help-28

[left creek]

ok, that makes a little more sense, I was trying to apply it like to everything

[plush egret]

you start at the end of a soccer field

and you run to the halfway point

then you run to the halfway point between the halfway point and the end

then you run half of the remaining distance .....

[left creek]

and so on

[plush egret]

where do you end up if you keep covering half the field forever

yea

this is justh

1/2 + 1/4 + 1/8 ....

intuitively if you do this forever you cover the whole field

[left creek]

yeah.. idk brain hurty

[plush egret]

but if you ever stop, youll land short

[plush egret]

just apply the formula

lets go back hcere

all i did was this

$\sum _{n=1}^\infty \frac{1}{2^n} = -1 + \qty( 1 + \sum _{n=1}^\infty \frac{1}{2^n} )= -1 + \qty( \frac{1}{2^0} + \sum _{n=1}^\infty \frac{1}{2^n})$

[glossy valve]

jan Niku

[plush egret]

seem okay?

[left creek]

ok

[plush egret]

now u can apply the formula

and be done with it

[left creek]

and then merge that in and change the index

[plush egret]

yup

then it matches your formula

[left creek]

idk why I don't have anything abt this in my notes...

[plush egret]

[left creek]

yeah, ok, that makes more sense. thank you for your help!

[plush egret]

np

[left creek]

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[scarlet flint]

i put the answer as 3 but i got it wrong

[hot herald]

how did you get 3?

[scarlet flint]

from 3-5 A2 = -4 and from 5-7 A3 is +1

so the final answer would be -3

[plain sable]

5-4+1-1

[scarlet flint]

but then there is the absolute value sign so that would make it positive 3 no?

[hot herald]

no

you're calculating something different

[plain sable]

the areas above the y axis are added and the ones under it are subtracted

[hot herald]

specifically, you're doing
$$\abs{\int_2^7 f(x) \dd{x}}$$

[glossy valve]

ℝαμΩℕωⅤ

[hot herald]

which isn't the same thing

[scarlet flint]

so do i need to do the reverse of what im doing so A2 = 4 and A3 = -1?

[hot herald]

no

should really involve - signs anywhere

to get what's being asked,
add up the absolute areas of each region

to make it easier to visualise you could first consider flipping stuff currently below the x-axis
for |f(x)|

[scarlet flint]

i dont think im following. So would i do 4+1?

[hot herald]

yes

[scarlet flint]

ok and then what do i do or is that the answer?

[hot herald]

that's it

[scarlet flint]

ok so when f(x) is inside of an absolute value i just take all values for A1 to A4 as an absolute value

is that the idea?

[hot herald]

you add the (unsigned) areas of the regions which are given to you

[scarlet flint]

ok i understand

ty very much for the help

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[lost raptor]

How would i do this?

[steel cove]

They both have the exact same range and domain

There’s probably some theorem about that

Isk

Idk

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[tiny terrace]

how do you handwrite aleph?

[glossy valve]

Stephen

[shrewd hamlet]

hm good question

[tiny terrace]

I've been wondering for a while
Cause every time I write Aleph in my notes, it comes out different

[shrewd hamlet]

https://youtu.be/CHhO-aS1bb4?si=O7CTOy3TIsoXuEkb

[tiny terrace]

nice

much easier than whatever I was doing

thx

[shrewd hamlet]

np

[tiny terrace]

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[shrewd hamlet]

btw google is much better for this kinda stuff

lol

[tiny terrace]

fair enough

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[trail rose]

#7 howd they get that

how did it seperate to 1/ln3

and 2x/x^2+6

[shrewd hamlet]

They just differentiated the numerator

[glossy valve]

Stephen

[shrewd hamlet]

@trail rose

[trail rose]

ohh

ok thanks

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[coral ravine]

can anyone help me?? do we use spherical coordinates to solve this?? I am stuck from the very beginning. I don't even know how to start..

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[odd olive]

pls explain de moivres thereom and polar form

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[odd olive]

sorry?

[fickle ruin]

,tex .demoivre

[glossy valve]

VishramKidPG123

[fickle ruin]

polar form:
x=rcos0
y=rsin0

(pretend 0 is actually a theta, I'm lazy)

[odd olive]

hahah thanks

Oh and @fickle ruin just wondering what the r means

[fickle ruin]

radius, which is usually defined as a function of theta

[odd olive]

Ok

also do you know what arcsin is

and arg(...)

[fickle ruin]

at least in polar calculus, I haven't done too much work with complex plane

[fickle ruin]

arcsin is another way to write inverse sine

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[crimson pawn]

hello can someone help me how to solve these two? i can't find it in the internet

[calm tinsel]

💀

5x5....

do some gaussian elimination?

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[golden shore]

can someone prove that PV=nKT from PV=nRT

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[pale shadow]

Hello

[pale shadow]

can someone explain the procidure and what exactly i have to do here

[floral widget]

U add by putting all the values mentioned for example the first problem will be 1/1+1/2+1/3+1/4+1/5

[full hinge]

This is Sigma notation. Essentially, for A it means to find the sum of that series starting at k = 1 and ending at k = 5. For B, it would mean starting at i = 1 and ending at i = 5. Substitute the values and add to find your sum

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[pale shadow]

so for k i just put in 1 and for b) i pit in 1 for i aswell?

[full hinge]

Starting at k = 1 and ending at k = 5 for part A

So it would be 1/1 + 1/2 + 1/3 + 1/4 + 1/5

[pale shadow]

and that for a) its K and for b) its i even tho they are different letters do they mean same thing

[pale shadow]

mhmmm so i have to basically sum the all up

[full hinge]

Yeah try A and then do B on your own and see what you get

[full hinge]

Yeah they mean the same thing

[pale shadow]

So basically this is the procedure

Now here it says write the first and the last three

so 50 51 52/ 100 99 98

but then it says calculate the sum again… just for those numbers or for all also it says using calculator

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[open minnow]

for a given set of data, the median is just basically the middle number right? then whats the difference between that and the median measured from the cummulative frequency graph?

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[torn jolt]

Would it be legal to write x_m instead of 100000 if y_m is discrete (either 0 or 1) and x_m is continuous?

I tried to do that in AMPL and it can't parse it if I use x_m instead of 100000, that's why I'm wondering

(Also I thought of this because no upper bound has been specified for x, 100000 is arbitrary but x might assume a higher value technically)

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@torn jolt Has your question been resolved?

[nova basin]

Then this is a question that has to be answered from context, which you didn't provide

[torn jolt]

Not really sure what to specify other than this is optimization and linear programming honestly lol

[nova basin]

then I can't argue in favor of doing that

unless indeed x < 10^5 is guaranteed somewhere

[torn jolt]

It isn't

The point of this is that if y is 0 then x should be 0

We chose (someone helped me in another channel) that value arbitrarily

But it is unspecified in the problem

So it should be valid to write?

Not sure why AMPL wouldn't accept it
Thanks btw

[torn jolt]

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[torn jolt]

Ocupied!!!!!!!

[viral shadow]

Last month. We got 63.000kr. from selling cakes, and today we got 107.217 kr., it is 70.22% increase since last time?

[viral shadow]

You rat.

[torn jolt]

Lol

Sorry.

MY question.

[acoustic forum]

If you increase by 70% you get to 170%

[torn jolt]

Can somone help me proof the Circle theorem?

[acoustic forum]

So check if the second number is 170% if first

[torn jolt]

Please in the right channel: https://discord.com/channels/268882317391429632/903486481929224232

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[torn jolt]

<@&286206848099549185>

[wheat ruin]

Which is?

[torn jolt]

Circle theorems?

[wheat ruin]

Ye

[torn jolt]

Angles in the same segment theorem
2:05 Angle in a semi circle theorem
2:52 Angle at the centre theorem
3:31 Opposite angles in a cyclic quadrilateral theorem
4:31 A tangent meets a radius theorem
4:55 Tangents from a point
5:55 Alternate Segment Theorem

[wheat ruin]

I've never heard of that

[torn jolt]

Woha rlly?

You veryyy bad at geometry.

You know equation of circle?

[wheat ruin]

Yes obviously

[torn jolt]

At least...

[wheat ruin]

Are those timestamps?

[torn jolt]

So what is your question today sir?

How Can I help you to learn and master the theorems?

[wheat ruin]

Nah

Let's start at the first one ig

I'm cracked at geometry trust

[torn jolt]

You said you dont know geometry?

[wheat ruin]

I have no idea what the theorem states

I do I just don't memorize names of theorems

[torn jolt]

You know the theorem of abc-s?

🤣

[wheat ruin]

Abc?!

I don't memorize names smh

Bruh

[torn jolt]

csq =asq+bsq

Anyways jokes aside.

[wheat ruin]

Pythagorean?

[wheat ruin]

Ohhh

[torn jolt]

Good. Yeah... memory... exactly names...

[wheat ruin]

For the first one

We have similar triangles

And

[torn jolt]

WHY?

[wheat ruin]

Okay wait

Simpler solution

They are binded by the same arc

Let me draw rq

[torn jolt]

So what?

[wheat ruin]

Oops

See how they share that arc

[torn jolt]

And?

They share the same cord too.

[wheat ruin]

Ok we can prove they are similar instead ig

Let's have arc AB

[torn jolt]

Find videos that proofs each please.

[wheat ruin]

https://youtu.be/nd46bA9DKE0?si=_rrTP9Q3_C_EQWIZ

[torn jolt]

Prooofs

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[errant bison]

v

[errant bison]

need help wtih part b to this question

<@&286206848099549185>

[torn jolt]

!15

!15m

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[torn jolt]

!status

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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

[errant bison]

mb

i have an answer but its wrong

ive found PQ

what are your pronouns

[minor crater]

show your work

[torn jolt]

Binary. 00110101 / 10100110

😂

[errant bison]

Here’s the work binary

[torn jolt]

A moment let me explain you.

[errant bison]

did u use ai?

[torn jolt]

Grammarly?

[errant bison]

lol

that does maths

[torn jolt]

No,

Just to formmat it.

[errant bison]

because the distance is 4/root(6)

not 2root4

[torn jolt]

OH.

!done

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[errant bison]

thats not right either

.close

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[slender pond]

is this not a binomial type question?

[frozen needle]

I'd say it's more of a geometric distribution. (A bounded one at that). Binomial distribution is a distribution of the number of successes, which is a bit different to "turns until first success".

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[slender pond]

im doing a level maths so i dont think im allowed to use it

but i do see it tho

[frozen needle]

You don't need to use it for this

Because it's the finite case, and they've bounded it by four, you'll just manually calculate each one

For example, P(Y=1) is 3/5 (i.e. the probability that the first spin is odd)

Then P(Y=2) is 2/5 * 3/5 (i.e. the probability that the first spin is even and the second spin is odd)

Then continue that logic

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[slender pond]

oh ok

thank you for the help

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[peak dome]

this is shell method i wanna know how to find the radius

[peak dome]

here too

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[sudden canyon]

I need help determining the area of a shadow projected onto a plane from a triangle

[sudden canyon]

I started a projection calculation but am unsure about the numbers used or the next step

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@sudden canyon Has your question been resolved?

[torn gust]

slightly different approach but if you get the normal of the triangle and the plane, the area will be the 3d area * cos(theta) of the angle between the normals

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[sudden canyon]

The cosine angle between the two normals is 0. Does that mean the 3d area of the triangle is the same as the shadow?

[torn gust]

yea

[sudden canyon]

Ok that should be easy enough

Thanks

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[torn jolt]

Hellooo, on the first image is the original question then it askes a modified question and i got the test statistic wrong. I tried replacing n by the women only.

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[torn jolt]

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[unique fossil]

Please help me with question 12 😭😭

What am I even doing wrong, I legit checked over it a million times

[hot herald]

you're supposed to use the values in the given table

[unique fossil]

I don’t get how

What do I do with the 7.46$

[hot herald]

read the text describing what the table represents

at the very top

[unique fossil]

Do I do 240000 / 1000

Times 7,46

[hot herald]

yes

[unique fossil]

Ohhh

But why can’t I do what I done in the working out ?

[hot herald]

the mechanics of the interest is unclear

how its compounded etc

[unique fossil]

Alrrr tysm

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[solemn garden]

when do you know if the integral yields to a natural logarithmic function?

[narrow ermine]

That's highly dependent on the integrand itself. If you mean when is the antiderivative ln(x), then that would be 1/x.
Do you have anything specific in mind?

[sharp flame]

I mean if you have a rational or irrational function

if the total degree is -1 then youre probably gonna get a logarithm

[solemn garden]

ok thank you...............

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[eager sparrow]

help please

[eager sparrow]

??

[nova topaz]

You know to add two fractions you have to make them all have the same denominator right

[eager sparrow]

yes

[torn jolt]

still need help?

[eager sparrow]

yes

[torn jolt]

why

Have you tried combining the fractions into one

[eager sparrow]

yes

[torn jolt]

what did you get?

[eager sparrow]

i don’t get it

[torn jolt]

what do you mean?

Show your work

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@eager sparrow Has your question been resolved?

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[real silo]

hello, how to do this?

[neat aspen]

what have you tried

[real silo]

actually i don't understand the question at all

[torn jolt]

here is a hint to guide your thoughts. Assume you want to find the probability of u managing to do something on the 5th try of you doing it

Wouldnt you agree that you need to fail the first 4 times, and then succeed at the 5th?

[real silo]

yess

[torn jolt]

right

so right here the probability you are trying to find follows something called a geometric distribution

a geometric distribution is basically "whats the probability that i will get this after exactly k tries?"

the probability then simply would be [
\6\P{X = x} = pq^{x-1}
]
where $p$ is the probability of success and $q = 1 - p$ is the probability of failing. $x$ represents the amount of trials

[torn jolt]

do you see how to connect this with your problem? @real silo

[real silo]

yesss i have better understanding abt it now

[real silo]

thank you so much!!

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[gaunt lance]

If x1,x2,x3 are linearly dependent, then x1 is a linear combination of x2 and x3

[gaunt lance]

Why is this false?

[foggy garden]

wait sorry i miss read your question

who says it is false?

[gaunt lance]

My teacher

[foggy garden]

it says the correct answer is this

[gaunt lance]

Its asking for false at the top tho

[foggy garden]

oh

great question I am unqualified to answer

[gaunt lance]

ah nws ty anyway

got it

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[torpid geyser]

i have no idea how to proceed

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@torpid geyser Has your question been resolved?

[torpid geyser]

<@&286206848099549185>

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[dim pumice]

"Calculate the double integral
Tip: Be sure to change variables"

[full forum]

@dim pumice Has your question been resolved?

[leaden ermine]

Are you sure

[dim pumice]

No but that is what he wrote haha

[leaden ermine]

How far are you

[dim pumice]

Not far at all

[leaden ermine]

Oh no

Tears

[dim pumice]

Do I switch out x and y to what he wrote?

Yes

[leaden ermine]

You can do that

[dim pumice]

or do i start with tranforming the limits from x and y to u and v?

[leaden ermine]

Yws that too

[dim pumice]

cuz we have 2|x| + 3|y| <= 6

then we do 2|x| + 3|y| = 6

2|3u| + 3|2v| = 6

6|u| + 6|v| = 6

|u| + |v| = 6?

[leaden ermine]

Do you notice what kind of shape that is

Yes

[dim pumice]

a square?

[leaden ermine]

Yes

Basically your bounds

[dim pumice]

does this represent the corners?

ohh okay

[leaden ermine]

In terms of x

Or u

[dim pumice]

Okay do we get (2|3u| + 3|2v|) * 6dudv then?

[leaden ermine]

From u = v+6 to u = v-6

[leaden ermine]

Yes

v = -6 to v = 6

[dim pumice]

Ohh okay

Don’t we get I = 6 and then the integral from 1 to -1?

Or is it easier to keep -6 and 6?

[leaden ermine]

Why from -1 to 1

[dim pumice]

Is it not because |u| + |v| = 1

So we get 1 and -1 or am I thinking wrong

[leaden ermine]

Oh yess

Mb

You are right like always

[dim pumice]

so we have this kind of

[leaden ermine]

No not twice

[dim pumice]

ohh okay sorry

[leaden ermine]

It's a square but the square is roatetd

Like parallogram

[dim pumice]

ohhhh

[leaden ermine]

Dont cry it's ok

[dim pumice]

Haha okay

okay so what do we do now?

[leaden ermine]

We figure out the functions

Im terms of u

[leaden ermine]

Also not dxdy

[dim pumice]

is it just dx?

[dim pumice]

Ohh okay

[leaden ermine]

Yes

From u = v-1 to u = v+1

And v =-1 to 1

Imagine u is x axis and v y axis

[dim pumice]

Ohh yeah okay

so we have this?

[leaden ermine]

😭 where is the other integral

Just left it

[dim pumice]

[leaden ermine]

It's dübbelintegrallo

You are first integrating between thosw two functions

[dim pumice]

Ohh okay

[leaden ermine]

,w Integrate[2abs(x)+3abs(y),{x,3y/2-3,3y/2+3},{y,-2,2}]

Anyway both should yield the same

[dim pumice]

Okay

How do we get to that?

[leaden ermine]

Cant you calculate it

[dim pumice]

No

[leaden ermine]

Ok

[leaden ermine]

We can factor here 6 again

[dim pumice]

Yes okay so we just get |u| + |v| in the middle?

And 6 outside of the integrals?

[leaden ermine]

,, 36\int_{-1}^1 \int_{v-1}^{v+1} \left ( \abs{u}+\abs{v} \right ) \dd u \dd v

[glossy valve]

𝔸dωn𝓲²s

[dim pumice]

Oh yeah and multiply with the other 6!

[leaden ermine]

Now |v| is a constant

When we integrate by u

Ok?

[dim pumice]

Okay

[leaden ermine]

The only thing that needs to be done is get rid of the absolute value

By splitting the integral

Thats the trick

,, = 36\int_{-1}^1 \left [ \int_{v-1}^{0} \left ( -u +\abs{v} \right ) \dd u + \int_{0}^{v+1} \left ( u +\abs{v} \right ) \dd u \right ] \dd v

That's it

For the negative part of the bounds you put a minus

[dim pumice]

Okay

[leaden ermine]

For the positive you can simply leave the absolute value

[dim pumice]

Right!

[leaden ermine]

Now you can do that for sure

[dim pumice]

Yes

[glossy valve]

𝔸dωn𝓲²s

[dim pumice]

Yes

[leaden ermine]

What you get

[dim pumice]

-u-v for the first one

And u+ |v| for the second

[leaden ermine]

✋

Hold

On

[dim pumice]

[leaden ermine]

Lemme check

[leaden ermine]

Wouldn't you get for the first one

,, = 36\int_{-1}^1 \left [ \left[ -\frac{u^2}{2} +u\abs{v} \right ]{v-1}^{0} + \left [ \frac{u^2}{2} + u\abs{v} \right ]{0}^{v+1} \right ] \dd v

[dim pumice]

Yes you are right

Do we need to simplify more or can we put in the bounds now?

[glossy valve]

𝔸dωn𝓲²s

[leaden ermine]

yes

put in the bounds what youget

[dim pumice]

We get (v-1)^2 / 2 - (v-1)u?

For the first one v-1 and 0

And for v+1 and 0 we get (v+1)^2 /2 + u(v+1)?

[leaden ermine]

not quiet

|v| remains as it is

,, = 36\int_{-1}^1 \left [ \frac{(v-1)^2}{2}-(v-1)\abs{v} + \frac{(v+1)^2}{2}+(v+1)\abs{v} \right ] \dd v

[glossy valve]

𝔸dωn𝓲²s

[dim pumice]

Oh okay

[leaden ermine]

Do you see your mistake

[dim pumice]

Yes

[leaden ermine]

it's okay

Now we need to do the same thing, splot the integral

to get rid of |v|

can you tell the bounds?

[dim pumice]

For the negative part from -1 to 0 and for the positive 1 to 0

[leaden ermine]

hell yeah*

,, = 36\int_{-1}^0 \left ( \frac{(v-1)^2}{2}+(v-1)v \right ) \dd v+ 36\int_0^1 \left ( \frac{(v+1)^2}{2}+(v+1)v \right ) \dd v

[dim pumice]

Yes

[glossy valve]

𝔸dωn𝓲²s

[dim pumice]

I’m sorry this is an occupied channel

[leaden ermine]

yummy will explain

[full forum]

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).

[leaden ermine]

yummy

[dim pumice]

Yes

Wait don’t we get 18 for the first part?

[leaden ermine]

I think the function is even too

[leaden ermine]

no

[leaden ermine]

yea it should be even as i did itmentally

[dim pumice]

Okay

[leaden ermine]

yea mb

[dim pumice]

Do we expand (v-1)^2?

[leaden ermine]

wait

if you know it's even

what can you do

Then you can integrate one integral

multiply it by 2

[dim pumice]

We can integrate one

Yes!

[leaden ermine]

ok i was too fast

i didnt know you were a critical thinker too!

[dim pumice]

Haha no I was too slow

Haha

[leaden ermine]

$= 36\int_{-1}^0 \left ( \frac{(v-1)^2}{2}+(v-1)v \right ) \dd v+ 36\int_0^1 \left ( \frac{(v+1)^2}{2}+(v+1)v \right ) \dd v\$
$\= 2 \cdot 36\int_0^1 \left ( \frac{(v+1)^2}{2}+(v+1)v \right ) \dd v$

[glossy valve]

𝔸dωn𝓲²s

[dim pumice]

Okay now we can expand (v-1)^2?

[leaden ermine]

You actually don't need to expand

[dim pumice]

Ohh okay

[leaden ermine]

(v+1)²/2 can be integrated using power rule because the argument is linear

But (v+1)v can be expaneded yes

[dim pumice]

Ohh okay

[leaden ermine]

So what would (v+1)²/2 be integrated?

using only power rule

[dim pumice]

No wait

[leaden ermine]

ok expand it then 😭

[dim pumice]

That was wrong

(V^2 +2v + 1)/2

[leaden ermine]

that is just expanded

I meant this yummy

[dim pumice]

V^2/2 + v + 1/2

[leaden ermine]

yea that is to be integrated now

but my point was

[dim pumice]

Okay hold on

1/6v^3 + 1/2v^2 + 1/2v

[leaden ermine]

yes...

but bee

you could have done this too

(v+1)²/2 integrated is simply (v+1)³/6

😅

[dim pumice]

Oh haha

[leaden ermine]

because v+1 is lienar

[dim pumice]

Right

[leaden ermine]

that's why it only work

but you like it the hard way is ee

what is (v+1)v integrated?

if you suffer lemme know

[dim pumice]

I don’t know

[leaden ermine]

Ok

NOW

somehow expanding is not an option...

😂

OH NOO it's (v+1)v it's so over

[dim pumice]

V^3/3 + v^2/2

[leaden ermine]

you did it

[dim pumice]

But I know we can write it in a simpler way

[leaden ermine]

ok i'll let you cook

$\= 2 \cdot 36 \left [ \frac{(v+1)^3}{6}+ \frac{v^3}{3}+\frac{v^2}{2} \right ]_0^1$

[glossy valve]

𝔸dωn𝓲²s

[leaden ermine]

No did we actually do some mistkae

[dim pumice]

V^2(2v + 3) /6

[leaden ermine]

FUCK

[leaden ermine]

that's wrong

[dim pumice]

No

[leaden ermine]

When we split this we just took the half part lMAOO

[dim pumice]

[glossy valve]

𝔸dωn𝓲²s

[leaden ermine]

Ok NOW

,,= 2\cdot 36\int_0^1 \left ( \frac{(v-1)^2}{2}-(v-1)v +\frac{(v+1)^2}{2}+(v+1)v \right ) \dd v

[glossy valve]

𝔸dωn𝓲²s

[leaden ermine]

We already integrated somethings yummyyyyyyyyyyyy

[dim pumice]

Okay

[leaden ermine]

It's not time to wipe tears yet ⚔️

[dim pumice]

Haha okay!

[leaden ermine]

what is (v-1)²/2 integrated?

quickly

[dim pumice]

This

[leaden ermine]

But

with

YUMMY

CMONM

[dim pumice]

What?

With -?

[leaden ermine]

yes

[dim pumice]

What is that?

[leaden ermine]

It's (v-1)^2/2

[dim pumice]

Ohhhh

[leaden ermine]

not (v+1)^2/2

[dim pumice]

I am dumb

[leaden ermine]

glory to the almighty

[dim pumice]

(V+1)^3/6

[leaden ermine]

ok

it's still (v-1)^2

[dim pumice]

Okay

[leaden ermine]

(v-1)^2/2 integrated is (v-1)^3/6

really surprising i know 🚬

[dim pumice]

Haha

[leaden ermine]

what is (v-1)v integrated?

[dim pumice]

V^2(2v+3)/6

-!!!

[leaden ermine]

you justdestroyed it

but i sse what you did

you try your very best

to keep up with this cruel world

[dim pumice]

Yes

[leaden ermine]

,,= 2\cdot 36 \left [ \frac{(v-1)^3}{6}-\frac{v^2(2v-3)}{6} +\frac{(v+1)^3}{6}+\frac{v^2(2v+3)}{6} \right ]_0^1

[glossy valve]

𝔸dωn𝓲²s

[dim pumice]

Okay yes

84?

[leaden ermine]

yes

times 2

[dim pumice]

Right

168

[leaden ermine]

times 2

[dim pumice]

Again?

[leaden ermine]

no

i tested you

[dim pumice]

Oh okay haha

[leaden ermine]

i tested your critical thinking

i thought you only say yes LOL

[dim pumice]

Haha nooo

[leaden ermine]

same integral without the hint

[dim pumice]

Are we finally done?

[leaden ermine]

What do you think?

[dim pumice]

Yes we are!

[leaden ermine]

Right!

[dim pumice]

[leaden ermine]

Have you managed the task from yesterday

[dim pumice]

No

But

[leaden ermine]

dübbel

[dim pumice]

I will take a break now

Haha

[leaden ermine]

you'reso funny

[dim pumice]

Dubbel

[dim pumice]

No you are

MAYBE I will do one more later

But we will see

[leaden ermine]

we will See

[dim pumice]

Yes

[leaden ermine]

👀

[dim pumice]

I have tomorrow to finish it as well

[leaden ermine]

well as!

i am gonna drink a coffee

[dim pumice]

Yes you do that!

Thank youuuu

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[torn jolt]

can someone tell me what are they trying to say
especially the equation?😭😭😭

[torn jolt]

u funny fr😭😭

[full forum]

@torn jolt Has your question been resolved?

[pliant dagger]

what is the question?

[torn jolt]

q was, if f(x) =2^x, solve f^-1 (x) = 5 what does this line mean btw

putting these together can be written like
f f^-1 (x) = (5)
which means functioning the inverse function of x will give us x
now putting the f on other side to find x
on thr right side they cancel out so
x = f(5)
f(x) also equal to f(x) = 2^x so, we will subsititue in f(5)
x = 2^5
x = 32
ye?

[torn jolt]

^

[strong leaf]

that line means "for what value of x, does f^-1(x) = 5"

this line tells us that f(f^-1(x)) = x

and we're given that f^-1(x) = 5

so they subsitute that

[torn jolt]

👍

okay

[strong leaf]

does that make sense do you know where to go from there?

[torn jolt]

yea

thank youuuu

[strong leaf]

okay cool

no worries

[torn jolt]

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[proper briar]

Can someone help me here...I can't understand the solution here from where t-1 and t-s came.

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@proper briar Has your question been resolved?

[proper briar]

<@&286206848099549185>

[proper briar]

.close

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[modest rain]

how do you learn unit circle by heart? or is just memory?

[somber niche]

you mean trig unit circle?

there's a trick to it

[wide sundial]

I just redraw the triangles with Pythagoras every time

[modest rain]

yh can you tell me?

[somber niche]

so for the sines you memorize angles 0 30 45 60 90, that's easy right?

as for the values of sine, they go, in order sqrt(0) / 2, sqrt(1) / 2, sqrt(2) / 2, sqrt(3) / 2 and sqrt(4)/2

so basically you just increase the number under the root

by 1 for each

for cos you do reverse order

and then you can derive tan as sin / cos

[modest rain]

what if i dont have a calculator at hand to compute the arccos for e.g?

[wide sundial]

Then you write the answer in terms of arccos

Whoever is reading can put it in the calculator themselves if they care to

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[tepid aspen]

If only someone could explain how to do a and b for me😔

[hardy snow]

Can you identify any special triangles

And transversals

Remember it’s a parallelogram

[tepid aspen]

Triangle ABE, Triale CDE and triangle BEC?

[hardy snow]

Yeah but any special properties

Ig we don’t need to know that right now but what angles do you know

[tepid aspen]

theres 90 degree angle in triangle ABE and EDC?

[hardy snow]

No…

[tepid aspen]

😢

[hardy snow]

Can you spot all the congruent sides first

[tepid aspen]

AB//DC , AD//BC

[hardy snow]

What other congruencies do we have

That are given

[tepid aspen]

AB=AE and CD=DE?

[hardy snow]

ok and what side lengths is given

[tepid aspen]

AB = 50 cm and DC= 50cm?

[hardy snow]

Ok can we find DE then

[tepid aspen]

I dont know 😦

[hardy snow]

mark the congruencies on your sheet

Do you know how to do that

[tepid aspen]

yea

[hardy snow]

Ok so looking at the markings

We need DE and we have DC

You’re so close

[tepid aspen]

am i

[hardy snow]

These are the markings right?

I hope this is what yours resembles too

[tepid aspen]

its the same

[hardy snow]

ok can you point to DC

And then look at DE

we know that DC is 50cm

[tepid aspen]

ahh

[hardy snow]

What 😭

[tepid aspen]

DC = DE... something something

[hardy snow]

Ok

So that means the length of DC and DE are equal right?

[tepid aspen]

yep

[hardy snow]

And what is DC…

[hardy snow]

Do you understand what these markings mean

It’s easy to keep track of your information by writing it down

[tepid aspen]

all i know is that they mean equal

[hardy snow]

Equal what

How are they equal

[tepid aspen]

like 1 markings means these sides are qual and two marking means these sides are equal and so forth

[hardy snow]

Yes so sides with the same amount of dashes means they have the same LENGTH

Now given that the length of AB is 50 what is DE

[tepid aspen]

DE = 50... given?

[hardy snow]

The length of DE isn’t given

But yes it is 50 do you see

[tepid aspen]

i can

[hardy snow]

Does it make sense

Can you make these markings on your paper

[tepid aspen]

i can make them

[hardy snow]

ok cool

can you solve for the perimeter

[tepid aspen]

lemme try

[hardy snow]

It helps a lot to solve when you have those markings so please do

[tepid aspen]

oh yeah i always make them but i just couldnt get where exactly they come in to play once i calculate

[hardy snow]

Did you calculate the perimeter?

[tepid aspen]

Im trying but i dont even know what it is

[hardy snow]

can I see your markings

[tepid aspen]

@hardy snow

[hardy snow]

Ok well you need to include the full shape

And all the given info

Can you mark that down as well

[hardy snow]

Ok mark all sides that are congruent to AB with 50cm

We are doing this because we want to find perimeter

[tepid aspen]

i marked it on my practise thing

[hardy snow]

ok so what is the length of the top side

[tepid aspen]

AE = 50
DE = 50

do we combine it for 100?

[hardy snow]

that is a question for you

would it make sense and why or why not

[tepid aspen]

it would because we have all the necessary lenght to get AD

but yeah maths can be wierd sometimes so im prolly werong

[hardy snow]

Correct

Another way to phrase it

Is that AE + ED = AD

Since E is a point on the line AD

so if you have the top side

What os the bottom side

[tepid aspen]

its 100

[hardy snow]

ok

What is the perimeter

remember to write down new information in the diagram

[tepid aspen]

300cm

?

[hardy snow]

I saw 300 degrees and I was shocked lol

[hardy snow]

Yes correct

Stunning

magnificent

[tepid aspen]

ahh danggg now i get it

teachers in primary usually skip over this so when we arrive in highschool we gotta relearn from scratch

but yeah thanks

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[torn jolt]

frrrrr

omg frrr

annoying bad trachers

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[charred escarp]

Say I have one point, and a circle that's been transformed to some other place in the x,y plane. How could I find the maximum distance between the first point, and a point on that circle?

[charred escarp]

Or to phrase it better, how can I find which point on that circle gives me the maximum distance to the first point

[grand yacht]

take the further intersection to the circle with that point and the center of the circle

[charred escarp]

Wdym?

Ohhh I get it, thanks

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[next basalt]

"determine the largest and smallest value of the function ... for ... hint: f is compact, i.e max and min is achieved through the critical points, the singular points or on the stripe.

[next basalt]

I've made some progress already but i need some help finishing up

It's solved with Lagrange

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[next basalt]

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[nova ravine]

Anyone mind assisting me with question 1?

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[fathom terrace]

Could someone plz help me put these two equations into standard form and y=format?

The highlighted ones, I mean

[viscid blaze]

ok

but they aren't equations

they are inequalities inside of the braces

[fathom terrace]

Those are just the domains/ranges

I mean, the other stuff

Everything before that

Yk what I mean?

[viscid blaze]

yeah

[fathom terrace]

So, could u help me out?

[viscid blaze]

bring all the ys and y^2 to one side

https://cdn.discordapp.com/attachments/903486480985489478/1234267583532896376/IMG_9218.jpg?ex=66301cc3&is=662ecb43&hm=77b38cff4f3886ebba4bece848b798ff3c6a6e96c70ff7935a25f5e2bfcde14d&

so you get 0.1y^2 + .2y = x - 7.1

then you bring the .2y to the other side

0.1y^2 = x - 7.1 - .2y

then you take the square root

[fathom terrace]

Is this correct for the y= format of the first equation?

@viscid blaze

[viscid blaze]

there are two answers

thats wrong btw

[fathom terrace]

Oh

[viscid blaze]

its + or - sqrt(10x-30) - 1

[fathom terrace]

Oh

[viscid blaze]

thats ok

[fathom terrace]

Where’d I go wrong?

I just wanna do it this way cuz I feel like it’s easier

[viscid blaze]

i have to go

srry

bye

[fathom terrace]

Ok, nw

Thanks

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[granite torrent]

trigonometric pythagorean theorem seems like a good way to continue

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[normal ginkgo]

Wait

.close

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[torn jolt]

how do i solve dis

what does it even mean

[silver lintel]

you plug in (x-5) in f

(x-5)^2 + 3(x-5) + 2

and then you expand that

[torn jolt]

how did u know

like why

[silver lintel]

this is called function composition

[torn jolt]

oh!

[silver lintel]

there is no reasoning it is just some operation

[torn jolt]

thank you!

🌹

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[coarse tulip]

just to double check but this is like undefined or something ritght

[umbral dome]

the integral is defined because ln(x) cos(x) is a continuous function

however, you can't write it in terms of elementary functions

[coarse tulip]

like "no solution" or is there one?

[craggy tapir]

not exactly no solution

it means you literally cannot express the result in terms of already known functions, like sinx cosx or anything

youd have to define a new function

[craggy tapir]

to elaborate on this further, this is because if you perform IBP on the function, youll end up with sinx/x, which is famously an integral that cant be written in elementary functions

[coarse tulip]

uhh

what about in terms of ap calc ab

[umbral dome]

what are the problem instructions for this?

[coarse tulip]

none

The title is "Practicing antidervatives for the AP Calculus AB Test"

[craggy tapir]

i dont think your prof is sadistic enough to give you these types of questions in ap calc ab

[coarse tulip]

she is

[craggy tapir]

is there DNE option?

[coarse tulip]

probalby

[craggy tapir]

but technically speaking the integral isnt a DNE though...

try to talk with your prof about this

[coarse tulip]

ok

but she never responds to emails

tf

troll sending gory images