help-23

[deft pivot]

Yes

[buoyant shadow]

that's not true

[deft pivot]

And I am nott satisfied with that kind of "stories" at all , because I can make them too which probably won't be right

[buoyant shadow]

actually yeah maybe

[deft pivot]

Yes it's correct but I need some kind of math here and not stories

[buoyant shadow]

there's only stories

this is wikipedia, it's the same number with a different story

[deft pivot]

I have seen this as well

And I don't like stories I like math

Can you explain it mathematically?

Because I can make my own stories as well

[buoyant shadow]

theres's nothing else

[deft pivot]

If it is story telling thing

For example I can say :
52C1 x 3C1 x 2C1 x 48C1
And the story goes like this : from 52 I choose any 1 , then from the second car I can only choose from 3 card ( the same card with I choose the first) ,then I choose from the remaining 2 card 1 and lastly I choose from the remaining 48 card any 1 card

Of course the answer is not correct but how about the story

[buoyant shadow]

if you're convinced, then other people will likely be convinced if they think about it

if you made it up they will likely sniff it out

that's all math and all science

[deft pivot]

I agree that it's math , that's why I am asking for a proof

These stories are not proof at all

[deft pivot]

Actually here I forgot to include one more car but this does not matters

Of course no one can explain anything here

Why I had any hope to get an answer

.close

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[stable spear]

The question is a little confusing

[tardy mango]

You found the area in part b. What’s another expression (involving AK) that’s equal to said area?

[stable spear]

Hm, uh im not really sure

[tardy mango]

Hint: you used BC as the base earlier

[stable spear]

is it 192 cm = Ak x 10?

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@stable spear Has your question been resolved?

[stable spear]

yea

i understand it now

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[lean otter]

hii I was wondering how you solve 23000 ÷ 1500 written in standard form eg. 1.533 x 10^1, when I type it into calc it gives me 15.33333333, I'm wondering how do I know how many 3's I put as the answer? The answer is 1.533 x 10^1, but how do I know that?

[tardy mango]

There’s infinitely many 3s, they only put 3 cause they felt like it or smt ngl

[lean otter]

so if I put the answer 1.5333 x 10^1 or 1.533333 x10^1 would that make it wrong

[buoyant shadow]

no

2,3,4 are usual amount people like

[lean otter]

ahh okay thanks that makes sense

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[spark epoch]

sqrt(x+1) - sqrt(x-1) = sqrt(4x-1)

[spark epoch]

seems simple to solve

you'll end up with x = 5/4 once you're done

but when you plug the value back in it doesnt satisfy the original equation

i just wanna know what's wrong with the normal method of solving these questions

[magic junco]

Have you plug it again?

[spark epoch]

yeah

[magic junco]

Double check the result

Alr

May I see your work?

[spark epoch]

give me a sec

sorry my working is not the neatest lol

[empty marten]

because of the fact that you squared both sides, youve created some false solutions

[spark epoch]

makes sense

[empty marten]

so unless your working has mistakes, the question has no solution

[spark epoch]

there'd be no other way to solve this without squaring both sides though, right

[empty marten]

yup

not that i know of

[spark epoch]

cool

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[knotty plume]

my math teacher made us do this without teaching it

i need help rn on how to solve it

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@knotty plume Has your question been resolved?

[winged flare]

for the first one, CE * CD = CB^2

thats the property of a circle

[knotty plume]

whats the measurement of CD tho? they only gave us how far it is from the circle

like how do i find it

[winged flare]

CD = CE + 2r

CE * (CE + 2r) = CB^2

you know exerything except r

[small epoch]

for the first problem, the way is to construct the radius from the center of the circle to point of tangency, then you notice you have a right-angled triangle and then you can solve for x

[small epoch]

for the second problem, i believe you just use the fact that angles subtended by the same arc are the same which you could prove

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[obtuse sonnet]

i get something else

i get 2r-2v/r^2

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@obtuse sonnet Has your question been resolved?

[obtuse sonnet]

<@&286206848099549185>

help me find this deravative

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[mortal sandal]

Are you not just given the inscribed angle theorem?

Well generally it also works when O isn't inside ABC

but yes

[mossy lotus]

You dont need to join anything. Look at the angles at A and C wrt the arcs DAB and DCB

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[woven quartz]

Not sure what to do for A. Did the same as B but with just the bag. Figured C was Ui=Kf but that came up wrong.

[safe radish]

@woven quartz Has your question been resolved?

[woven quartz]

<@&286206848099549185>

[woven quartz]

For c, the box ends up moving with the concrete doesnt it?

ok got c

[median vigil]

the friction on the box does some work here

[woven quartz]

So Im sure I understand right. The gravel is removed from the equation, so it's just the box and the concrete.

Kf is the Kinetic of concrete, the box, and the Work of kinetic friction.

Im guessing I need a better understanding of Kinetic and Potential, is it simply kinetic is moving and potential is it's "potential" amount of kinetic it can start once breaking into kinetic friction?

I still can't figure A out though. 😅

[median vigil]

for part A you just need to find a force equilibrium

[woven quartz]

Its the same as B just for gravel isnt it?

Just the box includes gravel since its on it.

$f_{friction}=.7 \cdot 48 \cdot 9.81$

[flat frigate]

Ousel

[woven quartz]

us * m * g

[median vigil]

that's not how you solve for the friction on the box, either

[woven quartz]

$f_{friction}=.7 \cdot (83+48) \cdot 9.81$ for box

[flat frigate]

Ousel

[woven quartz]

which is more than the tension

[median vigil]

we have that [ f_s \le \mu_s N ] which makes it a \emph{maximum value}

[flat frigate]

cloud

[woven quartz]

correct

[median vigil]

we are not guaranteed that the static friction is maximal in this situation

it takes whichever value it needs to to maintain equilibrium

[woven quartz]

f_s is tension of rope right?

[median vigil]

,calc .7*(83+48)*9.81

[flat frigate]

Result:

899.577

[median vigil]

let's draw a free body diagram on the gravel and box to confirm

[median vigil]

can you draw the free body diagrams for the 3 objects in this scenario?

[woven quartz]

?

[median vigil]

all 3 of them are separate objects, so they should get separate free body diagrams

[woven quartz]

Not sure if I should add mention on the cord as well

Um...ok then, I think I know for box and concrete then, not sure for gravel.

Think m2 has everything, m1 needs gravity and rope tension.

Would gravel just be weight, natural force, and static since nothing is technically pulling it?

[median vigil]

go ahead and draw them, then we can discuss adding anything missing

[woven quartz]

What I got so far. Still sorta checking it.

Can't think of anything else. @median vigil

Except I guess friction on m3?

or wait third law

[median vigil]

so we could potentially have friction of m3, yes

it should be noted that the "normal force" on m3 is coming from m2, so the force from m3 on m2 is equal-and-opposite by Newton's third law

[woven quartz]

Yeah, just realized that

[woven quartz]

got cut off

$f_s = u_s \cdot N$

[flat frigate]

Ousel

[woven quartz]

im trying to remember how those forces work against each other for N....

[median vigil]

that works for the normal force

we should remember that there are also two different friction forces: one between the box and the ground, and one between the box and bag

(similar to the normal forces)

[woven quartz]

theres a normal force on m3?

I have a friction on both as well?

its 0

[median vigil]

the force you drew "F AB" is a normal force (it arises from contact between them, and is normal/perpendicular to the surface of contact)

[woven quartz]

Im not 100% how I came to that conclusion, but I did lol

[woven quartz]

Yeah thats what I thought, is that not what the drawing is showing?

[median vigil]

yes but this seemed to be a question

so what are f and F1 representing in the diagram?

[woven quartz]

the bucket pulling on the rop is F1, f on m2 is static

[median vigil]

so for consistency we should use T for the tension force in both cases

and is f the static friction between the box and the ground or between the box and the gravel?

[woven quartz]

I guess I meant m2 (for option 1) and m3 (for 2)

I know m2 is static on the ground.

I was thinking m3 is on the box but Im now questioning it.

[median vigil]

so on the box we expect to see the following 6 forces:

• weight of the box
• tension in the rope
• normal force between box and ground
• static friction force between box and ground
• normal force between box and bag
• static friction force between box and bag

[woven quartz]

so there should be 2 frictions on m2's fbd?

[median vigil]

yes

[woven quartz]

Im assuming above the box by F_BA, pointing left as well, but that doesnt seem right and idk where else it would go?

So I know the answer is 0. is it because the box is technically carrying all the mass?

oh, wait, is it because there are no x (horizontal) forces on it

only other thing I can think is the 2nd (top) f_s is a vertical direction, but im not sure how that would work. Only seen friction horizontally.

[median vigil]

we should label the two different static friction forces separately

also the static friction by m2 on m3 should be equal-and-opposite to the static friction by m3 on m2

[woven quartz]

I don't understand this.

[woven quartz]

Like this?

[median vigil]

no, i just mean they are a newton's third law force pair, like FAB and FBA

[woven quartz]

oh so...

Oh wait reread it. Thought it just read weird, but I might actually not know what youre talking about.

Like that?

or is it maybe...

Realizing I don't have an example of friction on top of or within another object.

Think the arrows might be backwards actually

[safe radish]

@woven quartz Has your question been resolved?

[median vigil]

remember that m2 (box)is the object in contact with the ground and the bag, so it should have two friction forces on it, whereas m3 (bag) is only in contact with the box and only has on friction force on it

[safe radish]

@woven quartz Has your question been resolved?

[woven quartz]

???

[median vigil]

yes

so we can now figure out all of the forces using the fact that all 3 objects are in equilibrium

[woven quartz]

Are the arrows correct for m2 and m3?

[median vigil]

yes, looks fine

[woven quartz]

f_s2 = f_s3 right?

I cant wrap my brain around this...

all forces are vertical except the friction for the bag?

except friction which requires horizontal forces.

[median vigil]

we sum forces for all objects separately

so we can start with the bag

[woven quartz]

$F=mg, W=mg, -|N|=mg$?

so

[flat frigate]

Ousel

[woven quartz]

$F = W+-|F_{AB}|$?

for y
then f_s2 for x which is 0 since no acceleration?

@median vigil

[flat frigate]

Ousel

[median vigil]

yes

[woven quartz]

so it was what I said earlier xD

[median vigil]

yes

with only one horizontal force, the only way to have 0 acceleration is for that force to be 0

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@woven quartz Has your question been resolved?

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[zealous sable]

How would I solve this?

[austere cypress]

do you know the formula for the volume of a cone?

[zealous sable]

V=1/3πr^2h

i think

Then I would substitute V with 1234^3 right?

[austere cypress]

you can substitute V with 1234cm³

[zealous sable]

like this?

[austere cypress]

not 1234³

[zealous sable]

1234cm^3?

[austere cypress]

yeah, but you could just ignore the unit for now

[zealous sable]

kk

[austere cypress]

so just 1234

another substitution you could do is regarding the radius and the height, given that they are the same

[zealous sable]

so I could just assume they're the same since theyre both unknown?

[austere cypress]

you're assuming they are the same because that's what the question says

[zealous sable]

oh

I didnt read that 😭

so then i would do

[austere cypress]

yeah, but again not 1234³

[zealous sable]

oh

oops

multiply both sides by 3?

then divide by pi?

then lastly i would cube root right

[austere cypress]

yeah

[zealous sable]

ohhh

okok

thank you!!

.close

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[zealous sable]

.reopen

[safe radish]

✅

[zealous sable]

is it just pih^3

[austere cypress]

multiplying by 3 would just cancel with the (1/3)

[zealous sable]

okok thanks

okok i get it now

i got it

thanks !

.close

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[flat frigate]

gmdn

gmdn

gmdn

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[granite sentinel]

im in algebra 2, i need help on this question for my test review.

[spark epoch]

substitute value of x from eqn 1 in eqn 2

you'll get a quadratic in y

set the discriminant to 0

[granite sentinel]

one sec

wdym set the discriminant to 0

[spark epoch]

okay wait

[granite sentinel]

b-4ac=0?

[spark epoch]

b^2 - 4ac yes

[granite sentinel]

mk

[spark epoch]

but in this question you can try completing the square

[granite sentinel]

wbt q

[spark epoch]

?

your 'c' in this question would be 7-q

if that's what you're wondering

[spark epoch]

alternatively you could ask yourself 'what constant can I add on both sides such that the lhs can be factorised into a perfect square?'

[granite sentinel]

i got this

[spark epoch]

no no

look your quadratic would be y^2 -8y +( 7-q) = 0

[granite sentinel]

7-q?

[spark epoch]

bring q to the other side

leaving 0 in the rhs

[granite sentinel]

ah

can the q be with a variable or will it always be with a normal number

[spark epoch]

wdym

[granite sentinel]

why is the q linked to the 7 what if it was y^2 -(8y-q) + 7 = 0

[spark epoch]

wait

a general quadratic is in the form ax^2 + bx + c = 0

you agree?

[granite sentinel]

yes

[spark epoch]

here, q is not multiplied by y or y^2

so it's meaningless to group it with those terms

q will have to be grouped with a constant term

so 7-q is the 'c' of this quadratic

had it been q*y then you would have grouped it with -8y

[granite sentinel]

so whoudl it be like this?

[spark epoch]

-4q

[granite sentinel]

mb

ye

[spark epoch]

yeah

[granite sentinel]

liek so?

[spark epoch]

yep

[granite sentinel]

mk thx

[spark epoch]

or you can also think like this

adding 9 on both sides can help you factorise the rhs into (y-4)^2

[granite sentinel]

my teacher dose not like guess and hcek]

like pluging in random numbers\

but i understnad it now

[spark epoch]

no no I'm not asking you to sub in values lol

it's a method of factorisation

trust me they'll understand

[granite sentinel]

mk

thank you for your halp]

help*

[spark epoch]

yw

[granite sentinel]

have a good night

.close

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[haughty ruin]

Can someone help my aim get better

[haughty ruin]

My builds are good but

My aim is terrible

[austere cypress]

<@&268886789983436800>

[warped roost]

.close

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[warped roost]

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[fading hollow]

hey im confused.. how do i do y= 2x + 3 🤔 i didnt pay antetion in class..

[finite igloo]

what do u want to find with that

[fading solar]

What are you trying to solve for

[fading hollow]

IDK it just says to do it

[fading solar]

Yeah we need a little more context

[finite igloo]

can we see

😭

[fading solar]

There's like 5 different things you can do with this

At a base level

Yeah just send the question

[fading hollow]

Ok I will

[safe radish]

@fading hollow Has your question been resolved?

[safe radish]

@fading hollow Has your question been resolved?

[chilly phoenix]

like graph it?

[safe radish]

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[lean otter]

yo

[lean otter]

help me out

my teacher hasnt taught us anything about this

[opaque dew]

circumference of semicircle

[lean otter]

uh what 🙂

[opaque dew]

do you know how to find circumference of circle

[lean otter]

nope

its our assignment my teacher havent taught us anything about this yet

[austere cypress]

,tex .2d geom

[flat frigate]

kaue

[lean otter]

why the fuc k is there a pi

[lean otter]

p m oit

[subtle thorn]

Please is this correct

[lean otter]

HEY 🤬

THIS MY CHANEL

CHANNEL

[hollow gorge]

Hi

Someone help me with math

[lean otter]

omg.

[lean otter]

uys help me out

[hollow gorge]

[subtle thorn]

Sorry

[lean otter]

it ok ❤️

[winged flare]

say we need to make a square frame

with side length 3 m

how much wire would you need

[lean otter]

what ☹️

[winged flare]

like this

---------
|       |
|       |
|       |
---------

we are bending a wire to make it in this shape

for example

not the actual question

how much would we exactly need if the side length is 3 metres?

[safe radish]

@lean otter Has your question been resolved?

[keen nexus]

wiwi

@lean otter still need help?

[lean otter]

yas

[lean otter]

🙂

[winged flare]

wha

[keen nexus]

K, do you know the formular for perimeter of a circle

if yes, then what do you think perimeter signifies

[lean otter]

no

[keen nexus]

circumference?

[lean otter]

NO

my teacher doesnt teach us anything about this

[winged flare]

ok first of all

you are sure on what perimeter is?

[safe radish]

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[keen nexus]

.reopen

@lean otter

[lean otter]

no

[keen nexus]

.

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[primal shuttle]

find this

[primal shuttle]

find y

[vagrant ice]

!status

[safe radish]

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

[vagrant ice]

you just need the angles in the same segment theorem

[primal shuttle]

i got a quyestion

the angle inside looks like its of a mid part

like dude im so confuse

thats why i cant think of the angles at the same segement theoreom

like does a question have to explicitly state the middle for me to know that

[vagrant ice]

what?

[primal shuttle]

like the diameter is like at the center

so it confuses me

[vagrant ice]

you can't assume that

[primal shuttle]

i thoughts its a semicircle

yeah exactly

[vagrant ice]

yeah it didn't say so you can't assume

[primal shuttle]

does the question have to say it?

oh i see

[vagrant ice]

yes

[primal shuttle]

alr thanks

!close

?close

[vagrant ice]

.close

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[mental star]

can someone help me set up the partial fractions for

[mental star]

$\frac {x+1}{x^2-24}$

[flat frigate]

taf ☆

[past orbit]

Question?

[vagrant ice]

sure, so this is equal to $\frac{A}{x - 2} + \frac{B}{x + 2}$ right

[mental star]

uhhhh

[mental star]

Express as a sum of partial fractions

[mental star]

i dont think thats right? maybe?

is it?

[vagrant ice]

difference of two squares on x^2 - 24

[mental star]

OH

ROOT 4*

SORRY

I MEAN

-4

not -24

[vagrant ice]

ohhhhhhhhhh

[mental star]

$\frac {x+1}{x^2-4}$

[flat frigate]

taf ☆

south's secret twin brother

[vagrant ice]

okay that makes much more sense

[mental star]

thank you, thats perfect

ohhh

[vagrant ice]

yeah then x + 1 = A(x + 2) + B(x - 2)

[mental star]

that makes so much sense,

i was =thinking it was one of the ones that has like

A B, and C

scrap that question,

[vagrant ice]

cool

[mental star]

can i get help with this one

[vagrant ice]

okay then

[mental star]

sending it hold on!

$\frac{4}{(1+3x)(1+x)^2}}

uhhhh

[vagrant ice]

okay now this is $\frac{A}{1 + 3x} + \frac{B}{1 + x} + \frac{C}{(1 + x)^2}$

[flat frigate]

south's secret twin brother

[mental star]

thank you, what determines the denominator of c?

like

like why is it squared i mean

[vagrant ice]

so there's a rule that if you have say (1 + x)^n

you need fractions with denominators (1 + x), (1 + x)^2 .... (1 + x)^n

[mental star]

ohhh right

thank you!

thats all!

.close

$\frac{4}{(1+3x)(1+x)^2}}

[safe radish]

Channel closed

Closed by @mental star

Use .reopen if this was a mistake.

[vagrant ice]

no worries!

[mental star]

$\frac{4}{(1+3x)(1+x)^2}}$

oh oh!

also!

[flat frigate]

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

[mental star]

how do i fix the compile error?

$\frac{4}{(1+3x)(1+x)^2}$

[vagrant ice]

$\frac{4}{(1+3x)(1+x)^2}$

[flat frigate]

south's secret twin brother

taf ☆

[vagrant ice]

you had an extra } ye

[mental star]

ahh

i see

thank you!

thats all!

@vagrant ice

[vagrant ice]

it's ok I saw your message

[mental star]

how do i get it in

this format?

[vagrant ice]

what do you mean?

I just typed it

I didn't use LaTeX

[mental star]

no no i mean like

uhhh

like

4=A(1+x)^2…

[mental star]

not a on about latex!

just on about like

how do i put it into uhh

like writing it out

[mental star]

getting it from this,

to 4=A…

[vagrant ice]

oh okay so that means $4 = A(1 + x)^2 + B(1 + 3x)(1 + x) + C(1 + 3x)$ if you multiply it out

[flat frigate]

south's secret twin brother

[mental star]

thank!

wait

for c, why is there uhhh

[vagrant ice]

multiply everything by $(1 + x)^2 (1 + 3x)$

[flat frigate]

south's secret twin brother

[mental star]

thank you!

[safe radish]

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[hollow jewel]

hi, i am stuck in how we can conclude the statistical function for the parameter θ ,if we have a random sample of the exponential distribution , is a full distribution, more specific i can;t get why the φ(t) must be equall 0 for t>0 for the integral from 0 to infinity of φ(t)*e^(-θt)*t^n dt to equal 0

[safe radish]

@hollow jewel Has your question been resolved?

[hollow jewel]

$\int_{0}^{\infty}f(t)*e^{-ut}*t^n,dt=0$

[flat frigate]

hpodd

[hollow jewel]

.close

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[viral tiger]

need help with this integration (yielding inverse trigo functions) not sure what to do with the let u because of the 10
but i do know that a^2 = (ln 3x)^2, a= ln 3x

[viral tiger]

oh sorry

[gleaming hawk]

Hello @viral tiger

Substitute

Some a = ln3x

da= 3/3x * dx

From there on its simple integration

[viral tiger]

what about u

and the num 10

[gleaming hawk]

Give a second

I think it will be easier if I write it down

See if you can proceed from here

Sorry for the handwriting and the untimely response

I had to rush to find this thing loll

[viral tiger]

wait but it's supposed to yield an inverse trigonometric function

[gleaming hawk]

It will

You can ignore this

But even if you solved it

It would still yield sin^-1

You talking about this?

@viral tiger

[viral tiger]

wait i got confused

[gleaming hawk]

Hmm..?

With what

[viral tiger]

da= 1/3x 3dx right where will the 3 go

[gleaming hawk]

3/3x

Is 1/x

[viral tiger]

ohh

[gleaming hawk]

Yeah

It make sense now @viral tiger

?

[viral tiger]

yes but what happened to the 10-a^2

[gleaming hawk]

10a - a^2

= 25 - 25 +10a -a^2

Right?

@viral tiger

Which can further written as
25 - (25 - 10a +a^2)

Which is

(5)^2 - (a-5)^2

Exactly what I wrote

It's ok take your time

[tame sable]

Please help me graph this please..

I've been trying for hours now..

[gleaming hawk]

!occupied

[safe radish]

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).

[gleaming hawk]

We getting somewhere? @viral tiger

[viral tiger]

yes

thank u sm

[gleaming hawk]

You got it

?

This differentiation of sin^-1

Substitute a back

And you should get your answer

Welp good job @viral tiger

Anything else?

If not then please .close the channel

[viral tiger]

i'm gooooddd

thank u againn

.close

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[flat frigate]

ritam.in4k 🍡

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[chrome plank]

Q11.
I know that $b^2 - 4ac$ (discriminant) decides the nature of the root. But here, discriminant = +5 ! Why is the correct option (A) Irrational and distinct ??

[flat frigate]

ritam.in4k 🍡

Q11. 
I know that $b^2 - 4ac$ (discriminant) decides the nature of the root. But here, discriminant = +5 ! Why is the correct option `(A) Irrational and distinct` ??

[chrome plank]

@ me if reply

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[knotty jungle]

Is this how you calculate the adjunct?

[knotty jungle]

We didn’t see how to do it but I’m supposed to be able to do it so I’m not rlly sure

[delicate sphinx]

Are you finding the inverse of that matrix?

[knotty jungle]

no

im trying this formula

[delicate sphinx]

Your work looks correct

[knotty jungle]

okay yay

.close

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[spring oxide]

How does this make sense?

[spring oxide]

The second function in the inverse function

should have a domain between 3/4 and 1 right? not 1/4 and 1

because for x to be between 0 and infinity in the lower function, y has to be less than 1 as -log(0) = inf but larger than or equal to 3/4 as if you input 3/4 in the expression you get -log(1) = 0 but if its 1/4 you get -log(3)?

Even when i plot the functions its 3/4 not 1/4?

[inner tapir]

seems like a typo to me yeah
https://www.desmos.com/calculator/zlszpdnxsi

[safe radish]

@spring oxide Has your question been resolved?

[spring oxide]

Thx!

[safe radish]

@spring oxide Has your question been resolved?

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[rare abyss]

for B, i differentiated va and vb and set them equal to eachother but im not sure what to do next

[plucky elk]

!show

[safe radish]

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

[rare abyss]

i dont have my phone on me rn to take a picture of my book but what i have is 0.8pe^0.04t = -0.02pe^-0.02t

[plucky elk]

Simplify first then use logs to solve for T

[rare abyss]

i take ln of both sides

[junior smelt]

(be very careful as well, you don't set them equal "as is")

[rare abyss]

do i do minus dvb/dt?

i was just a bit confused bc when i differentiated it it was already negative so i thought that showed rate of decrease already

[junior smelt]

"decrease is negative increase" and all

[rare abyss]

4.8e^-0.02t = 0.8e^0.04t

solve for t using ln?

[junior smelt]

How'd you get the 4.8 tho?

[rare abyss]

because p = 240 from earlier part of the question

i forgot to sub it in

[junior smelt]

Ah I see, fairs

[rare abyss]

when i use ln does e^ positive become + and vice versa or is it multiply

[plucky elk]

,tex .log rules

[flat frigate]

riemann

[rare abyss]

thank you, got it now

@junior smelt @plucky elk

!close

.close

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[native robin]

Find the volume inside a sphere of radius 4 but outside the cylinder of radius 2 assuming both solids are centered at the origin.

[native robin]

guys is the answer 256pi/3-16pi sqrt3

[true horizon]

how did you arrive at that answer

[native robin]

i used cylindrical coordinates

[true horizon]

wait wait wait

inside a sphere of radius 4

ourside a cylinder of radius 4

[native robin]

sorry i meant to type radisu 2 for cylidner

[true horizon]

oh ok

[native robin]

the exact problem is
Measure the volume of the solid region that is inside the sphere
x^2 + y^2 + z^2 = 16 ,
but is outside the cylinder
x^2 + y^2 = 4 .
This solid is shaped like a napkin ring.

[true horizon]

ah, very famous problem

[native robin]

i got the plot of the solid

the top of the cylinder is at z=sqrt12

the bottom of the cylinder is at z=-sqrt12

[native robin]

dyk whats wrong

[true horizon]

hmmmmmm

i'm having a hard time reading your code but i suspect you haven't subtracted the caps on the top and bottom of the cylinder

[native robin]

should i paste the code instead

[true horizon]

i don't know the language so it wouldn't help

but i think

you took a whole sphere [r=4] and subtracted the cylinder [r=2, h=2sqrt2]

[native robin]

ye h=2sqrt12

[true horizon]

what about the round bits

[native robin]

which ones

[true horizon]

the caps

[native robin]

does that not get included when i do the volume of sphere calculation

[true horizon]

it does

but you're supposed to remove it

[true horizon]

no caps on this ring

[native robin]

ohhh i thought my plot was bugging or smth

so i have to remove the parts above the cylinder

[true horizon]

yeah

[native robin]

and below the cylinder

[true horizon]

well

strictly speaking the cylinder x^2 + y^2 = r^2 has infinite height
you say "above the cylinder" and "below the cylinder" like the cylinder was contained in the sphere to begin with

[native robin]

ye

[true horizon]

so when you remove the entire infinite cylinder

yeah

you remove the caps

[native robin]

are they a hemisphere

[true horizon]

very No

maybe try using cylindrical coordinates on the napkin ring

[native robin]

how should i go about calculating its volume

[true horizon]

i would ditch the sphere and treat the napkin ring as a volume of revolution

about the z axis

[native robin]

alr

[true horizon]

or

you could do an integral from -sqrt12 to sqrt12 of (big circle - little circle)

which is i think the standard soln

[turbid breach]

Pls what can i do in it (i m 14 yo) we study math with french pls how he can do that

Helps

[true horizon]

This help channel is occupied, maybe try posting in #help-49 @turbid breach

[turbid breach]

Ok

#help-26

[native robin]

.close

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[modest holly]

help?

[modest holly]

,rotate

[safe radish]

@modest holly Has your question been resolved?

[clear tiger]

are you able to take a clearer imedge please

[safe radish]

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[slender bane]

https://gyazo.com/4d04a851cba421b53413d7e0219cc3fc how do i find the end behavior of this?

[vapid folio]

I'm pretty sure that since e^(-x^2) dominates that function, and since that approaches 0, then the end behavior of g(x) is also 0.

end behavior of g(x) as x approaches infinity is 0

[slender bane]

yeah but what about the restricted domain

[vapid folio]

Idk about your class

But

just do g(-1)

[slender bane]

so a horizontal asymptote at y = 0?

right

[vapid folio]

Yeah

well no

[slender bane]

ok yeah

huh

[vapid folio]

the function is defined for y values 0

[slender bane]

ohh right

but

as it approaches negative or positive infinity the limits 0?

[vapid folio]

yeah

[slender bane]

ok thanks

[vapid folio]

as it approaches negative infinity, g(x) has no value tho

because of the domain restriction

[slender bane]

yeah but it would still approach zero till negative 1 no?

till where it's defined

idk this is a weirdly worded problem

[vapid folio]

maybe looking at the graph might help

[vapid folio]

g(-1) has a finite value

1/e im pretty sure

[slender bane]

ohhhhh

ok thanks

[vapid folio]

You should clarify with someone whether the end behavior for the left side is undefined or 1/e, I'm not really sure

[slender bane]

<@&286206848099549185> ^

[slender bane]

due today at 12 so can't ask my prof lol

[vapid folio]

lol

[slender bane]

unfortunately

we never even went over end behavior for graphs like these

[vapid folio]

In my pre-calc class at least, for functions like log(x) (which have a domain restriction x>0), we would have the end behavior as undefined and infinity

[slender bane]

i may just do DNE

but the thing is

-1 is included in the domain

[vapid folio]

Yeah

I would just say the end behavior is 1/e (as x decreases) and 0 (as x increases or approaches infinity)

Since the point of these questions is to guide you to graph the function

[safe radish]

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[zealous sable]

currently reviewing for a midterm

[zealous sable]

and for one of the questions im tryna get the gcf of 84 and 154

my prime factorization fo 154 is fine

but when for 84 I get 2^5*3

which is incorrect

and ive got no clue how or why

[solid shell]

Do you have your work?

[zealous sable]

yep

gimme 1 sec

[solid shell]

What do you get when you divide 84 by 2

[zealous sable]

oh shoot

whoops

im actually fried

thanks 😭😭

[solid shell]

Assuming you got 11x7x2 for 154

[zealous sable]

yep

i did

[solid shell]

Yeah just a dividing mishap lol you got it

[zealous sable]

yeah

thanks!

.close

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[solid shell]

[zealous sable]

.reopen

[safe radish]

✅

[zealous sable]

is it fine if i ask multip[le questions in one channel?

[solid shell]

Yup

[zealous sable]

alright'

is it fine if i keep this channel while im reviewing for my midterm?

just incase i have more questions

[solid shell]

The bot will ask you every once in a while if you still need help but I don't think it'll cause problems

[zealous sable]

kk

thanks

.close\

.close

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[zealous sable]

.reopen

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✅

@zealous sable Has your question been resolved?

[safe radish]

@zealous sable Has your question been resolved?

[safe radish]

@zealous sable Has your question been resolved?

[safe radish]

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[cunning path]

my r square has a discrepency to my excel r square, is there something wrong with my working:

[cunning path]

the excel r^2 is like 0.98556

formula for excel is

=CORREL(L2:L8,M2:M8)^2

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@cunning path Has your question been resolved?

[safe radish]

@cunning path Has your question been resolved?

[safe radish]

@cunning path Has your question been resolved?

[cunning path]

<@&286206848099549185>

[cunning path]

.close

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[errant obsidian]

Hey cant understand the formula

What is the n ?

I understand it with a and b being in power

But not when the power is a fraction or something like this

[strong oar]

The formulas are just stating that you can replace the root-dense expressions there with a-b and a+b respectively, but only so long as n is a natural number (a number that is both whole and positive)

that's what that $n \in \mathbb{N}$ means, it says that n needs to be in the set of naturals

[flat frigate]

aeiou

[errant obsidian]

How to i find n

Where to get it from exemple a+b

pr a-b

[strong oar]

it just means ANY n works, provided n is positive and a whole number, no finding required

[errant obsidian]

For exemple i need to factor

a+b

No n provided

[strong oar]

then you can use any n at all!

provided it obeys the formula

[errant obsidian]

like how

I dont really understand

[strong oar]

so you have a + b and you want to get to that root expression, then you can choose a whole number n

and write out that expression, replacing n with whatever number you chose

although this is usually used for getting to "a + b" and not the other way around

[errant obsidian]

And so how to implement that into formula?

Like n is 2

Then its (√a-√b)(√a+√b)

But what if its like 4

the n

Or nevermind

I will get into it

Thank you !

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[sand leaf]

Can everyone help me find angles 1,2,and 3

<@&286206848099549185>

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[desert pasture]

We wish to find $\lim_{x \to 3} \frac{x}{x-3} \int_{3}^{x} \frac{\sin(t) , dt}{t}$.
\
This is equivalent to $\lim_{x \to 3} \frac{x - 3 + 3}{x-3} \int_{3}^{x} \frac{\sin(t) , dt}{t}$,
\
which is equivalent to
\
${\lim_{x \to 3} \int_3^3 \frac{sin(t)dt}{t} + {\lim_{x \to 3} \frac{3}{x-3} \int_{3}^{x} \frac{\sin(t) , dt}{t}={\lim_{x \to 3} \frac{3}{x-3} \int_{3}^{x} \frac{\sin(t) , dt} $. This is as the first part evaluvates to $0$

Let $x - 3 = h$.

We thus have
$\lim_{h \to 0} \frac{3}{h} \int_{3}^{h+3} \frac{\sin(t)}{t} , dt$.

[desert pasture]

Looks good so far?

[mossy lotus]

You forgot to write anything about the leftover part after splitting the fraction

[desert pasture]

That's 0

oh , should have mentioned that

[mossy lotus]

And who will mention that?

[mossy lotus]

yea

[flat frigate]

A dense set(Ping when reply)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[desert pasture]

We wish to find $\lim_{x \to 3} \frac{x}{x-3} \int_{3}^{x} \frac{\sin(t) , dt}{t}$.
\
This is equivalent to $\lim_{x \to 3} \frac{x - 3 + 3}{x-3} \int_{3}^{x} \frac{\sin(t) , dt}{t}$,
\
which is equivalent to
\
$\lim_{x \to 3} \int_{3}^{3} \frac{\sin(t) , dt}{t} + \lim_{x \to 3} \frac{3}{x-3} \int_{3}^{x} \frac{\sin(t) , dt}{t} = \lim_{x \to 3} \frac{3}{x-3} \int_{3}^{x} \frac{\sin(t) , dt}{t}$. This is because the first part evaluates to $0$.
\
Let $x - 3 = h$.
\
We thus have
$\lim_{h \to 0} \frac{3}{h} \int_{3}^{h+3} \frac{\sin(t)}{t} , dt$.
\
Let $f(x)= \int_{3}^{x} \frac{sin(t) dt}{t}$
\
notice $f'(3)= \lim_{h \to 0} \frac{3}{h} \int_{3}^{h+3} \frac{\sin(t)}{t} , dt$.
\
We thus have $sin(3) =\lim_{h \to 0} \frac{3}{h} \int_{3}^{h+3} \frac{\sin(t)}{t} , dt$.
\

[flat frigate]

A dense set(Ping when reply)

[desert pasture]

How does this look?

detailed enough>

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[indigo seal]

i cant even do one question

please help

[lean otter]

For 1(a), first find 25% of 80 and then add it to 80

[indigo seal]

ohhh

wait I only needed the 1st

than you

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[sharp beacon]

Acting with a force F, always directed tangent to the trajectory, they raised a small 2 kg block from A to B. The necessary work developed by F for this purpose is (g = 10m / (s ^ 2))

[sharp beacon]

Please, help

[teal thicket]

Is the length ab given?

[sharp beacon]

no

only what is there, the 4 meter one and the 2 meter one

[teal thicket]

oh ok..

[sharp beacon]

Is there any formula that I can use?

[teal thicket]

you can use the work energy theorem…

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[bitter comet]

true

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[glacial raven]

.

[glacial raven]

I need help with this ex

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[sinful stratus]

What do I do next to find convergence?

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[plucky elk]

the photo is like 60% hand

actual math is too small

[lunar halo]

When you did the ratio test and changed n to n+1, 2n doesn’t change to 2n+1

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[cunning zephyr]

Hello I'd like some help with factorisation

x² + xy + ax + ay

I know this is dumb but I'm really stuck and don't know how to solve this.

[cunning zephyr]

elaborate

that's what I'm wanting to learn

two different things?

[mellow bobcat]

then you should see something they both have in common afterwards

[cunning zephyr]

so like x(x+y)

ooh wait a minute

x(x+y) a(x+y) ?

[mellow bobcat]

yes

[cunning zephyr]

and then (x+y) is the common

so (x+y)(x+a)

wow y'all smart wtf

or wait I'm dumb

thank you @mellow bobcat @hasty mason

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[cunning zephyr]

also quick question which i forgot

if a thing is raised to more than the power 2 ( x³ - x² + x -1 )

x ( x⁶ - 1) ?

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✅

[charred saffron]

ur asking for factorisation?

[cunning zephyr]

yes

[charred saffron]

any polynomial P(x) can be factorised into the form:
P(x) = a(x - x_1)(x - x_2)(x - x_3)....(x - x_n)

[cunning zephyr]

uhm

so is my assumption correct?

[charred saffron]

where x_1, x_2, x_3,... x_n are the solutions of the equation P(x) = 0

i mean if u expand x(x⁶ - 1), would u get x³ - x² + x - 1?

[charred saffron]

this rule is unreliable mostly when dealing with polynomials with degree higher than 2

[charred saffron]

just try to factor out the GCF of x³ - x²

[cunning zephyr]

yes

[charred saffron]

and find smt

[cunning zephyr]

what is GCF

[charred saffron]

well then smt is wrong

[cunning zephyr]

idk the English terms

[charred saffron]

Greatest Common Factor

[cunning zephyr]

ohh

the thing we do with the 5/6 * 5/4 ?

actually no

uhhh

fuck

[charred saffron]

eh

[cunning zephyr]

am i really that stupid wtf

this is like middle school

stuff

[charred saffron]

js try factorising x³ - x²

[cunning zephyr]

x ( x⁴) ?

[charred saffron]

how did u get that

im js wondering how u got that result

[cunning zephyr]

😭😭

idfk know bro

i guessed

[charred saffron]

wait no

this is the rule ur gonna have to follow @cunning zephyr