help-23

[dry sigil]

that manner

[bitter oyster]

oh like that yeah

[dry sigil]

yeah

[lean otter]

then solve for x

there's only one logical solution

[dry sigil]

^

[bitter oyster]

can you maybe explain to be sure

[lean otter]

you set 5x-28 = 0

5x = 28

x = 5.6

[bitter oyster]

oh ok

that makes sense

[lean otter]

do you understand

[bitter oyster]

thanks guys

[lean otter]

ok

good

[bitter oyster]

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[velvet kettle]

how do you do this q?

[dry sigil]

there's a neat trig identity

[lean otter]

is this the MAT?

[velvet kettle]

no

[dry sigil]

might be a cambridge entrance exam

[velvet kettle]

no

lol yeah but neat trig identity hmm

[lean otter]

Yeah, this appeared on the Oxford exam

[dry sigil]

ah

[velvet kettle]

hint pls

[vapid anvil]

What is sin^2(0°)

[velvet kettle]

0\

[vapid anvil]

Hmmm

Make pairs

sin^2(1°) = cos^2(89°)

sin^2(x) + cos^2(x) = 1

[velvet kettle]

wait lmao i never knew this

[vapid anvil]

sin^2(90 - x) = cos^2(x)

You didnt know this?

[velvet kettle]

oh wait no nvm lmao

yeah ik it

so is the answer 50

[vapid anvil]

Idk

You have to figure out the total number of

pairs

Let me see

[velvet kettle]

should be 50

1-cos^2(0) cancels with sin^2(90) so = 1 and 50 pairs

although i did it like really rushed may be wrong

45 pairs fk

actually i think 45 + sin^2(45)

wait no just 45

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[dawn carbon]

Calculate the probability of winning: Draw 2 cards from a standard deck of cards without replacement. You win if you draw a face card, then a number card in that order. A deck of cards has 12 face cards, 40 number cards and a total of 52 cards. Show as a percent to 1 decimal place with no percent sign.

[dawn carbon]

i'm pretty sure the probability for the face card would be 12/52, and then the probability of the number card would be 40/51, but i'm not sure what to do with the numbers now

[marsh walrus]

you probably want to find the probability in this way

find the total number of ways to win, and divide it by the total number of ways to play

whats the total number of ways to play?

like, how many different ways can you draw 2 cards

[dawn carbon]

2,652?

[marsh walrus]

yea

how many ways to win?

to me i think you can calculate in a similar way

[dawn carbon]

well like, it has to be a face card first then a number card

[marsh walrus]

yea

[dawn carbon]

i'm not really sure how many ways there are to win tbh

[marsh walrus]

so how many ways to draw the face card first

12 face cards, right?

[dawn carbon]

yeah

and there's 40 number cards

but just saying it's 52 ways wouldnt work right?

cause that's not taking into account drawing a face card first, then a number card in that specific order

[marsh walrus]

no, its not 52 ways

i usually think about it like a tree

you start at some original node

then you have 12 ways you can branch for face cards

at the end of each of those branches, is another 40 branches

one for each face card

[dawn carbon]

so would it be 40 x 12?

[marsh walrus]

yea

you can think about this another way, like

say you were curious about just uhh

the king of spades, and the 4 of spades

thats 1/52 * 1/51, right?

the probability

then, you can total up the king of spades and some other number card

so king of spades, 2 of spades, king of spades, 3 of spades, ....

thats 1/52 * 1/51 + 1/52 * 1/51 + ....

well idk how much sense this will make

[marsh walrus]

does it make at least a little sense lol

[dawn carbon]

i think so

so the final anwser would be 480/2652 = 18.0%?

[marsh walrus]

its not exactly 18

but yea

[dawn carbon]

yeah but the question just needs to be 1 decimal place

thanks!

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[solemn plover]

is 90 the right answer here

[hard crest]

show work

[safe radish]

@solemn plover Has your question been resolved?

[solemn plover]

7p4/2!2!

[hard crest]

you're either overcounting or undercounting... not all codes have duplicate letters in them

eg you're counting only a quarter for the code COEF

[solemn plover]

but can’t i just pick 4 from 7 and divide by duplicates

wait nvm

so do i get all cases

with no dupes

?

[hard crest]

you will have to consider cases here

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[lean otter]

I'm trying to figure out how much i could buy using 8.15 mil on just one thing, such as:
Pizzas
Cars
Houses
Refrigerators
and Washers

[obtuse plover]

What’s the price of each commodity

Unit price

[lean otter]

Pizza: $3.95
Car: 31,903
House: 486,800
Fridge: 500
Washer: 700
And like i said, all are bought with 8.15 mil.
Each

[stuck crest]

just do 8 150 000/(unit price) then always round down

[lean otter]

Like divide it?

[stuck crest]

yeah

[lean otter]

Pizza: 20,632, 091?

[stuck crest]

wait

2,063,291

sure u typed in a calc right?

[lean otter]

Yes, is that allowed?

[stuck crest]

yeah

8150000/3.95

[stuck crest]

try it again

[stuck crest]

yeah then just round down

= 2 063 291 pizzas

then u can do the same w the others

[lean otter]

Thanks

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[gritty hamlet]

ABCDE iss a regular pentagon how can I find the angle x

[ebon mesa]

try connecting EF

dur bekle duzgun cizip atayim

[gritty hamlet]

then

tamam bekliyorm

[ebon mesa]

uhhhhhh

trigonometrik cozum bulabildim

besgende altin oranlari biliyor musun

iki kenari ve bir acisi bilinen ucgen olusuyor ordan diger acilar da cikar

[gritty hamlet]

cok hakim deilim

[ebon mesa]

biraz daha bakicam geometrik cozum icin

[gritty hamlet]

tesekkurler cozumun icin cevap tam olarak ne oluyor :o

[ebon mesa]

soyluyorum hemen

96°

👍

[gritty hamlet]

👍

cozume kendim bakarim sagol

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[ebon mesa]

bulursam ozelden yazarim

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[delicate shell]

If I have 3 equations where equation 1 and 2 can be combined to give equation 3, does this form a dependent system?

[delicate shell]

im a bit confused on the idea of linear combinations and dependent systems

[lilac patio]

that just means 3 is a linear combination of 1 and 2

by dependent system do you mean linearly dependent?

[delicate shell]

as in the system has infinite solutions

[delicate shell]

oh. but if ( say equation x, y,z) x+y=z, then couldn't be it re-arranged to show each equation is a linear combination of the other

[lilac patio]

no cause its $xv_1+yv_2=zv_3$

[flat frigate]

Arctic

[lilac patio]

we are working with vectors

i guess if the vectors are one dimensional then they are always linear combinations

[delicate shell]

i dont think i'm far in enough because there's no mention of vectors

this was the definition I saw

[lilac patio]

what class are you taking?

[delicate shell]

i'm self-studying the intermediate algebra book, AOPS

[lilac patio]

algebra

not sure why this vocabulary is being used

needless to say i think its irrelevant

this is linear algebra vocabulary

i guess you are doing systems of equations?

[delicate shell]

yeah. i was expecting to just glance over and move on to other stuff, until I saw that

now I'm imprisoned, lol. i just can't move on until I get this right or I wont be able to sleep at night 💀

[lilac patio]

yeah i dont think its important

i never saw anything like that doing algebra

[delicate shell]

like say the 3 equations:

x+y=3
2x+3y=2
3x+4y=5

do they count as linear combinations of one another?

[lilac patio]

does your book have a definition for linear combination

[lilac patio]

alright its the same as linear combinations in linear algebra

[lilac patio]

i dont think so because the right column

3,2,5

you can get the same x and y values but 3,2,5 cannot be equal at the same time

[delicate shell]

oh

[hard crest]

I wouldn't normally call equations linearly dependent but whatever
yesh those are linearly dependent, the system is not fully constrained

[delicate shell]

perhaps I just don't know enough to make sense of this. I was thinking that it's enough that you can add/subtract each equation with another to get the other

[hard crest]

x=7, y=-4 satisfies all three...

[lilac patio]

oops

i was eyeing it

[delicate shell]

so then, that would make the system dependent right? that part is what I'm confused about.

[hard crest]

yeah since one of the equations can be removed and it'll still have a unique solution

[delicate shell]

but apparently online a dependent set has infinite solutions. this is what is messing with my head

am I just misinformed

[lilac patio]

it can have infinite solutions

not always

[delicate shell]

ahh

so dependent doesn't necessarily mean infinite

[hard crest]

if you imagine a third variable z (to go with the three eqns) then that could have infinitely many values

dependent system just means that the number of equations is misleading

[delicate shell]

so if I'm understanding correctly, dependent systems don't gurantee infinite solutions right?

[hard crest]

this is a dependent system:
x = 1
2x = 2
it only has one soln

if I told you that I had a system of 3 library equations and 3 variables, you wouldn't know how many solutions it had unless I also told you it was an independent system

every independent equation cuts down the dimensionality by 1 (plane -> line -> point)

dependent systems can be overconstrained or underconstrained depending on how many variables you start with

[delicate shell]

gotcha. thank you both for going through the trouble! i appreciate it

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[weary cliff]

Not sure where to start on this problem. I think I need to find the bounds of r to set up the integral, but I'm not sure how to find that.

But I know that mass is the triple integral of p * r dz dr dθ

[plucky elk]

your theta limits should be easy

do you know the maximum r can be?

[weary cliff]

I'm thinking r's maximum would be the cube root of 5

[plucky elk]

yes

0 < r < cuberoot(5)

what about z limits?

[weary cliff]

z is given so it would be lower bounded by r^3 and upper bounded by 5

[plucky elk]

i think you can write the triple integral now

[weary cliff]

ya i think so

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[brazen helm]

if f(x) is a polynomial is there a formula for the sum $\sum^n_{k=1}f(k)$

[flat frigate]

bigpufik

[plucky elk]

You sum the coefficients independently

ax+b + cx + d = (a+c)x + (b+d)

[split ether]

If you know the closed form of $\sum_{i=1}^li^n$ for any natural $n$ then shouldn't be a problem

[flat frigate]

alonelybean

[split ether]

What's the original question btw

[brazen helm]

nothing im just curious

I like asking myself questions, and writing small articles about them

writing one right now about function sums as integrals:D

so just asking

but thank you

[split ether]

Well, the $j$-th coefficient of $\sum_{k=1}^nf(k)$ should be just $\sum_{i=1}^ni^j$

[flat frigate]

alonelybean

[brazen helm]

Yeah okay, thank you :D

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[half gate]

Let $f$ be a non decreasing function on the interval $[0.1]$. Prove that for any $\alpha \in (0,1)$, $\displaystyle \alpha \int_{0}^{1}f(x) \mathrm{d}x \leq \int_{0}^{\alpha} f(x) \mathrm{d}x$

[flat frigate]

sotpau

[half gate]

Can I get a hint?

I tried breaking the LHS integral but nothing of value came up

The fact that it is non decreasing is probably key here, but I'm not sure how to use it

[plucky elk]

!show

[safe radish]

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

[half gate]

I mean, nothing of value came up. I said what I've tried. I got $\displaystyle (\alpha -1)\int_{0}^{\alpha} f(x) dx + \int_{\alpha}^{1} f(x) dx \leq 0$

[flat frigate]

sotpau

[plucky elk]

are you trying to work backwards?

[half gate]

yeah

it shouldn't be done this way though obviously

I tried to maybe morph it into some other known inequalities but I don't know any that have different limits

[plucky elk]

try breaking $\int_0^1 f(x) dx$ into a sum of two integrals

[flat frigate]

rie.mann

[plucky elk]

Say on [0, alpha], [alpha, 1]. then bound the second integral by the max of f(x) on that interval which you can find using f(x) being nondecreasing

[half gate]

oh so that's what "non decreasing" was for, I did it ty

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[dry sigil]

<@&268886789983436800> this user was posting nsfw content

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[steep pewter]

how tho?

[hazy elbow]

GIF i guess

[steep pewter]

he said "eureke"; its nsfw apparently

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[hazy elbow]

what have you tried?

[cobalt acorn]

No clue how to start

[hazy elbow]

do you know general formula of nth term of an AP?

[cobalt acorn]

The general term formula?

tn = a+(n-1)d

[hazy elbow]

Yes

so 7th term of an ap would be?

[cobalt acorn]

23

[hazy elbow]

and the formula?

t7=a+(7-1)d

[karmic garden]

arithmetic and geometric have a methodical formula you can follow

[cobalt acorn]

Yeah

[hazy elbow]

similarly you

you'll get an equation

with t11

so you have set of pair of linear equations in two variable

which can be solved for a unique solution

and then apply t28=a+(28-1)d

[cobalt acorn]

Ok I'll try it

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@cobalt acorn Has your question been resolved?

[safe radish]

@cobalt acorn Has your question been resolved?

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[tall bough]

test

[tall bough]

.open

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[heavy veldt]

Can someone help me understand this problem? I am confused on multiple points. First the question asks “Find the other solution…”, is this to be interpreted as there is a total of two solutions for the equation? These being x=r and the one represented in terms of our constants? Secondly I am confused about our solution in terms of p, q, and r. If the solution is s=p+q-r then our “r” term refers to a root and not the constant “r” term in the two binomials on the right side of our equation? So when the problem asks to find a solution in terms of “r”, that is to be assumed a root? I apologize for any vagueness if at all found and I will specify if need be.

[proper crypt]

@lean otter

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[proper crypt]

LOL

you are comparing two polynomials

one of them has root r

and if they are equal, the sum of roots is the same

well

wait im also a bit confused here let me think

x^2 - (p + q)x + pq = (r - q)(r - q)

[peak estuary]

if you were to subtract the constant term onto the other side, the cofficient of x doesnt change

[proper crypt]

oh yeah

you have x = r as one root

[proper crypt]

if the other is s, then it satisfies this equation

[proper crypt]

just saw that, yes thats what i was gonna say

so the sum of roots is still p+q

thanks

[safe radish]

@heavy veldt Has your question been resolved?

[heavy veldt]

What I believe was tricking me up was having a constant term in r and still representing the sum with r+s. Thanks for quick response.

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[glacial mesa]

Are you allowed to mess around with math like I did on this picture, or does it make it inaccurate?

[plucky elk]

holy shit that's small

[glacial mesa]

I’m sorry

But you can put it into pdf

For me it fits half the screen

Cuz phone

On image pdf it is big

I love how there are many answers to 1 + 1

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@glacial mesa Has your question been resolved?

[safe radish]

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[lean otter]

would 1/2f(x) imply horizontal shrinkage?

[zinc token]

let's suppose at x, f(x) outputs a

what does 1/2 f(x) output then

[lean otter]

a/2?

[zinc token]

great

now imagine that for all points

[lean otter]

x + y cords?

[zinc token]

x, y pairs yea

[lean otter]

like both of them

then it would look like this

if we are halving cords

yeah?

[zinc token]

yeah

seems right

[lean otter]

thank you

also

just a question for the future so i dont have to ask later

would 2f(x) imply streching then?

on both cordinates

[zinc token]

yep

[lean otter]

thank you

[zinc token]

[lean otter]

.close

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[high barn]

I'm trying to solve the following problem and I'm confused which of one of these methods is correct:
"Consider a standard 52-card deck. Find the total number of ways of drawing a flush (all 5 cards have the same suit)."

[high barn]

I think it would be simply 13C5 * 4 (bc there are 4 suits)?

[quasi bison]

do we maybe wanna exclude straight-flushes in accordance with poker rules?

which would make this (13C5 - 9) * 4

but if we consider straight-flushes as a type of flush, and not as the highest ranking hand in the game, then yes, 13C5 * 4 is correct.

[high barn]

ok gotcha

also what if I wanted to draw a two-pair (2 cards with one denomination, another 2 cards with a different denomination, and a fifth card with a third denomination)?

in that case, would it be just 13C2 * 11C2 * 9C1?

[quasi bison]

no

you choose which ranks the pairs are with 13C2 but then you choose which suits the pairs are going to involve with 4C2, and you do that twice (since you have two pairs), and then you choose one of the remaining 44 cards as your kicker

[high barn]

ohh so like (13C2 * 4C2 * 4C2) * (11C2 * 4C2 * 4C2) * (9C1 * 4C1)?

[quasi bison]

no

task failed successfully

[high barn]

uhh, not sure what I'm doing wrong

[quasi bison]

overthinking what i said

[high barn]

sorry I'm not really good at counting problems lol

[quasi bison]

again, your count will be:
13C2 for the denominations of both pairs (you are picking both of them at once, no need for any of that 11 shit)
times 4C2 for the suits of the first pair
times 4C2 for the suits of the second pair
times 44 for the last card in the hand

[high barn]

where does that 44 come again? is it bc we subtract 8 from 52 bc of the two 4C2 selections?

[quasi bison]

is it bc we subtrace 8 from 52

yes

bc of the two 4C2 selections?
no

44 is the number of cards in the deck that don't belong to either of the two denominations for the pairs.

[high barn]

ohh I see now, ok cool

ok last 2 quick questions

[high barn]

so if I were to select four-of-a-kind (4 cards have the same denomination in a 5-card hand)

would it be like 13C4?

or 13C1 (select a rank) * 4C3?

[quasi bison]

13 for selecting the rank, times 48 for the fifth card.

[high barn]

oh wow, ok

look simple but idk why I often confuse them so much lol

ok last question:

Find the number of length-7 sequences (so order matters) from the numerical digits {0,1,2,3,4,5,6,7,8,9} such that the sequence uses exactly 4 different digits.

umm, not sure where exactly to start on this one

[quasi bison]

looks ugly

[high barn]

is it something like (10C1 * (9C3) * 7P7) + (10C2 * 8C2 * 7P7) + (10C3 * 7 * 7P7)?

[quasi bison]

can't think of it off the top of my head

[safe radish]

@high barn Has your question been resolved?

[high barn]

if anyone could help me out getting started with this, it would be helpful!

[safe radish]

@high barn Has your question been resolved?

[delicate bobcat]

Maybe try analyzing it as a single set of four digits.

How many different sequences can you make with a specific set of 4, then determine how many sets of 4 there are

That would be my intuition to try and solve it

[safe radish]

@high barn Has your question been resolved?

[proper crypt]

this is probably inclusion-exclusion

first pick the 4 digits, so 10C4

and then you want sequences of length 7, with those 4 digits, and each digits is used at least once

thats the same as the number of surjective functions from {1, 2, 3, 4, 5, 6, 7} to {a, b, c, d} where a, b, c, d are your digits, so you can also just use it as a known result tho

the answer is ||1764000||. try it and check that you got the same answer

ping me if you need another explanation

@high barn

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[mild brook]

Help

[mild brook]

question 17

The book had a complicated working out

All i did was substitute d as 2r

is that still correct?

[split ether]

Yup, looks like you are already given V = 4/3 pir^3, so using it should be good

[cunning shard]

r=d/2

put that in and you good

[mild brook]

eh

that makes it longer

[mild brook]

so is my working out right?

[split ether]

Yes

[mild brook]

ah alright

thanks for the affirmation dude

[cunning shard]

$V= \frac{4}{3} \pi (\frac{d}{2})^3= \frac{4}{3} \pi \frac{d^3}{4}$

[mild brook]

?

[split ether]

No spaces right to the first $

[cunning shard]

which is what you wanted to show

[split ether]

You meant to say (d/2)^3 and d^3/8 ?

[mild brook]

wot

[split ether]

Why are you substituting r = d/3

[flat frigate]

dotdoc.

[cunning shard]

cancel off the 4 and you are done

[split ether]

Now it's right, except 2^3 is 8

[cunning shard]

typo

Obviously, radius has to be half of diameter

[split ether]

Yeah

[cunning shard]

I don’t see it

[mild brook]

i mean

[cunning shard]

but yeah both works

[mild brook]

ah

thanks guys

.close

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[fallen bluff]

, w x^3 + 2x^2 - 23x - 60 = (x^2 - 2x - 15) (ax + b)

[flat frigate]

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[fallen bluff]

uh

, w x^3 + 2x^2 - 23x - 60 = (x^2 - 2x - 15) (ax + b)

[flat frigate]

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[split ether]

You might as well use #bots

[fallen bluff]

my bad

but like

how do i get a and b

[split ether]

There is a quick way to get to those answers

[split ether]

When you multiply (x^2 - 2x - 15) by (ax + b)

[fallen bluff]

what does it equal

[split ether]

What will be the coefficient of x^3?

[fallen bluff]

ax^3?

[split ether]

Yup

And we want it to be x^3

[fallen bluff]

1

[split ether]

So we say a = 1

[fallen bluff]

alright

[split ether]

Similarly, the free coefficient will be -15b

Which we want to be -60

Meaning b = 4

[fallen bluff]

mhm

[split ether]

Though you would still need to verify whether (x^2 - 2x - 15)(x + 4) is actually equal to the given polynomial

[fallen bluff]

so how we verify it

[split ether]

Just expand the brackets, multiply term by term

And see if what we get is indeed x^3 + 2x^2 - 23x - 60

[fallen bluff]

so pretty much equations?

wait nvm

[split ether]

(As a side note, we do the verification because we haven't consider other terms, it might as well have been x^2 - 3x - 15 in there and we would still get to the conclusion (a, b) = (1, 4) despite it then getting us something other than x^3 + 2x^2 - 23x - 60)

[fallen bluff]

@split ether still reading side notes and i am still confused

is there a formula for this?

[split ether]

No, just plain logic

[fallen bluff]

is it alright if u solve it

[split ether]

Okay, so, when you multiply (x^2 - 2x - 15)(x + 4), you get x^3 + 2x^2 - 23x - 60, right?

[fallen bluff]

got it

whats next

[split ether]

But if we were given something else, like x^2 - 3x - 15

Then the multiplication would get us something else

[fallen bluff]

cant i be

[split ether]

Meaning there is no solution

[fallen bluff]

(x-5)(x+3)

[split ether]

Although, assuming we didn't do the verification, we would still think that a = 1 and b = 4 is a solution

[split ether]

Sure, whatever, but my point is that it's gonna give you something other than x^3 + 2x^2 - 23x - 60 in that case

[split ether]

This is why we should actually verify that a = 1 and b = 4 is a solution

[fallen bluff]

oh alright

[split ether]

The fundamental issue with our reasoning was that we didn't just say that a = 1 and b = 4 is a solution to the problem

Instead we showed that if there is a solution, it must be a = 1 and b = 4

But we aren't sure whether a solution exist, so we check ourselves

[fallen bluff]

by applying 1 and 4 into the equation?

[split ether]

Yup

[fallen bluff]

oh alrigh then

how bout for this

how we gonna find c

[split ether]

When we multiply (ax^2 + bx + c) by (x + 3), the free coefficient will be 3c and we want it to be -27

So we say c = -9

Similarly we get a = 1

To find b we could just expand the brackets and set the coefficients equal, this way we won't have to do another verification

[fallen bluff]

oh alright

wouldnt it just be 0?

or na

[split ether]

,w expand (x + 3)(x^2 - 9)

[flat frigate]

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[split ether]

Yup

Works out

[fallen bluff]

oh wow

,w expand x^3+2x^2-23x-60

wait wrong one

,w expand (x-5)(x+3)(x+4)

[flat frigate]

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[fallen bluff]

.close

[safe radish]

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[icy frigate]

can someone tell me where did the x go?

[glass carbon]

divide the top and the bottom by x

x/(x^2 + x) = 1/(x+1)

(for x non being zero)

[safe radish]

@icy frigate Has your question been resolved?

[safe radish]

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[peak wraith]

Understand the qn but don't understand what it wants

[peak wraith]

As long as u can find length SX which is longer than XR is the answer proven alr or is there more?

[pulsar pecan]

No

[pulsar pecan]

It requires you to show that X is the point on SQ closest to R

[peak wraith]

how would i do it?

i have length SX alreadly

same for XQ

[pulsar pecan]

You are asked to show that any point on SQ is farther away from R than X is

S is farther away from R than X is, and Q is too

[pulsar pecan]

If you don’t understand what the question wants, you don’t understand the question

[peak wraith]

yea i dont understand what the qn wants

[pulsar pecan]

It requires you to show that X is the point on SQ closest to R

[peak wraith]

but isint x alr on point SQ?

[pulsar pecan]

Yes, it is, you have to show that it is the closest point to R compared to the other points on SQ

[peak wraith]

if compared to point XQ and XS?

[pulsar pecan]

Yes, and also all other points on the line

[peak wraith]

isint XQ compared to XR

XQ is 18cm

[pulsar pecan]

Another way to say this is that you have to prove that XR is perpendicular to SQ

[peak wraith]

hmm

[pulsar pecan]

I meant QR vs XR, not XQ vs XR

[peak wraith]

oh

huh

Qr is 30cm

whereas XR is 24cm

doesnt it prove it for you alr

[pulsar pecan]

No, that’s like saying all prime numbers are even because 2 is even

Sure, it would be better to ask in #math-discussion

[peak wraith]

er im preety avg myself but er get help ig like how i am

ok erm i can prove triangle XR is perpendicular

since 1/2 x 40 x 30=600

area of triangle =1/2x base x height

SQ=32+18

[pulsar pecan]

you were supposed to use pythagorean theorem

[peak wraith]

1/2 x 24 x50=600

oh

[peak wraith]

yea i used it to find angle XQ

length*

[lean otter]

Alr! Thanks.

[peak wraith]

since ration is 16:9 i just went to find 1 unit and then 16 units

ratio

[pulsar pecan]

I’m not sure I understand how you used this to prove them perpendicular

[peak wraith]

SQR is a triangle so XR would be the height and SQ would be the base

i found XQ using phthagreon theorem

and then found SX with XQ

can i not prove it that way or?

[pulsar pecan]

When you use the pythagorean theorem, you must assume that it has a right angle

[peak wraith]

i assumed that X is a right angle

[pulsar pecan]

you are asked to prove that there is a right angle at X

[peak wraith]

ohh

i accl dk how to prove it if i cant use it

wait can u make a imaginary line from X to P and since they are cut equally in half XP =XR and then do 1/2 x base x 24=600 use algebra to find the base since area of rectangle is 40x30 1/2 x 40 x30=600 then once u have length SQ use the ratio to find length XQ and then u can prove X is a right angle

using phytygreaon theorem

[pulsar pecan]

by using the converse of pythagorean theorem

[peak wraith]

mine works?

[pulsar pecan]

not equally in half

[peak wraith]

oh

[pulsar pecan]

XP and XR are not equal

[peak wraith]

but they said a straight path cuts through the grass patch joins S and Q

[pulsar pecan]

it’s like this

[peak wraith]

er i dont understand

[pulsar pecan]

the point is, XP may not equal XR

as above

[peak wraith]

oh

[pulsar pecan]

so the way you’d prove it is by finding either XQ or XS, then use the converse of the pythagorean theorem

and you’d be pretty much done

[peak wraith]

but can you find either one using phytygreaon theorem

[pulsar pecan]

no, but it’s given in the question

as a ratio

[peak wraith]

but wouldnt you need either length to find the ratio?

use*

[pulsar pecan]

you can find the length of the diagonal

[pulsar pecan]

the ratio is given

[peak wraith]

using 1/2 x base x24=600?

[pulsar pecan]

That would give the area, not the diagonal

[peak wraith]

no 600 is the area

[pulsar pecan]

To find the diagonal, you can use the fact that the grass patch is a rectangle

[peak wraith]

oh wait

u cant do 1/2 x 30 x 40

u cant right?

or can u ....

[pulsar pecan]

you can carry out that calculation, but it won’t give you the answer

[peak wraith]

why? it gives you the area of the triangle which is 600 so 1/2 x base length x 24=600 and flip them over and you get base length=50 add the ratios 50 divided by it and then u can find the lengths? or

[safe radish]

@peak wraith Has your question been resolved?

[safe radish]

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[atomic jackal]

doubt

[atomic jackal]

i was looking at the solution of a question

this bit confuses me

if dy/dx is increasing that does not imply f(x) is increasing?

[quasi bison]

no

f itself is increasing when f' is positive

[atomic jackal]

so what does f' increasing imply

[olive fossil]

it means that f'', or the second derivative is positive

that is, the graph is concave upwards

[atomic jackal]

what does it imply for the function itself

[atomic jackal]

point of minima?

[olive fossil]

you mean minimal point? no, thats when f'(x)=0(and even that condition is not sufficient for minimal point)

[atomic jackal]

i meant minima as in local minima

[atomic jackal]

oh ok

[olive fossil]

yeah this is still a different thing

[quasi bison]

for f? it implies concave-up.

[atomic jackal]

hm

alright

makes sense

thank you both

[safe radish]

@atomic jackal Has your question been resolved?

[safe radish]

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[lusty spear]

can someone check whether I am right ?
Given 12 objects and two distinguishable bags how many possibilities are there to put 6 objects in each bag ?
My answer would be 2 * (12 C 6). Is that correct ?

[buoyant shadow]

no it doesn't make sense

[quasi bison]

nop

[lusty spear]

whats wrong ?

[twilit harness]

is ||2*(6!)|| the answer?

ok

[quasi bison]

why don't you tell us the logic behind your answer and we tell you where you fucked up

[lusty spear]

sure

so first I considered the case were both bags are the same

[quasi bison]

but why

[lusty spear]

then I thought it would be 12 c 6

[quasi bison]

the bags are distinguishable, as the problem says explicitly

it would be a different unrelated sitch if the bags were identical

[lusty spear]

because I thought I could simply count it as symmetric cases

[buoyant shadow]

that's also wrong

[quasi bison]

there's a red bag and a blue bag

[twilit harness]

oh wait || 4*6!|| ?

[lusty spear]

why is that ?

[lusty spear]

yes

[twilit harness]

ok i give up bye

[quasi bison]

each distribution of balls into bags is uniquely determined by what goes into the red bag

bc what doesn't go into the red bag goes into the blue bag

[lusty spear]

yes

thats why I thought 12 c 6

[quasi bison]

so the answer to your ORIGINAL problem is just 12C6,

but when you make the bags the same, it no longer is

[lusty spear]

wait

what is it then if the bags are identical ?

why is it not 12 c 6

[buoyant shadow]

it's too much

[lusty spear]

but what would it be instead

[buoyant shadow]

12 c 6/2

only half is unique

[lusty spear]

I see

[twilit harness]

oh wait is it just ||12C6||

[lusty spear]

thanks !

[twilit harness]

cuz the last 6 are fixed when you've filled the first bag

[lusty spear]

.close

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[lusty spear]

.reopen

[safe radish]

✅

[lusty spear]

oh now I see

ah nevermind

I thought there was this general formula

but it doesnt matter

.close

[safe radish]

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[safe radish]

@vale hollow Has your question been resolved?

[vale hollow]

<@&286206848099549185>

[plucky elk]

Correct

[high barn]

hey so I actually tried this and got 10C4 * 7P4 = 176,400. Is this right?

[plucky elk]

You need a new channel

[high barn]

oh sorry was asking a question previously but wasn't able to respond.

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[lean otter]

there are 3 red balls and 8 blue balls

[lean otter]

whats the probability your first ball was red given your 2nd ball was blue

is this a trick question?

oh it is

im an idiot

[quasi bison]

no it's an incomplete question

[lean otter]

it was on my quant OA

and im an idiot

[lean otter]

wdym

[quasi bison]

not told whether the balls are drawn with replacement or not

also might want to at least explicitly say you draw 2 balls

[lean otter]

w/o replacement

and you draw 2 balls

is it just 3/8?

@quasi bison

[quasi bison]

aeugh

no?

[rocky tulip]

list all the cases where the second ball is blue

[lean otter]

how do you solve it intuitively

[rocky tulip]

well, we have the fact that the second ball is blue, so we only have to care about the cases where the second ball is blue

so would you agree these are the cases?:

(1st = red, 2nd = blue) and (1st = blue, 2nd = blue)

[lean otter]

yes

so would you take the ratio

[rocky tulip]

so what would you say the probability of these cases are?

[lean otter]

idkt

[rocky tulip]

not quite

[lean otter]

if

1. (1st = red, 2nd = blue) and 2) (1st = blue, 2nd = blue)

Id say the take ratio of 1) / 1) + 2)

[rocky tulip]

yeah

I'd agree

so what are these probabilities?

[lean otter]

can i use bayes theorem at all?

or is it not usefull

[plucky elk]

probability tree is probably more appropriate

https://www.mathsisfun.com/data/probability-tree-diagrams.html

[lean otter]

1. 9 C 2 and 2) 9 C 3

?

[lean otter]

this is in the context of a timed 30 min exam so id like to have some sort of formulaik solution

[rocky tulip]

what?

9 choose 2?

[lean otter]

oh i meant those are the scenarios

[plucky elk]

it shouldn't take you 30 minutes to draw a probability tree

[rocky tulip]

there's probably a lot of questions on there

it would be more useful to give an average time per question

[plucky elk]

oh i thought he meant there was an avg of 1 question per 30 minutes

[lean otter]

it was an online quant trading assessment

with 20 q in 30 minutes

with those being the questitons

[plucky elk]

probability tree is how you learn how to do the problem

you don't need to draw it when doing actual problems

[lean otter]

ah

[plucky elk]

just for you to learn

[rocky tulip]

Once you do enough probability trees, your brain will do it automatically

that's the cool thing about brains

[lean otter]

so what would be the first branch? the probability the first ball is red?

[rocky tulip]

but thats the right idea

[rocky tulip]

yeah sure

[lean otter]

so P(1st is red| 2nd is blue)

= P(1st is red)*P(2nd is blue)

/

P(1st is red)

(3/11 x (3/11 x 8/10 + 8/11 x 7/10)) / 3/11

is this correct?

@rocky tulip

[rocky tulip]

I dont think so

so we are interested in when the second one is blue

so red blue and blue blue

so we get

(3/11)*(8/10) and (8/11)*(7/10)

but we care about red blue

[rocky tulip]

so we use what you said here

(3/11)*(8/10) / ((3/11)*(8/10) + (8/11)*(7/10))

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

wait

which of my assumptions is wrong

[safe radish]

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[dense solar]

how is the max 6.35 when if u plug in f(-2) into the equation you would get -6.35 instead of pos

so how did they get a positive?

because -2^2/3 is -1.5874 and if u mult that by the -1 u would get 1.5874 which you then mult by -4 since -2-2 is -4, which is then -6.35 (rounded)

so im confused on how their answer is positive

[queen parcel]

,w calc -(2)^(2/3) (-4)

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Requested by shenny17#0

[queen parcel]

That's how

[dense solar]

where did the negative go?

and its not just -2^2/3 its actually -(-2^2/3)

so it would be a pos 2^2/3 times -4 anyways

@queen parcel u there?

[queen parcel]

,w calc - (-2)^(2/3) (-4)

[flat frigate]

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[queen parcel]

Think about squaring (-2) to get 4, and then taking the cbrt of 4

[dense solar]

wow alr thx

idk why my calc didnt show me that tho?

.close

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[lean otter]

!help

[plucky elk]

[last heath]

^

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@lean otter Has your question been resolved?

[queen parcel]

@lean otter Just ask dude

And don't DM me "can you help with math"

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[gray swallow]

Answer key says A but I don't understand how. Isnt the distance between two objects the hypotenuse?

<@&286206848099549185>

[solemn vault]

No, dude

The distance is always measured in perpendicular terms...

[gray swallow]

so like horizontal?

[solemn vault]

If you've to measure distance between two lines you'll measure it by drawing a perpendicular line between them and measure the perpendicular line... and that will be your distance....

[gray swallow]

could you show me?

like draw it out for me as an example?

[solemn vault]

And... how ever the hypotenuse is also the distance between two poles, they aren't the shortest... and what I assume fron your question is that your question asked the shortest distance rather than any distance...

[gray swallow]

how do you know its asking for the shortest though?

[solemn vault]

Well that depends upon how you read it...

[gray swallow]

this is a question on the sat

i dont think its safe to assume like that tho

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[lean otter]

can smo check if dis question dat i js solved is right

[solemn vault]

How you got 61.3 in the third step?? Explain

[lean otter]

cus

4/3

each side

to get rid of 4/3

cus trying yo find

to*

r

but i feel like im incorrect

[lean thorn]

,rccw

[flat frigate]

Couldn't find an attached image in the last 10 messages.

Jul 13, 2023, 3:39 AM

[lean thorn]

,rccw

[flat frigate]

Couldn't find an attached image in the last 10 messages.

Jul 13, 2023, 3:39 AM

[lean thorn]

boo

[lean thorn]

,w (4/3)(pi)(4.42)^3

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[lean thorn]

yeah something wonky happened here.

To get rid of the 4/3, you need to multiply both sides by 3/4. So 735.6 * (3/4) = 551.7

not sure where you got the 61.3 from but yeah that's where the step went wrong

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[atomic jackal]

could someone explain these parts please

[pseudo scroll]

$(a \pm b)^2 = a^2 \pm 2ab + b^2$

[flat frigate]

neonperseus

[atomic jackal]

no

i meant the part where they introduced h

[pseudo scroll]

ah

x - 2 = h

[atomic jackal]

what about it in the second one

[pseudo scroll]

both use the same sub

[atomic jackal]

wouldnt it become x-4 in the second case then

[pseudo scroll]

$$(x^2 - 4)$$
$$h = x - 2 \implies x = h + 2$$
$$((h + 2)^2 - 4)$$

[flat frigate]

neonperseus

[pseudo scroll]

oh wait I think they're taking h --> 0^+

[atomic jackal]

but it shows (h-2)^2

[atomic jackal]

how did you figure that out

[pseudo scroll]

because if x approaches 2^- then x^2 - 4 has to be approaching 0^-
And this is only happening if h approaches 0^+, if it approaches 0^-, then the limit is approaching 0^+

Take a moment to digest that

[atomic jackal]

lost me at the second line

ill try reading it once more

[pseudo scroll]

(h - 2)^2

If h is approaching 0^- here then is it safe to say this equals 4^+

[atomic jackal]

ah okay

but how did we get that expression

(h-2)^2

[pseudo scroll]

Probably x = 2 - h

As h --> 0^+

[atomic jackal]

but then wouldnt we get (2-h)^2

[pseudo scroll]

same thing

[atomic jackal]

i get that since we're squaring it doesnt matter but then why not just write 2-h in the first place

[pseudo scroll]

looks nicer

[atomic jackal]

why create so much confusion

anyways

i think ive got it now

tysm for helping

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[harsh harbor]

Could anyone check if my answer is right?

[harsh harbor]

I have been unable to check this for 2 hours now so I would really appreciate if someone could let me know, as every website I have tried entering this into to plot has not worked

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@harsh harbor Has your question been resolved?

[harsh harbor]

<@&286206848099549185>

[harsh harbor]

Can I see your workings?

[final halo]

My workings?

[harsh harbor]

is that just my answer plotted?

[final halo]

Yes

[harsh harbor]

oh my bad sorry

what website is that?

[final halo]

Desmos

[harsh harbor]

Im think that means my answer is very wrong

I cannot tell where I have gone wrong tho

[final halo]

Your answer is very wrong indeed it seems

[harsh harbor]

Do you know how to do the question?

If so I could post my workings and see where I went wrong

[final halo]

Sure

[final halo]

You integrated from -π to π but the function isn't x+π on that whole interval

[harsh harbor]

I think I did that because it divided into each otehr

the 2π/2

[final halo]

You need to do $\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin(nx), \dd x$

[flat frigate]

ΣΑC

[harsh harbor]

oh it was odd too I forgot that

That changes Bn's formula

[safe radish]

@harsh harbor Has your question been resolved?

[harsh harbor]

I redid it with the correct formula but no idea what to do now

<@&286206848099549185>

[plucky elk]

your series shouldn't have a product of both sine and cosine. they should just be linear combinations of sines and cosines

maybe use product to sum formula

,tex .prod2sum

[flat frigate]

rie.mann

[harsh harbor]

does everything else seem to be correct?

[plucky elk]

no idea

[harsh harbor]

this is what the answer shold be btw

[plucky elk]

try this on your final expression

[harsh harbor]

do i multiply it out first

[plucky elk]

how else do you get a product if you don't multiply first

[plucky elk]

1st and 2nd rows here are what you should use

[harsh harbor]

I don’t see what I can do from here

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@harsh harbor Has your question been resolved?

[harsh harbor]

.close

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[flat perch]

What do they mean reduced to x

[flat perch]

Are they saying that the terms can be cancelled?

[glass carbon]

"simplified to x"

only

[flat perch]

Oh

[unique bison]

x=x.

[quasi bison]

good question, what the fuck do they mean??

[glass carbon]

e.g. using factorization or smth to get "x" as a final result

[flat perch]

Sorry

[quasi bison]

not an attack on you.

[flat perch]

Yeah I get it sorry

[quasi bison]

you have nothing to apologize for.

you presented the question exactly as it was given to you.

it is the question itself that is fucked up.

[unique bison]

Good

[flat perch]

Alr so the answer is false

Right

Cause u can’t cancel terms like that Lmaooo

That be amazing tho

Just getting to cancel terms

Whenever u like

Ty

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[lethal laurel]

Hello I thought I should use the compound interest formula A=P(1+r/n)^nt

[lethal laurel]

Is this the formula I should use for the problem?

[plucky elk]

yes

[lethal laurel]

P=$12,000, r=.11, n=4, t=7 is that correct?

Okay I got it. Thanks!

.close

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[crude vault]

i keep selecting (350,750) and ive tried (50,125) and several other points

i keep not getting any of these slopes

i also tried (225,500)

i wont let me select the points