help-33

[onyx sentinel]

I replaced

And then i attempted to solve

[stark trail]

where did the absolute values go?

that's the mistake

[onyx sentinel]

Ohhh

[stark trail]

here's what I would do

|1+(-1)^n/n| <= |1| + |(-1)^(n)/n| <= 1 + 1/n

the first inequality is true because of the triangle inequality, |a+b| <= |a| + |b|

from this, you can clearly see that as n gets larger, 1+ 1/n gets smaller

so it's bounded above by when n is at its minimum, n=1

i.e, it's bounded above by 1+1/1=2

However, you could be more strict with your bounds because the sequence is also bounded above by 3/2

[high tree]

another route for boundness would be lipschitz constant

or epsilon delta proofs

[onyx sentinel]

oh

[stark trail]

That's not at all needed here

[onyx sentinel]

epsilon delta is what my prof told me to use

at least in the examples she showed

[high tree]

yeee

[onyx sentinel]

i just couldnt understand

[high tree]

its not needed but better for general caess

[stark trail]

This has nothing to do with epsilon delta

[high tree]

are you 2nd year uni also?

[stark trail]

That is for proving continuity of functions

[onyx sentinel]

im only 1st

[high tree]

learned them a few months ago

[stark trail]

This is a sequence

[high tree]

dam unlucky

yeah>

you can use epsilon in sequences

[onyx sentinel]

one sec

[stark trail]

Not epsilon delta

[onyx sentinel]

oh its only epsilon i think

[high tree]

ahh

[stark trail]

It's completely unnecessary and overcomplicated for this problem

[onyx sentinel]

but if i needed to use it

how would i?

[stark trail]

It isn't used for this type of problem

[onyx sentinel]

oh ok

[stark trail]

If you wanted to find the limit of a sequence, you could do a formal proof using the epsilon-N definition

However this does not ask you to compute any limits

[high tree]

Truee trueee

mb mb

[onyx sentinel]

ah i see

[high tree]

we are just finding the sup ?

or is the condition not tha strict

[stark trail]

Determining if the sequence is bounded and/or monotonic

I'm unsure if we need to find the sup, or just determine if it's bounded

the question isn't worded clearly

[onyx sentinel]

i just need to determine if its bounded i think

what is sup?

[stark trail]

As we realized earlier

the sequence is bounded by 2

but also by 3/2

so when it asks for the sequences boundedness

it may be asking, is it bounded, answer yes or no

or it may be asking, is it bounded, what is the least upper bound

[onyx sentinel]

ohhh

[stark trail]

least upper bound = supremum = sup

[onyx sentinel]

ah i see

[stark trail]

just vocabulary

[high tree]

it approches 1

[onyx sentinel]

alright lemme take alook at ur solution again

[onyx sentinel]

how did you get the 1 out here? like shouldnt it be |1/n|?

sorry

nvm

i didnt read it right

for this part

even if i dont have an idea what its bounded by can i just start solving the question?

do i need to know the value 3/2

[stark trail]

you don't

[onyx sentinel]

oh ok

[stark trail]

but it would be a mistake to just start blindly solving aswell

simply looking at the sequence in most cases you should be able to tell

[onyx sentinel]

ah ok

[stark trail]

1+(-1)^(n)/n

it should be clear what this does, and if not find a few test values

[onyx sentinel]

could u explain why it would be a mistake to start blindly solving?

if i just needed to find if its bounded i mean

[stark trail]

Because what if the sequence is unbounded, and you start trying to do some complicated algebra to find a bound?

waste of time

[onyx sentinel]

ohhh

ok that makes sense

[stark trail]

or if the sequence is bounded and the bound is obvious by inspection, but doing the algebra might get messy

[onyx sentinel]

ahh

[stark trail]

Like

1+(-1)^(n)/n

just compute n=1, 2, 3

you get

1-1=0
1+1/2=3/2
1-1/3=2/3

can't you see what the whole sequence does now based on that?

it follows a very clear pattern

[onyx sentinel]

i dont get the first term though

like it starts at 0

increases

and then starts decreasing

[stark trail]

1+(-1)^(1)/1

=

1-1/1

=

1-1

=

0

[onyx sentinel]

i get that part

but like

why is it a clear pattern

[stark trail]

the sequence alternates

[onyx sentinel]

if it increases and decreases

[stark trail]

every term

you subtract 1/1

you add 1/2

subtract 1/3

add 1/4

subtract 1/5

add 1/6

see the pattern?

[onyx sentinel]

oh so i would need to plug in more values to see that though right

or

[stark trail]

I mean, if you personally need to sure

[onyx sentinel]

oh ok

[stark trail]

but the form of the sequence should be a giveaway aswell

(-1)^n alternates in sign

dividing by n will divide by increasingly larger n

[onyx sentinel]

ohhh

yeah that makes sense

[stark trail]

like

if we changed it to just

1+1/n

hopefully you can see immediately what that sequence is

1+1
1+1/2
1+1/3
1+1/4
....

right?

[onyx sentinel]

yeah

[stark trail]

all the (-1)^n does

is change the sign on half the terms

[onyx sentinel]

oh ok yeah that makes snese

[onyx sentinel]

so for this part

like this is to find what its bounded by right?

[stark trail]

yes

[onyx sentinel]

like to simplify it to make it easier to find?

[stark trail]

to find if it is bounded

yes

[onyx sentinel]

ok so when u get bounded by 2 from this equation as long as the question doesnt ask for the sup then i can just put its bounded by 2?

[stark trail]

I think you can just say the sequence is bounded by 2 yes

[onyx sentinel]

what if it does ask for sup?

[stark trail]

if you want to be thorough, you can note that it's actually bounded by 3/2

[onyx sentinel]

ah

[stark trail]

then you can prove it's bounded by 3/2

and it attains the value 3/2 at n=2

so there would be no smaller lower bound

[onyx sentinel]

but surely it wouldnt be this easy to just see for all equations

[stark trail]

You'd be surprised

[onyx sentinel]

lol alright

ok ill try to do some more questions thank you so much

[stark trail]

I mean sequences are basically like functions

except discretized

[onyx sentinel]

mmm okok

[stark trail]

you look at something like

sin(pi*n)

you can know it's behavior

by knowing sine's behavior

so with sequence questions it really just boils down to knowing how functions behave

and in this problem

the functions you needed to know

where (-1)^n

and 1/n

[onyx sentinel]

ah so i should study the different types of functions and how they behave

[stark trail]

it's not always this easy you're correct

but

they don't get impossibly difficult

[onyx sentinel]

ooo

[stark trail]

the functions just get a little more complicated

[onyx sentinel]

ok i guess i just need more experience

thank you

[stark trail]

Indeed

no problem

[onyx sentinel]

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[buoyant rover]

How many three-digit numbers can be formed with the digits: 0, 1, 2, 3, 4, 5, 6. If each one can only be used once.

[buoyant rover]

I got from this question 180 distinct digits

but now they ask me to find the amount of odd numbers

what could I do to find the amount? At first I thought it was 6 * 6 * 1, but it didnt give me the correct answer

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[daring blade]

relativistic collisions really confuse me, I get that we're generally supposed to find momentum before and after/total energy before and after but I get so lost with the formulas

[marsh citrus]

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[still temple]

I am in dire need of help for this problem

[small vector]

monte carlo? what class is this

what techniques have you used before

[still temple]

Numerical Analysis

this is the first module in our probability unit

[small vector]

do they want you to write code?

[still temple]

i mean its a part of our noes

notes

like he gives us what he wants

i guess can u just explain it for me if you can

ill send it

[small vector]

okay yeah they do indeed want you to write code

but for now imagine a square
and a circle touching all the edges of that square

[still temple]

ya

i mean is the gist of it like just finding n random points and checking how many fall into the circle

[small vector]

that's exactly what it is

[still temple]

so like

what is monte carlo method

bc the next question is asking to evaluate an integral using it

is it just like finding a certain amt of points and checking if it satisfies?

[small vector]

yeah

we just estimated the area of a circle

now we're going to estimate the area under a curve

[still temple]

ohhhhh

duhhhh

[small vector]

monte carlo lends itself nicely to visuals so here's $\int_0^1 x^2$

[elfin berry]

b

[small vector]

but yeah it's literally just brute force

[still temple]

so again u just generate points and check the probability of it landing under the curve?

[small vector]

yeah

[still temple]

so how does that translate to finding the value

[small vector]

in this example, if you made 1000 points, maybe 334 of them would land in the red area

because $\int_0^1 x^2 dx = \frac 1 3 \approx \frac{334}{1000}$

[elfin berry]

b

[still temple]

ahhh i seee

this helped a lot! thanks 🙂

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[rapid tangle]

Could I walk through a with somebody

The textbook for this chapter gave us the definitions for these

but never explained how to classify them

[marsh citrus]

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<@&286206848099549185>

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[gaunt gale]

no problem, when would you have a bit of time again?

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[arctic halo]

for part b

how did they find out the centre

[cunning jackal]

Angle subtended by a diameter on the circumference is 90 degrees

So essentially if a right triangle lies on a circle

It's hypotenuse is the diameter

[arctic halo]

oh like that

okay thank u

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[silver field]

If (x - 2)² is the factor of an expression of the form x³ + bx + c then the other factor is

[proud pawn]

So first, what is the degree of the other factor?

[silver field]

Only one factor is given in this question that is (x - 2)²

Degree is 2

[proud pawn]

So what must be the degree of the other factor?

[silver field]

One

[crude basalt]

ok so you have to expand this out

and see what factor, when you do all the multiplcations, gives you 0 for the x^2 term

[silver field]

Ok...then

[crude basalt]

ok show us what you do

[silver field]

Till now I have done is expand (x-2)² = x² +4 -4x

Don't know what to do next

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[deft karma]

Hello Guys can you help me ?

[novel juniper]

ask your question

if anyone can help, they will

[deft karma]

Can i send a photo

[novel juniper]

Yes

[deft karma]

Can you translate in englisch ?

From German ?

I have B

[novel juniper]

I don't know german, sorry

This isn't a. test, is it?

[deft karma]

No just Homework

[novel juniper]

cool

[deft karma]

Can you help ?

[novel juniper]

If you can translate it, maybe

[deft karma]

Okay

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[deft karma]

No

[whole compass]

This is a maths server.....I doubt many know German here

Use google translate

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[frigid lance]

For simultaneous equations, should we say
x= something OR x=other thing
or
x= something AND x= other thing?

[frigid lance]

OR implies AND/OR right?

[crude basalt]

you should use or

[low ruin]

or generally in mathematics is the inclusive or

[crude basalt]

or impliess that x can be either this or that

[frigid lance]

thanks!

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[naive sedge]

the topic is probability and compound probability the question is like idk number lower than 6 the given is 0-15 is the answer 1/3 or 2/5? like do u include the 0?

[whole compass]

Yes

If it's 0-15 then 0 is included

[naive sedge]

0 even right?

[whole compass]

Yeah

[naive sedge]

even number 0-15 8/15?

aight thanks

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[queen mantle]

Any good resources to learn everything about the desmos calculator? The SAT uses it and I heard its quite powerful so I want to be able to use it during the exam

[crude basalt]

indeed it is very powerful

perhaps too much 🤫

but the best way is to get familiar with it

and use it a lot

look up all the details and functions

everytime you need to figure how to do somehitng out, google it

[marsh citrus]

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[queen mantle]

👍

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[orchid oracle]

dont really know how to start this problem

[novel juniper]

is that. $3A-something = something else)$

[elfin berry]

ƒ(Why am. I here)=I don't Know

[orchid oracle]

No?

[novel juniper]

or -3A

3A or -3A

[orchid oracle]

Its just -3A

[novel juniper]

ok

\begin{bmatrix}
entry_11 & entry_12 & entry_13 \
entry_21 & entry_22 & entry_23
\end{bmatrix}

[elfin berry]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[novel juniper]

let this be A

wait, let me format the TeX

\begin{pmatrix}
a & b & c \
d & e & f
\end{pmatrix}

let this be A

[elfin berry]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[novel juniper]

can you do it from here

[orchid oracle]

So i prlly have to go over each row and col individually and solve

[novel juniper]

yeah

[orchid oracle]

okay

and why do u say let A be a 3x2 matrix?

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[timber hamlet]

Why does multiplying two vectors completely ignore all components they consist of?

[outer lodge]

what do you mean by multiplying vectors?

[timber hamlet]

This

[marsh peak]

That seems to be the dot product, next time call it that since "multiplication of vectors" is an ambiguous term

Anyway, why do you think the dot product ignores the components of the vectors?

[timber hamlet]

I mean if you have vector a=3i+4j and b=-i+3j and do a multiplied by b you get 3*(-1)+4*3=-3+12=9

(Scalar)

And you have a multiplication sign. I know what the dot product is, i know that it creates a new vector that is 90 degree angle to the area created

[marsh peak]

What you are describing is the cross product, not the dot product

[timber hamlet]

huh

[marsh peak]

Also, I don't see where the components of a and b are ignored in that part when calculating the dot product

[timber hamlet]

But scalar?

[marsh peak]

Perhaps you are talking about the fact that the information about the components is lost once given only the dot product?

[timber hamlet]

Yes...

?

[marsh peak]

Multiplication of real numbers behaves like that too then, you can't guess a or b given the product of a and b

[timber hamlet]

wdym? 3*4=12
like nothing dissapears

[marsh peak]

Say I am thinking of two numbers and their product is 12

[timber hamlet]

ok

[marsh peak]

Can you tell what those numbers are?

[timber hamlet]

No

[marsh peak]

So information is lost in the similar manner as with the dot product's case

If that doesn't answer your question, I need you to clarify more on what you mean by "components being ignored"

[timber hamlet]

Hmm... I dont think so. Cuz its still two numerical values that togheter make up x*y=12. There are no components.

[kind solar]

@timber hamlet Om du har finska som ett andra modersmål så kan du också försöka förklara frågan på finska. Asså jag tror att din screenshot är från ylioppilaskoe, ett finskt prov

vi kan också försöka på svenska men då kan det bli lite svårt för mig

[timber hamlet]

Toi on ylioppilas kokeest joo

[kind solar]

joo

[timber hamlet]

Siis ihmettelen ihan vaan miten toi vektoreiden kertolasku menee, ku ne i ja j komponentit vaan katoo ku kertoo kaks vektoria tolleen

Ja joo mä tiiän et se antaa luvun joka antaa tietoa niiden välisest kulmast, mut ei tunnu loogiselt, et sinne päästää ihan vaan kertomalla

[kind solar]

Kuten Lonely Bean sano nii kahen vektorin ristitulo on uus vektori, mutta kahen vektorin pistetulo on uus luku

Se miten se uus luku määritetään on että kerrotaan komponentit keskenään ja sit lisätään lopuks yhteen

Se on tosiaan aika yllättävä juttu että pelkästään kertomalla komponentit keskenään ja summaamalla vois saaha jotain tietoa niiden vektorien välisestä kulmasta, mutta se liittyy siihen että vektorin pistetulo itsensä kanssa, jos on vaikka xi+yj, on x^2 + y^2, joka on sen vektorin pituuden neliö (koska se pituus on sqrt(x^2 + y^2) ). Toi on se geometrinen yhteys mikä pistetulolla on

[timber hamlet]

Puhutko nyt siitä joka saadaan rastilla vai normi kertolaskun merkil

[kind solar]

pistetulo = normi kertolasku merkki

[timber hamlet]

Tai siis mun kysymys on siis se et miks komponentit katoaa laskutoimituksen jälkeen. siis sanotaan et vektori a = 2i + 3j nii tajuun mitä tapahtuu kun laskee 3a, mut sit vektori*vektori ei tunnu mahdolliselt

Vaik joo se antaa tietoa kulmasta, mut onks siis vaan päätetty että kertolasku vektoreilla meinaa tota pistetuloa

[kind solar]

Pistetulo on vaan määritelmä jolla lopulta sattuu olemaan kätevä geometrinen tulkinta. Mut se geometrinen tulkinta ei oo mitenkään ilmeinen ja vaatii ymmärrystä ortogonaalisista kannoista ja pistetulon bilineaarisuudesta

Kuten Lonely Bean sano nii ei yleensä sanota "kertolasku vektoreilla" vaan puhutaan joko vektorien pistetulosta tai ristitulosta

Niitä kutsutaan tuloiksi sen takia että ne toteuttaa joitain samoja ominaisuuksia kuin mitä tyypillinen kertolasku

[timber hamlet]

Eli muuten "vektoreita ei tarvii niin ku kertoa" muilla ku luvuilla

[kind solar]

mut ne ei oo kertolaskua siinä mielessä miten sä oot tottunu ajattelemaan kertolaskua

[timber hamlet]

Joo joo, siis mä tajuun sen et vektorit on vähä erilaisii

no kiitti.

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[kind solar]

ei mitää ^^

[marsh citrus]

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[dense steeple]

if something is 90% complete but only 80% successful, what would the final success rate be if the final 10% successfully completes

[dense steeple]

can you just add that 10%? meaning final success rate is 90%

[manic stirrup]

no

[dense steeple]

didnt think so

[kind solar]

yes, of course. x is the total amount of stuff, and 0.80x has succeeded, and you know there's gonna be 0.10x more that succeeds, so therefore the amount of success will be 0.90x, which by definition means that 90% of the stuff succeeded

[manic stirrup]

isnt this wrong

[kind solar]

Of course there could be some interpretations that I'm making that don't describe the original problem

so more context would be helpful

[kind solar]

I guess it depends on what that 80% refers to, does it mean that 80% of the total experiments succeeded or that 80% of the 90% which have already been tried succeeded

[dense steeple]

so .9 is the reference point meaning the actual success proportionality is 80% of 90

[manic stirrup]

80% succeess from 90% is bigger then 90% to 100%

if its supposed to be the same % of succeess rate it cant be that answer

[dense steeple]

.8*.9 = .72x?

[kind solar]

If 80% refers to the things that have already been concluded then the success rate would be

0.90*0.80 + 0.10

[dense steeple]

wait but that also doesnt make sense though

[kind solar]

at this point

!xy

[marsh citrus]

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.

[kind solar]

I mean, this part is definitely correct. You can absolutely add the final 10% to the success count

What's not clear is whether you start with 80% of ALL your stuff being successful or 80% of the 90% being successful

Upon thinking about it some more I would assume it's the latter situation

If this is an exercise then it's almost certainly the latter

but if this is some real world scenario then it's possible that someone is not doing a very good job at phrasing things and thus it could be either or

[manic stirrup]

how can you say its true when the % changes between the cases

80 90 =/= 90 100

it doesnt become more effective or less effective in the last 10% rather it should be constant

[kind solar]

It will be more effective in the last 10% since the entirety of the last 10% succeeds

[dense steeple]

A student has 3 categories that their total grade depends on: assessments, classwork, and homework. The weighting for each category is 45%, 45%, and 10% respectively. If the student's grade is an 80% after assessments and classworks were graded, what will the student's final grade be if they earn full credit in the homework category?

[kind solar]

that's a 100% success rate

[dense steeple]

you add the two total weightings together, so their current grade is 90% of what it should be

but they have an 80% in the class

but thats not .8/.9

[kind solar]

Yup so they lost 10% of all the credits when they got 80% from assessments and classwork but if they earn full credit in the homework category then they'll have 90%

So funnily enough my initial interpretation was correct

[dense steeple]

.45x + .45x + .1x = x?

but then that assumes the success rate is constant, no?

its not

its changing at two points

its more like .45x +.45x + .1y = ?

no no no not even that, i dont think that system works either

[kind solar]

If the student got up to 80% from assessments and classwork then they got 8/9 ~ 89% of all the points that you can get from those categories combined. So the 80% is referring to the TOTAL amount of credits available, not only to the first two categories

[dense steeple]

wait it would be

[dense steeple]

ahhhhh

yeah that makes sense

that makes sense

COOL

thanks

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[kind solar]

@dense steeple haha no worries ^^

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[still temple]

I forgot how to integrate 2x/3

[still temple]

pleasethanks

[proud ice]

Do you know $\int xdx$?

[elfin berry]

SWR

[still temple]

I have no clue what this is

oh I mean x2/2?

^*

I dont get how I would I tegrate 2x/3 tho, apart from partial fraction Integration I don't I derstand the rules of integrating a fraction

.Close

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[arctic halo]

for the width why isnt it 200-2(2x+20)/2

[pulsar trout]

The 200m fencing doesn't include the 20m wall

[arctic halo]

oh ok that makes sense

ty

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@grizzled bolt Has your question been resolved?

[marsh citrus]

@grizzled bolt Has your question been resolved?

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[mental pollen]

write the equation of the parabola that passes through the points -1,0 5,0 6,-14

[mental pollen]

i missed the lesson and now i have a test soon for it and need to learn how to solve it

[pliant siren]

oh these questions

do you know the principle on how to solve them?

[mental pollen]

somewhat but not really, the notes i got from other students were more confusing than helpful

[pliant siren]

what do the notes say is the first step to do it?

oh wait

nvm

[mental pollen]

give me a minute to find the notes again, i have a giant wad of papers here.

[pliant siren]

k

[mental pollen]

i dont believe i took the notes off of her

[pliant siren]

alr so the way I was taught to do this is to start with the general form of a parabola

which is?

[mental pollen]

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[pliant siren]

damn

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[still temple]

Need help, answer is 15cm^2

having trouble isolating y to equal something

If the rectangle's area is defined as xy, and based on the triangles dimensions the area can be concluded to be (12 - y)(5 - x). But I have no idea how to go about finding y so I can find the derivative for the area equation in terms of x

[drowsy breach]

ok assume the angle opposite to 12cm side is theta than tan theta =12/5=y/(5-x)=(12-y)/x

[marsh citrus]

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[still temple]

<@&286206848099549185>

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[wanton apex]

Sorry if this is messy but I’m not sure where I’m going wrong here

[wanton apex]

am I missing a minus somewhere that would make the p_2 term equal to 0 somewhere?

[stray field]

There is only helper right now so might take a while

[wanton apex]

ah

unlucky

[stray field]

Agreed as same here

[twilit grove]

I think you did the quotient rule incorrectly

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[opaque pendant]

How do I determine the y value on these types of graphs? are they all correct

[wide grail]

All except the 3rd

You see that G(x) has a discontinuity on x=2

[opaque pendant]

third is dne

[wide grail]

Noted by the white dot at 2

At x=2 the value is the colored dot above

[opaque pendant]

is this one also dne?

[wide grail]

here f(3)=1

It approaches 3 from the left

Approaches -2 from the right

But denoted by the full circle

Its value is 1 at 3

[opaque pendant]

so a full circle indicates a hole?

[wide grail]

I guess you can say that

More like the empty circle indicates a hole

And the full one indicates what's in the hole

[opaque pendant]

Is the 3rd then G(2)=4?

[wide grail]

Right

It is 4

[opaque pendant]

so when it is normal like that doesn’t break then the answer will be the same even if it wants what is going left or right?

will this one also be 4? or is it just 2

[wide grail]

Yes, if there was no hole then G(2) would be 2

No this one is 2, because its a limit

It approaches 2

But at x=2 the value is 4

[opaque pendant]

oh ok now i get it

[wide grail]

For x=1.999999999 y is 1.9999999

But for x=2 y=4

[opaque pendant]

so whenever is has x—>3 like this one it would be dne?

[wide grail]

This one does not exist because it approaches different values if we approach 3 from the left and from the right

[opaque pendant]

and if it were asking for f(3) the y will equal 1

[wide grail]

If we go from the left it approaches 4 and if we go from the right it approaches -2

Yes

Exactly

[opaque pendant]

now this is making sense to me

[wide grail]

This still means that lim x->3+ and lim x->3- do exist

But lim x->3 doesn't

[opaque pendant]

ok so x->3 will exist if the lines don’t break?

[wide grail]

Thats right

It must approach the same value left and right

[opaque pendant]

ok thank you, now this seems simpler

[wide grail]

It really is

[opaque pendant]

i think what really confused me was what to do with the holes

[wide grail]

Yeah its a weird notation when you see it first time

[opaque pendant]

thank you so much

if one of the limits is filled does that mean when g(-4) it will make y=2?

[marsh citrus]

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[opaque pendant]

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[north plover]

How do I do this?

[blissful mural]

so h(3)=-2. Then find 2h(3)-1

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[trail flint]

can someone help me understand why the endpoint 1 is divergent

I don't know of any test that works for (-1)^n besides ratio and alternating series and both don't work here

[turbid mica]

if x = -1 then you have (-1)^n * (-1)^n = 1

by the nth term test or divergence test the resulting series diverges

limit is infinity

when x = 1 it's alternating but the terms dont decrease so it's still divergent

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[safe knot]

hey i had a quick question about AIME. (AMC) do you guys think i can qual for aime next yr by self studying this summer? ive never done any math contests so far and i will be a junior next year. i do very well in school (4.0) and have a natural aptitude for math, typically scoring highest on math tests in my friend group. am i cooked? 🙏

[rough trellis]

Cant say for i have no experience, but goodluck 👍

[turbid mica]

AMC 12?

4.0 on a 4.0 scale?

zoinks

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[silk sable]

Post a screen capture of the math problem. Don't rely on people clicking on an unknown link.

[marsh citrus]

@still temple Has your question been resolved?

[safe knot]

Yeah

[safe knot]

Yep

[scarlet hearth]

this one

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[tall pike]

4r^2 - 20r + 25

can I factor this equation so that one of the brackets is (r -5)

[desert socket]

why do you think so?

you can try it yourself

let f(r)=4r^2-20r+25, then if (r-5) is a factor of f(r) then f(5) must equal zero

which in your case isn't

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[marsh citrus]

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[spice plover]

Let a and b be positive numbers and $A={a^n, n \in \mathbb{Z}}$ $B={b^m, m \in \mathbb{Z}}$ what are the conditions on a and b such that $A \cap B={1}$

[spice plover]

bruh

[bright geyser]

Use {}

[spice plover]

I did

[bright geyser]

\ { \ }

[spice plover]

aight

good enough

[marsh peak]

I think it wants you to wrap the first a and b in dollar signs

And the last a and b

[elfin berry]

The Prophet Of The Damned

[spice plover]

now

help

[bright geyser]

What condition is for 1 to be in A?

Oh it Z not N

[marsh peak]

The condition is the same though

Ah no

0 in Z

[bright geyser]

But 0 in Z

Yep

Every nonzero real works, no?

[spice plover]

ping me when you can figure it out because I'm stuck completely

[marsh peak]

No, let one of a or b be the power of another and it doesn't hold

[spice plover]

take 2 and 4 as a and b

[marsh peak]

So for all nonzero integers n, m we want a^n ≠ b^m

Say a and b are both different from 1

Since the other case is trivial

[spice plover]

hmm

[marsh peak]

So mlog_a(b) shouldn't be an integer

Neither should nlog_b(a)

[spice plover]

mmm

I'm with ya

[marsh peak]

Yeah looks like the only condition is that a and b are not integer powers of another

[spice plover]

how do I prove it

[marsh peak]

Use this

mlog_a(b) is not an integer iff log_a(b) is not an integer

[spice plover]

so I take 2 and 3 and the intersection is 1 and does that imply 2^n =/= 3^m

[marsh peak]

Yes, for nonzero n and m

[spice plover]

and log_a(b) is only integer when b's some integer power of a

aight thanks

.close

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[still temple]

help mehelphelphlp

[still temple]

send your question

[still temple]

fx is 1/x
and gx is x^2
so fgx will be

well gx is the input so..

(x^2)
now f it and f will just divide the input so..
fgx= 1/x^2

\begin{align}
f(x) = \frac{1}[x} \
g(x) = x^2
\end{align}
Find $f \circ g (x)$?

[elfin berry]

45
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[still temple]

hm

yes

am i right

[still temple]

.

yeah $f(g(x)) = \frac{1}{x^2}$

[elfin berry]

45

[still temple]

i hage another q

[spice plover]

convolution?!?!?

[still temple]

???

[still temple]

wat

what

[still temple]

send

[spice plover]

that dot is used for convolution

[still temple]

no

it means

f after g

ion no wat convolution is

it sounds complicated but watever

For a ,
f(x)..
Which is 1/x
And x > 0 and smaller than 1 ( 0< x < 1) so
Like..0.1, etc
And if we put 1/(0.1)^2
We will get 10.
or like 0.0001. Which would be 10000 etc so we can conclude that.. f(x) >1

For gx
X is greater than 1 so..
So.. g(x) = 2^2, 3^2.. etc.
So range for gx will be:
gx >1

Now, Does the fg(x) exist?
gx = x^2
f will divide it so
1/x^2
To know if it exisrs or not,
We know x is >1 therefore it exists as x is not 0.

Now range of gf(x)
f(x) is:
1/x
and g(x) = x^2
So, (1/x)^2
Now input a value as an example
1/1^2
Or 1/2^2
Or 1/3^2
So..
Range is f(x) >=1
Right?

^ the question (white pic)

question got three parts: a,b, and c

[still temple]

my answers in the text

@still temple

[spice plover]

that's f(g(x))

[still temple]

it represents dat too

[still temple]

someone help me check this shi

😭😭😭

pleasw

[spice plover]

notation

[still temple]

kepe

my hope

[open kayak]

$f(g(x)) = \frac{1}{g(x)}, 0 < g(x) < 1$.

Now, does g(x) fit between 0 and 1?

[still temple]

whats gx

x^2

why didnt u subsitute that

[open kayak]

$g(x) = x^2$ for $x > 1$.

[still temple]

yes

so

1/2^2 for example gives us 0.25 so it exists

[open kayak]

Sure, I could've done that. [f(g(x)) = \frac{1}{x^2}, \quad 0 < g(x) = x^2 < 1 \text{ with } x > 1]

But really the important thing is to look at the inequality here

[open kayak]

And for that, this is enough

What's the range of g?

[still temple]

0 < gx < 1

[still temple]

oh

gx >1

[open kayak]

That's our requirement when plugging it into f

For gx
X is greater than 1 so..
So.. g(x) = 2^2, 3^2.. etc.
So range for gx will be:
gx >1

Yep

[open kayak]

Ok, but to plug it into f(x), our requirement is 0 < g(x) < 1. And we know that g(x) is always > 1

[still temple]

this

[open kayak]

That means f(g(x)) doesn't exist

Does that make sense?

[still temple]

yea

hpw dod u get the requirement for fgx

[open kayak]

Well, it says [f(x) = \frac 1 x, \quad 0 < x < 1.] Now if I plug in $g(x)$ as $x$, that becomes [f(g(x)) = \frac 1 {g(x)}, \quad 0 < g(x) < 1.]

Does that make sense?

@still temple

[still temple]

indeed

thank u soooo much

[still temple]

are all my other answers except for the one that answered "does fgx exist?" right?

@open kayak

[open kayak]

@still temple

[open kayak]

I meant to say "Ok, 1/x ** > 1** is fulfilled when ...", sry

[open kayak]

Does this make sense? We got the "when 1/x > 1" requirement from g(x) and the "and 0 < x < 1" requirement from f(x)

[still temple]

yeees

dont be sorry
tytytyty

um

are all my other answers right?

[open kayak]

You mean to a?

f(x) > 1, yes, and g(x) > 1, yes.

[still temple]

what
bout fgx

finding the range of gfx

[open kayak]

It doesn't exist

It won't have a range

[still temple]

?

um dis one

sorry

[open kayak]

.

[still temple]

Hi

What are you doing?

[open kayak]

answering to questions about this, @still temple

[still temple]

Just a minute

[open kayak]

If you need help yourself, open a new help channel, this is occupied

e.g. #help-26

[still temple]

wait
range of gfx doe

[open kayak]

g(f(x)) > 1

[still temple]

oh

[open kayak]

But yeah, when you are given something like this, I'd write it down step by step, like here

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[dull sable]

How do I approach this problem

[broken dome]

I'd start by drawing a sketch of the situation

[dull sable]

This is what I think the problem looks like

[broken dome]

aight

Now how would you approach calculating this?

[dull sable]

Write down $$ \tau = Fr$$

[elfin berry]

Unknown

[broken dome]

Aight, now what?

[dull sable]

Maybe identify a force

[broken dome]

Wait, are you familiar with how beams are solved?

[dull sable]

Only force I can think of is Weight pulling down of the construction worker

[dull sable]

Not really

[broken dome]

Have you ever drawn a moment diagram?

for a beam

[dull sable]

No

[broken dome]

Well in this case you could make due without it I guess

anyway, what sources of forces do you have?

[dull sable]

Weight of construction worker and normal force exerted by beam?

[broken dome]

Yeah

so how would you turn those masses into weight forces?

[dull sable]

Well isn't W=mg

[broken dome]

Sure

[dull sable]

I can only see that for the construction worker force the normal force is something different

[broken dome]

hm?

[dull sable]

I don't think I can say N=mg is what I'm trying to say

[broken dome]

Why not?

[dull sable]

Maybe because I associate W as a downward vector not upward

[broken dome]

Wouldn't both of those forces be downward though?

Also pay attention to where you place the beam weight force

[dull sable]

I guess I'm pretending the beam itself is acting like the ground so the normal force would be up

[broken dome]

It should act from the beams center of gravity

[broken dome]

No, that's not the case at all

[dull sable]

To me that makes sense because of newtons third law

Unless that only applies to moving objects

[green cradle]

Don't ignore weight of 500kg

[broken dome]

Treat the worker and the beam as one body

they act on the bolt to the left

the reaction force is in the bolt

There is no ground here

[dull sable]

If true, can I add the masses up

[broken dome]

No, because their centers of gravity are at different places

Well you technically could, but that'd be a bit complicated

So you have the downward force of the workers weight

and the downward force of the beams weight

they cause a reaction force in the bolt

[dull sable]

Mostly every physics problem I've done is treat each object with it's own free body diagram

[broken dome]

as well as a reaction moment

you can if you want, but you'll get the same end result here

[green cradle]

Wouldnt the answer be
1.2×500g+ 2.4×70g

[dull sable]

This the part I'm confused at, why is it downward force

[green cradle]

@broken dome

[broken dome]

because gravity acts downwards

[broken dome]

ye

[dull sable]

Why can't I say the beam acts like the ground if I place and object on the beam it should direct the force back

[green cradle]

Beam is hanging in the air

It is applying force on the bolt as well

[dull sable]

So normal force isn't a thing for objects not in ground?

[broken dome]

you can, but then you have to observe the worker as seperated from the beam

[dull sable]

Would there be a normal force from the wall in the drawing

[broken dome]

yes

a normal force and a reaction moment

[green cradle]

But it internal so it will be cancelled?

[dull sable]

Going back to the weight force of the beam, is it from the center

I don't want to assume it's in the middle because, center of mass placement in my opinion has been weird

[green cradle]

Com of beam would be at its centre. But if look at beam and man together it would be shifted in the right

[broken dome]

it is in the middle of the beam

[dull sable]

So far I found the magnitude of the Weight of the beam and Weight of construction worker

[broken dome]

and you know where they are acting from right?

So you also know the distance from the bolt

[green cradle]

But why calculate that to make it complicated?

[dull sable]

So if like you said the weight is acting in the middle of beam then it should be half the distance I'd assume

2.40/2

[broken dome]

ye

[dull sable]

I’m still not sure of using torque but this what I have

[broken dome]

Good

the total torque will be the sum of the torque caused by the worker

and the torque caused by the beam's weight

So just calculate them individually

and add the together

to get the total torque

[dull sable]

My only issue would be, wouldn't it be -4900N

Vector points down

[broken dome]

They're both pointing down

so the sign will be the same

both -

or both +

depends on convention

[dull sable]

Would it make more sense for the total torque to be negative

[broken dome]

yes

depending on sign convention

[dull sable]

For the construction worker would the distance just be the entire 2.40m for the torque

[broken dome]

ye

[dull sable]

I got $$\tau_{tot} = -7526.4 Nm$$

[elfin berry]

Unknown

[dull sable]

The application wanted it to be positive and the answer was supposed to be 7500 for some reason

.close

[marsh citrus]

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[onyx sentinel]

Im kinda stuck here. I need to use lhopitals rule but idk where to go from here

[novel juniper]

don't use l'hopital's rule

\lim_{x\rightarrow0}\left(\frac{\tan\left(ax\right)}{az}\right)=1$

[marsh citrus]

@onyx sentinel Has your question been resolved?

[marsh citrus]

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[ocean salmon]

Hi, I'm confused on how to do this and my notes don't have anything like it 😊

[twilit grove]

solve the two inequalities

[ocean salmon]

Oh separately.?

okay

so would this be the answer?

[wispy lance]

Yes this is the right answer

[ocean salmon]

Thank you !!

.close

[marsh citrus]

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[desert prairie]

Can someone help me with 4

[dim zodiac]

Due to no movement, N1 applies.

Decompose the forces to x- and y components.

[green cradle]

Draw FBD.

@desert prairie Mg will act downwards. Take vertical component of Tension in both wire and equate it to mg.

[marsh citrus]

@desert prairie Has your question been resolved?

[desert prairie]

FBD?

[desert prairie]

How do I find the vert tension for P and B when I don’t have an angle for it?

[green cradle]

I meant Free body diagram. But it's actually not necessary here.
Just show all the forces and u r done

[desert prairie]

This?

The cos should be a sin

[green cradle]

Wait it is given

[desert prairie]

The angle QpB?

Is the equation I wrote correct?

[green cradle]

Both T1 and T2 should be same I guess.

Let me read the question again, 1 min

[desert prairie]

Kk

[green cradle]

Wait I think you r right and I made a mistake of assuming it to be in the middle

Yup. They have not even said whether the ropes are of same length or not

[desert prairie]

Kk

I found the length of BW

BQ

Not sure if that helps tho

[green cradle]

Eqn 2 T1cos45=T2costheta

[desert prairie]

Why are they equal?

[green cradle]

Yeah it does, then we can find the value of theta

Balancing horizontal forces

[desert prairie]

Oh ok

[green cradle]

But I am getting 2 eqns 3 variables

1 more relation is required to solve this

[desert prairie]

I got T1 as 6g over root 2

[green cradle]

How did you find length of ropes?

[green cradle]

Using only these eqns?

[desert prairie]

Can’t I sub in this

Into the equation I did before

[green cradle]

Oh it will be cancelled right

[desert prairie]

Yea

[green cradle]

I didn't notice

Yeah now find T2. That's what the question asked

[desert prairie]

Also when it comes to tension horizontal forces are always the same?

[green cradle]

In equilibrium condition they are

[desert prairie]

Ok thanks

[green cradle]

👍

[desert prairie]

I got 49 as t2

Does that seem right?

.close

[marsh citrus]

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[cedar palm]

hello

[cedar palm]

I think it’s correct up to the point where I apply L’hospitals rule

I don’t think I can apple it because the bottom (1/2x+1) approaches 0, but I don’t believe the top ln (2x-3/2x+5) is infinity over infinity?

How would I solve this?

[blazing pulsar]

The top goes to ln(1) = 0 so L'Hopital's rule applies

If you need convincing, just look at the limit as x tends to infinity of 2x-3 / 2x+5 then divide the numerator and denominator by 1/x and take the limit

[cedar palm]

Okayyy I think I got it 👍

I’m going to try to redoing it

[cedar palm]

Do you mind looking over this?

I don’t think it’s going the direction ?

[blazing pulsar]

You misunderstood what I was saying

I am just saying that if you divide the numerator and denominator by 1/x

and take the limit

It is obvious that ln(2x-3 / 2x +5) goes to ln(1).

Since 3/x and 5/x tends to 0

Don't actually do this, because it is going to make your differentiation for L'Hopital's a nightmare

You said you didn't think LHopital's applied and I showed you why it applies, you don't need to do that transformation prior to differentiation

[marsh citrus]

@cedar palm Has your question been resolved?

[marsh citrus]

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[chilly mango]

$\lim_{x \to 0} \left(\frac{\arctan(x)}{x}\right)^{\frac{1}{x^2}}$

[elfin berry]

Alex

[chilly mango]

$\lim_{x \to 0} e^{\ln\left(\left(\frac{\arctan x}{x}\right)^{\frac{1}{x^2}}\right)}$

[elfin berry]

Alex

[chilly mango]

$\lim_{x \to 0} e^{\ln\left(\frac{\arctan x}{x}\right)/x^2}$

[elfin berry]

Alex

[chilly mango]

is this a good start?

$e^{\lim_{x \to 0} \left(\frac{\ln\left(\frac{\arctan x}{x}\right)}{x^2}\right)}$

[elfin berry]

Alex

[chilly mango]

<@&286206848099549185> i am doing l,hopitsl twice already and still getting 0/0

[marsh citrus]

@chilly mango Has your question been resolved?

[marsh citrus]

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[dense lion]

I'm trying to solve for the part c of this question, i've plugged in multiple numbers but im still confused

[dense lion]

This is my work

[desert dirge]

why 77?

ah nvm

your times are the wrong way around

[dense lion]

ohhhhhhh, so what would it look like

[desert dirge]

where did your second value 10.63 come from

[dense lion]

thats when i did 80 instead of 77