help-13

[viscid ridge]

I need help on a project i need to do on math

[viscid ridge]

My school is about to close on 1st quarter, and on math is that i need to show a talent and use it to answer a math question, i don't have a talent worth mentioning, other than my perseverance, so i need some help

Forgot to mention I'm in Grade 11

[crimson sedge]

what's the highest level of math that you know

[viscid ridge]

General mathematics?

[crimson sedge]

hmmm what's the last topic your class has covered

[viscid ridge]

Logarithmic functions

[crimson sedge]

not sure how this idea might fare, but i'd imagine a way to show perseverance is to solve an extremely complicated logarithmic function

[viscid ridge]

Yea, I'm desperate at this point since the deadline is tommorow

And i still have another unfinished task that is also deadlined tommorow

[cedar kiln]

@viscid ridge Has your question been resolved?

[viscid ridge]

<@&286206848099549185>

[cedar kiln]

@viscid ridge Has your question been resolved?

[viscid ridge]

Rude

[cedar kiln]

@viscid ridge Has your question been resolved?

[obsidian coral]

You didn't really ask a question

[viscid ridge]

Fair point

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[stark turret]

just a lil something quick, how do i get 200 without trial and error?

[gilded granite]

240/(1+20%)

[stark turret]

where'd the 1 come from

[gilded granite]

Let x be the previous number of sales.
Then
x + 20%x = 240
x(1+20%) = 240

[stark turret]

big thanks man 🙏

[gilded granite]

You're welcome.

[stark turret]

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[restive apex]

Hey guys I tried to slove this question so many times. But i coulnt reach the solution yet. Could you help me please..

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[cloud glacier]

2 equations with two unknowns

y = 2x + 0
and
-12x+3y=-15

answers:
x = 2,5
y = 5

i am stuck with this question, the problem is the equation y = 2x + 0. we dont know, what it equels and i cant just put in to -12x+3*2x+0=-15, then it gives something else than already given answers

i just need the method to solve this, i havent been able to find examples, where they solves these

[cloud glacier]

well i tried, i may have calculated, wrong

-12x+3*2x+0=-15
-12x * 2x=-15-3-0
-24x=-12
x=-12/-24
x = -0,5

oh wait a second

arent you suppose to isolate the x's to the left

wait let me re calculate then

thx for the help seem like i keep making calculation mistakes

[cedar kiln]

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[woeful musk]

just solved the integral x^5(x^6+2019)^2020 dx. I was taught to solve them a specific way but can someone tell me why I could solve it without touching the x^5 part?

[woeful musk]

Like, same story with (x^3+1)^6.x^2

Without the x^2 part

I'm just confused, sorry if it's a stupid q

[dense wing]

Not sure what you're asking

Do you want to just do the integral without u sub...?

[austere kindle]

👀

[woeful musk]

I don't either

I'll just move on

question answered

[austere kindle]

@woeful musk type .close

[woeful musk]

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[zinc pagoda]

can someone help me with this question please?

[cedar kiln]

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[crimson sedge]

help

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[little stag]

sqrt(m) * x = cbrt(m)

so what's x?

[meager kettle]

for m=0, any x
for m=/=0, you can divide by sqrt(m) by both sides

and then find other value

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[turbid wagon]

Hey
Could you help me please with that :
Using Disjunction Elimination, prove that for every real number : sqrt(2x²-3x+3)-x+1>0

[cedar kiln]

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@turbid wagon Has your question been resolved?

[zinc finch]

huh whats disjunction elimination…

[cedar kiln]

@turbid wagon Has your question been resolved?

[turbid wagon]

it's proof by cases

[short magnet]

Use that sqrt(...) is always non-negative

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[open glacier]

is there a set that can be closed under division?

[open glacier]

The ratio of A and B is 3:4. The ratio of A and C is 2:7. The ratio of B and D is 5:3. If A, B, C, and D are whole numbers, what is the ratio of A:B:C:D?

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[sand plume]

Simplify (4−3𝑖)/(−4𝑖)

[sand plume]

i know the answer is (3/4) + i but i just need someone to explain how this is done

[dense wing]

do you know how to rationalize the denominator on something like $\frac{1+\sqrt{2}}{\sqrt{3}}$?

[wraith dagger]

Mosh

[sand plume]

no i dont think so

[dense wing]

ok...

[sand plume]

i prob do but im slow

[sweet dawn]

yeet

[dense wing]

ok well for your question you want to realise the denominator, what would you multiply top and bottom by to get rid of the i?

[graceful karma]

I think he just realized

[sand plume]

im so confused

to get rid of just the i?

[dense wing]

what is the definition of i?

[zinc mason]

why not just split the fraction into two?

[dense wing]

you still have to do what im trying to explain.

regardless of when you do it @zinc mason

[sand plume]

isnt it square root -1

[dense wing]

yes, i = sqrt(-1)

what's i^2 then?

[sand plume]

-1?

[dense wing]

yes

so if I multiply i by itself, I get rid of the imaginary"-ness" of it

so multiplying top and bottom by i (more generally the complex conjugate of the denominator) you "realise the denominator"

[graceful karma]

LES GOOOO

[sand plume]

so i multiply top and bottom by i?

[dense wing]

yes

then simplify

[sand plume]

wait now im confused

wouldnt the fraction then be 7/4

or am i being blonde

[dense wing]

being blonde

3 doesnt magically become 7

[sand plume]

it is 4 - 3(-1) for the top?

[graceful karma]

4i

[dense wing]

numerator is (4-3i)i
denominator is -4i^2

[sand plume]

oh

thanks i think i got it

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[graceful karma]

Try writing a series for the first number of each row first

Like

1,2,4,7,11

Si

Hmm idk I gtg

Just dont forget about modulo

[upper abyss]

The second difference is 1. That means this is T(n) = (1/2)n² + an + b

The first digit of every row, I mean.

Actually, nvm I have something easier. Consider the last digit of every row

1,3,6,10,15...

Pretty clear pattern

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[grizzled berry]

can anyone here help me with a calc 2 problem?

[grizzled berry]

https://cdn.discordapp.com/attachments/486844875632017408/906021032807252028/unknown.png

so i've tried to do this by integrating with respect to y

i've rearranged the equation for the black line to x = 2y^2 - 2

and the pink line to x = -y^2 + 1

so, i tried to calculate this integral:

and for some reason it gives me -4???

(and i just put it into mathway and i got 0, i did my calculation wrong anyways)

[upper abyss]

Remember you're integrating with respect to y. It looks like your bounds are set for x

[grizzled berry]

are they?

[upper abyss]

0 ≤ y ≤ 1

[grizzled berry]

how does that work

[upper abyss]

Nothing happens in negative-y region

[grizzled berry]

i got -2 and 1 by replacing y with 0 in both equations

oh wait a minute

i see what you mean

[upper abyss]

You got x = -2 and x = 1 haha

[grizzled berry]

so, all i have to do is set x to 0 for one of the equations to find the upper bound right?

[upper abyss]

Yeye, to find that point at the very top there

[grizzled berry]

alright ima try that and see how that goes

i got 2, doesn't seem right though..

i may be wrong

nevermind, its right

@upper abyss thanks

[upper abyss]

Np, feel free to ask if you have any others!

[grizzled berry]

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[tepid sail]

.

[crimson sedge]

Hi guys, could someone help me with this sequence?

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[wispy bluff]

Part c

[wispy bluff]

Do I complete a simultaneous equation?

[cedar kiln]

@wispy bluff Has your question been resolved?

[wispy bluff]

<@&286206848099549185>

[cedar kiln]

@wispy bluff Has your question been resolved?

[tropic oxide]

@wispy bluff do you still need help with this?

i have a feeling you've misunderstood part b of this problem.

[wispy bluff]

I think I have @tropic oxide

And I do still need help please

[tropic oxide]

here is a prettier graph of your sinusoid.

do you see that little arc in the middle (around t=3) which is below the blue line?

[wispy bluff]

Yes

[tropic oxide]

part b tells you this segment must cover 1 second on the time axis.

which, by the symmetry of the sinusoid, means it must go between t=2.5 and t=3.5 (since t=3, the lowest point, must be exactly in the middle)

do you understand this?

[wispy bluff]

Yes that makes sense

[tropic oxide]

so do you understand that the value of k is just the height of your sinusoid at t=2.5 (or t=3.5, which is the same)?

[wispy bluff]

Yes

Is there a reason that didn’t say h instead of k?

Since h is on the axis?

[tropic oxide]

hm?

k is a constant

h is the name of the function

[wispy bluff]

Nvm

Yep

Wait but

That constant is from the h axis

[tropic oxide]

??

[wispy bluff]

Could they have said find the value of h for which sin… is < during filming

[tropic oxide]

no, they couldn't.

h is the name of the function.

you can't reuse names like that.

[wispy bluff]

But aren’t we finding the point along the h axis when t is at 2.5 or 3.5

[tropic oxide]

the reasoning i laid out just above establishes that k = h(2.5).

[wispy bluff]

So we’re finding the constant at h(2.5)?

[tropic oxide]

you're overthinking it

[wispy bluff]

I hate trig 😦

Is k=sin(2.5pi/2)+1

[tropic oxide]

yes, k = h(2.5) just like i said

[wispy bluff]

Okay

[tropic oxide]

shall we move on to part c?

[wispy bluff]

Yes plz

[tropic oxide]

we are now told that the one-second interval has been shifted to [4, 5]

and we know from part b that before the shift, said interval was [2.5, 3.5]

how far and in what direction do we need to shift our graph?

[wispy bluff]

1.5 seconds right

[tropic oxide]

that's correct.

now, do you recall your graph transformations?

[wispy bluff]

I think so

[tropic oxide]

the new graph will have the formula $y = h(t - 1.5)$

[wraith dagger]

Ann

[wispy bluff]

Factor out the pi/2?

[tropic oxide]

you're jumping the gun.

[wispy bluff]

Oops sorry

[tropic oxide]

do you understand what i wrote here?

[wispy bluff]

T-1.5 means t=1.5

So 1.5 right

[tropic oxide]

no, t - 1.5 means t - 1.5

i'm not equating it to zero.

[wispy bluff]

Oh

[tropic oxide]

i repeat: do you understand what i wrote here, yes or no?

[wispy bluff]

Not fully

It’s negative 1.5 because that means right?

[tropic oxide]

yes, that's how we translate graphs to the right.

[wispy bluff]

I don’t understand what h is

[tropic oxide]

h is the function introduced in the beginning of the exercise.

[wispy bluff]

Ohh okay

Yep

[tropic oxide]

and now we actually take the formula for h that we have

thus y = sin(pi/2 (t - 1.5)) + 1

= sin(pi/2 t - 3pi/4) + 1

[wispy bluff]

Okay

One question

[tropic oxide]

yes?

[wispy bluff]

Can we not set that bracket equal to 0 and solve for t, like quadratics abs cubics?

[tropic oxide]

........

why would you need that?

[wispy bluff]

I don’t in context of this question, but could I set that bracket equal to 0 and find t when h is 0?

[tropic oxide]

what bracket?

the stuff inside of the sin()?

[wispy bluff]

Yes

[tropic oxide]

no, that wouldn't give you the solutions of h(t) = 0.

[wispy bluff]

Is it bc there is more than 1 solution

Or does it not give the right solution at all

[tropic oxide]

it doesn't give the right solution at all.

[wispy bluff]

Okay thank you

Would you help with an integration question please?

[tropic oxide]

well, while i'm still here, i don't see why not...

[wispy bluff]

Okay thank you

[tropic oxide]

and what is giving you trouble here?

[wispy bluff]

I’m struggling to start

Is first step to integrate the equation?

[tropic oxide]

no, the first step is to think.

[wispy bluff]

Wait

Can I use a definite integral from 0 to 10 then times that by 2?

[tropic oxide]

and what'll that give you?

[wispy bluff]

The area from 0 to 20?

But not the length ?

[tropic oxide]

the area of what exactly?

[wispy bluff]

Area of the speed bump

20cm in width?

Actually

If I find the area, can I then use length = area/width?

[tropic oxide]

the cross-section area of the speed bump.

length = area/width

our cross-section is not rectangular, so no, you cannot.

[wispy bluff]

What is the cross section exactly? I’m a bit unsure sorry

[tropic oxide]

the cross-section of the speed bump is that shape we have the graph of.

[wispy bluff]

Okay

[tropic oxide]

what i was trying to get you to say

is that once we know the area of the cross-section

we can use the formula Volume = Length * Cross-section area to get the length

since we will know the cross-section area, and the volume is given

[wispy bluff]

okay that makes sense

I am going to try the question now, I’ll show what I do

Does this look right?

[tropic oxide]

if i were grading this, i would be puzzled as to how you were able to do the integration in one step like you did.

you have shown no work whatsoever that went into calculating $\int_0^{10}2 \sqrt{\frac{2x}{5}} \dd{x}$, even though the problem has asked you to be explicit about it.

[wispy bluff]

You’re right, it says full algebraic working even though it’s a calc paper

[wraith dagger]

Ann

[wispy bluff]

I just put it in a calculator

[tropic oxide]

you put it in a calculator? why?

were you that unconfident in your own ability to work out the integral by hand?

[wispy bluff]

I will try to do it myself

[wispy bluff]

Im struggling to integrate it

I have factored our 2 so now have root(2x/5)

[tropic oxide]

if you factored out 2, why not also factor out sqrt(2/5)?

then you'll have $\frac{2\sqrt{2}}{\sqrt{5}} \int_0^{10} \sqrt{x} \dd{x}$

[wraith dagger]

Ann

[wispy bluff]

Didn’t think that, it helps heaps

Is this better?

Need to add cm^2 at the end of A

[tropic oxide]

i suppose this is fine now.

[wispy bluff]

Suppose? Wdym

Should I have kept going with the area working

[tropic oxide]

apologies for misleading you.

this is fine now.

[wispy bluff]

Okay

Thank you very much

You’re guidance was very helpful

.close

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[hallow stream]

How would I solve this limit

[cedar kiln]

@hallow stream Has your question been resolved?

[crimson sedge]

as x approaches -1/2

what does the y value approach?

observe the graph

[cedar kiln]

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[brisk drum]

hey id like to solve this with substitution

$2^{3x-2}-2^{3x-3}-2^{3x-4}$

[wraith dagger]

spool

[brisk drum]

$\frac{2^{3x}}{2^2}$-$\frac{2^{3x}}{2^3}$-$\frac{2^{3x}}{2^4}$

[wraith dagger]

spool

[brisk drum]

i get this, and i wanted to ask if i can subsitute as 2^3x = t ?

[regal sentinel]

is there an equation you're trying to solve? or just simplify?

[brisk drum]

id like to solve this for x

there would be a =0 after that i assume, i just got this problem as it was presented sorry ahahah

[regal sentinel]

hmm, it would matter where the equal sign is

could you check?

[brisk drum]

yeah sorry it was just given like this 😢

this is all i was given

[regal sentinel]

i wonder if one of it is supposed to be an equal sign instead of minus

[brisk drum]

hmm let me ask

[regal sentinel]

cuz technically, the first two terms just equal to 0?

[brisk drum]

yeah the second is wrong she wrote 2^3x-2 but its 2^3x-3

[regal sentinel]

oh

[brisk drum]

yeah xd i tried to solve this the other day when we talked and i was sure i had found the solution too cuz itd be 0

im gonna close and come back when i know more abt the problem

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[gentle patio]

could I please get help

[gentle patio]

Determine the equations of the vertical and horizontal asymptotes for the function f(x)=8x^2/81-4x^2 and describe the behavior of f(x) near the asymptotes.

[regal sentinel]

$f(x) = \frac{8x^2}{81 - 4x^2}$

[wraith dagger]

hyrflood

[regal sentinel]

Is this it?

[gentle patio]

yes

i need to get the vertical and horizontal asymptotes

and then describe the behaviour of f(x) near each asymptote

[regal sentinel]

yup what do you remember about vertical and horizontal asymptotes?

[gentle patio]

i need to solve for x in the denominator

and that would be the vertical asymptote?

[regal sentinel]

specifically for denominator = 0

[gentle patio]

oh wait yea

sorry

and for horizontal asymptote

[regal sentinel]

np just clarifying

[gentle patio]

it would be the coefficnts of the x's

so in this case

8/4?

8/-4

[regal sentinel]

well, that's kind of a shortcut that could lead to mistakes in some cases, just make sure you know why that works

[gentle patio]

but is that right?

[regal sentinel]

but yes, in this case it would trend towards -2 yes

[gentle patio]

oh okay

so for the vertical asymptote

what x value makes the denominator equal 0 right

[regal sentinel]

yup

[gentle patio]

what about the behaviours tho

idk how to explain that

[regal sentinel]

so there are some different cases you can consider for vertical asymptotes - and you'd have to check for both sides of each asymptote

one method is to check for the value of f(x) near the asymptote on both sides

if say the asymptote is at x = 1, then I might plug in 0.9 and 1.1 to see where it is (is it trending positive or negative)

[gentle patio]

i figured out the vertical asymptote

it is x=9/2,-9/2

so what value would I plug in to check?

[regal sentinel]

for -9/2 try something closeby on both sides

since -9/2 is -4.5, you could try -5 on the left and -4 on the right of it

[gentle patio]

sorry i am a little confused

so what do I plug it in to?

the actual equation?

[regal sentinel]

yes,

to find the y value of it

gimme a sec to draw it

[gentle patio]

okay

[regal sentinel]

notice how f(-5) and f(-4) are both defined since they're not on the asymptote (x=-4.5) itself, so if you plug it in to the function, the y value kind of tells you which direction it trends towards as it approaches the asymptote from left or right sides

if you wanted to be careful, technically you could find f(-5), then f(-4.9), f(-4.75)... and you will notice it gets more and more negative

[gentle patio]

ohh okay thank you so much

and what about horizontal asymptote

same thing?

like I use -3 and -1

and see which way it trrends

[regal sentinel]

well, that one is different, that is asking "what happens if I plug in x = very big and x = -very big"

which is why I discouraged the shortcut

I can show you how I think through it

[gentle patio]

im a little confused on this one

how is it different?

cant i just do the same thing I did for the vertical asymptote

[regal sentinel]

well, you wouldn't really need to

you already know what it looks like - horizontal asymptotes just mean your function gets close to that line/shape you found

in this case you found it to be y = -2

the red line is y=-2

[gentle patio]

yea

so how do I describe the behavoir?

of f(x)

so I know how to describe the behaviour for the vertical asymptote

but not the horizontal

[regal sentinel]

complete this sentence: as x approaches negative infinity, y approaches [ ]. As x approaches infinity, y approaches [ ].

[gentle patio]

is that for the horinzontal asymptote?

[regal sentinel]

yes

that is essentially the statement you are making every time you describe a horizontal asymptote's equation

[gentle patio]

as x approaches negative infinity, y approaches positive infinity . As x approaches infinity, y approaches positive infinity

is that correct or am I wrong?

[regal sentinel]

nope not quite

[gentle patio]

oh

[regal sentinel]

think of this - as x approaches negative infinity, I am looking to the left side of the graph

what happens if I keep zooming out

on the left, what happens to the y value?

[gentle patio]

it increases?

[regal sentinel]

how come?

[gentle patio]

on the left side

[regal sentinel]

refer to this

[gentle patio]

the line gets closer and closer to the horizontal asymptoe

meaning its slowly raising in y value

[regal sentinel]

yup! it's increasing but it approaches what value

[gentle patio]

-2?

[regal sentinel]

yup

[gentle patio]

ohhh

[regal sentinel]

so theoretically, if I could actually reach infinity

It would be -2

[gentle patio]

so then the answer would be

as x approaches negative infinity, y approaches -2?

[regal sentinel]

yes

👍

[gentle patio]

and

As x approaches positive infinity, y approaches -2

[regal sentinel]

you got it

just remember horizontal asymptotes are not always a number, sometimes it's a zero (still horizontal), but sometimes you could have a function as an asymptote

[gentle patio]

wow thank you sooo much

[regal sentinel]

no prob, i like rational functions

[gentle patio]

i have another question but its different from this

[regal sentinel]

yeah ask away, I have to be afk for 2 mins but I'll brb

[gentle patio]

okok

The concentration of, c, of a pain killer (drug A) in the bloodstream t hours after the drug was taken orally is given by the equation c(t)= , where c is measured in milligrams per litre. How would you interpret the meaning of the horizontal asymptote?

and the other part of the question is A new drug (drug B) has been introduced. The concentration in the bloodstream is given by c 2 (t)=. Given the
graph of drug A, sketch the graph of drug B on the same axes. Is Drug B a better pain killer than drug A?

[regal sentinel]

oh interesting question

did you happen to have the functions?

[gentle patio]

oh oops

[regal sentinel]

first thing that comes to mind is there is a half-life that defines these drugs and also kind of like where they "level off"

[gentle patio]

The concentration of, c, of a pain killer (drug A) in the bloodstream t hours after the drug was taken orally is given by the equation c(t)=18/7t+3 , where c is measured in milligrams per litre. How would you interpret the meaning of the horizontal asymptote?

A new drug (drug B) has been introduced. The concentration in the bloodstream is given by c2(t)=6/t+2. Given the
graph of drug A, sketch the graph of drug B on the same axes. Is Drug B a better pain killer than drug A?

[regal sentinel]

so $\frac{18}{7t+3}$

[wraith dagger]

hyrflood

[regal sentinel]

and $\frac{6}{t+2}$

[wraith dagger]

hyrflood

[regal sentinel]

? Just wanted to make sure the numbers in the right places

[gentle patio]

yes

this one really has me confused

[regal sentinel]

did you graph them yet?

[gentle patio]

yea i have it on desmos

[regal sentinel]

ok great, so let's take it step by step

maybe it would help to calculate the asymptote

[gentle patio]

for the first one right?

because that is part a

[regal sentinel]

yup, actually I'll tell you straight up both their horizontal asymptotes are the same

use your strategy for calculating it

[gentle patio]

how would I do it?

because if there was a t beside the 18

then i would do 18/7

[regal sentinel]

let's go back to the definition of horizontal asymptote

we're trying to figure out - what happens as x approaches infinity (gets larger and larger)

look at the function

what can you say about the denominator as x gets larger and larger

[gentle patio]

umm

it gets closer to 0?

as x approaches infinity, y approaches 0

[regal sentinel]

yup

so what is x and what is y in this context?

[gentle patio]

x would be the horizontal asymptote?

it is 0 right?

and im not sure about y

[regal sentinel]

wait that's not what im asking

im asking

in the situation, what does x represent

not the asymptote

[gentle patio]

oh

[regal sentinel]

the context

[gentle patio]

it would be the time?

in hours

[regal sentinel]

yes

and y?

[gentle patio]

i think the concentration

[regal sentinel]

yup

[regal sentinel]

so lets put those back into this statement

[gentle patio]

as time approaches infinity, concentration approaches 0?

or would I say something else

[regal sentinel]

"as time goes on" (after ingestion of druf)

drug*

probably

cuz we don't really experience infinite time xD

[gentle patio]

yea true lol

okay

so for the question

How would you interpret the meaning of the horizontal asymptote?

I would say

as time goes on (after the ingestion of drug), the concentration approaches 0

[regal sentinel]

yeah or the drug gets completely metabolized or leaves the body

[gentle patio]

so thats the answer to the question right

[regal sentinel]

yup I should think so

[gentle patio]

okay thank youu

[regal sentinel]

[gentle patio]

what about b)?

A new drug (drug B) has been introduced. The concentration in the bloodstream is given by c2(t)=6/t+2. Given the
graph of drug A, sketch the graph of drug B on the same axes. Is Drug B a better pain killer than drug A?

[regal sentinel]

what did you notice about their graphs

and also, I'm not a pharmacologist so I don't know what constitutes "better" pain killer

[gentle patio]

im not sure how to word it

but I think drug B is higher in concentration?

[gentle patio]

how would we know what the better pain killer is then?

[regal sentinel]

lol ask the person who created the question

[gentle patio]

okayy ima ask my teacher

but what about the actual solving part

[regal sentinel]

well reading the graph from left to right

[gentle patio]

i need to graph drug B

and I have it on desmos

[regal sentinel]

looks like drug A has a bigger concentration but then it dips quickly

[gentle patio]

but i gotta show m y work

[gentle patio]

yea

[regal sentinel]

and drug B dips less quickly than drug A (so it leaves the body slower)

[gentle patio]

yea

so maybe drug A would be better as it affects faster?

[regal sentinel]

well, it leaves the body faster, which probvably means it works for a shorter amount of time

[gentle patio]

oh

yeaa

then it would be drug B

[regal sentinel]

I guess that makes sense

[gentle patio]

how would I show myu work for graphing tho

[regal sentinel]

hmm are you allowed to graph using desmos for your assignment?

can you just put a screenshot on your assignment?>

[gentle patio]

no she said no technology

[regal sentinel]

oh lmao

what about calculator

[gentle patio]

hey wait

oh wait nvm

yea I can use a calculator im pre sure

[regal sentinel]

hmm, so maybe she wants to you graph by hand

have fun

[gentle patio]

she provided a graph of drug A

and said to graph drug B onto that

[regal sentinel]

I can;t think of any other strategy than plotting points and joining them for this case

[gentle patio]

hmm I just need to show my work

and then I already know what drug B would look like because of desmos

[regal sentinel]

you have to show you know what graph B looks like by plotting it by hand

[gentle patio]

do you know how I can possiubly show my work

[regal sentinel]

lol sadly

[gentle patio]

like wdym plotting points

[regal sentinel]

oh

make a table of values

and then plot them

literally do what the computer does for you on desmos

[gentle patio]

so I am only given the equation c2(t)=6/t+2

[regal sentinel]

for x= 0,1,2,3 etc find f(0), f(1), f(2) etc

then plot the pairs

[gentle patio]

and for y?

oh wait

[regal sentinel]

well f(0), f(1) f(2) are your ys

[gentle patio]

yea

sorry

im gonna do it rn

[regal sentinel]

theyre likely going to be fractions so, a bit annoying to plot

[gentle patio]

so I got

(0,3)

(1,2)

(2,1.5)

(3,1.2)

[regal sentinel]

oh not too bad

[gentle patio]

yea lol

im going to plot the points now

thank you so much for all your help

i understand everything

.close

[cedar kiln]

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[royal leaf]

how do i do this

[royal leaf]

idk what im doing

[meager kettle]

let's t=x^(1/3)

then it's just normal quadratic equation

[royal leaf]

oh okay

[meager kettle]

t^2 - 5 + 4=0

after you find t, you need to come back to t=x^(1/3)

sorry, i meant x^(1/3)

[royal leaf]

how does that work

[meager kettle]

for example

x^(2/3) = (x^(1/3))^2

so then for x^(1/3) we put t

and we do it for other one

and we get t^2 - 5 + 4=0

[royal leaf]

wouldnt that be x^1/9

[meager kettle]

no, why would it be?

[royal leaf]

oh mb

i did it on paper

u right

[regal sentinel]

$(x^\frac{1}{3})^2 = x^\frac{2}{3}\ x^{(\frac{1}{3})^2} = x^\frac{1}{9}$

[wraith dagger]

hyrflood

[royal leaf]

yes

so how would i know when to solve normally with indices stuff and when do introduce another variable and make it a quadratic

or cubic

or whatever

[regal sentinel]

you have to get used to recognizing quadratic structures

eventually you'd probably also need to solve quadratics with trig functions

[royal leaf]

i do that rn

in physics

[regal sentinel]

[royal leaf]

yeah

idk maths just doesn't work for me

anyways

thank you guys

[regal sentinel]

i would also say just get really good with your exponent rules, your mistake in the original prob was an exponent rule issue

[royal leaf]

likely yeah

im terrible with indices

.close

[cedar kiln]

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[crimson sedge]

.open

I need help with a simple question. Is the period in a sinusoid the same as the B in the particular equation?

[cedar kiln]

@crimson sedge Has your question been resolved?

[pliant shell]

what is particular equation here?

[crimson sedge]

I figured it out, I was just second guessing myself. Here it is though: y = C+AcosB(x-D)

@pliant shell

[pliant shell]

ok

[cedar kiln]

@crimson sedge Has your question been resolved?

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[cedar kiln]

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[dark lintel]

Is this really a rule or shoulld i multiply it to remove the negative from -5

[crimson sedge]

help pls

[dense wing]

wdym?

cos(x) is in fact an even function...

[dark lintel]

how does it being even mean that (-pi/6) can become (pi/6)

[dense wing]

definition of being even

f(-x)=f(x) iff f is even

[dark lintel]

hmm

would the same applly to sin?

[dense wing]

no

sine is odd

cosine and secant are even, all other trig functions are odd

[dark lintel]

so sin(-x) would become -sin(x) then right?

[dense wing]

yes

[dark lintel]

okok I get it now

thanks

[cedar kiln]

@dark lintel Has your question been resolved?

[cedar kiln]

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[cobalt parrot]

try plotting the points

like draw a graph

plot (7,4), (10,6) and (12,3) on a graph (just draw a simple one)

also: remember that the midpoint between the two ends of the diameter is the center of the circle

@inland tide

[inland tide]

Ohhh ok

[cobalt parrot]

i would jot a diagram first to visualize the circle better

[inland tide]

Yeah

[cobalt parrot]

(honestly, if i was in a test situation, I would do that)

because the question could be like:

a circle passes through points (7,4) (10,6) and (12,3)

find the equation of the circle

how would you do that?

the easiest way is to jot a graph and plot the points

and figure out what the diameter is

[inland tide]

Like the question or plotting

[cobalt parrot]

the question

i think question 6 u sent is trying to prepare you for the question i mentioned before

[inland tide]

Ohh

[wraith dagger]

IAMTHEFARMER

[cobalt parrot]

first step is plot the 3 points there

[inland tide]

Yeah

[cobalt parrot]

then, you will see which points are part of the diameter line

[inland tide]

Ok

[cobalt parrot]

then you find the midpoint of those 2 points

[inland tide]

Ohh ok

[cobalt parrot]

and that becomes your center

[inland tide]

Yeah

[cobalt parrot]

so then following the equation

[inland tide]

Oh ok

[wraith dagger]

IAMTHEFARMER

[cobalt parrot]

you plug in the midpoint points

[inland tide]

That makes sense

[cobalt parrot]

and then you can plug in one of the points mentioned in the question to find r^2

[inland tide]

Ohh

[cobalt parrot]

so yeah

but

thats a potential test question

that wasnt the question you gave

[inland tide]

Really

[cobalt parrot]

of course

when i did conics i once had a question like that

[inland tide]

Oh

It’s like practice questions for us

[cobalt parrot]

just gotta draw a graph its that simple

[inland tide]

I will keep that in mind

[cobalt parrot]

yeah keep that in mind

but dont over practice it just do what ur teacher gives

[inland tide]

Ty very much

[cobalt parrot]

np

[inland tide]

Ok

Ty ty

.close

[cedar kiln]

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[feral basalt]

Could someone please help me with this problem?

I have to get the inverse

[cobalt parrot]

?

unclear picture

[feral basalt]

Mathway says the answer is f(x)^-1=(1/x)+1

Is it my handwriting?

I got f(x)^-1=1/x-1

[cobalt parrot]

what uis the question

[cobalt parrot]

well i dont rlly see the question

whats the question

[feral basalt]

f(x)=1/x-1

[cobalt parrot]

find?

inverse?

[feral basalt]

Yup

[cobalt parrot]

ok

so when you find inverse, your x values become ur y values and your y values become your x values

therefore

[feral basalt]

Yup I got that

Idk how they got a +1

Is it because x is in the denominator?

[cobalt parrot]

$x = \frac{1}{y-1}$

[wraith dagger]

IAMTHEFARMER

[cobalt parrot]

cross multiply

you get:

$x(y-1) = 1$

[wraith dagger]

IAMTHEFARMER

[cobalt parrot]

divide by x

$y-1 = \frac{1}{x}$

[wraith dagger]

IAMTHEFARMER

[cobalt parrot]

bring over the 1

[feral basalt]

I see what I did wrong lol

[cobalt parrot]

therefore

[wraith dagger]

IAMTHEFARMER

[cobalt parrot]

or

[wraith dagger]

IAMTHEFARMER

[feral basalt]

How did you get “y-1=1/x”?

[cobalt parrot]

divided by x

x(x-1) = 1

divide each by x

[cedar kiln]

@feral basalt Has your question been resolved?

[cedar kiln]

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[cedar kiln]

@sinful bear Has your question been resolved?

[cedar kiln]

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[floral condor]

(240+200√2+120√5+100√10 )/-56

[floral condor]

can u simplify that

<@&286206848099549185>

;-;

[inner loom]

好

[floral condor]

?

[inner loom]

等一下

[floral condor]

好哒

[inner loom]

十分钟 hold up

[floral condor]

k

ima sleep ;-;#

[modest hull]

Yes, you can

[floral condor]

how

[modest hull]

Wait

I mean no

[floral condor]

k

ty

[inner loom]

不能

[floral condor]

.close

[cedar kiln]

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[floral condor]

oh oki

[cedar kiln]

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[earnest wedge]

I'm a little lost on why that sum of a_ij from i=1 to infinity equals (-1/(2^i-j))

[cedar kiln]

@earnest wedge Has your question been resolved?

[earnest wedge]

<@&286206848099549185>

[supple panther]

I’ve been looking at this for 15 minutes and I’m still confused lmao

I mean I haven’t learned infinite series yet but still

[floral thistle]

this is matrices and infinite series, so it’s pretty advanced

but you shouldn’t talk if you can’t help, because that pushes the question up, so people are less likely to see it

[supple panther]

Oops my b

[storm chasm]

An airplane is at a height of 1200 m when it begins its descent to land. How far is it from the runway, if a depression angle of 20 ° is applied to descend?

None of the above
6140 m
12280 m
3509 m

[cedar kiln]

@earnest wedge Has your question been resolved?

[cedar kiln]

@earnest wedge Has your question been resolved?

[cedar kiln]

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[languid void]

4x^(4)+25x^(3)+26x^(2)+28x+16

[languid void]

How woud I go about solving this without a graphic calculator

is it even possible?

without factoring?

[cedar kiln]

@languid void Has your question been resolved?

[supple panther]

Have you learned the rational root theorem?

[cedar kiln]

@languid void Has your question been resolved?

[languid void]

Yeah but it doesn’t help, I looked up the answer and none of the roots are rational

I learned it 2 years ago

[supple panther]

Oh yeah you’re right

[cedar kiln]

@languid void Has your question been resolved?

[cedar kiln]

Channel closed

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[cedar kiln]

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[versed lava]

Let there be a Set A {1,2,3}

[versed lava]

Set A is corresponding to x

now if there is a relation (a,a)

does the order within (a,a) matter

so is (1,2) the same as (2,1)

<@&286206848099549185>

.close

[cedar kiln]

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[pallid sapphire]

for a multivariable function, if the second derivative test fails what can be done to identify the nature of the curve?

[wraith blaze]

identify the nature of the curve
what exactly do you want to find out about it?

[cedar kiln]

@pallid sapphire Has your question been resolved?

[pallid sapphire]

if the stationary point on that curve is a max or min

is the only way to look at the value of the function close to the point?

[wraith blaze]

damn, i forgot that this bot was sometimes punctual

global extremum? or local?

[pallid sapphire]

just in general

[wraith blaze]

so global

[pallid sapphire]

this is just a hypothetical case, ive been pondering on for last 2 weeks

lets say global and local, what would you do?#

[wraith blaze]

what do you know about finding extrema in general for multivariate functions?

besides the SDT

[pallid sapphire]

you can look on either side of the SP

and find the value of the function

if its positive on one side and negative on the other its a saddle

if its positive on both sides the SP is minimum

if its negative on both sides the SP is a maximum

are the last 2 cases right or other way round/

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[wraith blaze]

nani

[pallid sapphire]

.reopen

[cedar kiln]

✅

[wraith blaze]

just 3 minutes and timeout

dang

[pallid sapphire]

lol

[wraith blaze]

wdym by SP?

[pallid sapphire]

stationary point

[wraith blaze]

i see

you don't know any theorems for finding extrema of multivariate functions

?

[pallid sapphire]

not really, but i can learn

are the statements that i made above true?

[wraith blaze]

not in general

[pallid sapphire]

for what cases are they true?

[wraith blaze]

basically you said that the sign of the function evaluated at the x-coord of an SP determines if the SP is an extrema or a saddle, yes?

are you sure you meant just "function" and not its derivative?

like, were you referring to the SDT?

[pallid sapphire]

my lecturer uses the function

not the derivative

[pallid sapphire]

evalutated at x and y coordinates

for example: if a stationary point occured at (1,0) i would look at the value of the function at (0.9,-0.1) and (1.1,0.1)

then depending on their sign determine the nature of the SP

this might be completely wrong, thats why i wanted to know for sure

i think it'd be easier to solve a qs and explaining

like this one

the SP is at (0,0) but the SDT is inconclusive

what would i do here?

cuz if i evaluate the curve at (-0.1,-0.1) and (0.1,0.1) I get positive values which means its a minimum

but if i evaluate the x-derivative at (-0.1,-0.1) and (0.1,0.1) i get negative values which would mean its a maximum

and if i evaluate the x-derivative at (-0.1,-0.1) and (0.1,0.1) i get a positive and a negative value which means its a saddle

<@&286206848099549185>

the topic this is, is called constrained optimisation

[wraith blaze]

take for example the function f(x,y) = x²y²-1

[pallid sapphire]

thats not a mutlivariable function tho

[wraith blaze]

graphically, it has a global minimum at (0,0) when f is projected onto the xz-plane

but f is negative at points below z=0

[pallid sapphire]

if its a minimum wouldnt the curve rise up, so there wouldnt be negative points? as long as x^2y^2 >= 1?

[wraith blaze]

,w plot f(x,y)=x^2y^2-1

[wraith dagger]

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[wraith blaze]

it does rise up from (0,0), but f is negative

below z=0

[pallid sapphire]

ok

but how did you know that its a global min if the sdt fails

and without looking at the graph

in a test im not gonna have a graphical calc or software

wait i was reading it

[pallid sapphire]

for point 3. by "largest values" you mean the positive largest right? and same for "least"

[pallid sapphire]

and finally how would i find g for this function?

[wraith blaze]

i meant "greatest" there

[pallid sapphire]

positive "greatest" right?

[wraith blaze]

it's possible, though, that evaluating f at any point obtained in (1) or (2) yields a negative value

being the "greatest" doesn't always mean f is positive at the global max

[pallid sapphire]

ah ok i see

and what about a saddle point?

[wraith blaze]

it doesn't help you identify saddle points, unfortunately

[wraith blaze]

anyhow, you mainly wanted to know when critical points can be considered as global extrema, right?

[pallid sapphire]

well i said general so that includes local and global

[pallid sapphire]

but this case is for both right?

[wraith blaze]

any absolute maximum of f is a relative maximum of f on any open region containing that point

similar for absolute minimums

[pallid sapphire]

ok

Thank you!

[wraith blaze]

there's a method that uses lagrange multipliers, and it's way nicer i suppose

[wraith blaze]

the advantage of that method would be, you wouldn't need to partition the restrictions of f into boundary curves

[pallid sapphire]

yes i know of it, but that only applies when you have been given a function f(x,y) and an intersection curve g(x,y) right?

or can it be used to on just a normal curve, f(x,y)

[wraith blaze]

yes, and this is similar to the previous method i talked about

in fact we find global extrema of a function whose partial derivatives are continuous

[pallid sapphire]

hmm, yes i understand that method but the problem question related the curve which is unbounded

but i do understand the lagrange method

[wraith blaze]

wdym

[pallid sapphire]

the question said on R^2

this was the example qs

[wraith blaze]

sorry, i meant to say here that $g_{h_a}$ is the restriction of f on each boundary curve $h_a$ (a function of $x$ and $y$), $a\in{1,2,\dots ,k}$ where $k$ is the number of partitions of the boundary

[pallid sapphire]

ah ok

last question:

[pallid sapphire]

you see on these boundary curves, how would i know which direction the curves are going?

[wraith dagger]

Configuration options for `keepsourcefor`

How many seconds to keep source for after compilation ('None' to never delete).
​ Current value: 0 seconds
​ Default value: Don't delete source (may be overriden by the guild)
Accepted input: A non-negative number of seconds, or None to keep forever.

[pallid sapphire]

an example of these is this:

[cedar kiln]

@pallid sapphire Has your question been resolved?

[pallid sapphire]

i'll just close it for now. If you find an answer please pm me. I have a mid-term next week
Thanks again

.close

[cedar kiln]

Channel closed

Closed by @pallid sapphire

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[fluid fulcrum]

Hi. If we have a derivative, d(sinx)/dx = cosx.. why are we able to say that: d(sinx) = dxcosx? I thought that the d(sinx)/dx was just notation, and it doesnt actually behave like a fraction?

[floral thistle]

@fluid fulcrum those are actually called differentials

and i think it’s usually written like y = sinx, dy = cosx dx instead

i just think of differentials as dy/dx but then you multiply both sides by dx

this is another reason why i think derivatives are fractions lol

when you get into integrals you’ll be using these a lot

[cedar kiln]

@fluid fulcrum Has your question been resolved?

[fluid fulcrum]

but, how, from first principles, does this work? I though dy/dx is just a form of notation to indicate 'the derivative'

[floral thistle]

by first principles, do you mean the limit definition of a derivative?

if so, i don’t think differentials have a limit definition

[cedar kiln]

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[wraith blaze]

is the bot malfunctioning? lmao

[zinc mural]

wdym

[wraith blaze]

the channel wasn't closed at all, this is sus

[zinc mural]

closed due to time out looks like

I can withdraw my question and he can reopen it

[wraith blaze]

yeah, but strangely it didn't say that

np

[zinc mural]

@floral thistle You still need this?

[zinc mural]

.

[floral thistle]

no, i wasn’t even the one who asked the question lol

[wraith blaze]

actually, it was @fluid fulcrum who was using this

[zinc mural]

oh oops sorry

MEH you still need this open?

[floral thistle]

they’re offline

although they could just be on invisible mode