help-33

[tired oxide]

if you want to show this for arbitrary polynomials, you have to show that x^k has this property

yeah

that's exactly right

[novel juniper]

I have all 4

[tired oxide]

which 4

[novel juniper]

product, quotient, sum ,, difference

[tired oxide]

yeah, but you don't have it for x^k where k is an integer

and this is where induction is used

because x^k = x^(k-1) x

[novel juniper]

ah

right

[tired oxide]

okay great

[novel juniper]

I suppose I could write an induction proof

[tired oxide]

and then you need to show this for sums of arbitrary length given that you have it for a sum of two terms

but that is also induction

you can just say "a similar induction argument shows this"

if you want to

[novel juniper]

We start of with the base case , which is $\lim(x_n) \to x$.By the inductive hypothesis, we have $\lim{x_n^m) \to x^m$. then $x_n^{m} \cdot x = x_n^{m+1}$ by the product rule of limits, and we're done

[elfin berry]

ƒ( wai ina teacup)= I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tired oxide]

i mean you could clean it up a bit if you were trying to write a really formal inductive proof but like yeah that's the idea

like the wording is a bit weird

are you really not at a point where you would just write "by induction on the degree, x_i^k -> x^k"?

and like they give you full credit?

[novel juniper]

I could yeah

I only have this course in my 3rd sem, so I'll surely be at that point by then

[tired oxide]

oh i see, this is self-study

[novel juniper]

yeah

[tired oxide]

tbh, i know that some people are gonna get mad at me for saying this, but i really don't think you ought to be so formal with proofs like this. the main thing that you want to be focusing on is "what facts do my assertions use? i should be citing the reasoning either from things i have already proved, or things that someone else has proved, or things that i could easily prove using [X, Y, Z principle, induction, X fact]

because you're just self-studying it

[novel juniper]

fair,yeah

Tbh, I'm also doing it early so that I can study multi var from shifrin

Atleast need knowledge of open sets, closed sets etc to study shifrin

thanks

.close

[marsh citrus]

Channel closed

Closed by @novel juniper

Use .reopen if this was a mistake.

[marsh citrus]

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.

[tepid hound]

The question is asking for the value of dy/dx

[tepid hound]

All options are constant numbers

[pseudo flax]

differentiate both expressions w.r.t t and find the ratio

[tepid hound]

I can't find a way to cancel out

The nominator and demon should be same

Living some constant behind

[main idol]

Show your whole work

[tepid hound]

.

[tepid hound]

This is my final answer

[fathom ridge]

you should substitute t = tan u

and then diff

[tepid hound]

Still stuck

[fathom ridge]

maybe not tan hmm

but there is probably some sort of trig sub

[tepid hound]

I get arcsin(tan2x) and arctan(sin2x)

[fathom ridge]

hmm

[marsh citrus]

@tepid hound Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

@buoyant jetty Has your question been resolved?

[night dew]

Can you translate it into English?

It looks like we have a line and 2 planes

But I don't see an L2

[buoyant jetty]

Let L = ..., Π1 = ..., Π2 = ... find lines L1, L2 such that

[night dew]

So we just need any line L1 in Π1?

[buoyant jetty]

is a subset

so its entirely contained

every point in L1 is contained in Π1

[night dew]

Mmhmm

[night dew]

Okay so to find L1, we need a point in the plane, and a direction vector in the plane

Any point will do, so let's just set x = 0, y = 0

Then we have 2z = 8

So z = 4

[buoyant jetty]

direction can be a multiple of (1,3,2) maybe

[night dew]

So (0, 0, 4) is in the plane

[buoyant jetty]

L1 ∩ L have a non empty intersection

[night dew]

We haven't finished yet

The equation of the plane includes a vector normal to the plane

In other words, (1, 3, 2) is normal to the plane

To be in the plane, we need to be tangent to the plane

[buoyant jetty]

the normal of the plane. is parallel to all the vectors residing in the plane

[night dew]

Which means perpendicular to the normal

[night dew]

perpendicular*

[buoyant jetty]

the direction of L1 has to be perpendicular to the normal of Pi1

[buoyant jetty]

yeah, sorry

[buoyant jetty]

for direction vector of L1

[night dew]

So we need the dot product with the normal vector to be 0

[buoyant jetty]

(a,b,c).(1,3,2)=0

(a,b,c) is direction of L1

we get a subspace of vectors orthogonal to the normal vector of Pi1

a+3b+2c=0

[night dew]

Yeah. I think the intention is to just pick one

So let's set a = 1, b = -1, c = 1

[buoyant jetty]

I was thinking of finding the intersection of L and this plane

but idk

[night dew]

Why?

L is just another line

We're finding a different line L1

There are an infinite number of options though

But we just need to make sure the one we choose is valid, so we need a point in the plane and a vector in the plane

The point in the plane can be found by substituting values into the plane equation

And the vector needs to be perpendicular to the normal vector

Those are the conditions we need to satisfy

So we have L1: r = (0, 0, 4) + λ(1, -1, 1)

Does that make sense?

[buoyant jetty]

sure

[night dew]

We don't have enough information to specify a single line

[buoyant jetty]

direction vector of L1 has to be orthogonal to normal vector of Pi1

we need L1 and L to intersect aswell and we need

L1 to be parallel to L2, I am trying

[night dew]

Why do we need L1 and L to intersect?

Oh wait we need all 3 at once?

[buoyant jetty]

ye

[night dew]

I was doing them as separate questions mb

[buoyant jetty]

yeah is my bad I didnt translated properly

[night dew]

Okay then so we need L1 to be in Π1, L2 to be in Π2, L1 needs to be parallel to L2, and L1 and L2 need to intersect with L

Okay so if L1 is in Π1, the direction vector of the line is perpendicular to the normal vector of the plane

Right?

[buoyant jetty]

ye

is a multiple of (1,3,2)

problem is what does L1 parallel L2 mean

they never intersect

yet they have same direction

[night dew]

It's not a multiple

The normal vector of the plane points out of the plane

We need a vector in the plane

So that is perpendicular to the normal vector

[buoyant jetty]

okay a vector that is orthogonal to the normal vector

[buoyant jetty]

yeah

(x,y,z).(1,3,2)=0

(x,y,z) is the direction of L1

[night dew]

We don't need that any more

If L1 is parallel to L2, then they both have the same direction vector

Right?

[buoyant jetty]

ohh okayy

,w (1,3,2)x(-3,-1,3)

[elfin berry]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[buoyant jetty]

I see what you mean now

[night dew]

Right so L1 and L2 both have the same direction vector

Which is perpendicular to the normal of both planes

Aka cross product time

Which is what you did there

[buoyant jetty]

what about non empty intersection L1∩L and L2∩L

[night dew]

So now we have the direction vector of L1 and L2

[buoyant jetty]

while they being parallel

[night dew]

But we need a starting point for each

[buoyant jetty]

yeah

[night dew]

We need L1 to obey the equation for Π1 and also L

And we need L2 to obey the equation for Π2 and also L

Yep that would work

[buoyant jetty]

L : X = k(2,0,1) + (0,2,5)

(2k,2,k+5) =(x,y,z)

x+3y+2z=8
2k+6+2k+10=8
4k=-8
k=-2

L1 : X = k(11,-9,8) + (-2(2,0,1)+(0,2,5))

L1 : X = λ(11,-9,8)+(-4,2,3)

[night dew]

And then similar for Π2

You seem to know what you're doing now

[buoyant jetty]

Π2 : -3x-y+3z=4
-6k-2+3k+15=4
13-4=3k
9/3=k=3
(2k,2,k+5)=(6,2,8)
L2: X = λ(11,-9,8) + (6,2,8)

[night dew]

I'd use μ for L2 personally

Since it varies independently of λ in L1

[buoyant jetty]

yeah

the course I bought made a mistake, is a course taught by undergrads from same university

should be different scalar

[night dew]

Do you have answers for this btw?

[buoyant jetty]

using same λ means they are the same and they are not

[night dew]

It's not terrible to do that

[buoyant jetty]

yeah, our stuff is correct

[night dew]

But it's not ideal

Okay then all good

If we didn't have answers it would be helpful to check our conditions

So for condition 1, we can sub in L1 into Π1 and L2 into Π2 and check they obey the equations

Condition 2 is easy since it just means the direction vectors are the same

[buoyant jetty]

being parallel also mean we need to show they dont intersect

[night dew]

If they have the same direction vector, they won't intersect

[buoyant jetty]

they could intersect at one point if they are perpendicular (same direction vector so they are perpendicular directions, no?)

[night dew]

No

[buoyant jetty]

,w (1,2,3).(1,2,3)

[night dew]

The direction vector of a line tells us the direction the line goes in

So the two lines are going in exactly the same direction

So they are parallel

Planes have normal vectors that are perpendicular to the plane. Lines have direction vectors that are parallel to the line

If the direction vectors are equal, then the lines are parallel

The only other option is that they are the same line

[buoyant jetty]

yeah

[night dew]

So we could check that they aren't

But I don't think that's necessary here

We're just checking our work

The direction vectors being the same is enough

For condition 3, we need to show that L1 and L intersect, so we could just put the equation of L1 into L and make sure it works, and similar for L2

But we have answers to check with here so maybe we don't need that

All good now?

[marsh citrus]

@buoyant jetty Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[patent fjord]

@still temple

[limber hearth]

,w henrycavillbutfat

[elfin berry]

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

Requested by yakubros

[patent fjord]

That isn’t like 1 the sum as you pointed out the function has finite discontinuities is the key

[marsh citrus]

@patent fjord Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[compact crescent]

Can I interpret logistic curve as a graph that grows and decrease exponentially and has a inflection point.

[marsh citrus]

@compact crescent Has your question been resolved?

[marsh citrus]

@compact crescent Has your question been resolved?

[glacial hedge]

It converges at an exponential speed towards both asymptotes

Increasing/decreasing exponentially sounds like the limit is infinite, not finite

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[honest creek]

f(2x - 6) is a horizontal compression by a factor of 2 and a shift right of 3 units right?

[honest creek]

and this is the logic i used for that^:
h(x) = f(2x) then h(x - 3) = f(2x - 6)

but then why is this wrong, starting with a shift of 6 units and then compression of factor of 2:

h(x) = f(x - 6) then h(2x) = f(2x - 6)

[odd orchid]

by order of operations was how i was taught

addition and subtractions

then multiplications and divisions

so translations before stretches

[honest creek]

okay but where am i going wrong?

[honest creek]

also uh this is right

i'm pretty sure

and so this:

h(x) = f(2x) then h(x - 3) = f(2x - 6)``` is probably right too

i just why a similar argument doesn't work for this:

h(x) = f(x - 6) then h(2x) = f(2x - 6)```

[marsh citrus]

@honest creek Has your question been resolved?

[marsh citrus]

@honest creek Has your question been resolved?

[honest creek]

.close

[marsh citrus]

Channel closed

Closed by @honest creek

Use .reopen if this was a mistake.

[marsh citrus]

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.

[elfin forge]

Hey

[elfin forge]

I need help

integration

I have this equation

IDK how to express it but

The intrgral notation with unknown letter b over 2 (3e^x+6e^-2x)=0

[odd orchid]

$\int_2^b 3e^x + 6e^{-2x} dx = 0$

[elfin berry]

ashy!

[elfin forge]

this

YES

this one

How do you do it

[odd orchid]

evaluate the integral

[elfin forge]

I know

[odd orchid]

latex 🙏

[elfin forge]

e^x remains the same

[odd orchid]

show your working

[elfin forge]

so 3e^x then -3e^-2x?

because we divide by the differential

[odd orchid]

yup

[elfin forge]

okay so

[odd orchid]

now evaluate it at the bounds

[elfin forge]

3eb -3e^-2b-3e^2+3e^-4

[odd orchid]

$3e^b -3e^{-2b}-3e^2+3e^{-4} = 0$

[elfin forge]

yes

[elfin berry]

ashy!

[elfin forge]

then what do you do?

[odd orchid]

first i would divide through by 3

[elfin forge]

did

that

then

[odd orchid]

multiply through by e^2b

then itll be a cubic

[elfin forge]

wait

[odd orchid]

$e^{3b} -e^{2b}(e^2+e^{-4})-1= 0$

[elfin forge]

where did u get e

[odd orchid]

-1 i meant

[elfin berry]

ashy!

[elfin forge]

u make x^2=e^1/5?

[odd orchid]

what

why

[elfin forge]

i mean 1/3

yo

[odd orchid]

wait wtf am i doing

b is just 2

$\int_a^a f(x) dx = 0$

[elfin berry]

ashy!

[elfin forge]

yeah?

[odd orchid]

so b is 2

since the integral is 0

[elfin forge]

wait what

[odd orchid]

you are solving for b?

[elfin forge]

yes

But how does this rule apply

[odd orchid]

the integral is 0

[elfin forge]

why

[odd orchid]

so the bounds are the same

[elfin forge]

oh

oh alr

Thank you

[marsh citrus]

@elfin forge Has your question been resolved?

[elfin forge]

hey

also

helloooo

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[elfin forge]

hey

[elfin forge]

Hello

so I have this integral AGAIN

and

Snake b over 3 (2x-6) dx=36

how do we find b

[devout mauve]

can you integrate 2x-6?

[elfin forge]

uh

oH WAIT it's in the square too

so snake b over 2 (2x-6)^2=36

[devout mauve]

$\int_2^b (2x-6)^2 dx = 36$ ?

[elfin berry]

Denascite

[elfin forge]

yes

[devout mauve]

do you know u subs?

[elfin forge]

what///

?

[devout mauve]

good then not

imagine you differentiated (2x-6)^2

what would you get

[elfin forge]

Wait

(2x-6)^3/2?

is this what you mean

[devout mauve]

do you mean $\frac{(2x-6)^3}{2}$ or $(2x-6)^{3/2}$?

[elfin berry]

Denascite

[elfin forge]

first one

[devout mauve]

good

[elfin forge]

okay I found the answer but

but

I have another equstion

Question

on another integral

sin^2 x times sin^x

What is its integral

[devout mauve]

$\int \sin^2(x)\cdot \sin(x) dx$ this?

[elfin berry]

Denascite

[elfin forge]

sin^x squared too

i mean AAAA

Sin^2x times cos^2x

this one

Sorry I'm just tired

[devout mauve]

$\int \sin^2(x)\cdot\cos^2(x) dx$?

[elfin berry]

Denascite

[elfin forge]

yes]

this one

[devout mauve]

use a few trig identities

[elfin forge]

like

[odd orchid]

sin double angle

[elfin forge]

oh ok got it

thanks guys

[marsh citrus]

@elfin forge Has your question been resolved?

[elfin forge]

guys

I got another question

again on integration

how do you solve (1+sinx)^2

[odd orchid]

integral?

just expand it

and integrate term by term

[elfin forge]

how

don't use any identity

[odd orchid]

$(a+b)^2 = a^2 + 2ab + b^2$

[elfin berry]

ashy!

[elfin forge]

ok wait

[odd orchid]

if u call that an identity

[elfin forge]

bruh it's easier than I thought

Okay

and now

my other question

snake ln2 over 0 (1/(1+e^x)

How do you find x?????????

[odd orchid]

$\int_0^{\ln2} \frac{1}{1 + e^x} dx$

[elfin berry]

ashy!

[elfin forge]

yea

[odd orchid]

u sub

[elfin forge]

what do you mean

[odd orchid]

do a substitution

[elfin forge]

pls Show

[odd orchid]

u = 1 + e^x

then itll turn into a partial fractions

[elfin forge]

wha?

[odd orchid]

ok scratch that

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd orchid]

i thought of a faster way

[elfin forge]

ok

[marsh citrus]

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.

[elfin forge]

uh

[odd orchid]

add e^x and subtract it from the numerator

$\int_0^{\ln2} \frac{1 + e^x - e^x}{1 + e^x} dx$

[elfin berry]

ashy!

[elfin forge]

yeah/

?

[proud arch]

$\int_0^{\ln2} 1 - \frac{ e^x}{1 + e^x} dx$

[odd orchid]

and now u = 1 + e^x

[elfin berry]

lex.in.a.teacup

[velvet cedar]

$\int_0^{ln2} \frac{e^{-x}}{e^{-x} + 1} dx$

[elfin forge]

then you reverse chain rule

[elfin berry]

StrangeQuarkAL

[elfin forge]

right?

[proud arch]

$\int_0^{\ln2} 1 dx - \int_0^{\ln2}\frac{ e^x}{1 + e^x} dx$

[elfin berry]

lex.in.a.teacup

[odd orchid]

yes

[elfin forge]

ok ty

Got it

[proud arch]

Ye

[marsh citrus]

@elfin forge Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[rotund violet]

I will just start off by saying that this is a pretty stupid question

Like they literally invent functions for how the ball moves in time which just don’t follow the rules of physics at all

Regardless let’s do this

Ok so first

It takes 1 second to go from 2 to 10 meters right

And then at the same speed

It goes down to 1 meter

So

Well it should be 1.125

9/8

Ok

Next it goes from 1 m to 5 meters

So it goes 4 meters up

Via a square root curve (which is stupid as gravity isn’t dependent on speed and it should be linear)

So this means that in order to go up a distance of t, it will take ct^2 time, where c is some constant

We know however

That if it goes a distance of 2 meters up, this will take 1 second however

Can you use this to determine c?

A constant is just a number

Basically it isn’t a variable which changes

It’s always the same

yeah, except it is now, every certain amount of meters = (a certain amount of time squared) times something

and im asking for what is that something

okay another way to think about this

if it takes 1 second to go up 2 meters

along a square root curve

how long will it take to go up 4 meters?

4

yes

so in total

it takes 1+1.125+4 seconds

right?

[marsh citrus]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

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.

[cold bloom]

this is my word

work*

am i correct

[rotund violet]

yes

you are correct

[cold bloom]

how can i be sure

[cold bloom]

this might cost me a mark on my exma tomorrow sir arnavutkoy

[rotund violet]

i mean you did the steps right? idk what else to say

im sure its correct. im just assuming by option c you mean the 3rd option rather than what is actually labeled as "c"

[cold bloom]

oh

yeah

why is it b d c e

ok 1 last hting

these are relatively easy

am i correct on these

[pliant maple]

25 is wrong

[cold bloom]

i know

it should be d

[pliant maple]

yes

everything else is good

[cold bloom]

ok what about this

i havent solved it

yet

cause i dont know

[pliant maple]

think about how you determine where a function has veritcal asymptotes

[cold bloom]

where the denominator of a function is equal to 0?

[pliant maple]

yes

where does the graph have vertical asymptotes?

[cold bloom]

-3

and

1

[pliant maple]

alright

[cold bloom]

so that must mean it has x+3 and x-1

[pliant maple]

yes

[cold bloom]

so its either c or d

and its c?

i think

idk

[pliant maple]

it's c

read d a little more closely

d isn't x+3 and x-1, but x-3 and x+1 instead

which is incorrect

[cold bloom]

oh cz the vertical asymptotes arent

yeah

sir 1 last problem pls

these questions r easily solved online but i wanna know how to solve them myself

also cause chatgpt is saying d and other classmates solved it as a

[pliant maple]

$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{cb}$

[elfin berry]

890s

[cold bloom]

stop edging it sir

is it a or d

[pliant maple]

my goal is to help you understand the process to solving it

i want you to know yourself if it's a or d without me telling you

[cold bloom]

youre right mb

so

where do i start

[pliant maple]

how can you use this information to help you solve the problem? ^^^

[cold bloom]

im not sure sir

[pliant maple]

let's look at the expression in the problem

it's a fraction of fractions

using the property i showed above

we can assign each part of the expression to a,b,c,or d

[cold bloom]

so we flip the fracitons

[pliant maple]

yep

[cold bloom]

andcross multiply

[pliant maple]

not cross, just multiply

[cold bloom]

sorry

[pliant maple]

but yea

[cold bloom]

multiply

yeah

[pliant maple]

and before you do foil and all that

i'd say factorize each of them

and try to cancel out common factors

[cold bloom]

at that point wouldnt there be no reason for foil

cause factoring and canceling

are done

[pliant maple]

you might have to foil at the end if you have leftover factors

like if you had (x-3)(x+1)/(x-5)

[cold bloom]

im doing of fuck this shit

sit

sir

would u perhaps tell me the answer

cause ive been sitting at mybtable solving this 80 question worksheet since like 1 pm

and its 10pm

[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.

[cold bloom]

the guy is edging me

[spark siren]

ignoring the servers rules is the best way to get help from volunteers. not.

[marsh citrus]

@cold bloom Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

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.

[finite scroll]

I need help with finding the derivatives! And what's chain rule and product rule and when to choose which one to use?

[finite scroll]

<@&286206848099549185>

[red nimbus]

You use the chain rule if you have a composition of functions (that means something like f(g(x)) one function is in the other). Product rule you use if you multiply two functions basically

[finite scroll]

Got it! I know it may sound a bit stupid but How do I know myself that the equation is going to be multiply two functions ?

[red nimbus]

f(x) * g(x)

for example

y = ln(x) * x²

f(x) = ln(x) and g(x) = x²

here you multiply ln(x) with x²

[finite scroll]

Alright thx
One more thing, my doctor wrote = cos (u) at beginning why?

[limber hearth]

To show the formula

Cos(u) where u is a function

d/dx that badboy gives - u' * sin(u)

[marsh citrus]

@finite scroll Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

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.

[still temple]

Could someone please check it

Also for d ,I didn't quite understand what they meant by linear combination for columns

And as for g, I know it's false but not sure how to prove it

Oh I was wondering who math rocks was

[novel juniper]

It's wai

[still temple]

Figured 😁

You changed your username too 🔥

[novel juniper]

fofr d they mean b lies in the span of the column vectors

(a),(b),c ,e are right

[still temple]

Could you rephrase that

are they saying b is a column vector ?

[novel juniper]

let the column vectors be v_1, v_2,\dots, v_n$
\
then $b = \sum_{i=1}^{n} a_i v_i$

for $a_i \in \R$

[elfin berry]

math_rocks

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

[still temple]

OHH

lol

It's true

Then

Could you check the rest please

[novel juniper]

okay

(h) ,(j),(k) are right

(l) required C to be invertibel

(m) is right

[still temple]

I assumed it to be invertible

So that we can multiply with C inverse on both sides

[novel juniper]

Can you do tha

[still temple]

Lol no

[novel juniper]

yeah, so pretty sure its false

in (n) won't the product have a row of zeros

[still temple]

1 min

[novel juniper]

oops

my bad

wiat

yeah

[still temple]

Columns

I took an example to be sire

Sure

[novel juniper]

hmm

one minute

okay

yeah

sorry

(o) is right too

[still temple]

(f),(g),(i)

Is the proof for g too big?

[novel juniper]

f , i are right

is g right though

[still temple]

It's false

[novel juniper]

yup

for a proof, a counter example suffices

[still temple]

Oh

Thank you

.close

[marsh citrus]

Channel closed

Closed by @ionic jewel

Use .reopen if this was a mistake.

[marsh citrus]

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.

[red nimbus]

Say we have two groups $G$ and $H$ such that $a \in G$ generates $G$ and $b \in H$ generates $H.\$
Does that imply that $(a,b) \in G \times H$ generates $G \times H$?

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

Probably $(a,b)$ can be expressed as $(g^n, h^m)$ where $n,m \in \mathbb{Z}$

[elfin berry]

𝔸dωn𝓲²s

[copper raven]

no

[devout mauve]

what have you seen for Z2xZ2 ?

[red nimbus]

😭

so right

Ok I will post the orginal and my progression of thoughts

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I have a stupid idea

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

If a+b=2 then n=5x+6y = 120 we get (x,y) = (0,20) hence there must exist a pair (a,b) with order 120?

[lucid zenith]

i shall be borrowing that notation for the modular arithmetic

[red nimbus]

I am kinda on the Schlauch right now arghhh

[lucid zenith]

why does n=120 mean that the order of the element is 120?

[copper raven]

can you give us a pair with order 120 then ?

[red nimbus]

😭

[red nimbus]

I wanted to find an element where it takes at least 120 additions withitself that lead to the neutral element

To show only the existence of a generator

[lucid zenith]

but the order could be any factor of 120, not just 120

[copper raven]

yeah no reason the order can't be lower than 120

[red nimbus]

Ok give me a last chance

Before you spoiler

[copper raven]

aight

[red nimbus]

There is no generator

lcm(10,12) = 60

[copper raven]

then you spoiler yourself

[red nimbus]

So one group is cyclic the other isn't so they cannot be isomorphic

[copper raven]

well yeah if we start with your thing, take an n such that n=0 mod 10 and n=0 mod 12, then you can guarantee n(a,b) = 0 whatever (a,b) are

i.e. all the orders are n or less

and the smallest such n is lcm(10,12)

60

[red nimbus]

so basically since 1 generates Z_10 and Z_12 we would see if (1,1) would also generate Z_10 x Z_12

[lucid zenith]

but how would you prove this

[copper raven]

essentially with the same argument I made above

[copper raven]

well you might have another generator who knows, just checking (1, 1) is very light

[copper raven]

here you cover all you bases, nothing can ever generate the whole group on its own

[marsh citrus]

@red nimbus Has your question been resolved?

[red nimbus]

I tried to prove it but I think I am missing something important

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I just claim that n = lcm

[copper raven]

you're just saying stuff you're not proving anything yeah

[red nimbus]

but like there are 2 cases which lcm covers all. That is ord(g) = k * ord(h) or gcd(ord(g),ord(h)) = 1

[copper raven]

ok but it still doesn't prove this

ord(g) = k * ord(h)
why so?

[red nimbus]

either this or ord(h) = ord(g) * l

I should have said with out loss of generality

(g,h)^n = (g^n, h^n) = (g^ord(g), h^ord(h)) = e

n = ord(g) = ord(h) implies they are equal but suppose they weren't

then n must be a multiple of ord(g) and ord(h)

n = ord(g)ord(h)

Ok I need also the condition that n should be minimal

therefore we take n = lcm(ord(g),ord(h))

[copper raven]

that's better yeah

[copper raven]

pretty much what I did here

[red nimbus]

so what I did rn should suffice

[copper raven]

yea

[red nimbus]

Is this fine now?

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I think the proof is shit

[copper raven]

main issue is this

how do you even justify that implication

[red nimbus]

g^n = g^ord(g)
h^n = h^ord(h)
there fore I thought n = ord(g) and n = ord(h)

I think I am being too rash and dont overthink which is badd

[copper raven]

(g^n, h^n) = e, means that g^n = e and h^n = e
therefore ... [what can you say about n from these two equalities ?]

n = ord(g) is too restrictive for example

[red nimbus]

n is a multiple of ord(g) and n is a multiple of ord(h)

[copper raven]

yea

[red nimbus]

say

n = a * ord(g) and n = b * ord(h)
then
n = a * ord(g) * b * ord(h)

[copper raven]

tfw n=n^2 always

[red nimbus]

haha noo

ok one step back

n = a * ord(g) and n = b * ord(h)

[copper raven]

well n is a common multiple of ord(g) and ord(h) then

[red nimbus]

hmm okay

and because of the minimality

lcm

[copper raven]

yea

[red nimbus]

😭

it took me so long and it was easy

[copper raven]

eh we all get stuck in weird proof ideas sometimes it's fine

[red nimbus]

thank you very much for your help aPlatypus ❤️

.solved

[marsh citrus]

Channel closed

Closed by @red nimbus

Use .reopen if this was a mistake.

[marsh citrus]

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.

[wheat trench]

I don't understand the definition of this line function. Isn't the result a set with a single point inside it?

[devout mauve]

why do you think there would only be a single point in the set?

[wheat trench]

I see an addition of a point and a vector, which yields a single point.

[regal coral]

You get a different point for each t

[wheat trench]

I would have expected that to have a colon then, like:

[devout mauve]

lots of ways to write the same thing

comma is unusual but its clear from context what its supposed to be

: and | are more common

[wheat trench]

ok, thanks for the help.

.close

[marsh citrus]

Channel closed

Closed by @wheat trench

Use .reopen if this was a mistake.

[wheat trench]

.reopen

[marsh citrus]

✅

[wheat trench]

Just want to mention that my confusion was because I interpreted that as saying something like "A point + a scaled vector, where the vector's scalar can be any number".

That syntax is ambiguous, and a ":" or "|" would have been better.

.close

[marsh citrus]

Channel closed

Closed by @wheat trench

Use .reopen if this was a mistake.

[marsh citrus]

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.

[burnt abyss]

why

[still light]

Just post the question

[wary kite]

<@&268886789983436800>

[marsh citrus]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

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.

[oak fog]

Anyone know's how to do this question

[oak fog]

I need it done in this exact way

[modern wigeon]

try setting y = 2^x

[oak fog]

gimme a sec

I have a very simular one already done

almost the same

[modern wigeon]

if you set y = 2^x you can rewrite it in the form a y^2 + by + c = 0 and solve for y

ah, yes, this is what you did here with t = 2^x

[oak fog]

i need it done in the same way though

[oak fog]

yeah..

[modern wigeon]

ok, so set t = 2^x and rewrite the equation in terms of t

[oak fog]

I genenly have no clue how to do this one

I'm really bad at math

[modern wigeon]

the only thing to figure out is how to rewrite 2^{2x - 1} in terms of t

[oak fog]

would you be able to do the steps

[modern wigeon]

what is 2^{2x} in terms of t, if t = 2^x?

[oak fog]

what 😭

[modern wigeon]

do you understand the solution to the other problem that you're mimicking?

[oak fog]

okay so

I have a test

and the practice question was

[oak fog]

this

[oak fog]

and in my note book i found tihs

which is almost to same

But i dont understand math at all

[modern wigeon]

yes, they are almost exactly the same

[oak fog]

especially in english

I just need someone to do the first question like the question that was done in the notebook

my proffesor has a specific way of doing steps

and if i do them any other way

[modern wigeon]

do you understand these steps from the notebook:

• 2^{2x} - 10 * 2^x + 16 = 0
• 2^x = t
• t^2 - 10t + 16 = 0

[oak fog]

it would not count

[oak fog]

uhm

kinda

[modern wigeon]

2^{2x} - 10 * 2^x + 16 = 0 is a quadratic in disguise because 2^{2x} = (2^x)^2

so if we create a new variable t = 2^x, then we can substitute t and get t^2 - 10 t + 16 = 0

[oak fog]

okay

[modern wigeon]

in the new problem, you almost have the same thing, except now instead of 2^{2x} it's 2^{2x - 1}. so first you want to rewrite it into the form a * 2^{2x}. do you see how to do that?

[oak fog]

uhh

okay

[oak fog]

okay

i just tried

to do the question

but im confused withsomething

2^2x -5 * 2^x(what happend to the +1 here)

@modern wigeon

[modern wigeon]

hint: 2^{2x - 1} = 2^{2x} * 2^{-1}

[oak fog]

I dont understand

why =

there's no = on the first

[oak fog]

this is what i got

@modern wigeon

[modern wigeon]

how did you get that?

[oak fog]

i just tried to copy

first question

man what

[modern wigeon]

the second line in the first question is wrong

[oak fog]

here?

[modern wigeon]

Here is a worked solution to the first question:

• We start with: 2^{2x} - 5 * 2 ^{x + 1} + 16 = 0.
• Since 2^{x + 1} = 2 * 2^x, we can rewrite the left-hand side: 2^{2x} - 5 * 2 * 2^{x} + 16 = 0.
• Simplifying: 2^{2x} - 10 * 2^{x} + 16 = 0
• Now note that 2^{2x} = (2^x)^2.
• So if we set t = 2^x, then we have t^2 - 10 + 16 = 0.
• Factoring, we find that (t - 2)(t - 8) = 0.
• Hence t = 2 or t = 8.
• Substituting back, we have 2^x = 2 or 2^x = 8.
• Hence x = 1 or x = 3.

[proud ice]

!nosols

[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.

[modern wigeon]

This is not a solution to his question. It's a correction of the notebook solution he linked.

[oak fog]

okay wait

how did we go from -5 to -10

[proud ice]

I retract my !nosols then

[oak fog]

is it because -5 *2 = -10?

imean

yeah

butjustmakingsure

[modern wigeon]

yes

[oak fog]

okay

think i got it

[modern wigeon]

once you understand this solution, you should be able to solve the new problem easily

[oak fog]

im

[proud ice]

Btw, sorry for the trigger-happy !nosols, I just have to be wary with new members. But I'm seeing your other posts and you seem to be acclimating here very well.

So allow me to reset, and say welcome to the server.

[limber hearth]

Can't wait to see his name in green

[oak fog]

i got

root68

but i cant get root from 48

68*

[modern wigeon]

I don't think that's correct

if you look a bit more closely, you will see that your practice exam problem is extremely similar to the notebook problem

[modern wigeon]

ah, you dropped the 2^{-1}

when you changed to t

[oak fog]

where

i just followed my note book

[modern wigeon]

everything looks good until 2^x = t. then you didn't substitute correctly

[oak fog]

so what should i change

i dont understand

[warm umbra]

Send me the problem

[oak fog]

.

[warm umbra]

No i meant type it

I dont see things clearly

[oak fog]

Its as clear as it can be

[warm umbra]

What s the goal ?

[warm umbra]

Maam im trying to help lol

[oak fog]

x = 1, x= 3

[oak fog]

ima a dude lol

you didnt even look at the photo

[warm umbra]

i did

[oak fog]

.

i men this

my bad

[oak fog]

need it done same as done in here

[warm umbra]

Lol i didnt see this one

[oak fog]

my badd

[warm umbra]

Im new here and first thing was the picture

So im not scrolling down lol

[oak fog]

oh lol

[warm umbra]

Anw waait

[oak fog]

i got

Root68

when im not supposted to

so idk what

uhh

@modern wigeon

helo

i think i got it

[modern wigeon]

as I said before, the mistake is in going from
2^(2x) * 2^(-1) - 5 * 2^x + 8 = 0
to
t^2 - 5 * t + 8 = 0

[oak fog]

c needs to be 16not 8

[modern wigeon]

you can't just change c to 16 lol

[oak fog]

oh

so what do i do..

[modern wigeon]

how did you get from 2^(2x) * 2^(-1) - 5 * 2^x + 8 = 0 to t^2 - 5 * t + 8 = 0

you need to understand the steps to spot the problem

[oak fog]

I mean

i just tried to follow my note book

what's it supposted to be like

[warm umbra]

I found 1 and 3

[modern wigeon]

will you have your notebook for the exam?

[oak fog]

probbobly

[oak fog]

can u flip

the

thing

[modern wigeon]

!nosols

[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.

[warm umbra]

Flip it using ur phone haha

[oak fog]

im on pc 😭

[limber hearth]

Return the pc

Jk

[warm umbra]

Hahahahha

[oak fog]

lmao

thank you

so much

im gonna add u

[warm umbra]

Lol nw

[marsh citrus]

@oak fog Has your question been resolved?

[oak fog]

uh

idk

Is this correct

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.

[lavish granite]

Is my solution correct for x?

[frosty thicket]

seems right

[lavish granite]

👍

.close

[marsh citrus]

Channel closed

Closed by @lavish granite

Use .reopen if this was a mistake.

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.

[red nimbus]

Hi, I am kinda hard stuck

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I literally have no idea how to disprove it

[main idol]

you can probably contradict Phi(0) must be (0, 0)

[upbeat sable]

Derive an expression for $\phi (p/q)$ in terms of $p,q\in \mathbb Z$ and $\phi(1)$

[elfin berry]

4573r01d|)d357r0y3r 45²

[marsh citrus]

@red nimbus Has your question been resolved?

[elfin berry]

4573r01d|)d357r0y3r 45²

[regal coral]

you can find an element of Q that generates both x and y but there's no element in QXQ that generates both (1,0) and (0,1)

[upbeat sable]

Yeah this is basically a basis argument

[red nimbus]

but i would then treat them as vector spaces, not groups

[upbeat sable]

Treat them as groups

[glacial hedge]

The argument is that a Q morphism is in fact Q-linear

[red nimbus]

wtf

[glacial hedge]

Did you read this?

[red nimbus]

no

[upbeat sable]

I wrote this because its similar to cauchys func equation which you have probably seen

[glacial hedge]

"if f: R -> R is continuous and verifies f(x+y) = f(x) + f(y), then f(x) = f(1)x"

[red nimbus]

How do we know Phi is linear?

[upbeat sable]

Because it is a homomorphism

[red nimbus]

huh

[upbeat sable]

With + on Q

I mean its by definition

[red nimbus]

no

[upbeat sable]

Yes

[glacial hedge]

The proof lies in the 3 lines of latex there

[red nimbus]

a homomorphism only shows that the operations are presevered of the structures

it's not necessarily a linear map

[upbeat sable]

Oh i see your issue

[glacial hedge]

Which is why we prove it here

[elfin berry]

4573r01d|)d357r0y3r 45²

[red nimbus]

ok i see sory

[upbeat sable]

First equality is by thr fact that phi is a homomorphism

Dont mind the misplaced )

[upbeat sable]

Yeah so after that you are pretty much done since p/q * phi(1) has a one dim range while QxQ is 2 dimensional

That is what Axe said by "you can find elements of Q that generate both x and y but nothing in QxQ that generates (1,0) and (0,1). "

Indeed say if x=m/n and y=r/s then z=1/ns generates both x and y. But there is definitely nothing in QxQ that generates (1,0) and (0,1)

[red nimbus]

Can you explain please how the 3rd line implies it's not surjective?

[upbeat sable]

phi(1) is a single vector in QxQ, so its span={ p/q * phi(1)} = range of phi is atmost 1 dim

While QxQ is 2 dim

As spaces over Q

[upbeat sable]

Honestly this is much more straightforward you should read this too

[regal coral]

then $(1,0)=\phi(x)=\phi\left(\frac{ms}{ns}\right)=ms\phi(z)$ and $(0,1)=\phi(y)=\phi\left(\frac{nr}{ns}\right)=nr\phi(z)$

[elfin berry]

Axe

[regal coral]

letting $(z_1,z_2)=\phi(z)$ we have $(msz_1,msz_2)=(1,0)$ so $z_2=0$. but then $nr\phi(z)=(nrz_1,0)=(0,1)$

[elfin berry]

Axe

[upbeat sable]

~~ => 1=0~~

[regal coral]

yeah it's a contradiction

[upbeat sable]

Im sure there are a few more ways to do this apart from these two

[spark otter]

have you seen (1,0) = phi(p1/q1) and (0,1) = phi(p2/q2), then (p2q1,-p1q2) = p2q1phi(p1/q1) - p1q2phi(p2/q2) = phi(0) = (0,0)

[marsh citrus]

@red nimbus Has your question been resolved?

[red nimbus]

guess that's what riemann meant, was too stupid to see it

[limber hearth]

||who said it was a contradiction afterall||

[red nimbus]

real 1 = 0 mod 1

[limber hearth]

Thats the math i like

[red nimbus]

but then I think we should say p1,p2 in Z\{0} and q1,q2 in N

I guess I am gonna close this, thanks to everybody for the help!

.close

[marsh citrus]

Channel closed

Closed by @red nimbus

Use .reopen if this was a mistake.

[marsh citrus]

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.

[surreal hatch]

Hello

[surreal hatch]

Hi I had a doubt in Jordan Normal Form theorem can you please surprise explain the proof of this theorem with voice clip

[marsh citrus]

@surreal hatch Has your question been resolved?

[hollow glen]

"surprise"

[gloomy merlin]

how does one "surprise explain" something

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[marsh citrus]

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.

[neat grove]

yo guys

[neat grove]

i have my mocks starting in a week

give me study advice please 😂😂 how do u guys cram for hard maths in a aeek

[proud ice]

Read your book and work on problems. Identify where you struggle and study at those points

[neat grove]

not just pastpaper questions?

[proud ice]

That's a start too, but you'll have a bias towards the problems you've already encountered.

[marsh citrus]

@neat grove Has your question been resolved?

Channel closed

Closed by @neat grove

Use .reopen if this was a mistake.

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.

[still temple]

Hi, happy new year. Just a little help if I may:

[still temple]

The graph shows a line going into the negatives

but I'm failing to understand how this function can ever be negative

because the absolute value will mean that for any x, the function will yield a positive result.

so |-8-1| + 1 = 10 for example.

What am I misunderstanding please?

[cunning fiber]

Where?

[cunning fiber]

Also, as a correction, absolute values return non-negative values, not necessarily positive (the absolute value of 0 is 0)

[cunning fiber]

2-x is negative for x>2

but the domain for the 2-x "part" is x<1, which doesn't include any part of x>2

[still temple]

I see, just gonna take a sec to process that

Sorry yeah I still don't get it, I get this part:

[still temple]

This makes sense

[cunning fiber]

[https://www.desmos.com/calculator/kdnsjgiiqk]

the red part is the part that's included in the graph

the blue part is where 2-x starts being negative

As I said earlier, none of that blue part is part of our original graph because of the domain restrictions

[still temple]

I see yeah

ooooooh

wait

Ooooh

I get it

yep its not negative, I was seeing it wrong

for g(-1) = 3

correct me if I'm wrong

that makes sense I think yeah

@cunning fiberThanks 🙏

.close

[marsh citrus]

Channel closed

Closed by @woven sinew

Use .reopen if this was a mistake.

[still temple]

god you actually went on to desmos and graphed it to show me

thanks so much lol