help-17

[raven owl]

right here your second step is good, simplify the -4+1

[sterile pecan]

yeah -3

[raven owl]

well ig you already did that

yeah

so can you tell me what the vertex is

It is in the proper form right now for you to extract that information

[sterile pecan]

OH YEAH IM SILLY (5, -3)

[raven owl]

yeah nice

You are basically just gonna do the same thing as in this image to find the x intercepts

pretend the problem started as 3(x-5)^2-3=0

[sterile pecan]

but that's where i'm stuck

because

idk why we had squared root of 4

i think it's because we were dividing by a negative number

but don't u have to add i to it too

oh wait nvm that's when your square rooting a negative number

[raven owl]

During step 2, they divided both sides by -2

yeah

[sterile pecan]

so i divide both by negative 3 and i just solve it normally?

wait

no

positive 3

wrong 3

[raven owl]

you first want to add 3 to both sides

Then you will divide by 3

[sterile pecan]

wait but then it would end up with a 3/0

[raven owl]

In this image the exact steps made were

subtract 8 from both sides

divide -2 from both sides

sqrt both sides

add 1 to both sides

[raven owl]

You will go from 3(x-5)^2-3=0 to 3(x-5)^2=3 first

[sterile pecan]

OH

wait

[raven owl]

does this make sense

[sterile pecan]

i get it now

[raven owl]

gg

[sterile pecan]

in the notes i thought the sqrt just came outta no where from doing -8/-2

but it's dividing -8 by -2 then square rooting 4

[raven owl]

Yeah when its written like that it gets confusing

[sterile pecan]

and sqrt the other side

k thanks

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[gilded summit]

I need help with Algebra 1 chapter 2 standard form

[hybrid wind]

which is..?

[gilded summit]

Find the equation of a line in standard form that goes through the point (5, 2) and is perpendicular to the line 3x - 2y = 7.

[soft walrus]

may I first suggest writing the equation 3x-2y=7 in slope-intercept form?

[gilded summit]

May I have some help on that?

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[valid socket]

what is meant by the last line about f(c) being the g weighted average

[flat whale]

That's just the definition

Just something you memorize

If you want you can solve for f(c)

Then try out different functions for g(x) like 1, x

[valid socket]

is there any intuition on understanding the second mean value theorem because right now, it just seems like an equation that I should memorise

[vocal sleet]

@valid socket Has your question been resolved?

[valid socket]

@flat whale if you don't mind having a look

[flat whale]

Just memorize

[valid socket]

thanks

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[terse grotto]

For question 3 I had equated 900+30x = 600^1.12x but I don't know how to solve

[vocal sleet]

@terse grotto Has your question been resolved?

[nova delta]

I’m not a helper but where are you getting these numbers from? You mean question 2 or 3?

[terse grotto]

Question 3

Oops wrong question

For 3 I got 456 +7x =2^x

But I couldn't resolve that either

[late ingot]

its slightly off still but i don't think this is solvable by hand or any method i can think of.

should be 449 + 7x = 2^x to match the totals for the first game.

where x = 1 for that initial game

i suppose you can use trial and error or a graphing utility

[nova delta]

I’m tired but I’ll try my best.

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@terse grotto Has your question been resolved?

[mild flower]

just make a table

[terse grotto]

Alright tnak you

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[potent delta]

Hey, I was trying to formulate this proof and was setting it up but I was struggling of like

[potent delta]

where do i go from this

Heres a better pic

[mild flower]

what does linearly dependent mean?

[potent delta]

It means this

[mild flower]

mmk

can you use this to show that u and v are multiples of each other?

[potent delta]

this is what I coocked up

this is a bit better and focused

[mild flower]

this is fine

when you have a structure like:

• suppose for contradiction not P
• stuff
• P, that's a contradiction
• therefore P

you can often rephrase that to just

• stuff
• P

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[rancid ibex]

can someone help me get started on this

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[unique path]

How does one do this:
I am fairly sure that its the bottom one but could someone show me some written working as I am not sure what I am doing is correct

[livid tapir]

Well it's multiple choice so u can do some tricks tbh

like 3 is possible if and only if 2 and 1 are both unbiased

So if you figure out one of them is biased then that's good enough to get u the answer based on the possible choices

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@unique path Has your question been resolved?

[livid tapir]

Honestly 2 is probably the hardest, 1 is possible to do inside your head if you remember E[X_i^2] - E[X_i]^2 = sigma^2

And use linearity properties

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[sudden compass]

$\int_{0}^{1} \frac{a}{{(xy - a)}^2} + \frac{1}{xy - a} dx$

[twin meteor]

ItzKraken

[sudden compass]

I have to evaluate this integral

this is what wolfarm alpha shows

I get $\frac{1}{a - y} - \frac{\ln{|y-a|} - \ln{|a|}}{y}$

[twin meteor]

ItzKraken

[sudden compass]

btw I am assuming log is equivalent to ln

[lapis marten]

Thats the same

lnx - lny = ln(x/y)

[sudden compass]

oh right 🤦‍♂️

[lapis marten]

Except the absolute values, but I think you are correct

[sudden compass]

kk

lemme check my work again

hmm seems correct

the absence of the absolute value might be related to this

what do the V and upside down V mean? something like and and or?

[hard atlas]

yes

V means or, the other one means and

[sudden compass]

I see

yeah makes sense why no abs value

okay lastly

$\int_{0}^{1} \frac{\ln{|1-\frac{y}{a}|}}{y} dy$

[twin meteor]

ItzKraken

[sudden compass]

I tried IBP, it made it tougher 💀

and no form of substitution comes to mind

[hard atlas]

just sub the numerator

[sudden compass]

1-(y/a) = u?

[hard atlas]

no the entire thing

[sudden compass]

wdym

[hard atlas]

with the ln

[sudden compass]

oh so ln((y-a)/a) = u?

[hard atlas]

yes

[sudden compass]

hmm k lemme try

dy = -(a-y) * du right

[hard atlas]

wait this isnt as nice as I thought

sry

[sudden compass]

no its fine

hmm if a=1 tho

then it does become nice, i think?

[hard atlas]

oh whoops

non elementary

[sudden compass]

?

[hard atlas]

,w integrate ln(1-x)/x

[twin meteor]

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[sudden compass]

whats Li_2

[hard atlas]

so yeah good luck

poly logarithmic function

[sudden compass]

wait so u mean dilogarithm?

[hard atlas]

probably

whenever WA uses these kinds of functions I just think to myself "nope"

[sudden compass]

well that means no sols?

btw the definite integral for this apparently is pi^2/6 (from 0 to 1)

how did wolfram calculate that

[hard atlas]

wolfram knows a lot of identities we dont

it doesnt mean no sols. it means that you cant write the antiderivative using any of the typical functions you know

[sudden compass]

I see

[hard atlas]

so you have to do some other trick. maybe some complex analysis trick. or series expansion

or this

or that

[sudden compass]

should I ping helpers? just to check if someone knows how to evaluate this one

[sudden compass]

[hard atlas]

sure. I just know that I cannot help further

[sudden compass]

Okay, thats fine, thanks for helping!

to evaluate: $\int_{0}^{1} \frac{\ln{|1-\frac{y}{a}|}}{y} dy$

[twin meteor]

ItzKraken

[sudden compass]

<@&286206848099549185>

pls ping when replying btw

[vocal sleet]

@sudden compass Has your question been resolved?

[sudden compass]

<@&286206848099549185>

[vocal sleet]

@sudden compass Has your question been resolved?

[sleek island]

If "a" is a constant, yes?

[sudden compass]

yes

[simple mason]

So firstly in my opinion
For this integration to be valid, either one of those should be true
a > 1
a < 0
Let's see why I say that
Now imagine that a is a value between 0 and 1 (the range that we want to calculate its area at) then when x = a the value in the ln will be zero which is not identified
So because of that it should either be greater than 1 or smaller than 0
For the first case all the values in the ln will be positive for the values for 0 to 1
For the second case also or the value inside the ln will be positive for values from 0 to 1

Which means that there is no need for the absolute values for that ln

The second point that you might have realized is that this function is not identified at y = 0

Which makes taking its integration a little bit tricky but it is still possible

$$\int_0^1 \frac{\ln(1-\frac{y}{a})}{y}$$

[twin meteor]

Sherif Player

[simple mason]

,wolf $$\int_0^1 \frac{\ln(1-\frac{y}{2})}{y}$$

[twin meteor]

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[simple mason]

It seams that they done it using integration by parts

$$\int_0^1 [y^-1 \ln(1-\frac{y}{a})] dx =
\ln(y) \times \ln(1-\frac{y}{a}) + \frac{1}{a} \int_0^1 [\frac{\ln(y)}{1-\frac{y}{a}}]$$

[twin meteor]

Sherif Player

[sudden compass]

well..after this?

[cobalt crypt]

is there context to this problem?

[simple mason]

I think that the right integration can be done by partial fraction but I am not sure how

[cobalt crypt]

or are you just computing a random integral

[simple mason]

Yeah this

[sudden compass]

right integration?

[simple mason]

Yeah there is two integrations in the equation
I am talking about the one on the right

[cobalt crypt]

you realise that wolfram alpha has already shown that the answer is expressed in terms of Li_2 right?

[sudden compass]

I know, how do I find Li_2 tho?

for some a

[cobalt crypt]

Li_2 is defined to be the integral you're calculating

[simple mason]

Yeah but this

[cobalt crypt]

because there exists no expression for it in terms of "simple" functions

[sudden compass]

exactly for a=2 how did wolfram get that

[cobalt crypt]

it is a known constant

[sudden compass]

yes ik its non-elementary

[sudden compass]

bruh 😐

[simple mason]

,wolf $$\int_0^1 \frac{\ln(1-\frac{y}{3})}{y}$$

[twin meteor]

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[simple mason]

LOL

[cobalt crypt]

https://mathworld.wolfram.com/Polylogarithm.html

check equation (16) here

wolfram is using that the value of Li_2(1/2) is known

you will certainly not get an answer as nice as that for general a

[sudden compass]

well is there a nice answer for specific a?

[cobalt crypt]

well for specifically a = 2, you've already seen it

for other values, who knows

go study what values of Li_2 are known

[sudden compass]

okay ig...

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[true lagoon]

What is in the subscript character of C?

is it i?

[outer warren]

1

[elder scaffold]

no its 1

its 5 combination 1

[outer warren]

poorly written notation

[true lagoon]

Oh

[elder scaffold]

true

[true lagoon]

So all of those are 1s?

[elder scaffold]

yeah

[true lagoon]

Ok sick

thanks

[outer warren]

$\frac{\combo{5}{1}}{\combo{26}{1}}$

[twin meteor]

ℝαμΩℕωⅤ

[elder scaffold]

but its kinda redundant as x combination 1 is always x

[true lagoon]

How do you read this?

[elder scaffold]

5 combination 1 over 26 combination 1?

[true lagoon]

I see

[true lagoon]

Ok, thanks so much

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[solid sage]

There are 5 students and 4 teachers. A group of 5 members are being selected at random. What is the probability for there to be at least 3 students in the group?

[solid sage]

i thought it was $\frac{^5 P_4+^4 P_4}{^9 P_9}$

[twin meteor]

annyeong

[solid sage]

but 1/21 doesn't seem right

[elder scaffold]

im not sure about the rest, but here you'd definitely use combinatorics

as order doesnt matter here

[solid sage]

oh true

[paper depot]

"combinatorics" is the entire field, you're looking for the word "combinations"

[elder scaffold]

oh true

[solid sage]

weird its the same if you replace P with C

[glad python]

By the way, complementary is easier here

[solid sage]

yeah i figured that but not sure what the complement is exactly

[glad python]

1-at most 2 students

[solid sage]

umm

[elder scaffold]

i'd say complementary is harder?

[glad python]

You only have two cases, as opposed to three

[elder scaffold]

as for probability we just have to find favourable outcomes over total outcomes

[glad python]

Yes thats the definition of probably

Probability*

[solid sage]

would the total be 9C9

wait no thats just 1

[glad python]

Total is 9c5

[solid sage]

total for all possible cominations?

[glad python]

Because we are picking 5 distinct people from 9 distinct people

[solid sage]

oh true

[elder scaffold]

i see what u mean kappa there will be a few cases here

u could have 3 students 2 teachers, 4 students 1 teacher and 5 students 0 teachers

[glad python]

If you don’t want to do complementary, we have to find the number of groups with 3 students, 4 students, and 5 students

[solid sage]

wouldn't it be 1, 2 and 3 studnets

[elder scaffold]

atleast 3 students

3+

[glad python]

The question asks for at least

3

[solid sage]

oh yeah

misread

i am so tired right now sorry

[glad python]

Now try it

[elder scaffold]

yeah

[solid sage]

ok

so students iwth 3 people

would be 5C3

[glad python]

Yep

[elder scaffold]

yeah

[solid sage]

so 5C3+5C4+5C5/9C5

[elder scaffold]

but u have to account for the teachers aswell

[solid sage]

wait what

am i doing

oh

how would i do that

[elder scaffold]

well u have 3 students AND 2 teachers

[glad python]

How many teachers do we need with 3 students if the group is of size 5?

[solid sage]

oh so multiply by 2C2

[elder scaffold]

almost

[solid sage]

wait no

4C2

[elder scaffold]

yuh

[solid sage]

wait i multiply?

[elder scaffold]

yes, when u have an AND case,its always multiplied where as an OR case gives a +

[solid sage]

so i multiply the whole fraction above with 4C2

[elder scaffold]

not exactly

[solid sage]

oh 5C3 x 4C2

[elder scaffold]

break the fraction into sections, so ignore the denominator for now

yeah

[solid sage]

and then subsequencty 5C4 x 4C1 etc

[elder scaffold]

yes

[solid sage]

Ah i think im starting to understand

[elder scaffold]

yeah combinatorics was a struggle for me when I first learnt it

[solid sage]

hmm yeah alot of practice im guessing

[elder scaffold]

nah i just winged the prelims

[solid sage]

lmaoooo ok

if there are 8 numbers to choose from, how many 3 letter codes can be made where order is important and letters can be repeated

tbh dont really know what order being important changes

oh wait i think i knwo what that means

its like 1 3 1 is not same as 1 1 3

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[wraith bear]

Something must be incorrect here right?

Ignore the numbers

[elder scaffold]

personally i would have it as (s-ut)/t^2 then times it by 2

[wraith bear]

Yeah thats what i did too

,rotate

[wraith bear]

Here is the formula by the book

[elder scaffold]

yeah 2s-ut is not the same as 2s-2ut

[wraith bear]

So i guess the solution is incorrect

Yeah

Bruh its such a pain when that happens

Thanks for the help

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[rugged orchid]

How do i describe the set of sequences that are longer than n

[rugged orchid]

im not sure how to do this without specifying how long my sequence is

maybe ${a_n}_{n\geq 1}$?

[twin meteor]

Frosst

[lapis marten]

{R^k | k > n}?

[rugged orchid]

but it's possibly infinite

[lapis marten]

Something similar might be here: https://math.stackexchange.com/questions/707705/what-is-the-cardinality-of-the-set-of-all-infinite-sequences

[blissful sentinel]

S = {s \in {0,1}* | |s| > 2}

[rugged orchid]

is that a thing, *

[blissful sentinel]

Oh possibly infinite

[rugged orchid]

also it'd be >=

[blissful sentinel]

You wanna allow infinite strings?

[rugged orchid]

well shouldn't it?

[blissful sentinel]

Kleene star

set of all finite strings

[blissful sentinel]

Well that depends on your situation

[mild flower]

there's a big difference between all finite strings of arbitrary length and infinite strings

[rugged orchid]

well i think it doesn't matter for this question

[blissful sentinel]

Most CS applications wouldn't allow infinite strings, I imagine

[rugged orchid]

but that definitely is my assignment question

[blissful sentinel]

Probably finite

[rugged orchid]

so directly helping that doesn't seem very ethical here

[blissful sentinel]

{0,1}* is the set of all bit strings

[mild flower]

you're right that that question would work in principle with infinite strings, but you can stick to finite unbounded

[rugged orchid]

well it says length 2 or more which to me says ≥2 and possibly infinite

an infinite sequence of 1's and 0's is most definitely longer than 2

[cyan talon]

It wouldn't really change much for this question tbh

[rugged orchid]

no but i was just curious about the notation

[cyan talon]

$$A^* = \bigcup_{n\geq 0} A^n$$ if we wanna get extra formal

[twin meteor]

aPlatypus

[rugged orchid]

what kind of union is this

what are the A's

ah

no i get it

[cyan talon]

A is just some set

[rugged orchid]

yeah no i get it

[mild flower]

infinite strings are 1) weird and 2) fundamentally different from finite strings, so your definition would probably have two pieces unioned together

[rugged orchid]

oh..

so like A* U A^inf

[mild flower]

yeah but A^inf might be something you need to define

[rugged orchid]

mhmm

why do you say infinite strings rather than sequences

well i guess what's the difference between a string and a sequence

since both can be finite and infinite

[mild flower]

they're the same thing really

[rugged orchid]

oh ok

[mild flower]

bc the question said strings lol

which like

[rugged orchid]

yeah makes sense

i mean i just dont really know what a "string" is

im assuming it means "sequence"

[mild flower]

id probably have defined string as a finite sequence tbh

[rugged orchid]

from here would i go $S = {s \in {0,1}^*| |s| \geq 2}$
$R = {(s_1, s_2)\in S^2| (s_1)_1 = (s_2)_1 \text{ and } (s_1)_2 = (s_2)_2}$

[twin meteor]

Frosst

[rugged orchid]

does that forst bit work really

the (s_1, s_2) \in S^2

ok thanks guys

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[vast shale]

.reopen

Hi

[vast shale]

two years ago,my age was four times the age of my son.eight years ago ,my age was ten times the age of my son .find the age of my son now​

Only one variable 'x' is used to solve this question.

I have not been able to solve using on variable

<@&286206848099549185>

<@&286206848099549185>

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[covert orbit]

so whats your unknown here

[vast shale]

x

[covert orbit]

as in

what do you want to put as your unknown x

[vast shale]

Father's current age

[covert orbit]

ok that would work

so what is your son's age two years ago

ok i think im skipping a few steps here

what is the father's age 2 years ago

[vast shale]

It is not mentioned in the question

[covert orbit]

no, but you can deduce

so now my age is x right? 2 years ago what would it be

[vast shale]

x-2

[covert orbit]

correct

so 2 years ago father's age is x-2, and also 2 years ago father has age that is 4 times that of son

so what is son's age 2 years ago

[vast shale]

(x-2)/4

[covert orbit]

great

now

rinse and repeat for 8 years ago

[covert orbit]

lets keep this equation in mind too

[vast shale]

Okay

(x-8)/10

[covert orbit]

ok

now, 2 years ago son's age is (x-2)/4 and 8 years ago son's age is (x-8)/10

how many years has passed between those 2 periods

[vast shale]

6 years

[covert orbit]

yeah

can you see the solution?

[vast shale]

(x-2)/4 = (x-8)/10

[covert orbit]

nope, remember they have a difference of 6 years

[vast shale]

I don't understand

[covert orbit]

(x-2)/4 - (x-8)/10 = 6

[vast shale]

I understand now.

[covert orbit]

so do you know how to solve for son's age?

[vast shale]

yes, we have to just substitute

father's age

Son's age = 11

[covert orbit]

yeah i didnt do the math there

[vast shale]

Fathers age = 38

[covert orbit]

lemme verify rq

how did you get 11?

ah yeah

my bad

11 is correct

[vast shale]

Thank you so much for your help!

[covert orbit]

np

[vast shale]

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[vast shale]

Im doing related rates and dont understand how to find the radius for this problem:
If the water level is decreasing at a rate of 3 in/min
when the depth of the water is 8 ft, determine the rate at
which water is leaking out of the cone.

[vast shale]

i get that r/h = r_0/h_0

but when i look online everything is saying r = 5/16(h) which makes no sense

2*8 = 16 so maybe thats where that is coming from

but where the heck is a radius of 5 coming from

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

even the books solution is a number

but it doesnt make sense

as theres literally not enough information

[elder forge]

Is there an illustration or something included with this problem?

[vast shale]

nope

[elder forge]

Can you post a picture of the complete problem

[vast shale]

yes

nvm

holy moly

sorry

.close

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[past shard]

can someone give me a little guidance on this question

[past shard]

i know the case of

2

and 3

but i cant seem to find a pattern to apply to the rest of them

ab = c, c is order 2 c must be a or b -> a or b = e contradiction

abc = d, d is order 2 d must be a, b, or c, wlog suppose c abc = c, ab = e, a = b-1, contradiction

abcd = e, e is order 2, e must be abc, wlog d, abcd = d, abc = e

and then im stuck, because ab = c-1, which igggg makes everything true idk? how do i turn ab = c-1 into a contradiction

<@&286206848099549185>

[vocal sleet]

@past shard Has your question been resolved?

[tepid plinth]

hmm

generalise...? it can have either zero or more than 1 elements of order 2. at zero we're done

[hard atlas]

do you know that you can decompose an abelian group into a product of cyclic groups?

[past shard]

uhhhh no i dont thinks so

supposedly there is an inductive solution?

and a neater one as well

[past shard]

yeaa, ive gotten the case where with 1 it must be the element of order 2, but it slips away from me when i try to go to 4 or more order 2 elements, im not really sure what to induct on

[vocal sleet]

@past shard Has your question been resolved?

[past shard]

<@&286206848099549185> anyone have any ideas haha , i completed the 1st and 3rd part of the problem, but im still stuck on the second,

[vocal sleet]

@past shard Has your question been resolved?

[vocal sleet]

@past shard Has your question been resolved?

[past shard]

.close

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[stone oasis]

OK I'm using Brilliant dot com to learn probability and stats. They have this pmf and ask for E(N)

[stone oasis]

But they forgot to plug in 2?

Or is there something I'm misunderstanding

I calculated the expected value to be 1/3

[flat whale]

send link

[stone oasis]

https://brilliant.org/courses/probability_ii/introduction-110/distributions/1/?from_llp=data-analysis

It's the free portion, you just need an account to see it

[flat whale]

oh i have to do the whole thing

nothx

[golden night]

they plugged in 0, 1, 2

[stone oasis]

Yeah, you don't have to

[flat whale]

they do plug in n=2 on line 4

[golden night]

but on line 3 its not fully shown

[flat whale]

the first n=0 term is 0

[stone oasis]

oh hmmm

Oh I see

[stone oasis]

They are just not showing it explicitly

The entire point of brilliant dot org was to teach it well for the normies

[flat whale]

they're leaving out 0+

[stone oasis]

Why even

[stone oasis]

I see it now

Thanks everyone!

[stone oasis]

Does desmos actually do calculations or did you type all of these

[golden night]

wym?

[stone oasis]

Also, while I'm here, can someone walk me through what happened here?

I see they expanded the (X-E(X))^2

But then the 2E[X]X turned into 2(E(X))^2

How

[golden night]

they distributed the E

[sly sierra]

you're taking the expected value of (-2E[x]x)

the -2E[x] part is constant, you can pull it out of the outer expected value

[stone oasis]

OH

Damn thats fire

I see it now

Thank you so much!

I should ask more questions here

This is really helpful

.close

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[vocal sleet]

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[spring sigil]

Need help with last question

[flat whale]

https://www.cuemath.com/calculus/asymptotes/

[spring sigil]

In my case the degree of numerator and denominator is the same so it would be just 1?

[flat whale]

it is yes

[spring sigil]

If it equals one, what does that tell me

[flat whale]

last sentence here

[spring sigil]

So it's the horizontal asymptote but there's also vertical asymptote too

I need two equations

[flat whale]

where does it say there's a vertical asymptote

[spring sigil]

It doesn't but in my question it says "equations' implying there's more than one

[flat whale]

there isn't more than 1

[spring sigil]

Not going to lie forgot about this but I also didn't really like graphing in math at all before

Thanks for the help, I'll save that screenshot you posted

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[icy bloom]

okay say uhhh

[icy bloom]

{ {}, {{1}}, {{2}}, {{3}}, {{1},{2}}, {{1},{3}}, {{2},{3}}, {{1},{2},{3}} } = some kinda set

would { {1}, {2}, {3} } be the union of that set?

[dark kiln]

yeah

in the sense where you can have a union of 1 set

[icy bloom]

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

how am i ever gonna prove dis then

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

i will never find a counter example t-t

[dark kiln]

show question

[icy bloom]

i can explain my thought process

take F = { {1}, {2}, {3} } (some family of sets)

then the power set of F would be

P(F) = { {}, {{1}}, {{2}}, {{3}}, {{1},{2}}, {{1},{3}}, {{2},{3}}, {{1},{2},{3}} }

and if i took the union, aka U P(F) i would get { {1}, {2}, {3} }

which friggin sucks cuz then theres no counter example

cuz every element in F is a subset of U P(F)

[dark kiln]

that's already a counter example i think

[icy bloom]

waht

[dark kiln]

A in { {1}, {2}, {3} }
so A is e.g. {2}

[icy bloom]

mhm!

[dark kiln]

and there's no 2

in U(P(F))

[icy bloom]

ummmm

if A is {2}

isnt {2} also in U(P(F))..

[dark kiln]

P(F) serves as a list of subsets

{ {}, {{1}}, {{2}}, {{3}}, {{1},{2}}, {{1},{3}}, {{2},{3}}, {{1},{2},{3}} }

[icy bloom]

mhm

[dark kiln]

there's no {2}

the closest is {{2}}

[icy bloom]

i understand that

but arent we looking at U(P(F)) and not P(F) tho?

[dark kiln]

U(P(F)) is F
so P( U(P(F))) is P(F)

we're searching P( U(P(F))) for {2}

[icy bloom]

dat is very interesting

[dark kiln]

again it's about counting brackets, they're just making sure you're paying attention

[icy bloom]

ah

T^T this seems very different than what the prof said on forums a few days ago

i will try to digest it on my own first

thanku tho

[dark kiln]

∈ would mean you're looking for {2} directly

and ⊆ is about one set being "smaller" than another

so {2} can be a subset of a set if that set contains a 2 directly

or it's an element of P() of that set, by definition

[icy bloom]

im gonna write what ur saying down

[dark kiln]

don't write smaller i guess

[icy bloom]

oke thanku

also frowny frog

i hope u frown less and smile more !!

[dark kiln]

it's pretty simple really, {1,2} is a subset of {1,2,3,4}
{1} is an even "smaller" subset
if there are more brackets
{{1},{2}} is a subset of {{1},{2},{3},{4}}
{{1}} is one smaller

[icy bloom]

i can see that :D

[dark kiln]

{1} is an element of {{1},{2},{3},{4}}
{1} ∈ {{1},{2},{3},{4}}
{1} is not ⊆ {{1},{2},{3},{4}}

and there's no quirks, or hidden rules, it's always about that

just there has to be the right amount of brackets

[icy bloom]

regarding {1} is not ⊆ {{1},{2},{3},{4}}

{{1}} ⊆ {{1},{2},{3},{4}}

right?

[dark kiln]

yeah

[icy bloom]

okie dokes

thanku

x2

!!

[dark kiln]

also {{1},{2},{3},{4}} ⊆ {{1},{2},{3},{4}}

the sign says "smaller or equal", that's what it also means

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@icy bloom Has your question been resolved?

[icy bloom]

okie

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[river spear]

im not sure where to start

[outer warren]

start with getting an exression for the area of the rectangle (in terms of x)

[river spear]

x*y=a ?

[river spear]

^

[outer warren]

(in terms of x)
and what are you told about y

[river spear]

so then x*sqrt(6-x)

[river spear]

^

[outer warren]

yes

now use calculus to optimise that

[river spear]

you mean take the derivative of it?

[outer warren]

yes, that's part of the optimisation process

[river spear]

how do i start it

[outer warren]

differentiating

applying the appropriate rules

[river spear]

how does multiplying get me that

wait nvm

i got 3(4-x)/2sqrt(6-x)

[river spear]

^

do i set it equal to zero and solve for x now

nvm i got it

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[river spear]

i dont know how to do this problem

[pale perch]

chain rule

[vocal sleet]

@river spear Has your question been resolved?

[river spear]

so just derivative of r^3 then solve for r in 3r^2=12

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@river spear Has your question been resolved?

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[vast shale]

ok so I found the average value

which is approx 22.9 cm

I am asked to find the lowest and highest bounds of the common interval

I am confused as to how to do it

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

anyone?

[vocal sleet]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[vast shale]

that's the problem I described it

3 measurments were given

I was asked to find the lower and higher bounds of the common interval of these 3 measurments

[thin vale]

don't describe it, show a picture or screenshot

[vast shale]

.close

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[warm crypt]

Hi, does anyone know the answer for this and how to solve it?

[vocal sleet]

@warm crypt Has your question been resolved?

[tidal umbra]

@warm cryptdo you know what is the case if any 2 lines are conincident to each other

[warm crypt]

ummm

isn't it infinately many solutions?

[vocal sleet]

@warm crypt Has your question been resolved?

[upper iron]

a consistent system means a system which has at least one solution

in order for your system to be consistent, the vector of the results must belong to the range space of the matrix of the coefficients

anyways, what you need to have is

Rank of matrix of coefficients = Rank of augmented matrix

$$r\begin{pmatrix} 1 & -1 & 8 \ -5 & 6 & -40 \ -7 & 8 & -56 \end{pmatrix} ,,,= ,,, r\begin{pmatrix} 1 & -1 & 8 & 3 \ -5 & 6 & -40 & -17 \ -7 & 8 & -56 & K \end{pmatrix}$$

[twin meteor]

Luna

[upper iron]

Find for which values of K this is satisfied!

[vale nexus]

you need to look at the row ranks

rows being horizontal

@upper iron whats the row rank of the left side?

are any rows a multiple of another?

[vocal sleet]

@warm crypt Has your question been resolved?

[warm crypt]

I still don't get it. im sorry

so what might be the answer?

[vocal sleet]

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[vocal sleet]

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[valid thunder]

Help, what's 2x²/x+2 =

[paper depot]

$\frac{2x^2}{x} + 2 = \frac{2x^2}{x}+2$

[twin meteor]

AnnGhost

[paper depot]

!original

[vocal sleet]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[paper depot]

@valid thunder please show the problem as it was given to you in full

[valid thunder]

Dividing polynomials using long division

[outer warren]

firstly, its ideal that you arrange your terms in order of descending powers

[valid thunder]

So change x³ and 2x² positions?

[outer warren]

-x^3, but yes

[valid thunder]

Oh right

[paper depot]

and then multiply x+2 by something that will turn the x (leading term) into -x^3

[valid thunder]

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[paper depot]

eh?

we got 2x^2 tho

[paper depot]

rude.

[outer warren]

oh nvm,fk

forgot about the alignment issue i myself mentioned

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[bleak remnant]

How do i find a and b

[vast shale]

can you send a picture where the writing is more clear

[bleak remnant]

Mb lol

After trial and error a=1 and b=4 lol

But no solution

[paper depot]

$\lim_{x \to +\infty} \paren{\sqrt{ax^2 + 2x} - \sqrt{x^2 - bx}} = 3$

[twin meteor]

AnnGhost

[paper depot]

this?

[bleak remnant]

Yeah

[paper depot]

ok, clearly a > 0 so that this limit makes sense in the first place.

i'd now multiply and divide by the conjugate.

[bleak remnant]

Oh i did that but

Do i like

Oh nvm

Do do it twice?

[paper depot]

you'll have $\frac{(a-1)x^2 + (2+b)x}{\sqrt{ax^2 + 2x} + \sqrt{x^2 - bx}}$

if i have not fucked up the signs

[bleak remnant]

Cuz im gonna get a square root at denom

[twin meteor]

AnnGhost

[paper depot]

no you don't need to do it twice

at this point you divide top and bottom by x

[bleak remnant]

Right

[paper depot]

leaving $\frac{(a-1)x + (2+b)}{\sqrt{a + \frac{2}{x}} + \sqrt{1 - \frac{b}{x}}}$

[twin meteor]

AnnGhost

[paper depot]

1/x = sqrt(1/x^2) of course etc etc

denom now approaches sqrt(a) + 1, a constant

[bleak remnant]

Written like this right

That's multiplied to lol

[paper depot]

what the fuck

[bleak remnant]

Wait

[paper depot]

also your twos are misshapen

[bleak remnant]

Hol up sorry

[paper depot]

but i mean sure yeah

[bleak remnant]

Can we go back to this

you'll have $\frac{(a-1)x^2 + (2+b)x}{\sqrt{ax^2 + 2x} + \sqrt{x^2 - bx}}$

[twin meteor]

7RgT

[bleak remnant]

Aint the numerator wrong?

[paper depot]

oh. no. yours is wrong

(p-q)(p+q) is p^2 - q^2 not p^2 + q^2 as you did.

[bleak remnant]

Then i multiply both numerator and denom by 1/x?

[bleak remnant]

I meant the (a-1)x²

[paper depot]

(ax^2 + 2x) - (x^2 - bx) = ax^2 + 2x - x^2 + bx

you are fucking up the signs

[bleak remnant]

Oh right

leaving $\frac{(a-1)x + (2+b)}{\sqrt{a + \frac{2}{x}} + \sqrt{1 - \frac{b}{x}}}$

[twin meteor]

7RgT

[bleak remnant]

Howd the denominator end up like that after multipling to 1/x

.close

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[tawdry wedge]

Find number of natural number solutions for a + b + c + d + e = 20 such that a < b< c < d < e

[tawdry wedge]

The answer lies between 0 to 9

not sure where to begin

[grand frigate]

You can just list it tbh, it's not that many.

[tawdry wedge]

So, just substitute random positive values of e and find it?

[mild sierra]

I’d start by finding an upper bound on e which is relatively easy to do

That way you massively cut down on the number of possible combinations, because to be honest I don’t know of a clever way to just get the number but it is a pretty easy thing to just compute in this case

[tawdry wedge]

alright, ill try

[mild sierra]

Finding an upper bound on e is the same as minimising the other numbers if that gives you a push in the right direction

[tawdry wedge]

.close

[vocal sleet]

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[open coral]

can someone help me with question 2?

[open coral]

I don't understand the process

of making a point into a vector

[frigid wolf]

what is x in 4^x=8

[open coral]

this channel is occupied, but the answer is 1.5

you can find this using log laws

[paper depot]

!occupied

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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).

[paper depot]

open your own channel!

read #❓how-to-get-help for instructions!

[frigid wolf]

thanks

[paper depot]

DON'T go into channels that have someone else's name on them!

[frigid wolf]

thanks

[open coral]

👍🏿

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[open coral]

<@&286206848099549185>

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@open coral Has your question been resolved?

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@open coral Has your question been resolved?

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[summer bronze]

why is there 8x^3

[keen flint]

expand (1-4x^3)^2

[summer bronze]

oh is it (1-4x^3)(1-4x^3)

i thought i need to distribute it

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[idle gazelle]

Can anyone help me write a proof for this i got the first two steps of writing the given statement. But im confused on what to do from there.

[raven owl]

The big triangle is similar to the ABD triangle

[idle gazelle]

Yeah but im required to write a proof

i got angle BAC congruent to angle CBD

as given

[raven owl]

Yeah just trying to guide you to the solution. If you consider what angles the similar triangles share, then in the next step you will be able to find angle CBD

[idle gazelle]

lemme send what i got so far

[raven owl]

To show in your proof theyre similar, well you know ADB is right angle because the other 2 angles in there are complementary, and obviously these 2 similar triangles share the angle BAC

[idle gazelle]

so my next step is to write angle ADB is congruent to angle CDB

because all right angles are congruent

[raven owl]

I would write before that, angle ADB is a right angle because sum of angles in triangle add up to 180 (it is not given that it is a right angle), then write that

[idle gazelle]

ok

[vocal sleet]

@idle gazelle Has your question been resolved?

[idle gazelle]

what should i do after saying angle adb is congruent to angle BDC

[raven owl]

triangle ABC and ABD share the angle BAC

so that means the triangles are similar

which ig immediately implies that the other angles are equivalent

[idle gazelle]

so angle ABC is congrruent to angle BCD

[idle gazelle]

wdym by share the angle bac

[raven owl]

lets think of it more like this, angle BAD, and angle BAC are equivalent, agree?

[idle gazelle]

yup

[raven owl]

that is what I mean by share angle

but they are the exact same angle so idk if I would say they are congruent

[idle gazelle]

i need to prove angle ABC is congruent to BCD right

[raven owl]

CBD

ok my bad

[idle gazelle]

?

[raven owl]

I was looking at the wrong smaller triangle

It is the same argument though I just got confused on what angle we were trying to say ABC is congruent to

Triangle ABC is similar to triangle BCD too

[idle gazelle]

so what do i write for angle abc is congruent to angle bcd

[raven owl]

Maybe you should write ABC=BCD because given (because we reach the conclusion by looking at the given chart)

I would say the reasoning is given

[idle gazelle]

mm

idk if my teacher would accept

[raven owl]

Actually im not sure what you mean by this

[idle gazelle]

beccause we only write given if it says that

so angle BAC is congruent to angle CBD

angle ADB is congruent to angle CDB

so im missing those two angles

right

AAA congruent to AAA

or is it something else

[raven owl]

Yeah thats basically what we are gonna finish with

[idle gazelle]

o

im on step 5

[raven owl]

this is the final step cuz this is what we are proving

[raven owl]

Ok I do nto think ABC is congruent to BCD though

[idle gazelle]

wait im solving for how the triangle is the same as the other triangle right

[raven owl]

The big triangle is similar to the triangle on the right

Ill try and draw on the picture to make this more clear

brb

[idle gazelle]

ok

ty

[raven owl]

@idle gazelle this is basically the entire argument we are making

from this we can clearly see BAC = CBD

[idle gazelle]

wait im pretty sure im supose to find out how triangle BAD is congruent to triangle BDC

?

[raven owl]

Nah the < looking thing means angle

[raven owl]

Also I think you are mixing up congruency and similarity

BDC is clearly smaller than BAC so they arent congruent

but all their angles are the same so they are similar

[idle gazelle]

i ment BAD

not BAC mb

[raven owl]

ok why do you think thats what your suppose to find

[idle gazelle]

i honesly have no idea

[raven owl]

Ok I should have payed more attention. When we have an IF THEN statement, we are trying to prove the THEN part

You wrote we are given BAC is congruent to CBD on your first step but that is what we are trying to prove

[idle gazelle]

o sht

it is?

[raven owl]

Yeah

The IF part is what we are given so that we can prove the THEN part

[idle gazelle]

aight lemme restart

we have to prove BAC CBD right?

by using the given

[raven owl]

yeah

[idle gazelle]

ohhhh

[raven owl]

Yeah that is one of the givens, we are also given ABC is a right angle

[idle gazelle]

So should i write both as the first two steps?

[raven owl]

yeah

[idle gazelle]

whats next

[raven owl]

next step would be showing angle ADB is a right angle.

[idle gazelle]

how?

Ohh

since Bac is ccomplementary to DBA, ADB has to be a right angles because they add up to 180

[raven owl]

Yeah

Then next we have BDC is a right angle because the angle of a line is 180 degrees

[idle gazelle]

ok

whats next

[raven owl]

this diagram shows how angle BCA and BCD are congruent

[idle gazelle]

?

i dont get this step

[raven owl]

Do you not get why its true, or why we are doing it

[idle gazelle]

so what do i write

[idle gazelle]

i dont get it because it doesnt really help

[raven owl]

Ok look at the triangles in the upper right. From those triangles we know BAC is congruent to DBC (AA similarity) which is what we want.

if we go back to the actual original diagram, we see that the top angles of the ones I drew are shared in the original diagram, meaning theyre the same

Basically we are just trying to show which angles of the triangles are congruent, then this will lead to us finding BAC and DBC are congruent (which would happen on the next step)

[idle gazelle]

so whats the next step

im still a little bit confused

[raven owl]

After this step, we have shown that both triangles have two congruent angles. This means the last angle must be congruent (which are the angles we have for our conclusion)

So the next step is just the conclusion,

[idle gazelle]

how to prove ABD is congruent to angle BCD

[raven owl]

we dont need to do that

[idle gazelle]

arent those the two angles we need

[raven owl]

its BAC and CBD

remember we are proving the part after "then"

[idle gazelle]

no im on the step after proving the right angles are congruent

[idle gazelle]

so we need to prove this right

[raven owl]

Nah, the triangle BAD doesnt have any of the final angles we are trying to prove are congruent so we dont really care about any of the angles inside it

[idle gazelle]

wait are we proving sas

aaa

or

wait can u write step 5 after that i think im good

[raven owl]

BCA is congruent to DCB

[raven owl]

AAA similarity is actually just AA similarity. If we have 2 angles of a triangle, we know the 3rd angle because they must sum to 180, so if the 2 angles add up to the same thing in both triangles, then the 3rd angle of both triangles must be the same

so basically we are using AA which is the same as AAA which means the third angle is the same

[idle gazelle]

whats the explanation

[raven owl]

I would say from the diagram which is basically the same as saying its given

because in the diagram those angles are the same thing

[idle gazelle]

why do we need to know Bca is congruent to DCB

[raven owl]

Because once we know this we can immediately jump to the conclusion

[idle gazelle]

how about BAD

do we not needto know

[raven owl]

In this picture, do you agree for the triangles I drew, the bottom right angles are congruent

[raven owl]

Nah

[idle gazelle]

how about ABD

[raven owl]

Nah that one is not needed either

[idle gazelle]

huh?

[raven owl]

Ok lets say we knew what ABD and BAD are, what would you want to do next

[idle gazelle]

for#5 i wrote A angle is congruent to itself

[idle gazelle]

we can know what triangle BDC is

like the angles are

[raven owl]

How

[idle gazelle]

by using the given info

like ABC is right

[raven owl]

Are you saying you want to try and find the exact angle of ABD and BAD

[idle gazelle]

yeah

[raven owl]

We do not have enough information to find that

That could work if we were able to figure then out, but the only thing we know about those angles is they add to 90 degrees

[idle gazelle]

so for step 6 we can conclude the answer

right

[raven owl]

Yeah

We dont need to know what the angles are, we just need to know that theyre the same

[idle gazelle]

so for step 6 i can write BAC CBD.

congruent

[raven owl]

Yeah

[idle gazelle]

what do i put for reason

[raven owl]

AA similarity makes the most sense but you could put other explanations