help-10

[upper fog]

pls help its 10pm and test is tomorrow morning

also our teacher doesnt let us use any sort of substitution LMAO

i cant take the constant out like normal

idk what to do

[spring trail]

is that 3y

[upper fog]

no

39

39/4

[spring trail]

by god

[upper fog]

the first step was complete the square

[spring trail]

i was soo confused

[upper fog]

sorry im really rushing

yeah

teacher wont check this , this is just f or practice

[spring trail]

i think u got ur question wrong

[upper fog]

??

the first part is wrong

its 1/5(x-3/2)^2 - 39/4

this is a better photo

[spring trail]

welp

ok

[upper fog]

i cant use symbolab because it uses substitution, but i have to do it manually

[spring trail]

so not even trig substitution

[upper fog]

nope, its a "simple" trig identity

the arctan one

i have to convert this shit to 1/(x+c)^2 + 1

which seems IMPOSSIBLE

[upper fog]

you cant trig sub this one btw

[spring trail]

like not allowed or CANT

[upper fog]

you cant

its a complete the square problem

not allowed u sub or integral sub

[upper fog]

for anyone coming into the channel, this is the equation i need help with. no u sub allowed

<@&286206848099549185> , its been 15 minutes

my friend gave up on it, maybe its impossible to do without sub?

but symbolab gives an answer

[spring trail]

well u can

but like

its kinda dishonest coz im just building back from the answer

apparently u can mutiply by 2/2

and take root(195) common from the denominator

take the 2/root(195) out of the intergral

multiply the 2 in the denominator

[upper fog]

?

yeah

[spring trail]

then try to get into the square + 1 form

[upper fog]

?

how do i get rid of the five

i cant

[spring trail]

wolfram spits this out

u can try building to this

[upper fog]

oh hell naw

im just taking the C

i dont even care anymore

[fierce lagoon]

Completing the square integrals bites ass

[upper fog]

.close

[obtuse pebble]

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[timid silo]

I don't understand why we divide it by negative sqrt x

[timid silo]

How do I know when to divide it by negative sqrt x?

and please ping me in your response, thank you very much in advance!

[royal basin]

not -sqrt(x) but -sqrt(x^2) @timid silo

In the numerator, let's divide by -sqrt(x^2), since for negative values [of x], x = -sqrt(x^2).

do you understand the bolded part?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

I’m sorry not really

Is it possible to explain it in simpler terms

Or give an example

[royal basin]

take x = -10

x^2 = 100, and so sqrt(x^2) = 10

so x = -10 and -sqrt(x^2) = -10

[timid silo]

Oh my god

Okay that clicked so well

Thank you very much

.cllse

.close

[obtuse pebble]

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[sharp hare]

its given that -2 is a root in the polynomial p(Z)

I need to find the multiplicity of -2

All I know is that I need to find out how many times -2 can be put into this root

[static marten]

do you know the method for finding multiplicity?

you differentiate and test the root

[sharp hare]

Can you give a super simple example?

[static marten]

f(x)=(x-2)^2

f'(x)=2(x-2)

f''(x)=2

so x=2 has a multiplicity of 2

f(x)=(x-1)^3(x-3)

f'(x)=3(x-1)^2(x-3)+(x-1)^3=(x-1)^2(4x-10)

[sharp hare]

hmm

[static marten]

f''(x)=2(x-1)(4x-10)+4(x-1)^2=(x-1)(12x-24)

etc.

x=1 has multiplicity 3

[sharp hare]

but in this case its complex numbers so idk

[static marten]

is it

it should still work for complex numbers

[sharp hare]

so you just differentiate every time? @static marten

because I already know 1 root

just need to know its multiplicity

[static marten]

yeah you keep differentiating and testing the root

as long as it keeps working is its multiplicity

[sharp hare]

How would you do it for one as long as the one in my picture? @static marten

Can we do it together?

[static marten]

uh P(z)=z5+4z4+4z3+5z2+20z+20

we know that z=-2

P'(z)=5z4+16z3+12z2+10z+20

test z=-2: 5x16-16x8+12x4-20+20=0

P''(z)=20z3+48z2+24z+10
test: 20x-8+48x4-24x2+10=-6

so it looks like z=-2 stops working for the second derivative @sharp hare

[sharp hare]

mhm

[static marten]

so what does that mean for the mulitplicity?

[sharp hare]

a multiplicity of 2?

[static marten]

yeah

[sharp hare]

okii

I really hope this also works for complex numbers xD

[static marten]

there shouldn't be any problem

if you have like (z-i)^2

that doesn't change how it works

[sharp hare]

wait

I think you miscalculated the second derivative

[static marten]

did i

[sharp hare]

nvm

its me who is wrong

but ok, this is a way simpler method

Do you have a way to do it with factorization ? @static marten

[static marten]

what do you mean

if you factorise it and find that it's like (z-2)^2, then yeah that's a shortcut

[sharp hare]

lemme send you a pic

this is how the teacher did it for the polynomium p3 i sent

[static marten]

yeah

that works too

[static marten]

but do you want to factorise this?

[sharp hare]

But their logic won’t work with large polynomials like the one I have right

[static marten]

it works, but it's harder

[sharp hare]

It seems too hard

Yeah I guess they want to factorise that one

[static marten]

this use of differentiation only works because of the factorised form

[sharp hare]

How do I use their method for large polynomials?

[static marten]

what do you mean? how to factorise large polynomials?

[sharp hare]

Yeah

[static marten]

just do it?

i mean you have different things like rational root theorem

[sharp hare]

Is there a method? Its extremely tedious

[static marten]

but that only works for... rational roots

[sharp hare]

How would you factorise that large one starting with z^5

[static marten]

so we know it's got one root at z=-2, so factorise that out

and then guess

z5+4z4+4z3+5z2+20z+20=(z+2)(z4+2z3+5z+10) it looks like

we found out it's got a double root at z=-2: (z+2)^2(z3+5)

and from here, z^3+5 is sum of cubes z^3+a^3=(z+a)(z^2-az+a^2)

so P(z)=(z+2)^2(z+cbrt(5))(z^2-cbrt(5)z+cbrt25)

$P(z)=z^5+4z^4+4z^3+5z^2+20z+20=(z+2)^2(z+\sqrt[3]5)(z^2-\sqrt[3]5z+\sqrt[3]{5^2})$

[warm shale]

chlamydia

[sharp hare]

okay, wow

[static marten]

and it looks like the last quadratic factor is irreducible

so yeah why would you want to do that

leave it for the low degree polys

[sharp hare]

you are totally correct

i guess they just wanted us to do the first step

kek

[obtuse pebble]

@sharp hare Has your question been resolved?

[obtuse pebble]

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[timid silo]

Hi

[timid silo]

I'm studying arcfunctions so

Is this the correct formula?

[royal basin]

yes all these are correct

[timid silo]

ok

bye

[balmy mortar]

.close on your way out

[obtuse pebble]

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[timid silo]

WAIT I FORGOT TO WRITE IT ON THE PAPER

NVM

[obtuse pebble]

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[twin stump]

just a quick question.. if im writing a solution set, do i have to write every possible answer or just the realistic ones

[twin stump]

like if it was a real world problem

negative answers dont make sense

[royal basin]

if it's a real world problem then you should filter out the ones that don't make sense

[twin stump]

so if im asked for the age, i should disregard negative values?

okay ty

[neon vector]

Yes.

Can't really be -3 years old 😄

[balmy mortar]

you have to write all the solutions, and only the solutions

no extra ones. Like -3 is extra when the solutions must be positive

its not a solution.

[twin stump]

i see

okay thx guys

.close

[obtuse pebble]

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[kindred moss]

I put the y value in but didn’t get the same x

so I think this answer is wrong

and I’m not sure where to go from here

[paper steeple]

Hey can I ask

I'm dumb at math

[kindred moss]

wat

[static marten]

which part is this

[kindred moss]

part a

[paper steeple]

That

[static marten]

#❓how-to-get-help

[kindred moss]

open ur own thing

[paper steeple]

I'm dumb

Like bro I'm literally dumb

[kindred moss]

OPEN UR OWN FORUM

[paper steeple]

What's that?

[kindred moss]

https://discord.com/channels/268882317391429632/903463351789694986

go there

[sage iron]

why did you calculate the second derivative

[kindred moss]

n paste it

[kindred moss]

cause it’s then in the form of mx + c

instead of a quadratic

we wanna find the line

[sage iron]

a tangent line has the same slope as the curve at a point

and when you hear the word slope, that’s first derivative

[kindred moss]

right

[sage iron]

second derivative is useless here

[kindred moss]

so what do I do with the first one

find x?

[nocturne nebula]

Substitute the value of X coordinate at point P as we are finding the slope of tangent at point P

[kindred moss]

we already have the x and y of that point

we need the equation of the line

[nocturne nebula]

Yeah to find equation of a line, we need to find its slope first

And slope of a line is found using the first derivative

[kindred moss]

kay so I put the x value into the first derivative?

[nocturne nebula]

Yes

[kindred moss]

I got 2

x is 5 y is 2

[nocturne nebula]

Nope, the value you got is just the slope not a coordinate, it is the value of m in the equation
y = mx + c

[kindred moss]

it is?

[nocturne nebula]

Yes

[kindred moss]

so it’s 2x?

y = 2x + c

now do I put x and y in

and find c

x and y of point p

[nocturne nebula]

Do you know the standard line equation, Y - y = m(X - x) ?

[kindred moss]

yeah

[nocturne nebula]

Just a sec

[kindred moss]

[nocturne nebula]

Yup just like that

[kindred moss]

tyyy

[nocturne nebula]

.close

[kindred moss]

waittt

part b

do I equal the equation of the line to the derivative

@nocturne nebula

[nocturne nebula]

Yup

[kindred moss]

okay ty

[nocturne nebula]

Wait, eqaute the equation of line to the equation of the curve

Not to the derivative

[kindred moss]

ah okay

and then find x?

how do prove it

[nocturne nebula]

You should be getting two values of X?

[kindred moss]

yeah

and do I just plot it?

like how do u prove c is on l

[nocturne nebula]

What values did you get?

[kindred moss]

I dunno i put my book away

im outside

I just want to know the steps

[nocturne nebula]

Wait

On equating both the equations you will get X = 0 and x = 5

The value x = 0 zero indicates that the curve and line meet on the y-axis

To verify substitute the value of X = 0 in the equation of curve and the equation of line, you will get the same value of y

@kindred moss got it?

[obtuse pebble]

@kindred moss Has your question been resolved?

[obtuse pebble]

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[elder nexus]

can anyone help me and teach me how to do number 15 pls

[neon vector]

,rccw

[neon vector]

Do you know how to find the inverse of a function?

[obtuse pebble]

@elder nexus Has your question been resolved?

[obtuse pebble]

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[timid silo]

I have no clue on this and ive attempted this ten times. DO NOT give me the answer.

[cinder lintel]

take 2 cases

both ardvarks at edges

and both armadillos at edges

youll get favourable outcomes

[timid silo]

I tried that

[cinder lintel]

divide by total permutations

what you getting

[timid silo]

1/20

which is wrong

[cinder lintel]

nope

you did it wrong

[timid silo]

all of the answers i tried

oh my

[cinder lintel]

retry that

i think

you didnt divide

because identical

[timid silo]

What probabilities did you get for ardvarks and armadillos?

i didn't really think of that...what i did once was take .6 for armadillos and .4 for aardvarks

[cinder lintel]

how many cases you getting each

no thats not how it works

[timid silo]

first i tried 6 cases cause i thought of overcounting

[cinder lintel]

do you know permutation and combination?

[timid silo]

i know permutations yes

[cinder lintel]

if u want i can explain the process just not do numerical

for armadillos right

[timid silo]

then i tried 5! cases

[cinder lintel]

if they are on the edge

[timid silo]

yeah

[cinder lintel]

then all spots have to be filled

by ardvark

so theres only 1 way

in this case

[timid silo]

yeah

[cinder lintel]

because they are identical

now if armadillo

on edge

you have to arrange

[timid silo]

yea

[cinder lintel]

1 armadillp

and 2 ardvark

[timid silo]

1/30?

[cinder lintel]

in 3 places

no

[timid silo]

oh

[cinder lintel]

how many cases

for this?

[timid silo]

4 successes

[cinder lintel]

4 total favourable

yes

1+3

[timid silo]

but we need the probability

not the cases

[cinder lintel]

yes

to get probability

u do

favourable by total

we got favourable

now what do you think

total outcomes

are

[timid silo]

total was the one i has issues

[cinder lintel]

its not 5!

[timid silo]

ik

[cinder lintel]

its 5!/3!2!

do you know why?

[timid silo]

no

[cinder lintel]

because all the 3 armadillos

are identical

[timid silo]

mhm

[cinder lintel]

say i have a1 a2

and arrange it as a2 a1

the word remains same

its still aa

so didnt count that

[timid silo]

i put 5/12

incorrect

[cinder lintel]

likewise

its not 5/12

its 4/10

[timid silo]

3! x 2! is 12

[cinder lintel]

yes

120/12 is 10

4/10

2/5

[timid silo]

i am kinda confused

[cinder lintel]

which part

[timid silo]

this is a new concept

for me at least

[cinder lintel]

im only dividing because identical objects

[timid silo]

yea

[cinder lintel]

theres a word dad okay

now

[timid silo]

can you reexplain it but go a bit slower

[cinder lintel]

only the division part?

or whole

[timid silo]

division tbh

[cinder lintel]

for division

aight

say you have a word dad

you should get 3! that is 6 permutations to this?

yes

[timid silo]

yeas

[cinder lintel]

but

[timid silo]

i know that

[cinder lintel]

if i exchange

both the ds

its still the same word

[timid silo]

but i thought if i did 5!

[cinder lintel]

yes?

[timid silo]

i was overcounting

[cinder lintel]

add

dad

dda

these are the only permutations

[timid silo]

yes but the word dad we get 3 cases

[cinder lintel]

ye

[timid silo]

because d gets overcounted

[cinder lintel]

what is this 3

its 3!/2!

[timid silo]

OH WAIT

I GET IT NOW WE OVERCOUNT THOSE TWO

[cinder lintel]

yes

because interchanging d

result in same word

[timid silo]

SO WE DO 120/12 = 10

[cinder lintel]

so divide by 2

yes

[timid silo]

AND THEN THE CASES ARE 4

[cinder lintel]

yes

[timid silo]

DIVIDE BY 2 AND GET 2/5

[cinder lintel]

yes

[timid silo]

thanks man

[cinder lintel]

np

pnc is a lovely topic

[timid silo]

yea

personally algebra is my game tho

.close

[obtuse pebble]

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[shadow lava]

Wouldn't it be infinity tho? it keeps going up forever if t is undefined

https://www.desmos.com/calculator/aednpwqkdb

right, so it's definately not infinity, but when I look at the graph it looks like it would be

the question is asking what the maximum height of the area under the curve is from 0 (lower limit) to t (upper limit)

however, the graph just just keeps shooting up forever so I'm not sure how to answer this

i also tried t = t but got the same result

[frosty spoke]

not sure that your graph that you posted is actually relevant

it's not asking for the local maximum of (x^2+13x+30)/(1+cos^2(x))

it's asking for the local maximum of the integral of that from 0 to t

so while you may be right that the area keeps going up, that's not super relevant, because it's possible for a function to have local maxima while the absolute maximum is infinity

[shadow lava]

oh

[frosty spoke]

I'm too lazy to actually plot the integrated graph to show you, but a natural thing to solve this problem would be using a first derivative test to find the critical points and a second derivative test to classify the critical points

[shadow lava]

you mean like here?

[frosty spoke]

no I don't mean like there

[shadow lava]

or again, graph not helpful

OK

just making sure

[frosty spoke]

do you understand the difference between the local maximum of f(t) and the local maximum of (x^2+13x+30)/(1+cos^2(x))?

[shadow lava]

no

[frosty spoke]

do you agree that f(t) and (x^2+13x+30)/(1+cos^2(x)) are two different things?

[shadow lava]

but from research, factoring is involved somehow

[shadow lava]

well, (x^2+13x+30)/(1+cos^2(x)) is not a function of t at all

t is the upper limit

meaning the right most point on the x axis

from 0 -> t

[frosty spoke]

suppose I write $f(t) = \int_0^t \frac{x^2 + 13x+30}{1+\cos^2 x} , dx$ and $g(x) = \frac{x^2+13x+30}{1+\cos^2 x}$. Do you see the difference between them?

[warm shale]

Saccharine

[shadow lava]

yes, one is f(t) and the other is g(x)

so g(x) equals that function

[frosty spoke]

do you see that g and f are not the same function?

[shadow lava]

f(t) I'm not sure lol

yes

but they are declared with different function names, f(t) and g(x)

[frosty spoke]

so does it make sense then that the local maxima of f and the local maxima of g are different?

[frosty spoke]

I can write f(t) = 3t and g(x) = 3x, and those would be the same function

[shadow lava]

right

[frosty spoke]

which of f or g do you think you've graphed?

[shadow lava]

but what is f(t) actually referring to? if not the function itself

[frosty spoke]

f is a function

[shadow lava]

oh so you just replace x with t

[frosty spoke]

did you not read the part of the question that says $f(t) = \int_0^t \frac{x^2 + 13x+30}{1+\cos^2 x} , dx$

[warm shale]

Saccharine

[shadow lava]

like a constant, f(0) you replace x with 0
f(t) you replace x with t

[frosty spoke]

no wtf

did you read the question or not

[shadow lava]

At what value of 𝑡
does the local max of 𝑓(𝑡)
occur?

[shadow lava]

yes, I read it

[frosty spoke]

if you're so hung up on the choice of letters, then write it as $f(x) = \int_0^x \frac{z^2 + 13z+30}{1+\cos^2 z} , dz$

[shadow lava]

oh

[warm shale]

Saccharine

[frosty spoke]

do you see that these mean the same thing?

[shadow lava]

I do F(t) - F(0)

[frosty spoke]

bruh I'm not even talking about what to do

I'm trying to get you to first even understand what the question is even SAYING

[shadow lava]

F(x) - F(0)

it's asking for the integral

[royal basin]

procedural understanding clashes with conceptual/comprehensional explanations

[shadow lava]

afaik this is what it's saying

[frosty spoke]

Question: "at what value of t does the local max of f(t) occur?"

it's asking for the integral

something doesn't match up here...

[shadow lava]

this on it's own, would be asking for the integral, right?

[frosty spoke]

that is not a question; that's just some expression written out

[shadow lava]

the end question is different, yes

but wouldn't I need to evaluate first? or no need

[frosty spoke]

I can define functions like $f(x) = \int_0^x 2 , dy$

[frosty spoke]

try to understand the question first before trying to churn out a bunch of useless steps

[shadow lava]

OK

I am trying

[frosty spoke]

there's a good reason why most people solve problems by thinking about them

[royal basin]

❌

stinky

[warm shale]

Saccharine

[royal basin]

there we go

[frosty spoke]

rip me I'm horribly jet lagged

[royal basin]

well some people are, or consider themselves to be, Not Good At Thinking

[shadow lava]

NGAT syndrome

my life with mathematics, until it finally clicks

[frosty spoke]

okay, ignoring the stuff after f(t) for a second

[royal basin]

NGAT

[frosty spoke]

do you recognize that the question is giving you a function f

[royal basin]

/ŋæt/

[frosty spoke]

and telling you to find its (presumably unique) local maximum?

[shadow lava]

yes

[frosty spoke]

so how do you find a local maximum of a function?

[shadow lava]

it's giving me more than a function tho

isn't it?

[frosty spoke]

no it's not

it's saying f(t) = something involving t

[shadow lava]

it's also giving me an integral with upper and lower limit, with respect to dx

[frosty spoke]

and asking you to find the local maximum of it

[frosty spoke]

that would fall into the category of "something involving t"

[shadow lava]

or would this also be considered a part of the function?

i look at the integrand as the only part of the function, maybe that is wrong to do so

[frosty spoke]

and I look at the moon and wish it were made of cheese

[shadow lava]

OK so the entire statement is a function

[frosty spoke]

do you agree that $\int_0^t \frac{x^2 + 13x+30}{1+\cos^2 x} , dx$ is "something that depends on t"?

[warm shale]

Saccharine

[shadow lava]

yes, but only for the upper limit of the area under the curve

[frosty spoke]

I don't care about the extra stuff you have to say about it, becuase it's mostly nonsense

[shadow lava]

something that depends on t: yes.

[frosty spoke]

do you agree that if I gave you a value of t, say t = 2, and then told you to plug it in, you would get a number out?

[shadow lava]

yes

it would become a definite integral and I would get a constant value for the result back

[frosty spoke]

doesn't that seem a lot like a function?

if I tell you that f(t) = 3t

[shadow lava]

it does seem like a function, yes

[frosty spoke]

and I tell you t = 2, you can plug it in and get a number out

[shadow lava]

f(2) = 6

[frosty spoke]

so what f(t) = [the integral that I won't keep bothering to reproduce here] means is that you can compute f(2) by plugging 2 into where t is in the integral

and likewise for any other number

[shadow lava]

but which number do I replace t with?

[frosty spoke]

like $f(3) = \int_0^3 \frac{x^2 + 13x+30}{1+\cos^2 x} , dx$

[warm shale]

Saccharine

[frosty spoke]

or $f(5) = \int_0^5 \frac{x^2 + 13x+30}{1+\cos^2 x} , dx$

[warm shale]

Saccharine

[sage dagger]

Yo am I tripping or is that texxit font different

[frosty spoke]

it is different

[shadow lava]

lol, yes I understand this. t can be anything

[sage dagger]

ok good thought I was going crazy

[frosty spoke]

so then do you understand that writing $f(t) = \int_0^t \frac{x^2 + 13x+30}{1+\cos^2 x} , dx$ defines the function f?

[warm shale]

Saccharine

[frosty spoke]

it tells you what f is

[shadow lava]

yes

[frosty spoke]

now that you know what f is, you're asked to find a local maximum of it

how do you find local extrema of a function?

that's complete nonsense

x is a dummy variable in integration

it could be anything EXCEPT t

[shadow lava]

i would do a sign test I think

[frosty spoke]

can you stop trying to incorporate what I'm saying into your completely messed up understanding of everything

[shadow lava]

factor to find roots

[frosty spoke]

like I'm trying to give you a reasonably complete explanation of the steps, and you're constantly saying non-sequiturs like you're on some parallel reality

[shadow lava]

dude, I'm just explaining how I misunderstood. it is wrong. i know that. sorry for speaking.

you know what I'm gonna close this. you obviously need to get some sleep from jet lag. thank you for your help.

.close

[obtuse pebble]

Channel closed

Closed by @shadow lava

Use .reopen if this was a mistake.

[balmy mortar]

🤨

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[shut pecan]

how Do I do this

[neon vector]

Do you know what f(c) is?

[shut pecan]

no

[neon vector]

Hm

Do you know what a function is?

[shut pecan]

yes

its like

f(x)

its a relation between y and x

[neon vector]

Yes

The function f takes the x, does something to it and outputs y

So let's say we have a function $f(x) = 2x$

[warm shale]

USS-Enterprise

[shut pecan]

Ok

y=2x

[neon vector]

This means the function f takes an x, multiplies it by 2 and returns y

So if we pick x = 3, what do we get as the output (y)

[shut pecan]

6

[neon vector]

Correct

Now it doesn't always have to be x

We can have $f(h) = h - 3$

[warm shale]

USS-Enterprise

[neon vector]

And if we pick h = 4 we get the output of 1

[shut pecan]

so h is just representing x?

[neon vector]

You can think of it as that

x is replaced by h

You can do it with any letter

But we just most often use f(x) because points on the cartesian coordinate system in a plane have this structure: (x, y)

[shut pecan]

ok

so

[neon vector]

So if we have a function f(c)

What is the value of the function at c = -1

[shut pecan]

y is -1

i got it

can we do the next one

[neon vector]

Why -1?

[shut pecan]

when the fuction is -1

means

that

y is equal to -1

so I found the points where y was -1

[neon vector]

OH nevermind

I misread

Yes you are correct

[shut pecan]

can we do next one

[neon vector]

What does f(-2) mean?

[shut pecan]

x is -2?

[neon vector]

You've got f(x) = y

Correct

[shut pecan]

oh

so

wait

no

i dont get it

oh wait

no

i get it

[neon vector]

You substite x for -2

[shut pecan]

is it 1

[neon vector]

It's close

But it's just somewhere there

I'd say $\frac{2}{3}$

[warm shale]

USS-Enterprise

[neon vector]

Imagine a line x = -2

And see where it intersects the graph

That's your y value

[shut pecan]

can we do 2 more

[neon vector]

Sure

[shut pecan]

I dont know what it wants me to compute

what does that last part mean

[neon vector]

h(h(7))

Think of it as h of h(7)

So once you get h(7) you do h of that

Get it?

[shut pecan]

ok

i get it

[neon vector]

So what's h(7)?

[shut pecan]

y

[neon vector]

How much

[shut pecan]

7y?

[neon vector]

Again, imagine the line x = 7 and see where it intersects the grapj

[shut pecan]

5

[neon vector]

Correct

[shut pecan]

yoo i got it

-1

[neon vector]

And now you need to do h(5)

Correct

[shut pecan]

I have 5 left are you ok with helping me with them

[neon vector]

Yes of course 🙃

[shut pecan]

tyyy

what does invertible mean

[neon vector]

Hm

You guys were given this but the teacher hasn't explained what an inverse function is?

[shut pecan]

I missed this class

im going to look in the transcript

[neon vector]

Ah

Pretty much

If you have a function $f(x) = x + 1$

[shut pecan]

yeah

[warm shale]

USS-Enterprise

[neon vector]

To find its inverse you first have to check whether it's bijective

[shut pecan]

bijective?

[neon vector]

It's bijective when it's both injective and surjective.

[shut pecan]

injective

surjective

hang on

[neon vector]

Very briefly explained, it's surjective if each element of the codomain is mapped to by at least one element of the domain

[shut pecan]

theres nothing in the transcript

is x+1 the codomain

[neon vector]

And injective if no two different x's go into the same y

[shut pecan]

well there is only 1x

so does that mean it is injective

[neon vector]

What I mean is, if for example you pick x=2 and x=9 and they both go into the same y

If this happens, the function is not injective and doesn't have an inverse

[shut pecan]

oh ok

wait so for this

do i find wehre x is 4

[neon vector]

Anyway

For this

[shut pecan]

when x is 4 y is 8

[neon vector]

All you need to know is that graphically, the inverse of a function is reflected in the line y=x

Which means its coordinates get switched

[shut pecan]

what

[neon vector]

If you have a point (x,y)

And reflect it in the line y=x

You get a new point with coordinates (y, x)

[shut pecan]

they have the same coordinates

[neon vector]

They switch places

[shut pecan]

ok

i get it

[neon vector]

So here you x=4 and y=8

The inverse will have x=8 and y=4

[shut pecan]

it doesnt go up to there on teh graph

[neon vector]

This graph won't show it

The graph of the inverse function will show it

[shut pecan]

yeah

the inverse function is

x=8, y=4 righty

[neon vector]

If you pick x=8 you will get y=4, yes

[shut pecan]

can I pick a diff point

like

2 and 1

i still dont udnerstand like

[neon vector]

(2, 1)?

[shut pecan]

what are we trying to find

[neon vector]

Yeah, it's pretty complicated

[shut pecan]

like for the answer

[neon vector]

Would take me quite a while to explain all of it

Your best bet is to ask your teacher if he can explain it to you or your classmates

[shut pecan]

this is like an online class

[neon vector]

Still

You first need to know what an inverse function is

[shut pecan]

ok

[neon vector]

Then this question seems easy to you

[shut pecan]

let me send the other ones

[neon vector]

Hm

[shut pecan]

we could do 8 maybe

[neon vector]

Do you know how to calculate when two graphs intersect

Oh

[shut pecan]

when they have a point that is equal to each other

like

honestlyu7

these are kind hard

[neon vector]

Yeah, it can suck when you miss class

If you have someone, I'd greatly suggest asking them to explain all of this to you in person

[shut pecan]

ok

thanks for help on the other ones

i got this one

[neon vector]

No problem

[obtuse pebble]

@shut pecan Has your question been resolved?

[obtuse pebble]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[civic bone]

in these equations

https://i.imgur.com/WxUmDEs.png

[civic bone]

I'm not following line 3

[royal basin]

AH = 1

AH/AC = cos(β)

=> AC = 1/cos(β)

do you understand this?

[civic bone]

yeah

where does sin BAC go though

oh nvm I get what it's trying to say

I was trying to get from lef thand side to rhs and I was like but you already need to know the equality it's trying to show

but you basically follow the left side vertically and the right side vertically

then it makes sense

unless I'm wrong lol

nope wait

going vertically from line 3 to 4 I still don't get how we get (2 x sin(angle BAC)) = sin (alpha + beta)

any helpers?

[obtuse pebble]

@civic bone Has your question been resolved?

[civic bone]

.close

[obtuse pebble]

Channel closed

Closed by @civic bone

Use .reopen if this was a mistake.

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dim bobcat]

Base*height

[native inlet]

V=bh

[wide vigil]

8x3?

[dim bobcat]

Area of base * height

[wide vigil]

?

[native inlet]

calculate the area of the triangle

[wide vigil]

72x3

?

is this correct?

72x3

[dim bobcat]

Nah

27×3

[wide vigil]

ok

[obtuse pebble]

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.

[wide vigil]

81?

@dim bobcat

[dim bobcat]

Yeah

[wide vigil]

ok ty

ur the best sir

[dim bobcat]

Nah

u just being too humble

[wide vigil]

ur still very helpful

i like people like u

ty

.close

[obtuse pebble]

Channel closed

Closed by @wide vigil

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[grim trout]

I am conpletely lost on this one, Im generally not good at functions so i could just be completely missing it, But how do i find the output when i only have a variable?

[wide vigil]

thats easy

u have to times by 4

or something

[grim trout]

why

[wide vigil]

7x4=28

[grim trout]

what do i times by 4?

ohhh

[wide vigil]

n

2x4=8

8x4=32

u dee the repetion

x4

[grim trout]

yeah

thank you

.close

[obtuse pebble]

Channel closed

Closed by @grim trout

Use .reopen if this was a mistake.

[wide vigil]

pls help

@dim bobcat

pls help

p.reopen

.reopen

[obtuse pebble]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

Available help channel!

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Remember:
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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[timid hemlock]

-3x+5y=-25 for y

[neon vector]

Solve for y?

[timid hemlock]

Yes

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[timid hemlock]

I need help starting if

It

[neon vector]

Right, do you know what solving for y means?

[timid hemlock]

We have to get everything y=max+b form?

[neon vector]

You mean $y = mx + b$

[warm shale]

USS-Enterprise

[timid hemlock]

Yeah?

[neon vector]

You said max 😄

[timid hemlock]

Oops

[neon vector]

This is the form of a linear function

[timid hemlock]

Autocorrect

[neon vector]

$f(x) = mx + b$

[timid hemlock]

Y=mx +b?

[warm shale]

USS-Enterprise

[neon vector]

Yes

f(x) = y

If I give you this equation

2 + y = 3

And tell you to solve for y

[timid hemlock]

We would move the 2 to minus 3?

[neon vector]

What do you mean

[timid hemlock]

Like to get f by it’s self we would move the number beside it

[neon vector]

Correct

And we would get what

[timid hemlock]

Y=1

[neon vector]

Correct

You've just solved for y

That's what solving for a variable means

Isolating that variable on one side and equating it to everything else on the other

[timid hemlock]

Ok

[neon vector]

So if we go to the original question

[timid hemlock]

-3x+5y=-25 for y

[neon vector]

$-3x + 5y = -25$

[warm shale]

USS-Enterprise

[neon vector]

Solve for y

We have to isolate y

What do we do first

[timid hemlock]

So we would want -3 to get next to the -25 right? So we would move it over

[neon vector]

Correct, but the x is hanging onto -3

So we move the entire term -3x to the right

[timid hemlock]

5y=-25+3x?

[neon vector]

Correct

Now it is best to use commutativity and put the terms with variables in front

$a + b = b + a$

[warm shale]

USS-Enterprise

[neon vector]

So we get

$5y = 3x - 25$

[warm shale]

USS-Enterprise

[neon vector]

Do you understand what we did

[timid hemlock]

Yes

[neon vector]

Okay

Now what do we do as the final step

[timid hemlock]

So do we divide 5y by everything

[neon vector]

Correct

And we end up with what

[timid hemlock]

Y=3/5x-5

[neon vector]

$y = \frac{3}{5}x - 5$

[warm shale]

USS-Enterprise

[neon vector]

Correct

You've just solved for y again

[timid hemlock]

Ok

Thank you

[neon vector]

No problem 🙃

[obtuse pebble]

@timid hemlock Has your question been resolved?

[obtuse pebble]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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[covert spoke]

what is the derivative of 3/x^2 and how do i figure it out

[royal basin]

the derivative of x^p is p * x^(p-1)

this works for all p

[covert spoke]

oh so it’s just 3/2x

[obtuse pebble]

@covert spoke Has your question been resolved?

[kind hawk]

no

remember that 3/x^2 is the same as 3x^(-2)

[obtuse pebble]

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[twin peak]

what is the difference between (y / x)^2 and y^2/x^2 in terms of cartesian plans

[dim root]

They are the same, no change between the two.

[twin peak]

.closed

[tepid mountain]

.close

[twin peak]

.close

[obtuse pebble]

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[cold brook]

how can we mathematically prove that 0,0 is not a tangent?

[cold brook]

i was thinking of

[kind hawk]

(0,0) is a point, can't be a tangent because a tangent is a line

or what do you mean

[cold brook]

i mean to find the point of tangency i did

-x = x^2 -6x^2 + 8x

and this is true when x = 3 and x=0. at x=3, the line y=-x is tangent to the curve but it just intersects the curve at 0,0 so i wanted to know how can we know without a graph that its not tangent at 0,0 and tangent at x=3

i was thinking of finding the deriviative

s

3x^2 - 12x + 8 = -1

and now we get x = 1 and x=3

so at x = 0, it doesnt have same -1 gradient so its not a tangent

but idk if thats the right way to do it

[kind hawk]

yes to be tangent they need to have the same gradient

tangent=intersection+same gradient at that point

[cold brook]

So in further questions like this

i should

equate the graidnet with the derivative

and the equation of the tangent with the curve equation

and find the points which are comman

that should be my answer right

since its fulfilling both the conditions

[kind hawk]

well you only need to solve one of the equations and then can plug into the other one

which is usually faster

but yes

[cold brook]

okayy thankyou so much

.close

[obtuse pebble]

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[split estuary]

What’s the initiative behind this question? I feel like it’s obvious but I can’t think

[vestal cape]

use the table bro it shows you the distance between the 2 places

lower number = close

high number = far

read it kinda like a multiplication table

[split estuary]

https://i.imgur.com/mPyNlg4.png

not sure really what im looking at

[vestal cape]

?

[vestal cape]

2nd column 1st row that square shows you the distance between EGVN and EGXC

[split estuary]

EGXU seems to be smallest compared to EGXV

[vestal cape]

yes

[split estuary]

So it could be 33 (EGXU) with 133(EGXV)

or

30 (EGXU) with 138 EGVN)

is it which is the smallest difference is better?

[vestal cape]

I would say the 2nd one

2nd one is closer and futher away

[split estuary]

how could u verify that?

[vestal cape]

cause the 1st number is smaller and the 2nd number is bigger

...

[split estuary]

ok, so it would be EGXU

[vestal cape]

yeah I'd assume so

[obtuse pebble]

@split estuary Has your question been resolved?

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[ruby flame]

How do i finish this?

[latent walrus]

what is (sqrt(2))^7

[ruby flame]

49

[latent walrus]

thats 7^2

[ruby flame]

oh

196?

[latent walrus]

$(\sqrt{2})^7=((\sqrt{2})^2)^3\cdot \sqrt{2}$

[warm shale]

AℤØ

[tall ore]

Hi

[latent walrus]

hi

[ruby flame]

11.3137085?

[latent walrus]

well.... yeah but you want it in the form a*sqrt(2)

[latent walrus]

thats why i separated it like this

[ruby flame]

Oh so you write it like that

[latent walrus]

its just a way to make it more clear

[latent walrus]

dont have to

[ruby flame]

So like this