help-23

[random rivet]

but how

do i figure it out

[sleek sentinel]

colinear: $v_1 = kv_2$ and perpendicular: $v_1 \cdot v_2 = 0$

expand these formulas
sometimes you'll have two variables, which means one can be whatever you want and then just solve for the other

[flat frigate]

b

[tardy mango]

^ For the second one, p and q aren't uniquely determined, so you can pick p (or q) to be whatever you want

then solve for the other variable

[random rivet]

where did the fomulas come from

[sleek sentinel]

do you know what a scalar multiple is

[random rivet]

ngl

no

[sleek sentinel]

we can multiply vectors, which gives you a "stretched" (or squashed) version of the original
something like this

[random rivet]

okay

[sleek sentinel]

can we agree these two vectors are colinear

*do you see why

[random rivet]

yes 👍

[sleek sentinel]

cool, so if you have colinear v1 and v2, they're both scalar multiples of each other

in other words, v1 = k * v2 for some k

[random rivet]

okay

[sleek sentinel]

is that enough info to do part a?

[safe radish]

@random rivet Has your question been resolved?

[safe radish]

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[idle schooner]

I know the answer is C to this question, but how am I supposed to find it. This is from my AP Precalc practice test

[safe radish]

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[idle schooner]

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[feral sonnet]

that is cos law

how can I find h

[river oriole]

just collect the like terms

[feral sonnet]

its so hard

[river oriole]

all of these have a real value except for h

ok then

[feral sonnet]

how can I do this on my test

I have to spend 1 hours

[river oriole]

open the brackets

see if that helps

[eternal carbon]

is this the original problem?

[feral sonnet]

there has to be a short cut

no

[river oriole]

yea

[river oriole]

cause h^2

[feral sonnet]

this the problem

[river oriole]

opening the brackets isn’t that hard

[feral sonnet]

but I might have 10 questions like this and 1 hour to finish

[river oriole]

yea

it takes 5 seconds

[feral sonnet]

really

can you please right it down then for me please

let me see how you solve it

[river oriole]

there you have your like terms

[feral sonnet]

now isolate for h

[river oriole]

i can’t do the whole thing kinda goes against server rules

[river oriole]

yea, that’s it

[feral sonnet]

please*

[river oriole]

nah sorry mate

you try and i’ll help you

[feral sonnet]

dude i knew that already

[river oriole]

yea

so isolate for h

what part are you stuck on exactly?

you don’t know how to isolate for h?

[feral sonnet]

time

there is no way you do the quastion under 10min

[river oriole]

well, then it just comes down to practice.

mate, i’ve done this question more than a million times, i can do it in less than 1 minutes

[feral sonnet]

I know there is a short way for it I just need that short cut

[river oriole]

just practice

[feral sonnet]

this is the quastion

ignore the blue 90

[river oriole]

what do you want me to do

[feral sonnet]

tell me the short cut

[river oriole]

shortcut method is just do the equation quickly

[feral sonnet]

are you sure

?

I think its somthing with cot

[river oriole]

cot would do the same thing

[feral sonnet]

ok then thanks but can you at least help me undrstad a nother quastion

[river oriole]

you just don’t have 1/tan

[river oriole]

send

[feral sonnet]

wtf is this quastion asking

[river oriole]

try drawing a diagram for your given knowledge first

[feral sonnet]

exactly

i dont know where to start

where is 12

12 digg

if I draw it I can solve it ezally

[river oriole]

ok then

first draw the two roads intersecting

[feral sonnet]

like this

[river oriole]

there should be two straight lines

[feral sonnet]

😂

thats not right

no ?

[river oriole]

:/

[feral sonnet]

it seems more like 120

[feral sonnet]

tf

how

see I always have problem undrstaning the quastions

maybe its because english is my second lanuage

just maybe

[river oriole]

ah that sucks man

think about a road

[feral sonnet]

one of them gose up and one stright ?

[river oriole]

intersection is two roads crossing each other

[feral sonnet]

like this

?

faster car gose stright and slower one is going up

somthing like this

?

[river oriole]

ur right this is worded weirdly

i got it wait

[feral sonnet]

other quastions are much easyer to undrstand

[river oriole]

yea this one’s a bit annoying

[feral sonnet]

see

the last quastion was number 8 and this one is number 7

[river oriole]

ok

so first we look at the road

[feral sonnet]

witch one

if I undrstand the 12dg I will be fine I think

[feral sonnet]

oooo the 12 is the angle btwenn the roads ?

[river oriole]

because the helicopter is in between them, we need to find the instance between them first

[river oriole]

yep

first can you find AB

[feral sonnet]

I thought its the slope

I can find the lengths now

20min * 90km/h and 20min * 120km/h .

right

that means divided by 3

[river oriole]

yep

[feral sonnet]

so 30km from start to point A and 40km from start to point B

[river oriole]

yep

[feral sonnet]

and we use cos law

no

wait

its 3 d

[river oriole]

cos rule yea

nono you’re on the right track we’re thinking about this just with a triangle rn

not 3d

[feral sonnet]

so then yes

cos law for finding A to B

c2 = a2+ b**2 -2abcosA

idk why it removed the **

for c and a

[river oriole]

it’s just a discord thing

[feral sonnet]

then what

[river oriole]

it bolds it

what did you find for it first

[feral sonnet]

do I really have to do it now .-.

lets just say c

[river oriole]

ok then

[feral sonnet]

I know that part trust me

[river oriole]

now we have a helicopter

[feral sonnet]

right

[river oriole]

this helicopter will be 1000 metres above somewhere on that line

[feral sonnet]

is it like this

[river oriole]

nah

cause we’re on line AB now

[feral sonnet]

o its 1d

2

[river oriole]

ye this questions weird

you got it from here?

[feral sonnet]

I think so

thx'

[river oriole]

sorry it ain’t 1-x

it’s AB - x

[river oriole]

gl man. if you’re stuck, just break the sections of the question up

[feral sonnet]

if

isnt suppose to be like this though ?

then we have c/2 for both left and right

[river oriole]

you can’t say it’s c/2

[river oriole]

cause it’s not right in the middle, it’s somewhere in the middle

nah the blue won’t work like that sorry nate

the bottom triangle is a flat plane, you can’t make the 12 degrees 3d like that

[feral sonnet]

ye . my teacher did a example in class it was 2d . you are right I think

thx'

[feral sonnet]

sure.

.close

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[golden pumice]

my soln is getting all over the place lol

after substituting cot with cos/sin what should be the next step?

[vale oriole]

cancel

[olive tinsel]

i don't like trig

[golden pumice]

oh it basically ended up at

sin^5/cos

but yeah

i kept getting the wrong answers lol

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[zealous dagger]

can pls someone help me with this problem

!help

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[olive tinsel]

Bro nobody's helping you with your test question

I saw your earlier post

[zealous dagger]

why lmao

[olive tinsel]

It's a test omg

[zealous dagger]

thats beacuse of you dont know it or smth

thats nonsense

[olive tinsel]

yes, i don't know how to do it but still cheating on your test ain't cool bro

[zealous dagger]

quiz finished like a while ago lol

i just want to know if i did it right

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[subtle ore]

hey there, can i get help with a topic i dont understand much

[brazen parrot]

😡😡😡

[subtle ore]

what

can someone help

[river oriole]

Post the question

[subtle ore]

i don’t understand this exercise much

i kinda forgot about the topic and the teacher still hasn’t reviewed yet

hello

[minor ermine]

U1= 3/8

I forgot the topic

But the second question I remember the method

You just verify u(0)
Then put Un verify and build Un+1

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@subtle ore Has your question been resolved?

[safe radish]

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[fiery otter]

Unsure if I am supposed to send the question in the help forum AND claim a help channel. In any case, here is my question: #1239163223685726210 message

[fiery otter]

.close

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[cloud mauve]

hello

[cloud mauve]

id like help w this question

i couldnt understand how to solve this as it isnt present in the standard form

that is ax+by+c

[final halo]

There is a typo in the question it is meant to be 25x² - 16

[cloud mauve]

o

oke wait lemme try solving it again tysm

[cloud mauve]

YOU'RE THE GOAT

I GOT THE ANS

tyvm!

answer is option a

[final halo]

Yep! Nice

[cloud mauve]

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[lean otter]

pls help my hw is due tomorrow and idk what this is 💀

[hard crest]

can you do (a)?

also did you do question 11?

[lean otter]

no.

[hard crest]

read i, and try to do at least that part of part a

[lean otter]

nope

ok tyy

i still dont get it 😭

oh wait

would it be y = 3/4 (x - 8) + 2?

[hard crest]

yep perfect

[lean otter]

oh ok

[hard crest]

now how can you convert that into the other form they want, for ii?

start by distributing the 3/4

[lean otter]

i have no idea

[hard crest]

do you know the distributive law? or sometimes called property?

it looks like this:

$\psq (\rsq + \bsq) = \psq\rsq + \psq\bsq$

[flat frigate]

hayley table

[lean otter]

oh yeah

ik that

[hard crest]

ok great

so let's apply it to: y = 3/4 (x - 8) + 2

we get y = 3/4 x - (3/4)(8) + 2

can you simplify that last bit?

[lean otter]

y = 3/4x - 8

[hard crest]

i don't think that's right

when you multiply the - 3/4 * 8 you should get -6 right?

so then we have: y = 3/4 x - 6 + 2

which will simplify to: y = 3/4 x - 4

[lean otter]

ohhh

ok

[hard crest]

can you do b? it's the same, just with different numbers

[lean otter]

i) y = -7/3 (x-15) - 6

ii) y = -7/3 - 41

[hard crest]

missing an x, and be careful of the negative signs again, i know there's a lot

y = -7/3 (x-15) - 6

y = -7/3 x + 35 - 6

[lean otter]

oh yeah

[hard crest]

so ultimately +29

[lean otter]

yup

[hard crest]

ok for the next ones, do you know how to calculate gradient? (also called slope)

[lean otter]

yeah i do

but what does it mean by lie on the line

[hard crest]

ok well i'm posting this anyway:

(to the tune of the chorus of YMCA)

you know that m equals y minus y
all over x minus x
watch your signs as you do the rise over run
and finding slope will be totally fun!

oh that they're on the line, so like the line goes through those points

[lean otter]

oh ok

[hard crest]

so once you've calculated the slope you're in the same position you were for the previous ones

[lean otter]

ohh ok tysm

and what abt g)?

[hard crest]

an x-intercept is the point at which the line crosses the x axis

like this

[lean otter]

would i do (4,0) (0,3) and then figure out the gradient?

[hard crest]

yep

[lean otter]

ohh ok

tysmm

[hard crest]

and then with that one

[lean otter]

ur a lifesaver

[hard crest]

i'd recommend using the y-intercept for the point-slope form thing

[lean otter]

ok

tyyyy

[hard crest]

good luck!

[lean otter]

tysm!

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[lean otter]

.reopen

[safe radish]

✅

[lean otter]

wait im so sorry

for c)

m = 1/4

so how do i put that into the y = m(x-h) + k equation

oh no wait

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[opaque bolt]

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[cosmic pier]

Hello?

[cosmic pier]

helllooooooooooooo

I have a math problem

very problematic

whatever

[spiral imp]

yes? share the problem

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[trail otter]

I need third order of total differential of a function $f(x,y,z)=xyz$. I know that $d=h_1\frac{\partial}{\partial x}+h_2\frac{\partial}{\partial y}+h_3\frac{\partial}{\partial z}$. Is third order of paritial dirrefential just $d^3=\left(h_1\frac{\partial}{\partial x}+h_2\frac{\partial}{\partial y}+h_3\frac{\partial}{\partial z}\right)^3$?

[flat frigate]

Slowaq

[trail otter]

pls help

[safe radish]

@trail otter Has your question been resolved?

[trail otter]

<@&286206848099549185>

or is it like $d^3f(x,y,z)(\vec{h},\vec{k},\vec{l})=df(x,y,z)(\vec{h})df(x,y,z)(\vec{k})df(x,y,z)(\vec{l})$?

[flat frigate]

Slowaq

[trail otter]

when $df(x,y,z)(\vec{h})=\left(h_1\frac{\partial}{\partial x}+h_2\frac{\partial}{\partial y}+h_3\frac{\partial}{\partial z}\right) f=h_1yz+h_2xz+h_3xy$

[flat frigate]

Slowaq

[trail otter]

<@&286206848099549185>

[meager shoal]

sorry bro Idk this I am only learning basics of calculus :d

wait

is this vectors?

[trail otter]

no problem

[trail otter]

yes

i figured it out now

[meager shoal]

good keep up yo work bro

[trail otter]

:)thx for encouragment i kind of needed that

[trail otter]

bye bye

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[storm ice]

Hi, quick question about linear algebra

[storm ice]

If I have four points

P = (2, 1, 0)

Q = (-1, 0, -1)

R = (0, 1, -1)

S = (2, -3, 2)

How would I go about proving they're in the same plane?

I can't get my original idea to work and well.. im not sure why

I was thinking that I could go from for example P to Q and R to get the direction vectors that would be in the plane and then take the cross product of that

If the points are in the same plane, to me putting in the coordinates should total up to zero, no?

So PQ = (-1-2, 0-1, -1-0) = (-3, -1, -1)

[plucky python]

you can find the equation of the plane passing through 3 of them and then substitute the fourth

using cross-product as you said

[storm ice]

How would I do that?

Why is it not enough to have 2 direction vectors?

PR = (0-2, 1-1, -1-0) = (-2, 0, -1)

[plucky python]

if you want to get the equation of the plane passing through P, Q, R for instance, you can take the cross product of vectors PQ and PR

to get the normal vector

[storm ice]

but it doesnt work for me :-;

cross product of -3 -1 -1 and -2 0 -1

[plucky python]

yes that is correct

[storm ice]

-3 -1 -1

-2 0 -1

1, 2-3, -2

aka

(1, -1, -2)

which should give x - y -2z = 0

oh

i may have forgotten about the constant

i realize now

when i tried plugging in for example P i get

2-1-0 = 0

1 = 0

but then i guess we might have -1 as the constant for the plane?

i think it breaks for R still

nvm i just cant do math

yeah i see the issue

i just forgot about the constant

that's why I got 1 = 0 and gave up and came here in frustration xd

thank you!

[plucky python]

oh ic lol

[storm ice]

just needed some reaffirming that i was on the right path :')

[plucky python]

np i didnt help lol

[storm ice]

well not that im here anyway

how would I calculate the area between these four points?

that's the last part of the question that I wasn't entirely sure about 🤔

[plucky python]

the area of the quadrilateral formed by them?

[storm ice]

yea

[plucky python]

you can first split the quadrilateral into two triangles

then find the area of each using cross-product

[storm ice]

will try that approach

thank you!

have an amazing day <3

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[plucky python]

you too! you're welcome

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[signal holly]

How could I calculate the point where the line intersects with the circle, given the red point, Radius, and blue point? I want a formula so i could move the 2 points and desmos would calculate it for me.

[brave wolf]

red point is center?

[signal holly]

yes

[brave wolf]

ah i see

we will call red point coords x1 and y1

and blue point coords x2 and y2

radius will be just r

[signal holly]

ok

[brave wolf]

this formula works for me @signal holly

the reason why is works is there are similar triangles

[signal holly]

thank you

[safe radish]

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[lean otter]

"How many integer pairs 'a' and 'b' are there for a < b, a + b < 100 that verify the equation (a/4) + (b/10) = 7?"

I'm not really sure what to do here. I'd prefer a solution avoiding trial and error and an actual logical response, please (if possible).

Thanks.

[frank horizon]

this is what's called a linear diophantine equation, i.e. equations of the form ax+by=c with everything being integers, if you multiply on both sides by 20 you get the equation 5a+2b=140 and this equation can be parameterized to get every single solution

[lean otter]

so, trial and error? I don't quite understand what you mean by "parametrized?"

[frank horizon]

you basically have a formula into which you plug in some value and get a corresponding solution

[lean otter]

so, you try a situation that satisfies both inequalities and the equation?

and keep doing it

[frank horizon]

no, the inequality isn't relevant right now, first you want to find some pair (x,y) which solves the equation, then it is a fact that if (x,y) is a solution then (x+kv,y-ku) is also a solution, with k being any integer and v and v being the quotients when a and b are divided by their gcd

so now you can find all k which satisifes the inequalities, and that set of k's should be easy to count

[lean otter]

the possibilities are endless without the inequality

are they not

[frank horizon]

yes, but now instead of counting the number of valid solutions, you count the number of valid k's which is much easier

it should just be some range of integers

[lean otter]

what do you mean "k's"

[frank horizon]

k is a parameter which you can use to get other solutions, so if you find a solution (x,y) to the equation, then (x+kv,y-ku) is also a solution

and u and v are a/d, and b/d

[lean otter]

where are you getting "a/d" and "b/d" from

just rational numbers?

[frank horizon]

no, d is the gcd of a and b, so necessarily d|a and d|b

you can verify just by computation that if (x,y) is a solution, then so is (x+kv,y-ku)

[lean otter]

why are you doing (x,y) and not (a,b) if that's what I'm trying to find

sorry if I'm asking too many questions

[frank horizon]

thats ok, sorry i changed variables because that's what im used to, in this case a=5, b=2, c=140, and x=a and y=b

so first try to find just any solution

[lean otter]

a=5 and b=2?

like the coefficients?

[frank horizon]

ye

[lean otter]

why is it a and b

[frank horizon]

well in what i was saying earlier, a and b were the coefficients, sorry if that was confusing

i can rephrase, in the equation we have 5a+2b=140, so if (a,b) is a solution, then so is (a+kv,b-ku)

[lean otter]

so,

(a,b) = (a+kv, b-ku)

yeah

[frank horizon]

right

[lean otter]

but then, what can I do from here?

guess k?

[frank horizon]

no, now find v and u

u and v are the coefficient divided by their gcd

[lean otter]

so, 2/10 and 5/10

[frank horizon]

err no, thats the lcm

the gcd is smaller than the numbers

[lean otter]

ohhh

so 1?

[frank horizon]

yes

in this case its simple

[lean otter]

2/1 and 5/1

[frank horizon]

right

so now what are the solutions?

[lean otter]

so

(a,b) = (a+2k, b-5k)?

[frank horizon]

yep looks right

[lean otter]

then I find k?

[frank horizon]

now see if you plug in that new solution, the equation is satisfied

[frank horizon]

now you can find all valid k

so first find a valid first solution

[lean otter]

one sec

2a + 5b = 140

now I plug in a+2k and b-5k??

[frank horizon]

err wait i think that's backwards

should be 5a+2b=140?

[lean otter]

does it even matter

ohh

[frank horizon]

yes of course it matters

[lean otter]

wait yeah

my bad

[frank horizon]

well it doesn't matter ultimately, but what you wrote down earlier would be wrong

[lean otter]

5a + 2b = 140

so actually

it should be 5(a+5k)?

and 2(b-2k)?

[frank horizon]

no what you wrote down earlier was correct

you can see why if you plug in the new solution

the k's cancel out

[lean otter]

5(a+2k) + 2(b-5k) = 140?

5a+10k + 2b-10k = 140

yeah, okay

but then what is the use of this situation

what can I do here

[frank horizon]

so first find any solution

[lean otter]

for k or for all the coordinates

[frank horizon]

just any valid solution

having the formula isn't helpful unless you have an initial solution

[lean otter]

but i dont get what you mean by solution

[frank horizon]

just any pair (a,b) which satisfies the equation 5a+2b=140

[lean otter]

oh, (2, 65)

[frank horizon]

cool

so now, you know that (2+2k,65-5k) is a solution for all k

so now the question is for what k is 2+2k < 65-5k, AND 2+2k + 65-5k < 100

[lean otter]

9

what do I do with this

just keep plugging values in

?

[frank horizon]

sorry, what do you mean, you should be able to get a range of values of k

[lean otter]

ohhh

i only did

[frank horizon]

i meant get every single k so that the inequalities are satisfied

[lean otter]

ok wait

-11 to 9?

is what I'm getting

[frank horizon]

wait let me check

yes i think

noninclusive right?

[lean otter]

yep

[frank horizon]

ok, so now just count how many are in that range

each one gives you a unique solution (a,b)

[lean otter]

oh my god

this is the greatest thing ive ever seen

it's 19

[frank horizon]

yep

[lean otter]

thank you so much

[frank horizon]

np

id suggest reading the wikipedia page on diophanine equations to read the proof of why what we did works

its basically just some number theory

[lean otter]

I will definitely do that

alr

[frank horizon]

👍

[lean otter]

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[static dune]

is this a to c + c to b property?

cuz where does the 0 come from

[tawdry needle]

Yes

[static dune]

but where does 0 come from

is that just part of it?

[tawdry needle]

0 is arbitrarily chosen and any number would work there

But ig 0 is good

[static dune]

so i could put like

200 in there

but it would make it harder to solve i suppose

[tawdry needle]

Yes 200 in both places

[static dune]

ok thx

[tawdry needle]

Ye thats why they chose 0

[static dune]

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[rigid sapphire]

Part e is the problem I'm having trouble with

I understand the normal distribution is symmetric, but can you express this mathematically or no?

cuz it says "Show that"

If I got this type of problem on a test I'd just say they're equal since the normal distribution curve is symmetric and draw the graph lol

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[nocturne blaze]

how do i integrate that

[gusty trench]

uh

like integrate it over some bounds to get a number? or find a generic antiderivative?

[nocturne blaze]

so its not easy to integrate?

[gusty trench]

no, it has a nonelementary antiderivative, so you can't really express the antiderivative in terms of standard mathematical functions

[nocturne blaze]

ok thanks

what if it was sin(pi/2 x)

$sin(pi/2 x)$

[gusty trench]

to clarify, is the x in the numerator or denominator?

[flat frigate]

Sprites

[nocturne blaze]

numerator

[gusty trench]

as in pi / (2x)?

oh

yeah that's easily integrable

[nocturne blaze]

how?

[gusty trench]

do you know the antiderivative of sine?

[nocturne blaze]

where does the x go?

yes - cos

[gusty trench]

one thing that could help keep track of chain rule here since it's not just sin(x) is u substituion, do you know that

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@dim cobalt Has your question been resolved?

[dim cobalt]

<@&286206848099549185>

[rough storm]

You could try to convert back to cartesian?

yeah that helps you visualize/graph it

[dim cobalt]

okkkk i definitely will give it a try, i think my proof will just be the graphs themselves

if i figure out geogebra

[rough storm]

i think it will help motivate your proof. Note specifically where the parts connect and disconnect. it's at a particular line

[dim cobalt]

okkkkk

ill give it a shot

thanks !

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[vast harbor]

I know that 0 is invalid

[vast harbor]

and 2 isn't either w the discontinuity

but is -3?

[magic tendon]

discontinous

[vast harbor]

so -3,2,0

[magic tendon]

yes

[vast harbor]

thx

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[paper phoenix]

can someone explain this parametric equation of stright line formula

[paper phoenix]

what is t

also what is a

what is at

i guess a, b and c are X2, Y2, Z2

but t

[mortal sandal]

if you set t to a specific value you get a point on the line

As you vary t over every value, the point moves across the line, defining it by sweeping it out

[paper phoenix]

gotcha, so in this case its just like a placeholder

[mortal sandal]

a b and c are components of the direction vector

[paper phoenix]

what about this

[mortal sandal]

Not points on the line, but they might be more like X2-X1

[paper phoenix]

what that in simple terms

ah

yeah this question states that

wait but a was 4 before

hmm maybe the question was solved wrong idk

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[tall drift]

Would this be correct?

[tardy pivot]

with logarithmic equations, always test your solutions with the original equation

oh well nvm it's fine here

it looks all good

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[paper phoenix]

i dont even understand the first equation he came up with

can someone explain

[final halo]

theyve taken the determinant of the matrix by expanding across the top row

[tardy pivot]

just another formula to remember sigh

[paper phoenix]

gotcha, why 6 - k tho, why the -

in formula?

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[paper phoenix]

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✅

[paper phoenix]

<@&286206848099549185> need some help

after 6 the negative i dont get

[mellow cypress]

(3)(-2) - (1)(k)

[paper phoenix]

oh negative is in formula?

3 (2) btw

[mellow cypress]

yeah, the determinant of a 2x2 matrix [[a b][c d]] = ad - bc

[paper phoenix]

got it

[mellow cypress]

( \begin{bmatrix} a & b \ c & d \end{bmatrix} )

[flat frigate]

shsgd

[paper phoenix]

thank you very much

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[safe radish]

@wheat ivy Has your question been resolved?

[wheat ivy]

<@&286206848099549185>

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[wooden sable]

.close

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[wooden sable]

.reopen

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[atomic pebble]

Guys I have this function and the question is: Find the largest interval containing the point 1/2 on which f is defined and injective. Find the regulation
for the inverse function of narrowing f to this interval and sketch the graph of this inverse function.

I don't really understand it

[grizzled fossil]

Know how to check if a function is injective?

[spice wing]

Yeah 🙂

For f(x) = f(y) if x=y then injective

[grizzled fossil]

are you OP?

[lean otter]

op?

[split fulcrum]

Original poster

[lean otter]

ok water beam

[split fulcrum]

OK YAJAT

[atomic pebble]

yes, what do I do then

[grizzled fossil]

Find the intervals where the function is injective

[solar hazel]

this isn’t even right

[spice wing]

:'')

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[manic shadow]

STEAL

[manic shadow]

Log(1-x)(5)=2

how

i got √5-1 apparently not

[spice wing]

?

can you write it properly ?

[manic shadow]

i tried finding x but i cant seem to

wait

[spice wing]

5 × log(1-x) =2 ?

Or log base (1-x) 5

?

[manic shadow]

(1-x)²=5

[spice wing]

Ok

Got it

log a base b

= ln b/lna

Know thiz ?

[manic shadow]

nope

i learnt logarithmic thru Youtube

[spice wing]

Wait

See the formula

so we have ln 5/ln (1-x) = 2

Or ln 1-x = ( ln5)/2

Now you can solve ?

Orln 1- x = ln √5

Or x = 1-√5

Okk ?

[manic shadow]

uh wait

so we substitute x

1-1-√5?

[spice wing]

Why ?

[spice wing]

We just remove ln

By raising them to the power of e

ln means log (base e)

[manic shadow]

i mean for this, if we substitute x, we get
(1-1-√5)²=5

?

[spice wing]

How will you substitute ?

And where will you ?

[manic shadow]

loga(b)=c
a^c=b
So
log(1-x)(5)=2
we substitute x?

i think im misunderstanding hold on

[spice wing]

Yeah

[spice wing]

Nah you did correct

If you subtitute there you will get 5=5

[spice wing]

But itz (1-(1-√5))

So its (1-1+√5)² okk

[manic shadow]

Oh ty, it says true

[spice wing]

Yeah

[manic shadow]

so this formula is called change of base?

[spice wing]

Yeah

[manic shadow]

Ight tyvm

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[flat frigate]

Akti

[grizzled fossil]

WLOG let x,y be negative. Simply just verify that x+y+z is not equal to 3

[sharp cedar]

Yh how?

[grizzled fossil]

x+y+z=x-z+z=x<0<3

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[sweet mica]

It's ilegible

[neat onyx]

Really? How is it showing up on your screen?

Would u mind showing me a screenshot

Of the illegible part

@sweet mica

[sweet mica]

The question text is ilegible.

[neat onyx]

My question was , the parts in white are the options and blue is me trying to figure it out but I’m not confident enough in the rules to know what to do next for certain to get the correct one

[neat onyx]

Resent

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[visual creek]

help

[sturdy hazel]

what can i help u with

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[versed wave]

is this a test?

[sturdy hazel]

ok sir, first u are going to substract the -2 to get the other side of the equation. Then the last number 3x is going to be divided by 6y then subtracted by -2. After that ur going to be looking at the second equation. First subtract 4 by 7y into the equation. Which leaves me with y= -26

[visual creek]

its on a learning website

for our school

[versed wave]

!nosols

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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[visual creek]

aight let me read through this

[trail locust]

Hello anyone can help with my problem

[visual creek]

its not even the right answer bro

[sturdy hazel]

sure what is ur problem?

[visual creek]

who is this guy

[versed wave]

!noadvert

[trail locust]

Sorry

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Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[visual creek]

rolex why are you on my channel

you have your own

[sturdy hazel]

math professer

[trail locust]

Come to my page sorry dear

Sorry

My professor also failed to solve this question

[versed wave]

what does the question ask?

[visual creek]

yu man buggin fam worda my motha you playin

[visual creek]

let me translate

it says

solve set of equations

[versed wave]

alright

the equations rn looking a bit ugly rn, why dont we clear it up?

convert the 2 equations into ax + by = c form would be a nice start

[sturdy hazel]

what in the confusion

u randomly got a a and b out of nowhere

[versed wave]

???

x and y are the variables, a, b are their coefficients

[sturdy hazel]

https://tenor.com/view/shinji-bankai-gif-13460197516612583169

[versed wave]

lets not drift off topic here, what matters is @visual creek 's problem

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[desert juniper]

ask the actual question?

[fervent meadow]

Help me

What’s the answer

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@lean otter Has your question been resolved?

[tardy pivot]

you can solve this algebraically or using symmetry

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[river oriole]

hi i’m not quite sure what to do with the information y=-4x +13.

[river oriole]

I know the -2a + b = 4

and 1=a-b+c

and that a is negative

what else would i be able to do

[brisk onyx]

Set up that 2ax +b = -4x for some x and that after you get that point you can set up a similar equation to this

[river oriole]

wouldn’t 2ax + b = -4, not -4x, as -4 is the gradient

or is this a rule, cause I thought the gradient would only equal -4 at that point

[brisk onyx]

So think about what the equation is trying to say rather than thinking about it like rules

[brisk onyx]

Over here why do we have the x on the lhs

[river oriole]

because of the derivative

[brisk onyx]

But what is the derivative telling you

[river oriole]

the gradient

[brisk onyx]

at a point

right?

[river oriole]

yep

[brisk onyx]

and so we know for example at the point 2 the gradient would be 2a(2)+b

[river oriole]

yes

[brisk onyx]

but when we are given that -4x +13 is tangent to the function

one big thing is we dont know where is it tangent to the function

so we know two things though

1. at that point this line and the function must have the same slope

2. the function and this line must intersect once

try putting those two conditions into equations

[river oriole]

this is what I thought would be 2ax + b = -4, and i honestly completely forgot how the second one would be put in an equation like form

[brisk onyx]

but here you are implicitly assuming that the point at which they share the gradient must be one cuz you subbed in one on the rhs already

[brisk onyx]

we can do the second one after we finish the first

[brisk onyx]

we know they must have the same slope at *some point * but we dont know that point yet

[river oriole]

wait, finding the equation for the tangent is like (y-k)= m(x-h) right? so if I had the equation as that, i thought -4 would immediately be the gradient since mx

[brisk onyx]

you know what i just realised

youre right

my bad

it should just be -4

[river oriole]

nw

[brisk onyx]

for the second eq

if they are equal at a point

just set the eq's equal to each other

[river oriole]

ax^2 + bx +c = -4x + 13?

[brisk onyx]

yeah

[river oriole]

and then should i make a tangent equation with the other one and use that as my third equation?

nah i’m getting ahead of myself

[river oriole]

after combining them, what would i be able to do?

[brisk onyx]

you should have a sytem of equations

and you can slowly hack away at them

do you wanna write them all out and as you solve them i can check for mistakes and all that

[river oriole]

yep that would be helpful

[brisk onyx]

-2a + b = 4
1=a-b+c
ax^2 + bx +c = -4x + 13
2ax + b = -4,

which one(s) seem like the least terrible

[river oriole]

1 and 4?

idk to go for the ones without x or the shortest ones

[brisk onyx]

so you can solve for either a or b in 1

then just sub that in somewhere else

[river oriole]

i feel like subbing it in would be kind of useless tho, since we don’t know x or c

[brisk onyx]

we dont know yet

but we have 4 eqs

and 4 unknowns

so its solvable

[river oriole]

oh damn ur right

[brisk onyx]

itll be a lot of algebra

which yk good practice

but not fun

[river oriole]

here’s where i’m at right now

is there any problems with this or am i good to go

[brisk onyx]

imma be honest

i havent seen a number above 5 in months

[river oriole]

lol

[brisk onyx]

i am no longer good at deal with this

it looks fine

to me

[river oriole]

nah there’s something wrong i did

[brisk onyx]

extra 4 i thnk

in the second ther,

4+2a+4

maybe it should just be 4+2a

[river oriole]

i wasn’t clear with that but that was supposed to be from the b+4 from 3rd equation

ah

the 13

[brisk onyx]

oh you mixed them i see

my bad

[river oriole]

nah i’m still a bit off lmao time to proof all my working with the answers

it’s 5+a whoops

[brisk onyx]

might i recommend, if you do have one, putting this into a calculator

if you have a TI ------ type thing i think i know how to put in the systems

[river oriole]

i have an old casio calc. the of but now forgotten plain looking sad calculator with an ink stain bleeding onto the screen :,)

i fixed my equation tho

i’m just gonna do the algebra to find a and then be a happy soul

aight i got a

thanks for your help mate

.close

[safe radish]

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[safe radish]

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[crisp gyro]

ik this isnt maths per se but ignore the context, how would i find the radius of the first curve? it says 1cm represets 10cm but like what do i mesure?

[cobalt sinew]

1cm of that diagram is actually 10 cm

[crisp gyro]

ye but like what do i actually measure

[cobalt sinew]

length i think

[crisp gyro]

that would be arc length

[cobalt sinew]

what is the formula to use in this question

because i dont know

[crisp gyro]

im only stuck on the measurement bit bc u have to mesure radius somehow

[safe radish]

@crisp gyro Has your question been resolved?

[crisp gyro]

<@&286206848099549185>

[gleaming field]

Who pinge me

[gleaming field]

Radius of a curve=? I assume youre supposed to stretch another curve to form a circle from it and then find the radius of said circle

[crisp gyro]

use a compass?

[gleaming field]

No a protractor i think

If you do use a compass that first curve would make a huge circle

If its to scale, then it really doesnt matter

Just complete both of those circles then find the radius

[safe radish]

@crisp gyro Has your question been resolved?

[safe radish]

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[icy coral]

.reopen?

[polar iron]

I need help

[polar iron]

@pine wind

All three questions need statement and reasoning

[safe radish]

@polar iron Has your question been resolved?

[polar iron]

A bit yes

[safe radish]

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[long fiber]

say i had a matrix [1,3,2,2]

[long fiber]

its 2x2

and the eigenvalues i got are 1, 2

which align with the trace

but the product of the eigenvalues should be equal to the determinant of the matrix

the determinant is -4 but the product is 2

nvm im stupid

.clse

.close

[safe radish]

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