help-23

[golden loom]

what is the rule for multiplying logs

log a x log b

[cedar void]

there is none

[golden loom]

then how do i multiply logs

[cedar void]

but that's not what you'll get

[cedar void]

it will be addition

not multiplication

[golden loom]

how

in the question it is multiplying

[cedar void]

.

[golden loom]

ok but i dont have log(ab)

or log a + log b

oh i get it

[cedar void]

good thing

[golden loom]

that log (a-b) and log (a+B)

im getting this

i dont think its right

[golden loom]

nvm i dont get it

[lean otter]

wow

logs

what is your questions

[lean otter]

uh

log both sides

[golden loom]

how to log rhs

[lean otter]

simple

[magic junco]

Is it really necessary?

[lean otter]

hum

maybe u can do some freaky substituion

looks like a formula

[magic junco]

Not necessary either

[warm fractal]

I'd say dont prematurely log, use factorisation and index laws to cancel as much as possible

[lean otter]

i mean its solving the exercise it doesnt matter if its neccesarry

[magic junco]

Sure

[golden loom]

i am studying log so how will i solve using log

my goal is not to solve the question but to learn log

[magic junco]

Alr, do you understand what 77 square and skill issue mentioned?

For example, this

[golden loom]

yes

i know that

i am stuck here

how do i proceed

[magic junco]

You can’t proceed further

Which implies you did something wrong initially

[golden loom]

what is wrong here

(a-b)^2x = 2xlog (a-b)

[magic junco]

As what skill_issue said, he assumed you made some mistakes between step one and step two

Have you checked it?

[golden loom]

this is the first step

and i think its not wrong

now what do i do

[magic junco]

Oh, mb

Let me check

[golden loom]

this is the rhs

[golden loom]

and i got this

[magic junco]

Step.1 is correct

I take that back, what is this?

[golden loom]

multiply

sign

[magic junco]

No, that’s not how it works

[golden loom]

these are in multiplication right

[magic junco]

yes they are

[golden loom]

then why is it wrong

[magic junco]

Why is it correct?

[golden loom]

huh

[golden loom]

why wont it be

[golden loom]

why do u subtract

they are multiplying

[magic junco]

A slower version

👍

[golden loom]

oh so the log is on the whole thing in the rhs and not seperately

[magic junco]

Affirmative

I gotta go in a few minutes

With confusions, leave them for the next helper 😂

Do you have any questions before I go? @golden loom

[golden loom]

wait

[magic junco]

Alr

Make it fast 💥

[quick crater]

what does swaamii mean

[magic junco]

Idk actually

It’s been one of my nickname for a year and a half

[golden loom]

i have to do trignometry also after this 😭

[magic junco]

I don’t find any problem with your work, but the result is different from what I got.

@quick crater Do you have any idea?

[quick crater]

um i havent tried it

gimme a minute

which question are we solving?

[golden loom]

.

[magic junco]

The one initiated an hour ago

Ah I found it

[magic junco]

A small hiccup, but you’re doing great as a whole

[golden loom]

whats the problem

[quick crater]

no idea sadly, not too experienced with logs

[magic junco]

Look at it carefully

[golden loom]

-2log(a+b)

oh

dang

didnt divide by 2

[magic junco]

All good, the answer is -1, ignore it

[quick crater]

a nice thing to maybe do is to simplify to a+b=n and a-b=m

[quick crater]

wtf it is?

[magic junco]

What did you get?

[quick crater]

nothing yet, but i swear i tried -1 and it didnt work

lemme recheck

[golden loom]

x = log(a+b) / log(a-b)

[magic junco]

This is my work without log

[plush geyser]

x² - 6x - 41 = 0

[magic junco]

….what did you do?

[magic junco]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[quick crater]

its (a^2-b^2)^(2x-2)

[golden loom]

correct that -2

[magic junco]

Is it? I thought the original question said x-1

Ah yeah, mb

[quick crater]

a^4-2a^2b^2+b^4=(a^2-b^2)^2

[golden loom]

so the answer is this?

[quick crater]

wait it wants it interms of a and b?

[golden loom]

idk

[magic junco]

Show what you have revised

[magic junco]

Correct

[golden loom]

finally

[magic junco]

1.5 hour work

XD

Alr, anyways

!done

[safe radish]

If you are done with this channel, please mark your problem as solved by typing .close

[magic junco]

Have a good one

[golden loom]

thanks

.close

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[crude shell]

guys can someone tell me how to find this

consider a line with slope k being tangent to a circle with center (-4,-6) at point (-7,-7) how am i supposed to find the slope of this line which is k?

[crude shell]

i remember that there was a way to do this but like i seriously don't remember what im supposed to do

[quick crater]

the blue line is the line with slope k

notice how its perpendicular to the line from the center to the tangent point

since you have the centers coordinate and tangent points coordinates, you can find the slope of that (yellow line), then find the slope of whats parallel to it

[safe radish]

@crude shell Has your question been resolved?

[crude shell]

ohh and then find their slopes using m_1 * m_2 = -1 bc they are perpendicular right

[quick crater]

yes!

[crude shell]

i actually thought of this but i wasnt sure whether they were perpendicular or not tbh

okay great

thank you!

im closig the chat btw have a great day

.close

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[quick crater]

you too!

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[gusty hound]

F(x) is an integral function for one of the given functions a-d

[oak tangle]

differentiate F(x)

[gusty hound]

what's e^2

derivate

[quiet plume]

Well it's a constant isn't it

[gusty hound]

i end up with

oh so e^2=0

?

ok thanks

how do i close this

/close

.close

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[cunning pasture]

how is C1 D perpendicular to C2 D

[cunning pasture]

bruv it was easy

half of 180 - theta

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[cunning pasture]

Hi

[cunning pasture]

please

how doth PD = BD + AD - AB / 2

PELASe

ok got it a+b - c/2

r = s-a

sic

s-cic

.close

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[twin atlas]

Hey, so

[twin atlas]

this is killing me

[spice furnace]

infimum is really easy

Consider the sign of the numerator and denominator

[stable axle]

Where did you get that it looks acient

[spice furnace]

Supremum can be done by inspection. I think you just need to think about the problem more. Understand what 2^x looks like and x^2

[stable axle]

Can i see how you tried to solve it?

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[floral sonnet]

Hi. I just finished Precalculus Honors last year at my old school and I'm now going to a different school this upcoming year. They put me in Precalculus Honors AGAIN because of my math placement score. I had to set up a meeting to get into AP Calculus AB and they finally did. However, they said to prepare for it.

Below, the Precalculus course on Khan Academy is shown. At my old school I only did the first 4 Units!!! Does this mean I am really not ready for AP Calculus AB? Or is the other stuff useless Precalculus stuff that doesn't get used in AP Calculus AB?

If my below question is true, and the other Precalculus units are really useless for AP Calculus AB... I am intending on taking Khan Academy's Get Ready for AP Calculus course...

Please tell me what I need to do, thanks.

[meager crow]

AP Calc AB doesnt get into complex numbers really or matrices. Focus mostly on algenrs (rational functions/trig are important)

Probability isnt really needed and I think part 10 is an intro to calculus

[floral sonnet]

So Units 5 through 10 in Precalculus are not needed to take AP Calc?

[meager crow]

limits and continituity are a big part of calc AB

[floral sonnet]

thats in the get ready course

unit 10 disregard that

5 - 9

do i need units 5 - 9 for AP Calc

[meager crow]

Matrices/vectors are seen in calc 3 and I think there is a unit on polar/parametrization

[floral sonnet]

so Calc I im fine?

i only learned first 4 units

of precal

[meager crow]

As long as you are good with algebra and trig you'll be fine

[floral sonnet]

so i should take the get ready for AP Calc course?

[meager crow]

Probably. Maybe study series/limits

[floral sonnet]

i dont need to go over units 5 - 9 on precal?

series?

[floral sonnet]

Okay ill do this:

I will do the Get Ready for AP Calculus course,
I will possibly look over series.

looks good?

the get ready course goes over limits

okay thanks man

goat

.close

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[molten flume]

they give different answers

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[slender frigate]

can someone help with b pls

[slender frigate]

its something to do with similar shapes but idk what to do

[hoary anchor]

wouldn't it just be 4x the volume?

because in the region given above 5-2 = 3 units

u double the height

wait

not 4x

2^3

[slender frigate]

why 2

[hoary anchor]

because

the height of the stool leg

is 6cm

6/3 = 2

3 is ur "height" of the volumes of revolution model in units

[slender frigate]

howd u get that

oh

cause its the distance nvm

i get it now

thanks

[hoary anchor]

no worries

[hoary anchor]

5-2

[slender frigate]

.close

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[haughty pendant]

Hello

[haughty pendant]

Im looking for a proof check so far

#3

So far, I have:

I was a bit sloppy in my language in some of the last lines. It would be more clear to have written, “if b is an integer and is greater than n+1, then [the case above],” since if b is not an element of N+, this means either b is not an integer, or b is an integer greater than n+1

If b is not an integer, then id use the result of exercise 1 with x=b and y=n+1 to show that there is some z greater than b which is also a lower bound for N+, again contradicting b being the GLB of N+

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@haughty pendant Has your question been resolved?

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[lean otter]

Can someone help me with my homework

[rigid inlet]

!da2a

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No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[lean otter]

Idk if my answer is correct

[formal ether]

i am getting 1/9 feet

convert inches into feet first and then apply unitary method

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[compact geyser]

How can I find whether this is convergent or divergent?

[compact geyser]

without using direct comparison test?

[obsidian oracle]

series/integral test

you have yourself a positive decreasing integrable function

[compact geyser]

how do you know that an integral test would work?

like does it always work for every series or something?

[obsidian oracle]

it doesn't hurt to try

[obsidian oracle]

shoot your shot when you're dealing with integrable and decreasing positive functions

like that's how you get that series 1/(n^p) converges only with p > 1

[compact geyser]

I see

[compact geyser]

okay wait i have another question

so i used d'alambert and this series does converge, but then they want me to calculate the sum of the series

which method do i use to calculate the sum of this?

[obsidian oracle]

d'alembert?

[compact geyser]

i think its called the ratio test in english

[obsidian oracle]

No i got that

but you really got that it converges with that

?

[compact geyser]

yes i believe its correct

i got 0

[final schooner]

where you from ?

[compact geyser]

would you have used the integral test for this?

[compact geyser]

belgium

[obsidian oracle]

mmmmmm not sure at all

show?

[final schooner]

doet ge industrieel ?

[compact geyser]

well uh its a bit hard to show it because my handwriting is bad but in the solution sheet its also convergent

[compact geyser]

ja man 😭

[obsidian oracle]

i'm not saying it isn't convergent

[compact geyser]

universiteit antwerpen?

[obsidian oracle]

I'm saying d'alembert is wrong

[final schooner]

hahah nee leuven

[compact geyser]

hm, what method would be best suited for this then?

[final schooner]

kben ook bezig met rijen en reeksen hahaha

[compact geyser]

moet je ook herexamens doen?

[obsidian oracle]

check the sign of each term

[compact geyser]

well its alternating right

[obsidian oracle]

yep

[final schooner]

ja herexamen van wiskundige basistechnieken

[compact geyser]

oh yeah there is an alternating test as well i forgot about that completely

[compact geyser]

basistechnieken is crazy man

das nog basis 😭

[final schooner]

veel herexamens of niet

[obsidian oracle]

for your info, if the ratio test is done properly the limit should be -1

[compact geyser]

yep, zo ong 8, gy?

[compact geyser]

-1?

[final schooner]

same hahaha

[compact geyser]

I don't know maybe I didn't do it correctly or something then i guess

[compact geyser]

wanneer begint die van jou?

[final schooner]

kbegin volgende maandag

[compact geyser]

damn ik volgende dinsdag man

ben er zo half op voorbereid ngl

heb je ook biochemie enzo?

[final schooner]

ja

heb je voor wisk ook mondeling gedeelte of niet ?

[compact geyser]

als ik mondeling had zou ik wiskunde nooit halen

[final schooner]

hahaha dat had me echt gefuckt ik haalde een 8

[compact geyser]

hoe werkt mondeling wiskunde eigenlijk

[final schooner]

grote vraag op 6 punten

krijgt zo max een 1h om die te voorbereiden en gaat dan samen met prof bespreken

[compact geyser]

damn

moet je dan zo opnoemen hoe je een probleem gaat aanpakken

[final schooner]

ja alles

van het begin tot einde hoe je het opgelost hebt

en stelt dan ook nog extra vraagjes

[compact geyser]

das wel crazy ngl

[final schooner]

ja nie fijn

welk examen ben je erdoor ?

[compact geyser]

van mijn minor, kinematica en fluiddummechanica

3/7

ik had echt telaat gestudeerd man

[final schooner]

hebben niet dezelfde vakken denk ik

[compact geyser]

waarschijnlijk onder een ander naam

[final schooner]

ja

[compact geyser]

@final schooner btw weet je toevallig hoe je de som van dit berekent?

[final schooner]

ff checken

[compact geyser]

aight

[final schooner]

heb je het resultaat ?

[compact geyser]

som is 1/2

[final schooner]

met welke testen hebt ge al geprobeerd ?

[compact geyser]

om convergentiegedrag te bepalen?

tot nu alleen d'alambert maar ik heb nog niet geprobeert de som van de reeks te berekenen want ik weet niet hoe het moet

[obsidian oracle]

Compute partial sum by finding Partial fraction decomposition

[final schooner]

yeah i did it but i dont think its right

[compact geyser]

right i thought about partial fraction as well, but then what's next?

[final schooner]

got 1/2 for both fractions

[obsidian oracle]

yes so $\frac 12\sum_{n=0}^\infty (-1)^n \left[\frac{1}{4n+1}+\frac{1}{4n+5}\right]$

[flat frigate]

rafilou2003

[final schooner]

yeah and i did than integral

[obsidian oracle]

integral?

we wanna find the exact value

not compare

[final schooner]

you can find it with integral also i think

[obsidian oracle]

what is there to integrate?

[final schooner]

wait but do we have to find the convergens or what

[obsidian oracle]

exact value

of the sum

[obsidian oracle]

also look at the first few terms, stuff cancels

overlooking the 1/2 coefficient we have:

1 + 1/5

-(1/3 + 1/7)

• 1/5 + 1/9

[compact geyser]

I don't quite see what cancels here

[obsidian oracle]

at this point it's 1+ 1/5

[obsidian oracle]

lemme check something

[compact geyser]

oh wait i'm stupid

n = 0

[compact geyser]

this is called the telescoping method, right? i didn't learn it yet

[obsidian oracle]

telescoping is a must here

[compact geyser]

how can you recognize that we can use telescoping to find the sum of a series? what gives it away?

[obsidian oracle]

oh wait I'm dumb yes it's telescoping

4n not 2

1 + 1/5

-(1/5 + 1/9)

+1/9 + 1/13

-(1/13+ 1/17)

[obsidian oracle]

here it was the fact that the series is alternating

[compact geyser]

so for every alternating series we can use telescoping to determine the sum

[obsidian oracle]

no it's like a possibility

whenever you have opposing signs in a sum

look at what cancels each other

[compact geyser]

oh i see

well, what other methods could we then use to determine the sum if the series isn't arithmetic nor geometric?

@obsidian oracle also, i was wondering how i could start with this specific series. How can you calculate whether its a convergent or divergent without doing a direct comparison test

[obsidian oracle]

wdym direct comparison

[compact geyser]

like checking its majorant

[obsidian oracle]

cause I would have done that

[compact geyser]

yea but apparently the formal way to solve whether this is convergent or divergent is not through direct comparison

[obsidian oracle]

formal way?

[compact geyser]

like the way its "supposed" to be solved

i'm sure if i did direct comparison in an exam, they'd mark it as correct

but i want to know how to solve it other ways

[obsidian oracle]

oh I wouldn't solve it any other way

why bother

especially when this series really has no other nice way

[compact geyser]

well fair enough i guess lol

So how i would have done it via direct comparison is that, when you expand the brackets in the denominator, you get n^3

and since 1/n^p with p > 1 is always convergent, this means that this series is also always convergent

did i get it right?

[obsidian oracle]

ln^2 grows really slow compared to n

[obsidian oracle]

slow down there's some ln^2 on the numerator

if you wanna compare with 1/n^p

the numerator should be 1

[compact geyser]

but isn't that irrelevant since ln^2(n) barely grows though?

ALSO before I forget

how do I find the sum of this geometric series where i = 100?

1/4^n(a^2 + a^2/2) with a = 1

[obsidian oracle]

grows faster than 1

so not a majorant

[compact geyser]

hm

[obsidian oracle]

use that it grows slowly

so compare it to n for example

ln^2(n) < n

[compact geyser]

o

n/n^3 > ln^2(n)/n^3

[obsidian oracle]

yep

[compact geyser]

i see i see

[compact geyser]

thanks for your help man, but could you also assist me with the geometric sum thing real quick please? 🙏

i tried to use (r^n - 1)/r-1 with n = 100 and r = 1/4 but it gives me 4/3

and the sum is supposed to be 1/2

[obsidian oracle]

this? 1/4^n(a^2 + a^2/2) with a = 1

[compact geyser]

yes

[obsidian oracle]

also I don't know if (a^2 + a^2/2) refers to 3a^2/2 or a^2

[compact geyser]

yea its 3a^2/2

[obsidian oracle]

ok so a = 1

3/2 * 1/4^n

and are you starting from n = 0?

[compact geyser]

from n = 1

n = 1 to 100

[obsidian oracle]

so

first term * (r^(100+1)-1)/(r-1)

first term is 3/2*1/4 = 3/8

[compact geyser]

wait what

why

O

right

oh my god

damn how did i not notice that

i put 1/4 as the first term

that's why i kept getting wrong answers

thanks alot

[obsidian oracle]

so yeah approximating r^101 as 0

you get approximately

[compact geyser]

1/2

[obsidian oracle]

3/8* 4/3

1/2

[compact geyser]

: D

thanks again bro

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[obsidian oracle]

you had put 1 as first term

to get 4/3

[compact geyser]

yeah i later put 1/4 as well and i still got it wrong

[compact geyser]

yep

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[pine wren]

I got 3/2 for the answer and tripple checked and dont see no problem at all, yet the solution should be -3/2 according to wolframalpha and my teacher, but what I solved also checks out. Can anyone see a mistake or is this an anomaly?

[ivory thicket]

i can't see the exponents in the denominator very well. is it to the 7th power?

[worthy estuary]

the roots you found are correct but that is not the correct factorisation of the quadratics

[pine wren]

No, to the power of 2

[mellow cypress]

Try just factorisingn the denominator

[pine wren]

Im not sure which part you mean by this

[worthy estuary]

the top and bottom quadratics

[pine wren]

Yes

[worthy estuary]

if u expand them you wont get the originals

[mellow cypress]

There is -4x^2 but your factorisation doesn’t lead to a negative coefficient for x^2 term

[worthy estuary]

maybe try factorising the numerator and denominator with a different method

[pine wren]

Oh ok, I see this now but I qm confused as to why this is since I followed the D = and x1,2 = formulas to the t

[pine wren]

But why is this method not working

If I didnt know the answer I would just go on with my day not thinking its incorrect

[worthy estuary]

the roots are correct

but those arent enough to define a quadratic

as there are an infinite number of quadratics that go through the same roots

like this

[pine wren]

Im slovenian so theres a bit of a language barrier in terms of math namings for me

What are roots?

[worthy estuary]

x-intercepts

[pine wren]

Oh ok

[worthy estuary]

the solutions x1,2

[pine wren]

I dont get this part though

Oh wait I get it

But I dont get how its related with my problem

[worthy estuary]

so for the quadratics in your problem, x^2 has coeffecients 2 and -4

but when you expand the factorised ones

[pine wren]

Ok

[worthy estuary]

you will get something like x^2 +.......

so they r not the same

[mellow cypress]

@pine wren an easier way to do it imo would be to first factor a -1

-(4x^2 -5x +1) = -(4x-4)(4x-1) = (1-x)(4x-1)

[pine wren]

So then I missed something when doing the steps, no? Aka when using the methods that I used

[pine wren]

But how would I know this beforehand

[pine wren]

For every negative number?

[mellow cypress]

From practising factorisation

I find it easier to factorise when the coefficient of x^2 term is positive

So if its negative I just factor it out and then it becomes simpler

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[full agate]

why does applying the chain rule normally not work here?

like, bring out 1/x in front, then -1 from the exponent, then times by the derivative of the exponent?

[torpid fable]

Chain rule is for functions of the form f(g(x))

The power rule is used for functions of the form x^(…) where (…) is constant

[full agate]

so for the power rule the exp must be a constant?

i see now

ig in retrospect

if a^x and x^a are different

it makes sense that x^x would aswell

thanks

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[turbid aurora]

Given two sets of positive integers, as well as the Collatz function $f(n) = \begin{cases}
\frac{n}{2} & n \equiv 0 \pmod 2 \ \
3n+1 & n \equiv 1 \pmod 2
\end{cases} \$

Whenever $f$ is applied to a number $k$ in the set, add $f(k)$ to the set.

The cost function $C(A \to B)$ is the sum of $k$ for all $f(k)$ used to create set B from set A.

[turbid aurora]

Given a set B, we must find the set A that minimizes $C(A \to B)$ under the restriction that $B \notin A$ initially

This feels like a trick question, kind of

Can't you just give a set A that's just 2B?

Or I guess the element's in the form 3n+1 you could put that

$B = [4, 7, 9]$, $A = [1, 14, 16]$, $C(A \to B) = 1+14+16$

[flat frigate]

7aman

7aman

7aman

[turbid aurora]

$B = [1,2,4,8,16,32,64], A=[64], C(A \to B) = 64+32+16+8+4+2 = 126$

[flat frigate]

7aman

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[wraith cave]

idk what am looking at all i know is that am supposed to a2 + b2 = c2 idk what that is help

[wraith cave]

a quick explanation would be fine

on like what it is why its necessary and how do i use it

also calculators r allowed in the exam

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[sinful wyvern]

guys,
are these both the same?

[cedar void]

yes

[sinful wyvern]

omg thank youuu

ive been struggling finding where i got wrong

😭😭😭😭

[sinful wyvern]

btww how do u know that it’s the same?

[cedar void]

in your work, try to take 3 out

from that square

[sinful wyvern]

ohhhh

but i will have 3 outside instead of 9

do i square the 3?

[cedar void]

obviously

to take that out, you have to square it

[sinful wyvern]

ooohhhh

so it becomes -9?

[cedar void]

3^2 is just 9
that negative is already present outside the parenthesis

[sinful wyvern]

yess

okayyy

holdonn

[sinful wyvern]

if we take out 3, what do we do with the -7/2?

[cedar void]

if we take 3 out, -7/2 will be divided by it

[sinful wyvern]

ohh

so -7/2 is divided by 3, and not 9?

[cedar void]

yeah

'cause it becomes 9 after it goes outside the parenthesis

[sinful wyvern]

ooohh

okayy thank you soo muchhh

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[sturdy girder]

I need help

[sturdy girder]

I need help on Propositional Logic

I am doubting every of my answer

I just wanna know If my answer is correct or not, If it is not, leave it up to me to figure out

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@sturdy girder Has your question been resolved?

[sturdy girder]

<@&286206848099549185>

[river steppe]

Hello everyone,

I'm a Mathematics and Statistics tutor. If you need any help with your studies, don't hesitate to reach out. I'm here to support you!

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[limber pine]

can someone guide me through this proof?

[lean otter]

uhhh

use binomial theorem

pretty much

[limber pine]

i know

but we have to prove it

[lean otter]

yes prove using binomial

[limber pine]

but

yeah but how

like i tried foiling it

and expressing things in terms of n

in hopes that they subtract out

but idk if its leaving me with x^n - y^n

[peak estuary]

show your work

[limber pine]

i made some notes

i know that x is going down after the dots but

[lean otter]

u have to have more then notes

[limber pine]

i dont know how to express these notes in mathematical form in terms of n because i dont kno wwhat n is

[limber pine]

i have calculations but theyre super messy and its based off the notes

[peak estuary]

well no wonder you dont get anywhere with writing it like that

[limber pine]

i dont know how else to express it though

[peak estuary]

if you cant deal with general n right now then do it for n=4

[pallid crystal]

I would prove by induction

[peak estuary]

well, foil and write a sum consisting of x^stuff y^stuff terms

[limber pine]

ill try with 4 and report back

[lean otter]

you have to take cases also

[limber pine]

can you explain?

[safe radish]

@limber pine Has your question been resolved?

[pallid crystal]

Standard induction proof

Show it’s true for n=1

Assume it’s true for n=k

Prove for n=k+1

[limber pine]

ah, alright

i might have to do that

i tried to make a case for when n = 7

but it didnt lead to x^n - y^n since all the other terms didnt cancel out. only some did

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[crude glen]

Would anyone be willing to hop on a voice call with me for a bit to help me take a basic algebra practice test? I just need help understanding how to problem solve. I am planning on taking the actual test tomorrow and want to be prepared.

[spiral saddle]

yeah I want the same can someone hop on vc for help me with linear algebra test

[empty gyro]

You'll get better help in chat

[crude glen]

Thank you. I'll copy past my message.

[torpid fable]

They meant you should ask your question in this chat

[crude glen]

Ohhhh

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[inland dust]

Hi

Can someone help me with this math exercise please 🙏

A medicine bottle contains 250ml and needs to be administered in doses of 20ml every 8 hours. How many days will the bottle last if the patient takes one dose per day?

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[potent summit]

alr yeah

[potent summit]

here it is again

[twilit forge]

the integer one is probably where you should start

[potent summit]

alright

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[olive vault]

so im reading about linear transformations

[olive vault]

this is supposed to prove that given T : V -> U, T(0_V) = 0_U

but i dont understand how this proves anything

[steep lily]

subtract T(0) from both sides

[olive vault]

what sides

[steep lily]

of T(0) = T(0) + T(0)

[olive vault]

oh

lol

thanks

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[proven yew]

I need help solving these, have absolutely no idea, and just need the answer and help

[sleek plank]

it might help if you individually define x^2, y^2, x+y, x-y, and xy

for example, you can define x+y as 6, which is easier to sub in than (3+sqrt(3)) + (3-sqrt(3))

[proven yew]

I have tried and tried and I need the answer in like 2.5 minutes or else I don’t get credits so am desperate for answer

I tried those already, but haven’t worked

[sleek plank]

are you sure ur not supposed to simply verify if the equations are true or not?

[proven yew]

No

I have to plug the x= stuff and y= stuff in too

[sleek plank]

the question also says expression when these are all equations

could you send a screenshot of the whole question

[sleek plank]

im getting the same answer as you for 4

[proven yew]

Ok

[sleek plank]

and im getting 12sqrt(3) for 5

[proven yew]

So how would you write that out

[sleek plank]

$x^{2}-y^{2}=(x+y)(x-y)=(6)(2\sqrt{3})=12\sqrt{3}$

[flat frigate]

890s

[proven yew]

So would it be the 12 sqrt sign 3

I got it thank you very much

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[old basin]

I found the speed

which is 9.270... knots

but i dont know how to calculate the direction

This is my graph so far

[icy lance]

could use the sine rule if you fancy

[old basin]

i havent learned that yet soooo...

is it possible to use cosine rule instead?

if not then ill have to learn the sine one then

[icy lance]

sure you can use the cosine rule

[old basin]

I get how to use the cosine rule and all

but when im deciding the c value

how do i know which side to pick

[uncut star]

cosine rule always works i think 😄 no picking sides

i guess hmm

you gotta think of the angle as an eye

it looks at the corresponding skde

side*

that’s how you ”pick”

[old basin]

ahh

[uncut star]

no opp, adj or hypo

[old basin]

so then would it be 6 in this case then?

[uncut star]

let me see the pic i didnr acrually check it

relative to the angle 135?

what did you call 135

if 135 is an angle C then it is staring at side c which would be 9 * 270

[old basin]

because it said that the boat way going west and got hit with a northwest wind

northwest is 45 degrees right?

[uncut star]

oh wait

i didnt read the question

crud i have confused you unnecessarily and myself in the process

dump all the information received by me as fast as you can

[old basin]

done

[uncut star]

throwir OVERBOARD

[old basin]

LMAO

[uncut star]

wait okay i read the question now

i heard cosine rule and i was happy

[old basin]

lol

[uncut star]

okay so you are on a boat doing things mmm yes okay

hit by wind mm ok

[uncut star]

can you talk to me about this picture you drew

[old basin]

yes

the 4 knots side is where the boat is going

the 6 knots is the wind it is hit by

the 9.27 is the speed of the boat after being hit by the wind which i calculated

now i wanna see the direction of the boat

[uncut star]

ah

well okay

maybe i my info wasnt totally useless before then anyway

so cosine rule

[old basin]

now i know how to use the cos rule equation, but when im choosing the c value how to choose?

[uncut star]

gives us the whole triangle

yeah ok

so my silly way to describe it earlier

actually still works but

it got confusing because of the question

[old basin]

ngl id rather get the math way instead of the silly way, i have a lot of hw i need to do thats due VERY soon

[uncut star]

if i say each angle in the triangle looks at an opposing side

damn aight nvm nvm

tbh i am sure there is a proof right

but i swear they just teach you it is the opposing side of the angle ig

[old basin]

so c = 4?

[uncut star]

so for arbitrary angle a

[old basin]

dude atp i think i figured it out, sorry

i have more work to do

i gtg

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[uncut star]

gl

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[iron wolf]

how do i integrate dx/dt dx ?

[iron wolf]

please @ me if you come back to me, thanks :)

[quiet plume]

Set dx/dt = u and then you have dx = u dt. Then you can plug that back into your integral to get an integral with respect to t.

That is, assuming that you do mean $$\int_a^b \frac{dx}{dt} dx$$

[flat frigate]

Azyrashacorki

[quiet plume]

@iron wolf

[iron wolf]

Yes i do mean that

I see

Ive tried this and gotten (dx/dt)^2 times t ? is that right

[quiet plume]

Yeah it should look like $$\int_{a = x(t_1)}^{b = x(t_2)} \frac{dx}{dt} dx = \int_{t_1}^{t_2} u^2 dt $$

[flat frigate]

Azyrashacorki

[iron wolf]

I see

Thank you

Though its a shame because im trying to solve for t in this one question

And that just cancels it out

T-T

thank you anway !

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[ancient solar]

ABC is an isosceles triangle. AH ⊥ BC. M is the midpoint of AH. HK ⊥ MC. HK ∩ AC = P, BK ∩ AH = Q.
Prove that ∠BAC = 2.∠HKQ

[ancient solar]

my progress:
PE // BC
MC ∩ PE = I
BK ∩ IP = F, therefore PF = PI
PQ ∩ BC = J
now im trying to prove E is the orthocenter of triangle MCJ, so that: JE // PH, and EPHJ is a parallelogram then QH = EQ = 1/2 EH
EP ⊥ AH (PE // BC) therefore △HEP ᔕ △HKM therefore HK.HP = HE.HM = HQ.HA therefore △HQK ᔕ △HPA therefore ∠HKQ = ∠HAP = 1/2 ∠BAC

"." means multiply

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@ancient solar Has your question been resolved?

[sturdy girder]

E is the orthocenter of △MCJ, EPHJ is a parellogram, and QH = EQ = 1/2 EH, satisfying all the given condition

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[ancient solar]

thats what im trying to prove

[ancient solar]

im trying to prove E is the orthocenter of MCJ

and im stuck at proving it

[mellow wren]

Nice approach with that construction. But there’s an easier solution using similar triangles. Let me quickly write that down and post a photo here

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@ancient solar Has your question been resolved?

[ancient solar]

thanks

[ancient solar]

ah thank you

[mellow wren]

No probs

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[lilac osprey]

i have a question about identifying polar curves specifically how to distinguish a circle from a limacon

[lilac osprey]

in my question r=4sin(theta), its supposed to be a circle but couldnt you argue its supposed to be a limacon where r=asin(theta)+b where a=4 and b=0 like how would you know that its not actually a limacon and is a circle

or can a limacon not have a value b=0 with the form r=asin(theta)+b

[median vigil]

for that formula, you get a circle when b = 0 and a limacon if b ≠ 0

[lilac osprey]

oh ok, so you just have to note that b cant be 0 for a limacon

does that reasoning also apply to distinguishing between a cricle and a rose

like r=asin(btheta) for a rose, if b=1 does that mean it cant be a rose

cause it would just be 1 petal and i guess it would be shaped like a circle not like a longer petal shape

[median vigil]

well b = 1 also gives a circle in that case

[lilac osprey]

yeah

i guess ill just pay attention to these things

ty

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[strange plaza]

Not sure how I will approach this

Im thinking of looking for turning points?

[safe radish]

@strange plaza Has your question been resolved?

[prisma mauve]

Try to see what its derivative should look like

It should have two zeros and one which is of particular interest

[strange plaza]

What do you mean by that?

[prisma mauve]

So the sign tabular of the derivative should look like that

| + 0 + 0 - |

Thats not a sign tabular of a polynomial of degree 2 right ?

Because the first zero with same sign of both sides can only be a root of even multiplicity, i.e. something like (x-a)^2 or 4 or 6 ...

So the derivative is a priori of minimal degree 3

Another argument : You can see two inflection points, one very early on the graph, one between the two extremas, so the second derivative is of degree at least 2, so the derivative of degree at least 3

Can you do the rest ?

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[brazen parrot]

what is invertible

[brazen parrot]

this is a past test

;-;

[proud dust]

invertible means that if you mulitply a matrix by its inverse you get the identity matrix

[brazen parrot]

is it this

[proud dust]

not sure tbh

[brazen parrot]

why is it not equal

[proud dust]

turns out lmao

for a matrix to be invertible

[drifting knot]

Cz

Ok he is writing

[proud dust]

its determiant can'tbe 0

[drifting knot]

Good

Sir

[proud dust]

yuh im cookin 🧑‍🍳

[drifting knot]

🥚

👞♂️

Sheesh

[proud dust]

who let this man cook

[safe radish]

@brazen parrot Has your question been resolved?

[brazen parrot]

what does invertible mean..

[proud dust]

invertible means if you multiply a certain matrix by this matrix you get the identity matrix

A * A^-1 = In

and in order for it to be invertible its determiant must be different to 0

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[indigo mason]

how to type fractions in a graphign calculator

[prisma karma]

depends on your calculator, but you should read the manual (or find a tutorial somewhere, hopefully)

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[edgy charm]

this is for spectral decomposition. I don't understand how we know the operation is equivalent to a projection onto u_1

[edgy charm]

.close

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[edgy charm]

.reopen

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✅

[edgy charm]

<@&286206848099549185>

[kind tinsel]

can you give me the formula for projection onto $u_1$? (the one that you’re familiar with)

[flat frigate]

smaych principle

[edgy charm]

(x . u_1)/(u_1 . u_1) * u_1

[kind tinsel]

okay, so we can rewrite x•u_1 as u_1^T x, and u_1•u_1=1

[edgy charm]

oh

oh ok that makes sense

thanks

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[native robin]

how do u do 10 and 11

[native robin]

i started 10 but idk how to get the last z

[idle frigate]

so, what do you use for z to get a tube?

you need an additional parameter

(x(s, t); y(s, t); z(s, t)) = (2cos(s); 2sin(s); t)

with s between 0 and 2pi and t between -oo and +oo

[native robin]

why do you need a

s*

[idle frigate]

but you want to center it on the curve too

well you want an additional parameter

do you know about cylindrical coordinates?

you need to draw every single circle, for all values of z

[native robin]

ye

[idle frigate]

so you want a parameter that goes through all the values of z

[native robin]

so that’s the z=t part right

[idle frigate]

no

well, yes

if you call the first parameter s

you need two parameters

if you do {2cos(t); 2sin(t); t} you will draw a spiral

not a cylinder

[native robin]

ohhhhhhhh yea cuz it’s a surface

which needs 2 parameters

i forgot

[idle frigate]

yes

but that's not the full answer

you want to center the circle on the curve

[native robin]

ye idk what that means

i didn’t understand that on the question

oh wait

i get it

so u need to have

[idle frigate]

well the question is a bit weird indeed

if z(t) is not increasing/decreasing it doesn't work

or strictly monotonous

although no I guess it works for any z

it's just a weird looking tube

[native robin]

${x[t,s],y[t,s],z[t,s]}={2\cos s+x,2\sin s +y,t+z}$

[idle frigate]

you want to use s for the z coordinate

[flat frigate]

Jash

[idle frigate]

or that yes

but what do you give to x y z?

it should be (2cos(s)+x(t); 2sin(s)+y(t); z(t)) I think

you can check with geogebra

[native robin]

is it similar to this

[idle frigate]

yes

same thing as what I said

$$(2cos(s)+x(t); 2sin(s)+y(t); z(t))$$

[flat frigate]

suut

[native robin]

we don’t know what x(t), y(t), and z(t) is tho

ohhhhh

it given

[idle frigate]

yeah

[native robin]

so the bounds would be -inf<t<inf and 0<s<2pi

right

[idle frigate]

yes

well t is between 0 and 4

according to the exercise

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[timid scroll]

A n B'

When I draw it, it looks like a B'

Can someone go through the steps of how to do it

I just really really really dont get it

[lime nexus]

in simply terms the ‘ means not
so in the middle box they’re asking for A and NOT B, the left circle has only A, and there’s no B in it, whilst the middle has both A and B (in which they do not want thus y it’s not coloured), and the right is only B and it’s not coloured bc we don’t want B

[glacial meadow]

Which one are you struggling with?

[lime nexus]

so essentially the middle box says we only want A and no B at all, where the left part only has A

[timid scroll]

they share a B

[lime nexus]

and there is no B in it

[gentle sun]

It's A intersect everything that isn't B

[timid scroll]

Isnt it apart of it?

Like to my logic it would be this