | "a_1 on left matches: a_1 in A on the right", "a_2 on left matches: a_2 in A on the right"}

2. {(a_1, a_2) | "a_1 on the left matches 'a_1 in A, a_2 in A, ...' on the right; then "a_2 on the left matches 'a_1 in A, a_2 in A, ...' on the right}

[sharp lagoon]

Well, it's (a_1, a_2) on the left.

As far as matching, there isn't really any matching.

You can think about it as manufacturing.

You start on the right side.

You get something like a_1 is in A and a_2 is in A.

Then, you fill in the variables with all possible combinations of two things.

So, like a_1 = sandwich, a_2 = 5.

If that combination fulfills the condition on the right of |, then the variable values get filled in on the left.

So, let's say that A = {1, 2, 3, 4, 5}.

The condition is a_1 is in A and a_2 is in A.

We fill in a_1 = sandwich and a_2 = 5.

a_1 is in A fails, so since and requires them both to be true, the whole condition fails and that combination won't go to the left side.

We fill in a_1 = 2 and a_2 = 3.

The condition is that both of them must be in A, and they are.

So, those variables go to the left.

(a_1, a_2) is the form of the elements on the left, so you get (2, 3) in your resulting set.

[manic bramble]

Yes, this all makes sense to me.

At this point,

Am at a point that I'm not even sure if I'm still confused. But, if I still am, I'm not sure what I'm confused about. I know that may not make sense.

I'm going to go over the chat, and review what we discussed.

I wont take more of your time

[sharp lagoon]

Well, there's one more thing to go over. With {(a_1, a_2) | a_i is in A, i is in {1, 2}}, it's sort of not well-formed.

[manic bramble]

You've been very kind

[sharp lagoon]

i can be 1 or 2.

Let's say i is 1.

Then, the other part of the restrictions says that a_i is in A, so a_1 is in A.

But then it doesn't have a condition on a_2.

Similarly, if i is instead 2, it doesn't have a condition on a_1.

It's sort of fuzzy.

It doesn't clearly state whether the restrictions have been satisfied.

[manic bramble]

thinking...

[sharp lagoon]

I guess you could have like (sandwich, 5) and (2, kitchen sink) where either one had to be in A and the other one could be anything, but it doesn't make it clear that that's what is desired.

[manic bramble]

I think I get it

without the forall, then it would only "select" one variable (a_1, a_2, ...) on the left. Like it could be "a_1", or "a_2", but not both. Then, that selected a_i variable, say a_1, could be anything in A. Where all forall "selects" all possible "a_x" of the left side tuple, and then "sees" if that's in A.

[sharp lagoon]

Right.

For all loops over each i value.

If you've done programming so far.

[manic bramble]

yep, I'm a professional c++ programmer. 🙂

[sharp lagoon]

Oh, OK.

[manic bramble]

Ok, I think I'm not confused anymore.

I really appreciate the help. You've been very kind.

[sharp lagoon]

You're welcome.

[manic bramble]

see you around.

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[analog sand]

Stop doing this. Don't guess straight out solutions and give them.

[analog sand]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[primal horizon]

I tried to do it ina a way but it did not work

[analog sand]

I see, can I see your work?

[primal horizon]

I dont know how to approach

[primal horizon]

I just expanded

[pallid sorrel]

thats a pretty reasonable approach

u mightve expanded slightly wrong?

it looks like a lot of terms should cancel

the ugly stuff

[primal horizon]

Is there a way to do it without expanding? maybe

[pallid sorrel]

uhm, i think expansion would be most straightforward

[analog sand]

Take x-2y = t, but you'll still have to expand.

[pallid sorrel]

can you see how some of the stuff will end up cancelling? like, when u expand those brackets you get a x^2 term, and theres a x^2 at the very start

[onyx glen]

(x-2y-1)^2 = -4y(x+y) - x^2

is this anything?

[pallid sorrel]

same with the y^2 on the right, etc

[primal horizon]

I thought we could bring the RHS to the LHS and do some (x+y)^2

here y = x - 2y - 1 (the both y are different)

[tame cradle]

So?

[primal horizon]

I dont think it is possible

[onyx glen]

how did you get here

[primal horizon]

y1 = x - 2y - 1

x1 = x

(x1 +y1)^2

[onyx glen]

y = x - 2y - 1 (the both y are different)
btw if you have to clarify that two instances of the same letter are different then you are definitely doing things wrong

[primal horizon]

mb

[tame cradle]

What if you try subbing x-2y=t

and like converting in terms of t and x

i mean like simplify

[primal horizon]

Need to try that

1 min

[tame cradle]

@primal horizon

Look

(x^2)+(x-2y-1)*2=-4yx-4y^2

@primal horizon

Bro look carefully

question is done

try shifting terms on other side and simplify

[primal horizon]

ye

1 min

[tame cradle]

Bro

[primal horizon]

not getting

[tame cradle]

Look

take terms on rhs to the lhs and rearrange

you will see what i mean to say

[primal horizon]

I dont see it

[tame cradle]

look what did you get after you took all terms on rhs to lhs

after expanding on rhs

[primal horizon]

(x^2) + 4xy + 4y^2 + (x-2y-1)^2

[tame cradle]

now try to look

what do the first three terms form together

[primal horizon]

yes

omg

I see it

x-2y = 1

thx

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[mental tree]

am i starting off this problem right? i feel like it went to smooth and finding the horizontal tangent line of the graph doesn’t seem right to me

[sharp lagoon]

No, the chain rule is done wrong.

You need sin^2(x) -> 2 sin(x) (sin(x))'

[mental tree]

wait shouldn’t it be multiplied by cos?

[sharp lagoon]

(sin(x))' is cos(x).

[mental tree]

since i’d take the derivative of the inside

[sharp lagoon]

' means you still need to take the derivative.

[sharp lagoon]

Looks good.

[mental tree]

then would i multiply that sine and cos

6cos + 2sincos

[sharp lagoon]

Well, I'd factor out the 2 cos(x).

Then you can use the zero product property.

[mental tree]

then i’d set both to 0

[sharp lagoon]

Almost, the cos(x) leaves the first term.

[mental tree]

they both share 2cosx so i’m not taking out the 2cosx? just cosx

sorry that’s a bit hard to understand

like this sorry

[sharp lagoon]

Yes, that's right.

Then factor out the 2 as well.

Then you have a product with 3 factors.

[sharp lagoon]

Looks good.

[mental tree]

now if set both to 0 to find the horizontal tangent line

or is there more that i’m not seeing

[sharp lagoon]

Well, set the whole thing equal to zero to show what you're doing, then do the separation.

[sharp lagoon]

OK, looks good. When is cos(x) = 0?

[mental tree]

pi/2

[sharp lagoon]

Where else?

[mental tree]

3pi/2

[sharp lagoon]

OK, where else?

[mental tree]

could i answer as pi/2+pi k k is an integer

[sharp lagoon]

Yes.

OK, so we have the cos(x) = 0 part done.

What about the sin(x) = -3 part?

[mental tree]

then would the sin be invalid since we’re only between 1 and -1

[sharp lagoon]

Right, so that doesn't have any solutions.

[mental tree]

i do have a question on that

if no given restriction would i still include that at all or whenever its above 1 or below negative 1 it’s a “no solution”

[sharp lagoon]

Well, the only way to get solutions for sin(x) = -3 is to use complex numbers.

You're probably only dealing with real numbers here.

So, you'd just write no solution for that part.

[mental tree]

ok i see

[sharp lagoon]

And then the solutions for 2cos(x)(3 + sin(x)) = 0 are x = k pi + pi/2 where k is any integer.

So, that's the answer to the question.

[mental tree]

ok ok i see now thank you just learned about the chain rule so the beginning had me for a turn

have a good day

[sharp lagoon]

You're welcome.

You too.

[mental tree]

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[worn aurora]

,rotate

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[worn aurora]

2 ig

[analog sand]

Show your work

[worn aurora]

i just drew that line and found it was 54

i havent done much

[analog sand]

What is 54?

Which angle?

[worn aurora]

bda

and bad

[hidden harbor]

why is it 54 tho?

[worn aurora]

isosceles

o my teacher said change it to 72 mb i forgot to say

[analog sand]

wait nm

Yeah i got confused lol the measurements dont match the diag

Yeah ok ur right so far

[worn aurora]

im not surew what to do next tho

@analog sand u got any ideas abt it?

[analog sand]

Yeah sorry was away

Take DBC as y

And take BDC as z

Then you'll see that since opposite angles of a parallelogram are equal

54+z=72+y

Then you can also see by the same property

That BCD is 54

And we know the sum of interior angles of a parallelogram is 360

So 72+y+54+54+z+54 = 180

Two equations, two variables

find y and z

[worn aurora]

how do we know its a parralelogram

[analog sand]

Hm yes opposite sides don't seem parallel, I apologise.

I'm gonna let someone else help you with those lol, bit fuzzy rn.

[worn aurora]

alr algs ill try find an aswer as well

[brisk helm]

is ABC 90?

[strange basalt]

oh

[analog sand]

yuh

[brisk helm]

and is DCB 90 too?

[analog sand]

what?

[worn aurora]

not told unless u found smt?

[analog sand]

wasnt responding to you lol sry

[brisk helm]

dw

[worn aurora]

how did u get 90?

[strange basalt]

?

why is it writen 72 , corrected?

[worn aurora]

here

[strange basalt]

mb

[brisk helm]

iwas askin

btw how old r u rice

[worn aurora]

nah unless ur told it prolly isnt

[brisk helm]

cuz im still forteen

[worn aurora]

same

[brisk helm]

ow

where r u frrom

[worn aurora]

australia

[worn aurora]

u got any ideas im rly stuck 😭

wait i think ifound the answer

[brisk helm]

54?

[worn aurora]

i got 36

[strange basalt]

yeah

i got 36 too

sorry

net

work

omg

my net died

are we back

we so back

[strange basalt]

what process you used , your contruction was not needed that much here

[worn aurora]

yeah, i made dbc as y so bdc was (180-y)/2 and then acb would be (108-y)/2 and then (108-y)/2+x=(180-y)/2 and x=36 cuz the y cancels out

[strange basalt]

ahhh ic

i just wrote a lot of equations and looked into it to find 2 which worked

💀

[strange basalt]

thats good

[worn aurora]

which equations did u find?

[strange basalt]

a here i used for showing same side , the double bar clutters my brain

[worn aurora]

wait why does 2a+ o thing equal 180 and squigly line + a + 2x-180?

[strange basalt]

lemme rewrite

squigly line = del

so

del + phi + x = 180 from triangle pdc

thumbsup if you following

[worn aurora]

which one is phi

[strange basalt]

this is phi

[worn aurora]

alr

yeah

[strange basalt]

right

then

phi = theta +x

isocelles triangle

[worn aurora]

wihich triangle??

[strange basalt]

bdc

bd=bc

[worn aurora]

ohh yeah

[strange basalt]

so
from these two i can say theta+del + 2x = 180

now consider triangle apb

theta + 72 +del = 180

[worn aurora]

yeah

[strange basalt]

theta + del = 180-72 = 108

[strange basalt]

( theta + del )+2x = 180
108 +2x = 180
x=36

[worn aurora]

js wondering what abt 2 theta + phi =180 at the top there?

[strange basalt]

mainly triangle property used and one vertical opposite angle on apb = dpc = del

[strange basalt]

i told you i just wrote whatever i saw , and tried to work out x , that one was not needed

[worn aurora]

ye but can u explain how it equals 180? just for like general knowledge next time or smt

[strange basalt]

sure lemme check

Oh wait my bad that was from the old attempt when I tried your construction one

Different labelling here

Mb

Too many lines so I decided to restart lmfao mb

[analog sand]

,rcw

[worn aurora]

O lol alg what’s the root 2a from tho?

[worn aurora]

@strange basalt ?

[strange basalt]

Oh

That was a thought check

If abc was a right angled triangle

Its my rough space 💔😔

[worn aurora]

O alg I thought I was like missing something

[strange basalt]

Yeah my bad 😔

Just consider the below diagram this one is all weird

I was high

[worn aurora]

Nah dw abt it tysm for helping as well

[strange basalt]

You got it yourself first , I just showed you a diff aproach

[worn aurora]

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[worn aurora]

Thxx

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[torpid plover]

hey

[torpid plover]

idk how to do this

like do I find b, a, or a + c?

[analog sand]

For what?

To find the angle you just need to take the dot product and divide by product of mods

[torpid plover]

to find the angle between i and the acceleration

[torpid plover]

i dont understand this

or why it works

[analog sand]

It's just a formula

[torpid plover]

i dont think a formula (like that) is needed

its just finding an angle in a triangle

right?

[analog sand]

Well you can just do tan^-1 (2/3) then

[torpid plover]

so im finding angle A

[analog sand]

yes

That's the x axis

[outer lotus]

i is just a positive x-axis direction. You need to find the angle a

[analog sand]

The unit vector i lies in that direction

[torpid plover]

ok makes sense

what about if it was with j

which angle would i find

[outer lotus]

Then c

[analog sand]

Then find a+90

[analog sand]

thats with -j

[outer lotus]

Right

Sorry

[analog sand]

since j is along the positive y axis

[torpid plover]

why

wait

this is how the vector would look on a graph right

[analog sand]

yes

[torpid plover]

so why dont we find this angle

[analog sand]

Thats the angle with -j

-j lies along the negative y axis

with +j you need to find with the positive y axis

[torpid plover]

hmm

how do we find the angle with +j then

[analog sand]

90+ angle with the positive x axis

[torpid plover]

why does this make it go into the +j

[analog sand]

Phi is with +j, theta is with -j

[torpid plover]

makes sense

niceee

is this the same with +i and -i

[analog sand]

yuh

[torpid plover]

cool cool

thanks

[untold oasis]

can anyone help me understand the difference between finding an angle of the longest length within a cuboid, and finding the angle which the diagonal makes with the length edge, width edge, base length, height etc?

[torpid plover]

.close

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[analog sand]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[torpid plover]

post now @untold oasis

.close

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[autumn raptor]

Can sb explain why the derivative of fx is C

[thick hedge]

the derivative of $\sin^{-1}(x)= \frac{1}{\sqrt{1-x^2}}$

[glossy valve]

wai

[autumn raptor]

Could u like explain?

I forgot how to to do it

Oh wait

Is that the formula?

[sudden condor]

y=arcsin(x)
sin(y)=x/1
y'(cosy)=1
cosy=sqrt(1-x^2)/1 [using a right angle triangle with side x, hyp 1 and angle y]
y' sqrt(1-x^2)=1
y'=1/sqrt(1-x^2)

[sudden condor]

yea

[autumn raptor]

Oh I got it thx

[autumn raptor]

Thank u

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[thick hedge]

So I'm trying to prove every convergent sequence is cauchy convergent.
\
Let $(a_n)$ be convergent to $L$. So $\forall \varepsilon >0, \exists N \in \N : n≥N \implies \abs{a_n-L}<\varepsilon$.Let $m≥N$. so we have $\abs{a_m-L}<\varepsilon$
\
We then have $-\varepsilon+L< a_n < \varepsilon +L$ and $-L-\varepsilon<-a_m<\varepsilon-L$.
\
We thus have on adding these two inequalities, $-2\varepsilon<-\varepsilon<a_n-a_m<\varepsilon<2\varepsilon \implies \forall \varepsilon>0 \exists N \in \N : n,m≥n \implies \abs{a_n-a_m}< \varepsilon$

[glossy valve]

wai

[thick hedge]

Does this proof suffice

[leaden ermine]

If you add the inequalities you get -2ε < a_n - a_m < 2ε but I don't understand how you derive -ε < a_n - a_m < ε.

I think choosing ε/2 might help

[full forum]

@thick hedge Has your question been resolved?

[thick hedge]

Epsilon is positive

So -2e<-e

And e<2e

[leaden ermine]

|x| < 4 doesn't always imply |x| < 2

[thick hedge]

I know

But here it does i think

Because eps is positive

Ah

I see the problem

-e> a_n-a_m is possible

Got it

Thanks

Sorry

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[urban olive]

a and b are both true right

[urban olive]

im not tripping??

[nimble crane]

[urban olive]

thank yer

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[nimble crane]

wait

.reopen

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✅

[nimble crane]

sorry, b isn't necessarily true

the inclusion should be the other way around

[urban olive]

what

[nimble crane]

f(f^-1(B)) \subseteq B

[urban olive]

but foesnt f(f^-1(B) give Y?

[nimble crane]

what is Y?

oh, I see

why would it give Y?

[urban olive]

idk..

cuz f-1(b) gives x and then f(x) gives y

[nimble crane]

it can only equal Y if B = Y and f is surjective

[urban olive]

hmm

so why is a true

[nimble crane]

in general, f(f^{-1}(B)) \subseteq B with equality only when f is surjective

[nimble crane]

well, you can prove it directly if you want to

if x is in A, then f(p) is in the image of f, {f(x) : x in A} = f(A). but by the definition of preimage, we have f^-1(f(A)) = {x in X : f(x) in f(A)}. since f(p) is in f(A), we have x in f^-1(f(A)), so A is a subset of f^-1(f(A))

that would be the argument

[urban olive]

oh lord

thank you

[rapid rain]

uhm

f^-1(f(A)) = {x in X : f(x) in f(A)}.

?

[nimble crane]

oops, should be X

I don't think the argument fails though

[rapid rain]

it doesn't

[nimble crane]

that was just me being dumb

it was because the last time I proved this, I did it when the domain was A

not X

so my brain malfunctioned

[urban olive]

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[lyric narwhal]

Need help with question 10
i understand its a stoke theorem question
but how do i find the normal
i saw u can take the gradient and then a different method using the parametrisation

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@lyric narwhal Has your question been resolved?

[lyric narwhal]

<@&286206848099549185>

[waxen flax]

I'm not sure I understand
you're saying it you want to do with without Stokes' theorem?
Or you are saying “I know how to do it, I just wonder what the normal is”?

[lyric narwhal]

im guessing that we would be using stokese theorem, but i dont know how to find the normals, so i can use it

[waxen flax]

Do you know what Stokes' theorem says?

[lyric narwhal]

that the double integral of a closed positivly oriented surface is equal to the integral of its boundary?

[waxen flax]

Yes but which integrals specifically, can you write down the formula?

[lyric narwhal]

is this what u meant?

[waxen flax]

So if you want to compute what's on the left by using the formula, you're moving to the integral on the right

But you don't need the normal to D to compute the integral on the right

[lyric narwhal]

i just realise the way i was doing it was without stokes theorem, okay so i dont need the normal for the integral of the boundary

for my question do i split it into two parts, the cyclinder part and the half sphere

for the sphere should i use that boundary idea with stokes theorem and then for the cylinder just compute it normally?

[waxen flax]

Why do you want to split it?
You have 1 surface, regardless of how it's built, which has a simple boundary

You might want to draw a picture

[lyric narwhal]

am i wrong to say its a cyclinder connected to a half sphere centred around the x axid?

[waxen flax]

It is, now forget that it's originally described by the union of 2 surfaces, can you tell me what the boundary is?

[lyric narwhal]

i honestly dk what its boundary is

[waxen flax]

that's why a picture would help

[lyric narwhal]

okay

can its boundary be a surface or is that wrong

[waxen flax]

When you take the boundary you're decreasing the dimension
the boundary of a solid is a surface
the boundary of a surface is a curve

[waxen flax]

What would you say the boundary is?

[lyric narwhal]

because what u said above where the boundary of a surface is a curve, i want to say a curve but i dont know how to describe that curve

[waxen flax]

look at it this way
take any point “in the middle” of this surface and draw a little circle around around it all lying in the surface, can you do it?

[lyric narwhal]

sorry im so lost, not sure i understand what u mean

[waxen flax]

like this (hold on)

[lyric narwhal]

by middle do u mean miuddle of the volume?

[waxen flax]

The subject is the surface and a surface has no volume

pardon my mouse art

I picked that point in green and I was able to draw a little circle around it (imagine the circle is curved so that it really lies on the cylinder and sphere part)

Do you agree that this is possible for that point?

[lyric narwhal]

yes i agree

[waxen flax]

Do you think it is possible for every point?

[lyric narwhal]

yes apart from the boundary

[waxen flax]

then you can identify the boundary then

what part is it?

[lyric narwhal]

the edge of the cyclinder

[waxen flax]

great, we found our boundary

[lyric narwhal]

u jjust u nclocked a new part of the brain for me

very thankful

[waxen flax]

it is the circle on the left side of the cylinder

what's its equation?

[lyric narwhal]

its equation is y^2 + z^2 = 4, for x = 1

[waxen flax]

Now you just have to compute the work of the field through this circle

[lyric narwhal]

the field is quite complex is there an easy way to do this

do i want to switch to polar

[waxen flax]

you might want to first write down the integral

[lyric narwhal]

I normally wouldn’t use notation like that for the bounds, it ,like this

[waxen flax]

also x=1

Then what does that integral look like?

[waxen flax]

Maybe it looks nice in polar coordinate
(btw did you compute the curl of F? Maybe that also it looks nice, but idk yet)

[lyric narwhal]

the curl of F is 3*(a-1)x^2

[waxen flax]

so for x=1 it's constant

that's very nice

[lyric narwhal]

good point

[waxen flax]

you might want to specify the coordinate

(0,0, 3(a-1)x²) ?

[lyric narwhal]

sorry im lowkey consufed again

i get that the curl of F is (0,0,3*(a-1)x^2)

i thought we were using stokes theorem which uses the integral of the field dotted with dc

and i thought we can only use the x = 1 when we use the stokes thorem, because x isnt constant for the curl of F integral

[waxen flax]

that is true, what I am suggesting can be formulated in two different ways

• Way 1) Start with Σ, use Stokes' theorem to go on ∂Σ = γ, note that γ lives on the 2-d plane at x=1 so we can think of γ as a planer curve and completely forget about the x coordinate, then once we are on the plane we can use Stokes' theorem in 2-d (or Gauss-Green formula, whatever you call them) to move to the integral on the filled up circle
• Way 2) Stokes' theorem tells us that the flux through Σ is the same as the work over ∂Σ, but ∂Σ is also ∂S where S is the filled up circle in x=1, so we can use Stokes' theorem again to move from the work over ∂S to the flux over S, and here the curl is costant

These are completely equivalent and basically the same thing

[lyric narwhal]

thank you i think that makes sense\

[waxen flax]

You might want to write the computations down

I can see how one would get confused without some tangible calculations

[lyric narwhal]

way 2 makes more sense to me
but i somehow got lost, my bad

so i can compute the flux over the surface to just be the flux over a cirlce, y^2 + z^2 = 4, where x = 1
then isnt my normal (-1,0,0) and the curl is (0,0,3a-3) so i get 0

[waxen flax]

oh yeah lol, that's quicker than I thought it would

[lyric narwhal]

so its just equal to 0?

[waxen flax]

well yeah, unless you got some computations wrong

[lyric narwhal]

thank you, so theres two ways to use stokes

go from a flux integral to the work integral might be simpler

or go from flux to work to flux where the shape is easier and then it simplfys

[full forum]

@lyric narwhal Has your question been resolved?

[waxen flax]

Note that this can also be paraphrased using the divergence theorem
if you add the circle S at x=1 to Σ you get a closed surface S∪Σ, so the flux through it is the divergence on the inside, but the divergence of a curl is 0
conclusion: flux through Σ = flux through Σ∪S - flux through S = 0 - flux through S
(+ and - have to be adjusted for orientations)
all these theorems are really just one big theorem with a lot of formulations

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[onyx haven]

No idea how to approach this tried something with AM-GM but it’s ab1 and not ab

[echo marsh]

isnt this AGM

[green merlin]

variables are shifted

[onyx haven]

?

<@&286206848099549185>

[full forum]

@onyx haven Has your question been resolved?

[uncut skiff]

This is a very arbitrary and weird substitution but put b=acosy

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@onyx haven Has your question been resolved?

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[torn jolt]

What are these type of matrixes called ?

[urban jacinth]

Where is it given that b is bounded?

[torn jolt]

that person closed the thread

[urban jacinth]

Yes my bad

[torn jolt]

<@&286206848099549185>

[shy bane]

can you share the whole question?

[torn jolt]

sure

[torn jolt]

"Find the determinant of :"

just in place of A=(matrix) , it should be Delta=

[urban jacinth]

Are a1, a2,a3 a4 in ap gp hp or just a random sequence

[shy bane]

do you have your formula list for this?

because this is not any type of matrix

[urban jacinth]

Exactly

[glossy zodiac]

if a1= a2 =a3

and so on

[torn jolt]

They just variables

[glossy zodiac]

its like identity matrix

[urban jacinth]

Id say square matrix but that is too trivial

[shy bane]

umm no?

[urban jacinth]

No????

[glossy zodiac]

like a1=a2=a3=....

like that

[shy bane]

it will be identity matrix when a1=a2=a3=.....=an = 1

[glossy zodiac]

no ig

[urban jacinth]

Still no?

[glossy zodiac]

yeh

[shy bane]

1

[glossy zodiac]

i understood

my bad

[urban jacinth]

Bro what

[glossy zodiac]

yeh

im wrong

it should be 0

srry

[urban jacinth]

It can never be identity

[torn jolt]

the solution is basically a summation to approximate the determinant

[shy bane]

oh yea

sorry

[shy bane]

exactly

[glossy zodiac]

srry,im dumb

[shy bane]

do you know any other formulas to find determinant

or are you givem

[urban jacinth]

Ohk

[torn jolt]

the solution of this is very long , but its an approx. formula for the determinant

its basically

a_1 product (i=2 to n) (a_2-1) + sum (k=2 to n) * product (i=1 1=/k to n ) (a_1-1)

[shy bane]

umm what

[torn jolt]

this is supposed to be the solution

[torn jolt]

hold on

[shy bane]

just saw your tag TT i dont think i can help you sorry

[torn jolt]

what is TT ?

[shy bane]

its crying emoji lol

[torn jolt]

Yeah I am cooked

[urban jacinth]

Oh no is this linear algebra theory

[shy bane]

tag helpers someone might come

row reduction?

[sand escarp]

are you trying to find the determinant?

of A?

[torn jolt]

YES

this is the solution

[sand escarp]

laplace expansion?

[uncut skiff]

it's given a>b>0 and you can prove that the sequence does converge

[sand escarp]

guys find a different help channel

[uncut skiff]

mb

[sand escarp]

this ones miyamotos

[torn jolt]

idk, my teacher started expanding it by a column or a row , is that what is called ?

[sand escarp]

Yes

[torn jolt]

here is another "easier one"

[sand escarp]

horrific

but yeah row reduction also works

[torn jolt]

ok laplace reduction it is , thanks

.close

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[shy bane]

i thought one of you was gonna explain laplace reduc lol

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[hot kernel]

I have a question

[hot kernel]

What's wrong in this solution

[indigo ore]

You have 2 equations for two variables

Find them individually and then solve

It would be easier

[hot kernel]

Look is there any mistake?

[indigo ore]

Btw your mistake is in (a-b)^2 expansion

you wrote a^2 - b^2 - 2ab

[analog sand]

(alpha-beta)^2 expanded wrong

[hot kernel]

Whats the difference?

[indigo ore]

💀

Its a^2 + b^2 - 2ab

Use this

[hot kernel]

I found the value of α²-β² = 192

Then I put it in the equation like it is shown

[indigo ore]

Do you know how to solve a system of linear equations

The answer will be way quicker if you use this

[hot kernel]

Tell me what it is?

[indigo ore]

add the two equation given in the question

LHS + LHS, RHS + RHS

What did you get

[hot kernel]

I am wrong

My expansion is wrong

[indigo ore]

What

[hot kernel]

I am now crystal clear

[indigo ore]

Alr

Nice

[hot kernel]

Look what I have done

[plain oasis]

meow meow

[hot kernel]

My brain is dumb

.close

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[rare pine]

translation: 'Let triangle ABC have I, O, I_A as its incircle, circumcenter, and excircle opposite A respectively. Let D be a point on BC such that DA = DI_A. Let the radius of its incenter be 8, and the radius of the circumcircle be 18. Suppose that IO// BC. Then whats the length of AD?

[rare pine]

i geniunely have no idea

sigh

wait wait

A,I,I_A are colinear right>>

<@&286206848099549185>

[grizzled otter]

how can you prove it 😮

[rare pine]

is that not by uh

construction

[grizzled otter]

could u elaborate

[rare pine]

incenter is constructed by the intersection of angle bisectors

so is excenter

also more accurate diagram where IO // BC

i have a strange feeling A,O,D collinear from the img..

[grizzled otter]

yea right

a o and d?

[rare pine]

yea

probably just a coincidence tho

[grizzled otter]

yeah

what exactly is the q

[rare pine]

find length of AD

if incenter radius is 8, circumcircle radius 18

and D exists such that AD = DI_A

[rare pine]

oh duh

im so dumb

they are collinear

[grizzled otter]

why dont you try taking 1 vertex as origin

[rare pine]

im NOT coord bashing this, sorry

[grizzled otter]

it would make the calc simpler

[rare pine]

i just dont want to

[grizzled otter]

oh okay

[rare pine]

i wanna practice my geo thingy

[grizzled otter]

solely based on geometry?

[rare pine]

yea

[grizzled otter]

that would be a bit tough for me 😮

i was thinking of using the formula io^2 = R^2-2Rr after coords

[rare pine]

ohh

geometry gods please bestow upon me thy knowledge 🙏

AD = 18AC/AF = 18AB/AE

thats all ive got vro

[full forum]

@rare pine Has your question been resolved?

[rare pine]

anyone..?

[tame cradle]

Yo

[rare pine]

hii

[rare pine]

^

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[lunar grove]

CAn u help with part b

If i sub X=the x Co ord of M and Y = y cord of M

Into eq bottom

Is that allowed

[hasty sail]

<@&286206848099549185>