help-27

[desert fulcrum]

Then divide that by 2023 for the probability

[weak cove]

Wait

again

[desert fulcrum]

correct

[weak cove]

say we had two solutions like

0 and 10

their difference

would not satisfy this

that cos(0-10)>= sqrt(3)/2

[desert fulcrum]

you don't have to do anything with 0 to 10

[weak cove]

so, the number of solutions to sqrt(3)/2 <= cos(P)

[desert fulcrum]

it doesn't matter at all

[weak cove]

just as an example

[desert fulcrum]

Ok

[weak cove]

how do we know that because 2 solutions satisfy this

[desert fulcrum]

see

[weak cove]

that their difference, satisfies

what we want

[desert fulcrum]

Let's make it more simple

x⁷²-1 is 0

72 solutions

5° angle

[weak cove]

yes

[desert fulcrum]

So you have 0<=5n<=30

0<=n<=6

[weak cove]

we would count n=0 and n=1 as solutions

right

[desert fulcrum]

yes

[weak cove]

but if we used those as our theta

0-5

it wouldn't work

[desert fulcrum]

that is the number of solutions

Not rheta

Theta

[weak cove]

but I'm saying that that isn't the number of solutions

because some of them don't work

[desert fulcrum]

It will all work, it's an inequality

[weak cove]

I don't think you get my confusion

[devout snow]

@weak cove Has your question been resolved?

[weak cove]

ty anyways notkavin

[devout snow]

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[restive river]

take x^2 as the factor in the radical

that should be the clue

[sterile wind]

is this supposed to have a - between the terms?

[dense jay]

most likely

[wise plank]

should be 2-3x

[sterile wind]

in which case the denominator of this should have a + correct?

I see

there is a +

but the limit goes to -infity

[wise plank]

$\frac{2-3x}{2x-\sqrt{4x^{2}+3x-2}}\to\frac{\frac{2}{x}-3}{2+\sqrt{4+\frac{3}{x}-\frac{2}{x^{2}}}}$

[woven radish]

Combustion

[sterile wind]

in the rationalization the numerator should be -

[wise plank]

you use |x| = -x to get here

factor out x^2 out of the root

[sterile wind]

now this is going to be similar to the first probelem we did where you want to factor a sqrt(x^2) from the square root in the denominator

so that you can "cancel" an x in the numer ator and denominators

but for x going to -infinity

sqrt(x^2)= -x

mostly, you ended up getting rid of the extra terms in the square root, which works out in the limit, but not technically equality

but yeah, your function has the same limit as this

from here you can use the fact that |x|=-x when x is negative, so sine you are taking a limit to negative infinity you can rewrite it

yep! and how does that let you simplfy the whole expression?

yeah! the reason why we do this is because when we look at and expression like $\frac{3x^2-x+6}{6x^2-7x}$ directly

[woven radish]

Willow

[sterile wind]

then in a limit at infinity all of the terms get bg

*big

so we have a lot of inf-inf indeterminate forms

but if we instead make terms get small, its much easier to deal with because instead we have a lot of +- 0

so in the above expression, dividing out the highest order term gives $\frac{3-1/x+6/x^2}{6-7/x}$

[woven radish]

Willow

[sterile wind]

where anything with an /x or /x^2 all go to 0

so in essense, for limits at infinity of rational expressions we only care about the leading terms

again its for the same reason, after you factor out the highest degree of x from the square root, the rest is either constant or goes to zero

$\sqrt{x^3+x+1}=\sqrt{x^3}\sqrt{1+1/x^2+1/x^3}$

[woven radish]

Willow

[sterile wind]

something like this might be preferable if instead it led with a x^3

yeah, for the expression you had, if you factored out sqrt(x^3)

you would end up with a inf*0 indeterminant form

yep

to some extent it depends since there are some complicated real analysis limit laws happening when you do this

if you can assert it to be 0 without having trouble in the rest of the limit then it is fine

but if it was x*sqrt(1/x)

you cant assert it to be 0, because then you would get x*0, which has limit 0, where as the original function had limit inf

yeah, thats an important case of something similar

yo uend up losing information on how the h in the numerator and denominator interact

for 93 its a limit as x goes to infinity

which means you want to show that for every $\varepsilon>0$, there exists an $M>0$ such that whenever $x>M$, you have that $|f(x)-k|<\varepsilon$

[woven radish]

Willow

[sterile wind]

your observation that $|f(x)-k|=0$ everywhere is helpful here since then for ANY $x$ we have that $|f(x)-k|<\varepsilon$ already

[woven radish]

Willow

[sterile wind]

so you can really choose any value for $M$ since choosing larger $M$ doesnt make you closer to 0

[woven radish]

Willow

[sterile wind]

yep

[devout snow]

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[wicked egret]

Can someone explain why they removed all the denominators

[wicked egret]

After they all became common why did they remove it in the next step

i know 5 /12 = 60 / 12 = 5 but why the rest is gone

[devout snow]

@wicked egret Has your question been resolved?

[upbeat prawn]

they just multplied by 12 throughout the equation

and in this equation, since every term is divided by 12, it cancels out 12 throughout the thing

[devout snow]

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[light cradle]

Hello

[light cradle]

can i get help with this problem

I have no clue how to start it

[wise plank]

substitution

for the first one u = 8t

change the bounds accordingly

[light cradle]

ok

so

du=8dx

dx=du/8

so would I plug in u to f(8t)?

[wise plank]

yeah

also the bounds would change

[light cradle]

bounds are now 0 to 64

[wise plank]

so it'll be $\frac{1}{8}\int_{0}^{64}f\left(u\right)du$

[woven radish]

Combustion

[wise plank]

and we know that $\int_{0}^{64}f\left(t\right)dt=-25$

[woven radish]

Combustion

[light cradle]

so it's 1/8 0 to 64 because it's 0 to 8 which is 8 times smaller

it's equal to -25

i'm stuck

[wise plank]

$\frac{1}{8}\int_{0}^{64}f\left(u\right)du=\frac{1}{8}\left(-25\right)$

[woven radish]

Combustion

[light cradle]

hmmm ok

so would that be the answer?

oh yeah it is

ok that makes sense

[wise plank]

for the second one you do the same

[light cradle]

ok not bad I will attempt it thank you for your help

[wise plank]

[light cradle]

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[hollow void]

Hi

[hollow void]

How do we go about this? Im kind of lost

[winter torrent]

um one thing you could do is suppose you had two ways to express it, then show that those are actually the same way

[devout snow]

@hollow void Has your question been resolved?

[midnight sluice]

assume a set of coefficients k_i and another m_i for 1 <= i <= n, then {m_iv_i}-{k_iv_i}=0 because they represesnt the same vector, divide by the vector of v_i's, you get m_i=k_i

horrible wording and horrible notation, but in essense write two possible representations, put them in an equation, divide by $v_1,\dots,v_n$ and you've shown they're the same representation (same coefficients)

[woven radish]

metnal

[midnight sluice]

or not divide since i dunno if you've defined division for vectors that way if at all, show that the corresponding cells are equal

[devout snow]

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[scarlet peak]

is there anyone who can help with this lean code? I don't understand what goals aren't being solved on line 72

[desert arch]

Spaces

[scarlet peak]

?

[devout snow]

@scarlet peak Has your question been resolved?

[sullen island]

h (m+1) tells you f(m+1) + f(m+2) = f(m+3) right @scarlet peak

you don't have f(m+1) + f(m+2) in your goal though, how do you expect that rewrite to work ? (as lean says, the tactic failed)

you can do it in two steps though

take f(m) + f(m+1), from the hypothesis h (m) it's f(m+2)

then you have f(m+2) + f(m+1)

and then you can use h (m+1)

[lethal pulsar]

ig

[devout snow]

@scarlet peak Has your question been resolved?

[devout snow]

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[graceful folio]

AB = BC, KD = ME
Prove that the yellow triangles overlap

[radiant anvil]

they don't

[graceful folio]

they do tho lol like its stated in the question that they do

[frozen aurora]

i think you mean equal

[graceful folio]

yeah sorry just google translate bugging

in my language the word is overlap

[radiant anvil]

congruent?

[frozen aurora]

OOOH i see

yes he means equal

[radiant anvil]

is ACMK a rectangle?

[frozen aurora]

overlap as in if you put one on top of the other they would overlap completely

[split gulch]

Yo am I tweaking?

[frozen aurora]

notice that B is midpoint of AC

[devout snow]

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).

[cosmic gust]

I'm assuming that's a rectangle, put alpha and beta in the left and right bottom triangles, you will find that they are exactly the same therefore, CE=AD

[graceful folio]

ohh then congruent

[cosmic gust]

And also since the C angle is 90 degrees too, the inside angles of them and 2 sides match, therefore BE=BD, therefore they are the same triangles

[graceful folio]

can you elaborate a little?

[cosmic gust]

Let me draw it

1sec

[cosmic gust]

1. Red
2. Yellow
3. Blue

You can ask me the step you don't get*

Also I flipped it to show it easily

[graceful folio]

Ohh yeah I thought about it now, now I realize that ADB and BEC are congruent

ty

but how do I prove that BE and BD are the same?

[cosmic gust]

Since EC and BC are same, and the angle between them are the same, the third side should be same too

It's a law

It can be somewhat proved with cosine law

[graceful folio]

oh ok

thanks

thanks a lot

$close

.close

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[devout snow]

@upbeat crest Has your question been resolved?

[upbeat crest]

no it has not

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[upbeat crest]

1

[frozen aurora]

this looks like squeeze theorem to me

[upbeat crest]

dont have that in course

[frozen aurora]

especially the definition of g(x)

what do you have in the course, then?

[upbeat crest]

unless they want us to derive it ourselves

[upbeat crest]

for functions

and

nothing related to squeeze theorem

integral calculus

[frozen aurora]

well, $\lim_{x\to 0} g(x) = 0$

[woven radish]

artemetra

[frozen aurora]

do you see why?

2sqrt(x) goes to 0

the integral goes to zero too (the bounds both converge to the same value -> the area under the graph is 0)

f(x) seems to diverge to infinity

i would use l'hôpital but i have no clue how here

[upbeat crest]

whats that

[frozen aurora]

nvm then

[upbeat crest]

i found the solution on a website

[frozen aurora]

it's a very useful property of limits but you haven't studied it so i don't think you are supposed tl ise it here

[frozen aurora]

that's nice

[upbeat crest]

something like that

dont know why they put this up before uni

anyway

.close

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[hexed willow]

hi, is this correct?

[dense jay]

that looks like a quadratic

not a cubic

[hexed willow]

oh yeah

wait

[radiant anvil]

feels off

[hexed willow]

give me a sec fixing

[cosmic gust]

If a polynomial has a root at some x point, what does that mean?

[hexed willow]

I don't know.

[cosmic gust]

Let's say you have a polynomial: x^2 + 3x + 2, you can factor it as (x + 2) (x + 1), which then you will realise that its roots are -2 and -1

What does it mean if the root was x=1

[hexed willow]

you would use it to find the vertex?

[cosmic gust]

Yes and also you will know that it has a (x - 1) factor in it because when you plug in 1, it gives you 0

[hexed willow]

meaning the vertex would be (1, 0)

[cosmic gust]

So therefore this third degree polynomial has a (x - 1) in it

So therefore you can do polynomial division

To find the other factors, to find the other roots

[hexed willow]

ok, thanks for the help!

.close

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[digital mirage]

i need help to proof this formula by using the given formulas. it did only work with different formulas but i have not been able to use sin^2(x)+cos^2(x)=1

[upper schooner]

You might want to combine both $\sin^2(x) + \cos^2(x) = 1$ and $\tan(x) = \frac{\sin(x)}{\cos(x)}$ together to figure out how to change $1 + \tan^2(x)$...

[woven radish]

@upper schooner

[digital mirage]

but i dont know how to do that

i dont get what to do after that

[devout snow]

@digital mirage Has your question been resolved?

[devout snow]

@digital mirage Has your question been resolved?

[devout snow]

@digital mirage Has your question been resolved?

[devout snow]

@digital mirage Has your question been resolved?

[devout snow]

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[stark flax]

Need help w this

[devout snow]

@stark flax Has your question been resolved?

[devout snow]

@stark flax Has your question been resolved?

[devout snow]

@stark flax Has your question been resolved?

[devout snow]

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[shadow rose]

Can anyone help with 15?

[devout snow]

@shadow rose Has your question been resolved?

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[restive river]

whyd she break this into two parts

[weak cove]

because you can

it makes it easier to evaluate

[restive river]

how did she know she can

[weak cove]

you always can with finite sums

split them apart at addition/substraction

[restive river]

so since their is two different terms in the parenthesis

she does the sum of each one right

[weak cove]

the parenthesis aren't even really doing anything here

it's just making it clear that the -3 is inside the sum

[restive river]

oh okay i see, this my first time learning sums so it's a bit of a learning curve for me

[weak cove]

so you don't accidentally read it as, sum 2i^2 and then -3

[restive river]

ah i see

[weak cove]

it is saying sum 2i^2-3, and uses parenthesis to make it clear

and sums are linear so you can split them apart at subtraction/addition

(finite sums)

[restive river]

oh okay i see

thanks!

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[weak cove]

no problem

[devout snow]

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[twilit estuary]

lets say i measured an angle of 90 degrees

[twilit estuary]

i have to measure the other two angles

to know what special traingle im using

right?

[dense jay]

pardon?

[lusty sapphire]

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[dense jay]

measure the other two angles?

i can tell you how to construct those triangles if you want, but im not sure exactly what you mean

[twilit estuary]

acc

what i mean is

yk how the triangle has 3 angles

[dense jay]

all do, yeah

[twilit estuary]

i was measuring a triangle from a random image

it had 90 degrees

so its gona be a special triangle

but isnt there two diff ones?

[dense jay]

wrong

[twilit estuary]

💀

help

[dense jay]

not all right angle triangles are those two triangles

having an angle of 90 degrees just means its a right angle triangle, nothing else

[twilit estuary]

so how do ik if its special

[dense jay]

you generally dont just find those two triangles, you tend to just make them yourself if you need them

[twilit estuary]

i measure the other two angles

and then ik?

[dense jay]

i think youre a bit hung up on them
if you need the triangles then you create them

any enlargement of those triangles will also be the same thing

but i dont see a situation where you would ever need to measure the angle to confirm its one of them

one is just a right angle triangle with 45-45-90 angles
the other is an equilateral triangle of side length 2 that was bisected down one of its axis of symmetry

[twilit estuary]

okay i need to find sin cos and tan

of 90 def

degrees

i made it into radians

which is pie over 2

[dense jay]

not necessary here, but sure

[twilit estuary]

and pie over teo

two

is in the special traingle?

[dense jay]

the triangles wont be so much helpful for the 90 angles, theyre generally only useful for 30, 45, and 60

[twilit estuary]

so if its 90 degree

its not special?

[dense jay]

it should be easy to memorise that sin90=1 and cos90=0 though

if in doubt just draw a sin or cos graph

[twilit estuary]

yea

so it is a special angle

[dense jay]

not really, the triangles dont help

cos 90 would be adj/hyp, but two sides are adj to 90

[twilit estuary]

hm

so i dont use the special triangle

when i have 90

[dense jay]

yes, theyre unhelpful

use them for 30,45,60

[twilit estuary]

so would i refer to this

[dense jay]

looks like a work in progress unit circle

[twilit estuary]

yea

i mean

no need to continue u it

bec

its pie over two

[dense jay]

that graph isnt actually telling you anything though

[twilit estuary]

u split the top into 2 and bottom into teo

then how am i supposed to show my work

for

sin cos and tan for 90

without a unit circle

[dense jay]

draw a graph of sin and cos

no more really would need to be done

[twilit estuary]

she didnt ask for graphs tho

[dense jay]

what did she ask for

[twilit estuary]

she just find trig rations

and bec its a special angle

she said we have to give exact values

if not exact u estimate

[dense jay]

the special triangles tell you nothing about 90, at least not cos(90) anyway

sin 90, maybe

[twilit estuary]

hm

this is confusing

[dense jay]

i dont think there would be any real issue with directly quoting that sin90=1 and cos90=0

[twilit estuary]

graph does make sense

[twilit estuary]

i mean she wants us to show how we got what we got

[dense jay]

ill default back to draw a graph then, seems the simplest

itll help with 180,270,360, etc too

[twilit estuary]

kay

[devout snow]

@twilit estuary Has your question been resolved?

[devout snow]

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[sly galleon]

in this definition of a spherical sector, i was a little confused by the phrase near the end: "but which contains no point inside the sector" like how does that work if the axis passes through the center of the sector?

[sly galleon]

im pretty sure this is how the sector is formed right?

then clearly the axis has some points contained within the sector

[devout snow]

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@sly galleon Has your question been resolved?

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[covert pebble]

wait k=0.5

i think

[devout snow]

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[covert pebble]

.clear

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[dry ivy]

idk bro

[cerulean quarry]

let u = 9 + x^2

[dry ivy]

yeah

oh man

oh shoot

i added an exponent on accident my fault

i got to

(1/2) ln |12| + 1/2 ln |9|

oh i got it

.close

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[dim prawn]

,rotate

[dim prawn]

This graph is a periodic graph for y = x in Fourier series

For this graph of semicircles, i made the equation as stated above

,rotate

[dim prawn]

Can someone let me know if I’m correcr?

[devout snow]

@dim prawn Has your question been resolved?

[devout snow]

@dim prawn Has your question been resolved?

[devout snow]

@dim prawn Has your question been resolved?

[dim prawn]

<@&286206848099549185>

[thin inlet]

sqrt(1-x^2) instead

it makes sense when you think about the domain

[dim prawn]

Oh yes i made a mistake there

But shouldn’t it be sqer(pi^2 - x^2)

[thin inlet]

yes

well

wait

they aren't technically semicricles if they only go up to 1 right?

[dim prawn]

Umm why not?

[thin inlet]

if they were semicircles this distance would also be pi

[dim prawn]

Ohh yes

So

How about a half cosine wave

Piecewise

[thin inlet]

yeah, that could work, that's probably what they want yes

[dim prawn]

Ohkayy

[thin inlet]

you could do some ellipse thing but a cosine wave seems most apparent/easiest to do a fourier series for

probably what they want

[dim prawn]

Thanks alot

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[bright juniper]

ok so there is a factory that makes 7200 toys each worker make the smae amount of toys but in the factory 3 workerss left so other workers needed to make 400 more each for there plan hom many workers are there

[devout snow]

@bright juniper Has your question been resolved?

[bright juniper]

<@&286206848099549185>

[regal blade]

Let the number of workers be w and the number of toys produced per day be n. Write out the equation.

[devout snow]

@bright juniper Has your question been resolved?

[bright juniper]

wrong

[dapper tiger]

there are n workers, each worker does 7200/n toys, but when 3 workers leave, there are n-3 workers left, and they need to make 7200/(n-3) = 7200/n + 400 toys
solve for n

[devout snow]

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[bright juniper]

Thank you

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[restive river]

asked to prove the identity above

got stuck near the end, last step was a desperate attempt lol

most likely wrong

[potent nebula]

which identity exatcly

[restive river]

nvm wrote it wrong

thanks anyways

.close

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[quartz pond]

hi

[devout snow]

Please don't occupy multiple help channels.

[quartz pond]

I am no

t

can someone explain this weird angle thing to me

it uses the sin law but what is this

does it want me to use different combos depending on obtuse or acute

[lavish radish]

right so have u heard of small side big side angle and big side little side angle?

probably haven’t cuz that’s the way my teacher calls it and i’m not sure if that’s the way it’s taught normally

basically when given SSA, you have a chance to get an ambiguous case

to discern whether it’s sSA or SsA you have to look at the side opposite the angle that ur given

if it’s bigger than the other side ur given then u have SsA or big side little side angle

there should only be one possible triangle when it comes to this

but when you have sSA u can have either 0 or 2 possibilities

so just calculate the angle as usual using law of sines and if the value u get is greater than 1, there are no triangles possible

[quartz pond]

thanks bro

let me read it

sorry

[lavish radish]

if u get a number less than 1 then plug it into inverse sin to get the final angle value

then subtract that final angle value from 180 to get the second solution

[quartz pond]

there were 2 triangles here!

[lavish radish]

well yeah that fits a possibility

[quartz pond]

how do u know 1 or 2

[lavish radish]

since ur problem is sSA

[lavish radish]

i explained the sSA

SsA is 1 case

sSA is 0 or 2 cases

[quartz pond]

oh ok

then what would I do here in this scenario

I solved the sin angle normally

angle*

now for the other one do I do 180-angle

and its just other variables=same value

[quartz pond]

is that right

[lavish radish]

yes

[quartz pond]

oh ok

[lavish radish]

chill lil bro

[quartz pond]

wait what

[quartz pond]

what did I say

[devout snow]

@quartz pond Has your question been resolved?

[quartz pond]

HEY BROO

I did one part

I think its right

but how abuot the obtuse side

thats confusing

FUCK HELP ME MAN

FUCK

fuck

fuck

fuck

this math shit makes me wanna fucking jump off a cliff

I swear whoever tf fcreated this

@lavish radish

FUCK HELP ME MAN

[devout snow]

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[wicked egret]

Step 2 is wrong right?

[wicked egret]

Isnt it supposed to ve x^2-x-2x-2

[dense jay]

mhm mhm

[wicked egret]

i did this and got A = 1, B = 1, and C = 4

Is that right

i got

[dense jay]

2(x+4)(x-4) 2(x-4) 1/(x+4)
seems alright

[wicked egret]

1/x+4

ye but do they try to confuse u with the B

[dense jay]

how so

[wicked egret]

well cuz of the power

cause its just supposed to be 1

[dense jay]

1 is a power

i wouldnt say its a trick or attempt to confuse people really

knowing that x=x^1 should be fairly standard

[wicked egret]

ye ik but like they never write 1 so thats what i mean

How do I do this @dense jay

[dense jay]

-2/3 and 7/4 are excluded from the domain

this means they are roots of ax^2-bx-c

when x=-2/3 or x=7/4 then ax^2-bx-c=0

[wicked egret]

Ohhh so like its like how it would be (x+4)(x-2) ors omething like that right?

[dense jay]

sure, they probably want a b and c to be integers, that my impression anyway

so keep that in mind

i could be wrong, feel free to try not doing so

[wicked egret]

@dense jay ty for ur help i just got a few more questions like this right

so i did it

and got 3(x-4)(x-2)

but it says (ax-b)

whats a here?

[dense jay]

you can factor the 3 into the other bracket

its valid

[wicked egret]

so it would be

(3x-4)?

[dense jay]

not like that

3(x-4)(x-2)=(3x-12)(x-2)

like that

[wicked egret]

oh so like foil the 3(x-4)?

[dense jay]

kinda, i just distributed it into that bracket

[wicked egret]

oh mb ye i mean that

@dense jay i got x^2-14x-6 = 0

-(-x^2+14x+6)

or did i do somethn wrong

ty for responmd to me btw i relly appreciate it

[dense jay]

9x-x+6=x^2-6x
x^2-14x-6=0

[wicked egret]

wait i forgot to give u the options

[dense jay]

you seem to be fine

[wicked egret]

8x-7 = 0

8x+5 = 0

x^2-14x-6 = 0

x^2-14x+6 = 0

those options

[dense jay]

you have one of those

[dense jay]

distribute the -

[wicked egret]

yup

x^2-14x-6

[dense jay]

indeed, =0

[wicked egret]

im still rlly confused ab this tho

like that question confuses me alot

[dense jay]

if x+2/3=0 then 3x+2=0

that could be one of the factors for example

then youll have integer coefficients when you expand

[devout snow]

@wicked egret Has your question been resolved?

[devout snow]

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[royal laurel]

what is the alternative word for "interview"

[royal laurel]

the situation is not a face-to-face meeting, instead, I'm going to send somebody a test to see if he/she is appropriate for this job position

what word can I use here

[wicked egret]

what does this gotta do with math 😭

i mean u could just ask chat gpt like i just did and it said:

In a situation where you are sending someone a test to assess their suitability for a job position, you might refer to it as an "assessment," "evaluation," or "test." So, you could say something like "I will be sending you an assessment/test to determine your suitability for the job position."

[royal laurel]

ohhh

[wicked egret]

ez chatgpt

[royal laurel]

assessment

[wicked egret]

yes

[royal laurel]

shit, i didnt know chatgpt is that good at answering this type of questions

[wicked egret]

chatgpt is good at everything except math

[royal laurel]

thats lit

thank you so much!

[wicked egret]

but genuine question tho

why would u come here

in this math server

to ask that

lol

[restive river]

they're not good when humans have the ability to shut them off

[wicked egret]

na they gonna become self aware soon

and then they will be unstoppable

[restive river]

possibly not, we are not in skynet

[wicked egret]

im kidding 😭

[restive river]

nuh uh, u said what u said, terminator is coming down

[devout snow]

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[leaden bramble]

I need help

[leaden bramble]

A triangular banner has an area 729 square feet. Find the measures of the base and height of the triangle if the base is two-ninths of the height.

I konw that you gotta do

A=1/2bh

but

I dont know how to implement it in this one

[wooden wraith]

"the base is two-ninths of the height"

can you express that in an equation?

[leaden bramble]

729=1/2(2/9)h

[wooden wraith]

almost

you replaced b with 2/9

but it doesn't say the base is 2/9

it says the base is 2/9 of the height

[leaden bramble]

729=1/2b 2/9 ..........?

[wooden wraith]

no, don't focus so much on the area formula just yet

"the base is 2/9 of the height"

b = (2/9)h

so you can replace the b in the area formula with (2/9)h

[leaden bramble]

729=2/9h so i multiply by 2 then divide by 9 ?

[wooden wraith]

wait, hang on

you had

A = 1/2 bh

if b is equal to 2/9 h

then

$$A = \frac{1}{2}bh$$
becomes
$$A = \frac{1}{2} \cdot \frac{2}{9} h \cdot h$$

[woven radish]

tatpoj

[wooden wraith]

since b is (2/9)h

notice nothing else changed, b just became (2/9)h

oh but yea of course you can put 729 in for A

so then if you simplify the right side, you should get $\frac{1}{9}h^2$

[woven radish]

tatpoj

[leaden bramble]

hmmm

i did something similar like that but got 2/18

let me try again

[wooden wraith]

no that's good

2/18 = 1/9

if you simplify the fraction

[leaden bramble]

ohh

I was confused there

[wooden wraith]

so then we have $729 = \frac{1}{9}h^2$

[woven radish]

tatpoj

[leaden bramble]

hence why i came here

[wooden wraith]

oh i gotcha lol

[leaden bramble]

0=1/9h^2-729

then i do the ( )( ) thing?

or am i missing something

[wooden wraith]

that'll work

but you can also just get h^2 by itself and then take the square root

which is probably simpler

[leaden bramble]

3h^2-729?

then do ( )( )?

[wooden wraith]

nah what I mean is

we had

729 = (1/9)h^2

multiply both sides by 9

to get h^2 by itself

so then 6561 = h^2

[leaden bramble]

aha

[wooden wraith]

and you can just take the square root to get h

[leaden bramble]

ohhh

[wooden wraith]

no need to factor

[leaden bramble]

so 81

[wooden wraith]

factorizing would work too but kinda hard to do with numbers this big

yeah h = 81

[leaden bramble]

while b is 6561?

[wooden wraith]

not quite

what did the problem say about the base again?

[leaden bramble]

oh u right

2/9

?

[wooden wraith]

2/9 of the height

so (2/9) * 81

since we just found the height was 81

[leaden bramble]

gotcha

let me write all this

and get a new question

so i can do it my self a couple of times

ty

[wooden wraith]

no problem 👍

[leaden bramble]

aight ty very much

[devout snow]

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[primal tusk]

so im doing some basic trig

and it seems my formula was wrong

but i cant figure out why

my math works out to be just wrong

but as far as i can tell the formula is right?

its invers cos

im a dumbass

/close

.close

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[carmine hill]

LOLL

sorry

[steady pivot]

It's all good

[carmine hill]

i need help finding the consumer and producer surplus

ive literally asked like 6 tutors and no one could help

this is what i have done

and theres no way the number can be that big

and i tried 1979 and it was wrong

<@&286206848099549185>

[devout snow]

@carmine hill Has your question been resolved?

[carmine hill]

PLEASE HELP MEEE

[devout snow]

@carmine hill Has your question been resolved?

[devout snow]

@carmine hill Has your question been resolved?

[carmine hill]

.close

[devout snow]

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[proper island]

How did they simplify it?

[grand siren]

$\sin(2x)=2\cos(x)\sin(x)$

[woven radish]

bee [it/its]

[grand siren]

if you didn't already know that identity i'm not really sure what to say other than uh... yeah that's just true, you can google "double angle formula"

[proper island]

Oh is that a rule

Alright thank you! I will Google it

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[frozen spoke]

?

[final dune]

PQRS is a cyclic quadrilateral, so the opposite angles add up to 180

Can you use that to solve the question?

[frozen spoke]

Uh

2x+45 + 3x +8 = 180

?

Or is that wrong

@final dune

[final dune]

Thats not right

[frozen spoke]

What would it be?

I mean

[final dune]

PQS+PRS=180

[frozen spoke]

2x+45 +x = 180?

[final dune]

PQS+VQS=180

[final dune]

But PQS doesnt have a value of x°

[frozen spoke]

Oh yeah

VQS does though

[final dune]

Thats not given in the question

Yes

[frozen spoke]

Yeah

So what's after?

😭

Sorry I'm stupid

[final dune]

Nah its ok to not know the answrr

PQS+VQS=180
PQS+PRS=180

Notice something strange?

[frozen spoke]

Yes

PQS in both

[final dune]

And 180 in both

So can you convince yourself that VQS=PRS?

[frozen spoke]

Oh

Yes

Oh yeah

Oh definitely

Oh i GET IT now

Easy.

Thanks

One more thing

Why is this not 50?

[devout snow]

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[rain apex]

Two particles A and B with initial positions (in m) 5bı−13b and 2bı− b have
constant velocities (in m/s) 2bı− b and 3bı−5b respectively. Find the closest
distance between the two particles.

[devout snow]

@rain apex Has your question been resolved?

[rain apex]

<@&286206848099549185>

[pastel pasture]

I think the formatting is messed up

Can you fix it

Plain text is fine

[earnest dirge]

yea

[rain apex]

Two particles A and B with initial positions (in m) 5i−13j and 2i− j have
constant velocities (in m/s) 2i− j and 3i−5j respectively. Find the closest
distance between the two particles.

the i's and j's are supposed to have hats on them

[earnest dirge]

45/root17

[rain apex]

what did you get for B's position relative to A's

@earnest dirge

[earnest dirge]

yea

[rain apex]

what did you get for B's position relative to A's

[earnest dirge]

cant you figure it out bruv

its easy

how old are you

im 15

from india

[undone gull]

shut upp man

unban eamonn004

right now

Hes truly humble under god

and Im the one that typed that

[devout snow]

@rain apex Has your question been resolved?

[devout snow]

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[weary token]

Need help with those two questions

[remote ember]

Try 100 for the first one

Shall I explain why that's the answer?

And for the second one, you need to use the formula for the area of a circle multiplied by ¼ to get the upper left quadrant, then subtract the area of that triangle to get the shaded region

[weary token]

no i figured why

[remote ember]

Okay

[weary token]

i dont fully undertsand this one tho

[remote ember]

Imagine you have a full circle, now because the angle including the area that you want is 90° (as is evident by the corner) you only want to focus on that quadrant. If the area of a full circle is πr^2, and you only want 90° out of the full 360°, you have to use the area of a circle multiplied by 90/360 which simplifies to 1/4

Now you've got the area of that quadrant

The only thing that you don't want in it is the area of the triangle, so subtract it

[weary token]

ohhh

so will the answer be D?

[remote ember]

No

Because ¼ times π(4)^2 is 4π

Then subtract that from the area of the circle, which is ½ times 4 times 4 which is 8

[weary token]

Ohhh makes sense now, thanks!

but one more question

[remote ember]

Yeah?

[weary token]

i tried solving this one

but it doesnt make sense

cause like

i was writing

x+2y=x^2-4y^2

and i couldnt find the right solution using that

i dont know if the formula im using is wrong

or if it is somethign else

[vapid fjord]

damn what hard ass math is this

is this college

[remote ember]

If you rearrange the first equation for x, you'll get x= 3-2y, then substitute that into the second equation

[weary token]

no 🥹

[remote ember]

By doing this, you'll get the y's on there own, then you can rearrange then solve for y

Once you have y, substitute that into the first equation to solve for x. Now you have x and y, substitute that into the third equation to get your answer

[weary token]

I did that, but I ended up with this

[remote ember]

(3-2y)^2 is not 9-4y^2

Because (3-2y)^2 is the same as (3-2y)(3-2y)

[weary token]

Let me try that

One second

[remote ember]

Okay

[weary token]

I was stuck on like putting the -4 to the other side

[remote ember]

No, you don't add (3-2y) to (3-2y) you multiply them together, but you have to multiply each term by the other. Do you know how to expand double brackets?

[weary token]

Oh yeah

I do know how to but that rule left my brain for a moment

[remote ember]

Okay, try that

[remote ember]

12 divided by -12 is -1

[weary token]

Oh

added it

[remote ember]

Now substitute your answer for y into equation one to solve for x

[weary token]

Ok

I do not know how to make the equation

[remote ember]

I will show you

[weary token]

Pleasee

[remote ember]

Equation one is : x + 2y = 3

Now, replace the y for a -1

So, you'll get : x+ 2(-1) = 3

[weary token]

Oh

so x = 5

[remote ember]

Yes

[weary token]

Ahaa

[remote ember]

Now you have x and y, put them into the third equation

[weary token]

I got 5 for the thrid equation

but its not on the choices

[remote ember]

The third equation should look like this 5 - 2(-1) = ....

[weary token]

ohh

i wrote 3 instead of 5

So the final answer is 7

correct?

[remote ember]

Yes

If you want to look more into this, the topic is called simultaneous equations or systems of equations

[weary token]

Ok, I'll do that later

Thanks tho

[remote ember]

No problem

[weary token]

.close

[devout snow]

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[onyx grove]

How is the solution integrating something like that?

Surely you would have to do some substition or something

[celest tide]

yea, you can sub 2 - 3x = t

[devout snow]

@onyx grove Has your question been resolved?

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[devout snow]

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[lime dome]

does this proof look good?

gotta fix the syntax

Maybe I should an include a line which says since n is in Z, and n is in that set, it shows Z = the set

updated ^

how does this look

[static pumice]

The last statement is not correct: for instance let n = 2. It is true that n is in {2,3,4} and in {2,3,4,5}. But that does not imply that {2,3,4}={2,3,4,5}

[lime dome]

right

little unsure how to finish this last statement, gonna take a few minutes to think about it

[static pumice]

You’re on the right track. First you show {12a +25b | a,b in Z} is a subset of Z. But to show those sets are equal you also need to show that Z is a subset of {12a +25b | a,b in Z}

Also use $\subseteq$ instead of $\subset$

[woven radish]

Intrer

[static pumice]

If you want to get fancy, you can invoke the axiom of extensionality https://en.wikipedia.org/wiki/Axiom_of_extensionality

[lime dome]

I did show {12a +25b | a,b in Z} is a subset of Z. What am I missing to show Z is a subset of {12a +25b | a,b in Z}

[static pumice]

Can you say that every member of Z is in 12a +25b | a,b in Z

In other words, for any n in Z, can n be expressed as 12 a + 25b for some a,b

Which you already kinda did

[devout snow]

@lime dome Has your question been resolved?

[devout snow]

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[steady stratus]

not rlly sure how to solve this one

[steady stratus]

Specifically how to check the boundary, I found the only critical point to be (1, 1/3)

so they say this... I understand why the volume would equal 0 if x or y = 0, but how do I know that the volume equals zero for 1-1/3x-y=0?

[devout snow]

@steady stratus Has your question been resolved?

[steady stratus]

<@&286206848099549185>

[devout snow]

@steady stratus Has your question been resolved?

[devout snow]

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[devout snow]

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[barren frigate]

Can someone please help me

[barren frigate]

do we calculate height first?

[versed vault]

is this not just ambigous case

[versed vault]

not really needed

[barren frigate]

I dont know Im trying to answer it

[barren frigate]

ohh

so what do I do?

[versed vault]

So this is ur triangle right

[barren frigate]

yes

yep

[versed vault]

so set up ur law of sines

rule

[barren frigate]

nani

[versed vault]

do u know law of sines

[barren frigate]

yes

[versed vault]

so set it up

[barren frigate]

theres 3 tho

formulas

[versed vault]

dont memorize the formula goober

memorize what it means

[barren frigate]

😨

[versed vault]

its just a ratio

not a formula

[barren frigate]

39/10

B/14

C/c

[versed vault]

well sin39/10

=