help-0

[grizzled zephyr]

I was thinking of doing integral from 1 to 9 - integral from 9 to 7

[vale crag]

you have 1 to 9 and 7 to 9

integral from 1 to 9 - integral from 7 to 9 tho

but yes that's the idea

[grizzled zephyr]

thanks

for b

i dont have g(u)

what do I do?

a is -√2

d too

I only was able to answer a and c , both = to -√2

@vale crag 😅

[vale crag]

you have a damn lot of questions lol

[gray isle]

its a dummy variable

[vale crag]

the fact that the variable is called u or t doesn't matter at all

ye

[gray isle]

you could replace every u or t variable with a banana and it won't make a difference to the value of the integral

[grizzled zephyr]

sorry

I am a starter in this lesson

[grizzled zephyr]

how come

f(x) is not like g(x)

[gray isle]

$\int_a^b f(x) \dd{x} = \int_a^b f(u) \dd{u} = \int_a^b f(\smiley) \dd{\smiley}$

[grizzled zephyr]

really?

[ocean seal]

ℝαμΩℕωⅤ

[grizzled zephyr]

thats weird

so g(x) = f(x)

?

[gray isle]

no

that's not what i said at all

[grizzled zephyr]

f(2) is not like f(5)

right?

then it matters?

[gray isle]

not sure what you're getting at

f(2) by itself is different from f(5) by itself yes

but that doesn't really say much

[grizzled zephyr]

mhm ok thanks I understand now

moving to a new lesson now 😄

fundemental theorm of calculus

lets see

ill close this

thanks

.close

[lone heart]

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[opal merlin]

Thinking either c or d

[storm ridge]

What have you tried

[opal merlin]

not sure what to do

[storm ridge]

$\frac{r}{s} - \frac{r+1}{s+1} < 0$

[ocean seal]

Lorentz

[storm ridge]

You can rewrite it like this

Generally it's best to not cross multiply for inequalities

In usually leads to loss of a solution

[mellow grail]

hmm true

[mellow grail]

u will do LCM here?

[storm ridge]

Ye

[mellow grail]

aight got it

[raw stream]

and u can't do it always

[storm ridge]

It would be fine for this I think, but I meant generally

[storm ridge]

Hmm

[opal merlin]

hmm

[raw stream]

cross m after transposition or before?

[mellow grail]

before

[proper tangle]

imho, using cross multiplication might be simpler for OP

[opal merlin]

So I’m thinking a)

[storm ridge]

Ye

Coz it becomes $\frac{r-s}{s{(s+1)}}<0$

[ocean seal]

Lorentz

[storm ridge]

The denominator is always positive so it's the numerator that matters

[opal merlin]

I’m on another problem one question

f(x) cannot be 0 right?

[mellow grail]

wat is f(x)

[storm ridge]

Is the pic loading?

[opal merlin]

yh

[storm ridge]

f(x) can be 0, x cannot

[opal merlin]

can’t be 0

[proper tangle]

instead of considering difference of two fractions, you can using cross multiplication, so that you only have plus sign, and you don't have minus sign

it's given that both r and s are positive numbers, so you can multiply both sides by rs, expand and cancel out rs on both sides

[opal merlin]

gotcha

[gray isle]

question seems messed up

[storm ridge]

This one no?

Ig the only value f(x) cannot be is 3

But that isn't one of the options

[opal merlin]

It is

Got cut off

of the pic

[storm ridge]

Oh

[opal merlin]

how’d you get that

[storm ridge]

For the options you sent, i just equated f(x) to each of the options and saw if there was a value of x or not
But then I found 5/x cannot be 0, so 3 + 5/x cannot be 3

[opal merlin]

so it’s not 0, it’s 3

[raw stream]

Yeah the value of function is undefined for x=0

[lone heart]

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[opal merlin]

.close

[lone heart]

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[willow mist]

ok so what I did first was split 2^x+2

so i got

(2^x)^2 - 12 x 2^x = 64

and then I replaced the 2^x with y

so y^2 - 12y -64

and I got the two values -4, and 16

but what do I do after that?

also, this question

Here I don't even know where to begin,
I did
8x^-3/2x^-1/3
2x^-6/2x^-1/3

[vale wigeon]

how did you turn $8x^{-3}$ into $2x^{-6}$...?

[ocean seal]

Ann

[vale wigeon]

don't use the letter x as a multiplication symbol.

[vale wigeon]

you got that the possible values for y are that. now, given y = 2^x as you yourself said, solve for x.

[lone heart]

@willow mist Has your question been resolved?

[lone heart]

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[long panther]

Have they integrated tan x wrong here?

I thought the integral of tan x is -ln cos x

since cos x differentiates into -sinx x, so the negative signs cancel and it becomes sin x/cos x = tan x

How did they get sec x

[hot swallow]

integration tan x can be represented by various formulas

and i mean

VARIOUS

one of them in ln sec x

[long panther]

oh ok thanks

.close

[lone heart]

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[lone heart]

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[dawn crest]

Trying to use Taylor’s to get to the result 2 but it’s not working at all

[lofty creek]

is it not possible to simplify the equation first?

that expression looks horrible

[dawn crest]

It really does

I changed it to be able to use Taylor's approximation but it didn't get me anywhere

[lofty creek]

!occupied

!occupied

[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).

[alpine sable]

Dude 2 mins ago it wasn't

[crystal thistle]

Calculate ln(3n+1)/ln n, and also ln(n^2+n)/ln n

first

which should be easy enough

[lofty creek]

it is what it is

[alpine sable]

Chill bro :/

[crystal thistle]

then n^2/ln(e^(n^2)+1)=n^2/(n^2+ln(1+e^(-n^2))=1/(1+ln(1+e^(-n^2))/n^2)=1/(1+0)=1

@dawn crest

[dawn crest]

So I should start like this?

[crystal thistle]

yeah

[lofty creek]

how did you get log(n)log(n)

nevermind

[dawn crest]

Now I get this instead but the result is supposed to be 2

[lofty creek]

shouldnt this be log(n^2)=2log(n)?

[dawn crest]

Omfg you’re right I’m a dumbass

[lofty creek]

then you should get the exact same thing but with a 2 instead of a 1 there

[crystal thistle]

well you could've just used lhopi

[lofty creek]

for this awful expression?

[dawn crest]

Yes exactly

[crystal thistle]

for the ln terms

[dawn crest]

I didn’t think of it because I would rather die than integrate the full thing but it would’ve made sense for the separate terms…

Sorry I meant differentiate

[crystal thistle]

yeah i meant using lhopital for the seperate terms

instead of taylor

[dawn crest]

Thank you sm for helping me solve it I was struggling for no reason

[crystal thistle]

but they are basically the same thing

np

[dawn crest]

.close

[lone heart]

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[wild kindle]

Does this seem right?

Its on the terminal side, however, it lies outside the unit circle

[raw stream]

👍

[wild kindle]

Thanks!

.close

[lone heart]

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[dry ibex]

Help

[dry ibex]

Okay so

Is this tangerine double organic?

[lone heart]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[dry ibex]

I'm asking a question

[chrome tiger]

It has a literal embryo inside it

Stop trolling

[lofty creek]

what...

[dry ibex]

I'm gonna fail English if I don't find the answer

[gray isle]

is it math related though?

if not it doesn't belong here

[dry ibex]

Everything in life can be math related

Am I wrong?

[lofty creek]

#discussion

[dry ibex]

Flappie why are you so against me

[lofty creek]

???

[gray isle]

are you doubling down on your intentions to post questions not related to math curriculums in channels intended for them?

[dry ibex]

What are you on about

Stop bullying me

.close

[lone heart]

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[dry ibex]

I'm done

[gray isle]

because that isn't a good idea if you intended to use this server as a resource for actual math questions in the futhre

[dry ibex]

I study math in a different language so it's not really helpful when I ask math related questions

Because idk the mathematical terms in English

[lofty creek]

you can just ask in your language

[umbral karma]

"Given three positive real numbers a, b, c satisfying a + b + c = 3, prove that:

[lofty creek]

!original

[lone heart]

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

[lofty creek]

see command

[umbral karma]

Can someone help please

[lofty creek]

!occupied

[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).

[umbral karma]

oh okay thank you

[lone heart]

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[ebon torrent]

ABCD is a rectangle and I need to find B , C , D

And I don't know where to start

[lofty creek]

draw the point directly below C on the x-axis

[ebon torrent]

ok I did, what do I call it

[lofty creek]

C'?

just something else

[lofty creek]

yes

now look at triangles OAB and DCC'

[ebon torrent]

these?

[lofty creek]

no

O is the origin

ah wait

you swapped B and D

[lofty creek]

here look at AOD and DEC

[ebon torrent]

O is (0,0) ?

[lofty creek]

yes

[lofty creek]

now what do you see about these triangles

they look kinda alike dont they?

what can we say about them

[ebon torrent]

how do I know that they are the same for fact?

[lofty creek]

look at angle OAD

call it x

AOD is a right angle

[ebon torrent]

oh

[lofty creek]

and AOD+OAD+ODA = 180

90+x+ODA=180

so ODA=90-x

right?

[ebon torrent]

yea

[lofty creek]

so now what is angle DEC?

[ebon torrent]

90 degrees right angle

so those 2 triangles share the same ratio

[lofty creek]

exactly

so we have OB/CE = BA/CB = AO/BE

i think you know how to go from here

[ebon torrent]

this will help me find B C and D ?

[lofty creek]

yes it will

[ebon torrent]

do I need to calculate the distances first?

[lofty creek]

because if we know the length of BE

then we can know the x value of C

right?

[ebon torrent]

oh yea makes sense

[ebon torrent]

so how do I get the values for this equation

[lofty creek]

have you had crossmultiplcation?

[ebon torrent]

I know what it means if thats what you ask

[lofty creek]

so, lets call the x value of B, x

so B = (x,0)

so we have OB

[ebon torrent]

ight

[lofty creek]

we also know CE

we also know OA

so what can we solve for?

[ebon torrent]

when you say we know CE what exactly do you mean?

[lofty creek]

well, what did we define point E as?

[ebon torrent]

ce/ao?

[lofty creek]

point E is the point C straight down

so it has the same x value as C

but y=0

[ebon torrent]

oh

but I dont know its X worth

[lofty creek]

doesnt have to be

how do you calculate the distance between two points?

[ebon torrent]

d formula

[dry ibex]

Hey flipflop message me if you need any help in math

[ebon torrent]

@lofty creek so CE/BO

[lofty creek]

distance = sqrt((x2-x1)^2+(y2-y1)^2 right?

but here x2 and x1 is the same by definition

so x2-x1=0

thus the distance is y2-y1

which are both given

[dry ibex]

Duh

Obv

[ebon torrent]

so this is the distance?

[lofty creek]

yes

that is EC

so we have OB, OA, and EC

so then what is BE?

[ebon torrent]

@lofty creek when you say we have OB what do you mean?

I did not calculate the distance there

[lofty creek]

we have the distance OB

we called it 'x'

[lofty creek]

here

[ebon torrent]

oh ok

[lofty creek]

you will see later

that it has to be one value

but right now

[ebon torrent]

then I write OB = x

[lofty creek]

we dont know that yet

[lofty creek]

yes

[ebon torrent]

and the distance of OA is 1 since it's Y = 1

[lofty creek]

exactly

OA = 1
CE = 6
OB = x
BE = ?

[ebon torrent]

I don't know

perhaps EB/OA ?

[lofty creek]

lets look at it this way

how many times bigger is triange BCE than OAB?

[ebon torrent]

without knowing EB I can't tell you that

or OB

I would say 6(ce)/x(ob)

@lofty creek is this correct^

[lofty creek]

that looks correct, yes

so with this multiplication factor

we can go from OA to BE

right?

[ebon torrent]

yes

[lofty creek]

so do that

[ebon torrent]

OA/EB ?

@lofty creek I'm not sure how to do it

[lofty creek]

ffs man

you said

you knew how to crossmultiply

we have CE/OB = EB/OA

[ebon torrent]

so x = 0

[lofty creek]

???????

[ebon torrent]

6/x = 0/1 ?

[lofty creek]

why tf would EB be 0?

what are you even typing

go think

[ebon torrent]

I dont have B's X so how would I know

I just know that OB = x by what you determined

[lofty creek]

so then write EB as a function of x

???????????

[ebon torrent]

did you say earlier that EB is 3?

[lofty creek]

i said assuming

for an example

not that it actually is 3

BE is what we want to solve

[ebon torrent]

can you help me a bit with how to solve for BE I'm getting lost

[lofty creek]

make sense now?

A = (0,1)
B = (x,0)
C = (?,6)
E = (?,0)

CE = 6
OB = x
BE = ?
OA = 1

yes?

[ebon torrent]

in the middle its bc/ab right

[lofty creek]

yes

[ebon torrent]

ok hold on I'll write it down and come up with an answer

[lofty creek]

$\frac{CE}{OB}=\frac{BC}{AB}=\frac{BE}{OA}$

[ocean seal]

Flappie

[ebon torrent]

is this enough information to use this equation?

[lofty creek]

yes why not??????

[ebon torrent]

you seem to be upset with me... I'm not trying to waste your time, I'm genuinely struggling with this

I forgot most of this material

[lofty creek]

but thats not what youre telling me

[ebon torrent]

my issue here is that I don't have the distance for be and ob

[lofty creek]

youre telling me you know one thing

then i ask you that one thing

and you dont know it

[ebon torrent]

I know cross multiplication

[lofty creek]

we have OB

its x

so then BE = OA*CE/OB

[ebon torrent]

so i wrote down 6/x =

[lofty creek]

BE = 1*6/x

so BE = 6/x

[ebon torrent]

yes but after that I don't know how to continue

[lofty creek]

what would be the x coordinate of C

[ebon torrent]

I don't know what you mean by that

[lofty creek]

we have

OB

and we have

BE

[raw stream]

he is asking abscissa

[lofty creek]

so then we must have OE

which is the x value of OE

and we defined E as having the same x value as C

so x value of C is=?

[lofty creek]

whats that

[raw stream]

P(x,y) , x-> abscissa , y-> ordinate

[lofty creek]

i have never heard of abscissa

but sure

[ebon torrent]

C = x + BE ?

[lofty creek]

yes and BE = ?

we just calculated it

2 seconds ago

[raw stream]

fr?

[lofty creek]

i legit have no clue what abscissa is

[raw stream]

Ok

[lofty creek]

never used it

never heard of it

well, probably used it

[raw stream]

u have used it but never knew the name

[lofty creek]

interesting

[ebon torrent]

did we calculate the value of BE ?

[lofty creek]

WE ALREADY HAVE BE

[raw stream]

.

[lofty creek]

sorry

[raw stream]

memory loss?

[ebon torrent]

alright so its 6/x

[lofty creek]

yes

so fkin plug that shit in

and bits bats boish

[raw stream]

u forgot the equation now didnt u?

[ebon torrent]

look its really simple for you and a lot more complicated for me. Can you tell me how to write down the equation or formula or how to plug that fkin shit in because that might help me understand it for the next rectangle that I'll solve

[lofty creek]

i already gave you the formula

[raw stream]

C = x + BE , you have be = 6/x

[lofty creek]

and the stuff oyu need to plug in

but youre not doing it

[lofty creek]

.

[raw stream]

substitute better word

[lofty creek]

.

[ebon torrent]

let me write it down

[raw stream]

c = x + 6/x

[ebon torrent]

ok thats what I wrote

[raw stream]

nice

[lofty creek]

you continue

ty

im taking a big break

[raw stream]

i gtg srry

[ebon torrent]

its ok... I appreciate the help

[raw stream]

Oh np, any doubts u could dm him

[ebon torrent]

its just not for my level..

[raw stream]

What grade are you in

[ebon torrent]

It's not about grade.. I'm completing mathematics after 10 years of not doing it

[raw stream]

What

How old are you?

[ebon torrent]

27

[raw stream]

What the hell

I am just dam

[ebon torrent]

I'm in a class with a bunch of people my age struggling with this shit too funny enough

you tend to forget most of this stuff after 10 years of not doing it

.close

[lone heart]

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[rose mulch]

find the zero of the polynomial - p(x) = 2x^2 + 3x - 2

[rancid jay]

use the quadratic formula

[raw stream]

no need

just factorise it

by splitting the middle term

it will be faster 💯

[rancid jay]

u just need to plug in values into the quadratic formula
much simpler

true

its faster

[raw stream]

really?

I betcha

[rancid jay]

i meant factorizing is faster

[raw stream]

Oh

np

both of em work

[rose mulch]

I am getting an introductory lesson to polynomials...so in my exercise the highest degree of the polynomial was 1 in the questions for finding the zero of the polynomial...so I was wondering if we can find the zero of a polynomial where degree is more than 1?

so I guess cubic formula would work when degree is 3. but what if the degree is 10?

or a really large number....Does it mean that for a degree of 10 there will be ten soln as zeroes of the polynomial....and is there any way to find out the zero of it?

[hushed locust]

an important skill when working with polynomials is to factor them (split them into lower degree polynomials multiplied together)
for example
3x² - 2x - 8 = (3x+4)(x-2)
the zeroes of the degree 2 polynomial on the left correspond to the zeros of the two degree 1 polynomials on the right.

There are general formulas for the zeroes of polynomials of degrees 1-4. polynomials of degree 5 and higher can only be solved exactly if they can be factored (which they can't always), but there are many techniques to find approximate zeroes.

[rose mulch]

got it

.close

[lone heart]

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[steel garnet]

I have no idea how to solve this

[buoyant saddle]

is it equal to 4?

[steel garnet]

yeah

[buoyant saddle]

id move one of the fourth roots to the other side

[violet urchin]

one way you could approach this is to move one radical to the right

[buoyant saddle]

then raise it to the fourth power

[violet urchin]

then put it to the fourth

but thats kinda bad

[steel garnet]

yeah, I need to solve it by replacing something with a variable

but I don't know what

[buoyant saddle]

u mean like some sort of u sub

[violet urchin]

yeah but for not calc i assume

[steel garnet]

I don't know how to explain this, some parts of equation

[tacit arch]

See if each term can both be 2

Or if it can be 1+3, 0+4, etc.

Raising both sides to the fourth power should work, but it's very difficult

,w solve (18+5x)^(1/4)+(64-5x)^(1/4)=4

[ocean seal]

Results provided by WolframAlpha

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Requested by rie.mann

[tacit arch]

Nevermind my suggestion

Good luck on the algebra

[steel garnet]

yeah there is a solution with replacing something but I can't find it

[buoyant saddle]

yea ik an algebraic sub

[lone heart]

@steel garnet Has your question been resolved?

[lone heart]

@steel garnet Has your question been resolved?

[remote escarp]

replace first root with a, second with b, you'll get system of equations

[steel garnet]

okay, I will try

[steel garnet]

I don't understand, I just get a+b = 4

[remote escarp]

also notice that $a^4+b^4=82$

[ocean seal]

Alisia

[steel garnet]

wait what if I replace 18 + 5x with A and make a 18 + 5x from the second root also and also replace it with A

then I get ⁴√a + ⁴√(82-a) = 4

[remote escarp]

it would be harder to work with that

[steel garnet]

wait how you got this?

[remote escarp]

$(18+5x)+(64-5x)=82$

[ocean seal]

Alisia

[steel garnet]

and how this?

[remote escarp]

what do you mean?

[steel garnet]

I don't understand where it came from

[remote escarp]

you have two roots of degree 4

and $(\sqrt[4]{x})^{4}=x$

[ocean seal]

Alisia

[steel garnet]

yes I know that but I still don't understand

[remote escarp]

what happens when you put each root in 4th degree?

[steel garnet]

Do I have to move one of them to other half to do this?

[remote escarp]

no

just see what happens when you put each root in 4th degree, then sum the 2 results

[steel garnet]

so I just multiply both of them by itself?

[remote escarp]

no, put in 4th degree

[remote escarp]

like this

[steel garnet]

wait so I get 18 + 5x + 64 - 5x then?

[remote escarp]

yes

[remote escarp]

here it is

[lone heart]

@steel garnet Has your question been resolved?

[lone heart]

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[vernal kite]

Is there some type of rule for this stuff ?

[golden viper]

3rd i think

[tardy stag]

technically none of these are right

well the first one is most correct i think

since anything can be written as itself over 1

[limpid turret]

When they say "fraction" I presume they mean rational number

[tall topaz]

[limpid turret]

but hayley still brings a good point

[golden viper]

irrational numbers can't be written as fractions

[limpid turret]

$\frac{1}{\sqrt{2}}$

[tall topaz]

One ought to define what a fraction is.

[ocean seal]

SWR

[tardy stag]

sure they can, $\pi = \f\pi1$

[ocean seal]

hayley

[vernal kite]

I think they meant rational function

It’s not that deep

[golden viper]

I meant the root of 2 can't be written as a fraction

[vernal kite]

We haven’t done pi or radical fractions or whatever

[tardy stag]

you know what they say abt presumptions, they make a pres out of ump and me

[tall topaz]

anyway

[tardy stag]

they clearly do, but the third one isn't right either on the surface

[tall topaz]

When you multiply an irrational number with a rational number, what do you get?

[tardy stag]

in terms of a rule: rational numbers can always be written as terminating or repeating decimals

[limpid turret]

$\frac{\sqrt{2}}{1}$

[ocean seal]

SWR

[vernal kite]

Irrational?

[tardy stag]

or at least potentially irrational

[lofty creek]

is it not always irrational?

[tall topaz]

So, what do you know about irrational numbers?

[tardy stag]

nope, exercise to you to figure out why

[lofty creek]

[\scalebox{0.2}{\text{bully}}]

[ocean seal]

Flappie

[tall topaz]

Flappie, stop using mathmode and \text

[lofty creek]

never

[tall topaz]

Wait hayley

Am I being dumb

Isnt it always irrational

[limpid turret]

yeah I'm with pure and flappie

I think it's always irrational

[lofty creek]

PROOF BY COUNTEREXAMPLE OR RIOT

[tall topaz]

Suppose it's not irrational then you'll get a contradiction that the irrational number can be written as a fraction.

[vernal kite]

I chose the 3rd one

[tardy stag]

#math-discussion

[vernal kite]

Is that right

[tardy stag]

that's the one that seems the most correct yeah

[buoyant saddle]

if u multiply by zero

[vernal kite]

Ok thx

[tall topaz]

It says non-zero rational

[tardy stag]

I'd just change that to "potentially nonrepeating"

[buoyant saddle]

oh

[tardy stag]

oh does it

ah yes reading has failed me once again

[buoyant saddle]

^

oh yea

the question says nonzero

[vernal kite]

.close

[lone heart]

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[violet urchin]

isosceles triangle because 45 45 90, then pythagorean

[spare fern]

Oh ok, thanks.

L = 212m

.close

[lone heart]

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[crude birch]

I don't understand how:-

[alpine sable]

what do you not understand

[crude birch]

How this product applied to independent events?

[alpine sable]

are you asking why $\m\P{A\given B} = \m\P A$ given an independent event $A$?

[crude birch]

No, that I understood.

[alpine sable]

so?

the last equation?

[crude birch]

I don't understand how this works for indendent events?:-

[alpine sable]

[
\m\P{A\cap B} = \m\P A \cd \m\P B
]
basically

[alpine sable]

following [
\m\P{A\given B} = \f{\m\P{A\cap B}}{\m \P B}
]

[crude birch]

I see. One more question.

Could you please explain me this?:-

From what I understood, shouldn't this be just P (A or B) = P(A) + P(B)?

Why we need this part?:-

[alpine sable]

no

that means the two probabilities are disjoint

being disjoint is NOT the same as being independent

common mistake you should be careful of

[crude birch]

Now it makes! Thank you 🙂

.close

[lone heart]

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[cosmic acorn]

hi

[alpine sable]

hi

[cosmic acorn]

does this work

no ig

[orchid quartz]

sorry i cant read your handwriting

[alpine sable]

that doesnt work

[cosmic acorn]

what cant you read

[cosmic acorn]

why

[alpine sable]

try expressing both sides of the equation using the same base

[cosmic acorn]

does that mean i can do log(x^3) = ... and because of the fact that all the logs have the bases 10 we must not write the logs and can solve the equation after x

?

just minus and plus

[alpine sable]

all the logs dont have base 10 in your equation

not even one in fact

[cosmic acorn]

If I write log(x) then the base its actually 10

normally

[alpine sable]

yeah

okay

change the basea

base

[cosmic acorn]

into?

[remote escarp]

no, but you can use that $\log(3x)=\log(3)+\log(x)$ and it will work

[ocean seal]

Alisia

[alpine sable]

that's one approach

[cosmic acorn]

Are you sure with the ln?

ln(3x)

into log(3) + log(x)

[remote escarp]

why ln? you went from base e to base 10 as i see

[cosmic acorn]

oh that. Yes ofc sorry 😅

[lone heart]

@cosmic acorn Has your question been resolved?

[lone heart]

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[alpine sable]

what's the problem?

[lethal dock]

we have to find x

i am getting 3

book says 2

[lofty creek]

is $hg(x) = h\circ g(x)=h(g(x))$?

[ocean seal]

Flappie

[lofty creek]

!show

[lone heart]

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

[lethal dock]

oh ok

[lethal dock]

I don't know this circle notation

[lofty creek]

did you write x-3=x-3 no solutions?

[lethal dock]

yes

[lofty creek]

isnt it all x?

for all x this equation is satisfied

also

g(h(x)) is not correct

g(x)=x-3

h(x)=x^2

$g(h(x))=g(x^2)=x^2-3$ right?

[ocean seal]

Flappie

[lethal dock]

yes

[lethal dock]

that is also correct though

[lofty creek]

yes but right now what you did is that you plugged h(g(x))=(x-3)^2 into both sides

you need to do: $\g(h(x)) = h(g(x))\x^2-3=(x-3)^2$

[ocean seal]

Flappie

[lethal dock]

hmm

[lethal dock]

this has no solutions though

I've tried it

[lofty creek]

then it has no solutions

$x^2-3=x^2-6x+9\6x=12\x=2$

[ocean seal]

Flappie

[lofty creek]

not sure how you got that though

[lethal dock]

oh wait 💀

[lofty creek]

💀

[lethal dock]

damn

i solved it wrong before

[lofty creek]

haha

happens

[lethal dock]

but well what's wrong with my other method

[lofty creek]

what other method?

[lofty creek]

this one?

[lethal dock]

yeah

it gives 3

[lofty creek]

here you wrote h(g(x))

but you never actually inserted it

you just ignored it and replaced it with g(x)^2

which is g(h(x))

[lethal dock]

hmm but then why does it lead to a solution 😭

[lofty creek]

it doesnt lol

what youre doing there is (x-3)^2=(x-3)^2

and the taking the root

which is quite obviously nonsense

[lethal dock]

yeah true

but when i take the root

it makes one of them plus minus right?

i feel like this concept of adding plus minus with root is messed up in my brain

[lethal dock]

$x^2=64$

$\sqrt(x^2)=\sqrt(64)$

$x=\pm8$

right?

[ocean seal]

Jug

[lethal dock]

but then

[lofty creek]

this is like saying $\1=1\1^2=1^2\1=\pm 1\1=1 \lor 1=-1$

which obviously makes no sense

[ocean seal]

Flappie

[lethal dock]

💀 true

[lethal dock]

why can't we say $\pm x=\pm8$ here btw

[ocean seal]

Jug

[lethal dock]

im probably asking the dumbest question ever

[lofty creek]

you mean here?

[lethal dock]

yeah

[lofty creek]

if you mean this, this is then always true

assume x=12

then -x=-12

because (-1)*x = (-1)*12

?

[lethal dock]

yeah

[lethal dock]

but here if we got -x=-8 then x=-8 wouldn't be a solution

so why is it that only 8 is plus minus and not x?

even though x had the root as well

[lofty creek]

but thats not what youre saying

youre saying

$\pm x=\pm 8$

[ocean seal]

Flappie

[lofty creek]

so this says $x=8\lor-x=-8$

[ocean seal]

Flappie

[lofty creek]

if you said $x=\pm8$

[ocean seal]

Flappie

[lofty creek]

then itd be

$x=8\lor x=-8$

[ocean seal]

Flappie

[misty mauve]

this is correct in the sense that when x is positive, 8 also is stricly positive and same for the negative sign

$like the other guy explained $1^2 = 1^2 wont imply 1 = \pm 1 but rather implies \pm 1 = \pm 1 $

[lofty creek]

damn, im just "the other guy"

💀

[misty mauve]

sorry if that hurt, i shall call you flappie from now on

and what happened to latex?

[lethal dock]

oh

[misty mauve]

$1^2 = 1^2 wont imply 1 = \pm 1 but rather implies \pm 1 = \pm 1 $

texit does not seem to be working

[lethal dock]

but for x there, +8 and -8 are solutions right?

[misty mauve]

yes

[lethal dock]

if we do $\pm x=\pm 8$ then x=+8 always

[ocean seal]

Jug

[lofty creek]

Lol

Your latex broken?

[lofty creek]

Its okay 😔

[lethal dock]

nova left a gap at the end

[misty mauve]

no wait hold on, if x^2 = 8^2, then x = $ \pm 8$

[lofty creek]

XD

no wait hold on, if x^2 = 8^2, then x = $ \pm 8$

[misty mauve]

+x = + 8

[lofty creek]

Its broke

[misty mauve]

and -x = -8

[lofty creek]

Wtf

$\text{nova is a bozo}$

[ocean seal]

Flappie

[lofty creek]

Nvm, still works

[lethal dock]

which results to x=8

[misty mauve]

yes

[lethal dock]

so no x=-8

[misty mauve]

sometimes another notation is used where we write -+x = +-8

in that case -x = +8

and +x = -8

[misty mauve]

yep

[lethal dock]

which case is correct though

[lofty creek]

$\pm = -\mp$

[ocean seal]

Flappie

[misty mauve]

look closer, both equations essentially mean the same thing -x = +8 is same as +x = -8

[lethal dock]

yeah

[misty mauve]

$In summary \pmx = \pm8 states that x = 8 whereas \mpx = \pm8 states that x = -8$

[ocean seal]

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

[lethal dock]

but im saying $\pm x=\pm 8$ results to $x=8$ and $x=\pm 8$ which is commonly written results to $x=+8$ and $x=-8$

[lofty creek]

$\pm x=\mp 8 = \left{x=-8\-x=8$

[ocean seal]

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

Jug

[misty mauve]

$x = \pm 8 is correct for the quadratic equation x^2 = 8, but when \pmx = \pm8 it strictly means x=8 or -x=-8 and thats it$

[ocean seal]

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

[misty mauve]

oof latex aint gonna work today

[lethal dock]

$x = \pm 8 is correct for the quadratic equation x^2 = 8, but when \pm x = \pm8 it strictly means x=8 or -x=-8 and thats it$

[ocean seal]

Jug

[lofty creek]

...

Still going on about this?

[lethal dock]

$x = \pm 8$ is correct for the quadratic equation $x^2 = 8$, but when $\pm x = \pm 8$ it strictly means $x=8$ or $-x=-8$ and thats it

[ocean seal]

Jug

[lethal dock]

ohh thanks

[misty mauve]

hope it makes sense now

[lethal dock]

so we just do the plus minus on one side for the quadratic equation solution?

[misty mauve]

yes

only and only on one side

if you do on both side then there will be only one solution which is not true for a quadratic equation unless the equation has equal roots

even in the equal roots case you are still writing $\pm$ but this time it will be operated on 0 and +0 and -0 are the same anyways

[ocean seal]

nova

[lethal dock]

I see

thank you so much nova

and flappie

.close

[lone heart]

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[hallow mortar]

Having an issue with part b, i have no working and dont know what the division identity is. Idont know how to link the remainders of each factor together

[hallow mortar]

Just need a hint and a point in the right direction

[modest idol]

bruh why is ur text rainbow

[tall topaz]

[sour verge]

FashionTex

[tall topaz]

wait i think thats too much info

We have that
[\wrb*{\t{Dividend} = \t{Quotient}\cd\t{Divisor} + \t{Remainer}}
]

[ocean seal]

Pure

[hallow mortar]

I dont understand how to link those components together, im a bit stuck

[tall topaz]

For example

12 divided by 5 gives quotient of 2 and remainder of 2.

[\unb{\77{12}}{\t{dividend}} = \unb{\7Z{2}}{\t{quotient}} \cd \unb{\7R{5}}{\t{divisor}} +\unb{\7P{2}}{\t{remainder}}]

[ocean seal]

Pure

[modest idol]

this is all you need for the answer

[hallow mortar]

okay

[tall topaz]

Yeah

[hallow mortar]

thank you

[modest idol]

@tall topaz hwo do you get the rainbow

text

[tall topaz]

magic

[lone heart]

@hallow mortar Has your question been resolved?

[hallow mortar]

Hi pure, I still cant figure it out can ou give me another hint

please

[waxen flame]

Do you know what the Division Identity is?

[hallow mortar]

no

or not by that name

[waxen flame]

It's basically what Pure showed.\\

$P(x) = Q(x)\cdot D(x) + R(x)$

[ocean seal]

Kookiemon

[hallow mortar]

Oh yes i know that but i just dont know how to use that in this question

unless does it just want me to work back and find the quotient

and original poly

[waxen flame]

So let's refer back to the question real quick.

Notice that it states that P(x) is divided by (x-1)(x+2). Where does (x-1)(x+2) go in the Division Identity?

[hallow mortar]

Divisor

[waxen flame]

Correct.

Do you understand why part (a) is true, that R(x) = ax + b?

[hallow mortar]

Yes

[waxen flame]

So you know that R(x) = ax + b

Do you know the Remainder Theorem?

[hallow mortar]

yes

to both

[waxen flame]

So given that P(1) = 2, how can you write out the Division Indentity using the information you are given?

[hallow mortar]

P(x)= Q(x)(x-1) + 0x+2

[waxen flame]

Not quite.

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

[hallow mortar]

i think ive been approaching the whole problem wrong, ive been treating P(1) as dividing by (x-1) and P(-2) as dividing by x+2.

[waxen flame]

Yeah, that would be a problem.

Let me give you a hint on how to think about the problem.

[hallow mortar]

so P(1) would be written as Q(1)(1-1)(1+2)+a+b=1
P(-2) would be written as Q(-2)((-2-1)(-2+2) -2a+b = 5

oh i get it

thanks b=3

[waxen flame]

You almost have it.

[hallow mortar]

a = -1

R(x) = -x+3

Is that it

[waxen flame]

You need to replace x in each one, not a.

[hallow mortar]

whoops

Thank you for your help

[waxen flame]

Looks like you got the correct answer anyways.

[hallow mortar]

.close

[lone heart]

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[vernal kite]

Do I look at the table side that’s X or f(x) for -1

[modest idol]

f(x) takes the input of x

so f(-1) is 0

[vernal kite]

So I would look at the x side right

[elfin fiber]

yes. what's f(9)?

[vernal kite]

-3?

[elfin fiber]

Yes

[vernal kite]

Tysm

[modest idol]

ok type .clsoe

,close

.close

[vernal kite]

.close

[lone heart]

Channel closed

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[vernal kite]

.reopen

[lone heart]

✅

[vernal kite]

This is similar and i would do the same thing right?

I looked at the X side domain

For 4

[violet urchin]

f(x) is an element in the range here

[elfin fiber]

so think of f(x) as something where you input a value of x and it spits something out

[violet urchin]

so we want the element that is mapped to 4

not the element that 4 is mapped to

[lone heart]

@vernal kite Has your question been resolved?

[raw stream]

The range is 4 so the domain is 2

@vernal kite

[violet urchin]

!nosols

[lone heart]

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.

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[crisp hill]

7.43

[crisp hill]

This is like

Boggling my mind

Twisting it in knots

[split mantle]

Do you mind trying to take a clip of the question, instead of a whole side?

[crisp hill]

It's gonna be sideways tho

😭

[split mantle]

All fine, throw it up.

[crisp hill]

Oh I didn't knowni could just switch thst

Thats awkward

Uhm

It's the suppose y is jointly

Porportional

Thst thing

ye

[split mantle]

This might be something that I haven't been into. So you might want to wait for another person.

[crisp hill]

alrighty

uhm

[elfin dust]

Hi again

[crisp hill]

oh hello

[elfin dust]

when y is jointly proportional to x_1 ~x_100

[crisp hill]

uh

[elfin dust]

It means that y is directly proportional to each of the x_i's

[crisp hill]

how would you write that down

😅

[elfin dust]

so y=k * x_1 * x_2 * ... * x_100 for some constant k

[crisp hill]

ahaha......

i cant read text format

[elfin dust]

hold on

lemme write it in latex

[crisp hill]

ok ty

[crisp hill]

hm

i dont really understand that too well

why

😭

[elfin dust]

That's the definition on "jointly proportional"

of*

[crisp hill]

isnt uhm

jointly orportional like uhh

z=kxy

?

[elfin dust]

Yeah

when z=kxy

z is jointly proportional to x and y

[crisp hill]

ah right

[elfin dust]

so if y is jointly proportional to xi

then y=kxi

[crisp hill]

ohhh

so then y is uhm

y/xi

right>

?

[elfin dust]

you mean k?

[crisp hill]

uhm

y/xk

right?

[elfin dust]

well anyway

[crisp hill]

lol

so uhh

[elfin dust]

like jiri did

[crisp hill]

oh wait

so

uhm

like

idrk

😭

i dont understand this at all

[crisp hill]

i honestly have no clue all ik is that its like (x_1 .....x_25)3

[elfin dust]

You follow?

[crisp hill]

3 to the power of 25?

why

not rlly

[elfin dust]

Yes because jiri tripled EACH of the 25 intergers

[crisp hill]

Alright then

I'm following now

so

uhm

im confused now

like

what am i supposed to do with this lol

[elfin dust]

Let's say

Vlatko multiplied a

to each of the 100 intergers

then

Right?

[crisp hill]

i dont understand no

[elfin dust]

It's the same thing

[crisp hill]

ah