help-33

[marsh citrus]

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[hollow basin]

can someone please explain the first part of how to start doing this

[waxen dust]

!noclopen

[marsh citrus]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[hollow basin]

wha

i wait like 4 hours for a human being to even respond

and the only time people respond is to say im being impatient

lol what

[lost merlin]

@hollow basin There's a formula. Your stats book should have a table to get the zscore value. Grab that and plug and play.

[marsh citrus]

@hollow basin Has your question been resolved?

[hollow basin]

thank you

[marsh citrus]

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[marsh citrus]

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[rocky goblet]

i'm confused

[young owl]

in u = cos(2theta), take the derivative on both sides to get du
then rewrite it so there's only sin(2theta)dtheta on rhs

[marsh citrus]

@rocky goblet Has your question been resolved?

[rocky goblet]

thank you

[marsh citrus]

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[tawny dragon]

does anyone know if the college board accepts synthetic division

[main idol]

Wot

Do whatever method you want as long as you get the solution right

[tawny dragon]

yeah but the cb doesnt teach synth division

(not 100% sure) don't some questions require showing work?

[main idol]

Who cares as long as your work is right

Just do problems from practice tests

[tawny dragon]

thats the thing tho if work is graded and the cb doesnt know wth im doing im not getting full credit (prolly)

[main idol]

...

Graders know synthetic division if the question can be solved using it

[tawny dragon]

probably, but the rubric could say to use long division

[main idol]

...

Then you need to follow instructions

[tawny dragon]

(if work is graded)

[main idol]

Just do problems

[tawny dragon]

rubric isn't provided for test takers (as far as i know)

[main idol]

Too many ifs to be a good use of time

[tawny dragon]

i mean there is a free response section

so it most likely requires work

[main idol]

.

[blazing pulsar]

I don't think you realize how irrelevant long division vs synthetic division is. You may as well ask if the graders will care if you cross your t's from left to right or right to left.

[tawny dragon]

makes sense now that you put it like that

but my teacher told us that the cb doesn't like synth division and that's why they don't teach it (or something like that)

[main idol]

Then listen to your teacher

[tawny dragon]

yet my teacher still taught it to us and we use it on tests

[main idol]

Great listen to your teacher

Do .close

[tawny dragon]

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

A typical ring has charge $\dd Q$, inner radius r, and outer radius $r + \dd r$. Its area is approximately equal to its width $\dd r$ times its circumference $2\pi r,$ or $\dd A = 2\pi r \dd r$

[still temple]

here is a sketch for visualisation

(reopened this because my previous channel closed apologies)

but how is area = width*circumference?

also im sure this is meant to be derived from the proper equation [
A = \pi(R^2 - r^2)
]
where $R$ is the outer radius and $r$ is the inner radius of a ring

[still temple]

but its a bit odd, do we take the derivative with respect to r here?

[still temple]

oh i mean this is a seperable differential equation that ends up getting you A = pi*r^2

but, isnt that the area of a circle? we are dealing with a ring here doe

[rigid sequoia]

since the width is very small, if you cut it open it's like a thin strip of length 2pi*r and width dr

btw you get the area formula by integrating 2pi*r dr

[still temple]

elaborate on "cut it open"?

[still temple]

yes it does give me pi*r^2 but my confusion here still remains

[still temple]

.

[rigid sequoia]

cut it like that so it becomes a strip

[still temple]

i get what you are going for but its still kinda iffy

[rigid sequoia]

that equation gives you pi r^2 if you integrate it 0 to r and if you integrate it r to R you will get the area of a ring

since your ques is talkinng about a differential ring this ring has very small width

so you can just do 2pi r dr

[rigid sequoia]

also if you put R = r+dr in this you will get 2pi r dr + (dr)^2 and (dr)^2 is very small so we can neglect it

[still temple]

darn differentials are weird

but that explanation makes sense i think

[rigid sequoia]

this is kind of the same thing as a derivative instead of h we have dr

but we keep the dr and neglect (dr) ^2 for the differential equation

[still temple]

i see...

thanks

.close

[marsh citrus]

Channel closed

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[marsh citrus]

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[marsh citrus]

@tender mantle Has your question been resolved?

[rigid sequoia]

RUR is just R
RnR-1 will have only the symmetric and reflexive elements of R, so if (a,b) and (b,a) are in R (a,a) and (b,b) will also be in R because its transitve hence RnR-1 is also transitive

for c) if R has (a.b) (b,c) (a,c) then RUR-1 will have (a,b) (b,c) (a,c) (b,a) (c.b) (c,a)

and you see (a,b) and (b,a) but there is no (a,a)

[marsh citrus]

@tender mantle Has your question been resolved?

[marsh citrus]

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[still temple]

.rotate

uhh

One sec

Oh it’s a comma

,rotate

[still temple]

Am I doing this right

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[neon wedge]

whoever wrote this is a genius

probably handsome as well

[marsh citrus]

@radiant mountain Has your question been resolved?

[marsh citrus]

@radiant mountain Has your question been resolved?

[marsh citrus]

@radiant mountain Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[modern belfry]

I’m kind of at a loss

Can I get some direction?

[marsh citrus]

@modern belfry Has your question been resolved?

[modern belfry]

<@&286206848099549185>

[modern belfry]

<@&286206848099549185>

.close

[marsh citrus]

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[olive aspen]

https://tenor.com/view/ownistcord-cord-calculus-schmibby-maz-gif-20893296

[marsh citrus]

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[still temple]

Is this right?

[still temple]

$$\int \tan(x) dx$$

[elfin berry]

help me

[still temple]

$$\int \frac{\sin(x)}{cos(x)}dx$$

[elfin berry]

help me

[still temple]

$$\int \frac{\sin(x)cos(x)}{cos^2(x)}dx$$

[elfin berry]

help me

[still temple]

$$\int \frac{\sin(x)cos(x)}{1-sin^2(x)}dx$$

[elfin berry]

help me

[still temple]

u= sin(x)
du = cos(x)

$$\int \frac{u}{1-u^2} dx $$

[elfin berry]

help me

[still temple]

$$ s= 1-u^2$$
$$ds = -2u du$$

[elfin berry]

help me

[still temple]

$$-\frac{1}{2}\int \frac{1}{s} dx$$

[elfin berry]

help me

[still temple]

$$\int \tan(x) dx = -\frac{1}{2}ln(1-\sin^2(x)) + C$$

[elfin berry]

help me

[still temple]

What did i do wrong?

[late geode]

a few issues in your Intermediate steps

[late geode]

du = cos(x) du

[late geode]

\int \frac{u}{1-u^2} du

technically you should've used abs val bars for the log when integrating the 1/s,
what you currently have can also be simplified

[still temple]

simplfication is for simpletons

[still temple]

I could had just did

$$\int \frac{sin(x)}{cos(x)} dx $$

u = cos(x)
du = -sin(x)dx

[elfin berry]

help me

[still temple]

ty

.close

[marsh citrus]

Channel closed

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[marsh citrus]

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[gloomy salmon]

cant get this one

[gloomy salmon]

isnt that just quotient rule

[still temple]

!status

[marsh citrus]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[gloomy salmon]

so i need to find the derivative

and then use point slope formula..

[still temple]

Yes.

[gloomy salmon]

wait no

wait yes

ok lemme retry

um

so how do i rewrite it in terms of y

i cant take derivative unless i do that

y(ysqrt(e)^x)?

[still temple]

what?

implicit differentiation

[gloomy salmon]

wait i think i got it

omg it worked

ok sorry lmao

thx

.close

[marsh citrus]

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[burnt ingot]

I need help

[burnt ingot]

I get f'(x)=-(4x)/(sin(2x^2)^2)+1/(sin(x))-(xcos(x))/(sin(x)^2) and plug in pi/2 for x, getting -(2pi)/(sin(pi^2/2)^2)+1

But it's not right

[gloomy salmon]

we have the same homework

[burnt ingot]

Dang, you having any luck?

[gloomy salmon]

yes i got that one but idk how to solve this

did u get that one

[burnt ingot]

I did

[gloomy salmon]

omg

[burnt ingot]

Can I dm you?

[gloomy salmon]

we can trade 😄

yes

[burnt ingot]

Bet

[still temple]

what?

[burnt ingot]

I am guessing that's very off based on that reaction

.close

[marsh citrus]

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[marsh citrus]

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[humble crescent]

can someone help me with this

[marsh citrus]

@humble crescent Has your question been resolved?

[marsh citrus]

@humble crescent Has your question been resolved?

[humble crescent]

<@&286206848099549185>

[humble crescent]

<@&286206848099549185>

[marsh citrus]

@humble crescent Has your question been resolved?

[marsh citrus]

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[tender mantle]

588=2^2×3×7×7
14=2×7

how to solve this?

[marsh citrus]

@tender mantle Has your question been resolved?

[marsh citrus]

@tender mantle Has your question been resolved?

[marsh citrus]

@tender mantle Has your question been resolved?

[tender mantle]

.close

[marsh citrus]

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[marsh citrus]

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[slow stratus]

Looking to solve [1] through any other way than using [2]. The wished result is 2^6

[slow stratus]

[1]

[2]

[untold nimbus]

noo

[slow stratus]

What

[untold nimbus]

( a b ) ^ x : a ^ x , b ^ x

and decompose the 4

by 2 ^ 2

mb its wrong btw 😭

[slow stratus]

It’s 2 in the root

Not 4

[untold nimbus]

yep sorry

the result is the same btw

( 2 ^ 2 ) at the end

just to change 4 by 2

its better

( by using your [ 2 ] )

sqrt 2 = 2 ^ 1/2

[marsh citrus]

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[slow stratus]

Thanks boss @untold nimbus

[marsh citrus]

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[untold nimbus]

np ^^'

[slow stratus]

i MUST use this then?

[untold nimbus]

yup

but the " m "

is not in the root

maybe the same thing

[slow stratus]

the exponent multiplies only the numerator?

[untold nimbus]

yep

a * 1 / b

a / b

XD

[slow stratus]

Huge help.

[marsh citrus]

@untold nimbus Has your question been resolved?

[marsh citrus]

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[wraith ledge]

when i have a metric d(f,g) = max| f(x) - g(x)| where x is in [a,b], what should convergation mean of a sequance in this metric space

[wraith ledge]

should i think about it as we are {fn} from 1 to inf we are reducing the distance and getting closer

or i have a wrong view

[marsh citrus]

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[storm zodiac]

Help me

[storm zodiac]

4^x - 10^x - 25^x=0

Solve for x

[stoic saddle]

!status

[marsh citrus]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[rapid escarp]

what had you try?

[storm zodiac]

Change of base

2^2x - (2^x)(5^x) - 5^2x =0

Then I let a = 2^x and b = 5^x,
that gives me a^2 - ab - b^2 = 0

but theres no solution

[rapid escarp]

umm... try to div 4^x

in both side

[storm zodiac]

Ah I see 1 - 2.5^x - 6.25^x = 0

then we can sub in a = 2.5 and a^2 = 6.25?

[rapid escarp]

just let a is ok

[storm zodiac]

1 - a - a^2 = 0

[tall umbra]

but... the x...

[rapid escarp]

i think just typo

[storm zodiac]

a = 2.5^x

Got it thanks

Its difficult to know the 1st step to divide by 4^x

and 2.5 and 6.25 just happens to be square of each other 🤔

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[toxic flame]

how can I find the Fourier serios of something like this?

[toxic flame]

something like this?

[marsh citrus]

@toxic flame Has your question been resolved?

[marsh citrus]

@toxic flame Has your question been resolved?

[toxic flame]

<@&286206848099549185>

[marsh citrus]

@toxic flame Has your question been resolved?

[toxic flame]

.close

[marsh citrus]

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[inland geyser]

is my approach valid?

[brave marsh]

It seems good. Just be careful when taking the square root; you're missing the solution t=-sqrt(5/6).

[inland geyser]

right got it

thx

[marsh citrus]

@inland geyser Has your question been resolved?

[marsh citrus]

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[whole minnow]

Problem with logarithms

[whole minnow]

This is solved

I understand turning the logs into natural logs fractions

I dont understand how is (ln3/ln2)(ln4/ln3)....(ln32/ln31) equal to ln32/ln2

[glass perch]

That's like a telescope product

For every 32 > x > 2 there is one ln x in the numerator and one ln x in the denominator

And they cancel. e.g for the 3 there is ln 3 in the first term numerator and ln 3 is in the second term denominator

[whole minnow]

Ohhhh

I got it

Thanks

Its a cool problem

[glass perch]

You're welcome !

[whole minnow]

[glass perch]

Telescope sums/products are always satisfying

[glass perch]

Do you want another one?

[whole minnow]

Oh sure

But Ill do it after I finish this sheet

Its 20 really hard problems from stewart calculus book

[glass perch]

Try to calculate the value of the infinite sum 1 to infinity of 1/(4n^2-1) (not hard, but it involves telescope sums). Had to do this during an oral exam

[whole minnow]

Thanks

Ill let you know when I finish it

✅

.close

[marsh citrus]

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[glass perch]

I'll probably be sleeping then, but sure

[marsh citrus]

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[lethal bridge]

Does the disk formula for solids of revolution vary depending on what axis the solid is being rotated around

[marsh citrus]

@lethal bridge Has your question been resolved?

[marsh citrus]

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[lyric bay]

Hello

[lyric bay]

I don't understand the first step

Where did 6b go?

I understand how they divided into three factors

[main idol]

Where's the question

[spark otter]

the 6b went nowhere

we're just using a different result from polynomials

[lyric bay]

Oh sorry here

[spark otter]

On one side, $p(z) = 3z^3-az^2-bz + 6b$

[lyric bay]

But we have to consider the leading coeff of the whole polynomial?

[elfin berry]

rafilou2003

[spark otter]

On the other, since 2+2j, 2-2j and w are the three roots of p

[lyric bay]

Ohhh

Why was I confused

I understand

[spark otter]

p is divisible by (z-...)(.....)

[lyric bay]

I also don't understand the last part

Where the factored polynomial help us to solve b

The -b = 12w + 24

This part

[spark otter]

what's the coefficient of p(z) in z^1?

[lyric bay]

12?

[spark otter]

look again

p(z) = 3z^3-az^2-bz + 6b

[lyric bay]

-b

[spark otter]

yes

[lyric bay]

Ohhh

I seee

That makes more sense

Thanks so much

.close

[marsh citrus]

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[toxic flame]

How do I proceed with this An?

[marsh citrus]

@toxic flame Has your question been resolved?

[main idol]

Do cases when T_p is less than T and greater than T

[toxic flame]

but how do I solve it since I got A and not an actual number?

[main idol]

A is a constant

Like 5

You can and should factor it out with 2/T

[toxic flame]

which I did in An

[main idol]

So what are you asking then

[toxic flame]

sorry, so should I just move it outside ?

[main idol]

That's what factoring out means yes

[toxic flame]

right?

[main idol]

Where did T_p go

[toxic flame]

tbh I dont know how to solve it

would it be at the denominator of both sin?

[main idol]

When you integrate, you find the antiderivative and evaluate it at the endpoints

if the antiderivative was sin(t), then with the same endpoints, what do you get

[toxic flame]

cos(t)

[main idol]

...

[toxic flame]

-cos sorry

[main idol]

No

sin(t) is the antiderivative

Endpoints are what

[toxic flame]

tp/2

[main idol]

That's one yes

What's the other

[toxic flame]

and -tp/2

[main idol]

Evaluate this at the endpoints

Also this reverse chain rule / u sub is wrong

Little t shouldn't be in the denominator

[toxic flame]

is it Tt/2npi ?

[main idol]

Is what this

I can't read your mind what you're referring to

[toxic flame]

instead of T/2npit should it be Tt/2npi ?

[main idol]

No

Do you know u subs for integrals?

int cos(ax/b) dx = ?

[toxic flame]

I dont remeber

[main idol]

Oof

[main idol]

https://tutorial.math.lamar.edu/classes/calci/SubstitutionRuleDefinite.aspx

Review it there

[toxic flame]

ok ty

.close

[marsh citrus]

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[humble river]

hi

[humble river]

so

for h^-1(-3)

they want inverse?

like should i find h(-3) first

then do inverse of that one

thnxs in advanc

[twilit grove]

no, it's the value of x for which h(x) = -3

[marsh citrus]

@humble river Has your question been resolved?

[marsh citrus]

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[sullen ice]

how do i show a constant function and the identity function are continuous

[native frigate]

constant function being f(x) = a and identity function being f(x) = x?

[sullen ice]

yes

[native frigate]

Is this a past calc class or a before calc class

[sullen ice]

past

wait

year 1

year 1 calc

so no real analysis i dont think

but we've used epsilon-delta

which is real analysis but isnt advanced

[native frigate]

yeah this is supposed to be an e-d proof

do you know the e-d definition of continuity?

[sullen ice]

im def making this problem more complicated than it is then lol

not rlly

i basically have a peicewise function with the top half being a polynomial and the bottom another polynomial

i need to show that there is some a <= polynomial 1 and a>=polynomial 2 that makes the function continuous

[native frigate]

i think this is a different problem than what you were asking, but I don't really understand the second part you said

[sullen ice]

im supposed to use IVT which implies that if i just prove all polynomials are continuous, it'll make the theorem work

[native frigate]

what is the a here

[sullen ice]

a is any value you just have to prove there is a value a that makes the function continuous

[native frigate]

I'm pretty sure you can just assume that polynomials are continuous, unless there was some reason to suspect you couldn't

[sullen ice]

if thats true im all good

which i think it is

[native frigate]

i mean the e-d proofs of continuity aren't that hard if you for whatever reason felt the need to include them, but assuming continuity of those is surely fine

[sullen ice]

ty o/

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

How is this 4x+6

[hasty ruin]

show your work

[uneven zephyr]

help

[hasty ruin]

#❓how-to-get-help

[still temple]

wait bruh i realized was i doing wrong

[uneven zephyr]

no one reacting me

[still temple]

i was doing f(g(x)))

not g(f(x)))

[uneven zephyr]

i got exam in 3 mins

[still temple]

4(sqrt(x+2))^2-2

thats what i got

and now im wondering

how would i simplify

(sqrt(x+2)^2) further

(sqrt(x+2))(sqrt(x+2)) = ?

[hard gull]

square cancels sqrt

it isnt too hard

[still temple]

oh lol

ok thanks

.close

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[tawny lava]

not sure what it means by expression

[tawny lava]

ping when reply thanks

[marsh citrus]

@tawny lava Has your question been resolved?

[nova totem]

Write the slope as an equation

Expression means equation

Well

You gotta evaluate that equation at x = -3

[tawny lava]

using the non derivative equation?

[nova totem]

Using the derivative

It's asking for the slope of the tangent line

the derivative is legit the slope of a tangent line at some x location

,tex \linearization

[elfin berry]

Umbraleviathan

[tawny lava]

but i tried 1 and it was wrong

[nova totem]

Oh maybe it just wants the equation for the slope in general then

Try that @tawny lava

[marsh citrus]

@tawny lava Has your question been resolved?

[tawny lava]

.close

[marsh citrus]

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[tender mantle]

How to solve this?

[marsh citrus]

@tender mantle Has your question been resolved?

[tender mantle]

any hints?

[tender mantle]

.close

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[marsh citrus]

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[still temple]

Just want some verification. This calculation seems wrong to me for some reason

[still temple]

I keep getting $1.152 \cdot 10^{-18}$ in my calculator

[spark otter]

Did you factor in the ^2 ?

or bad parentheses maybe

just compute 8.99 * (1.60)^2/(2.0)^2

and then compute $10^9 * \frac{(10^{-19})^2}{(10^{-10})^2}$

[elfin berry]

rafilou2003

[still temple]

oh yes i see my mistake now

thank you!

.close

[marsh citrus]

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[wary vortex]

This might be a no-brainer for a lot, but if I've got 19 dollars. I've spent 60% to get down to 19 dollars; how much did I spend?

[wary vortex]

I know it's just reverse calculation of percentages but for some reason I've completely fumbled the bag

[frank orbit]

I might be worse by suggesting u cross multiply
Ur trying to find the 100% of the original
But after spending 60% u left 19

[wary vortex]

Yeah pretty much

[frank orbit]

U wanna try cross multiply see whr that takes u? 🤣

[wary vortex]

I remember doing this stuff in school, but I've lost complete touch of how to do it

[frank orbit]

100 = x
(after spending 60%) = 19

[wary vortex]

That would be 31.6 wouldn't it?

solving for x

100x19 = 60 x X

X= 100x19/60

[frank orbit]

No

Raed again

[wary vortex]

Don't know how to use the smart formatting in here

[frank orbit]

Ik what ur talking abt

U got the format right
Just needa read again and readjust the numbers

[wary vortex]

I've read it like five times and my brain is not processing anything

[frank orbit]

spent 60% to get down to 19 dollars

[wary vortex]

Oh

it's 40, not 100

Cause 40 is the left over so 40 is the 100?

I feel like I am so wrong

wait wait

100 x 19 = 40 x X

[frank orbit]

Ye

[wary vortex]

x = 100 x 19 / 40

x = 47.5 yes?

[frank orbit]

I mean that's what I got

Unless I'm tripping by using cross multiplication

[wary vortex]

Seems reasonable

Thanks man, it actually made sense the way you explained it once I thought about it. It's easy to forget sometimes.

🙏

Thanks again :D

[marsh citrus]

@wary vortex Has your question been resolved?

[marsh citrus]

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[humble river]

should i plug in a number

like 1

[desert dirge]

could just draw it

[humble river]

Ok

oh yeah

this not not one-toone no chance

wait

it is

cuz its only half of it

[desert dirge]

its only for 0<=x<=7

,w graph x^4+4, 0<=x<=7

[elfin berry]

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[humble river]

yes

Perfect ok

thnxs

.close

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[gloomy salmon]

first line right side

d/dx 2xy

obv d/dx x is 1

but how does d/dx 2y = 2y

oh wait its product rule

ok nvm

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[tender mantle]

Please check my solution answer is A but my answer is D

[marsh citrus]

@tender mantle Has your question been resolved?

[magic trellis]

log t= 2i/π should be -2i/π

may be wrong tho

[gilded cape]

D is the correct answer

,w (-1)^(2i/pi)

[elfin berry]

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[tender mantle]

I see thanku

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[still temple]

How did they solve the marked part?

, rotate

[sullen ice]

wdym?

let u =r^2

then u have u^2+u-90

then just quadratic formula/factoring

and replace u with r^2 at the end

[still temple]

Yeah!!

That helps

Thx

[sullen ice]

np

[still temple]

.close

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[queen prairie]

i need help doing all of them , it is what we'll do tmrw at the test , we've done this at class but a revision of what i know and also learning what i dont know could greatly help me , i'll provide the neccesarry translations as well for the test

[gritty spruce]

Not everyone understands spanish

[queen prairie]

as i said , i'll provide proper translations for everything in there

also , it is romanian

[gritty spruce]

Mistake on my part

[queen prairie]

so you're willing to assist me ?

[gritty spruce]

In the test?

[queen prairie]

nono , learning for the test

we have an exampale for the test tmrw

[gritty spruce]

Sure let's if i can

[queen prairie]

alright

so first

a number that's not part of Q is

i mean ik that Q is rational numbers

aka fractions

[gritty spruce]

Root 5

[queen prairie]

why

[gritty spruce]

It's irrational

Can't be expressed in p/q form

[queen prairie]

cus it cant be made as a fraction nor includes one ?

p/q ??

[gritty spruce]

Yeah not a fraction

[queen prairie]

i mean 1 , and sqr 4/9 evidently include or can be expressed as a fraction

but what about -1

is it just -x/x?

[gritty spruce]

A rational is a number which can be expressed in p/q form where p and q belong to z and q not equal 0

[queen prairie]

Z is full numbers , right ?

like -3 , 6 , 2 ?

[brittle vessel]

integers, yeah

[queen prairie]

ah

mb

so sqr 4/9 includes a fraction so it is part of Q

right?

[brittle vessel]

no

[queen prairie]

so it is not rational?

[brittle vessel]

it is, but not because of your reason

its equal to 2/3

[harsh jungle]

Sqrt(3/1) also involves a fraction and is not part of Q

[brittle vessel]

which satisfies the p/q criteria

[queen prairie]

so p < q for it to be true ?

or

[brittle vessel]

no

any integer

so the A B C is rational, so we know D is not (and luckily its not asking you to prove it)

[queen prairie]

so 2/3 satisfies the q/p criteria

but why 3/1 doesnt ?

nvm

since it would just be an integer at that point

3

or

[brittle vessel]

root 3

[queen prairie]

sqr 3/1 isnt cus it is sqr 3

[brittle vessel]

not 3

[queen prairie]

ye

but sqr 2/3 doesnt give an integer luckily

so it satisfies the p/q criteria?

i mean the 2/3

isn't an integer

but a rational number ?

[brittle vessel]

you are changing up stuff 😄

[queen prairie]

so im not saying stuff right

[brittle vessel]

its sqrt(4/9)

[queen prairie]

OH

[brittle vessel]

so sqrt(4)/sqrt(9)

which is 2/3

[queen prairie]

damn

[brittle vessel]

ok, what is the next that you dont know?

[queen prairie]

and 1 and -1 can be represented as a fraction

so yeah

here

im not sure

[brittle vessel]

every integer is rational

[queen prairie]

we have the following numbers

the highest number is

we just put everything under sqrt

right?

[brittle vessel]

yeah

[queen prairie]

so both integers and fractions are rational , right?>

[brittle vessel]

if the fraction has only integers

[queen prairie]

so if the fraction has full numbers ONLY it is considered rational

so at 2

at b

we put the 4 under the square

by squaring it up

sqrt**

16x2 = 32

so sqr 32

and at c

9 x 2 = 18

at d

it is 25

so the highest number is 4 sqr 2?

[brittle vessel]

yes

see? you knew the answer

[queen prairie]

eh

lets try and do 4 first

since it'll take less time

the full part of number 3 - sqr 6 is

whole part of a number*

we put the 3 under square

right>?

and we get sqr 3 out of it

then we just find the closest full number

our candidates aree

[brittle vessel]

no, i dont think thats the way

[queen prairie]

why not

[brittle vessel]

between what 2 numbers will root6 will be

[queen prairie]

2 and 3 ?

[brittle vessel]

yes

[queen prairie]

but it is closer to 2

since 6 is closer to 4 than to 9

[brittle vessel]

yeah, but we dont need it

[queen prairie]

what do we need

[brittle vessel]

only this

3- (a number between 2 and 3)

what will it be?

[queen prairie]

idfk

a number between 1 and 0?

[brittle vessel]

yes

[queen prairie]

so what's the naswer

[brittle vessel]

so the answer is?

[queen prairie]

1?

0?

[brittle vessel]

its less then 1, so it cant be 1

[queen prairie]

oh

so we doin

ok wtf we doing rn..

[brittle vessel]

or im not understanding the question?

[queen prairie]

we're doing 4

[brittle vessel]

if we have 1.7 it will be 1 right?

[queen prairie]

the whole part of the number..

[queen prairie]

since we approximating to the lower part

[brittle vessel]

ok

then now

[queen prairie]

so why cant we just put the 3 under the square and do it

[brittle vessel]

we have 0.something, right?

[queen prairie]

yeah

so it'll be 0

[brittle vessel]

so 0

[queen prairie]

.

[brittle vessel]

do what?

[queen prairie]

sqrt

like

do sqrt 9 - sqrt 6

NVM

NVM

NVM

[brittle vessel]

😄

[queen prairie]

damn

tf did i just say

[willow veldt]

Hey I need an immediate help with a math rn please someone help me

[main idol]

!help

[marsh citrus]

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

[willow veldt]

Good music taste

[queen prairie]

can we redo 4 again tho

[brittle vessel]

xd, big brother is watching us

[queen prairie]

so we got the whole part of the number 3- sqrt 6

[willow veldt]

Somebody help me thoooo 😭

[main idol]

!help

[marsh citrus]

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

[willow veldt]

The c number

[willow veldt]

I did these

[main idol]

!occupied

[marsh citrus]

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).

[queen prairie]

and sqrt 6 x , x being 2 <_ x < 3 ?

[willow veldt]

@queen prairie can I get ur help

[brittle vessel]

open a new channel

[queen prairie]

u asking the wrong guy ngl

[queen prairie]

so we got 3 - 2 <_ x < 3 ?

[willow veldt]

Idk but I got a test in few hours and needa solve quick 😭

[queen prairie]

ik this is pretty badly written but it's alr

[queen prairie]

i got a test tmrw and i'll fail it fr

[willow veldt]

Help me and lemme pass fr

[queen prairie]

so x is < 3 , so it cant be 3

[brittle vessel]

Ari, ping helpers then, but dont bother the ppl in occupied channels pls

[queen prairie]

so our number that'll be out wont be 0

but

since it is higher and equal to 2

it can be 2 as well

so we get 1

[brittle vessel]

it cant be 2

[queen prairie]

why not

it is only higher than 2 ?

[brittle vessel]

cuz we dont have sqrt(4)

[queen prairie]

so 2 < x < 3 ?

[queen prairie]

ah yeah..

we got sqr 6

so it be 2 since that would be sqr 4

[brittle vessel]

pretend its 2.5 or something

[queen prairie]

or

we could write it as

2 < sqr 6 < 3

[brittle vessel]

yes

[queen prairie]

sqr 4 < sqr 6 < sqr 9

so that way we're also proving that it CANT be 2

so it is LOWER than 3

so out answer form that question is 0< n

but it is also < 1

since it CANT be 2

so if we get the full number out of it

it'll be 0

since yeah we go down baby

now the part wehere we'll die

ah it aint that bad

[brittle vessel]

indeed

[queen prairie]

we get 3 - 2 * 5 + |1/2 -1 |

the numbers in the module is my problem

it is 0.5 - 1

but it'll be negative as an outcome

so we change the signs ?

-(1/2) + 1?

[brittle vessel]

or just change whatever you got as result to positive

[queen prairie]

• 0,5

-0,5

yeah it's the same thing tf

[brittle vessel]

thats the abs value definition

[queen prairie]

what is the abs value def

[brittle vessel]

absolute value

nvm

[queen prairie]

i'll just put it under abs value since it's way easier

we get 3 - 2 * 5 + 0,5

3 - 10 + 0,5

-7 + 0,5

so it is D

right..?

so

the fractional part of -2,76 is

it is -0,76 ?

nvm

so since fractional part of a number is always positive

we have to divide it by the next integer

[brittle vessel]

divide?

[queen prairie]

minus it

not divide

divide is : ?

fuck knows

but

for example

if we have -3,23

[brittle vessel]

subtract

[queen prairie]

we have to substract the next full number

right?

[brittle vessel]

it will be -4 +0,77

[queen prairie]

ye im just dumb and trying to explain things the hard way

but on short

we can

take the 23 , aka the fractional part

and just substract it out of 100

and we'll get the fractional part?

[brittle vessel]

yeah

[queen prairie]

it aint that hard

[queen prairie]

since the fractionar part of a number cant be negative

it'll be 0,24

the intersection of intervals I and J

intersection aint that bad

what's in both of the intervals

so what we got is [0;5) ?

tf

i mean

at a

the sqrt b^2

wont it just be put under | |?

[brittle vessel]

yea

[queen prairie]

so it is |a|+|b| ?

my teacher really be tricky and wants us to fail

or make us want to die

both works

[brittle vessel]

or want you to understand

[queen prairie]

or want me to get the exam in 4 years and go to a good univ

so

we have to put them under ||

|2sqr3 - 4 |

| 3 sqr 3 - 6|

and cus we cool we'll put the numbers under the square root

so sqr 4*3 - 4

and sqr 9 * 3 - 4

we get |sqr(12) - 4|

and |sqr(27) - 6|

right?

[brittle vessel]

yes

[queen prairie]

so we just do what we did before for the sqrt?

i mean we couldd also try to just take them out of the module

[brittle vessel]

we need to know if those are neg or pos

[queen prairie]

ahah

so we'll put the integers under sqrt

we get |sqr(12) - sqrt16|
and |sqr(27) - sqrt36|

so shit they both be negative

and we cant just change the sign of what's left

we have to change the signs here for them all

we get -sqr(12) + sqrt16
and -sqr(27) + sqrt36

tf why do we still have them in the sqrt

-sqr(12) + 4 - sqr(27) + 6)

but it is + -

so it'll be negative?

since ++ and -- equal to ++

the rest is -

[brittle vessel]

??

[queen prairie]

wbhat

[brittle vessel]

what will be negative?

the sum of two positive number?

[queen prairie]

what

wdym

no

[brittle vessel]

u add these

[queen prairie]

we had to do |a|+|b|

yeah

but the first number from the second

is negative

wont that interfere?

[brittle vessel]

both are not negative

[queen prairie]

wdym

[brittle vessel]

so the sum cant be either

[queen prairie]

WAIT WHAT

What

wdym

[brittle vessel]

they have ||

[queen prairie]

wait im

[brittle vessel]

so they will be positive

[queen prairie]

they dont?

here we had them ||

and when we got them out of the module

we changed the signs

[brittle vessel]

yes, and by that those became positive

like a good | | sign makes it for us : )

[queen prairie]

we get -sqr(12) + sqrt16
and -sqr(27) + sqrt36

ye

the outcome is positive

and we must add them

the a and b

[brittle vessel]

yes

[queen prairie]

-sqr(12) + sqrt16 + -sqr(27) + sqrt36

but we got + - sqr27

and that's what i was talking about

wont it just get rid of the + and replace it with a -?

[brittle vessel]

• (-5) is -5, yeah

if thats the question

[queen prairie]

ye

so it'll be

-sqr(12) + sqrt16 -sqr(27) + sqrt36

we taking those 2 numbers out of the sqrt

-sqr(12) + 4 - sqr(27) + 6