help-39

[pearl pond]

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[gilded hollow]

A student must answer seven out of eleven questions on an exam. In how many ways can the student answer the questions if at least four of the first six questions on the exam must be answered

[toxic fractal]

divide it in cases.
a) ways when he answers exactly 4 of the first six.
b) ways when he answers exactly 5.
c) ways when he answers exactly 6.

[gilded hollow]

$\binom{6}{4} \binom{5}{3} + \binom{6}{5} \binom{5}{2} + \binom{6}{6} \binom{5}{1}$

[jolly parrot]

Merineth

[gilded hollow]

This is the answer but

I have literally no idea what i'm even reading

First it states that he must answer 7/11 questions? Then it says he must answer at least four of the first six questions?

[toxic fractal]

okay, so i'm gonna help you with case (a)

you have the first 6 questions, and you answer 4 of them. So you have 6C4 ways to answer those questions, correct?

[gilded hollow]

Yes

[toxic fractal]

okay. So you now need to answer 3 additional questions. Since you're not answering any of the first six, you're left with the other 5, so for each of those 6C4 of the first block, you got 5C3 ways to answer the second block of questions, correct?

[gilded hollow]

Yeah

[toxic fractal]

that would be your first term. Is this enough information for you to do (b) and (c)?

[gilded hollow]

I think so

So for b) is when he answers 5 of the first 6 questions

6C5

[vestal tapir]

that's not realistic

that the student wants to not do to well

[toxic fractal]

hm?

what do you mean?

[vestal tapir]

this way you;re not counting answering 9 questions right

maybe that's the intetion, just it's not realistic

[toxic fractal]

there's many kinds of exams where you just get to answer a determined amount of questions, not all of them

[vestal tapir]

oh it's not about getting it right, i see

yes i see it now

[gilded hollow]

I get at most 20 out fo 50 right on every math exam so far :) so it's realistic :(

[gilded hollow]

But yeah i see what you mean when it comes to this

The sum of all the three outcomes where he answers 4, 5 or 6 of the first 6 questions

[toxic fractal]

that would be the number of ways to answer at least 4, yes

[gilded hollow]

It's just crazy to me that i'm meant to solve these myself. Clearly something wrong with me

[toxic fractal]

combinatorics problems are like 95% reading extremely slowly and 5% knowing the few formulas

[gilded hollow]

I wish it was that easy

I can't solve a single one lol

"How many 4 digit numbers can you create of the numbers 1,2,3,4,5,5,6,6

You say it's about reading it slow but

It really doesn't matter in my case

[toxic fractal]

are the 5 and 6 supposed to be repeated?

[gilded hollow]

Yes

[toxic fractal]

okay, divide it in 4 cases

[gilded hollow]

My thinking is

1,2,3,4,5,6

[toxic fractal]

(a) you dont repeat any number
(b) you repeat the 5
(c) you repeat the 6
(d) you repeat both

can you solve (a)?

[gilded hollow]

6P4

should be a)

[toxic fractal]

order matters

[gilded hollow]

typo

But yeah 6P4

[toxic fractal]

can you solve (b)?

[gilded hollow]

b) should be 7P4

[toxic fractal]

how do you get that

[gilded hollow]

1,2,3,4,5,5,6 i repeat 5 once which means we have total of 7 but we still choose 4. 7P4

actually nvm

[toxic fractal]

nPk and nCk have more requirements than having n and taking k

[gilded hollow]

it should be

/ 2! ?

[toxic fractal]

seems to me like you need to pay more attention to the details of the definition

i'm gonna go and detail (b) a bit more

we're saying that we're taking both 5's

you have a 4-digit number

in how many different positions can the 5's be?

[gilded hollow]

4

[toxic fractal]

how do you get that

[gilded hollow]

_ _ _ _

i have 4 spots to place them?

[toxic fractal]

that would be if you had only one "5" to place

but you have 2

[gilded hollow]

Well the first one would be 4C1 and then 3C1

so 4 x 3

[toxic fractal]

that is not correct

you have 4 slots

you have 2 coins

the coins are equal

how many ways can you distribute the coins in the slots, if you can only put one coin per slot?

[gilded hollow]

1P4 then 1P3?

[toxic fractal]

permutations require the items to be distinct

[gilded hollow]

1C4 then 1C3?

[toxic fractal]

there's no "then" in combinatorics

[gilded hollow]

1C4 * 1C3

[toxic fractal]

55__
5_5_
5__5
_55_
_5_5
__55

[gilded hollow]

Oh

[toxic fractal]

4C2

[gilded hollow]

Ok

[toxic fractal]

you should revise the definitions

you're overlooking important details that make you unable to distinguish when to use each correctly

[gilded hollow]

It just doesn't make sense to me

I've spent nearly a year trying to learn this. I just dont get it

[toxic fractal]

Permutations of n elements:
Ways to order n distinct elements.
Pn = n!

Permutations of n elements, we take only k:
Ways to order the first k out of n distinct elements.
nPk = n! / (n-k)!
n! is the ways to order all of them.
(n-k)! is the ways to order the last elements that we dont care about

[gilded hollow]

I'm aware the permutations is used to rearrange items that are distinct.
Combinations is used to select/choose items that are indistinct

[toxic fractal]

Combinations nCk:
Ways to select k items, among n distinct items, ignoring the order.
nCk = n! / [ k! (n-k)! ]

[gilded hollow]

I just dont understand how you could so easily determine that you are supposed to divide it into a) b) and c)

[toxic fractal]

see that nCk = nPk / Pk

or in other words, the ways to order the first k elements, and then ignoring the order

[toxic fractal]

the more you do, the more clear it becomes

the thing is, i assume you were given 6 different formulas

[gilded hollow]

We aren't given any formulas

[toxic fractal]

permutations with and without repetition, variations with and without repetition, and combinations with and without repetition

(names might vary in your place)

the three formulas i've given you would be permutations, variations and combinations without repetition

[gilded hollow]

On our exam we aren't given any formulas that regard permutations. He expects ut to know them all by heart

[toxic fractal]

i'm not talking about the the exam. I'm talking about your lectures.

[gilded hollow]

We dont have any lectures, we only have the book for reference

[toxic fractal]

the book surely has the formulas for combinatorics explained

you need to know where they come from. Knowing the formula by heart without knowing how to obtain it yourself is almost completely useless

and if you know how to obtain them you dont need to memorize them, cuz you can get them back on the spot

[gilded hollow]

Oh i see so i have to read the entire book + memeorize each page with each theorem to be able to get htem back on the spot meaning i wont need to memorize them?

[cobalt monolith]

You need to understand the principles... Not memorize. Don't be so dramatic, you are not the first student taking combinatorics class

[gilded hollow]

Dramatic?

[cobalt monolith]

Yes

[gilded hollow]

It's a sincere question?

[toxic fractal]

well, you usually do need to read once the full text of the material you're covering

[gilded hollow]

Considering there are people here who are able to answer every single question, means i have to be doing something wrong to not be able to solve any, no? How is it dramatic to ask a sincere question of how to actually be able to solve the questions at hand?

[toxic fractal]

we are able to answer them because we have studied it back in the day, same as you will at some point

[cobalt monolith]

This sounds cynical to me

[gilded hollow]

It wasn't meant as cynical, sorry

[toxic fractal]

but you wont magically learn to do the problems if you just memorize formulas

you NEED to understand where formulas come from and WHEN they can be used

[gilded hollow]

I see. It's just so hard to learn them. I'm not really that good at reading university textbooks

[cobalt monolith]

You should try to find some lecture videos online or something

Shouldn't be hard to find...

[gilded hollow]

Yeah i've watched the entire series of Khan academy about Combinatorics and the organic chemistry tutor. And i understand everything they say. I just can't solve problems on my own

[cobalt monolith]

OK, continue solving problems, we are here to help 🙂

[pine jay]

The sort of reasonable memorization to have here is to be able to recall definitions when needed, however understanding these are as important but the trouble is thinking you can re-derive most definitions

[gilded hollow]

I see

[pine jay]

Most proofs are just by definition

[cobalt monolith]

Let's get back to concrete questions...

[cobalt monolith]

Here you should understand that when you multiply you treat the 5's as distinct

[gilded hollow]

Yeah

[cobalt monolith]

Here you need to try and solve it with the tools you are familiar with and see where you get stuck. Then try to come up with a solution

[toxic fractal]

that's why i separate it in cases, for example. Each case corresponds to one of the elemental definitions

[cobalt monolith]

One way it to break it into distinct cases and use the "Sum rule"

[toxic fractal]

^important here, the cases must never be able to coexist

[cobalt monolith]

all the basic "formulas" are simply based on the "sum rule" and the "product rule"

you need to understand these principles well

[gilded hollow]

And that applies only to the sum rule?

[toxic fractal]

yes, because if they can coexist, you cant just add them

[gilded hollow]

and if they do exist together it's the product rule?

[toxic fractal]

no

[cobalt monolith]

Product rule is for "and" and sum rule is for "or"

The "or" can be simply summed if the cases do not overlap

How many ways can you do A or B?

#(ways to do A) + #( ways to do B) - #(ways to do A and B)

if A and B is nothing

then you simply add...

[gilded hollow]

1,2,3,4,5,5,6,6

If i'm going to make a 4 digit combination how many possibilites are there
(a) No repeating numbers
(b) you repeat the 5
(c) you repeat the 6
(d) you repeat both

a) If no reapting numbers we jsut remove 5 and 6 from the pool meaning 1,2,3,4,5,6 all are distrinct so it's permutation. 6P4
b) If i repeat the 5 _ _ _ _ like you mentioned they are the same so it's combinations. 4C2 but then we have to add all the other numbers? 1,2,3,4,6 Shouldn't it be 5C2 ?

[gilded hollow]

I remember this specifically being in the book but it's crazy to me how you are able to remember it. Feels like it's mandatory to be able to memoerize formulas like that to be able to solve it?

[cobalt monolith]

You should simply think of a venn diagram

In the last part, with the numbers 1,2,3,4,6 the order is important since the numbers are distinct

So 5C2 is not the right answer

[cobalt monolith]

You should internalize the logic behind it. Not memorize

the basic building blocks (like the sum rule, and the product rule) should be understood really well... And everything follows from that

[toxic fractal]

if you repeat the 5's, you start by placing the 5's on two of the slots. I put the bruteforce method up there, where you could see 6 distinct ways to do it (4C2)

now, for each of those six positions, you'd still have to put the remaining numbers on the two slots remaining

[gilded hollow]

Trying to draw one but i dont even know what supposed to be included haha

[toxic fractal]

now since we explicitly say that we do not repeat the 6, you're left with five options: 1, 2, 3, 4, 6, and two slots, where the order is important

[gilded hollow]

Yeah that makes sense

So we have 2 spots remaining with 5 different numbers

[toxic fractal]

which means that for those 6 cases of the positions of the 5's, you got 5P2 ways to put the other two numbers

so (b) would be 4C2*5P2

[gilded hollow]

Yepp that makes sense

(c) should be the same method as (b).
4C2 for placing the two 6s. Then we have 5P2 left
4C2*5P2

[toxic fractal]

now try (d)

[gilded hollow]

well that essentially means we are placing 4 digits 5,5,6,6into 4 spots. This should be 4*3*2*1 possible locations?

[cobalt monolith]

again, you why did you multiply?

you have elements that are the same

[gilded hollow]

My reasoning for multiplying is _ _ _ _ i have 4 spots. and 4 digits 5,5,6,6 first i take the 5 and i have 4 spots to place it, then i take the other 5 and i have 3 spots to place it ... all the way to the last digit

[cobalt monolith]

yes, so you count, for instance, 5 5 6 6 multiple times

for instance, you place the first 5 in the first place then the second 5 in the second place and so on

you would get the same if you place the first 5 in the second place and the second 5 in the first place

so you can't do this.

[gilded hollow]

I see hmm

_ _ _ _ 5 5 6 6

[cobalt monolith]

You already did something similar in the previous cases

how did you treat the 5's?

in case (b)

[gilded hollow]

I placed them and then the remaining options?

[cobalt monolith]

...

So what do you get...?

[gilded hollow]

In past questions i solved double letters with dividing with them
4! / 2!2!

[cobalt monolith]

This works

[gilded hollow]

Oh it works?

[cobalt monolith]

But you need to understand what you are doing

the simple way to do it is to choose the spots for the 5's

and the remaining spots are for the 6's...

so it's simply 4C2.

or 4C2 * 1 (since there is one way to place the 6's after you place the 5's)

[gilded hollow]

is it 4C2 because we have 4 spots but only 2 distinct choices?

[cobalt monolith]

4 spots, and we want to choose 2 spots to place the fives

So we choose 2 out of 4

which equals 4C2

[gilded hollow]

4C2 is only for the 5s?

[cobalt monolith]

yes

but once you place the 5s, you have one option to place the 6s

So its 4C2 overall

[gilded hollow]

I thought it was both for 5 and 6

I had a problem where i had algebra and the solution was 7! / 2!

[cobalt monolith]

it is, as i said...

you choose the spots for the 5s say 5 X 5 X

(I don't know how to do underscores lol)

now you place the 6s, but there is one way to place the 6s once you place the 5s

5 6 5 6

[toxic fractal]

you have four slots.
You have two 5's, and two 6's.
You start positioning the 5's, which can be placed in any of the 4C2 = 4!/2!2!=6 ways.
Now you got two slots remaining, for two 6's.
Since they are identical, there's only one way to place them

[gilded hollow]

Yeah okay i see what you mean

so (d) would only be 4C2

[toxic fractal]

because at that point, you were asked to reorder all the letters.
Assuming the letters were distinct, you got 7 letters.
So you have 7! ways to order them.
Since the "a" is repeated, it doesnt matter where you place each of them, so you divide by the ways to order the two "a"s that are not, in fact, distinct, 2!

[gilded hollow]

6P4 + 4C2*5P2 + 4C2*5P2 + 4C2 = 360 + 6*20 + 6*20 + 6 = 360 + 120 + 120 + 6 = 606?

hmm yes it's 606 according to the book also

I understand when you explain i jsut can't create it myself

Or solve it rather

Dead inside I’ll pick it up after a break, thanks tho

.close

[pearl pond]

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[cloud oyster]

Hello I need help in this problem,not in the problem itself is just that I need to calculate the area of the side of the house, I thought it was just multiply 35 by 10 but it is not and I do not really don’t know what to do

[pearl pond]

@cloud oyster Has your question been resolved?

[cloud oyster]

<@&286206848099549185>

[pearl pond]

@cloud oyster Has your question been resolved?

[pearl pond]

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[void storm]

Hello i need help with determining the traversal of given binary trees

[void storm]

what is going to be the inorder and postorder traversal of the tree

what i know so far:

for inorder its gonna start at the top first and traverse all the way left down to 4 8 and then go back up to '-' and down right to '%' the on the left to 7 then back up to '%' and then down to 2

and then i think its gonna go to the root node? and then im not sure how

[heavy snow]

for inorder you can just think of it as left to right

for postorder you follow the perimeter in the clockwise direction

[void storm]

so im confused what happens when it goes back to the root node after its done with the left side of the tree

for inorder^

[heavy snow]

starts with the leftmost node in the right subtree of the root

[void storm]

so 1 8?

[heavy snow]

yep

[void storm]

then it goes to - ?

[heavy snow]

try tracing the recursive algorithm yourself

[void storm]

hmm i was watching a video

[heavy snow]

inorder(node):
    inorder(node.left)
    payload()
    inorder(node.right)

payload() is just writing down the node data in this case

[void storm]

i see

let me try doing that

[pearl pond]

@void storm Has your question been resolved?

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[daring valley]

Connor has a mass of 76.0kg and is competing in the state diving championship. He leaves the springboard from a height of 3.00m above the water with a speed of 5.94m/s in the upward direction. Use Work-Kinetic Energy theorem to solve the following quesiton.
a. Determine Connor's speed when he strikes the water
b. Connor's body plunges to a depth of 2.15m below the water before stopping. Determine the average force of water resistance experienced by his body

[daring valley]

so for a. i figured out that his speed is 9.70m/s (same as answer key)

but for b. my answer is rounded to 1660N when the answer is 2410N

Using the work-kinetic energy theorem, WDtotal = change in KE

so force x displacement = 1/2mv^2

1/2mv^2 = 1/2 * 76.0 * 9.70^2 = 3575.42J

force = 3575.42/2.15 = 1660N

[pearl pond]

@daring valley Has your question been resolved?

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[gilded hollow]

$x_1+x_2 = 6$ (0.6)(1.5)(2.4)(3.3)(4.2)(5.1)(6,) why doesn't the formula for stars and bars work? $\binom{n+k-1}{k}$

[jolly parrot]

Merineth

[cursive wraith]

here k being the bottom number is the sum, 6
n is the number of variables, 2

7C6 = 7

[gilded hollow]

n = number of numbers 6
k = x_1 and x_2 = 2

?

[cursive wraith]

no... you got it wrong

if n is your sum

then the formula is $\binom{n+k-1}{n}$

[jolly parrot]

rafilou2003

[gilded hollow]

Hmm, that formula isn't listed in the book. How do they differ?

[cursive wraith]

$\binom{\text{total sum}+\text{number of variables}-1}{\text{total sum}}$

[jolly parrot]

rafilou2003

[cursive wraith]

it's exactly the formula you got in your book

except you probably mixed up k and n

[cursive wraith]

if you're using this formula

then it's $x_1 + .... + x_n = k$

[jolly parrot]

rafilou2003

[gilded hollow]

$\binom{k+n-1}{k}$ This is the one listen in my book

[jolly parrot]

Merineth

[cursive wraith]

ok

[cursive wraith]

so it's for this

[cursive wraith]

if you're using this formula

[cursive wraith]

then it gotta match with this

x_1 + x_2 = 6

k = 6

n=2

[gilded hollow]

k = size
n = total objects

says the book

[cursive wraith]

yep

how many "x" are there?

[gilded hollow]

4

[cursive wraith]

huh?

[cursive wraith]

how many "x_..." appear in your sum?

[gilded hollow]

2

[cursive wraith]

ok

so how many total objects?

2

[gilded hollow]

yes

[cursive wraith]

n = 2

k is the sum or "size", it's 6

k+n - 1 = 7

k = 6

7C6 = 7

7 ways

[gilded hollow]

i was under the impression that k are how many boxes there are and n how many stars?

[cursive wraith]

no, you got them reversed, k is the number of stars

[gilded hollow]

oh...

[cursive wraith]

could you send a picture of where it's mentioned in your book?

we could clear things up

[gilded hollow]

but for normal combinations for example 5C3 $\binom{n!}{k!(n-k)!}$ where n = total and k = 3 for which n stands for the total and k stand for choices?

[jolly parrot]

Merineth

[cursive wraith]

you're confusing things

$nCk = \binom{n}{k} = \frac{n!}{k!(n-k)!}$

[jolly parrot]

rafilou2003

[gilded hollow]

Yeah that's the one i have

[cursive wraith]

yes

and not \binom{n!}{k!(n-k)!}

[gilded hollow]

"The number of combinations of size k of n different objects" Is the translation from the book

[cursive wraith]

can you send pic of the stars and bars formula tho

i'm interested

[gilded hollow]

Sure

One sec and i'll translate

"The number of combinations with repetitions of size k among n given objects is equal to ..." (where the ... is the formula)

And the formula for combinations is

"The number of combinations of size k of n different objects"

So the problem i've had all along is that i assumed for example $x_1+x_2 = 10$ where n = 10 and k = 2 but in fact it was n = 2 and k = 10?

[jolly parrot]

Merineth

[cursive wraith]

yeah

[gilded hollow]

WELL FUCK ME

I'm gonna have to write that down lol

[cursive wraith]

the thing is

https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) on wikipedia they have this

which is the correct thing

(but they swap k and n)

[gilded hollow]

oh right i see that

[cursive wraith]

so it might be complicated. But just in case, screenshot this

it's not using k and n

just what it's supposed to be

[gilded hollow]

Okay that does clarify a lot tbh

Let me test it

Determine the amount of integer solutions to the difference \$x_1+x_2+x_3+x_4 < 40$ which holds the following condition :
a) $x_i \ge 0 $ for $i = 1,2, \dots , 5$

This seems to be a stars and bars problem?

Where k would be 40 and n = 4

$\binom{k+n-1}{k} = \binom{40+4-1}{40} = \binom{43}{40}$

[jolly parrot]

Merineth

[cursive wraith]

uh

[gilded hollow]

Hmm wait that might not be right because of the "<" ?

[cursive wraith]

it's not =

and wait x_i >= ?

[gilded hollow]

my bad , sorry

[jolly parrot]

Merineth

[cursive wraith]

ok so

[gilded hollow]

if it was "=" would i have the right asnwer?

[cursive wraith]

yes

but

notice that we can make this into multiple problems

[gilded hollow]

one where it's 39, one where it's 38 .. etc?

[cursive wraith]

or actually

well this COULD work

but there's a prettier way

imagine you do have 40 stars

oh well actually

let's say we have 39

since x1 + x2 + x3 + x4 <= 39 right?

[gilded hollow]

Yeah

[cursive wraith]

so

for each star, you can either place it in one of the 4 boxes

or you can place it outside

and make "outside" become a 5th box

[gilded hollow]

Yeah someone mentioned earlier "outside" but i'm having a hard time digesting that

[cursive wraith]

and the problem becomes x1 + x2 + x3 + x4 + x5 = 39

[gilded hollow]

How would adding a new x5 determine the amount of solutions?

[cursive wraith]

look at what we've been doing

each of the 39 stars that didn't get into one of the 4 boxes

we just group them together and we call them a new box, x_5

[gilded hollow]

Ok

[cursive wraith]

and the problem becomes x1 + x2 + x3 + x4 + x5 = 39

with xi >= 0

[gilded hollow]

ok

[cursive wraith]

you know how to solve "=" problems like this

it's stars and bars

[gilded hollow]

Yeah

$\binom{43}{39}$

[jolly parrot]

Merineth

[cursive wraith]

yep

[gilded hollow]

the right answer should be 44C39

there is a hint given in the book

"The differences can be formulated as the equation x1+x2+x3+x4+x5+x6 = 40 where x6 has to be x6 > 0"

[cursive wraith]

uhm

doesn't make a lot of sense to me

where does x5 and x6 both come from

[gilded hollow]

it doesin't say

[cursive wraith]

I would have agreed if the original problem was x1 + ... + x5 < 40

but it's x1 + .... + x4 < 40

[cursive wraith]

oh wait... did you not copy it correctly?

it says xi >= 0 for i = 1,2,3,4,5

[gilded hollow]

yeah sorry damn

it was an x5 also

but the same method applies like you said

by introudcing a new box

Why does it work by introducing a new box?

Does the new box just represent when there are leftovers?

5 + 5 + 5 + 5 + 5 + 15 = 40
Where 15 is x_5 which makes it possible for us to calculate all the possible solutions including those that are > 40 ?

How did you come up with that idea?

[scenic aurora]

I think changing the inequality f(x) <= k to f(x) + r = k is commonly known as using slack variable

because when r is positive integers, you are consider the case of k - 0, k - 1, k - 2, ...

[gilded hollow]

I’ve never heard of that before hmmm

[scenic aurora]

It is common in optimization field

[gilded hollow]

Optimization field?

I Will just have to remember how its solved I guess😞

Thanks rafil for the help

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[gilded hollow]

How many different arrangements can be formed from the letters PARALELLOGRAM if
a) the four L cannot be placed next to each other

[gilded hollow]

I dont want to the solution i just want to know if i'm thinking right

First i want to find out how many possible arrangements of the L i can make, right?
_ _ _ _ _ _ _ _ _ _ _ _ _ _ Since we have 14 spots in total and the condition is that they aren't next to each other means we have 7 spots right? Meaning that the placement of the L's are 7C4 ?

[fringe robin]

let me know if you believe otherwise but I don't believe that that is correct since you don't know where the Ls are to start with

so you don't know which spots to avoid

[stone warren]

this does not seem right and you can do an easy sanity check for this, just fix the first 2 L's as 1st and 3rd position and count the number of positions available for the 3rd L

[gilded hollow]

ok.. i'll rethink brb

The absolute total amount of permutations that can be made are 14P14 = 14!

Since i know this, it might be possible to find out the conditions for when the L's are next to each other and subtract those from the total?

nvm

[gilded hollow]

$\frac{14!}{4!3!2!} - \frac{11!}{3!2!}$

[jolly parrot]

Merineth

[gilded hollow]

Why does it become 11! And not 10?

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[sullen leaf]

I got this answer

what are your thoughtd?

thoughts*

[pearl pond]

@sullen leaf Has your question been resolved?

[sullen leaf]

<@&286206848099549185>

[sullen leaf]

<@&286206848099549185>

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[fair oasis]

For trapezoidal/midpoint rules, how am I supposed to find K for the error bound formulas?

[fair oasis]

I understand that it has to do with f''(x) but that's really all I know

take something like this for example

i find the 2nd derivative right

which would be this

but then i don't understand where to go from here

nvm, i somewhat figured it out

actually

no

I still don't understand the a <= x <= b

I'm plugging in |f''(a)| or |f''(b)| for k but it isn't correct

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[summer hearth]

im not sure why these zeros aren't working, i even checked on a graphing calculator and the only zeros are -4 and 2

[quiet tendon]

enter all answers, including repetitions

-4 is a root that's repeated twice

so they want the answer as 2, -4, -4

[summer hearth]

ooh i see

thank you

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[dark ingot]

Coefficent of x⁹⁹ in (x-1)(x-2)....(x-100)

[dark ingot]

Is it 5050 or -5050?

[inland ivy]

How can we know for sure?

[dark ingot]

?

[inland ivy]

First tell me how you got 5050

[dark ingot]

(x-1)(x-2)...(x-100) is a polynomial whose roots are 1,2,...,100
So 1+2+...+100= -coefficientof x⁹⁹/1
Coefficient of x⁹⁹=-5050

But answer given is 5050

[inland ivy]

It’s wrong

[inland ivy]

This is correct

[dark ingot]

Ok

[inland ivy]

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[dark ingot]

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[vernal grove]

How is P(Z <= 3.33) = 0.9996?

[eternal tulip]

what are you confused about?

[cerulean ridge]

According to the table, there are 311 places where z ≤ 3.33, and the total times are 315

[pearl pond]

@vernal grove Has your question been resolved?

[vernal grove]

I dont know how one would solve P(Z <= 3.33) using that specific table

[vernal grove]

how did you even find that out though? there are no rows for 3.3

[eternal tulip]

note this table represents P(Z>a) where a is a given value in the table

so you would first find P(Z>3.33)

do you see why this would help?

[vernal grove]

oh but even then how would you find P(Z>3.3) I am still lost

like there is no row for 3.3 in that table...

[eternal tulip]

you would estimate it i suppose

[vernal grove]

how would I do that

[eternal tulip]

halfway point between 0.001 and 0.0001 is

,calc 0.00135+0.000233

or i gues it would be

233

[jolly parrot]

Result:

0.001583

[eternal tulip]

,calc 1- (0.00135+0.000233)

[jolly parrot]

Result:

0.998417

[eternal tulip]

hm

[vernal grove]

oh shit wth

[eternal tulip]

still not close

i mean it is

but

[vernal grove]

tysm

that shld be good enough

[eternal tulip]

this would be for more for 3.25

but its all approximation anyways

¯_(ツ)_/¯

[vernal grove]

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[kindred flame]

Hi, im trying to solve this problem and I understand that the vertical asymptote is x = -4 because x cannot be -4 in the denominator. However, how would I know which of these graphs is the correct answer?

[gleaming musk]

is the graph negative at any point?

[kindred flame]

am i supposed to know that based on the function?

[gleaming musk]

i would look at the denominator

[kindred flame]

i used the denominator to find the vertical symptote

but i forgot how i would know what the graph looks like

[gleaming musk]

consider h(x) = (x+2)^2

what can you say for h(x) in terms of whether it is ever negative

[kindred flame]

its never negative

[gleaming musk]

what can you say for 3/h(x)

its also never negative

[kindred flame]

wait

can you break that one down

3/h(x)

why is it never negative?

[gleaming musk]

3 is a positive constant right?

[kindred flame]

i know h(x) is never negative because anything sqaured is positive

but not the 3/h(x)

correct

[gleaming musk]

a positive divided by a positive still yields a positive

[kindred flame]

ahh ok

[gleaming musk]

so if we apply that logic here

if (x+4)^2 is always positive

then 3/(x+4)^2 is always positive

[kindred flame]

OHHHH

i see

so it would be the bottom left one,

because it's always positive

[gleaming musk]

👍

[kindred flame]

gotcha

can you help me with 1 more thats similar

this one is a bit more complicated

[gleaming musk]

sure

[kindred flame]

cool ill send a pic one sec.

so for this one its a bit more complicated

i have no idea where to start

like i was gone for a few days from class so im not sure if its in my notes tbh

[gleaming musk]

can you first try factorising that

[kindred flame]

ok

-4(x^2 + x - 2 ) / (x + 4) (x - 4)

[gleaming musk]

you can still factorise the top

but it wouldnt make too big of a difference

[kindred flame]

-4 (x -2) (x + 1)

[gleaming musk]

👍

[kindred flame]

(x+4) (x-4)

[gleaming musk]

ok so we know where the asymptotes are

which is x = +-4

more importantly, have you dealt with thinking about the graph as they approach x = +- infinite?

[kindred flame]

i somewhat recall that, but its hard to rememeber

[gleaming musk]

ok we can do this step by step

so both the numerator and denominator contain x^2 term right?

[kindred flame]

correct, so to find horizontal asymptote it's the ratio of the coefficients

[gleaming musk]

and we know that x^2 is much larger than x as x approaches infinite, regardless of which direction

yep

so then we only care about -4x^2/x^2

[kindred flame]

so the horizontal asymptote would be y = -4 ?

[gleaming musk]

yep

[kindred flame]

gotcha

so wouldnt that just make the graph in the top left the answer?

[gleaming musk]

correct

[kindred flame]

since its the only one with a line through y = -4

i see, i guess my teacher is mostly just testing us if we know

H.A / V.A

with the graphs

instead of actually graphing it

atm

thanks for your help sir

[pearl pond]

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[sharp smelt]

Write out the lower and upper sums for f(x)=1/x in the interval [1,3] where the length of each small interval is 1/2

[sharp smelt]

so just wanted to confirm that it's $\frac{1}{2}(1+ \frac{2}{3}+ \frac{1}{2} + \frac{2}{5}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

right?

[rough forge]

which is that lower or upper sum?

if it's upper looks right

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[vast berry]

For (c)

Why cant i let x = -3y

and substitute that in here

[pearl pond]

@vast berry Has your question been resolved?

[pearl pond]

@vast berry Has your question been resolved?

[vast berry]

<@&286206848099549185>

[pearl pond]

@vast berry Has your question been resolved?

[chrome chasm]

I haven't done matrices in a while so forgive me for any mistake

i think its is because y = -3x is an eigenvector, meaning that while the line remains invariant, the vectors need not

so like the point (-3, 1) could change under the matrix to (-6, 2), or (3, -1) and the line would still be invariant, but the vector muliplies by some constant number, lambda. In this case tho, I think it just so happened that the eigenvalue, lambda is equals to 1.

[pearl pond]

@vast berry Has your question been resolved?

[vast berry]

how do i know y=-3x is an eigenvector? ty

[last moth]

it's because that line is invariant under the transformation M

meaning when you apply M, the vectors along the line y=-3x only get multiplied by a scalar

[vast berry]

ohh

[vast berry]

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[royal frost]

how do I get y by itself here

(-1/y) = x^3 + C

[full agate]

Inverse and multiply by -1?

[royal frost]

idk what that means, in layman's terms pls

like what do I do, describe in actions.

[full agate]

I can give a solution but try first yourselfff

[royal frost]

I have the solution

im just asking why

or how i mean

[full agate]

Idk where they got y=0

[royal frost]

ignore that

[royal frost]

bet

[full agate]

Its answer not solution

[royal frost]

potato, tomato

[full agate]

Potato and tomato are different af

[warm sonnet]

i disagree

[full agate]

You whaaat?

[royal frost]

depends

[full agate]

You are guys geniuses

[full agate]

Anyway

understand it

[pearl pond]

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[inner galleon]

was trying to solve this by a different way -i already solved it using riemann sun and integration xsinx....- but i said why not take a different approach and use tylor summation of sin(x) and approximating kpi/n+1 =~ kpi/n -idk if its possible or not -

[inner galleon]

would it be possible ?

i tried taking first three termes from tylor series , and transformed them to integral through riemann sum

i noticed a pattern getting reapted, can we turn it into summation

[inner galleon]

im jst playing around, idk if what im doing is mathematically correct but when solving integral with integral calculator im getting a close answer to (1/pi) which is the correct answer

[inner galleon]

this integral by taking first 5 termes in tylor series, so can we make a summation of all termes ? and integrate it

if dominated convergence theorem verefied or smthing like that

<@&286206848099549185>

i kinda thought of smting like this, but uhhhh i dont think it works

[pearl pond]

@inner galleon Has your question been resolved?

[inner galleon]

[pearl pond]

@inner galleon Has your question been resolved?

[inner galleon]

<@&286206848099549185>

[silk venture]

nah i dont think you can create a sum like that

[inner galleon]

alright, do u think this approach is correct or nop. im jst playing

[silk venture]

hmmm not sure you should keep exploring if that interests you, I only did that usin riemann sums

but if you skip riemann sum I guess there wont be any integration

its definetly solvable w/o integrals

[inner galleon]

ye i did with riemann sums too , but i wonder if i can reach to the same result by expanding sin(....) using tylor series and do some bunch of stuff

jst wanted to play with riemann sums and integration since im not quite used to them, and i wonder if m other approach is valid or nop

[pearl pond]

@inner galleon Has your question been resolved?

[inner galleon]

[pearl pond]

@inner galleon Has your question been resolved?

[last moth]

it is correct yeah

using the dominated convergence theorem as you said

[last moth]

although shouldn't it be x+1 rather than x in the denominator?

it definitely seems harder to evaluate than using the fundamental theorem of calculus on x sin(pi x)

[pearl pond]

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[hallow condor]

need answer asap

[hallow condor]

whats the answer

been tyring for hrs and need to sumbit asap

<@&286206848099549185>

[west sapphire]

did you write the integrand as a power series first?

[last moth]

please wait 15 minutes before pinging Helpers, in the future

[hallow condor]

yes as a series expansion

then integrated the series

[west sapphire]

what did you get for your integrand after expressing it as a power series?

[hallow condor]

i just need the answer

i got 9 mins before it done matter

[west sapphire]

well no one's gonna just give you the answer, that's not how this server works

[hallow condor]

.close

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[teal hare]

How does this become 60x^2 y^8

[teal hare]

I understand the correct formula for the fifth term but I don’t know how it becomes 60x^2 y^8

Can someone show me the steps?

[prime bramble]

iirc, Pascal's triangle is really useful here

[teal hare]

I was using his triangle and it doesn’t make sense here

60x^2 y^8 ain’t even on here

I don’t understand

[pearl pond]

@teal hare Has your question been resolved?

[quiet tendon]

the triangle just gives you the binomial coefficients, so you’d use it to find 6C4

the rest of the factors comes from 2² and then the x and y comes form the rest of the formula

[pearl pond]

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[stray scroll]

I have a doubt in this solution that how do we know before solving that only one squared term will have x? Is there any way to know this or is it just a matter of hit and trial until we get the correct pair of squares?

[stray scroll]

Is there any common format for the expression that we will get by squaring and adding 2 terms which are Ax + By + C form so that we can use that to know if x is there in both terms on only one of them?

[pearl pond]

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[stray scroll]

Or can anyone please tell me where else should I ask this question?

[pearl pond]

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[pearl pond]

@stray scroll Has your question been resolved?

[pearl pond]

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[teal hare]

I know how to find it if it's variables such as m but Is it the same with numbers?

[naive hemlock]

Yes

like consider (9-2)^7

So 5th term will be 7C4 9^7-4 2^4

If you do the binomial expansion and add every term it will eventually equally to 7^7

[teal hare]

One moment imma try

[naive hemlock]

See take (4-2)^2
So the expansion will be 2C0 4^2 (-2)^0+ 2C1 4^1 (-2)^1 + 2C2 4^0 (-2)^2

Now see first term is equal to 16 second is equal to -16 and 3rd is equal to 4 so
16-16+4 is 4

Which is in fact (4-2)^2 i.e (2)^2=4

@teal hare

[pearl pond]

@teal hare Has your question been resolved?

[teal hare]

Sorry I was trying my best but I didn't get it, I'll take a look at your exampkle

What do the c's mean exactly?

its okay Ill come back in a little bit

.close

[pearl pond]

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[pearl pond]

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[dry wren]

Is my diagram right

And if so how do I find h

[sage escarp]

I can do it for u wait a sec

[dry wren]

Thanks

[sage escarp]

Let D be the intersection of the height in C

AD=x, BD=50-x

Then you can make two equations with two variables

With tangent

[dry wren]

Show d is on the ab line?

What with tangent

[sage escarp]

Yeah, it’s just for the facility of naming it

[dry wren]

Opp/adj

[sage escarp]

Yeah

[dry wren]

Oh do I like factor out the x

[dry wren]

That’s what I have

[sage escarp]

Well don’t know what u mean

[dry wren]

Ye

Wait

[sage escarp]

Then u already have it

[dry wren]

Wait

I times 50 and x

Both by tan 70

[sage escarp]

I mean it is a simple x, y equation

Y can do it anyways,

[dry wren]

I think I made a mistake

[sage escarp]

U have problems with parenthesis

Tan70(50-x)

[dry wren]

How is the answer 93

Do I first make both equations equal to h

[dry wren]

Oh

[sage escarp]

Note that u have not divided entirely by tan80

So it should not be done that way, RHS should be x(tan70+tan80)

[dry wren]

Oh

Where do I get 86 from

[sage escarp]

Where did u get 86, I thought it was 80

[dry wren]

Ye

I think that was the error

[sage escarp]

Well yea, if u typed in 86 then u have double error

[pearl pond]

@dry wren Has your question been resolved?

[pearl pond]

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[pearl pond]

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[dense iris]

This has me lost

7b

the right hand side is simple

but the left hand side is like

oh wait never mind

.close

[pearl pond]

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[dense iris]

.reopen

[pearl pond]

✅

[dense iris]

Ok I am extremely lost in 7b

ive gotten to the point

that ln(p) - ln(10000-p) = 1/200t + c

[cerulean ridge]

So you have ln() - ln() = stuff

You can simplify the ln()s into one ln()

[dense iris]

ye

and then I raise it the base e

idk how it helps tho

[cerulean ridge]

It helps because we're getting closer to "p"

[dense iris]

Oh I need the p value?

Like that throws me off cause it keeps it in terms of p in the question

[cerulean ridge]

Yeah, it even said in the question: "give the answer in the form p = ..."

[dense iris]

uh ok

[cerulean ridge]

Basically, instead of solving for x, we're solving for p

[dense iris]

so I got

p + 2500p = 2500000

idk if i can factorise p to (2500+1) tho

that seems wrong

[cerulean ridge]

You definitely can

[dense iris]

oh ok

[cerulean ridge]

1 thing + 2500 thing = 2501 thing

[dense iris]

I get a rly long decminal number tho

if I am to divide through

yh

i mean

I get p = 250000/2501

and have no idea what to do with that

[pearl pond]

@dense iris Has your question been resolved?

[dense iris]

.close

[pearl pond]

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[sterile osprey]

Anyone ever done a question like this?

[glass meadow]

Are these marks your attempts?

Never mind it works, though it's quite a bit more than necessary

What's your question exactly?

[pearl pond]

@sterile osprey Has your question been resolved?

[pearl pond]

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[round tide]

hello fellow helpers

[round tide]

I have a calculus question

I am studying integrals, but before moving forward I need to resolve a question about the dy/dx notation

what I came up with is the following:

The formula for the slope of a line is 🔺y / 🔺x. Now since the derivative is the same equation only when 🔺x tends to 0, I concluded that dy = 🔺y when it approaches 0 and that dx= 🔺x when it tends to 0. Basically, dy/dx is the same as saying 🔺y/🔺x only that by saying dy/dx we are implicitly saying that they are VERY small values. Therefore, since they are finite values, they can be treated as numbers and can be solved (for example, by solving dx of dy we get the differential of the function)

is this correct?

[calm wing]

yes, by the limit definition of a derivative

[round tide]

tell me more about it

[signal bison]

delta x and delta y are really small numbers yes

[round tide]

I really need to master this topic

[calm wing]

$f'(a) = \lim_{b\to a} \frac{f(b)-f(a)}{b-a}$

[jolly parrot]

artemetra

[calm wing]

$\approx \frac{\Delta f(x)}{\Delta x}$

[jolly parrot]

artemetra

[round tide]

there you go

must be tricky to type with that

[calm wing]

like it's the same thing but taking the limit of the difference of both things

[calm wing]

lol sorta

[round tide]

so is this correct?

at least for now

like could you define with your own words what a differential is

[calm wing]

yes

[calm wing]

it's an informal way of saying the same thing as this

like one to one

[pearl pond]

@round tide Has your question been resolved?

[pearl pond]

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[pearl pond]

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[polar steppe]

I'm confused by what they mean by formula here?

[polar steppe]

It's not a matrix is it?

for x y and z

I followed the note and got
$\begin{matrix}
XA & 0\
0 & ZC
\end{matrix}$

[jolly parrot]

unbearablefrequentist1

[sharp quest]

that's not the product of the left side

and by formula, they mean that each entry of the correct product will equal the entries of the matrix on the right.
So you'll have XA = I, for example.

[polar steppe]

really? Do I have the form incorrect?

[sharp quest]

your bottom left entry should not be 0

[polar steppe]

Like X = XA = I ?

where I the Identity matrix

[sharp quest]

No, X does not need to equal XA.
If you do the product. You'll get XA in the top left spot. That should equal the matrix on the right, which has the identity matrix in the top left spot. So those are equal. i.e. XA = I

[polar steppe]

Oh interesting

If I understood, you're saying XA = the whole Identity matrix on the rhs?

[sharp quest]

What do you mean by 'whole identity matrix'

[polar steppe]

Okay it's kind of weird because the rhs looks just like the identity matrix. You're saying XA = I or XA = rhs?

[sharp quest]

$\begin{bmatrix} a & b\c& d\end{bmatrix} = \begin{bmatrix} w & x\y& z\end{bmatrix} \implies a=w, b=x, c=y, d=z$

[jolly parrot]

Zybikron

[sharp quest]

I'm saying the top left entry of lhs equals the top left entry of rhs,
XA = I

[polar steppe]

okay

hmm

makes sense

[sharp quest]

So they want you to find X, Y, and Z in terms of A, B and C

[polar steppe]

Before I continue then, what part of the matrix that I computed was incorrect?

[sharp quest]

bottom left

[polar steppe]

since we established that the top left element was XA, would X = XA = I?

would that be the formula

[sharp quest]

No

[polar steppe]

darn

[sharp quest]

why would X = XA?

[polar steppe]

idk it's in the lh spot

I think I'm still unsure of what I'm being asked

[sharp quest]

All you have is XA = I, from that you should be able to work out how X and A are related.

[polar steppe]

oh

wait

X = I / A?

[sharp quest]

almost. You can't divide by a matrix, you'd use the inverse.

[polar steppe]

I did that because you mentioned "..how X and A are related"

ohhh

how would you get the Inverse of A

[sharp quest]

X= A^-1

[polar steppe]

I have a theorem,
$A^-1 = \frac{1}{ad-bc}
\begin{matrix} d & -b\ -c & a {matrix}\end$

[jolly parrot]

unbearablefrequentist1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[polar steppe]

I give up on making that

I think I'm gonna close this

clearly I'm missing a lot more than what I was given

.close

[pearl pond]

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[pearl pond]

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[lethal stag]

For integers n

[lethal stag]

Can someone help me with this?

[spare lark]

Induction

Theyre might be a chad way to do it

[lethal stag]

I just need to simplify it I think

[spare lark]

Maybe the limit in +infinity and that sequence is decreasing will show that its roofed by the first term which is 9/11 < 1

[spare lark]

How would you simplify ?

[lethal stag]

Turn it to a normal looking quadratic and factorise

Idk what to do after tho

[spare lark]

Like 3n < n^2 +2
-n^2 +3n -2 < 0 ?

[lethal stag]

Yep

[spare lark]

And study when this is negative ?

[lethal stag]

Yes but I'm not sure how

[spare lark]

Will give us $-(n-1)(n-2)$

[jolly parrot]

YakuBros

[lethal stag]

I divided by -1 to get $n^2-3n+2>0$

[jolly parrot]

DapperJaguar197

[spare lark]

Works too

[lethal stag]

Idk how to apply the 2 factors to get a solution for n tho

[spare lark]

You have 1 and 2 as roots

So the quadratic is positive on [2,+inf) but n > 2 so its strict positive

[lethal stag]

Is n>1 also true? Or do both have to be satisfied?

[spare lark]

n has to be n > 2 as said in the exercice statement

[spare lark]

The quadratic is positive for all n > 2, and the exercice supposed it so we have proved that its true

It wont work for n >1 cuz for n=2 we have 6/6 which is equal to 1 and not striclty less

As asked

[lethal stag]

Okay understood

Thanks for your help

.close

[pearl pond]

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[spare lark]

Can you show the whole work please ?

Ok ok

What is px ?

So 20 - 2x - Lpx = 0
-Lpx = 2x -20
L = (20 -2x)/px , indeed

I hope too,

[glad rose]

hello, i'm new here

[spare lark]

Welcome then !

[glad rose]

Thx!!

[pearl pond]

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[spare lark]

If you need help with maths problem, you can ask in the free-channels to get help

[glad rose]

free-channels? where is it, you mean help-34 ...etc?

[spare lark]

The not-occupied one/avaible

[glad rose]

alright

[pearl pond]

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[wispy quail]

Need help with matrices
Need to understand the manipulations and algebra

[pearl pond]

@wispy quail Has your question been resolved?

[wispy quail]

@feral sedge
a(bx+cy)^2-b(ax+by)(bx+cy)+(ax+by){b(bx+cy)-c(ax+by)}

[tawdry wing]

is this laplace?

[wispy quail]

matrix, co-factor expansion

[tawdry wing]

yup yup

[wispy quail]

factor out ax+by?

[feral sedge]

can't be done

[wispy quail]

oh shoot.

i'm hallucinating

[feral sedge]

I'd say, distribute it all out and see if you can look for factors of b^2-ac

[wispy quail]

pain

[feral sedge]

could try to get wolfram to distribute it lol

maybe there's something better I don't know about, I'm not a huge linear algebraist

[wispy quail]

idk why this problem exists in my textbook

it's supposed to be high-school stuff

i learned matrix operations

to make zeros

this is unfair

btw, it's getting nasty in the distribution

@feral sedge
ac2y2+ab2x2-a2cx2-b2cy2

[tawdry wing]

do you need solution

i just solved it

[wispy quail]

ax2(b2-ac)-cy2(b2-ac)

[wispy quail]

damn

nice

explain please

[tawdry wing]

just 2 minutes

you will get solution in the details

[wispy quail]

(b2-ac)(ax2-cy2)

WHERE'S THE 2BXY

[tawdry wing]

i dont get this

i get something similiar, but not exactly.

[wispy quail]

do you have the 2bxy?