help-19

[iron bear]

this is false btw

b must be even

[lofty sigil]

.reopen

[odd edge]

✅

[lofty sigil]

Wdym

[iron bear]

any number ending in 18 is not divisible by 4

you have to care about the 8

for instance, you cant just take a = 9, b = 9

which would be the largest possible pair if you ignore the condition on b

[lofty sigil]

If b = 1 that means a has to be 2,5,8

[iron bear]

b cannot be 1

this is disallowed by the divisibility by 4 condition

[lofty sigil]

Wait oh ●_● yea cus of divided by 3

[iron bear]

no

by 4

[lofty sigil]

No I mean cus after it's divided by 3

[iron bear]

divisibility by 3 does not prohibit b from being 1

20224218 is divisible by 3

but not by 4

[lofty sigil]

Anyways a + b has to be a odd number

Besides 1

[iron bear]

no, it has to be divisible by 3

[lofty sigil]

Yea so an odd number

[iron bear]

6 is divisible by 3

[lofty sigil]

Oh oops I forgot 5 and 7 and etc

[iron bear]

and not odd

[lofty sigil]

Yea ur right it had to be divisible by 3

A + b ÷ 3 has to have no remainder

So a + b has 3 , 6 , 9 , 12 , 15 and 18

[iron bear]

you should determine b first

[lofty sigil]

How tho

[iron bear]

b is constrained to be even by the divisibility by 4 condition

[lofty sigil]

Hm

A + b = divisible by 3 and ab8 has to be divisible by 4

Since 20224000 is divisible by 4

[iron bear]

just b8 is enough

do you know the divisibility by 4 test?

[lofty sigil]

Why? We don't know a value yet v:

[iron bear]

that is due to the divisibility by 4 test

[lofty sigil]

Oh the divisible by 4 rule

[iron bear]

last numbers must be divisible by 4

because 100 is divisible by 4

[lofty sigil]

Oh yea ur right then b8 is enough

[iron bear]

so b must be even

[lofty sigil]

So b = 2, 4 , 6 ,8

And a + b has to be divisible by 3

So a is odd

[iron bear]

no

[lofty sigil]

?

[iron bear]

again, multiples of 3 do not have to be odd

[lofty sigil]

Yea sorry

Since b = 2,4,6,8

Let's just take the biggest number

Making it 20224a88

[iron bear]

you should map out the largest choice of a for each b

[lofty sigil]

Yea and since a is less then 10

[iron bear]

taking the largest b is greedy, you are not guaranteed to succeed a priori

[lofty sigil]

9 or 8 won't work

So a = 7 b = 8

[lofty sigil]

I mean yea

[iron bear]

yes but also when b = 6, you can take a = 9

[lofty sigil]

But then 6 times 9 = 54

And 8 times 7 is 56

[iron bear]

6*9 is only less than 7*8 by 2

[lofty sigil]

It says maximum

[iron bear]

well yes

but my point is

taking the largest b is too greedy

you should be checking other b as well

this time it was less by 2

[lofty sigil]

I mean like if b = 6 largest number a can be is 9

[iron bear]

maybe next time it wont be

[lofty sigil]

If b = 4 then largest number a can be is less then 10

[iron bear]

2 is not a very big margin

[lofty sigil]

So doesn't matter what a is ab is less then 40

[iron bear]

well yes but now you are doing the actual checking

[lofty sigil]

It says maximum 😭

[iron bear]

which was my point

you should be doing this

well

without checking

how do you know

you only conclude after the fact that 56 is the largest

[lofty sigil]

No cus I knew if b is under 5 that means it's wrong

Since a isn't larger then 10

[iron bear]

okay but when b = 6, its not immediately clear

[lofty sigil]

So only have b = 6 or 8

[iron bear]

until you do the actual calculation

[lofty sigil]

It is if u just take b8 and plus everything together basically

[iron bear]

thats after you do a check

[lofty sigil]

Yea I alr check b = 6 in my head ●_●

[iron bear]

im saying you should not just greedily take b = 8, and not check anything else

[lofty sigil]

Cus for that means the max a can be is 9

So I know it's js 9 x 6 = 54

And 8 x 7 = 56

Not that hard to check

[iron bear]

thats not what it sounded like here

i know its not hard to check

but im saying that you ultimately need to check

[lofty sigil]

👍

Alright

Thanks for helping me

[iron bear]

👍

[odd edge]

@lofty sigil Has your question been resolved?

[odd edge]

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[steep mantle]

for the first its just the area of the triangle minus the area of the incircle

do you know how to find the incircle's radius of a triangle?

[odd edge]

@reef zinc Has your question been resolved?

[reef zinc]

No that's what I'm stuck on

[reef zinc]

<@&286206848099549185>

[signal yacht]

this one is such a rare formula lol

[warped glacier]

it's not it's very standard

[signal yacht]

i dont see it often 🤷‍♂️

then again i dont do geo often

[warped glacier]

yeah must be why

[signal yacht]

lol

[reef zinc]

am I just supposed to know this?

bec I've never taken this so I'd assume there's another way of doing this

[steep mantle]

no clue

[signal yacht]

probs

[spark lintel]

have you taken quadrilaterals

[reef zinc]

not good w topic names what exactly does that cover

what about the second question

[spark lintel]

shapes with 4 sides

[reef zinc]

the 3D trig one

[reef zinc]

yeah I know but like

what

their properties or

idk

[spark lintel]

you must know that

[reef zinc]

but yeah I'd assume so

[spark lintel]

all the r's are equal

in there

so you should be able to distinguish a quadrilateral from there and try to get the area yourself

[reef zinc]

but how do I find the radius

[spark lintel]

from the quadrilateral or the rule sent

[reef zinc]

okay and the second question?

[frosty badge]

!status

[odd edge]

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

[reef zinc]

1 💀

ik which angle I'm tryna find

but other than that idrk

[frosty badge]

my approach would be like so

find the height of the prism

and let's call the point of orthogonal projection of J onto GH J'

[viscid shadow]

🥲

[frosty badge]

we also want to find AJ'

[viscid shadow]

Help....!!!!!!

[frosty badge]

pythagorean theorem

since in that picture it looks like GJ' = J'H = 12/2

once you find all that

you can use tan to find the angle that you're looking for

tanx = O/A = JJ'/AJ'

lmk if u find trouble

[frosty badge]

open a help ticket in available

[odd edge]

@reef zinc Has your question been resolved?

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[frigid grove]

The altitudes of an acute-angled scalene triangle ABC with center H are BE and CF. Given that BC ∩ EF=P and the midpoint of BC is M, prove that AP ⟂ MH.

[frigid grove]

tried some angle chasing but didnt lead anywhere

<@&286206848099549185>

[odd edge]

@frigid grove Has your question been resolved?

[celest acorn]

this is not helpful, but usually when something involves midpoints, angle chase is not enough

[frigid grove]

using midpoint to say BM=FM=EM

if it went unused i’d say the samw

[celest acorn]

let us reframe the qn

you have a cirlce with center M

and two diamterically points B and C

and you take a line through M

and you take any point on it

H

you take interesections of BH and CH with the circle

and you call them F and E

[frigid grove]

how is that supposed to help

[celest acorn]

rephrasing the qns many times brings insights

do you know inversion?

[frigid grove]

kinda

[celest acorn]

hm

i have a feeling

that the intersection of MH and AP is the inversion of H around circle BFEC, i think that was a theorem (about this sort of construction), that might be hlpful

[frigid grove]

how to show that

or can you link if you know the theorem

[frigid grove]

also i think this is false

[celest acorn]

something about complete quadrilaterals

i just checked on ggb

it is true

[frigid grove]

my head hurts trying to find a solution

[celest acorn]

i think you need to cinstruct some more cyclicities

[frigid grove]

let AH∩PC=T

FHTB is cyclic

EHTC is cyclic

H is the intersection of angle bisectors in the triangle FTE

[celest acorn]

let MH cap AP = K

try showing a cyclicity involving K

[frigid grove]

PKHT is going to be cyclic

but i cant show

[celest acorn]

also there are some cool stuff if you invert this around cirlce BFEC, lots of nice stuff happens

[frigid grove]

can you show please

[celest acorn]

P goes to T

for instance

[odd edge]

@frigid grove Has your question been resolved?

[odd edge]

@frigid grove Has your question been resolved?

[frigid grove]

choose point G on FC such that MG//PE
HMDE is cyclic
ACP-F and CEM-G cevas are equal since 4 angles are equal
<EHK+<KAE=180, therefore AKFE is cyclic and MH is perpendicular to AP

i didnt mention the angle chasing

my head finally blew

.close

[odd edge]

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[opaque spindle]

Don't know how to approach this beyond what I have

[opaque spindle]

currently I have that C is (28; 24.248) and D is (7; 12.124)

but maybe there's a smarter way to do it without manually calculating the lines and algebraicly solving for x

.close

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[potent mulch]

Struggling with equations how do I get rid of a variable from one side without getting rid of the other/how do you know which side to put the variable on?

[ember oak]

it doesn't matter which side

you can start by dividing both sides by -2

[amber schooner]

i mean in this one the answer for y should be obvious because everything cancels here pretty much

[potent mulch]

okay well im still struggling to understand that given a different equation

[hot harbor]

you could try to get all of the things with y into one side and the rest into the other side

[potent mulch]

yeah how do you do that without getting rid of both variables

i cant take away the variable from jsut one side i have to do it to both

[hot harbor]

well, divide both sides by 4+3y or 4+y

[potent mulch]

i thought you can only divide by a whole number

[ember oak]

!original

[odd edge]

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.

[ember oak]

you can divide by any non-zero number

[amber schooner]

what

[hot harbor]

what ?

[amber schooner]

why would you ever do that

[hot harbor]

yeah, true, didn't think that much when i said it

what i wanted to say is usually, you try to get all of the variable on one side fo the equation

[potent mulch]

yeah i get that but i dont understand how you do that without getting rid of both

[ember oak]

Show the original

[potent mulch]

-2(4+3y)=-2(4+y)

[ember oak]

Both sides are multiplied by -2. So, if you divide both sides by -2, you can get rid of the -2, making the problem slightly easier

[hot harbor]

show all of the exercice, not just what you want us to help you on

that is showing the original

[ember oak]

How are you "getting rid of both"?

[potent mulch]

i geuss i more so dont understand how you get both variables on one side

[ember oak]

consider problem 2:
$$7+9d=7d+3$$

[clever fjord]

SWR

[ember oak]

subtract $7d$ from both sides:
$$7+9d-7d=7d+3-7d$$

[clever fjord]

SWR

[ember oak]

simplify, giving you
$$7+2d=3$$

[clever fjord]

SWR

[potent mulch]

what if both variables are the same

if 9d was 7d instead that wouldnt work

becuase then both variables are gone

[ember oak]

Then there is either no solution, or infinitely many solutions

For $2x+2=2x$, there are no solutions. For $2x+2=2(x+1)$, there are infinitely many solutions

[clever fjord]

SWR

[ember oak]

Just some examples

[potent mulch]

so how do you end up with the right answer for those?

[ember oak]

You say that there are either no solutions are infinitely many solutions

If you can end up with the same exact thing on both sides, then there are infinitely many solutions, otherwise, there are no solutions

Btw this is why i kept asking to see the original problem you were looking at. I figured this was what was bothering you

Next time, please, actually show the problem you are worried about

[potent mulch]

okay so for 2 (3 - h) -6 = -5h

everytime i try to solve i end up with nothing on the right side of the equal sign

[hot harbor]

5h-5h = ?

also, you added 5h to the right side, but added -5h to the left

[potent mulch]

would it transfer because its -5h?

[hot harbor]

well, if you want the transfer -5h to the other side, you have at add 5h so the -5h from the left "disappear", but you'll still have a +5h on the left side

[potent mulch]

okay so in trying to solve -2q -3 = -2(2q+1)

i ended up with -2q = 2q +1

asnwer sheet says i should get 0,5=q

[hot harbor]

how does -2(2q+1) = 2q-2 ??

-2×2q≠2q

[potent mulch]

i think this is just confusing me more lol

[hot harbor]

wait, let me grab something to write

[hot harbor]

is that better ?

[potent mulch]

i dont understand what you did in the second line

on the right side of the equal sign

[hot harbor]

you did it right there when you did 2(3-h) = 6-2h

left side first to second line

[potent mulch]

i geuss i more so struggle with understanding what to do first becuase it seems like bedmas has gone out the window

or pedmas which ever you prefer

[hot harbor]

well, did you understood what i did now ?

[potent mulch]

not entirely 🙁

[hot harbor]

what did you not understand ?

like which line

[potent mulch]

idk im jsut getting more confused you can close this channel so someone else can use it

[hot harbor]

well, if that's fine with you

[hot harbor]

you have to close it tho, do .close

[potent mulch]

.close

[odd edge]

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[odd edge]

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[mystic saffron]

hi

[mystic saffron]

i need help

can you guys helps me with this – 2 –3(-5)

[ember oak]

$-2-3(-5)=-2+(-3)(-5)$

[clever fjord]

SWR

[mystic saffron]

huh

wdym?

[mystic saffron]

?

[ember oak]

subtracting is the same as adding a negative

[mystic saffron]

ok

[ember oak]

I'll write it like this, instead
$$-2-3(-5)=-2+(-(3(-5)))$$

[clever fjord]

SWR

[mystic saffron]

so where did you get negative 3 from

[ember oak]

Because if you are subtracting $3(-5)$, that is the same thing as adding $-(3(-5))$

[clever fjord]

SWR

[mystic saffron]

ok

so would the answer be -13

[ember oak]

No, it would be positive

[mystic saffron]

how?

[feral halo]

can you show us your working out?

i wanna see how you would solve this question

[mystic saffron]

alr

so i do (-5) and chage it inti 5 becuase of its absolute value

[feral halo]

what

[mystic saffron]

*into

[feral halo]

why would u change -5 to 5

why are you talking about absolute values

[mystic saffron]

doesn't the parentces mean absolue vanue?

why am i spelling so much things wrong...

[feral halo]

absolute value is represented by
abs(x) or |x|

brackets just show you that it is a term

so your question was
-2-3(-5), right?

[mystic saffron]

yes

[feral halo]

-2 - 3(-5)

-2 - (-15)

-2 + 15

= 13

do u get that?

[mystic saffron]

the last part is comfusing

[feral halo]

which part

[mystic saffron]

-2 + 15

[feral halo]

you can rewrite that as 15 - 2

[mystic saffron]

oh

[feral halo]

(-2) + (15)
(15) + (-2)
15 - 2
13

[naive pond]

Nice

[mystic saffron]

i need help on another problem too

8 (-2) - 2 + (-5) – (1 – 7)2

[feral halo]

let's rewrite it to make it look better

8(-2) - 2 - 5 - 2(1-7)

is that correct?

i need to verify before i calculate

[mystic saffron]

the 2 is a exponent

[feral halo]

then use ^

[mystic saffron]

ok

[feral halo]

8(-2) - 2 - 5 - (1-7)^2?

[mystic saffron]

yup

[feral halo]

-16 - 2 - 5 - (1-7)^2

-23 - (1-7)^2

-23 - (-6)^2

-23 - 36

= -59

[mystic saffron]

i got -57?

[feral halo]

then show me your working out

i can't help if u just say that

[mystic saffron]

1-49=-48

-5-(-48)=43

43+-2=41

-16-41=-57

that is my work

[feral halo]

bro

start from the beginning

idk where u r

where does 1-49 come from

[mystic saffron]

7x7=49

[feral halo]

where did 7x7 come from

[mystic saffron]

the exponet

*exponent

[feral halo]

isn't it (1-7)^2?

= (-6)^2

= 36

[mystic saffron]

ohhhhh

alr

now i get it

<@&286206848099549185>

[odd edge]

@past otter Has your question been resolved?

[odd edge]

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[scenic lodge]

what am i supposed to do if it says don't completely solve?

[pastel orbit]

a system of equations is consistent if there's at least one solution, and inconsistent if it isn't

you can find the RREF of the matrix that represents the system of equations and see if any zero rows equal a nonzero constant in the last column

if you get something like [0 0 0 0 | 1], then you have an inconsistent system

because it corresponds to 0x_1 + 0x_2 + 0x_3 + 0x_4 = 1

which is impossible

[sand horizon]

||not OURfallenstars anymore¿||

[pastel orbit]

but if you have [0 0 0 0 | 0], it's okay

because 0x_1 + 0x_2 + 0x_3 + 0x_4 = 0 means there are infinitely many solutions to that equation

[scenic lodge]

so do ineed to just reduce it as much as possible

i have reduced the first 3 ros.

rows so far

[pastel orbit]

you need to get it into reduced row echelon form

[scenic lodge]

so that is 1 for each pivot and 0's above and below all 1's

or do pivots not need to be 1?

like i mean that pivots = 1

in each row

[scenic lodge]

you dont have to check my work but ilk i had to do all that? i couldn't have stopped earlier to asay its consistent ? like it seemed obvious earlier idk y

[odd edge]

@scenic lodge Has your question been resolved?

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[hallow kraken]

i need help with this

i don't know if i'm doing it wrong but the way i'm doing it would be practically impossible to answer the question without running out of time

[green elm]

how much time are you given? and do you have to do it by hand or with a computer or calculator?

[odd edge]

@hallow kraken Has your question been resolved?

[hallow kraken]

90 mins and the whole test is 100 marks

so a little less than a mark a min

for the first question how would i get the characteristic polynomial in under 3 mins

and the SVD in under 7 mins

[hallow kraken]

and by hand and i can use calculator

[odd edge]

@hallow kraken Has your question been resolved?

[hallow kraken]

no 😭

<@&286206848099549185>

[odd edge]

@hallow kraken Has your question been resolved?

[odd edge]

@hallow kraken Has your question been resolved?

[odd edge]

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[quasi inlet]

if I wanted to graph this transformed sin function, could I/should I split the transformed x-values into slices of pi. Something like n(pi)/8?

[cerulean vortex]

you already have points you can plot

(4,-1)
(9/2, -4)
and so on

[odd edge]

@quasi inlet Has your question been resolved?

[quasi inlet]

yeah I'm aware of that, might not but that useful here, but I was just wondering if it's actually possible. I want to scale it into evenly slices of pi/(some number).

I feel like I should be able to, but my brain is just farting. Especially considering the x values increment by 0.5

[cerulean vortex]

i don't get it

the points you have already will be multiplied by pi inside the sin function so it will give you slices of pi

[quasi inlet]

yeah maybe I'm just overthinking/over complicating it, anyway thanks for the help

.close

[odd edge]

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[compact nymph]

aye

[compact nymph]

could someone explain what happend here

how the formula of the circle would later be (x-(a+1)^2+y^2=(a+1)^2

[warped badger]

The guy just added (a+1)^2 on both sides

[compact nymph]

yh but i dont understand how did he get (x-(a+1)^2+y^2=(a+1)^2 as the circle formula

like where did the -2 and x go

oh i got it

(a+b)^2= a^2+2ab+b^2

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[cyan sky]

Find x.

[cyan sky]

( i have made some lines on that shape)

[mellow tundra]

The sum of all the exterior angles in a polygon is equal to 360 degrees.

[cyan sky]

I thought that only apply on quadrilateral

I guess that works

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[carmine sentinel]

Help please

[carmine sentinel]

I got 17.5 for the lower bound 18.5 for the upper bound

But the answers say 17.75 and 18.25

[carmine sentinel]

So they have a distance of 0.5 between them?

[austere panther]

however, so far as i can tell on the internet this terminology is not really standardized. I would consider this a poorly phrased question

[austere panther]

really seems like it

[carmine sentinel]

Ohh okay

The answers also say this 1 sec

[austere panther]

in practice i would always just use $\pm$

[clever fjord]

Eagle-From-Aut

[carmine sentinel]

but idk what that means-

[carmine sentinel]

alright

[austere panther]

i mean thats not really practical advice for the situation you found yourself in, but if you yourself are noting it down you can avoid ambiguity that way

[carmine sentinel]

Alright then

thank you

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[carmine sentinel]

.reopen

[odd edge]

✅

[carmine sentinel]

what does b mean?

[warped glacier]

can you take a picture of the map?

you basically need to compare distance on the map : distance in real life

then simplify that ratio by dividing by the same thing on both sides

[warped glacier]

which means you need to measure the distance AC with your ruler, methinks

[carmine sentinel]

okay

[carmine sentinel]

Here

[carmine sentinel]

1:200?

[warped glacier]

ah that's even easier

[warped glacier]

no, because cm and metres are not the same units

[carmine sentinel]

Wait yeah sorry

1:20000?

[warped glacier]

yes that's correct

[carmine sentinel]

is that all i need to write down?

[warped glacier]

yes, also note it's 1 mark

1 mark questions do not require you to write down any working

although you might want to write down a bit if it helps

[carmine sentinel]

okay thank you<3

also

[warped glacier]

yeah?

[carmine sentinel]

for h

b*

do i need to use a multiplier or something

can't remember-

[warped glacier]

yes, this is a working backwards problem

it's easiest to write an algebraic equation

(house price in 2017) * (1 + 7.7/100) = (house price in 2019)

x * 1.077 = house price in 2019

[carmine sentinel]

why did you add the 0 to 1.077?

[warped glacier]

you mean why did I add the 1 to 7.7/100?

[carmine sentinel]

is it the same way when calculating percentage increase we take off the 1?

no the 0-

[warped glacier]

well 1 + 7.7/100 = 1.077

you can move the decimal point of 7.7 two places to the left

then add 1

[carmine sentinel]

OHH

thank you<3

[warped glacier]

nw again!

[carmine sentinel]

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[stable needle]

What's a border hessian

[woven pier]

https://letmegooglethat.com/?q=What's+a+border+hessian

[stable needle]

Ok man

Thanks

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[tepid quarry]

$x^2 +4x + 4 = 0$ so our delta here = 4^2 - 4 * 1 * 4 right?

[clever fjord]

Simon James B
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[tepid quarry]

delta = 16 - 16 = 0 that means we have no real solutions

[sand horizon]

no need

[tepid quarry]

but there is a solution of -2 tho

?

[sand horizon]

whats x² + 4x +4 ?

[tepid quarry]

i know but i need to solve it using delta

[sand horizon]

ok

but delta = 0

means unique sol

-b/2a

[tepid quarry]

oh delta = 0 is one solution?

[sand horizon]

yes

[tepid quarry]

i do not understand why we did -b/2a

[sand horizon]

whats quadratic formula ?

[tepid quarry]

idk math terms well in english

it the (a+b)?

^2

[sand horizon]

the (-b +- sqrt(delta) ) / 2a

[tepid quarry]

oh is that what that is called

here we just say find x roots

wait what sign do we choose + or - if we have one sol

[sand horizon]

if delta = 0 what happen to quadratic formula ?

[tepid quarry]

it does not change it's value

ohh so it would be -b/2a yes i get it

ty

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[tepid quarry]

what is this symboll

[woven pier]

Which symbol?

[sand horizon]

a in ???

this ?

[tepid quarry]

yeaa

i never seen that Y shaped symbol before

and then i see a is part of that? it broke me into tears lol

[sand horizon]

i don't have any idea

looks like a little gamma that got long legs

oh

or upsilon

[tepid quarry]

i think it is upsilon

[sand horizon]

with the variant graphic

[tepid quarry]

but what meaning does it have in math

[sand horizon]

03D2

[tepid quarry]

?

[sand horizon]

thats the unicode

[tepid quarry]

oh

[sand horizon]

you can check ifw

but its not a math thing

i believe

[tepid quarry]

then what am i supposed to do? I need to find a with the condition that it has values in upsilon

like if it would bee Real numbers sure but upsilon?

[sand horizon]

maybe its a typo and they meant real

[vale vapor]

is that set not defined somewhere earlier?

[tepid quarry]

(a-1)x^2 + 7x + a = 0 Find a with values in upsilon knowing that x is -3

this is what i gotta do

[sand horizon]

yeah prob means R

if upsilon nowhere else

its probably a typo

[tepid quarry]

they never do typos tho

it's clear that the they are the same symbols

[rapid silo]

Could it be a symbol for integers?

[tepid quarry]

we write integers with Z

Rationals with Q

Real with R

[rapid silo]

The usual symbol is Z, I know, but an older symbol that was used at one time was J

[tepid quarry]

don't think it is interger either

So i guess real?

[rapid silo]

I guess so... it's not a notation I've seen before.

[tepid quarry]

me neither

they never showed such thing in class and they give it for homework lol

does this mean that a has values in the interval [0;400]?

[rapid silo]

I doubt it

[tepid quarry]

this letter makes me hate math😭

[rapid silo]

just ignore it for now?

[tepid quarry]

it is my last homework problem

[rapid silo]

And have you solved it, except for this upsilon thing?

[tepid quarry]

a = - 3

[rapid silo]

then maybe move on to something else?

rather than spending time on something obscure that they haven't even taught you about

[tepid quarry]

because we got a horror teacher

if i won't solve this we might get a F

[rapid silo]

That's really unair 😦

[odd edge]

@tepid quarry Has your question been resolved?

[odd edge]

@tepid quarry Has your question been resolved?

[odd edge]

@tepid quarry Has your question been resolved?

[tepid quarry]

I'll take it as no one knows an answer so I am going to close the channel. Thanks anyways

.close

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[mystic saffron]

how do we get these values

[mystic saffron]

how to derive them?

our teacher told us to just remember them for future derivations

this is from ellipses

is anyone there?

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[mystic saffron]

.reopen

[odd edge]

✅

[final anchor]

Ping helpers :)

they can most definitely help you

[mystic saffron]

i'm new , how to do that?

[final anchor]

@ helpers

you are allowed to do that after 15 minutes of no response

[mystic saffron]

ok

[final anchor]

do you want me to do it?

[mystic saffron]

yep, why not

[final anchor]

<@&286206848099549185>

[mystic saffron]

thnx

[final anchor]

you welcome

hopefully they can help you cuz i sadly cant sorry

[mystic saffron]

no worries

[mellow tundra]

use basic deifnition of ellipse : PS/PM = e

and plot it

[mystic saffron]

how to do it algebraically?

[mellow tundra]

doing it graphically will give u a better understanding

[mystic saffron]

ok

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[lethal spoke]

This channel is closed

[neat zodiac]

o

oops

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[fair briar]

am i on the right track with this?

[mellow tundra]

yo we meet again

i shall say u are on the right track 🗿

[fair briar]

should I keep multiplying by 1/x / 1/x?

[mellow tundra]

yes

theres actually a standard result on that

[mellow tundra]

but thats the way of approach

[fair briar]

should I have done this instead?

highest power 5

[mellow tundra]

yes

as u can see 1/x as x tends to inf is equal to 0

fyi

[fair briar]

what Go and bo?

oh thats an a

hmmmm

[mellow tundra]

lmao

[mellow tundra]

u understood by what i meant here right

@fair briar

[odd edge]

@fair briar Has your question been resolved?

[fair briar]

sorry for late reply..
hmmm... somehow i gotta get this to evaluate to 1/6 tho

[wooden gorge]

No

[fair briar]

hmmmm

[wooden gorge]

Once again, I'm completely flabbergasted with your affection for FOIL

[fair briar]

[wooden gorge]

@fair briar what's the highest power of x in the numerator

[fair briar]

5

[wooden gorge]

And in denom?

[fair briar]

5

[wooden gorge]

Divide both num and denom by x⁵

[mellow tundra]

am i correct this time 😭 arya

[wooden gorge]

no.

[fair briar]

like this?

[wooden gorge]

@fair briar You're left with $\frac{2(1+\frac{4}{x^2})^2}{(3+\frac{4}{x})(2+\frac{1}{x^2})^2}$

[clever fjord]

Arya

[wooden gorge]

nO

[mellow tundra]

where did i go wrong?

[wooden gorge]

Alright.. the idea is to use the fact that 1/x -> 0

[fair briar]

aren't I just multiplying by the fraction?

[wooden gorge]

No, we DONOT plan on using FOIL

[fair briar]

thats not what i mean

[mellow tundra]

foil = ?

[fair briar]

when i divide by 1/x^5

i'm just multiplying, so why would i multiply all terms

[wooden gorge]

Yes we are, but look at this

Do you see how it's much different from what you have

[fair briar]

can you show the work to get there?

im not following

[wooden gorge]

In this, I can use the fact that 1/x -> 0, to get limit [(2•1)/(3•2²)]

[fair briar]

all I did was multiply the leading terms

by 1/x^5

[wooden gorge]

No, you need to break x⁵ in x • x⁴

Look

What's power of x in 2x

[fair briar]

1

[wooden gorge]

In 3x + 4?

[fair briar]

hmmm, i thought i was following this

[fair briar]

1

[wooden gorge]

So what power of x⁵ should be adjusted with these?

[fair briar]

wdym?

it might be easier if I see the steps, sorry I'm just not following

[mellow tundra]

FOIL=?

[fair briar]

it sounds like this is basic pre-calc tho

[wooden gorge]

Yes it is

[fair briar]

i didn't think i was using FOIL here, I'm just dividing both numerator and denominator like asked

[wooden gorge]

Hmm it'd be a lot of LaTeX hmm

[fair briar]

are you asking what it stands for?

[mellow tundra]

ye

[fair briar]

but just a quick question, how are you giving advice on limits if we don't know this acronym yet?

maybe it's not taught in all schools?

[wooden gorge]

I have an easier assignment for you

[fair briar]

wait before we change it

[wooden gorge]

Put x = 1/y in the given limit and evaluate

[fair briar]

where did i go wrong with dividing num and den by 1/x^5

I thought I did what you asked of me

[mellow tundra]

u guys blocked me or smtg

[fair briar]

now I'm really confused

[wooden gorge]

I'm tryna run from typing so much LaTeX

Okie I'll give u a quick glance at it

[quasi sparrow]

use words. avid can understand do do his own math

[vale vapor]

you will never escape.

[wooden gorge]

When I asked you to divide by x⁵, I wanted you to do something like: $\frac{\frac{2x}{x}(\frac{(x^2+4)^2}{x^4})}{(\frac{3x+4}{x})(\frac{(2x^2+1)^2)}{x^4})}$

[clever fjord]

Arya

[fair briar]

ah

hmmmm

i've never seen it broken up like that before

[wooden gorge]

Phew, bs LaTeX skills but that's a quick one

[fair briar]

i guess it's still x^5

[wooden gorge]

Hehe

[fair briar]

ty for writing it, it helps to see what you meant

[mellow tundra]

bro tell me the full form of FOIL

[fair briar]

so what i did was still correct, but it doesn't help

[wooden gorge]

Yes, you divided by x⁵ correctly but it's no help

[fair briar]

OK, phew

[wooden gorge]

As it does not make use of 1/x -> 0

[fair briar]

I thought I was hallucinating for a sec

[fair briar]

OK

[wooden gorge]

Btw, your next assignment, and one which might make your life easier: put x = 1/y in the given limit

Tell me what you got

[fair briar]

I don't follow 😛

Oh like this?

[wooden gorge]

And you have to replace all x with 1/y ... Also please remove the (1/x⁵)s, this one's a seperate assignment

[wooden gorge]

Which simplifies to?

[quasi sparrow]

1/y should not be going to infinity

oh wait yes it should.

but y should be going to 0+ under the limit

[fair briar]

oops

[wooden gorge]

$\lim_{y \to 0^+} \frac{2(1+4y^2)^2}{(3+4y)(2+y^2)^2}$

[quasi sparrow]

this is poor notation

[fair briar]

i did multiplication before power, mb

[cerulean vortex]

please use () for 1/y squared 🙏

[clever fjord]

Arya

[wooden gorge]

@fair briar ping me when you arrive at this

[fair briar]

getting close?

[fair briar]

I thought you wanted 1/y

[quasi sparrow]

stop writing lim 1/y -> inf

[wooden gorge]

Nuuuuu, I want you to simplify that so we can use y -> 0 on the simplified version

We started with one hole and now the pot's leaking a river

How did (1/y² + 4)² becomes (1/y⁴ + 16)

[fair briar]

i need to review powers

I thought multiply num and den with power

nvm

that's when no + or - involved

[wooden gorge]

Actually

[fair briar]

here i can use difference of squares i think

since FOIL is wrong direction

[wooden gorge]

Do you get how I got this

[fair briar]

from this? no i don't see it

[wooden gorge]

Mmkay, do you see how

1/y² + 4 = (1+4y²)/y²

3/y + 4 = (3+4y)/y

2/y² + 1 = (2+y²)/y²

[fair briar]

oh

you are multiplying a term by 1, in any form

y/y
x/x
etc

to get common den

[wooden gorge]

That's basic algebra

Taught precalc =_= taking LCM, getting common denom stuff

[fair briar]

OK, I just don't see it for some reason

why does 1 = y^2

[wooden gorge]

Nvm, so do you get this now?

Ig this is what I get for skipping steps

[fair briar]

and why is it y -> 0+

where did (1/y) -> Inf go

[wooden gorge]

Because 1/y -> inf

That's why y must be approaching 0^+

[fair briar]

^

[wooden gorge]

For 1/y to approach inf

[fair briar]

i dont get the context here
it needs brackets? or 1/y is wrong

[wooden gorge]

Hmm, you should ask Riemann that ✓

[fair briar]

@quasi sparrow when you get a chance, we would like to know what you mean by this

[cerulean vortex]

i guess 1/y -> inf is maybe poorly defined

we have a definition for limits of a variable approaching a value

i haven't seen a definition for an expression approaching a value

though it's pretty clear in this context

[quasi sparrow]

don't invent new notation because you're gonna confuse yourself. limit statements should be $\lim_{\alpha \to c}$ for $c = \y, -\y$ or some real number. and $\alpha$ is a variable name like $x, y, t$.

[clever fjord]

riemann

[quasi sparrow]

the more notation you make up, the more likely you're gonna confuse yourself and make a mistake

[wooden gorge]

Btw I gotta go but @fair briar , you can use this and the fact that y -> 0+ to plug y as 0 in this expression and get your limit 1/6

[odd edge]

@fair briar Has your question been resolved?

[odd edge]

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[sturdy crane]

What's the formula for image of a point with respect to a line in 3D of the form r = a + Ux where U is a variable.

[odd edge]

@sturdy crane Has your question been resolved?

[sturdy crane]

<@&286206848099549185>

[feral scaffold]

$proj_r(v)$?

[clever fjord]

Sepdron

[sturdy crane]

not sure what you mean by that

[feral scaffold]

it's the notation for vector projection
it's a1 in this picture

[sturdy crane]

Oh, no i didnt mean this

something like this

just in 3D

[odd edge]

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[burnt lion]

Hello! Currently stuck on solving the vertex form for this

[grave roost]

Do you know the vertex

[oak crown]

recall that vertex form is y=a(x-h)^2+k, where (h,k) is the vertex

[burnt lion]

yep i just dont know how to solve the equation y = 3x^2 + 8x + 4

[oak crown]

y=3(x-h)^2+k

[burnt lion]

I got 3(x^2 + 8/3x) +4 i dont know how to solve the parentheses part of the equation

[oak crown]

recall that h=-b/2a

[burnt lion]

so will y = 3(x + 4/3 )^2 - 4/3 be the answer?

[oak crown]

yeap

[burnt lion]

thank youu our teacher didnt explain these parts properly

[oak crown]

lol

[burnt lion]

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[indigo salmon]

i have a question

[steady tide]

you may ask it

[indigo salmon]

how come on the right

there are to 1-

two

but the asymps are 1 and 1

is it because x and y v alues are decreasing

[steady tide]

well recall the definition of horizontal/vertical asymptote

you are just examining the behavior of the function from left to right/from right to left

[indigo salmon]

well

i had them at 1+ on my answer cuz

i thought it means y goes very close to positive 1 but not touching it

oh wait

is it basically

when its -infinity, the number will be negative? and vise versa

Uh anyone

[oak crown]

let me see

@indigo salmon at least to explain the 1+ and 1- for the limits as x->+-infinity, it’s cause some rational functions can have different H.As as they tend to infinity and negative infinity, respectively

[indigo salmon]

ok i thnk i understand

i did another question and i got it right this time

[odd edge]

@indigo salmon Has your question been resolved?

[oak crown]

👍

[odd edge]

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[prime solar]

I didnt mean to select it but

Am i over thinking it or is it seriously just as easy as it looks

[cerulean vortex]

looks good

[oak crown]

yeah, lol

go with ur instinct 99% of the time

[prime solar]

Ok lmaoo

So im guessing its A

wait no

i just said that for easy answer ngl

Oki doke thank ya

[cerulean vortex]

oh i thought that was your answer

[prime solar]

Wait

it is

OH

[oak crown]

do it term by term lol

[cerulean vortex]

??

[prime solar]

i see it now

My bad lol

was looking at it weird

[oak crown]

6-(-2), -9-(-1), 10-3

[prime solar]

Yeah i forgot about the - in the middle

"polynomial" What am i looking for to start with

[copper quarry]

expand the ones on the left

and see if they match with the right

[prime solar]

Sorry what do you mean by "expand"

I prob understand just dont know what you mean by expand

[oak crown]

distributive law

[vapid kayak]

just multiply the polynomials

[oak crown]

a(b+c)=ab+ac

[cerulean vortex]

and then you can use the FOIL method for (2x-3)(4x+5)

[prime solar]

Okay ima havge a go at it and we will see if i understand what yall say lol

What is the FOIL method?

[vapid kayak]

a(c+d)+b(c+d)⇒ac+ad+bc+bd

First: Multiply the first terms: a×ca×c
Outer: Multiply the outer terms: a×da×d
Inner: Multiply the inner terms: b×cb×c
Last: Multiply the last terms: b×db×d

[cerulean vortex]

why did you write a×ca×c twice?

[prime solar]

Wow ive never even heard of this method

[vapid kayak]

i guess you have to remind this too x^m×x^n=x^(m+n)

[prime solar]

Ooooh wait i know this method but it wasnt called foil method Good to know

[vapid kayak]

The FOIL method is a technique used for multiplying two binomials

The FOIL method is a shortcut for multiplying two binomials. It's based on the distributive property of multiplication, which states that you multiply every term in one expression by every term in the other expression.
Theoretical Explanation:

When multiplying two binomials, say (a+b)(a+b) and (c+d)(c+d), we use the distributive property to ensure that every term in the first binomial gets multiplied by every term in the second binomial. The FOIL method is a mnemonic to remind you of the steps in the distributive process:
(a+b)(c+d)
(a+b)(c+d)
Step-by-Step Breakdown:

First: Multiply the first term of the first binomial by the first term of the second binomial.
    In mathematical terms: a×ca×c
    This results in: acac

Outer: Multiply the outer terms (the first term of the first binomial and the second term of the second binomial).
    In mathematical terms: a×da×d
    This results in: adad

Inner: Multiply the inner terms (the second term of the first binomial and the first term of the second binomial).
    In mathematical terms: b×cb×c
    This results in: bcbc

Last: Multiply the last term of the first binomial by the last term of the second binomial.
    In mathematical terms: b×db×d
    This results in: bdbd

Combining the Products:

After multiplying each pair of terms (using the FOIL steps), the results are added together to form the final expression. In the case of (a+b)(c+d)(a+b)(c+d), you have:
ac+ad+bc+bd
ac+ad+bc+bd

This result can then be simplified further if there are like terms.
Generalization:

For any two binomials (x1+x2)(x1​+x2​) and (y1+y2)(y1​+y2​), multiplying them using the FOIL method involves:
(x1+x2)(y1+y2)
(x1​+x2​)(y1​+y2​)

First: x1×y1x1​×y1​
Outer: x1×y2x1​×y2​
Inner: x2×y1x2​×y1​
Last: x2×y2x2​×y2​

Thus, the result is:
x1y1+x1y2+x2y1+x2y2
x1​y1​+x1​y2​+x2​y1​+x2​y2​

is it helpful?

[oak crown]

what in the

@cerulean vortex is there a channel for helpers to discuss things?

[prime solar]

slowly but surely getting there but for (2x-3)(4x+5) i think i keep getting 8x^2-10x-12x-15

What am i messing up

[oak crown]

its +10x, not -10x

[vapid kayak]

you have to keep in mind + - barriers

[prime solar]

i got help from my friend and they showed me how to get this i wanna understand it fully

[oak crown]

i think the 3rd on the left is fairly obvious.

[prime solar]

Well i already had that one before the help it was easy but the rest i know how to read it but dont know how to continue with it

[vapid kayak]

I will suggest to watch a video of khan academy . it can be hard to understand it from text . or you can read the text i sent before .

[prime solar]

I already understood the FOIL method ive learned it before i know how to do it i just got it mixed up slightly

[oak crown]

try to only distribute on the first one

[prime solar]

What do you mean

[odd edge]

@prime solar Has your question been resolved?

[prime solar]

Is this correct?

[vapid kayak]

If the last one -15 at the end . then yes all are correct

[prime solar]

Oki doke then this can be closed

Okay since the bot hasnt asked me for a close i do need help with something

What is it meaning and where do i start

[oak crown]

solve the equation

[prime solar]

Yeah... how

like

[oak crown]

x^2-4x=140, x^2-4x-140=0

im assuming you dont know how to factor

[prime solar]

Its getting late im not remembering much so probably not at the moment

[oak crown]

k

[prime solar]

why is the =0 at the end needed

[oak crown]

subtracted 140 from both sides

[prime solar]

oooh

[prime solar]

x=14,−10 ??

[vapid kayak]

yes

[oak crown]

what is an extraneous solution here

[prime solar]

I hate english class no clue what that word is

[oak crown]

one that shouldnt exist

can ft be negative

[prime solar]

yeah i dont have a clue what your saying sorry

[oak crown]

can a measurement be negative

[prime solar]

No

Technically no it shouldnt

Wait so the solution is 14,-10?

makes no sense

[vapid kayak]

so thats the illogical solution in that case . you have to make dismiss it . actual practical answer is 14

[prime solar]

So for me to get the answer right i just put in 14

[vapid kayak]

yeah , the solution of your equation is right . but the practical one is 14

[prime solar]

What in the world is a domain

[vapid kayak]

-3

Domain is the set of all possible values of xx for which a function is defined.

To determine the domain, look for restrictions like division by zero, square roots of negative numbers, and the argument of logarithms being non-positive.

[prime solar]

I even did it in a actual calculator and the calc said none of these options

[prime solar]

Why are they giving me assignments about things ive not once been told about

[vapid kayak]

may i know your class?

[prime solar]

Im in geom atm never done alg 2

[vapid kayak]

who is giving that ? your school/

?

[prime solar]

Yes

[vapid kayak]

i can feel your pain .

this is option 2

[prime solar]

What is

[vapid kayak]

the answer of your last question. make it as product of functions .

you get it/