help-19

[clever fjord]

Cyrenux

[spring maple]

yes

[ebon harness]

Well to answer that, first you need to know why n is power to the one when the difference between two consecutive terms is a constant

[spring maple]

how am i supposed to find the value of c

a, b, c could be anything couldnt they?

[ebon harness]

Here

[ebon harness]

Pretty sure there is only one specific ordered (a,b,c) which satisfies this but we'll see

let n=1 then what does
c_n = an² + bn +c
become?

[spring maple]

$c_2 = 13 = an² + bn + c$ a = 2 b = 2 c = 1

[clever fjord]

tokyo

[spring maple]

idk

[spring maple]

$c_n = a^2 + b + c$

[clever fjord]

tokyo

[ebon harness]

You mean c_1

[spring maple]

yeah

[ebon harness]

And is it n² or a²?

[spring maple]

uhm

[ebon harness]

Coefficient of n² is a

Not a²

[spring maple]

oh

true

$c_1 = a + b + c$

[clever fjord]

tokyo

[spring maple]

thats correct?

[ebon harness]

Ye

What was c_1 equal to?

Its given at the start

[spring maple]

1

[ebon harness]

So a+b+c = 1

[spring maple]

yes

[ebon harness]

Now let n=2 to find c_2 in terms of a,b and c

[spring maple]

okay

$c_2 = 4a + 2b + c$

[clever fjord]

tokyo

[ebon harness]

This is equal to?

[spring maple]

13

[ebon harness]

Yes

And what about c_3

In terms of a,b and c

[spring maple]

$c_3 = 9a + 3b + c$

[clever fjord]

tokyo

[spring maple]

= 37

[ebon harness]

Now notice coefficient of c is always 1 in every equation

[spring maple]

yes

[ebon harness]

So you can extract two equations to get rid of c

Now try to get rid of c by extracting two equations

[spring maple]

$c_3 - 9a - 3b = c$

[clever fjord]

tokyo

[spring maple]

wait

$c_3 - 9a - 3b = 1$

[clever fjord]

tokyo

[ebon harness]

Uhh maybe using c wasnt the brightest idea

You confused it with the sequence c_n

[spring maple]

oh

[ebon harness]

Yeah poor choice of letter from me but whatever

[ebon harness]

c_1 = 1 not c

[spring maple]

so what do we do now

okay

[ebon harness]

Find c_3 - c_2

$c_3 - c_2 = ?$

[clever fjord]

Cyrenux

[spring maple]

24

[ebon harness]

And what about in terms of a,b,c ?

[spring maple]

uh.

[ebon harness]

You typed them above

[spring maple]

5a + b

[ebon harness]

What is this equal to in terms of numbers

[spring maple]

24

[ebon harness]

Ye

[spring maple]

can b be a negative?

[ebon harness]

Now evaluate $c_3 - c_1 = ?$

[ebon harness]

Of course

[clever fjord]

Cyrenux

[spring maple]

$c_3 - c _1 = 8a + 2b$

[clever fjord]

tokyo

[ebon harness]

What is this equal to numerically?

[spring maple]

uhm

36

oh i get it now

so now we can find

the values

or a and b?

[ebon harness]

$8a + 2b = 36$
\ $ 5a + b = 24$

Yeah

[clever fjord]

Cyrenux

[spring maple]

okay so uh

[ebon harness]

Before we evaluate, we can simply first equation

Do you see it

[spring maple]

uhm

yes

4a + b = 18

[ebon harness]

Yes

[clever fjord]

tokyo

[ebon harness]

Use double \

[spring maple]

oops

uhm

[ebon harness]

Discord only prints 1

So you have been decieved by text

[spring maple]

$4a + b = 18$
\$5a + b = 24$

[clever fjord]

tokyo

[spring maple]

damn

so a = 6

[ebon harness]

Yes

[spring maple]

amd b =

uh

-6

$a = 6$
\$b = -6$

[ebon harness]

Now you can find c as well

[clever fjord]

tokyo

[spring maple]

yes

uh

[ebon harness]

$c_n = an² +bn + c = 6n² -6n + c$

[clever fjord]

Cyrenux

[ebon harness]

Simply let n = 1

[spring maple]

oh

i was just going to use this

$c_2 = 4a + 2b + c$

[clever fjord]

tokyo

[ebon harness]

That works too

Any works

[spring maple]

okay so

$13 = 24 - 12 + c$

[clever fjord]

tokyo

[spring maple]

c = 1

then

yeah?

[ebon harness]

$$c_n = 6n² - 6n + 1$$

[clever fjord]

Cyrenux

[ebon harness]

You found the sequence formula you have been searching

[spring maple]

ayy

🥳

oh gosh

i need to type this all out for explanation

😭

[ebon harness]

My explanation is kinda wack but whatever

[spring maple]

idk we got there

so what do we do next

btw is there any chance you can help me type it out or no

because im gonna find it hard to get the right pieces of information

i've been kind of all over the place in this

[ebon harness]

Dont have pen with me currently

[spring maple]

oh i meant as in digital

im doing this on laptop

[ebon harness]

Well you can find (d) by the formula you found currently

[spring maple]

yes i've seen that

[ebon harness]

Have you done (c)

[spring maple]

i have almost no idea how to type out my method of finding the formula

[spring maple]

i have not done any of them

[ebon harness]

Well let me explain it better

[spring maple]

alright

[ebon harness]

Do you know what a 'linear' graph means

[spring maple]

yes

[ebon harness]

Like graph of y = x or y = 2x

[spring maple]

yes

[ebon harness]

Tell me your idea

[spring maple]

uh wdym

[ebon harness]

You said you know what it means

What makes it different than y = x² , sequence wise

First be aware that input of sequences are positive integers

You DONT calculate $c_{-1}$ or $c_0$ or $c_{\frac{1}{5}}$

[clever fjord]

Cyrenux

[spring maple]

uh

theres other numbers like

b, c

not js n^2

[ebon harness]

sequence of y = x
Should be like:
1,2,3,4,5,6,...

[spring maple]

?

[spring maple]

yeah

[ebon harness]

1 and 2 are consecutive terms right

And similiarly 2 and 3 are consecutive too

[spring maple]

mhm

[ebon harness]

What is the common difference between them

1 right

If you have 5th term, you add common difference to find 6th term

5th term here is 5, to find 6th term you add 1 to obtain 6

Common difference as you see is 1 which is a constant value

On your question common difference is not a constant

1st term is 1
2nd term is 13
3rd term is 37

Difference between first and second term is 12

While difference between 2nd and 3rd term is 24

Then later you saw that difference between 3rd and 4th term is 36

[spring maple]

yes

[ebon harness]

So from here you should see that common difference changes by 12 every time

[spring maple]

mhm

[ebon harness]

First diff is 12, then 24 and then 36

[spring maple]

yeah

[ebon harness]

So ( the sequence) is not linear

But change of difference is linear

Because it changed the difference of terms by 12 each time

Thats why variable with highest degree is n² and not n³

[spring maple]

okay

wait

im going to type out how i found the formula

ill be back after a while

[ebon harness]

For a better explanation you can search
'How to find general term of a non linear sequence'

Its kinda hard to explain to a middle schooler considering derivates arent even taught yet

[spring maple]

alright

i have typed my explanation for how to get the formula and i have completed 6d.

i am going to eat dinner but may i ping you when i'm done

[spring maple]

@ebon harness how can i approach 6a

[ebon harness]

Do you know what a triangular number is

@spring maple

[spring maple]

uhm

numbers like

x^3

?

[ebon harness]

Nope

[spring maple]

numbers which can be root to 3?

[ebon harness]

Amount of dots it wil take to form a triangle union 1

[spring maple]

ohh

i get it

i found an image that explains it

pretty much this?

[ebon harness]

Well googling it is the betyer way yeah

Yep

You need to know general formula for this

[spring maple]

mhm

[ebon harness]

Do you know sum of positive integers from 1 to n

1 + 2 +3 + ... + n = ?

[spring maple]

i learnt it

i forgot though

[ebon harness]

Yeah forget now

I will show you how formula is found

[spring maple]

n(n+1)/2

[ebon harness]

Do you know how to find it though

[spring maple]

nope

that was off google

[ebon harness]

Let $1 + 2 + 3 + ... + n = k$

[clever fjord]

Cyrenux

[ebon harness]

Wouldnt this from reverse also be equal to k?

[spring maple]

yes

[ebon harness]

$ n + (n-1) + (n-2) + ... + 1 = k$

[spring maple]

mhm

[ebon harness]

$1 + 2 + 3 + ... + n = k $
\ $n + (n-1) + (n-2) + ... + 1 = k $

Latex hello

I mean texit

Anyday now...

[spring maple]

well..

[ebon harness]

$1 + 2 + 3 + ... + n = k$
\ $n + (n-1) + (n-2) + ... + 1 = k$

@clever fjord

[spring maple]

..

the bot just disappeared smh

[clever fjord]

Cyrenux

[ebon harness]

So 1 matches with n here right

[spring maple]

mhm

[ebon harness]

When you sum them upside down

You obtain n+1

2 matches with (n-1) here

Since we are adding these two equations together we get :
2 + (n-1) = (n+1)

Similiarly 3 matches with (n-2)
Which will make their sum n+1 again

Now lets evaluate how many (n+1) values we will obtain

Our original equation was 1 + 2 + 3 + ... + n = k

How many numbers are there between 1 and n

(Including both as you can see)

If you dont know, do a more simple example like amount of numbers between 1 and 3

[spring maple]

uh

just n?

[ebon harness]

Yes n amount of numbers

Doesnt that mean we will get n amount of (n+1)

So n(n+1)

[spring maple]

mhm

[ebon harness]

Which will be equal to 2k since we summed these equations upside down

[spring maple]

i get it

[ebon harness]

k + k = 2k

[spring maple]

yep

thats why u then divide by 2

to make it just k

[ebon harness]

2k= n(n+1)

Yep

K was the value we were searching for

[spring maple]

mhm

[ebon harness]

Now you will ask why i showed you this

[spring maple]

yeah

what does this have to do with triangular numbers

[ebon harness]

consider a sequence where A_n = n(n+1)/2

Evaluate A_1 , A_2 and A_3

If you noticed, top of triangular figures have 1 dot at the very top

[spring maple]

yeah

[ebon harness]

Then 2 dots below the top

3 below 2 dots

4 below 3 dots

So on...

[spring maple]

ohh

do the formula for a triangular number

is the formula for

1 + 2 + n = k

[ebon harness]

T_n calculates amount of total dots so
T_n = 1 + 2 + ... + n

[spring maple]

mhm

[ebon harness]

Make sure to state 3 dots though

[spring maple]

wait

let me write this down

[ebon harness]

Otherwise it can be interpreted as 1 + 2 + n = 3 + n

[spring maple]

yes

i was just a bit lazy with that

my brain is hurting lmao

[ebon harness]

You know the formula of T_n

[spring maple]

yes

[ebon harness]

What is T_3

[spring maple]

this is basically all

uhm

T_3 would be 6

[ebon harness]

Yeah now do what it asks to do you in (a)

[spring maple]

what does it mean by illustrate though?

i dont really understand what the question is even asking

do i just show how the values of

$C_4 = 12T_3 + 1$

[clever fjord]

tokyo

[spring maple]

is equal to the 4th pattern with the general formula

?

$C_4 = 6 x 4² - 6 x 4 + 1$

[clever fjord]

tokyo

[spring maple]

oh

how do i make multiplication

sign

im not sure though

[ebon harness]

I kinda didnt get either but i think it wants you to evaluate 6(T_3) + 1 and say its equal to C_4

[ebon harness]

Pretty sure you dont need the formula

[spring maple]

okay

[ebon harness]

It says use the pattern

[spring maple]

i'll ask one of my friends rq though

[ebon harness]

So counting

[spring maple]

this is what i've been told

[odd edge]

@spring maple Has your question been resolved?

[odd edge]

@spring maple Has your question been resolved?

[odd edge]

@spring maple Has your question been resolved?

[odd edge]

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[grave tartan]

how to solve this question

[odd edge]

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[sleek falcon]

im pretty confused about what steps you need to do to do the following

[sleek falcon]

i mean i think a is = R^3

[tall veldt]

why do you think its R^3

[sleek falcon]

because the zero vector is in R^3

[tall veldt]

ok?

the zero vector is in every vector space

[sleek falcon]

do i need to use RREF

[tall veldt]

for what?

[sleek falcon]

to find a basis for it

[tall veldt]

what would you be putting into RREF?

do you understand what $S^{\perp}$ is

[clever fjord]

ΣΑCu

[sleek falcon]

orthogonal = perpendeicular

meaning dp = 0

[tall veldt]

yes but what is $S^{\perp}$?

[clever fjord]

ΣΑCu

[sleek falcon]

The supsace containing orthogonal vectors?

[tall veldt]

orthogonal to what?

[sleek falcon]

like any combination of the 2 vectors in the subspace

[tall veldt]

S^perp is the space of all vectors that are orthogonal to all elements in S

[sleek falcon]

yes

[tall veldt]

so, which vectors in R^3 are orthogonal to all elements in {0}?

[sleek falcon]

if they are linearly independent?

[tall veldt]

not sure what you mean

[sleek falcon]

well

any vector whose dot product = 0

[tall veldt]

you cant just say "whose dot product", you always need to talk about the dot product between two things

[sleek falcon]

let a, b in R^3 if a dot b = 0 then they are in the set

[tall veldt]

i dont understand why you think that is the condition to be in S^perp

[tall veldt]

you were happy with this

[tall veldt]

and so elements of S^perp for (a) should satisfy this

i.e. which vectors in R^3 are orthogonal to the zero vector

[sleek falcon]

anything that has a solution of 0??

[tall veldt]

what does that mean

[odd edge]

@sleek falcon Has your question been resolved?

[odd edge]

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[odd edge]

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[nimble plaza]

The problem:
The experiment is that we draw four marbles consecutively
(i) Find the probability that the first marble is green ant the second marble is golden
(ii) Find the probability that NOT all marbles will be Green (* Hint remember 𝑃𝑟(𝐴) = 1 − 𝑃𝑟(𝐴̅)
)[what is the opposite of NOT all marbles will be green?
The focus: on (ii) for h
My question: Can someone walk me through how to solve (ii)? what steps do I take?
I'm completely stuck on how to approach this and also between 2:15 to 3:30 I won't be able to respond due to a class

[odd edge]

@nimble plaza Has your question been resolved?

[odd edge]

@nimble plaza Has your question been resolved?

[odd edge]

@nimble plaza Has your question been resolved?

[odd edge]

@nimble plaza Has your question been resolved?

[nimble plaza]

<@&286206848099549185>

[glass hazel]

any other info in the problem?

[nimble plaza]

nope, just that

why is it missing something?

[nimble plaza]

<@&286206848099549185>

[nimble plaza]

<@&286206848099549185>

[drifting valley]

it is missing from what we draw the marbles

no calc can be done if we don't know the probabilities

we need the number of marbles for each color

btw, the opposite of not all marbles will be green is all marbles will be green

so calc the proba that all marbles will be green and take 1-that

[nimble plaza]

so due to the number of green marbles not being specified, I would calculate the probability like this?
Pr(4) = |A|/|S| = 4/4 = 1? |S| is the sample size and |A| is the odds

[drifting valley]

you can't calculate

that's just not possiible

[drifting valley]

^

[nimble plaza]

What if we use part one to do part 2?

[drifting valley]

I can't be clearer than I've already been

[nimble plaza]

Ok

.close

.close

[odd edge]

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[spiral roost]

Hey, could someone help me understand this? I'm not really seeing how the top inequality gives that absolute value part

[gilded hemlock]

Okay so lets say I have f(x) = 1/x^2+2. So we know x^2 > 0, for all real numbers in x. We also know x^2 + 2 >= 2. So why when we get to 0 < 1/x^2 + 2 <= 1/2. Why is the <= assigned to 1/2 instead of 0.

[spiral roost]

Hey sorry this channel is occupied

[gilded hemlock]

Oh mb

It said open on the thing idk

Is it okay if you can help me with this quick question, this little thing is stumping my brain

[spiral roost]

Umm well in general if a<b then 1/a> 1/b

[gilded hemlock]

hmm

okay I will ponder on that ty

[drifting valley]

|q'-q| is just q'-q since q' > q

-q is smaller than -(x-e/2)

since q' < x+e/2
and -q < -(x-e/2)
q'-q < x+e/2 - (x-e/2)

since q'-q is positive, so is the RHS, you can just put the absolute values

[odd edge]

@spiral roost Has your question been resolved?

[spiral roost]

Ah and since x+e/2 > x - e/2 we can say that x+e/2-(x-e/2) =|x+e/2-(x-e/2)|

[drifting valley]

yes

[spiral roost]

Oh, great. Thank you

[odd edge]

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[sweet chasm]

Hello, I did part a and b(i), but are now stuck on b (ii)

[odd edge]

@sweet chasm Has your question been resolved?

[sweet chasm]

@rancid moat

[sweet chasm]

@rancid moat

[odd edge]

@sweet chasm Has your question been resolved?

[sweet chasm]

.close

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[willow anvil]

.close

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[mystic saffron]

Factor by grouping x^2+2x+5x+10

[mystic saffron]

X is common to both on the left so would it be x(x+2) 5(x+2)

[north crow]

so (x+5)(x+2)

[neat loom]

Can anyone tell me if Id have to use product rule for this

[wooden python]

missing a plus sign in the middle

[mystic saffron]

Ok thank you

[wooden python]

!occupied

[odd edge]

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).

[nimble blaze]

$x(x+2) \red{+} 5(x+2)$

[clever fjord]

ℝam()n()v

[mystic saffron]

That’s the right answer? Because I got that

[nimble blaze]

including that red + sign,
that's only the first step

[wooden python]

it's an intermediate step

[mystic saffron]

So what’s step 2

To show your work on how you got it

?

[wooden python]

well how do you further factorize x(x+2) + 5(x+2)?

[mystic saffron]

Multiply the outside of the bracket by the inside?

[nimble blaze]

identify the common factor in your terms
similar to what you already did

[mystic saffron]

Oh ok thanks

[nimble blaze]

that just undoes what you just did

[mystic saffron]

(x+2) (x+5)

?

[wooden python]

lowercase x.

[mystic saffron]

Is that the full answer

?

x+2 (x+5)

With the brackets on both sides

(x+2) (x+5)

[nimble blaze]

yes

[mystic saffron]

Ok thank you

Appreciate it a lot

Both of you guys

🙂

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[timber bronze]

Hello
I have some difficulties in trying to compute the distribution of some binomial probabilities.

I have n1 events that could result in a success with probability p1 each
and n2 events that could result in the same success but with a probability p2 each

I know how to get the probability of getting K successes from each group of events, but how can I know the probability of getting K successes from all the events (n1 + n2) ?

Some sources tell me to look about Poisson's law, but I struggle to see if it can solve my problem, should I look further into it ?

[timber bronze]

I tried to simplify my problem, in reality I have 9 different events with multiple probabilities and different number of occurences

[upbeat drum]

!occupied

[odd edge]

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@timber bronze Has your question been resolved?

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@timber bronze Has your question been resolved?

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@timber bronze Has your question been resolved?

[raven grotto]

if i understand correctly, the probability of getting K successes in all n_1 + n_2 events is the sum from i = 0 to K of the probability of getting i successes in the first n_1 events and K - i successes in the last n_2 events

and they are independent trials (so the "and" splits into a product of probabilities)

since the only way you get K successes in the n_1 + n_2 events is if there is some i from 0 to K such that you get i successes in the first n_1 events and K - i successes in the last n_2 events

then you sum over all the cases

[timber bronze]

That makes a lot of sense, I didn't see it that way !
However, since my real problem involve 9 events (as you can see on my screen, 1 time 6.5% chance, 183 times 0.5% chance, ...), wouldn't it be complicated to find all the combinations of the probabilities ? (would it be 9! probabilities to compute only one K ?)

[raven grotto]

yeah; to find the probability of K successes, you would have to sum over all (i_1,...,i_9) such that i_1 + ... + i_9 = K, and you would sum the probability of getting i_1 successes in the first n_1 trials, i_2 successes in the second n_2 trials, ..., i_9 successes in the last n_9 trials.

it can be done, you just need to write a computer program i think

[timber bronze]

It looks like a cool challenge to do, thank you, it's much more clear for me now

.close

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[mystic saffron]

I NEE DHELPPP

[mystic saffron]

How do I determine this slope

[tribal ore]

divide both sides by 5 so that the LHS is Y only

[mystic saffron]

u can use y = mx + c

yup

[tribal ore]

and the slope would be 2/5

[mystic saffron]

y intercept 2 ie slope meets y axis at (0,2)

So I just divide by the y?

o

[tender gulch]

You have to isolate the y

So you need to get rid of the 5

How do you get rid of it?

[mystic saffron]

dividing it by the number next to it?

yup

divide both lhs and rhs by 5

so lhs will only have y

hold on, ill fetch ur ans from ms paint

It’s alr I actually should try to learn it

yeah just explaination

nw

going downwards

using the formula y=mx+c since m is negative, the slope will go down, and c is the y intercept, meaning where the slope meets the y axis. since c=2, the slope will meet y at (0,2) coords.

How do u find the slope of that

Would it be like 1/2 for number 5

just plot the ordered pairs

x,y => (5,10) then (10,20) like that

on a graph

No like how do I find the slope on a table

uhh like wdym by "on a table"?

Is the slope 10/5?

.close

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[low orchid]

Hey guys, I'm trying to solve this differential equation, and I'm stuck on integrating sinx/(1 -cosx)

[low orchid]

Sorry it's kind of messy

[mystic saffron]

Let cosx= u

Well 1 - cosx = u substitution works as well

I’d choose the second sub

[low orchid]

ok cool ill try that

and im on the right track right, for solving?

[stoic cosmos]

hey im new on the server are there any rules for asking help

im a french 1st year med student and theres a lot of maths in the program which i have trouble understanding

its really the basics such as limits and graphs and stats but i cant get to understand it

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[mystic saffron]

what other methods exist besides base*height/2 and heron's formula for calculating the area of a triangle?

[mystic saffron]

wait nvm

.close

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[mystic saffron]

graphical, right?

[mystic saffron]

reopen

i got it thanks

my dumbass forgot google exists

.close

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[fading snow]

yo

[fading snow]

so I solved a question and have a question

I put x^1/2 in the denominator section. Does this make sense?

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[heavy yacht]

i want to prove the bijectivity of a translation, but the example i have is very vague- what should a proof look like, even?

[low locust]

you could find the inverse function

[heavy yacht]

that's a part of the question, yeah. but the example i have starts with $(x_1+1, y_1+2)=(x_2+1, y_2+2)$ without much explanation as to why we're assuming that

[clever fjord]

WayneTundra

[low locust]

presumably they want to show injective and surjective separately

[heavy yacht]

i get that it needs to be one-to-one, i just don't know what that proves

[low locust]

bijective=injective+surjective. so if you have shown both of those then you are done

[heavy yacht]

ok. surjective= onto, right?

[low locust]

yes

[heavy yacht]

ok, cool. i still don't grasp what this algebraic argument is achieving though, or rather, how it's proving what it's claiming it proves

[low locust]

it starts with f(a)=f(b) and shows that this implies a=b

which is the def of injective

[heavy yacht]

isn't that basically just saying "x is x because it's x"?

[low locust]

no

[meager kelp]

Sorry to chime in and if I create more confusion.
What f(a)=f(b) => a=b means that if you propose two inputs to a mapping, that if they have equal outputs. Then your two proposed inputs are actually equal. Which is to say then that it maps distinct elements of its domain to distinct elements of its codomain. Or each element of the codomain is the image of at most one element of the domain.
Does this make sense?

[heavy yacht]

i think i follow the logic, yeah

and to prove the surjectivity i'd compare the image set?

[low locust]

show that every vector in R^2 is in the image set

[meager kelp]

Or for every possible output you can find some input that maps to it

[heavy yacht]

ok, sweet. thank you both for the help

.close

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[north moth]

can someone help me out with this related rates question

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[north moth]

<@&286206848099549185>

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[inland rain]

Hello I have problem with mixing some fluids.

I normally have 8 bottles of 10 ml, with 3mg nicotine. They cost, 39 each.

I can get 1 bottle of 0mg/10ml for 20.
I can get 1 bottle of either 12, or 9, or 6, or 3 mg/10 ml for 39 each.

Wouldn't it be cheaper to mix some bottles of 0mg with some bottles of 12 mg to get a final end result similar to my normal 8 bottles?

[odd edge]

@inland rain Has your question been resolved?

[inland rain]

I'm assuming 3 bottles of 0mg to 1 bottle of 12mg would result in 4 bottles of 3mg when mixed?

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@inland rain Has your question been resolved?

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@inland rain Has your question been resolved?

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@inland rain Has your question been resolved?

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@inland rain Has your question been resolved?

[odd edge]

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[jovial cape]

im tryna factor

[jovial cape]

Is there a number that adds to -2 and multiplies to -80

Im thinking no but idk cuz this is the third question with no answer

[upbeat drum]

no yeah

try solving a+b=-2 and ab=-80

[wooden python]

do you mean a pair of numbers like that?

[jovial cape]

yeah

[jovial cape]

no like there is no answer?

im honestly so out of it ive been doing hw for a min so im confused

[upbeat drum]

there's an answer

[jovial cape]

Could yall help me out im not seeing it

Cuz

If i want to get -80 i need 2 negatives or 2 positives

but if i do that then i cant add to get -2

[wicked kestrel]

the product of two negatives or two positives is positive

[wooden python]

no?

quite the opposite

to get -80 as a product you need one pos and one neg

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@jovial cape Has your question been resolved?

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[static lance]

Couldn't solve Q27,28

[static lance]

For question 27 number of outcomes wouldn't it be 9C6 + 9C3?

[wicked kestrel]

It would not. Where is that coming from?

[odd edge]

@static lance Has your question been resolved?

[static lance]

I thought it would be 9C6 would give all possible out comes of getting 6 heads and we add it to 9C3 which is all possible outcomes of tails, but I'm not sure

[wicked kestrel]

9c6 indeed gives the number of ways of getting 6 heads. However, if there are 6 heads, there are already 3 tails

[static lance]

That was dumb by me.

What about question 28?

[wicked kestrel]

Have you heard of the binomial distribution?

[static lance]

I took it in previous stat course, but we always used excel for it.

[wicked kestrel]

Well 28 is solved by directly using that distribution

[static lance]

Yeah I found the formula, thank you so much.

[wicked kestrel]

[static lance]

.close

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[languid drum]

Hi, for rational inequalitities, my question nis (x+1)/(x-1) is greater than 0

[languid drum]

we set the denominator to 0 in order to prevent the denom from being 0

and the numerator to 0 because the zeros of the numerator is the zeros of the functions right>

?

(I want to know if my reasons are correct)

[pallid tinsel]

r u trying to find values of x such that (x+1)/(x-1) > 0 ?

[languid drum]

yes

[pallid tinsel]

yeah u must look at where ur denominator becomes zero

and at the zeroes of the function

and look at the behavior of the function within all these points

[languid drum]

YESS
T

THANK YOU

.close

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[modest iris]

hi. i solved it but i am not sure about it. is it correct? and if there is an easier way i hope you tell me 🤍

[modest iris]

the question is to prove algebraically that each one is inverse function to the other

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[mystic saffron]

0=(3x+2)^2

[mystic saffron]

there are two ways to solve this

one is to just put -2/3 as x

and the other is a binomial formula

where one gets x1=0 and x2=16/9

which is right and which is wrong and why?

<@&286206848099549185>

[void yew]

what are $x_1$ and $x_2$

[clever fjord]

992qqoloy

[mystic saffron]

I said

0 and -16/9

but I could also just put in -2/3

that's the question

which method is right?

[void yew]

No but how do they show up

[mystic saffron]

binomial formula

0 = 9x^2 + 12x + 4

[void yew]

Oh ur saying x1 and x2 are the roots

[mystic saffron]

0 = x ( 9x +12) + 4

x1=0 and x2=-16/9

[mystic saffron]

yup

but I could also just put -2/3 as the root

so I'm wondering whether both are correct?

[void yew]

no

[mystic saffron]

good

how

[void yew]

No idea how ur getting that the roots are 0 and -16/9 from this yet

Oic

OK so

[mystic saffron]

what

I literally said

[formal coyote]

Δεν καταλαβενω τιποτα

[mystic saffron]

0 * (9 * (-(16/9))+12)+ 4

[void yew]

Hold on I understand now

That's 4

And the other one is 4*16/9 + 4

[mystic saffron]

how

[void yew]

0 cancels out everything but the 4

[mystic saffron]

ah

so that's why it isn't correct

[void yew]

Well now ur putting both roots in

[mystic saffron]

so what you're saying

is that -2/3 is the correct solution

[void yew]

Yeah

[mystic saffron]

because 0 * (16/9*9-12)

equals 0

+4

equals 4=0

which is false

[void yew]

Yeah

But

It should be 0*(0*9 - 12) + 4 = 4

[mystic saffron]

why

lol

the question is to find the spot where the tangent is horizontal

aka f'(x)=0

[void yew]

And for the other "root" not root you'd evaluate (-16/9)(-16/9*9 - 12) + 4 and see that that's also not 0

[void yew]

cus u pkug in for $x = 0$ to evaluate $f'(0)$

[mystic saffron]

yeah I remembered now that the rule only applies if what is in the parenthesis equals 0

[clever fjord]

992qqoloy

[mystic saffron]

which it doesn't

so it would technically be correct if the 4 weren't there but then with -12/9

0 * (9*-12/9+12)

totally valid

but with the +4 no

[void yew]

No

Wait

[void yew]

OK at ur expression, it's x*(9x - 12) + 4 originally

[mystic saffron]

bro

why are you making problems when there are none 😭

I understand this fully

let's move on to the next one can we?

[void yew]

I guess this would equal 0 but

There are tho bro

[mystic saffron]

exactly

0 =0

[void yew]

Ur failing to understand how to substitute in values into variables

[mystic saffron]

we're done

[mystic saffron]

Satz des nullprodukts

dunno what that is in English

but that's the rule I'm using

it totally makes sense

[void yew]

No the problem is that 0*(9*(-12/9) + 12) isn't the original expression but without the 4 and with 0 substituted into the x's

[mystic saffron]

I know that...

I said that if it were that

it would be correct

but it isn't, so it isn't

[void yew]

Now idk why you're putting in both a 0 and a -12/9

Ugh nvm can't get thru u just fail basic algebra and let's move on

[mystic saffron]

bro what

I think you just completely misunderstood me

I said this:

if the statement

WERE

[void yew]

idc

Move on

[mystic saffron]

x ( 9x - 12)

WERE

[void yew]

We're done

[mystic saffron]

it ISN'T

if it WERE that

then 0 and -12/9 would be correct

but it IS NOT

NOT

⛔

[void yew]

YES BUT YIU DIN'T PUT JN BOTH AT ONCE IDIOG

[mystic saffron]

so it isn't correct

got it?

[mystic saffron]

WDYM

WHY DO YOU HAVE SO MANY TYPOS

TYPE SLOWER

[void yew]

U either do 0(9*0-12)

[mystic saffron]

bro satz des nullprodukts

[void yew]

Or (12/9)(9*12/9 - 12)

Oml

[mystic saffron]

no

you're wrong

lol

[void yew]

<@&286206848099549185> help me find a way to explain to him how to plug in numbers into expressions

[mystic saffron]

the original statement is f'(x) = 1/3 (3x+2)^2 * 3

aka (3x+2)^2

making that 0 = 9x^2 + 12x + 4

which I could turn into 0 = x (9x +12) + 4

and if that +4 WEREN'T there

then I could just do x1 = 0 and x2 = -12/9

[void yew]

Idc about the original statement im referring to this

But u also made the mistake with the original statement too

See how u put in 0 in one instance of x

And - 12/9 in the other instance of x?

[mystic saffron]

yes

satz des Nullprodukts

[void yew]

You're supposed to put in the same number into both instances

[mystic saffron]

no I am not

for the third time

[void yew]

🤦‍♂️

[mystic saffron]

are x1 and x2 the same here?

[void yew]

Yes those are the roots

No but when u pkug in to check if they are root in theory

U should be plugging in the same number into each instance

Look, imagine if instead I told u to find p(4) when p(x) = x^2 + x + 1. I put in the 4 into x^2 and the 4 into x

Ur supposed to evaluate p(x_1)

And p(x_2)

It just happens to work out the way ur doing it anyhow but

I'm pointing out that ur using the wrong reasoning to get there

If instead I said p(x) = ax^2+ bx + c

And found x1 and x2 to be roots

I wouldn't calculate a(x1)^2 +b(x2) +c would I?

[mystic saffron]

I have no idea what you're saying rn

instead of doing that

explain to me how the zero product property works

[void yew]

🤦‍♂️

[mystic saffron]

this

[void yew]

Ur stupid I'm done bye

[mystic saffron]

💀

you really shouldn't be teaching anyone anything

[void yew]

Lmao

Ur by far the stupidest person I've ever helped in this server

I'm sorry

But it's true

[mystic saffron]

oh no

I just got called stupid by some random dude on a math server

😨

[hoary burrow]

Whats goin here

[mystic saffron]

what am I gonna do now?

[mystic saffron]

dude can't explain so he just started calling me stupid 😭

[void yew]

Ugh just forget it this dude can't understand how to calculate f(x)

[mystic saffron]

and he can't type

[hoary burrow]

Those are possible vaule of x

[void yew]

Not just stupid btw, I'm seriously not joking I've helped hundreds of people by now and none of them have been as obstinate

[hoary burrow]

Because if each one of term is divided to zero then two possible values come

[mystic saffron]

get a life man

[mystic saffron]

yeah

that's the logic I used

[mystic saffron]

all you are is some math nerdneck who lives off his ego because of 2 roles on a discord server

is that really what you wanna be?

[hoary burrow]

Then on what condition argue is going

[void yew]

lmao idc about my roles

[mystic saffron]

he's saying that what I'm saying is wrong

I don't get why

[void yew]

OK look

[mystic saffron]

so he just starts going stupid

[void yew]

I'll show u bob

[mystic saffron]

"in all of my years of being on this discord server" bro idc

instead of just explaining the issue he just starts going off

[hoary burrow]

[mystic saffron]

signs of childhood trauma or something

[lethal bronze]

wtf

[void yew]

This is the first place he makes the mistake

[mystic saffron]

solving homework with dad went south it seems

[mystic saffron]

I literally said that that was a mistake and explained why

🤦‍♂️

[void yew]

The expression is

[hoary burrow]

0+4=4

[void yew]

$x(9x + 12) + 4$

[clever fjord]

992qqoloy

[mystic saffron]

yes and I literally said that

[void yew]

But he puts in 0 for the first $x$

[clever fjord]

992qqoloy

[void yew]

And $-16/9$ for the second x

[clever fjord]

992qqoloy

[hoary burrow]

Wait can you show me real question

[mystic saffron]

right here

[void yew]

The question is fine we resolved that part

[mystic saffron]

there is no question...

I got it already

that's what I'm saying

this dude is literally just tryna argue

[void yew]

But for some reason he can't understand how to calculate p(x1) or p(x2) properly

Or p(y) or any substitution probably

[mystic saffron]

4x^2 +8x can be represented by x ( 4x + 8)

that's all I'm saying

using this

[hoary burrow]

Yes taking common

[mystic saffron]

x would be 0

and the other x -2

just using pure logic

[hoary burrow]

Yes

[mystic saffron]

🤯

almost like that's what I've been saying all this time

crazy stuff man

but hey I'm stupid and all that

[void yew]

I odnt have an issue with that part

[mystic saffron]

I don't care anymore

you ruined my mood

I'll just solve this alone

thanks for nothing

[void yew]

The issue is that u write stuff like $0(9*(-16/9) + 12) + 4$

[clever fjord]

992qqoloy

[void yew]

U don't put x1 into one x and x2 into the other x

[mystic saffron]

you really don't get it do you?

then what do I do?

[mystic saffron]

that's literally what I did here

and it's correct

[void yew]

U write x1 into both instances of x

Yeah I also mentioned that

Ur getting the right answer but the wrong way

It only works out here because

[mystic saffron]

the right answer is -2/3

we said

[void yew]

Ur multiplying by factors and it happens to be the root of a factor

Yeah I agree with that much

[hoary burrow]

Another bracket if forget to close

[void yew]

OK look say we had like $p(x) = (2x +3)(5x +6)$. Then $x = -3/2$ is one solution right?

[clever fjord]

992qqoloy

[void yew]

So if I put in -3/2 it'll evaluate to 0

But that's because the 3x + 2 factor evaluates to 0

So on the second x u can put whatever u want and it'll still evaluate to 0

For instance u could write $(2(-3/2) + 3))(5(84746366374744) + 6)$ and that equals 0

[clever fjord]

992qqoloy

[mystic saffron]

but that's not what the zero product property says

the goal isn't just equaling 0

the zero product property says that both sides must be equal to 0]

that's why

[void yew]

Yeah? Both sides would be equal to 0

Cus u have 0 times a number on one side

[mystic saffron]

no no

I don't mean 0 = 0

I mean the x * (x)

the parenthesis must equal to 0

which it doesn't here (clearly)

[hoary burrow]

Abosoluty by factor theorem

[void yew]

ok what's the point of having both parentheses be 0?

[mystic saffron]

I repeat

zero product property

have you never heard of it

[void yew]

not the way ur saying it

Oh

Well

I googled it and it says the product of two nonzero elements is nonzero?

That doesn't explain why ur plugging in $x1$ and $x2$ simultaneously but

[clever fjord]

992qqoloy

[hoary burrow]

You mean this question has said to find factor of p(x)

[void yew]

OK so the zero product property says that for the product to be 0 either factor must be 0

That doesn't mean both factors have to be 0 simultaneously

[mystic saffron]

bruh

what does it mean then

[void yew]

That either one factor or the other is 0

Maybe both if they're equal fsctors

[hoary burrow]

So this is the question

Lol

[mystic saffron]

yes 😭

[hoary burrow]

Only one solution will come

[void yew]

No we already resolved that part

[mystic saffron]

-2/3

[hoary burrow]

Yes

Then question you guys are arguing on

[mystic saffron]

bro honestly

idk anymore

seriously idc

.close

[odd edge]

Channel closed

Closed by @torn swan

Use .reopen if this was a mistake.

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[amber fable]

Vikas Gupta?

[woeful ravine]

Yes

[upbeat drum]

reverse product rule

[woeful ravine]

By parts?

[upbeat drum]

not really, just to notice that derivative by product rule

[amber fable]

Integration by parts may be one way to do this question. But there is another way too.

We know integral e^x ( f(x) + f'(x) ) dx = e^x f(x). Something similar in this question.

[odd edge]

@woeful ravine Has your question been resolved?

[amber fable]

Did you get it? Just substitute
x lna = Y and do integral with respect to Y. Or you may also use integration by parts.

You get same answer.

Or in worst case during exam, if nothing comes to mind just differentiate the options.

[woeful ravine]

I am thinking of using integral f(x)+xf'(x)= xf(x)

[amber fable]

That may also be a way, but kind of hard to see it this way. The other way is little easier to see. You can try to do something like this if visible.

[woeful ravine]

.close

[odd edge]

Channel closed

Closed by @woeful ravine

Use .reopen if this was a mistake.

[heady dragon]

i need help

[royal herald]

#❓how-to-get-help

[odd edge]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[solar burrow]

in MIT classes they call this taylor formula, on kahn academy they say it's MacLaurin formula, because it's centered at zero.. but here it's also centereed at zero because f(0)

[solar burrow]

can anyone clarify this?

[solar burrow]

that's what i said, so why they call it taylor?

[upbeat drum]

taylor series can be found at any x value

[solar burrow]

is the formula different then?

[upbeat drum]

slightly

[solar burrow]

oh, ok so it's not really correct what they say in MIT

ok

[upbeat drum]

well a maclaurin series is still a taylor series

but not all taylor series are maclaurin series

[solar burrow]

and, just so i understand the idea of this

i can transform any function in a series passing it throught this formula, right?

ideally

i'm a bit confused they way they use the formula for exmaple for find the series for sin and cos