help-10

[errant sonnet]

so 7^log(n)=-1/2 ?

@restive acorn

[restive acorn]

If you change the left you have to make same change to the right

Making it an exponent is an action

So like usual you make sure to do it to both sides

[errant sonnet]

so instead = -1/2 i do 7^-1/2 ?

[restive acorn]

Yea

[errant sonnet]

but how does that help me

what do i do after

[restive acorn]

Because you have 7^log7

That part cancels

You're left with n on the left side

[errant sonnet]

oh but how do i get an answer eve after that

7^-1/2

or is that the final answer

@restive acorn

[restive acorn]

You can simplify that

[errant sonnet]

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[oak cape]

can someone tell me if my solution is the correct way to do this or not, or if the solution is even correct?
we know $$y(t) = e^{4t}$$ which means
$$y'(t) = 4e^{4t}$$
plugging that into the given differential equation
$$4e^{4t} = 3e^{4t} - 4t + g(y)$$
$$g(y) = e^{4t} + 4t$$
but since $$e^{4t} = y$$, $$t = \frac {1}{4} \ln {y}$$ i can sub those into the equation to get
$$g(y) = y + \ln {y}$$
which would bring me back to the original differential equation:
$$y' = 3y - 4t + (y+\ln {y})$$
??

[warm shale]

TheRuleOfEngineering

[oak cape]

i have a feeling this solution is wrong because when I try to solve the differential equation i got in the end, im not able to, its weird

[eager niche]

What do you mean you’re not able to solve it?

[oak cape]

idk how to solve for
$\frac {dy}{dt} = 4y + \ln {y} - 4t$

[warm shale]

TheRuleOfEngineering

[oak cape]

and i tried putting it into wolfram alpha and its a weird solution

shouldnt I be getting y(t) = e^4t back as the solution when i solve the ODE

[eager niche]

$$y’= 3y-4t+\left(y+\ln{y}\right)$$
With
$$y=e^{4t}$$
So
$$\frac{dy}{dt}=4e^{4t}$$
So substituting from above
$$4e^{4t} = 3\left( e^{4t}\right)-4t+\left( e^{4t} +\ln{e^{4t}}\right)$$
Or
$$4e^{4t} = 4e^{4t} -4t +4t$$
So that is true.

[warm shale]

Pixelius

[eager niche]

Wolfram alpha has trouble with functions sometimes, and inputs can be finicky

[oak cape]

i see

thanks a lot!

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[vestal stag]

Idek what it’s asking tbh. Do I just find the absolute value or something

Of both

[river falcon]

yes

KN is 3

[tired skiff]

whats the distance from i to k

[vestal stag]

Oh damn so it’s just 6…?

[river falcon]

what, no?

IK isnt 3

[vestal stag]

Oh

0 then

[river falcon]

what 0

[vestal stag]

The answer

[river falcon]

how

[vestal stag]

Idk tbh

[river falcon]

IK is 2

[vestal stag]

Oh

Alt bet I got it right

Alr*

Thanks bro

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[errant sonnet]

what do i do? i got x=-3,2 but its wrong

[unreal musk]

How did you try to solve it?

[errant sonnet]

x^2-6+x then tried to factor

[unreal musk]

(plugging in either one of those points will give at least one side of the equation to be undefined)

Yea, as above then, make sure to check the points you found work in the original equation

x = 2 doesn't work, because the logs of both sides are of negative numbers

Whereas x = -3 does work, left hand side if log_4(3), right hand side is log_4(3), which are both allowed

[errant sonnet]

but 2^2-6=-2

so why isnt x=2 right?

[unreal musk]

You can't take logs of negative numbers

[errant sonnet]

so right side of the equation would not allow it?

since log4(-2)

[unreal musk]

Both sides, the left hand side is also log_4(-2), both aren't valid

[errant sonnet]

oh yeah

i understand now Thank you!

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[wintry oracle]

Can someone verify the highlighted result. I don’t think it’s right but I can’t find where I went wrong at the same time. I remember there was a result that looked something like this: $\sup \alpha A \le \alpha\sup A$ for $\alpha\le 0$

[warm shale]

harshul

[obtuse pebble]

@wintry oracle Has your question been resolved?

[wintry oracle]

<@&286206848099549185>

[wintry oracle]

<@&286206848099549185>

[tardy epoch]

That's some hard to read handwriting

[obtuse pebble]

@wintry oracle Has your question been resolved?

[wintry oracle]

😅

<@&286206848099549185>

[wintry oracle]

Do you have any idea? I can walk you through it if you want

[tardy epoch]

Hurts too much to read

[wintry oracle]

For $x\in[x_i,x_{i+1}]$:
$$\inf {\alpha f(x)} \le \alpha f(x)$$
$$\frac{1}{\alpha} \inf {\alpha f(x)} \le \alpha f(x) \le \alpha \sup {f(x)}$$
And for $\alpha < 0$:
$$\sup{f(x)} \ge f(x) \implies \alpha\sup{f(x)} \le \alpha f(x) \implies \alpha\sup{f(x)}\le\inf{\alpha f(x)}$$

[warm shale]

harshul

[tardy epoch]

Unclear what you're even trying to do

[wintry oracle]

For $x\in[x_i,x_{i+1}]$:
$$\inf {\alpha f(x)} \le \alpha f(x)$$
$$\frac{1}{\alpha} \inf {\alpha f(x)} \le f(x) \le \sup {f(x)}$$
And for $\alpha < 0$:
$$\sup{f(x)} \ge f(x) \implies \alpha\sup{f(x)} \le \alpha f(x) \implies \alpha\sup{f(x)}\le\inf{\alpha f(x)}$$

[warm shale]

harshul

[wintry oracle]

So this means $\alpha\sup{f(x)} = \inf{\alpha f(x)}$ for $\alpha < 0$

[warm shale]

harshul

[wintry oracle]

But this does not make sense

[wintry oracle]

Man

Wait actually it does make sense

Alright never mind

It’s correct

I can use it like this right?

Seems right to me

I have my mid-sem tomorrow I can’t be spending time converting these to latex 💀

[heady turtle]

K...Good luck

[wintry oracle]

Thank you 🙏

I believe these are correct though

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[lone egret]

if f(c) is defined it cannot be an open circle right

[lone egret]

i see

ty

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[vital bay]

Cleveland is sooooo stupid 415 city by the bay always on top never gonna drop we beat Cleveland in their own turf don't ever mess with the big city boys especially from the Bay Area 415 we beat LeBron in Cleveland we beat LeBron in Miami and we beat LeBron in the Lakers nomatter where he goes he gets beaten we got much more history in the Bay Area 415 than Cleveland we got 49ers 5 super bowl champion we got history in Giants baseball we won 3 world series in 5 years and we got many great legends in giant baseball we got the Warriors we got many record including 73 and 9 and after we won 4 NBA finals champion were just that damn good and S.F city by the bay is 100 times better than Cleveland we got more history more fun In the city by the bay nothing ever happens in the sleepy town Cleveland we went to your home turf beat u got a big wooping took the championship and brought it back to the city by the bay nobody cares about Cleveland we got world wide fan base just admit it 415 city by the bay S.F bay area teams are just that damn good if u ain't form the Westcoast you guys ain't nothing Let's go Warriors

[vital bay]

pls no bann

[obtuse pebble]

@vital bay Has your question been resolved?

[stiff leaf]

?!?!

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[fallow sail]

i need help for the question b
i do untill here then dont know hot to continue alrd

[last pilot]

multiply with the inverse of 111 mod 1001

[fallow sail]

Like this?

[obtuse pebble]

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@fallow sail Has your question been resolved?

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[azure wraith]

Not sure

But split into two fractions

And integrate

[restive gorge]

My guess is also you set it equal to dz = ... then integrate both sides, and then to find such a so that it is continuous

[obtuse pebble]

@heady turtle Has your question been resolved?

[heady turtle]

What is continuous

[restive gorge]

Your function z = f(x,y)

[heady turtle]

But it doesn't have to be continuous to be differentiable iirc

[restive gorge]

what

but if it is differentiable

that implies what for z = f(x,y)

[fringe nebula]

so a = 2 and C(x) is just a constant.

[heady turtle]

You spoiled the answer😭

But thank you anyways

[fringe nebula]

sorry. but you still have to do the integrations and differentiations to get through the steps.

yes

[heady turtle]

Ok thank you

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[coarse orchid]

Hi

[coarse orchid]

Am I correct?

[light raft]

working seems okay

[coarse orchid]

I'm confused, because the answer key is -8

The answer key might be wrong?

[light raft]

maybe check with wolfram alpha

[marsh geyser]

,w lim x->2 (4-4x+x^2)/(sqrt(cos(2-x))-1)

[warm shale]

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Upgrade to WolframAlpha Pro!

Requested by _arkimond_

[marsh geyser]

@coarse orchid

Your answer is correct

Book is wrong

[coarse orchid]

Ohh, alr then tysm

[marsh geyser]

!done

[obtuse pebble]

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[coarse orchid]

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[jolly radish]

derivative of a function

for item 1 do i do the foil method first?

[restive gorge]

You can

You can also use product rule

Also are you calculating the first derivative of f'(x) or the second derivative of f(x)?

[jolly radish]

uhh theres a second one?

[restive gorge]

You wrote f'(x)

But I think you mean

f(x) = (x+2)(x²-3)

And then want to calculate f'(x)

[jolly radish]

f'(f(x))?

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[coarse swift]

hov i fix this

[shadow dagger]

lol what r u asking? when does the analytic continuation of this function equal 0

[coarse swift]

i cant under stand the logic i am confused

[jolly radish]

just curious what topic is this

[coarse swift]

sorry this i the image of integration

[obtuse pebble]

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[warm shale]

l'agit

[obtuse pebble]

@pliant adder Has your question been resolved?

[civic kettle]

set A = {4,5,6}

[safe ravine]

What

[obtuse pebble]

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[plain tulip]

hello does anyone have a reviewer for rational functions

[plain tulip]

in 11th grade rn, just trynna advance study

[obtuse pebble]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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[slow prawn]

what are they asking here

[ripe flower]

is there more context?

[slow prawn]

nope

[cursive forge]

where are A and B

[ripe flower]

there has to

[slow prawn]

oh

yeah

im

stupid

LOL

anywya ty

[rotund ice]

AB = BA ?

[slow prawn]

thats why i was so confused

they did give matrices

[rotund ice]

Oh lol

[obtuse pebble]

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[unkempt lotus]

@wary badger would you know how to do this one ?

[polar pond]

you have to find the linear function that goes throught the points A(0, 65.9) and B(206.6, 0)

Do you know how a linear function typically looks like?

[unkempt lotus]

we're doing kinematic equations

for physics

[wary badger]

you can find the acceleration

[unkempt lotus]

yeah i already did

[wary badger]

using vf = 0

[unkempt lotus]

but then i got stuck

[wary badger]

where did you get so far

[unkempt lotus]

i got accel is -10.51018877

i usesd this

[wary badger]

,w (-(65.9)^2)/413.2

[warm shale]

Results provided by WolframAlpha

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Requested by c.knief

[wary badger]

yup

checks out

[unkempt lotus]

now im stuck

[wary badger]

so how far would 32.9 m be before it stops

[unkempt lotus]

173.7

[wary badger]

so 206.6 - 32.9

yup

now you use the same formula

just with d = 173.7

instead of the full d

and solve for v

you know a and vo

[unkempt lotus]

what is this formula for? anything? lol

[wary badger]

it’s a kinematic equation that doesn’t use time

[unkempt lotus]

that's what threw me off cause the last one asked us to find time

[wary badger]

the most used i would say is x = xo + vot + 1/2 at^2

[unkempt lotus]

i thought we would have to find time for some reason

[wary badger]

you’ll get there in a bit

[unkempt lotus]

ive used that one

[wary badger]

it’s definitely the most used

in my physics experience

but atm you don’t need it

[unkempt lotus]

so v0 is 65.9?

[wary badger]

yup

[unkempt lotus]

and we're solving for the other v

ahh

[wary badger]

mhm

[unkempt lotus]

makes sense actually

okk

this is just 3.4+0) / 2 ya?

for average speed

[wary badger]

there’s no time interval

what’s the context

[wary badger]

yea sure

[unkempt lotus]

the next question asks if it took the baot 5.0 seconds to reach that speed. what is the distance

[wary badger]

i guess it’s speed

[unkempt lotus]

is that just D = RT?

[wary badger]

to reach 3.4?

[wary badger]

this is for constant speed

not constant acceleration

[unkempt lotus]

oh i thought you could use the average

okk

makes sense

[wary badger]

v = vo + at

x = vot + 1/2 at^2

but vo is 0

so it’s really x = 1/2 at^2

where a is v/t

so you get 1/2 vt

or i guess that actually works out the same lol average speed times time

yea

[wary badger]

you can

just not the 3.4

[unkempt lotus]

but the boats average speed is 1.7

[wary badger]

yea

you use that

[unkempt lotus]

so could you do d = 1.7(5)

[wary badger]

(1.7)(5)

mhm

[wary badger]

this is why

if you look at the derivation

[unkempt lotus]

so thats like the same thing as using that

[wary badger]

yes

you’re just using the end result formula

it’s good to know why it works

and being able to derive it

better than memorizing a bunch of formulas

[wary badger]

make sure you understand how i got this

[unkempt lotus]

wait

so

i get how to do it like v = v0 + at

but if we do that then its 3.4 / 5 = a

[wary badger]

don’t rely on numbers

[unkempt lotus]

and a = .628

[wary badger]

no numbers

get used to using just variables

get rid of number dependence

[unkempt lotus]

i just wana see how it works that way

[wary badger]

yea use algebra

let’s start with two equations

v = vo + at

x = vot + 1/2 at^2

since vo = 0

v = at

and x = 1/2 at^2

then we can say x = 1/2 (at)(t) = 1/2 vt

[unkempt lotus]

oh

[wary badger]

or if you prefer you can solve for a here to get a = v/t then substitute that in the x = 1/2 at^2 equation

[unkempt lotus]

i just had to do that distance formula

[wary badger]

to get 1/2 (v/t)(t^2)

which is 1/2 vt

[wary badger]

yup

notice we don’t need to calculate a at all

[unkempt lotus]

well i need an answer

eventually

[wary badger]

you do

[unkempt lotus]

yeha

[wary badger]

but the point is having the value of the acceleration is irrelevant

[wary badger]

because this is all you need

and it’s your answer

[unkempt lotus]

ok

i see

[wary badger]

(1/2)(3.4)(5)

[unkempt lotus]

which was the same as 1.7 * 5 lol

so much less work

but yes i see what you are doing

[wary badger]

you’re missing the point

[unkempt lotus]

simplyfying equations

no./..

im not at all

[wary badger]

you’re using the equation that’s derived from what i did

[unkempt lotus]

i get that

[wary badger]

you should know how these are derived

[unkempt lotus]

i just said that

[wary badger]

instead of memorizing everything

because in the event you forget

how will you solve it?

[unkempt lotus]

well

i want to solve it each way to see that they are all the same

that will help me

[wary badger]

yea that’s fine

i was just showing you why it works

[unkempt lotus]

i know why it works

thank you

the easiest part would be to realize that d = rt and you can use the average velocity for that

which is good to know since we have the ave velocity

and the time

lol

ok makes sense

i like how you did that

thanks @wary badger

my question is how do you do d = rt when you have constant acceleration as the rate @wary badger

like this one here

[wary badger]

d = rt is just v = d/t

it’s analogous to v = at if you want the rate to be acceleration

[wary badger]

this is for when vo = 0

it comes from (v - vo)/t = a

[obtuse pebble]

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[steep flax]

i cant figure out how this was wrong, was i supposed to square -4 and THEN multiply it by -3?

[steep flax]

because -3(-4) is 12
and 12^2 is 144
so what the heck

ahhh

okay

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[gusty portal]

Can someone help me w the diagram of this question please?

[polar pond]

so you got this, but where is point E?

i got it

[gusty portal]

thats what im not sure how to draw it

prolly somewhere in the triangle ADC?

[polar pond]

it has to look like this, but f needs to be twice as long as h

[gusty portal]

😭

[polar pond]

you know, how to find the length of AC don't you

[gusty portal]

ah

yes

[polar pond]

with this, you know, how long CE needs to be

[gusty portal]

I find AC okay, but how do I find CE now??

[polar pond]

the task says, that AC : CE = 2 : 1, wich means, that AC is twice as long as CE

[gusty portal]

ah yes

!

ohh

somewhat close to the answer

but

now my brain isnt working

[polar pond]

i think the best way from now is to find the angle between h and c, which is equal to the angle between f and d + 90 degrees

you know, wich one i mean?

[gusty portal]

yes yes

[polar pond]

Do you know about sinus cosinus and tangens

[gusty portal]

yes

[polar pond]

so it shouldn't be a problem for you

[gusty portal]

ah thanks!

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[polar pond]

wait, it doesn't works as i said, because the triangle DCE has not angle with 90°

[polar pond]

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[tawdry vortex]

ik it not rlly math but kind of 😭

[tawdry vortex]

has to be B right?!

like not accurate not precise

[restive acorn]

Yes

[tawdry vortex]

Ok thnx

[restive acorn]

"Accurate" means close to bulls eye

"Precise" means a low spread between shots in this case

So not accurate not precise means

Not close to bulls eye

High spread between shots

[tawdry vortex]

Yes

Thnxs very much

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[unkempt lotus]

i have the distance for half the height which is 34.4424 in meters

and the velocity is 26 m/s

at that point

<@&286206848099549185> im just having trouble setting up the kinematic equation

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[wind flame]

$\lim_{x \to 3^-} \frac{1}{3 - x}$

[warm shale]

Quantie

[wind flame]

we get $\frac{1}{0^-}$ right

[warm shale]

Quantie

[wind flame]

Why does that equal to $+\infty$ instead of $-\infty$

[warm shale]

Quantie

[daring rock]

No, the denominator is approaching 0 from the positive side

[cursive forge]

cuz that is approaching 3 from the left

so it increases

[wind flame]

Does $3^-$ mean it approaches 3 from 0 to -3 or from -infty to +3

[warm shale]

Quantie

[cursive forge]

3- means it is apporaching from -infinity

[wind flame]

So 0+ means it approaches from +infty? So the values become 1/-0.001 and so on until -infty

[daring rock]

As x is approaching 3 from the negative side, think of x as a number slightly less than 3

3-x will be 3-(something slightly less than 3), like 3-2.99999

which is positive

[wind flame]

Yeah makes sense

[daring rock]

So your denominator is 0+, not 0-

[wind flame]

On this?

[daring rock]

yeah

[wind flame]

Ok i see it now

Thank you

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[daring rock]

np 👍

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[tough geyser]

Here's a proof I organized. (thank you to everyone who helped me)
I don't think this proof is the first of it's kind, but I think the situation for the proof is rare
I couldn't find anything online so I organized it myself. Please do let me know if you know anything similar or relevant!

[opaque mesa]

what's your problem?

what's your question?

[tough geyser]

Is there a proof simmilar to this?

Or a another theorm organizing this specific instance?

(2nd page repost sorry)

[alpine bison]

I get it

[tough geyser]

It's understandable?

[alpine bison]

sorry I was thinking about another method, I hope it could help you

[tough geyser]

OH WAIT NO THANK YOU

ILL LOOK INTO IT

[alpine bison]

theta and alpha are strictly smaller than 90 as there is already an anlge equal 90 in the riangles

the one in H

in the last example H= P as P is the tangency point

[tough geyser]

OH YEAH

[alpine bison]

sin is increasing in the interval (0,90) so the inequality is preserved for angles

[tough geyser]

So the part where

the proof is proving

" Regardless of where point P is on circle O,
OAP < 90° is always true. "

Is simplified down by a lot right!?

[alpine bison]

mmhh I think so

[alpine bison]

I didn t notice that the construction of the point H also prove this fact with ease

you can always construct H as

[tough geyser]

Wait I thought you were purposlely constructing the H to cut down that specific part

[alpine bison]

if AP is secant then H is the projection of O in the line PP'

if AP is secant then H=P

[tough geyser]

Were you trying to cut down the part where I maxmize the value of Sine?

I don't understand how that would work then or am I getting them mixed up 😭

[alpine bison]

yeah in that draw you can see that the angle in A is always smaller than 90 ...and also by the pythagora theorem you can also check that A is maximum when AP is secant

[tough geyser]

Huh wait sorry I'm lost at the pythogarm theorem part

I get that Angle A has to ALWAYS be smallr than 90, but how do you know A is maximum when AP is secant?

[alpine bison]

which point cause you some problems ?

[alpine bison]

ohh sorry it is not pythoagorm theorem ...I have only used the property of the cosine

[tough geyser]

OH WAIT I SEE IT NOW

YEAH I UNDERSTAND.... THATS ACTUALLY SO SMART OF YOU

THANK YOU SO MUCH... REALLY

[alpine bison]

hahah tahnks

[tough geyser]

that is actually so mind opening

Thanks again

[alpine bison]

np I really appreciated this (apparently) simple problem

[tough geyser]

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[drifting helm]

What are the referring to here? I know how to calculate the determinant now, but I still don't know what they are trying to say here.

[wary badger]

https://youtu.be/Ip3X9LOh2dk?si=pkiDH6xwXAnaSuF3

[worn yoke]

given matrices A, B of the same size, then det(AB) = det(A) det(B)

[wary badger]

in the derivation of cramers rule it’s also the denominator for each individual xi

[drifting helm]

Do you think it's possible to just, clarify the wording they are using before linking a 3Blue1Brown video?

If you happen to know what they mean yourself.

[wary badger]

which words? determinant? scalar valued function? square matrix?

if you want an intuitive understanding of the determinant then watch the video lol

[wary badger]

this is a bit condescending sir

[drifting helm]

No, literally. If you know what they mean yourself, then I'd gladly hear you out.

[wary badger]

you didn’t specify which words confused you

[worn yoke]

if you are referring to the highlighted part then it's this

[drifting helm]

I know what each word means, I just think maybe it could be rephrased, that usually clears things up to me.

[wary badger]

all they did was list properties

[drifting helm]

When they say product of matrices, do they mean a product of 2, 3, 4 matrices? Do they refer to something that is already done when getting the determinant?

[wary badger]

^

[tardy epoch]

@drifting helm again, it's just this

You can extend that to ABCD... as many matrices

[drifting helm]

Can we just quickly define what we mean by size here?

[wary badger]

nxn

same dimensions

[drifting helm]

Okay gocha

[tardy epoch]

determinant is only defined for square matrices

[worn yoke]

we would require them to be the same size in order to multiply them (obviously we can multiply their determinants regardless)

[tardy epoch]

so it makes sense to have one number defined as "size"

[drifting helm]

I think I realized another thing that didn't feel very clear. What do they mean by this first sentence here?

They name a few attributes, but thing is, I don't know how to interpret that. Do they mean: Determinants always follow the rule det(AB) = det A * det B?

[drifting helm]

I see, they meant size both in rows and columns

[worn yoke]

that means that if you set out to make a function that takes a matrix as input and outputs a number, and you specify it should have these two properties:

• det(AB) = det(A) det(B)
• determinant of a triangular matrix is the product of its diagonal entries
  then the only function you could come up with would be the determinant

[drifting helm]

So basically, they did mean "Determinants always follow the rule det(AB) = det A * det B" ?

Was I right, is there any sign that I understand what they mean here?

[worn yoke]

it does say that, but it also says that it's a unique enough property that we can use it (alongside one other property) as a definition of the determinant

[drifting helm]

I mean, I can just verify it for myself, but it'd be easier if I could just understand basic facts about them that I read on the internet.

[drifting helm]

Okay, gocha. That sounds very neat and clear.

Now, I found an example of the determinant, and I honestly have no clue, why do they suddenly take the determinant here? Is there any reason you'd try to use the determiant trying to find the intersection point for a line and a circle?
https://mathworld.wolfram.com/Circle-LineIntersection.html

I can't think of any.

[obtuse pebble]

@drifting helm Has your question been resolved?

[obtuse pebble]

@drifting helm Has your question been resolved?

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[vagrant vapor]

what is the value of sin x and cos y? and what relationship do they both share? please help thank you!

[sturdy arch]

you forgot the formula or smth?

[vagrant vapor]

I working it out rn I’m stupid slow with geometry ☠️

[chrome thunder]

To find sin(x), make a fraction of the length of the side opposite to the angle and the length of the largest side

$\sin(x)=\frac{\textnormal{length of side opposite to angle}}{\textnormal{length of largest side}}$

[warm shale]

747244351179980930

[chrome thunder]

What’s the length of the side opposite to the angle x?

[vagrant vapor]

this is 12 i think

[chrome thunder]

The one that isn’t touching angle x

[vagrant vapor]

oh oh oh okay im spacing i forgot how to find this

im so sorry i just got back with school after awhile 😭

[chrome thunder]

It’s fine

[vagrant vapor]

is it 13?

[obtuse pebble]

@vagrant vapor Has your question been resolved?

[vagrant vapor]

is the answer for both 0.3846153846?

sorry that took me way too long

[obtuse pebble]

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[wary badger]

@hidden valve

crazy

good attempt

[glossy basalt]

Hello there @hidden valve 👋
Do you know what do adjacent angles mean?

[hidden valve]

no

[glossy basalt]

ok, do you have any lecture notes, or handwritten notes that you've learnt about them?

[hidden valve]

no because i only had one class then we just got this quiz

[glossy basalt]

two angles are adjacent to each other if they share a line, and to usually see this, we can see the angles signs are exactly next to / stick to each other

[glossy basalt]

e.g. for this question, there are no given angle indicators, so we will use the first thought

i.e. "if they share a line"

[glossy basalt]

as we can see, angle DOC and angle COB are adjacent to each other since they share the line OC

[hidden valve]

so c & d share one line and b & a do too?

[glossy basalt]

careful, when you are indicating angles, you should use names with 3 points (except named cases)
e.g. DOC where D is the starting point, O is the point where the angle is at, and C is the ending point
it's like drawing the angle itself, but using words

[glossy basalt]

so, I'm not sure about what are c&d and b&a here, sorry 😦

[hidden valve]

i'm confuesd

[glossy basalt]

do you know how to name an angle? e.g. angle DOC?

[hidden valve]

pretty sure i do

[glossy basalt]

that's great!

[glossy basalt]

so, for this question

[hidden valve]

[glossy basalt]

we have
angle COD (blue)
angle DOB (green)
angle BOA (orange)

e.g. angle COD and angle DOB shares a line, which is line OD

so angle COD and angle DOB are adjacent to each other

[hidden valve]

okayy

[glossy basalt]

I hope that helps!

[obtuse pebble]

@hidden valve Has your question been resolved?

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[cosmic fulcrum]

i am confused on what it's asking me to do here for this "multipliers" table
is it asking for the given points at the respective point
and the multipliers are the "amplitude"

[cosmic fulcrum]

is this right

[surreal forge]

yes

it seems the point of the exercise is to get you to notice that a * f(x) multiplies the value of f(x) at x by a scaling factor a

[cosmic fulcrum]

i can see now, thank you. i know how to graph trig functions however these notes confuse me and make me believe i dont lol

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[surreal forge]

first, factor denominator

[unkempt lotus]

why denominator, not numerator?

[surreal forge]

the numerator is already factored

[unkempt lotus]

oh

i mean

dont i need to expand numerator

[wet moss]

No

[unkempt lotus]

and then do long division

[surreal forge]

it may be helpful to factor the denominator before expanding the numerator

you might see why in a minute

[unkempt lotus]

oh i am allowed to cancel out?

[wet moss]

I would highly recommend not doing that

[unkempt lotus]

i didint know you can simplify

lol well i have already started lol

[wary badger]

nonsense long division sounds like a great idea!

[unkempt lotus]

i just needed help with the long division part lol

ok so i can just cancel out

[wary badger]

brother

[surreal forge]

when integrating, yes

[unkempt lotus]

ok

[wary badger]

just cancel first

[unkempt lotus]

let erase half my page

jk

but not really

let me erase half my page

[wary badger]

did you foil the entire numerator

💀

then divide that by x^2-9

[unkempt lotus]

thath part wasnt hard at all lol

[wary badger]

using long division

[unkempt lotus]

yes

[wary badger]

yea ik pascal’s triangle but still

[unkempt lotus]

i am in the middle of it

but i am cancelling that

[wary badger]

oh brother

[snow moth]

why would you do it like that

[wary badger]

who knows

[unkempt lotus]

because im so much smarter than you why else would i come here to brag about how i did it that way

[wary badger]

crazy bio

[snow moth]

is ur calc class up to u-substitution?

[snow moth]

lol mb mb

[snow moth]

its my alt instagram lol

[glossy basalt]

so.... shall we come back to the question?

[wary badger]

he’s working on it

[unkempt lotus]

ok so after you cancel it out then you still get $\frac{(x-3)^2}{x+3}$

[wary badger]

parentheses

but yes

[warm shale]

wakamole

[wary badger]

you can let u = x + 3 to make the division easier

what is x-3 in terms of u

[unkempt lotus]

can we just do it not like that for the sake of my class

i have $(x^2 -6x+9) \div{(x+3)}$

[wary badger]

huh

brother do it my way

it’s simple

[snow moth]

err im not sure if the long division for this problem is gonna cut out smoothly if ur gonna integrate without u-sub ur prob gonna use partial fraction decomp

[wary badger]

sure it does

but my way is better

[unkempt lotus]

yes plz

[wary badger]

wrong order

[unkempt lotus]

true

[snow moth]

presumably their class hasnt gotten to usub yet

[wary badger]

do you know what u sub is?

eugene

[wet moss]

Well it depends if you can integrate A/(x+3)

[warm shale]

wakamole

[wet moss]

Since doesn’t that use a u sub in itself?

[glossy basalt]

knief is helping 🙂 thanks knief~

[wet moss]

Is that question directed at me?

[wary badger]

yes

[wet moss]

If so then yes

[unkempt lotus]

ok i need to do it the long division way plz

[wary badger]

you could also use synthetic division

[unkempt lotus]

NEED

[wary badger]

but yes long division works too

[unkempt lotus]

to

[wary badger]

nah i was just saying your name

[wet moss]

Fair enough

[wary badger]

do you know how to dive polynomials using long division?

i’m not sure what the latex command is for the long division house

whatever you’d like to call it

[wet moss]

But I’m wondering how you would eventually integrate A/(x+3) without a u sub (unless you are fine with using a u sub for the trivial parts)

[unkempt lotus]

i needhelp with this part

[wary badger]

how many times does x go into x^2?

[unkempt lotus]

1

[wary badger]

cap

x times

[unkempt lotus]

tru

1 times x is x

[wary badger]

right just x^2/x

x times x is x^2

thus x goes into x^2 x times

[unkempt lotus]

yea but 1 times

is x

[wary badger]

stop trolling

[unkempt lotus]

ok

im not

trying 2

[wary badger]

so then multiply (x+3) by x

[unkempt lotus]

sry

[wary badger]

then subtract it from (x-3)^2

[unkempt lotus]

x^2 + 3x

[wary badger]

do you remember long division with numbers?

[unkempt lotus]

yeah i just forgot if i can do the entire divisor at once

i know you can sometimes

we did in class but it was like 6x^3 + 4x^2 / x^3 - x^ 2

so it was perfectly equal basically

[wary badger]

you’re getting sidetracked

[unkempt lotus]

ok

so just go 1 by one

[wary badger]

^

[unkempt lotus]

-3x + 9

[wary badger]

are you sure about that

[unkempt lotus]

oh

[wary badger]

(x^2 - 6x + 9) - (x^2 + 3x)

[unkempt lotus]

-9x - 9

[wary badger]

right

now we keep dividing

how many times does x go into -9x

[unkempt lotus]

ahh now we can divide the whole thing

-9 times and 3 goes into -9 -3 times

[wary badger]

true

[unkempt lotus]

can i do that

since theres no polynoomoials in between

or should i continue doing it 1 by 1 term

[wary badger]

you mean get x - 3?

[unkempt lotus]

ya

[wary badger]

no

[wary badger]

this

[unkempt lotus]

ok

so -9x / x = -9

[wary badger]

^

yup

so now our quotient starts with x - 9

[unkempt lotus]

ye

[wary badger]

what’s -9(x+3)

[unkempt lotus]

-9x - 27

[wary badger]

subtract that from ^

[unkempt lotus]

-18x - 36

[wary badger]

no

[unkempt lotus]

shit

just 36

[wary badger]

what’s -9 - (-27)?

[unkempt lotus]

18

[wary badger]

good man

[unkempt lotus]

now?

[wary badger]

so can x go into 18?

no.

so 18 is the remainder

which means we’re left with remainder/divisor

thus our quotient is x - 9 + 18/(x+3)

i recommend you watch some khan academy on polynomial long division

[wary badger]

now integrate

[unkempt lotus]

yeah good idea

wait

its x-9 + (18/(x+3)

dont i need to write that in a form now

[wary badger]

bro

what

[unkempt lotus]

q(x) + r(x) / dividend

[wary badger]

it is in that form

[unkempt lotus]

oh you did

[wary badger]

🤔

[unkempt lotus]

oh ok cool

so that is the new integral now?

like how do i write it

[wary badger]

huh

it’s an equivalent form of what you had before dividing it

just integrate

term by term

[unkempt lotus]

$\int x-9 \frac{18}{x+3}$

[warm shale]

wakamole

[unkempt lotus]

like this?

[wary badger]

$\int x - 9 + \frac{18}{x+3} dx$

[unkempt lotus]

are both integrals

[warm shale]

knief

[unkempt lotus]

so $\int x-9 + \int\frac{18}{x+3}$

[warm shale]

wakamole

[wary badger]

yea

[unkempt lotus]

so $\frac{x^2}{2} - 9x + 18 ln |x+3|$ ?

[warm shale]

wakamole

[unkempt lotus]

did i do the secod one wrong?

[wary badger]

yep +C though

it’s indefinite

[unkempt lotus]

this is my teachers answer key

y do they have 36

i think we did math wrong

[wary badger]

,w integral (x-3)^3/(x^2-9)

[warm shale]

Results provided by WolframAlpha

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

Requested by c.knief

[wary badger]

what

,w (x-3)^2/(x+3)

[warm shale]

Results provided by WolframAlpha

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

Requested by c.knief

[wary badger]

am i bugging

[unkempt lotus]

oo

i see

where

for the first division

when we multiplied

x (x+3)

i wrote x^2 - 3x

[wary badger]

oh

[unkempt lotus]

nevermind thats right

cause we're subtracting it

[wary badger]

no its x^2+3x

[unkempt lotus]

wait

no its wrong

ya

so it should be -3x

[wary badger]

subtracting it from x^2-6x+9

gives -9x+9

oh that’s your mistake

you said -9x-9

[unkempt lotus]

oo

[wary badger]

then the rest is right just change -9 for 9

[unkempt lotus]

so its 9-(-27)

SO 36

[wary badger]

you’ll get 9-(-27) which is 36

yea

[unkempt lotus]

ok now i have to explain all this to my mom so she gets it

[wary badger]

to your mom?

huh

is she in the class too

[unkempt lotus]

no cause if i explain it to her then it helps me remeber it

[wary badger]

true

i used to do that with history

not because i wanted to just because of remembering it

i also just found it interesting and wanted to tell them

ap world

[unkempt lotus]

nice

[wary badger]

nice

[unkempt lotus]

.close

[obtuse pebble]

Channel closed

Closed by @unkempt lotus

Use .reopen if this was a mistake.

[obtuse pebble]

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>.
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[nova vale]

ques no 6

[obtuse pebble]

@nova vale Has your question been resolved?

[nova vale]

<@&286206848099549185>

...

[daring mortar]

What happens when you plug in a for x

[nova vale]

hi

are u still here

[daring mortar]

Sure

[nova vale]

Its 0/0 form

[daring mortar]

Ok so what do you do when your limit is in indeterminate form

[nova vale]

l hospital

[daring mortar]

ok so apply l’hopital

And see what that gets you

[nova vale]

i tried

[daring mortar]

What did you get

[nova vale]

wait

a^x.loga - x^a log x

for numerator

[daring mortar]

No

What’s the derivative of x^2?? Not x^2*log2

[nova vale]

nx^n-1?

[daring mortar]

Check the second part again

Yeah

So do that with a instead of n

[nova vale]

but should i apply that in a^x or x^a?

x^a right

[daring mortar]

Yeah only x^a

[nova vale]

so it will be a^xloga-ax^a-1

what will for x^x

[daring mortar]

This one’s toigher

[nova vale]

a^a will be 0

[daring mortar]

Here’s a hint though

[nova vale]

yeah

[daring mortar]

$x^x=e^{x\ln{x}}$

[warm shale]

za

[nova vale]

hmm

so is that the derivative

[daring mortar]

No

You use that to get the derivative

[nova vale]

e^xlnx /x

[daring mortar]

No that’s not proper chain rule

[nova vale]

is this correct

[daring mortar]

No

[nova vale]

how

[daring mortar]

$\frac{d}{dx}e^{x\ln{x}}=e^{x\ln{x}}\cdot(\frac{d}{dx}x\ln{x})$

[warm shale]

za

[nova vale]

Is this uv rule

[daring mortar]

This is chain rule

[nova vale]

ok

so it should be lnx+ 1

[daring mortar]

No

Now you have to use product rule

Actually

Yes sorry

I misread oops

[nova vale]

np

so now

[nova vale]

btw what is this

is this a formula

[daring mortar]

It’s just algabraic manipulation using exponent rules

[nova vale]

Oh alr

[daring mortar]

You use it to help you get the derivative easier

[nova vale]

alr

now what

[daring mortar]

Well rewrite it after applying l’hopital

[nova vale]

ok

a^xloga- ax^a-1/ e^xlnx . lnx+1

=-1

@daring mortar now?

[daring mortar]

Simplify the denominator

Remember $e^{x\ln{x}}=x^x$

[warm shale]

za

[nova vale]

ye

[daring mortar]

And then plug in a and solve

[nova vale]

can i put a=x

[daring mortar]

Yeah just replace all X with a

And set equal to -1

And solve for a

[nova vale]

its hard to solve when log and e are around

[daring mortar]

Well just try it

And say if you get stuck

And we’ll work through it if you do

[nova vale]

ok

[nova vale]

i cant solve it

gimme a hint

[daring mortar]

Can you show me your work so far

[nova vale]

i tried taking denominator to left hand

[daring mortar]

Why is x still there

You haven’t plugged in a yet

[nova vale]

alr

[daring mortar]

Do that first it’ll simplify some of the weirdness

[nova vale]

alr

what

wait

sorry

whats wrong here @daring mortar

[daring mortar]

You didn’t expand the denominator correctly

x^x(lnx+1)=x^xlnx+x^x

[nova vale]

it will be the same no?

i forgot the brackets

[nova vale]

if u take x^x common

log a + ln a = 0

so log a= 0

ln a= 0

[daring mortar]

Ok so what’s a then

[nova vale]

answer is 1

[daring mortar]

Yep

I’m not 100% how you got that but yeah a=1

[nova vale]

thnx

[nova vale]

then how do u solve it

[daring mortar]

No no I think you solved it right

[nova vale]

log a = lna

so a = 1

thanks @daring mortar

have a good day