help-10

[sinful falcon]

x^e

[subtle sinew]

Nope

Think power rule

[sinful falcon]

ex^e-1

[subtle sinew]

Yes exactly

[sinful falcon]

How was this right then?

[subtle sinew]

Because the derivative of e^x is e^x

[sinful falcon]

Ok

[subtle sinew]

Instead of x's, replace it with r's again

[sinful falcon]

They would be re^r-1

[subtle sinew]

Where is that from?

e^r + r^e is what you have

[sinful falcon]

Yes

[subtle sinew]

What's the derivative of e^r?

[sinful falcon]

e^r

[subtle sinew]

Then r^e?

[sinful falcon]

er^e-1

[subtle sinew]

Yes

[sinful falcon]

Is that my derivative?

Just that

[subtle sinew]

Yes

[sinful falcon]

.close

[obtuse pebble]

Channel closed

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[sinful falcon]

.reopen

[obtuse pebble]

✅

[sinful falcon]

Sorry for keep reopening this concept is hard for me

Problem 42, could someone guide me?

[modern onyx]

We need to rewrite the function first to make it easier

So u have v^1/3/v - 2ve^v/v

[sinful falcon]

Yes

[modern onyx]

We can simplify this

[sinful falcon]

I don’t know how I could take out the v

Since it’s raised to 1/3

[modern onyx]

V^-2/3 - 2e^v

[versed pier]

just rewrite

1/v = v^{-1}

and apply exponent rules

[modern onyx]

A^b/A^c = a^b-c

[versed pier]

or use that

same thing

[sinful falcon]

So just move v to the top?

[versed pier]

mhm

and then simplify with exponent rules

in particular,$ v^{a} * v^{b} = v^{a+b}$

[sinful falcon]

This doesn’t look right

[subtle sinew]

You did the same mistake again, saying 1/v is adding v^-1

[sinful falcon]

What is it then?

Sorry

[subtle sinew]

Try doing what you did in 25

[sinful falcon]

But isn’t this what I’m suppose to do

Like the other person said

[subtle sinew]

Moving it to the top isn't adding v^-1

It's multiplying

For example, (x +1)/x = (x + 1)x^-1

[sinful falcon]

Yes

[subtle sinew]

If that made sense, then why did you add v^-1 here?

[sinful falcon]

I forgot parenthesis

Do I

Multiply everything

now

[subtle sinew]

Sure

[sinful falcon]

That seems like so much work though

Is there a easier way

[subtle sinew]

It'll be the same amount of work still, you need to simplify it to make it easier

[sinful falcon]

ok

[modern onyx]

I feel like rewriting the equation and splitting the fraction would be easier

[sinful falcon]

I did not understand this

Sorry

[modern onyx]

4th one

[sinful falcon]

Here uh did I do this right

[modern onyx]

Wait

Whys there e^v-1

[sinful falcon]

Because I added 2v to v from previous line

Which is 2v^2

[modern onyx]

What happened here?

[sinful falcon]

I moved v up from denominator but I forgot to put parenthesis

[modern onyx]

What abt the -v^-2?

[sinful falcon]

I branch down the -1 which makes it negative and -2 because you subtract 1

Bring

v^-1 because it’s from denominator then it becomes -v^-2 because -1(v) and -1-1 becomes -2

[obtuse pebble]

@sinful falcon Has your question been resolved?

[obtuse pebble]

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[sweet bear]

Hi, wondering how I can solve this, this comes from surface area of revolution of a cubic function (ax^3+bx^2+cx+d)

[obtuse pebble]

@sweet bear Has your question been resolved?

[thorn summit]

Disgusting integral damn

[obtuse pebble]

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✅

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[delicate lava]

Hello! I'm currently doing IGCSE preps, I'm so stuck with this question. Does anybody have resources or if possible can you teach me? You can use the other examples to help me understand :]

[obtuse pebble]

@delicate lava Has your question been resolved?

[analog creek]

this one is p simple

i can explain it if you havent figured it out yet :)

[delicate lava]

that would be amazing!

[analog creek]

basically ab and cd are parralel

and bc they intersect with the other one

you can solve this just by finding the scale

so AB is 3.0, and cd is 7.5

thats a 1:2.5 ratio

so every side of the big one is 2.5 times the same side of the small one

lmk if this helps :)

(first step is to find XC if you need a hint)

[delicate lava]

o

[civic kettle]

Dini is 100% correct however

Do you know the concept of similarity?

Of triangles

[delicate lava]

I only know about the 180 degrees for every triangle things, sorry I'm not so good in math.

[civic kettle]

Then I'll explain from the basics

[delicate lava]

Sure thing!

[delicate lava]

Thank you :]

[civic kettle]

First we'll know what we mean by when we say

If two triangles are 'similar'

That means that their ratio of the side will be constant

Let's say we have a triangle of sides 5,10,15 and another triangle of sides 15,30,45

We'll find their corresponding ratio

First side of triangle 1/First side of triangle 2

5/15

[analog creek]

its basically just finding the ratio

i think we just worded it differently

[civic kettle]

Yes absolutely I'm just teaching the basics because it is necessary I feel

[civic kettle]

The ratio of first sides will be 1/3 upon simplification

[analog creek]

fair enough

[civic kettle]

Now we'll do it for second sides

10/30

Upon simplification it's 1/3 as well

[delicate lava]

1/3 as well

Mhm!

[civic kettle]

And so will be the case for the third one

15/45

[analog creek]

so you only need to know one

which is what your problem is asking for

so multiply BX by your ratio, and thatll be XC

then u just add them and your set

[civic kettle]

I appreciate your help,We'll get there

Let me tell her how to know

If triangles are

Ratio-able

[civic kettle]

1/3 as well

All the side's ratio is 1/3

Hence it is a similar triangle

[analog creek]

sure ill let u take it from here

lmk how it goes :)

[civic kettle]

Thank you

[civic kettle]

Sure thing

[civic kettle]

Understood?

[delicate lava]

understood !

[civic kettle]

Noicee

Now you know the property of similar triangles

Now how to know how to know triangles are similar

[civic kettle]

In this question

We can see both the triangle touching each other

At centre

[delicate lava]

Yup

[civic kettle]

So both the angles of triangles

Gonna be equal

Because

[delicate lava]

I see

[civic kettle]

They are vertically opposite angles

They'll always be the same no matter what

If they are vertically opposite angles

[delicate lava]

got it

[civic kettle]

We got 1 angle common on both the triangle

To prove they are similar we need to get another 1

We are proving they are similar by AA similarity (angle-angle)

When two triangles have 2 same angle

They are similar also

[delicate lava]

[civic kettle]

You're getting everything right?

Do tell me if something bothers you

[delicate lava]

will do!

3/7.5 = 2/5 when I simplified it

so does the means the rest are 2/5 as well?

feel free to correct me tho!

[civic kettle]

You are absolutely correct

But you gotta know

For sure

That the triangles are similar

Only then it'll work

[delicate lava]

o

oke oke

[civic kettle]

In this case it is

But I'll teach a last thing

Which'll help you to understand

It mostly

When the sides are parallel like that

AB and CD

[delicate lava]

Yup yup!

[civic kettle]

A diagonal line intersecting them making those angles in the triangle I marked

Are same

Angle XCD and Angle XBA

Is equal

[delicate lava]

I see

[civic kettle]

We can do this for the other side also

Hence we now have 2 angles equal

Of the 2 triangles

They are similar

And we're sure of it

[civic kettle]

One is a vertically opposite angle and one is a parallel one (it's called alternate interior angles)

Now that we're sure

We can do how you were doing it

Putting the sides on ratio

3/7.5

2/5 that is

[delicate lava]

Okay!

[civic kettle]

Now equate the ratio with other sides

[delicate lava]

So now I know that it has to be 2/5 I have to find the denominator?

[civic kettle]

Exactly

Correct

BX/XC should be equal to 2/5

Or 2.7/XC = 2/5

[delicate lava]

Yup

[civic kettle]

So what will be the value of XC

[delicate lava]

6.75?

[civic kettle]

Exactly

Now do the same with other side

[delicate lava]

5

[civic kettle]

2/XD = 2/5

[civic kettle]

Yesssss

There you go

You got the values

[delicate lava]

BC = 9.45

AD = 7

O

WAIT I GOT THE ANSWER

[civic kettle]

Yeahhhh

BC is 9.45

That's the answer

[delicate lava]

Thank you so much! That's really what i'm looking for, I like to get the right answer but understand it as well. Thank you so much both @civic kettle and @analog creek you guys are very helpful :]

[civic kettle]

no problemo

glad to help

[delicate lava]

Do you mind if I dm you for future guidance?

[civic kettle]

Not really I'd love to help you but like I barely come online on discord

I might not be that great of a help

But I'm always willing when I come!

[delicate lava]

That's okay :]

[civic kettle]

Okay then

Farewell

Have a great day ahead

[delicate lava]

.close

[obtuse pebble]

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[obtuse pebble]

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[twilit nimbus]

did i mention everything?

[rocky goblet]

think about $\lim_{x\to 0} \frac{x^2}{x^2}$

[warm shale]

bee [it/its]

[twilit nimbus]

that would just be 1

i was going to mention the lim of sinx/x

[rocky goblet]

yeah obviously that's the correct answer, but what answer do you get if you apply the formula in the poster?

[twilit nimbus]

yeah true

[rocky goblet]

because when x=0 it's 0/0 so this is a case where you can apply l'hopital, if for some reason you don't just cancel it

[twilit nimbus]

yeah i see

but how about the conditions that i mentioned in that description?

do u see anything that i missed in terms of conditions?

[rocky goblet]

well you missed the condition i'm trying to hint you towards, which is that in the case of x^2/x^2 the poster gets the wrong answer :)

but aside from that it all seems correct

[twilit nimbus]

I just realised that that example would work

[rocky goblet]

i mean the incorrect claim isn't that l'hopital applies, you can apply l'hopital here

you just shouldn't do it the way the poster does

the correct formula is $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

[warm shale]

bee [it/its]

[rocky goblet]

they missed the limit on the right so applying l'hopital to $\frac{x^2}{x^2}$ gives $\frac 00$ instead of $\lim_{x \to 0} \frac{2x}{2x}$

[warm shale]

bee [it/its]

[twilit nimbus]

oh lol

i didnt see that

[obtuse pebble]

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[obtuse pebble]

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[sharp star]

this is my first proof i’ve ever done, i just want to know if i approached it correctly and if there’s anything i should improve on my latex. thanks for any help 🙏

[pseudo swift]

3^7b = 3^a, how do you end up w/ that ?

[sharp star]

i can write out the math, one second

[pseudo swift]

so log_3(7) = 7 ?

[sharp star]

no, i raised both sides to power of 3 which makes this 3^(log_3(7)) * 3^b, 3^(log_3(7)) is equal to 7

[pseudo swift]

ok so you don't have 3^7b

you have 7^b

[sharp star]

no i can show you

nah u right

i mixed it up in photomath calculator

it’s my first time doing math after 3 months tbf 😭

[pseudo swift]

so 7^b = 3^a, here FTA makes more sense

you can't have a power of 7 equal a power of 3

[sharp star]

yeah i was just gonna say ill try to fix it with that because that would work better

do u have any tips on the latex or do you think it looks good

[pseudo swift]

it's fine to me

maybe one line per statement is a bit overkill but I can see the appeal

[sharp star]

yeah i just didn’t know because couldn’t find anything online 😭 thanks for ur help 🙏

[pseudo swift]

don't forget the \log on the 1st line

[sharp star]

thanks i just noticed

[obtuse pebble]

@sharp star Has your question been resolved?

[obtuse pebble]

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[tulip zodiac]

Hello. Yesterday I made the following question:

A student takes a multiple choice test with 3 questions. Each question has 5 possible answers, only one of which is correct. what I'm asked is: If a student answers each question at random, what is the probability that they will answer one correctly

Why is nCr(3,1)*(0.2)^(1)*(0.8)*(2) not equivalent to nCr(3,1)/2^(3)? I'm trying to figure this out without the binomial distribution

[tulip zodiac]

This is what I get from using a calc

[plain stag]

what's your rationale behind the second expression?

[tulip zodiac]

My reasoning: Since I have to calculate the probability of students answering 1 question correctly random, I can just use nCr(3,1) because each set corresponds to a valid question + 2 incorrect ones divided by the sum of combinatorics from k = 0 to n = 3 which is my sample space cardinality. I thought this was a valid answer because I can interpret the problem as the questions being chosen are the ones answered correctly. So, it'd just be a matter of adding up sums of combinatorics

[plain stag]

using combinatorics like this gives you the probability of selecting the question, but it doesn't also give the probability of then answering it correctly

it's not like the usual combinatorics questions with balls in hats or decks of cards or such, where you get a jack if you get a jack

[tulip zodiac]

how so?

[plain stag]

how so what?

where in your second expression is the information about the number of answers and/or the probability of answering correctly encoded?

[tulip zodiac]

like the sets in nCr(3,1) are not the possible combinations in which a question can be correct?

[plain stag]

3C1 is just the number of ways to choose 1 question from the 3

"sets in nCr" doesn't make sense. nCr just outputs a number

[tulip zodiac]

but it could be interpreted as the number of ways each question can be correct if order does not matter no?

[plain stag]

what

"number of ways each question can be correct"?

each question can be correct in one way

[tulip zodiac]

if I have 1,0,0, it means the first question is correct, and the other 2 are incorrect no? so, nCr(3,1) would just be counting 1,0,0 0,1,0 and 0,0,1 no?

[plain stag]

if I have 1,0,0, it means the first question is correct, and the other 2 are incorrect no?
sure
so, nCr(3,1) would just be counting 1,0,0 0,1,0 and 0,0,1 no?
again, nCr outputs a number, not a list/vector/however you want to call this

i really want to stress that while 3C1 allows you to choose what question you want to call correct, it does not further provide any information on the probability that the question is actually correct

[tulip zodiac]

yeah exactly, nCr(3,1) would be outputting 3 because its the number of ways one can select a correct question + 2 in correct ones among 3 questions

[plain stag]

indeed

[tulip zodiac]

exactly, that's why I add sum nCr(n,k) from k = 0 to n = 3 in the denominator

to count the cardinality of sample space

[plain stag]

the probability of each outcome isn't equal, so you can't do that

[tulip zodiac]

hmm could you please elaborate? i think i get it but i dont know how to put it in words

[plain stag]

what do you think your sample space is?

[tulip zodiac]

(0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), (1,1,1)

[plain stag]

you only do the "divide by the size of the sample space" bit when each of your elements has the same probability

what you're basically saying is that the probability of getting a question correct is 1/8, and then since you have 3 questions, the probability of getting "the first or the second or the third" correct is 1/8 + 1/8 + 1/8 = 3/8

but that's not true. The probability of getting no questions correct is

,calc .8^3

[warm shale]

Result:

0.512

[plain stag]

the probability of getting the first (or second or third) question correct is

,calc 16/125

[warm shale]

Result:

0.128

[plain stag]

etc

[tulip zodiac]

1/8 would be the probability each combination in the sample space has, right? so, yes I would be saying 3/8 is the probabilty of the first or second or third question being correct

so, what's wrong is that they do not have the same weight? as in for instance (1,0,0) in reality represents 4 combinations?

[plain stag]

1/8 would be the probability each combination in the sample space has, right?
no. 1/8 would represent the probability of obtaining each outcome if they were equally likely, but they aren't

[tulip zodiac]

ahhh I get it

[plain stag]

(0,0,0) has a different probability than (1,0,0) which has a different probability than (1,1,1)

[tulip zodiac]

im not taking into account that each outcome has different weight

and by ignoring that I'm saying they all have the same probability

if the problem said that each question had 2 answers, then it would be equivalent because, for instance, (1,0,0) would appear just once as the there's only one way in which the other 2 questions can be wrong, right?

[plain stag]

right

[tulip zodiac]

ahhh so, what im lacking is that im not accounting for that each combination has different weights

weight as in frequency

ok, noted

thank you!!

[plain stag]

[tulip zodiac]

.close

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[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

hello i need ur help

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[hexed knoll]

I believe this is true

I have down that we let x be in the set A u B u C

And we have x can be in a , b or c

By def of set union

So we have three cases

If x is in a, then x cannot be in (B and C) by our given statement

I’m not sure where to go from there

I want to say x in a and x not in b so get one the statements to the right

But I feel like I can’t split the statement up because since we know x is not in the intersection (B and C)

It could be the case that x exists in B and not in C

I belief

Believe *

[mighty geyser]

just try splitting it

so not in B, you are done (how?)
in B implies not in C, which means you are done (how?)

then there's also the other direction to show

which should be clear

[hexed knoll]

Oh ok

So we have like sub cases almost

So if it’s not in the intersection it can be in one or the other

It could also be outisde both sets

Wouldn’t that make a lengthy proof for this question because this was a practice test question for my class a few years ago

We could use demorgans I think !

So we would end up with x in A and x not in B or x in a and x not in C

But we only have one of those in the right side… would thsi make it not equal since we are not accounting for (A-C)?

[mighty geyser]

what I was showing was splitting by cases on whether x is in B

In the first case, x is not in B. In the second case, x would be in B.

[hexed knoll]

Isn’t the intersection of (B \cap C) included in B though?

I guess in the first case x is in b and x no in c, so we have (B-C) when x is not in b x we have (A-B)

Would it be valid to apply demorgans and say x in A x not in B or x in A and x not in C?

[obtuse pebble]

@hexed knoll Has your question been resolved?

[mighty geyser]

We start from x in A.
Split by whether x is in B.
x not in B -> you should be able to see you are done
x in B -> x not in C -> you should be able to see you are done

[mighty geyser]

this looks like the idea

[hexed knoll]

Yeah it seems so obvious now lol

Thanks a lot !!!!!!

.close

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[thin plover]

hello i have a calc 3 question

[thin plover]

so i have this so far

i could find a line thats perpendicular to the given line through dot product

[obtuse pebble]

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[thin plover]

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[warm shale]

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

[marsh blade]

Okay have a look at (A◇B)

Do you know what exactly this means?

@pliant adder

[obtuse pebble]

@pliant adder Has your question been resolved?

[marsh blade]

Yes

Yeah but what exactly does that expression mean?

Yes

So C intersected with that being the empty set, means that C is partly in the intersections of A and B or outside of A and B

Or its inside both

Like it has elements that are neither in A nor in B

But it could also have elements that are in both A and B

And now here is the thing

Yes

With P(A U B) = 0.9

Meaning P(O/(A U B)) = 0.1 with O being Omega

I wish I can express myself better but idk how to use the Tex bot nor can I stream cos Im.on phone

Oh nice du kannst deutscv

Wollen wir weiter auf deutsch?

Also Omega ohne A und B hat 0.1

Und Durschschnitt A und B ist 0.2

Aber jetzt kommt der Knackpunkt

C hat 0.3

A Durchschnitt B hat 0.2

Und Omega ohne A und B hat 0.1

Genau

Aber hast du jetzt erkannt was C genau ist

Nicht nur mindestens

Es muss 0.1 sein

Ja

C ohne den Durschschnitt ist wie Omega ohne A und B

A und B beide haben 0.9

Ohne ihren Durschnitt wird das zu 0.7 da A Durschnitt B 0.2 hat

Nach Vorraussetzung hat C 0.3

Hab mir die Mengen etwas bildlich vorgestellt

So etwas hilft bei so einem Mengenschlacht immer wieder

Gerne, keine Garantie dass ich sofort antworte aber

[obtuse pebble]

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[spice isle]

Hey there I feel like this is supposed to be simple or something so I feel stupid for asking but I genuinely don't know what to do here, can someone help?

[glossy basalt]

which part?

[spice isle]

Do I have to ask a question like, seperarely each time? If so just uneerstanding a) would help

[glossy basalt]

nah, i mean like I'm not sure if you have started anything

cuz sometimes when i help people from the start and after typing for 5 mins, they say they only need help in part (e) or something

[spice isle]

Oh ok, yeah If possible I'd like someone to help me with both, I've forgotten how to do a lot and need help

[glossy basalt]

anyways for part (a), you just have to construct similar triangles

[spice isle]

I didn't realise you could do that with 3D shapes, thank you I'll try to work with this

[glossy basalt]

I'll be waiting, do ping if i don't reply in time

[spice isle]

I won't lie I still have no idea how to find the radius with this, I think it might be just me forgetting basics, thanks for the help I'll just go back and revise some stuff and try again

[glossy basalt]

i can draw something to help

[spice isle]

If that wouldn't be too much trouble then I would still appreciate that a lot

[spice isle]

Oh, I forgot that rule was there, thanks, I think I know how to do this now plus I know to revise that too

[glossy basalt]

good luck! do tell if you need more help

[obtuse pebble]

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[weak meadow]

Hello

[weak meadow]

How to answer please

I can't remember

what do you call this?

[sage dagger]

call what?

an integral?

[weak meadow]

nevermind that, sorry

I forgot how to answer it

[high lily]

since the degree of the numerator is greater than the denominator , start with long division

[weak meadow]

2x + 4/(x-3)

am I right?

[high lily]

yes

[weak meadow]

I did synthetic division

[high lily]

and then integrate that

[weak meadow]

How to do it?

[high lily]

basic integration rules, like power, log

[weak meadow]

I don't remember it

How to do 4/(x-3)?

It's 2x²/2 right?

The first term

[high lily]

yes

[weak meadow]

How about the second one

[high lily]

do you know how to integrate 1/x

[weak meadow]

what?

oh wait

ln |x-3| ?

[high lily]

yes

[weak meadow]

That's correct?

[high lily]

yes, don't forget the +C

[weak meadow]

So, 2x²/2 + ln |x-3| + C?

[fallow sky]

Hello can anyone help me? Answer a question my friend gave me?

The question in there

[weak meadow]

Wait, what? I'm not finished yet?

How did you

[fallow sky]

Wdym?

[high lily]

!occupied

[obtuse pebble]

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).

[weak meadow]

Please check

[high lily]

2x^2/2 can be simplfied

[weak meadow]

Ohhh

Rightt

x² + ln |x-3| + C?

Correct, yeah?

[high lily]

yes

[weak meadow]

Okay, thanks

How do I close this again?

.close

[obtuse pebble]

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[weak meadow]

Can I ask one more?

.reopen

[obtuse pebble]

✅

[weak meadow]

Is it the same?

I divide or something

[high lily]

same approach

[weak meadow]

2x - 1/(x+1)?

x² - ln |x+1| + C?

[high lily]

your division is wrong

[river falcon]

it should be 2x - 2 + 2/x+1

[weak meadow]

Ohh rightt

so

x² - 2x + ln |x+1| + C?

[river falcon]

yes

[weak meadow]

Okay

Thank you so much everyone

.close

[obtuse pebble]

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[river falcon]

wait

[weak meadow]

What?

[river falcon]

2ln|x+1|

[weak meadow]

Oh you do that

Okay

Thank you

[obtuse pebble]

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[timid silo]

does z^3 = 1 imply z^{3k} = 1?

[timid silo]

k belongs N

ofc

[upbeat nymph]

Idk if I'm right but to make z^3 = 1 => z = 1 and 1 to the power of any N is always 1

[timid silo]

that's an real solution z = 1

but remember it has 2 more complex solutions

and i dont know if those are still 1 if raised to ^6

lets see

ok yeah they are still one

thankss @upbeat nymph

.close

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[iron juniper]

$$\int \frac{x^2}{(xcosx-sinx)^2} dx$$

[warm shale]

RadMeerkat62445

[iron juniper]

Can't figure this out

Taking $t = x cos(x) - sin(x)$ and hence $dt=-xsin(x)dx$, I took the original integrand as $\int (-xcosec(x))(\frac{-xsin(x)dx}{(xcos(x)-sin(x))^2})$

[warm shale]

RadMeerkat62445

[iron juniper]

I did integration by parts, taking $u=-xcosec(x)$ and $dv=\frac{-xsin(x)dx}{(xcos(x)-sin(x))^2}$

[warm shale]

RadMeerkat62445

[iron juniper]

Hence $du=-cosec(x)-x(-cosec(x)cot(x)) dx = cosec(x)(xcot(x)-1) dx$

[warm shale]

RadMeerkat62445

[iron juniper]

ok no i found my error

.close

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[dim cedar]

Yo guys how do I simplify this I’m trying to find its domain

[river falcon]

you dont simplify a function before finding its domain

[dim cedar]

So how would I find its domain from that expression

[river falcon]

x is under square root

and whatever is inside square root cant be less than zero

so 7x -7 >= 0

[dim cedar]

So wait I only focus on 7x-7?

[river falcon]

no

iwas typing further

[dim cedar]

Ok

@river falcon yo

My friend where did you go

Can someone else come

[river falcon]

back

sorry

so i was saying

just start from outside

so 7(sqrt(7x-7))-7 >= 0

now solve this

@dim cedar

[obtuse pebble]

@dim cedar Has your question been resolved?

[dim cedar]

Yea I got it

Domain is infinity and minus infinity anyways

Can you help me out with finding this domain

Not totally sure I got it right

The first line is the equation

[obtuse pebble]

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[warm shale]

l'agit

[alpine bison]

https://tenor.com/view/roll-dice-roll-the-dice-gif-7337987

[bronze mica]

it seems reasonable

[obtuse pebble]

@pliant adder Has your question been resolved?

[bronze mica]

share it

[warm shale]

l'agit

[obtuse pebble]

@pliant adder Has your question been resolved?

[obtuse pebble]

@pliant adder Has your question been resolved?

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[pastel berry]

is anyone here good with accounting? it has math and i guess im doing it wrog

[lunar rover]

just ask the question

[pastel berry]

why is my 2nd number wrong. i thouoght the credits always have to match/balance the debits (already did E, and that was correct. even though the debit still didn't match the credit for that part either?)

[obtuse pebble]

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[obtuse pebble]

@pastel berry Has your question been resolved?

[pastel berry]

.close

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[golden sphinx]

idk how to find the area of the shaded region

[latent walrus]

youll have to split it into two integrals

[golden sphinx]

how would i go aheaded and do that

[latent walrus]

what does integrating a function between two bounds give you?

[golden sphinx]

i got 40/3 but it says its wrong

[tardy epoch]

!show

[obtuse pebble]

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

[golden sphinx]

Base: 2 - (-2) = 4
Height: 4 - 0 = 4
Area = (1/2) * base * height = (1/2) * 4 * 4 = 8 square units
Triangle 2:

Base: 2 - 0 = 2
Height: 6 - 4 = 2
Area = (1/2) * base * height = (1/2) * 2 * 2 = 2 square units
Parabolic Region:

To find the area under the parabola y = x^2 from x = -2 to x = 2, we can use the definite integral: ∫(-2 to 2) x^2 dx = (1/3) * x^3 | from -2 to 2 = (1/3) * (2^3 - (-2)^3) = (1/3) * (8 + 8) = 16/3 square units

then add all units to get 40/3

also

just did

∫(-4 to -4/3)[(8 - 2x) - x^2] dx + ∫(-4/3 to 2)[((2/3)x + 16/3) - x^2] dx

Evaluating the First Integral:
∫(-4 to -4/3)[(8 - 2x) - x^2] dx
= [8x - x^2 - (1/3)x^3] from -4 to -4/3
= [(8(-4/3) - (-4/3)^2 - (1/3)(-4/3)^3) - (8(-4) - (-4)^2 - (1/3)(-4)^3)]
= 128/27

Evaluating the Second Integral:
∫(-4/3 to 2)[((2/3)x + 16/3) - x^2] dx
= [(1/3)x^2 + (16/3)x - (1/3)x^3] from -4/3 to 2
= [(1/3)(2)^2 + (16/3)(2) - (1/3)(2)^3) - (1/3)(-4/3)^2 + (16/3)(-4/3) - (1/3)(-4/3)^3)]
= 128/27

Adding the Areas:
Total area = 128/27 + 128/27
= 256/27

but 256/27 is not right

[surreal forge]

[golden sphinx]

😭

[obtuse pebble]

@golden sphinx Has your question been resolved?

[obtuse pebble]

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[restive coral]

hey quick question: I am following calculus for computer science and these are the topics in the syllabus. Until what videos do I need to watch in the professor leonard playlist? https://www.youtube.com/watch?v=fYyARMqiaag&list=PLF797E961509B4EB5&index=1

[restive coral]

I want to use leonard as supplementary material

[obtuse pebble]

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[unique shuttle]

can anyone confirm if this is correct

ignore the yr2 bit at the end

[latent walrus]

should be (-3x)

[unique shuttle]

oh shit yeah

[unique shuttle]

thank you man

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[golden tartan]

i need help on this question, finding the final answer

[alpine bison]

what about the first answer?

[golden tartan]

noo, i just wanna know the answer to this question

[alpine bison]

maybe you need to read a little bit the rules of this discord server ... firstly we need to know some of your attempts

[obtuse pebble]

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[ancient vessel]

Need some help with these problems right here, if anyone is able to link help videos in DMs that would help me out a lot too

[obtuse pebble]

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@ancient vessel Has your question been resolved?

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[wet ridge]

Looking for some help with this question, I'm really not sure what it means when it says the first positive value of x or what my interval for bisection is supposed to be

[wet ridge]

I graphed part a using demos which I guess is supposed to help in telling me what I should be doing but I'm really lost

[graceful rapids]

more numerical analysis?

[wet ridge]

yep, this is the next section of the homework

[graceful rapids]

say there's a solution in (0, 10) or whatever

is there a solution in (0, 5)? if not, the solution must be in (5, 10)

if you're familiar this is just binary search but looking for a real number

[obtuse pebble]

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[proud fiber]

Here the functions input are not given in terms of x

so how do I do this?

[broken gorge]

I dont know exactly but I would try substituting A with 90 and B with -(A+B)

[river falcon]

just start differentiating with respect to x

[proud fiber]

hehe, but then you wouldn't have to differentiate at all

[proud fiber]

You would directly get the formula I think

[broken gorge]

true that

[proud fiber]

so I am assuming that A+B is a function of x?

[river falcon]

d(sin(a+b))/dx would be cos(a+b).d(a+b)/dx

chain rule

[proud fiber]

I would have to assume A and B as functionsS of x

[river falcon]

!occupied

[obtuse pebble]

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).

[thorny iris]

oh

my bad

[proud fiber]

but what would I put for d(a+b)/dx

[river falcon]

nothing

it will cancel out later

differentiate this whole relation with respect to x

[proud fiber]

ok I'll try

[tardy epoch]

<@&268886789983436800> spam

[proud fiber]

Got it

thank you

.close

[obtuse pebble]

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[gloomy vector]

find the largest natural number n such that $2^n$ divides $(3+\sqrt{11})^{2021}$

[warm shale]

Skill_Issue

[gloomy vector]

status 1

[pallid basin]

My guess is 0 because that number is irrational

Even if it was rational, 3 and 11 are prime and do not have 2 has a factor

[haughty stream]

It's false

[haughty stream]

We will still count the rational part

[haughty stream]

Do you know about lifting the exponent lemma?

[gloomy vector]

no

[haughty stream]

It's the most essential thing to do this...

[timid silo]

use binomial expansion

[gloomy vector]

ans choices are in the thousands lmao

[haughty stream]

It would be so hard

[timid silo]

not really

[haughty stream]

And I mean, even if we use binomial expansion, we can't say that their sum might be divisible by some power of 2...

[haughty stream]

If you have any solution by binomial expansion, please tell him!

Otherwise, @gloomy vector I would recommend to go through the wikipedia of Lifting the exponent lemma. It's a really important topic

Oh wait, but I think that LTE is just for integers...

We would have to make some changes too...

[obtuse pebble]

@gloomy vector Has your question been resolved?

[obtuse pebble]

@gloomy vector Has your question been resolved?

[timid silo]

is the answer1010

[gloomy vector]

its 2021 apparently

[timid silo]

hmmmm idk then,im getting 1010

[obtuse pebble]

@gloomy vector Has your question been resolved?

[obtuse pebble]

@gloomy vector Has your question been resolved?

[obtuse pebble]

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[languid sphinx]

does a unit vector times an area scalar equate to the cross product of the vectors that form said area?

[languid sphinx]

this question comes from this image, just wondering if conceptually my thinking is correct or not

the unit vector is perpendicular to the plane, just like if you were to take the cross product.

[past sand]

If you use the correct order, yes

[languid sphinx]

right, otherwise the area vector would have the opposite direction

[past sand]

The length of the cross product corresponds to the area of the parallelogram formed by the two input vectors

[languid sphinx]

thats still so cool to me lol, alr thx

.close

[obtuse pebble]

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[obtuse pebble]

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[ancient saddle]

60,6*10 in the power of negative three standart expression

[ancient saddle]

can anyone help me with my homework

i have to write number in standart expression

[slow prawn]

do you mean three significant number?

maybe we have differnet ways of saying it

[ancient saddle]

i translated it from lithuanian it could be wrong

[slow prawn]

if 3 significant number then

[ancient saddle]

yeah okay

can you help

[slow prawn]

just move the

comma

3 times in front

if its 10^-3

aka 60.6 becomes 0.0606

you move the comma

3 times to left

[ancient saddle]

and if negative

to the right

[slow prawn]

well it is negative

positive youd move to the right

[ancient saddle]

oh wait the other way around

if its positive

[slow prawn]

if it was 10^3

then if would be 60600

[ancient saddle]

so if 17*10^3

then 1.7?

[slow prawn]

but three significant figure it would be 60.6 x 10^3

[slow prawn]

uh

not quite

you move 3 to left

17 = 17.000

if you move three comma to right

it would be 17000,0000

right

so 17000

[ancient saddle]

okay so lets say

15 * 10^-3

i go to the left?

[slow prawn]

yes because its negative

now what do you think it is

[ancient saddle]

so its 0.015

[slow prawn]

yup

[ancient saddle]

okay

one more to make sure

501 * 10^-5

is 0.0501?

[slow prawn]

almost

remmeber 10^-5

what you did there is 10^-4

[ancient saddle]

ohh so there has to be 5 numbers after the comma then?

to the right

[slow prawn]

nope negative

so move left

[ancient saddle]

0.00501?

[slow prawn]

yes

[ancient saddle]

okay

yo thanks alot i would have stayed up late

ac theres another one

0.03 * 10^6

[slow prawn]

try to do it

ill tell you if its good or nto

remmeber positive so moving right

[ancient saddle]

postive so to the right

so instead of decimals it becomes a whole number right?

[slow prawn]

yup

[ancient saddle]

is it 3000?

[slow prawn]

yyp

you got it

[ancient saddle]

omg lets go

finally

thanks alot man

[slow prawn]

gotchu 👍

[obtuse pebble]

@ancient saddle Has your question been resolved?

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[sullen spruce]

is (√15)/(4) the same as √(15)/(4)

[gaunt helm]

does the root affect the whole fraction in the second

[sullen spruce]

yh

[worn yoke]

,, \sqrt{\frac{15}{4}} \qeq \frac{\sqrt{15}}{4}

[warm shale]

cloud

[timid silo]

no

[sullen spruce]

ok

[timid silo]

this is not true

[sullen spruce]

.close

[obtuse pebble]

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[obtuse pebble]

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[frail depot]

Can someone hint me where to start?

[frail depot]

Or what I should read up on?

So, I get that the deviation from the true mean and the sample mean should be at least 0.5 with probability 100%

And then I'm lost 😄

[sullen spruce]

bro wtf is this level of math

[frail depot]

Machine learning A in a master course at my university and I’m still at my bachelor and do I regret it, yea

Mostly because I have an injury so it’s really hard to study the appropriate time

[obtuse pebble]

@frail depot Has your question been resolved?

[obtuse pebble]

@frail depot Has your question been resolved?

[shadow dagger]

i mean it might just be me being dumb rn

but can't you literally just set X1 = X2 = X3 = ... = Xn

and set X1 ~ Bernoulli(1/2)

[obtuse pebble]

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[foggy epoch]

What I have filled out are the correct answers, but it's confusing to me

[foggy epoch]

Why does the chance of catching a fish at the ith cast continue to decrease if he doesn't get a fish
AND YET, the probability of catching a fish at any given cast is independent and stays at 0.4

This just doesn't make sense to me and seems contradictory

<@&286206848099549185>

Quick correction
p(c_2) is 0.24

[marsh blade]

Do you mean that he isnt catching a fish at all up until the ith attempt?

Because if you are talking only about the event of him not catching a fish until the ith attempt then yes.
Because for that specific event to occur in all of the previous he mustnt catch a single fish

If he catches only one fish before, then this path isnt part of what you described earlier before

Im other words if you look at your task description, if he catches a fish earlier on, he will stop fishing at all and leave

So in order to catch a fish at the ith attempt he musnt catch a fish beforehand

[foggy epoch]

Makes sense to me n ow

thank u

.close

[obtuse pebble]

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[undone sentinel]

what did i do wrong

[tranquil geyser]

You're meant to use a perfect cube number close to 9

For x

[undone sentinel]

what? but i still plug in 9 itself. i only need to do that to find dx

distance between 8 and 9 is 1

not that one nvm

[tranquil geyser]

OK

[undone sentinel]

so my x is 8

what abt here

in this example problem

they do 624-625

ohhh

but they still plug in 625

not 624

i plug in the number close to 9

is that it?

so i would have this

[tranquil geyser]

Yes

[undone sentinel]

so this would be correct

[tranquil geyser]

Let me check

[undone sentinel]

well?

[tranquil geyser]

Yes I am checking

;-;

[undone sentinel]

sorry i didnt mean to be pushy 😭

i just used the examples in my textbook to guide me

[tranquil geyser]

NP

[tranquil geyser]

This is correct

[undone sentinel]

okay

ty

one more question btw

[tranquil geyser]

Ok

[undone sentinel]

how do i know if delta y is greater or less than dy

im not given values so i cant compare it

how do i infer it

[tranquil geyser]

It says it is increasing

When a function is increasing dy/dx>0

[undone sentinel]

well i know its positive

deltay>0

i know im right on that

but is the rate of change of y smaller or bigger than the instantaneous rate of change of dy

[tranquil geyser]

Ok

The only thing I could think of is dy/dx =lim as Delta x -> 0 of delta y/delta x

I'm not too sure on this one sadly 😞

[undone sentinel]

someone else told me to use the secant line but idk how when im not given values

[tranquil geyser]

Wait

Yes the secant line is a nice idea

Look at it

[undone sentinel]

yea like how its used to approximate the derivative

well deltaY is bigger than dy

its a bigger interval

is that what u mean

[tranquil geyser]

YUP

[undone sentinel]

okay it was right

but

why?

like it makes sense from just looking at it

but does that mean this will always apply

[tranquil geyser]

As far it's increasing

And concave up

That's concave up I think.

[undone sentinel]

it is

concave up is given

[tranquil geyser]

So Delta y should be greater than dy

[undone sentinel]

but why does it matter if it was concave down*

like if it was decreasing and concave down why would it change

i cant think of an example

[tranquil geyser]

Gimme a sec

, rotate

[tranquil geyser]

[undone sentinel]

what are you trying to show

i see delta y is still bigger than dy

[tranquil geyser]

Yes it is

You asked for decreasing and concave down

[undone sentinel]

oh true

well

what would be a situation where dy is bigger than delta Y

[tranquil geyser]

Concave up and decreasing maybe.

I'll draw it

[undone sentinel]

or can i move forward with the idea that delta Y is always bigger than dy

im happy with accepting that lol

[tranquil geyser]

No

Delta y is smaller than dy for Concave up and decreasing maybe and concave down and increasing

[undone sentinel]

oh

thanks for all the help

[tranquil geyser]

No problem

[undone sentinel]

.close

[obtuse pebble]

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[tranquil geyser]

https://youtu.be/2ooWs_8hzxQ?si=0j4I0eBkqNmzBsOL

[obtuse pebble]

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[sacred grail]

Can anyone verify if this is correct?

[obtuse pebble]

@sacred grail Has your question been resolved?

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[fathom kindle]

whats the domain of this

im confused because how is it XER

if theres a vertical asymptote

i think im going crazy

[uneven flame]

There's not a vertical asymptote?

[fathom kindle]

wait

rlly????

WHAT

[sage dagger]

you mean horizontal?

[fathom kindle]

oh

wait

hold up