help-0

[eager kraken]

Ohhhhhhhhhh

wait so we start from

from the line in the start on the left

to the 2

[gray isle]

well you don't really care about what's going on that far back

[eager kraken]

Oh alright

[gray isle]

only what's around a bit to the left of 2

[eager kraken]

Wait since its lim f(x) we see at the Y axis for the answer

But im confuse which one

[gray isle]

what you're given is the graph of
y = f(x)

the value of f(x) at a certain value of x is represented by its y-coordinate

[eager kraken]

Ohhh

I remmeber now my teaher said that

Wait so it will be at -1 right

since its closest to 2

[gray isle]

yeh

[eager kraken]

Ohh ok that make sense

but b im confused sry

[gray isle]

same idea)
just now you approached the location where x=2 from the left
the + in superscript indicates approaching from the right

[eager kraken]

Oh ok

like this ?

[gray isle]

drawing the direction of the arrow is important

and there's no need to include the stuff far away from 2

[eager kraken]

oh ye sorry forgot

[gray isle]

and regardless, no

you extended your curve

[eager kraken]

Oh so we have to look at the curve that is close to 2 from right side?

[gray isle]

yes

[eager kraken]

Ohhh

like this right

[gray isle]

yes

[eager kraken]

-1 again?

[gray isle]

yes

[eager kraken]

Thx so much

Now for the one without notation Im confused in it also

Bc im not very sure where we have to look

[gray isle]

you consider what you just calculated for a) and b)

they are the same value, the limit will be that value

[eager kraken]

Ooooooo

Wait so

If they were both different value lets say

The limit will be DNE?

[gray isle]

yes

[eager kraken]

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

Thanks

I never looked at this that way

Oki for d

Like this?

[gray isle]

yeh

i wouldn't bother circling that 0

[eager kraken]

Oh ok

I have question please

What if there is case like c

without notation

But there is no values before it

Like its just asking for lim 2 not 2+ 2-

[gray isle]

usually you'd still have to go through the one sided limits

[eager kraken]

Oh is it fine if you can explain it please

How to find

[gray isle]

if you know the function is continuous there, you can just plug in the value / read if off the graph

well its just doing what you did in a) and b) again

[eager kraken]

Ohh

But still im little bit confused how to find it

where do I have to look

[gray isle]

can you give a specific example

c)?

[eager kraken]

Yes

[gray isle]

though i answered that already

[eager kraken]

Ik its -1 but

I just am asking like if they never even asked for 2+ 2-

How could I find 2

Finding 2+ 2- even tho they dont ask ?

[gray isle]

you'd consider what's happening when you approach from both directions

yes

[eager kraken]

OHHH

I get it now thanks so much

Oki for d

That means it will be -3 right

[gray isle]

following the curve towards x=2 from the left/right
are you approaching the same y - value

[eager kraken]

Yes

[gray isle]

yes

[eager kraken]

But I have question, why its not 3?

Is it because theres no curve from the left to the 3

[gray isle]

wdym

why's what not 3

[gray isle]

well the y-coord at the head of your arrow is -3

[eager kraken]

So we dont consider the top right?

[gray isle]

no

[eager kraken]

Only where arrrow ends

[gray isle]

yes

[eager kraken]

Oooh

Okay

for e like this?

[gray isle]

yes

[eager kraken]

ooooooooo okay

so its -3 and so f is also -3

[gray isle]

yes

[eager kraken]

can i say limit exist instead of saying -3 for f

Or I have to say -3

[gray isle]

well it does exist,
and if it exists they'd expect a value

[eager kraken]

okay

thx so much

[gray isle]

unless the ask in the form:
do the following limits exist: yes/no

[eager kraken]

I see

oki and for g

Same process as 2- earlier

[gray isle]

yes

but again, you don't need to draw the arrow that long

[eager kraken]

Ohh

Im just confused where to start it tbh

[gray isle]

you only really care about what's happenign around x=1

i'd recommend drawing the head first

and then a short tail

[eager kraken]

Is it fine if you can please show sorry for inconvenience

So I can use for future reference

[eager kraken]

Oooooh okay

tysm

[eager kraken]

Ahh okay

It will be -2 agin

[gray isle]

yes

[eager kraken]

Oki

am confused for j

is it like this?

[gray isle]

yeh

[eager kraken]

Wait so I have to look at this line?

[gray isle]

no

the dashed line is to help you read coodinates

[eager kraken]

Ohhh

wait so its 0 then

since theres nothing else

[gray isle]

you pretty much only care about the head of your arrow

so answer to j) is 0

[eager kraken]

ah Okay

Uhhh

Im confused with k 😂

The line on top of it

Is also at 3

and bottom line

[gray isle]

well pay attention the the sign for the direction

for k), ^+ you only care about the right side

[eager kraken]

Ohhh

So we focus on the first curve on the right side

[gray isle]

wdym by first curve

[eager kraken]

Not the other one

[gray isle]

i wouldn't number the curves/ call that the first

[eager kraken]

Oooh alright

[gray isle]

for the right side, you don't care about what's going on at the left of 3

[eager kraken]

Ohh Okay

So it will be 3

[gray isle]

yes

[eager kraken]

Thx so much

and since we have different values

Its DNE

[gray isle]

yes

[eager kraken]

thx bro :D

[gray isle]

np

[eager kraken]

Sorry for wasting ur time

Have nice day

[gray isle]

You learned quite a bit so no time wasted. all g

[eager kraken]

my teacher just upload another one same thing but it has some extra things that im not sure familiar with 😭 😭 😭 😭

is it okif u just help me with the first few bc the graph is weird tbh

[gray isle]

sure

[eager kraken]

thx so much bro u really helped me a lot today

Is the picture clear?

[gray isle]

yes

[eager kraken]

Oki

But tbh my main questions are what are those extra lines and how to find +infinity and -infinty

[gray isle]

those extra lines at x=-2, x=2, y=1
are vertical and horizontal asymptotes
they should've been dashed

[eager kraken]

Oh

So we odnt consider it right

its like the lines in the other graph that were dashed

[gray isle]

they aren't part of the main graph,
they give an indication of what values are being approached / undefined

[eager kraken]

Oooo alright

Wait for +infinity we look at which quadrant?

[gray isle]

you consider what happens as x is getting very large in the +(positive) direction

[eager kraken]

Likethis?

[gray isle]

yeh

[eager kraken]

Oooh alright

[gray isle]

same idea, you'd draw the arrow/tail
and here the asymptote indicates the value being approached

[eager kraken]

tbh idk where it starts from 😂

[gray isle]

wdym where it starts

[eager kraken]

like the tail

or is this right

[gray isle]

doesn't matter too much, this is fine

the full graph would extend infinitely, but this is sufficient to identify the limit

[eager kraken]

Alright

wait but if its inifnity how can you determine how much the number will be

[gray isle]

and here the asymptote indicates the value being approached

[eager kraken]

using the asymptote?

[gray isle]

yes

[eager kraken]

Ohh

So we kinda have to consider it this time

but when we look to find the infinity we look at the main graph

[gray isle]

https://www.desmos.com/calculator/nylktdel9i

for a better graph

[eager kraken]

Ohh

Dang u made it fast

Wait so basically what ur saying is that the x will infinitily keep going near the 1 on the y axis?

[gray isle]

as x keeps increasing the y value approaches 1

[eager kraken]

Oooh alright so a) is 1

[gray isle]

yes

[eager kraken]

Alright

and for b

x apporaching negative infinity like this?

[gray isle]

yeh

[eager kraken]

Ohh so it'll be 1 again

[gray isle]

yes

[eager kraken]

tysm

I think i got the infinity idea now

is C like this?

Same thing as demos

[gray isle]

yes

[eager kraken]

Oh so it will be infinity?

• infinity

[gray isle]

yes

[eager kraken]

since no intercept at Y

Ahh ok

for d like this?

[gray isle]

yes

[eager kraken]

ohh okay

so it will be

-infinity and

i mean e dne

[gray isle]

yes

[eager kraken]

thx

I think I got it now I really appreciate ur help

[gray isle]

np

[eager kraken]

I will do it and ask if its right if ur available

Thx a lot bro talk to you later

I finished but im unsure about i) @gray isle

I wrote -infinity for i

[gray isle]

is incorrect

[eager kraken]

Oh

[gray isle]

for i), refer back to what i told you in the previous question set

(draw those arrows)

[eager kraken]

Oh

I tried doing 1 apporaching left and right

I got -inf for both

ill try it again

This is for

1 from left

Oh wait

WAIT

I just realized that its more than 1 if i do that

so it has to be 0

or still wrong

[gray isle]

redraw your arrows

[eager kraken]

1^-

[gray isle]

yes

and for 1^+?

[eager kraken]

tbh idk how to draw the arrow for the 1+

but it touches here at the curve 😭

[gray isle]

same idea, arrow head facing the 1, tail going in the right direction

[rain marsh]

lim f(x)_ x \to 1+ = 0

[gray isle]

but yeh,the limit will be 0

[eager kraken]

like this?

[gray isle]

yes

[eager kraken]

lets go

bro

[rain marsh]

lim f(x)_ x to 1+ = lim f(x) _x to 1-=0

[eager kraken]

thanks so much u saved me

i appreciate all ur help

now ill come to class with an idea of what is happening at least

[gray isle]

and if the curve is clearly continuous or has a removeable continuity at the specified location,
you can just take the point/value that's there

[eager kraken]

wait whats

continuity

the teacher said he will teach us that but im not sure tbh

[rain marsh]

that is theorem

[gray isle]

in simplest terms, curve is unbroken
can be drawn without lifting your pen

[eager kraken]

I think i get it lmao

so basically what we did for i) right

[lone heart]

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[dull trail]

I know this is physics but could anyone help with 3?

[glad root]

Try using X = 1/2 . a.t^2

Here we know x and t cuz of the graph

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[alpine sable]

I have a math problem

[plucky orbit]

Take all numbers except v to the right side

You should get the intervals

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Hmm

I know that bit

but how would it be writen for and?

[fresh wedge]

Did you find v> x and v<y values ?

[alpine sable]

I'm kinda struggling to under stand these types of questions

v>6, V<1

[fresh wedge]

Yup that's the range

[alpine sable]

but who would it be written

1, 6?

[fresh wedge]

(x,y) are used when there are no equals
Else [x,y]

Oh you have 2 intervals

[alpine sable]

if you can vc to show that would help more

can I dm vc?

[fresh wedge]

No , i am on mobile and I don't have pen paper unfortunately

[alpine sable]

ok

[fresh wedge]

See

If one solution is
x<a that means the interval is (-infinity,a)

x>b that means the interval is (b,infinity)

[alpine sable]

is it this?

[fresh wedge]

Yup

[alpine sable]

Ah so that is how alright

[fresh wedge]

Yes if equal sign was also there then it would have been
,1][6,

Like this

[] are used for inclusion

[alpine sable]

Oh it says No solution

I do dislike these problems

So new one

(y, -2) (y, 5)

[fresh wedge]

Yeah should be

[alpine sable]

Yes

[karmic harbor]

hi

[alpine sable]

I want to get better at those so I'm gonna do a few more

(x, 5] [2, x)

Wait it over laps

so NS?

or can it overlap?

[fresh wedge]

What overlap?

And write in ≥ ≤ type not (x,5] and [2,x) type 😅

[alpine sable]

because it passed 2 for the second

[fresh wedge]

x≥5 and x≤2

[alpine sable]

Ok then

[fresh wedge]

So again 2 intervals, but with inclusion so use []

[plucky orbit]

Shouldn't x<=5?

Since it has a - sign

[fresh wedge]

Both side has - should not it get cancelled? I might be wrong

[plucky orbit]

When - gets cancelled the inequality still changes (for example -2>-3 but 2<3)

[fresh wedge]

But I think you are correct

[fresh wedge]

@alpine sable ^^

My bad

[alpine sable]

hmm

[alpine sable]

So it's this?

[plucky orbit]

The answer?

It should be x belongs to (-inf.,2] U (-inf.,5] which results into (-inf.,5]

[karmic harbor]

gutys

i have a question

[alpine sable]

I do dislike these prolems

[karmic harbor]

what is -root-36

[alpine sable]

v<=6 U v >3

[plucky orbit]

Yea

[alpine sable]

how is that wrong?

nvm

I saw

[plucky orbit]

Never use closed brackets "[]" on infinity

Also 3 has no equality sign

[alpine sable]

OMG

So this is my weakness

[fresh wedge]

Its actually (-infinity,6] u (3, infinity) which can be written as
(-infinity, infinity)

[alpine sable]

I may take a break and come back later to this

So I'm more calm

[fresh wedge]

[lone heart]

@alpine sable Has your question been resolved?

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[daring tree]

Hello i just started linear algebra, and my book introduces the word parameters, rather loosely. From mult. calculus i know, that if a function f(x,y), where x(t), y(t) can be described as being parametrized by t. However is that how it's meant to be understood in linear algebra aswell, simply that t would be an independent variable, where x,y... are dependent on t ?

[naive valley]

can you show a screenshot from your LA book?

[daring tree]

I mean it does describe it as independent variables. But is that how im meant to understand theorem 1.2.2?

[mortal trellis]

well here the variables x1,x2,x3,x4 are all functions of s and t

just like in your example x and y are functions of t

[daring tree]

So theorem 1.2.2 states that there are n-r independent variables?

[mortal trellis]

yes

[daring tree]

So everytime i read parameter i just think independent variable

and in this case r dependent variables, n-r independent ones,

[mortal trellis]

yes

[daring tree]

Okay neat thank you again Denascite - I confused myself again.

.close

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[fickle musk]

here n and m are natural number and x is rational then it is 0

[fickle musk]

i wanna understand the behaviour of it when x is irr

[naive valley]

the inner limit won't exist, right?

due to equidistribution

[fickle musk]

yea

nvm

guess its too complicates

[vapid shuttle]

Hello sir

[naive valley]

good morning/afternoon, welcome to doubt clearing time

[vapid shuttle]

No need to welcome me to a time I’m constantly in

[naive valley]

real

[lone heart]

@fickle musk Has your question been resolved?

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[gentle jay]

I need help with the problems that has the lil white dots

[buoyant saddle]

inverse trig functions essentially ask for the angle which when substituted into the normal trig function gives the input

[gentle jay]

Yeah

[buoyant saddle]

so for 9

it’s asking "what angle of cotangent makes cotangent equal to 1"

[gentle jay]

Yeah

[buoyant saddle]

and there is of course a restriction

on the domain

for these functions

because there are infinitely many such angles

[gentle jay]

Right

[forest marsh]

What knief is saying will be the same for all dotted problem

[gentle jay]

Ohh alright

[buoyant saddle]

do you know which angle makes cotangent equal to 1?

[gentle jay]

No 😭

[buoyant saddle]

what about tangent

what makes tangent equal to 1

[gentle jay]

Hollon wait lemme see

[buoyant saddle]

we know tan(x) = 1/cot(x)

so if tangent is 1

then so is cotangent

[gentle jay]

45?

[buoyant saddle]

yes or pi/4 radians

[gentle jay]

Ohhh

[forest marsh]

Same for all co-something

[gentle jay]

Ohhh alright

[buoyant saddle]

best not to confuse bom

she may interpret that as cosine is 1/sine

[gentle jay]

What do we do after we know this?

[buoyant saddle]

the co has more to do with the derivative no?

and the pi/2-x

identity

not the reciprocal identities

[buoyant saddle]

you solved it

[gentle jay]

Ohhh okay okay

[buoyant saddle]

arccot(1) = pi/4

[gentle jay]

3pi/4?

Oh wait

[buoyant saddle]

are you doing the next problem?

i have to go

perhaps @forest marsh can help you

[gentle jay]

Oh alright

[gentle jay]

I’m still doing 9

[forest marsh]

9,ok

[forest marsh]

Sure

[gentle jay]

I forgot how to solve it icl

[barren nymph]

how do i find the range or is the range always +infinity ,-infinity

[forest marsh]

!occupied

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

[forest marsh]

Please

[forest marsh]

You have arccot(1) and you know cot(pi/4) = 1

[gentle jay]

Yeah

[forest marsh]

So it will be rewrited as arccot(cot(pi/4))

Since they are inverse

It will last pi/4

[gentle jay]

Ohhh

Alright alright

[forest marsh]

Do the next one

[gentle jay]

I think I got it thank you

Idk how to end this channel

[forest marsh]

!done

[lone heart]

If you are done with this channel, please mark your problem as solved by typing .close

[forest marsh]

You're welcome

Well explained @buoyant saddle tho

[gentle jay]

.close

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[dawn quail]

why is this allowed?

for example let's make this equal to 0 to find the roots

[vapid shuttle]

Why is what allowed? There's kind of multiple things going on in that image

[dawn quail]

It seems that I can mix these brackets ()

lemme try to explain

see how the second factor is (6x - 5)

[vapid shuttle]

yes

[dawn quail]

but here the second factor is different

they both get the same result

[vapid shuttle]

They divided by 6

they don't get the same result

[dawn quail]

but only one side

[vapid shuttle]

you have to then multiply through by 6 afterward

[dawn quail]

aren't we supposed to do same operation to both sides?

[vapid shuttle]

I'm not seeing any equal signs in the image you posted, just looks like scratch work

[dawn quail]

let's say its = 0

[vapid shuttle]

If you take the 3rd line on the right hand side, and multiply by 6 it will be the same as on the LHS

[dawn quail]

i'm now dividing only one factor by 6, shouldn't i divide both?

[vapid shuttle]

Also, this wont affect the roots

if y=0 then y=0/6=0

and 0*6=0

[dawn quail]

ah, so this could be why?

[vapid shuttle]

scaling doesn't matter at the roots

yes

[dawn quail]

but if it equal something other than 0, this would fail right?

to mix the factors like this

(hope my question makes sense)

[vapid shuttle]

yes

they're not equivalent as functions

but they have the same roots

[dawn quail]

right

i think that's why i assumed you can't mix factors

normally you cannot

but if it's for the root(s), doesn't matter i guess

interesting.. dunno if this ever becomes useful

i have used the diamond method here to find the roots

but you can go about it different ways for the same result

here is another way, using diamond method

where you divide a by the factorsum you find

my method is the inverse

but end result is the same

[vapid shuttle]

I think it could be useful

[dawn quail]

how so?

[vapid shuttle]

Say you have a polynomial you want to factor but all of the coefficients are fractions

[dawn quail]

oh

[vapid shuttle]

but all you care about is finding the roots

the fractions might be annoying

so if you just mulitiply by their least common multiple

you clear all of the fractions

and can factor more easily

[dawn quail]

but first you have to make sure the polynomial = 0, right?

[vapid shuttle]

Yea you have to just be looking for the roots

it isn't true that you can just multiply the polynomial by something and it will remain the same

but the roots will

[dawn quail]

hmmm, so let's say the polynomial is 5x^2 + 3x + 5 = 4

first you would have to subtract 4 from both sides

5x^2 + 3x + 1 = 0

[vapid shuttle]

https://www.desmos.com/calculator/ri2f5yfk8i

Here is an example

starting with the red polynomial, the blue I mutliply by the least common multiple

they have the same roots

but different graphs

[dawn quail]

oh

[vapid shuttle]

And this is just because, at any point that is non-zero, multiplying by something will change where that point lands

this much is obvious

but if the point is zero, i.e, a root

then multiplying by something, it stays at 0

[dawn quail]

but you can multiply by any number, the roots are the same

[vapid shuttle]

Yes

exactly

[dawn quail]

why did you choose 5, 2, 7, 9?

[vapid shuttle]

the reason I multipled by the least common multiple was because then the coefficients become the simplest integers

instead of fractions

so to make it easier to factor

hypothetically

[dawn quail]

oh

but would that work? do you just cancel them out? i thought you gotta distribute each multiplication into all of it

[vapid shuttle]

Yes, distribute the multiplication in and you will see everything cancels

[dawn quail]

oh, i was not aware of that

kinda crazy

but wait

[vapid shuttle]

This is how it simplifies

[dawn quail]

ahh, I see now

yeah the numerators still change

lots of steps involved here

i thought maybe only the denominators cancelled out, but i see now

this almost seems like more work just to remove fractions

but that's interesting

is there a shortcut method to this?

how would you tackle it

distribute into each?
so first 5
5/5 - 5/2 + 5/7 - 5/9
then 2
10/5 - 10/2 + 10/7 - 10/9
then 7
70/5 - 70/2 + 70/7 - 70/9
then 9
630/5 - 630/2 + 630/7 - 630/9

= 126 - 315 + 90 - 70

this is really cool actually, to remove any fractions from a polynomial, just distribute the value of the denominator(s) into the polynomial

[dawn quail]

@vapid shuttle you would agree with this statement? or some caviats to this

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[rustic coral]

You can think of it as factoring out the least common multiple of the denominators

[rustic coral]

that's rlly it

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[alpine sable]

m is a natural number, $\mathbb{R}^m$ means m copies of $\mathbb R$

[ocean seal]

uop

[alpine sable]

you can think of $u=(u_i)_{i=1}^m$ where $u_i\in \mathbb R$ are reals.

[ocean seal]

uop

[rustic coral]

(you could also think of it as the space of all m-dimensional vectors with real components)

[alpine sable]

yes, or as a 1-by-m or m-by-1 matrix, if it helps with the arithmetic

You can use $\langle x,x\rangle =||x||^2$ with $x=u-v$.

[ocean seal]

uop

[alpine sable]

and also $\langle x,y\rangle = \langle y,x\rangle$

[ocean seal]

uop

[alpine sable]

yes, with this, both statements I mentioned above are trivially true

you will also need $\langle x+y,z\rangle=\langle x,z\rangle + \langle y,z\rangle$. Also from this, it follows trivially

[ocean seal]

uop

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@final oak Has your question been resolved?

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[green wave]

https://www.desmos.com/calculator/uvgg4z8kla

[green wave]

what is the correlation of this data set

is it no correlation or negative correlation?

should I not use a histogram

[lone heart]

@green wave Has your question been resolved?

[green wave]

<@&286206848099549185>

or how would I be able to view it from the scatter plot

because the way I was given the data the scatter plot doesnt represent much

[green wave]

.close

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[light carbon]

I need help finding the y(t) for this equation, i know that I have two plug in both of the initial solutions to find not a and b. But I am not sure how to differentiate the equation . Il post my work as well

[rustic coral]

Chain rule

[light carbon]

Ahh ok , how would I use the chain rule for this? I search it up online but I am not sure how I would use it with this. Maybe I’ve done before but with the exp I guess it’s throwing me off

[rustic coral]

Rewrite it as e^x instead of exp(x)

I’m sure it’ll be very vicious after that

[light carbon]

Ah ok

Let me give it a shot

I am not sure if I differentiate it correctly?

[light carbon]

👍🏻

Nice now for the answer I got b is that correct

[rustic coral]

,w a+b=5, 6a+9b=5

[ocean seal]

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[light carbon]

Ahh ok cool Il try out the a see if can get the same answer, but I see what I did wrong the exp= e^x which was my problem.

@rustic coral thank you again civil service pigeon

Your service is very appreciated

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[broken river]

How would I solve this: To reach its intended destination on time a cruise ship needs to travel north at 70 mph. However, it is traveling in the Gulf of Mexico where there is a Gulf Stream current of 12 mph in the direction S25 degreesE. At what speed and direction does the captain need to aim the boat so that the result takes them due north at the required 70 mph.

[fickle musk]

resolve the vector diagram

and see if u can figure it out

[broken river]

Ok I got it

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[soft pagoda]

hi

[soft pagoda]

how do i go about finding the width and length given area?

and knowing length is

4x width

l = 4w

[flat roost]

so 4w*w = 230400

[soft pagoda]

hmm yes so should i

let me write it cownd

down

could i

add the 4w * w

4w^2 would it be?

[rocky copper]

Yes

[soft pagoda]

okay

so 4w^2 = 230400

so say i root

to get rid of ^2

[rocky copper]

No

Divide by 4 first

[soft pagoda]

oh okay

okay

so w = 240

so we know that w is 240

l is 4x w

so

960

960x240

230400

ah

okay thanks

[soft pagoda]

considering youve been helping me alot do u mind if i post another one here for a bit of guidance

for you to take a look at if ur not busy

not entirely sure how to deal with the 9/4

or

maybe i can

i gotta figure out ways to get rid of these fractions

frick

[rocky copper]

No problem

Multiply both sides by 4

[soft pagoda]

ohhh

[rocky copper]

0*4 = 0 so it stays the same

[soft pagoda]

and the 9/4 becomes

-36/4

so

-9

tes

[rocky copper]

You could’ve just canceled the 4

[soft pagoda]

true

but im so bad at fractions it helps me when i

actually do the math and think about it that way

so after factoring i believe i get

2x-1 and 2x+9

[rocky copper]

I think so

Yes

[soft pagoda]

okay

x = 1/2

x = -9/2

okay thank you

@rocky copper

so for this one

i first did -x - 19 = -4x - 7

and got x=4

but im not sure how to the 2nd solution

[rocky copper]

x+19=-4x-7

There are technically 4 solutions but a=-b is the same as -a=b, and a=b is the same as -a=-b

So there are only 2

[soft pagoda]

wait so

hmmm

x+19 = -4x-7

and x+19 = 4x+7

do i just solve both

[rocky copper]

Yes

[soft pagoda]

ok last one for now is this

im thinking

that i can root

or maybe move all to 1 side

[rocky copper]

Factor out x^2

[soft pagoda]

would that just beee

x(4x)

[rocky copper]

Ok first

Move everything to one side

[soft pagoda]

okay

x^4 - 4x^2

[rocky copper]

=0

[soft pagoda]

yes

soz

[rocky copper]

Now

Factor out x^2

[soft pagoda]

by x^2 you mean the

4x^2?

or just the x^2 alone

is that possible

[rocky copper]

What

No

Factor it out, x^4 has x^2 as a factor and 4x^2 has x^2 as a factor

[soft pagoda]

ooh all of it

okay

hm

would

x^2(x^2-4)

right

[rocky copper]

Yes keep going

[soft pagoda]

oh do we get to factor more

[rocky copper]

yes

[soft pagoda]

x+4 x-4?

[rocky copper]

Yes

Well

[soft pagoda]

okay

wait

2

mb

[rocky copper]

Its x+2 x-2

[soft pagoda]

x+2 x-2

okay

so final answer is

0, -2, 2

[rocky copper]

Yes

[soft pagoda]

ty

ty

[rocky copper]

Im gonna go eep so i cant help anymore

[soft pagoda]

np im done

ty

[lone heart]

@soft pagoda Has your question been resolved?

[lone heart]

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[still herald]

.

claimed

[still herald]

help - 0

16y/7 < x < 37y/16

Where x and y are integers (postivive)
Find the minia of x.

minima*

[patent lantern]

@ashen basalt

[still herald]

bro what

[zinc haven]

[still herald]

no one know s you

who are uu?

[severe portal]

Isn't this just finding the lowest integer value of 16y/7?

Idk if you meant something else

[still herald]

lowest integer valure of x

posirtive

[severe portal]

Right so look at the lower bound of x

[still herald]

for any positive x

ok

y*

[still herald]

yeaj

[severe portal]

What's the smallest positive integer that can be?

16y/7

[still herald]

what

what? what?

[severe portal]

Hmm?

[still herald]

I am going to something for 10 minutes bye

[severe portal]

Oh alright

[still herald]

hi

.close

[lone heart]

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[still herald]

bye

I got it

y = 10

and x = 23

[severe portal]

Huh?

[lone heart]

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[weary stag]

U = {1,2,3,4,....}
B = {2,4,6,8,...}
C = {1,3,5,7,...}
D = {1,2,3,4,5,6,7,8,9,10}

Solve:
{(C∪D)'\B} ∩ B

[weary stag]

is chat gpt right for saying an infinite set combined with a finite set = finite set

[dark meadow]

its practically an infinite set

{1,2,3,4,5,6,7,8,9,10,11,13,15,17,19,21,23,25,27,...}

[lone heart]

@weary stag Has your question been resolved?

[carmine reef]

No

Chat gpt is not reliable for math

[dark meadow]

sometimes it is

but idk what chat gpt was doing here

[carmine reef]

No, it never is

[dark meadow]

maybe basic math only

[carmine reef]

That's the definition of reliable

It's sometimes right, never reliable

[dark meadow]

yes

it will help on basic algebra

[weary stag]

so is my given just messed up or is there an actual answer to this?

[lone heart]

@weary stag Has your question been resolved?

[meager karma]

complement is only defined if you are given universe

Oh i guess that's what U is

Well then
CuD = {a in U | a <= 10 or a is odd}
(CuD)' = {a in U | a >= 12 or a is even}
(CuD)'\B = {}
{(CuD)'\B} = {{}}
So the result is {}

[lone heart]

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[fallow galleon]

Hi can someone help me with Q14 pls

I have attached the solution to the Q which I don't understand what on earth they're doing

And sorry I accidentally uploaded this on the preuni chat because i just joined and idk how to do this

[azure needle]

The variable point is (3t, 2t²)

Where t can be any number

[fallow galleon]

Ohh

But then how do they create an equation with X equals ... And y equals ...

Because I thought you must have a function like y=X squared

And that X and y is just a co-ordinate in that

So it wouldn't make sense to form a Cartesian equation from that

[azure needle]

A point is of the form (x,y)

[azure needle]

If (a,b) = (c,d) then a=c and b=d

[fallow galleon]

Ohh because it is a 'variavle point' so that it adjusts to any coordinates

[azure needle]

Here we have (x,y) [which is how we write every point on ] and (3t,2t²)

[azure needle]

Yes

(3,3) is not variable

It is a fixed point with a certain location

But (t,t) is

Because t could be any value

[fallow galleon]

So then how would you form a cartesian equation from the 'variable coordinates'

[azure needle]

What is the way you represent a point in rectangular coordinates

[fallow galleon]

Rectangular coordinates?

[azure needle]

Yes like Cartesian

Just google it

[fallow galleon]

OHHH

Omg I actually get it

The whole changing the t thingy is to make t the subject so that you can plug it back into y=x

Just like a function

[azure needle]

🎉

[fallow galleon]

Thank you so much dear!!!!

[azure needle]

Nvm

[plucky forge]

Hey guys. Can someone please explain how to solve this?

[fallow galleon]

OMG IS THAT COMBINAOTIRCS

[lone heart]

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).

[fallow galleon]

Are you in Australia?

[fallow galleon]

Are you doongbyr11

Maths ext1

[plucky forge]

what

[fallow galleon]

Are you an aussie

[plucky forge]

Can you explain this to me please?

[lone heart]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[fallow galleon]

Ok I can try

[azure needle]

Please open your own channel

[fallow galleon]

What yr are you in?

Is anybody in Australia!I live in australian

@plucky forge

[plucky forge]

11

[fallow galleon]

SAME

Are you in Australia!

What country?

[azure needle]

.

[plucky forge]

ok done

[fallow galleon]

David are you from australia?

Im so confused because I thought everybody here is an aussie

[fallow galleon]

Where is the channel? I also would like to see the explanation for that Q because im doing the same thing as well

.close

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[plucky forge]

in the help forum thing

[lone heart]

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[lone heart]

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[valid pendant]

hello

[valid pendant]

why ao x ob = co x od

anyone

[junior kraken]

Intersecting chords Theorem

[junior kraken]

I advice you to look for the proof

[valid pendant]

that the the proof

here the ques

@junior kraken illuminate further

[junior kraken]

Google? It's a theorem bro

Search on wiki

[valid pendant]

ok

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[red spade]

hello, i really only need the domain and range, ive already solved it and i got

-3 for domain
-9 for range

my friend has a different answer from mine and now we’re confused (i cant read their handwriting so yeah)

[sterile trench]

,w x^2 + 6x range

[ocean seal]

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[sterile trench]

,w x^2 + 6x domain

[ocean seal]

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[sterile trench]

it's a polynomial so it's domain would be R

[red spade]

we dont put r

we put the numbers associated with it,

example

axis of symmetry is -3
then vertex is (-3,-9)

now, the domain is (-3) right?
and the range is (-9)?

[pallid scarab]

???

[sterile trench]

uh?

[red spade]

ah

[pallid scarab]

the domain is the set of all possible "x"

[red spade]

wait

[pallid scarab]

the range is the set of all possible "y"

[sterile trench]

yup^

[red spade]

this is how we were told to do it

[pallid scarab]

unfortunately not correct

domain is all possible x values, so "x is a real number"

Domain is $\bR$ or $(-\infty,\infty)$

[ocean seal]

rafilou2003

[pallid scarab]

Range are all the possible values of y

since you're looking at a quadratic with positive x^2 coefficient, the possible y values lie above the vertex

so $y\geq -9$ or $(-9,\infty)$

[ocean seal]

rafilou2003

[sterile trench]

see the graph would be parabola upwards (like a U) who most minimum point would be -9 and then it would tends infinity

[red spade]

oh i see

[sterile trench]

,w graph x² + 6x

[ocean seal]

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[pallid scarab]

yeah

after graphing the function, finding the domain is kinda like flattening the curve on the horizontal axis

[red spade]

got it ! tyvm

[pallid scarab]

and seeing which parts are coloured

finding the range is the same but you flatten it on the vertical axis

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[alpine sable]

To get the amplitude of a tan graph sin graph or cos graph we will always do (max - min) / 2 ?

[alpine sable]

I m trying to solve this problem

but kind of unsure about the formulas

[jagged cobalt]

tan graphs dont have amplitudes

try comparing it to a regular tan(x) graph

the vertical translation is probably the easiest to start with

then the horizontal, then the stretch

[alpine sable]

But like I did compare to regular one

I feel like it’s moved one unit up

It’s 1/4 pi forwards?

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

How to find value of a

Guys help pls

<@&286206848099549185>

It’s been 15 mins

[nimble fern]

oh hi there

ah, typical atan(x-b)+c

[nimble fern]

do you know the period of tan(x)?

[torn remnant]

Hi, can u help me with a simple math pls? I can post it here if u are free

[nimble fern]

sorry, this channel is occupied

[torn remnant]

oh okay, thank youu

[nimble fern]

please read #❓how-to-get-help 😆

[torn remnant]

thank u

[alpine sable]

Hi

Yeah so

I got b and c

how to get a

[nimble fern]

heyyy

oh nice

[alpine sable]

Do I just like

Substitute

I wanna know any proper way of doing this

[nimble fern]

substitute yea, for tan, I'll usually check the (π/4,1) point

since it's a*tan(x-b)+c
I'll first check where did the (0,0) point of tan(x) moved to this new graph