help-33

[marsh citrus]

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[sudden sky]

im not able to get to the answer

[slender socket]

For the reply that was just deleted: It was correct, but also unhelpful without more labels on it

[knotty trellis]

it was not helpful, because we dont know whether the segment ending at P extends towards center

and it doesnt I think

[sudden sky]

yeah it probably doesnt

[slender socket]

I think it's implied that it does otherwise it's unsolvable

[knotty trellis]

I got a new sketch

just labelling it...

[sudden sky]

i got here

[sudden sky]

ohh alright

[slender socket]

Nevermind, I missed the arc length constraint

[knotty trellis]

maybe this could help

[sudden sky]

i dont know how to proceed after this

[knotty trellis]

and the angle must be right, because it's 1/4 of the arclength

so quarter circle

Like this

[sudden sky]

yepp

[knotty trellis]

with this, we could at least find the radius

[sudden sky]

ohh i somehow got the radius as sqrt(19) but I'm not sure about it

[knotty trellis]

let me check that

[sudden sky]

okayy

[knotty trellis]

,w 32 + 4 - 2*4*sqrt(2)*2*cos(135deg)

[elfin berry]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by methisalwaysright

[knotty trellis]

...

32+4 is 36

[sudden sky]

😭

[knotty trellis]

I think it's not sqrt(19)

,w 32 + 4 - 24sqrt(2)2cos(135deg)

[elfin berry]

Results provided by WolframAlpha

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Requested by methisalwaysright

[knotty trellis]

huh

oh, yeah

[sudden sky]

i got XY as sqrt(38) but idkk

by using loc

[knotty trellis]

I tried using it here

[sudden sky]

ohh

[knotty trellis]

,w (4sqrt(2))^2 + (2)^2 - 2*4sqrt(2)2cos(135 deg)

[elfin berry]

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Requested by methisalwaysright

[slender socket]

The final answer is ||sqrt(10)||, working on the drawing now

[knotty trellis]

yeah, im getting sqrt(52) instead of sqrt(38)

[sudden sky]

ooo

waitt

i did something wrong in my calculations its not sqrt(38)

then R= sqrt(26) ?

[knotty trellis]

yeah, I think so

Unless I made a mistake, I got a different answer

than sqrt(10)

@slender socket Do you have a drawing?

[slender socket]

I just did it numerically to a ton of digits and it matched, I might be measuring the wrong thing

[knotty trellis]

kay, i might be doing something wrong then

[knotty trellis]

3.13785816221
This is my result (approximately) lol

extremely close

[sudden sky]

ohh

[knotty trellis]

I might have made some tiny error

by a unit or sth

[knotty trellis]

Can you show your solution btw?

if you have one

other than the numerical

[slender socket]

I don't yet but I have to go for now, so someone else can take this and use it to figure it out:

[knotty trellis]

,calc sqrt(26)

[elfin berry]

Result:

5.0990195135928

[knotty trellis]

,calc sqrt(52)

[elfin berry]

Result:

7.211102550928

[knotty trellis]

,calc 65/13

[elfin berry]

Result:

5

[knotty trellis]

oh

im getting sqrt(10) too now

my approach was setting up this x and y

annd setting up 2 equations

[sudden sky]

ohhh

[knotty trellis]

with these 2 triangles

(26-y)^2 + x^2 = 32

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

[sudden sky]

oh i got it!!

thankk youu so much!

[marsh citrus]

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[marsh citrus]

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[obtuse plume]

So I'm trying to get an easy way of calculating big numbers

It's a part of my physics course, but the equation is k * 10^n where k is a number between 1 and 10.

so from what i understood is that if I were to go from Tera to Nano I could do this here to make the conversion between the two. (k*10^12) x (k*10^-9) = k*10^12+(-9) = k*10^3

Now my other question is that let's say k = 0.69GwH, we'd need to make 6.9 out of it since k > 1 < 10. How could I convert this the easiest way? if I move the comma from 0.69 to 6.9 do i need to increase the exponent?

[obtuse plume]

Please tell me if I've explained it in a bad way

[marble grove]

if you move the comma from 0.69 to 6.9 then you've multiplied by 10. You have to cancel it by dividing by 10 so you substract 1 from the exponent

[obtuse plume]

thank you

.close

[marsh citrus]

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[marsh citrus]

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[lapis shale]

how do I solve this?

[lapis shale]

I tried to open the binomial but then do I need to take cases for when the power of x is 3 and 4?

because then wouldn't I end up with 3 cases for each?

[calm harbor]

please don't

[lapis shale]

please don't?

I meant opening in the sense of writing the general term

[raw hawk]

Don't you only need the first five terms of (1-2x)^18

[lapis shale]

but then I'd have sit and calculate all of them right?

[raw hawk]

?

[lapis shale]

I'd have to find out 18 C (1 - 5) right

[still temple]

yes

[lapis shale]

the solution given here is solving it using the general term

[raw hawk]

Yes

[lapis shale]

honestly I don't even know hwo its solving it

[still temple]

can you send it ehre

[still temple]

thats the same thing the multiplied 1 to coeff of x^3 a to x^2 and b to x

and the sum is zero

ax when multiplied to the 2nd degree term gives a third degree term,bx^2 when multiplied to the 1st degree term gives a third degree term

[lapis shale]

ah so basically we're taking the three cases summing them up and equating them to 0?

[still temple]

yes

[lapis shale]

ah gotcha

[raw hawk]

Yes

[lapis shale]

I thought there was a simpler way of solving this

well thanks

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[marsh citrus]

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[wide sage]

general question, for a 2 variable function, does the gradient vector represent the direction that produces the maximum rate of change

[wide sage]

yup, thanks!

will do

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[marsh citrus]

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[marsh citrus]

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[clever tangle]

Why do we use this specific vecotr field

[marsh citrus]

@clever tangle Has your question been resolved?

[marsh citrus]

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[proper locust]

Integrate f = y = √x over the triangle region cut from the first
quadrant of the xy-plane by the line x + y = 1.

[proper locust]

can someone solve this in both x and y

i get different answers in both

<@&286206848099549185>

[elder crag]

is this what you need to find?

@proper locust

blue line being x+y=1

green line being sqrt(x) = y

[proper locust]

umm sorry but i dont actually know

[elder crag]

huh, lemme check online as im not sure either

[proper locust]

the ans is 4/15

[eager tinsel]

!show

[marsh citrus]

Show your work, and if possible, explain where you are stuck.

[grim hemlock]

Are you sure there aren't any z's involved in the question?

[proper locust]

yeah

no z's

[silent marten]

I am sorry can you tell me what is the question?

[grim hemlock]

I assume are are currently busy with multivariable calculus?

[proper locust]

its under the topic of double integration

yeah

[grim hemlock]

I think the question was supposed to be f = z = sqrt(x) and if so then I agree with this answer

[silent marten]

That makes more sense

I was a little bit confused too

[proper locust]

maybe

i not sure either

but i get the same ans on chatgpt

for both the variable x and

y

if the function is y = x^1/2 , we should be able to integrate it as both dA = dxdy and dA = dydx

but i get the wrong answer solving it as dydx

[grim hemlock]

Assuming they meant f = z = sqrt(x) then this should change if you want to solve it as dydx

[proper locust]

hmmm

why?

[grim hemlock]

because when you change the order of integration, you don't change the actual function you integrating. How did you get from sqrt(x) to y in the first place? From the question saying f = y = sqrt(x)? I feel like that must be a mistake and in my answer I said that I assumed that what they meant was f = z = sqrt(x)

[proper locust]

hmmm

ok

i think there is a typo

thank you

.close

[marsh citrus]

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[marsh citrus]

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[near elbow]

Hi guys, i don’t understand how they get that (highlighted yellow)

[late geode]

factorisation

same as factorising quadratics

[dapper aspen]

1+tan2x=sec2x

just sub tanx=y or something

and solve ig

then when u get y values

solve tanx

!occupied

[marsh citrus]

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).

[near elbow]

Oohhhh

Okok thankuuu sm guys

.close

[marsh citrus]

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[marsh citrus]

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[red phoenix]

can someone help with this?

Permutations and combinations

[knotty trellis]

Do you need examples? Or do you have a specific question?

[red phoenix]

i dont understand the topic

like what they mean and what it is about

[dapper aspen]

u dont understand the entire chapter?

[knotty trellis]

hmm maybe it would help if you just continued reading

[thorny bison]

It's about counting things

[red phoenix]

yeah

[red phoenix]

oh

[red phoenix]

i did

[thorny bison]

with permutations we count arrangements considering that order matters meaning we take (a, b) and (b, a) diff arrangements
With combinations we do the same but considering order doesn't matter

[knotty trellis]

Let's say that we have letters
A, B, C, D, E

Then A, B, C is a permutation of 3 objects.
This permutation is different from let's say B, A, C. Because in permutations, order matters.

On the other hand, if we take A, B, C as a combination of 3 letters, then it's the same combination as B, A, C or C, A, B

[red phoenix]

ok i get both

from display and math is always right

[thorny bison]

Il suggest you to read combinatorics from ctpcm

[knotty trellis]

okay, it would be helpful if you asked some more specific question if you have one

[red phoenix]

can u explan this page:

both

most ly this

[knotty trellis]

what exactly do you not understand about it?

the concept itself, the formula, or what?

[red phoenix]

the 2nd line

"so if we could choose 4 times then the arrgentmet would be..."

[knotty trellis]

If we have n=3 objects (say digits from 1 to 3), then we can make 3 x 3 x 3 x 3 different arrangements of length 4 of those objects, given that we can repeat them.

Here are some examples:
1, 2, 3, 3
1, 1, 1, 1
3, 2, 3, 1

[thorny bison]

Think about it you have to make all 3 digit numbers let's say so you have 3 spaces _ _ _
And you can use any number from 0-9 on that space so each space has 10 choices which are independent of each other meaning if I choose 4 for the first place it won't affect my choice of the 2 nd or 3rd place so total choices = 10×10×10 = 10³

[knotty trellis]

this explains the concept, and display's message should explain the formula

[thorny bison]

[red phoenix]

https://tenor.com/view/ok-mickey-mouse-okay-good-gif-18768605

but it says:

"So if we can choose 4 times, then the arrangement would be 3x3x3x3"

[knotty trellis]

You are making 4 choices of 3 objects

[red phoenix]

does that mean is there are 4 at a time?

[knotty trellis]

for the first choice, you can choose 1 of 3 objects

for the second choice, you can also choose 1 of 3 objects

so for the first 2 choices, there are 3 x 3 arrangements

[red phoenix]

so Im choosing the arrangemet of 4 possible things

so i have 4 choices??

[knotty trellis]

you are choosing length 4 arrangement of 3 things

such as the example with digits

there are 3 different objects, 1, 2 and 3

and you make 4 choices

and the result is a sequence of length 4 of those digits

[red phoenix]

lenth?

[knotty trellis]

if you choose 1, then 2, then 1 and then 3, you get 1, 2, 1, 3

and that's one possible arrangement

[red phoenix]

oo

[knotty trellis]

4 choices from 3 objects

repetition is allowed

[red phoenix]

so this means 4choices at a time

[knotty trellis]

and you have 3 options for each choice, so the total number of arrangements is 3^4

[red phoenix]

like if its a pass word is a 4 diget password with the options of 1,2, or 3?

[knotty trellis]

it means that you will make 4 choices from the set of 3 objects

[knotty trellis]

yes, that would work

[red phoenix]

ok

can you explain this part?

[thorny bison]

If u can't understand for 4 take smaller cases
Btw the question is saying that you have 3 Choachlates A B C and u need to arrange them in 4 spaces
_ _ _ _ each place has 3 choices independent of each other meaning
If I choose a for first it won't affect what I choose for second so total cases will be 3×3×3×3

[knotty trellis]

"Choachlates"? lol

[thorny bison]

Chocolate*

[thorny bison]

Yea mb

[knotty trellis]

Yeah, so it says that if there are n objects to choose from, and you are supposed to make r choices, then there is total of n^r arrangements

[red phoenix]

it will effect the 2nd right cause then u cant use athe 1st on agoins?

*again?

[thorny bison]

They have generalized the chocolate prob - so now you have n chocolates and r spaces so each soace will have n independent choices so total choices is going to be n^r

[thorny bison]

Space*

[red phoenix]

so its nuber of objects^choices?

[thorny bison]

Yes
If it is given that repetition of choices is allowed

[red phoenix]

ok

i get that part

[thorny bison]

[red phoenix]

ok so now im on permutions without repetition

so after u use an option it cant be used again right?

[thorny bison]

ok now let's take a previous problem u have 3 chocolates A B C
And u want to arrange them in 2 spaces first space has 3 choices but now since repetition is not allowed and we already have chosen one chocolate for the first space we are going to have 2 choices for the second place = 3×2 choices = 6

[thorny bison]

Yes

[red phoenix]

yea ok

oh ic that its like this

6x5x4x3x2x1

=720

[thorny bison]

yes

[red phoenix]

"a factorial funtion (!) means to mutilply all whole numbers from a given number down to 1"

so

5! = five factorial=5x4x3x2x1

[thorny bison]

Yes

[red phoenix]

ok

[thorny bison]

What will be 0!

[red phoenix]

I thogt 0

but it says 1

[thorny bison]

Ok let me explain
Can i say that 5! = 4! ×5

[red phoenix]

um

no?

oh yeah

u can

[thorny bison]

Ok so can i say 4! = 3! ×4

[red phoenix]

yes

[thorny bison]

So generalizing this argument can i say that n! = n× n-1!

[red phoenix]

so 0!=0x-1

=0

[thorny bison]

Factorials arnt defined for -ve numbers so u can't say 0! = 0× -1!

[red phoenix]

oh

[thorny bison]

Rather what u can say is that 1! = 1×0!

And we know 1! = 1

uh?

[red phoenix]

oh yeah

[thorny bison]

So 0! = 1
U can also do it with help of gamma function

[red phoenix]

ok

[thorny bison]

This is the gamma function and it is used to define factorials

[thorny bison]

Yup that's pretty much it

[red phoenix]

wait

[thorny bison]

Yes

[red phoenix]

how does thiswork

[thorny bison]

It is the no of arrangements of n objects into r places
So thinking about this no of choices for the first place is n
Then for the second place it's n -1
Then for the 3rd place it's n-2 .......
For the rth place it's n - (r+1)
So total ways of arranging = n × n-1 × n-2....... n-(r+1)
Which is as good as writing n! /(n-r) !

[red phoenix]

what?

[thorny bison]

U have n object and r places ok

[red phoenix]

i kinna get it

its ok

but look

[thorny bison]

Ok

[thorny bison]

Yea what's the problem?

[red phoenix]

so why its 5x4x3 and not 5x4x3x2x1

nvm

[thorny bison]

Because u only have 3 spaces and 5 objects so by my original argument we will have 5×4×3

[red phoenix]

my way worked

[thorny bison]

[red phoenix]

so now

wait

is combinations with out repetiion possible? @thorny bison or @knotty trellis ?

[thorny bison]

Combinations are without repetition

[red phoenix]

i thoght with were with

[thorny bison]

?

[red phoenix]

i thoght

[red phoenix]

combination are with repition

opps

[thorny bison]

Nope

[red phoenix]

"a combination i s a group of objict in which orber does not matter."

oh ok

[red phoenix]

how does this work?

@thorny bison

<@&286206848099549185>

[thorny bison]

Ask @lucid heron I'm a little busy rn sry

[red phoenix]

ots ok

@lucid heron

<@&286206848099549185>

[spark siren]

stop pinging to often.

your question is?

[red phoenix]

sty

*sry

id get this

,, _n P _r

[elfin berry]

Pl@nnΣd∀φd⋔MDCXXI

[red phoenix]

permution formula

[spark siren]

can you be a little bit mor specific? what do you net get?

[red phoenix]

how is it 5x4x3

[spark siren]

you have 3 positions, right?

[red phoenix]

yes

[spark siren]

you have 5 ice cream flavors?

[red phoenix]

yea

[spark siren]

how many options of ice cream do you have for the first position?

[red phoenix]

5

so then

[spark siren]

well. how many for the second position?

[red phoenix]

4

[spark siren]

and for the thrd?

[red phoenix]

3

5x4x3

[spark siren]

thats it.

[red phoenix]

what does that have to do with this?

,, _n P _r

[elfin berry]

Pl@nnΣd∀φd⋔MDCXXI

[red phoenix]

how its thr formula

[spark siren]

this is just a notation - its a sequence of characters. the meaning of this notation depends on how you have defined it.

[red phoenix]

so how its used?

[spark siren]

again. its a notation. the meaning depends on hwo you ave defined it. in life insurance nPr means premium at age r for period of n years.

look at your book, where this notation is defined

[red phoenix]

so how do i use this?

[spark siren]

ok, assume you have a choice of ice flavors: strawberry - vanilla - hazelnut. is this - in the context of your example - the same as vanilla - hazelnut - strawberry?

[red phoenix]

no its combintaion made when 4 numbers are selected out of 8 numbers

[spark siren]

ok, you are switching the example. not very helpful. anyway. you need to understand if the order matters or not.

[red phoenix]

ok

[spark siren]

in the ice cream example: you have a solution 5 tines 4 times 3., so strawberry - vanilla - hazelnut is different to vanilla - hazelnut - strawberry:

If the order does not matter, which means strawberry - vanilla - hazelnut should be counted as the same as vanilla - hazelnut - strawberry then you need to take into account on how many ways you can arrange 3 options. In this case you get 5x4x3 divided by 1x2x3.

[red phoenix]

Ok

[spark siren]

choosing 4 numbers out of 8 is just the same as the ice cream example. 8 options for the first position, 7 for the second .... gives you 8x7x6x5 - if the order matters. if the order does not matter ist 8x7x6x5 divided by the number of posibilites you have to count as one .reorder 4 -> 1x2x3x4.

[red phoenix]

Oh ok ic

Thx everyone

.close

[marsh citrus]

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[quiet tree]

do the logarithmic identities cause equations to be implications if they're used on logarithmic expressions that don't contain variables?

[silver mantle]

i think it's just equivalent

so $\Updownarrow$

[elfin berry]

pixel

[quiet tree]

Well from what I've heard, when we apply these kinds of identities to equations, we may end up losing answers since they only apply for x values above 0

[silver mantle]

yeah, but 3>0 right?

so we don't lose solutions

[quiet tree]

Your explanation does make sense

Okay, so unless we apply these logarithmic identities on logarithms with variables, we shouldn't lose any values and we get to use the equivalent arrow

[silver mantle]

yes

[quiet tree]

This is a case where it wouldn't be valid. In fact, it'll produce wrong answers

[silver mantle]

no

we have x^2=9

so $x=\pm3$

[elfin berry]

pixel

[quiet tree]

yeah i've emphasised that it's wrong

[silver mantle]

oh i'm sorry

[quiet tree]

i also seem to have messed up the application of the identity, too

[silver mantle]

i read it wrong

yes

[quiet tree]

we may actually forget about this as it's utterly wrong to the point. I didn't even add correct things

[silver mantle]

so is your problem resolved?

[quiet tree]

yeah the problem is solved, thanks a lot for clarifying

.close

[marsh citrus]

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[marsh citrus]

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[lilac light]

which one do you want help with

[stiff hare]

the first one

[lilac light]

first isolate the variable

so -5k

2/5+8k-1=-3/5

[stiff hare]

right

[lilac light]

did you get stuck when you got 8k=0

[stiff hare]

yea

[lilac light]

k=0

[stiff hare]

oh

ok

[gritty hawk]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:solve-equations-inequalities/x2f8bb11595b61c86:linear-equations-variables-both-sides/v/solving-equations-with-the-distributive-property-2

you should practice these

[stiff hare]

alright i’ll check that

thanks

.close

[marsh citrus]

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[indigo karma]

how do i integrate this

[acoustic bramble]

I would first split it up into 2x/cosx and (x^2*tanx)/cosx

[indigo karma]

then

[acoustic bramble]

Simplify it and use the integration by parts method

[naive fox]

Hint: You don't need to integrate 2x/cos(x) directly. It would be almost impossible to do so. But you can try using by part and see something nice. Another hint would be d/dx tan(x) = sec(x)tan(x). You already have sec(x) in there somewhere

[marsh citrus]

@indigo karma Has your question been resolved?

[marsh citrus]

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[steel musk]

can anwo help me w my homework:>

[marsh citrus]

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

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

[steel musk]

1

our teacher wanted us to study word problems in advance

[still temple]

For 1, let's do the problem from back to front

So, the final result is 4/3, ok?

[steel musk]

yeahh

so = 4/3?

[still temple]

Ok

[steel musk]

this is supposed to be rational equations btw

I just need help for number 3

[still temple]

I've just realised that solving it from back to front is not such a good idea here

Ok, initially we have $\frac{x+2}{x}$, right?

[elfin berry]

ScaleOk

[steel musk]

yeaa

[still temple]

What did the problem do next?

[steel musk]

I think its = 4/3 then cross multiply

[still temple]

No, I meant, the problem says clearly that they take the fraction and sum the numerator and denominator by 3

So, we need to do that with x+2/x, right?

[steel musk]

yeah...

[still temple]

So, what'll we get with it?

[steel musk]

x+2/x = 4/3?

[still temple]

No

Because, before it, there's the sum of 3

[steel musk]

so we add 3 to x+2/x first?

[still temple]

We need to sum the top and the bottom of x+2/x by 3

What'll we get?

[steel musk]

3x + 6/3x?

[still temple]

It is not multiplication

It is sum

[steel musk]

oop mbmb

[still temple]

$$\frac{x+2+3}{x+3}$$

[elfin berry]

ScaleOk

[steel musk]

its just x+5/x+3?

[still temple]

Yes

[steel musk]

then can we put = 4/3

[still temple]

Yes

The problem says to us they are equal

[steel musk]

okai

the lcd is 3(x+3) right?

[still temple]

lcd²

?

what's that?

[steel musk]

like

oh wait shoot

[still temple]

oh, i've got it

[steel musk]

i forgot i had to cross multiply it

[still temple]

Do you know why do you cross multiply?

[steel musk]

because its easier-?

[still temple]

Because understand it is more important than just memorize it

[still temple]

Yes, but how does it work?

Why is that true?

[steel musk]

idk..im just following what our teacher does

[still temple]

I'll strongly suggest you to watch the Khan Academy lessons

They'll teach you the motives behind everything in math

[steel musk]

notedd

ive tried cross multiplying..

[still temple]

Understand is superior to memorize

So, I'll show you how to do it now

[steel musk]

but it gave me 3.857142

i dont think i did it right lol

[still temple]

We have $\frac{x+5}{x+2}=\frac{3}{4}$

[elfin berry]

ScaleOk

[still temple]

So, if we do the same operation to both sides of an equality, will we still get an equality?

[steel musk]

No?

[still temple]

In other words, if X is equal to Y, and I apply the same operations in the same order to X and Y, they will be still the same?

[steel musk]

yes

[still temple]

Excellent

The x+2 term correspond to which math operation?

(My english grammar is bad)

[steel musk]

addition

[still temple]

No, I meant what's this term doing on the fraction x+5/x+2

Addition? Subtraction? Division? Multiplication?

[steel musk]

multiply

[still temple]

The denominator is multiplying?

[steel musk]

oh wait no

its dividing then?

[still temple]

Yes!

So, what's the inverse operation of it?

[steel musk]

of what

[still temple]

Of the x+2 on the fraction x+5/x+2

[steel musk]

x+2/x+5?

[still temple]

So, x+2 is dividing x+5

So the inverse of division is multiplication, right?

[steel musk]

yeaa

[still temple]

So, if I multiply both sides by x+2, will I still get an equality?

[steel musk]

yea

[still temple]

So, multiplying both sides by x+2, we'll get $(x+2) \cdot \frac{x+5}{x+2}=(x+2) \cdot \frac{3}{4}$

[elfin berry]

ScaleOk

[still temple]

Right?

[steel musk]

mhmm

[still temple]

So, as we have x+2 multiplying x+5 and x+2 dividing x+5, so we have the same term doing opposite operations

So, I can cancel out x+2 on the left, can I?

[steel musk]

yea

[still temple]

So, we get $x+5=(x+2)\frac34$

[elfin berry]

ScaleOk

[still temple]

We can do the same process with the 4

[steel musk]

how?

[still temple]

(4 is dividing -> inverse operation is multiplication -> I can operate the same thing in the same order to both sides of an equality and still have an equality -> multiplying both sides by 4 -> 4 is both multiplying and dividing on the right -> Canceling out 4 on the right. On the left, I'll still have the 4 multiplying the left)

Have you gotten it?

[steel musk]

yep

i think ive solved for no 1 alr

can we focus on no 3

[still temple]

So, we'll have $4(x+5)=3(x+2)$

[elfin berry]

ScaleOk

[still temple]

What do we do next?

[steel musk]

then we'll equatw it so its 4x - 3x = -20 +6

[still temple]

Yes

We're done

[steel musk]

so x = -17?

[still temple]

-20+6 is equal to -17?

[steel musk]

-14*

[still temple]

Yes

[steel musk]

its 6/7

if we substitute -14 into our first equation

so i think its right

anw

thanks for all the help^^ defo checking out KhanAcademy maybe after school

.close

[marsh citrus]

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[spiral plank]

im having a massive brain fart but how do you find the taylor series of 1/(x+1) centered at 1?

[pallid tapir]

you can repeatedly take derivatives and evaluate them at 1

[spiral plank]

ugh right lmao

[thorny bison]

Use the formula

[spiral plank]

got lazy lol forgot you can take derivatives

[pallid tapir]

that's so real lol

[thorny bison]

tf is there another way to do it?

[spiral plank]

no, but for other questions involving "important" functions like sinx cosx etc you can generalize all sums into a summation

instead of repeating derivatives

[pallid tapir]

there is like, an nth derivative formula for x^k

[thorny bison]

Ahh ok
Or u could do e^ ln( 1/1+x)

[spiral plank]

hmm that is an interesting though

ill try taking derivatives first though cause it seems like the question just needs the third polynomials

nvm im lowkey stupid

i forgot you can just divide out a coefficient in the denominator and that its fine if the resulting x is not over 1

[marsh citrus]

@spiral plank Has your question been resolved?

Channel closed

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[marsh citrus]

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[crude slate]

Hello, I was wondering if someone could help me with this, I'm not sure what explicitly in terms of x means?

[crude slate]

<@&286206848099549185>

[elder crag]

you want to get y = something with x

[lavish sparrow]

hi

[elder crag]

try getting 0 on one side of the equation

and then use the quadratic formula

[crude slate]

can you tell the answer so i can compare please

[elder crag]

!noans

[marsh citrus]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[elder crag]

you can show us your steps so far

if you only want the answer use wolfram alpha

[crude slate]

So far I’m here, how should I apply the quadratic formula when its not ax^2 + bx + c format?

[elder crag]

it is in the ax^2 + bx + c format

but x is y in this case

[hidden dawn]

this is useless but

sure

[elder crag]

?

[hidden dawn]

u want y = smt x yes ?

[elder crag]

yea

the quadratic formula will give you y = ...

[hidden dawn]

notice that y^2 + 6y + 9 is a remarkable identity

can be factorized

then simply find y

but i have no hate towards quadratic formula

[elder crag]

i think the quadratic formula is the quickest and best way to find it, as finding y after factorizing can be a pain

[hidden dawn]

wdym

its straight forward

cmon

[elder crag]

a is 1, b is 6 and c is -x+1

[hidden dawn]

js a question

[elder crag]

ye?

[hidden dawn]

we talking abt real values for x and y here or?

cause if so, a quadratic formula would be false

[elder crag]

yep

oh does it not work?

i didnt try it out

[hidden dawn]

i mean

look

if real values

who told u the discriminant is positive or negative

or equal to 0

[elder crag]

i mean yea you need to check and if that doesnt work than factorising it the way you suggested is ofcourse the only option

[hidden dawn]

that b squared - 4ac

[hidden dawn]

yep

in this case ydk x and y

[elder crag]

oh wait you mean cause c is x

like c depends on x

i mean it doesnt matter how you solve it you will have x under a root and the domain/image wont be R

cause you have a y^2

[hidden dawn]

no

look

do yk the whole point of a discriminant?

[elder crag]

yea?

[hidden dawn]

it has many points tbh

but

talking about real numbers

it determines the amount of roots yes?

[elder crag]

0,1,2 (in the reals)

[hidden dawn]

yes

discriminant is always D = b^2 - 4ac

[elder crag]

yes i know how quadratics work

[hidden dawn]

if u sub the values of x and y in the discriminant

u wont know

wether D is negative

positive

[elder crag]

but that doesnt matter

[hidden dawn]

it does

do u want undefined answers?

[elder crag]

i mean? a functions domain can be restricted?

[hidden dawn]

oh shit yeah

[elder crag]

its not like y = sqrt(x) is not a valid answer to y^2 = x cause maybe x is negative, we just restrict the domain of the function to be R+

[hidden dawn]

this isnt a regular equation

yo mb

ur right

[elder crag]

yea ahaahah

[hidden dawn]

CONTINUE

[elder crag]

cause its just finding a curve not really values

[hidden dawn]

IM A BASTARD

[elder crag]

nooo why?

[hidden dawn]

js playfully, bud

not serious

lol

[elder crag]

ah ok lol

[crude slate]

I plugged it in and solved it out, my answer is a little different than mathway, can i square root the 4x?

[elder crag]

write 8+4x as 4(2+x)

and now remember that sqrt(a*b) = sqrt(a) * sqrt(b)

[crude slate]

ah ok so would the end answer be y=-3+sqrt(x+8) then?

also what happened to the initial y's are those not needed for explicitly in terms of x?

[elder crag]

why +8

[crude slate]

not 8 i mean 2sqrt(x+2)

so y=-3+2sqrt(x+2)

[elder crag]

remember!

that its not plus

its plus or minus

$\pm$

[elfin berry]

woomy

[elder crag]

this is important as the next question is: "Is this a function?"

And can it be a function with the plus or minus symbol?

[crude slate]

I forgot about the plus or minus would it be y=-3+/-2sqrt(x+2)?

[crude slate]

no right?

[elder crag]

👍

it cant be a function cause for any given inputs you can have 2 outputs, one with + and one with -, but functions cant have 2 outputs for the same input

,w y^2 + 6y + 9 = x + 8 solve for y

[elfin berry]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by woomygfx

[crude slate]

so why is it x+8 here

[elder crag]

maybe a mistake in your steps

let me look at them

[elder crag]

seems like you are writing c as "x"

but its -x+1

[crude slate]

where

[elder crag]

Notice that your equation is:
$y^2 + 6y - x + 1 = 0$

[elfin berry]

woomy

[elder crag]

at the beggining

so "a" is the term infront of y^2, which is 1, b is the term infront of y, which is 6, and c is everything else, in this case -x+1

[crude slate]

ok so what are you referring to

[elder crag]

this

[crude slate]

I did -x+1

[elder crag]

yep

but look at the final equation

thats the quadratic we are working on

here when doing b^2 - 4ac

you plugged in "x" for c instead of "-x+1"

[crude slate]

thats after i simplified it because it was 36-4(-x+1) which turns into 32+4x

[elder crag]

ohh my bad then

[crude slate]

wait so why doesnt sqrt(x+8) turn into 2sqrt(x+2)

[elder crag]

i think its this step then

you factored the 2^2 out from 32 but not from 4

[crude slate]

yeah earlier you said it should be 4(2+x), so that wold make 3+/- sqrt(2+x) then right? but the answer says -3 +/- sqrt(x+8)

[elder crag]

starting from the mistake it would be:
$\frac{-6 \pm \sqrt{32+4x}}{2}\$
$\frac{-6 \pm 2\sqrt{8+x}}{2}\$
$-3 \pm \sqrt{8+x}\$

[elfin berry]

woomy

[elder crag]

the 4(2+x) step we did cause what was in the sqrt was wrong

sorry for all the edits im kinda tired, its 3.40am here

[crude slate]

its all good

Ok I understand

Thank you

[elder crag]

yw :)

[crude slate]

.close

[marsh citrus]

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[marsh citrus]

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[halcyon scaffold]

How do I draw a vertical asymptote on a CAS

[halcyon scaffold]

Like how do I type out the asymptote x = -2

[grave dove]

something like f(x) =kx+2k

with k being very large

try f(x)=10000 x + 20000

[marsh citrus]

@halcyon scaffold Has your question been resolved?

[marsh citrus]

@halcyon scaffold Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[night solstice]

as

[night solstice]

Can sb help me pls

[void elm]

[night solstice]

GIven x, y is positive integers. Find x, y so that 3^x - 2^y = 1

prove it pls 😦

$3^x - 2^y = 1$

@sharp wing

[devout pebble]

Prove it or solve it?

Cuz it looks like solve

[night solstice]

solve it pls

[elfin berry]

Nguyễn Tuấn Minh

[devout pebble]

I think you need to use logarithms to get the variables on their own

[night solstice]

but we haven't studied logarithms

[proud basin]

https://sci-hub.ru/10.2307/2371997

ive asked this question before, might help

[night solstice]

but i'm only a 8-graders

[cunning oak]

I’d recommend you study logarithms first, I’m 8th grade as well

[coral remnant]

bài quen quá =))

[night solstice]

maybe because we're in different countries and + i'm starting 8th grade

[night solstice]

m VN ak

[coral remnant]

ừa mà quên cách giải r chắc chờ ai giúp ik

[night solstice]

cô t sắp mang máy chém cổ ra đây r 💀

[night solstice]

lục lại sách vở đi

bruh

[marsh citrus]

@night solstice Has your question been resolved?

[night solstice]

<@&286206848099549185>

[proud basin]

wait, do you want to find all?

[night solstice]

yes 🙂

[marsh citrus]

@night solstice Has your question been resolved?

[lucid turret]

3-2 = 1
so x,y = 1 is a solution

also x = 2 and y = 3 is also a soluton

I used python for x, y= 1000
and couldn't find a soltuion

[night solstice]

but can you prove it??

[wheat moth]

hi

[vernal mantle]

Have no idea but I found this video that might be revelant

https://youtu.be/jQnIvegVhXg?si=taXuIPNWgq920E5N

[marsh citrus]

@night solstice Has your question been resolved?

[calm harbor]

Hint: Consider two functions, 3^x and 2^y

[night solstice]

hmmm

can you give me another hint??

.close

[marsh citrus]

Channel closed

Closed by @night solstice

Use .reopen if this was a mistake.

[night solstice]

.reopen

[marsh citrus]

✅

[night solstice]

.close

[marsh citrus]

Channel closed

Closed by @night solstice

Use .reopen if this was a mistake.

[night solstice]

.reopen

[marsh citrus]

✅

[night solstice]

.close

[marsh citrus]

Channel closed

Closed by @night solstice

Use .reopen if this was a mistake.

[night solstice]

.reopen

[marsh citrus]

✅

[night solstice]

.

.close

[marsh citrus]

Channel closed

Closed by @night solstice

Use .reopen if this was a mistake.

[valid cape]

!noclopen

[marsh citrus]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[calm harbor]

Please be patient….I wasn’t available

[night solstice]

!

[marsh citrus]

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[marsh citrus]

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[copper pebble]

How do I do q14

[rotund raptor]

is it x^2-1 or x^3-1

[copper pebble]

X^2

-1

[eager tinsel]

Hint

?

[copper pebble]

Wdym

[eager tinsel]

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

[copper pebble]

Yh ik that idk what to do after that 😭

[rotund raptor]

P(x)=(x+1)(x-1)Q(x)+(3x-1)

and you're dividing (x+1)(x-1)Q(x)+(3x-1) with x-1

[copper pebble]

So do I multiply

Then polydiv the remainder?

[rotund raptor]

(x+1)(x-1)Q(x) is divisible by x-1

[copper pebble]

Is the remainder 2

[rotund raptor]

yeah

[copper pebble]

Ohh ok

.close

[marsh citrus]

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[marsh citrus]

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[novel juniper]

So I'm trying to graph arcsin(sin(x))

[novel juniper]

I realise it's just x

but just wanted to be sure, it's just x on [-1,1]

right?

[dusky viper]

uh nope

[vivid chasm]

shouldn't the domain be x is an element of all real values cause the range of sin goes between -1 and 1

[novel juniper]

I mean arcsin(x)'s range is [-1,1]

[leaden monolith]

look at the domain and range of arcsin

[novel juniper]

oops

[leaden monolith]

the graph of arcsin(f(x)) is always going to be a subset of arcsin

[novel juniper]

I mean [-pi/2, pi/2]

[dusky viper]

still not the case

its domain is all real x

[leaden monolith]

well this is not trivial to see it's true

care should be taken to make sure this is in fact the case

[knotty trellis]

Keep in mind that sin(x) maps R to [-1, 1]

[leaden monolith]

that's important to consider yes

[novel juniper]

Ah yeah

true

thanks

[knotty trellis]

it might be easier to sketch the graph rather than describe it algebraically btw

[hushed egret]

the domain of a composition is always the domain of the first function though

[novel juniper]

This is what I was thinking

[knotty trellis]

The first function being the one inside or outside?

[novel juniper]

so the domain is [-1,1]

now the range of that function is -π/2,π/2

wiat

no

in this case the domain would be [-pi/2,pi/2]

my bad

[hushed egret]

the first function to be applied to the input

[knotty trellis]

so arcin(sin(2pi)) would be undefined?

[novel juniper]

That would just out put 0

not 2π

I guess it will be periodic then?

[knotty trellis]

Then that would be true only as long as the domain of the 2nd function is subset of range of the 1st function, no?

[knotty trellis]

yep

[crystal lintel]

hi snow :waves:

[novel juniper]

Ooh

okay

[knotty trellis]

Try to actually prove / show that it's periodic

[hushed egret]

then there is no composition

[knotty trellis]

and also find the period

[novel juniper]

so It's a saw-tooth function

[knotty trellis]

yeah, it looks like that

but it's not this sawtooth function

[novel juniper]

Yeah

[knotty trellis]

can you actually sketch it and send it here

[novel juniper]

yeah

[hushed egret]

[knotty trellis]

But we can still talk about a function such as f(x) = sqrt(x^3)

[novel juniper]

,rotate

[knotty trellis]

yep

[novel juniper]

Now to actually prove periodcity

[knotty trellis]

the pi reffers to this point, right?

[novel juniper]

Yeah

[hushed egret]

but it would be an ill-defined function unless you've restricted the domain

so really its x^3 with its domain restricted composed with sqrt

[knotty trellis]

yeah, the domain would have to be restrictd

[hushed egret]

so there is no composition for the unrestricted functions

[novel juniper]

I still feel the domain should be [-π/2,π/2]

[hushed egret]

the ol arrows between ovals diagram

[novel juniper]

OK. cool

Thanks

[hushed egret]

you're probably thinking of the range of arcsin

which is the domain that sin(x) must be restricted to to take the inverse

[novel juniper]

Yeah

[hushed egret]

its okay to compose the unrestricted sin with arcsin though

[novel juniper]

Got it

so sin(arcsin(x)) on the other hand will be restricted

right?

[knotty trellis]

it's domain will be [-1, 1] if you mean that

but you wont have to restrict domain of arcsin(x) in order to compose them

[novel juniper]

Yeah

[knotty trellis]

(Btw, this is how function composition was defined in enderton, the domain was restricted automatically)

[hushed egret]

ew

[knotty trellis]

it's not ew, i like it this way

[hushed egret]

the category theorists hate you

[knotty trellis]

and I hate the category theorists

[hushed egret]

a function should be the data of its domain, codomain, and the underlying mapping

two functions with different codomains but the same underlying mapping are different functions

[knotty trellis]

they do the same thing

[hushed egret]

one might be bijective, and one might not be

its a huge difference

you can claim the bijective one to be invertible

but the other, not so

[knotty trellis]

just restrict the domain

of the inverse

[hushed egret]

no, this is a corestriction of the codomain

otherwise you only have a one-sided inverse, the existence of which might even depend on choice

[knotty trellis]

Hmm how would it depend on choice?

[hushed egret]

constructing a one-sided inverse depends on choice

in the general case

[knotty trellis]

if one is bijective, then the other must be at least injective, no?

[hushed egret]

okay but where are you sending the elements that arent mapped to

[knotty trellis]

either nowhere, and I just restrict the domain

or just pick a single element

choice is needed for inverse of surjective function i think

[hushed egret]

true

other way round then

[knotty trellis]

I'll probably just accept choice then

[hushed egret]

but this is silly, because you want the distinction between bijective and injective things

[knotty trellis]

possibly

I mean, yeah, it is kind of silly. But the idea of composing functions and then restricting the domain isnt that bad imo

if i say sqrt(x^3), everyone will know what kind of function i mean

[hushed egret]

thats because the notation is convenient

but when you're working with function notation, that would be bad

the situtation is possibly worse when you have more structure in consideration

an injective homomorphism might not admit an inverse, but a bijective homomorphism automatically becomes an isomorphism for algebraic things

[marsh citrus]

@novel juniper Has your question been resolved?

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[novel juniper]

I was thinking $x = \frac {f^{-1}(h(x))}{c}$

[elfin berry]

f(why am i here )= I don't know

[knotty trellis]

if it's the inverse, then when you apply h on it, it should return x

try it

[still temple]

1/c*(f-1(x))

[marsh citrus]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[still temple]

😂😂

😂😂😂😂😂

[leaden monolith]

no? look at arcsin(sin(x) + 1)

you're restricted to the preimage of the domain of the 2nd function

[knotty trellis]

the point is that composition of functions arcsin(x) and sin(x)+1 isn't even defined (according to the formal definitions)

unless you restrict the domain of sin(x)+1

[leaden monolith]

no but my point is that when you say oh sin(x) has all reals as its domain, so arcsin(sin(x)) also has that

[hushed egret]

nooo you're doing HS definitions

[leaden monolith]

huh?

[hushed egret]

thats how they define it in highschool

[knotty trellis]

I was brainwashed for too long to forget HS definitions

[leaden monolith]

im very confused now

[hushed egret]

anyway this is not right

that equation doesnt even define a function

[crystal lintel]

where is the function at

[novel juniper]

I started with h(x)=f(xc)

then apply f inverse on both sides

$f^{-1}((h(x)) = xc$

[knotty trellis]

The manipulation is not wrong, but what is the inverse function? All that shows is some random equation, which doesnt define a function

[elfin berry]

f(why am i here )= I don't know

[knotty trellis]

make it so that there is h^-1(x) on the LHS, and the rest on the RHS

[novel juniper]

what

[knotty trellis]

this is just a random equation, it does not directly define a function

you want to reach something of the form
h^-1(x) = .....

[novel juniper]

so I have $x= h^{-1}(f(cx))$

[elfin berry]

f(why am i here )= I don't know

[novel juniper]

I'm confused

[knotty trellis]

How would you find inverse of f(x) = (x+3)^3

Where would you even start?

Would you start by doing $x = f^{-1}((x+3)^3)$?

[elfin berry]

MæthIsAlwaysRight

[novel juniper]

no

I would start by taking the cuberoot of both sides

[knotty trellis]

okay, that would give $\sqrt[3]{f(x)} = x+3$

[elfin berry]

MæthIsAlwaysRight

[knotty trellis]

now I suppose you would subtract 3

[novel juniper]

yeah

[knotty trellis]

$\sqrt[3]{f(x)}-3 = x$

[elfin berry]

MæthIsAlwaysRight

[knotty trellis]

now what is the inverse function?

how do we extract it from this?

[novel juniper]

wait

f(x)=y

[knotty trellis]

okay, this could work

[novel juniper]

$(y)^{1/3}-3=x$

[elfin berry]

f(why am i here )= I don't know

[knotty trellis]

what is x in terms of y?

[knotty trellis]

using this