help-23

[safe radish]

@crude vault Has your question been resolved?

[dry sinew]

What is your answer

2.14 something

[crude vault]

2

for 225,500

2.08 for (50,125)

2.5 for (300,625)

what points should i be picking?

[crude vault]

?

[dry sinew]

This all means that this is not a straight line

[crude vault]

how do i get the slope then>?

[dry sinew]

What are those red dots

[crude vault]

they arent possible to be clicked

[dry sinew]

Ig t1= 300 and t2=325, v1=650 and v2=700

Now try these points

[crude vault]

wym double x?

i see

is that 325,700

[dry sinew]

V1=650

V2=700

[crude vault]

why did you choose those?\

[sage shore]

Guys guys, it passes through origin (0,0)

One point to choose only xd

[crude vault]

wtym

[sage shore]

(V-0) / (T-0)

directly V/T

of some point

[dry sinew]

That's the problem

[sage shore]

it is a straight line, and since it passes through origin, you do not need to "guess" 2 points

you already have one point, the origin, (0,0)

[crude vault]

so its just 350/750

[sage shore]

The inverse you mean, V/T

750/350

[crude vault]

yes

750/350

[sage shore]

But wait

[crude vault]

thats none of the ans

[sage shore]

Yes

Look, the message in red

[crude vault]

its a jpg

[sage shore]

mhmmm :|

We need the precision :p

[dry sinew]

1. This line has changing slope
2. Those red dots are the points. Ig not sure

It's written that you can get cordinates by clicking on the points

[crude vault]

yes but i cant

[sage shore]

it's really necessary, otherwise how would we know whether it is 2.178 or 2.18

i guess you're gonna have to choose one of them randomly D:

You got it from here right?

[crude vault]

can you click them,

@sage shore

[sage shore]

Nooo :(

I guess you're gonna have to randomly choose one

[crude vault]

thanks anyways though guys!

.close

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[hard veldt]

hi hT I 2 +2

[half bane]

huh?

[tall bough]

Most comprehensive math question

[half bane]

lol

[karmic hedge]

Is it not 6?

[half bane]

how did you even understand the problem ngl

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@hard veldt Has your question been resolved?

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[plain cosmos]

By the first equation:

(x+1/x)^2019=(-1)^2019

Expand the LHS

[trim wigeon]

LHS is going to be -1

[plain cosmos]

No

In terms of x

What do you get in terms of x on the LHS

[trim wigeon]

what do you mean in terms of x?

[plain cosmos]

What is (x+1/x)^2019

[trim wigeon]

(x^2+1)^2019/x^2019

im not sure

so then would it be x^2 + 2x + 1 / x^2

[broken yew]

???

[hasty wagon]

sorry for interrupting:
i think this really depends on part (i) of the question

[broken yew]

Thats a very good point - definitely include that

lmao

[karmic hedge]

Maybe find a recursion and find it’s closed form

[rich laurel]

it s just 2

x+1/x=-1 and this was a nice suggestion

[trim wigeon]

i don't understand

[rich laurel]

x^3+1/x^3+3(x+1/x)=-1

x^3+1/x^3+3(-1)=-1

yeah mb confused

there

so x^3+1/x^3=2

now consider y+1/y=2 and find what is y^3+1/y^3

keep in mind 2019=3*673 and u can solve it from here if u need help tag

[trim wigeon]

@rich laurel im still confused, i dont know what to do

i dont get it

[rich laurel]

what is y^3+1/y^3

[trim wigeon]

1?

[rich laurel]

no

do it again if u reach 1 show ur work

[trim wigeon]

is it y < 0

[rich laurel]

what does this have to do with anything?

[trim wigeon]

idk

[rich laurel]

you have : y+1/y=2 , cube both sides and tell me the result

[trim wigeon]

y=2?

[rich laurel]

wdym y=2? i m not asking you to find y... im asking you to find y^3+1/y^3

and no y=2 doesnt work since we get 2+1/2=2

[trim wigeon]

no u wrote y=2, so i was asking if that was a typo

is it meant to be y=2?

[rich laurel]

do you know how to expand (a+b)^3?

i wrote y + 1/y

=2

no it s not it s (y + 1/y) =2

[trim wigeon]

a^3+3a^2b+3ab^2+b^3

[rich laurel]

nice now expand (y+1/y)^3

[trim wigeon]

y^2 + 3y^2 + 3y + 1/ y^3

[rich laurel]

that s not correct

[trim wigeon]

(y+1)^3 / y^3

[rich laurel]

no...you can just replace a with y and b with 1/y here

[trim wigeon]

can u help

im stuck

[rich laurel]

it s (y)^3 +3 (y^2) (1/y) +3 (y) (1/y^2) +(1/y)^3

just by replacing a with y and b with 1/y

doing some algebra we get y^3+3y+3/y+1/y^3

[rich laurel]

you were almost there with this attempt

[trim wigeon]

ohh yes that makes sense

[rich laurel]

so we started with y+1/y=2 and cubing both sided we get y^3+.....+1/y^3=2^3

correct?

[trim wigeon]

yes

[rich laurel]

this is y^3+1/y^3 + 3 (y+1/y)

but y+1/y=2 so y^3+1/y^3 + 3*2

good?

[trim wigeon]

yep got it

[rich laurel]

to be clear this can also be written as a^3+b^3+3ab(a+b)

[trim wigeon]

yes

[rich laurel]

so we get y^3+1/y^3+6=8 so y^3+1/y^3=2

[trim wigeon]

yep

[rich laurel]

and that about concludes our problem...almost

so what do you think will happen if i take y^3+1/y^3=2 and cube both sides again

(keep in mind we started with y+1/y=2 and concluded y^3+1/y^3=2)

[trim wigeon]

3/2?

wait no

2/1?

so just 2?

[rich laurel]

yeah...do you get what is happening tho

you didnt have to calculate anything to realise it will be 2

[trim wigeon]

yep but can i ask why we are using y? the question only uses x

[rich laurel]

just a moment

so what if u cube again?

[rich laurel]

the "nice" way to think about this is: you have y^3+1/y^3=2 so just set z=y^3 so we have z+1/z=2

and we are back at the original problem

[trim wigeon]

ohhh

right

[rich laurel]

so no matter how many times we i want to cube i can just set k=z^3 or something and know the answer immediately (which is 2)

[trim wigeon]

yep

[rich laurel]

now back to our problem we got x+1/x=-1 and we found out that x^3+1/x^3=2

just to be sure you understand...what is x^9+1/x^9?

[trim wigeon]

x^9 + 9(x^2)(1/x) + 9(x)(1/x^2) + (1/x)^9

[rich laurel]

just set y=x^3 and we are back at y+1/y=2...

y^3+1/y^3=2 so x^9+1/x^9=2

[trim wigeon]

yep

[rich laurel]

i might have to leave in a bit so i ll try to explain it a bit fast maybe someone else can take it if you need more help

x^3+1/x^3=2 => (x^3+1/x^3) * (x^3+1/x^3)=4

x^6+1/x^6 + 2=4 so x^6+1/x^6=2

doing that again we get (x^3+1/x^3)*(x^6+1/x^6)=4 we get x^9+1/x^9+ (x^3+1/x^3)=4 so x^9+1/x^9=2

if you keep doing this (multiplying by x^3+1/x^3) you will be getting all multiples for 3 as powers... meaning that you will get x^12+1/x^12=2 x^15+1/x^15=2.....

if you do that 673 times you will get x^2019+1/x^2019=2 since 2019=3*673

[trim wigeon]

do i need to include all that in my lines of working out? is it necessary to include everything?

[rich laurel]

eehh i dont know how to reply to that...i m here to make you understand, if you do you can judge whether something is necessary to include or not

[trim wigeon]

okay

[rich laurel]

the aim is if the teacher asks you how you got something you will be able to reply

[trim wigeon]

yep

thanks

[rich laurel]

pretty hard problem for your class i think...np cya

[trim wigeon]

.close

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[cobalt acorn]

Can someone check if this is correct

<@&286206848099549185>

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[cobalt acorn]

Ok

I have a test soon so can someone help quick

[wise quest]

it seems correct

you can just put it in desmos or smthn to check 100%

[cobalt acorn]

Ok Thanks

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[pale bloom]

I’m almost done with this, I don’t know what I’m doing wrong, I’m supposed to have log |x| in the answer but I’m getting log|2x|

[safe radish]

Please don't occupy multiple help channels.

@pale bloom Has your question been resolved?

[obtuse plover]

Did Hayley not answer ur question in #help-16 ?

Stick to one channel

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[tall bough]

yes

[obtuse plover]

Have u ever graphed piecewises before

Watch this video

https://youtu.be/Uzw9tsGq2Pw

[thin bridge]

there's usually something like "if" or "when" in between

[thin bridge]

the second half u follow what it says
poor description

don't really know what you mean by that

[safe radish]

@lean otter Has your question been resolved?

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[lean otter]

im trying to solve $(A-AX)^{-1}=X^{-1}B$ for $X$ but im definitely messing up somewhere

[lean otter]

this is definitely very sus

[quasi bison]

nyeh?

feels like a roundabout route

why not invert from the getgo

get $A - AX = B^{-1}X$ and a lot less headaches

[flat frigate]

Ann

[lean otter]

oh

why did that fly over my head

okay so [
A-AX = B^{-1}X \
A= B^{-1}X+AX \
A=(B^{-1}+A)X \
(B^{-1}+A)^{-1}A= X
]

this still feels fucked though

oh wait i can just add them cant i

[lean otter]

okay

that still seems wrong

OH

[
A-AX = B^{-1}X \
A= B^{-1}X+AX \
A=(B^{-1}+A)X \
(B^{-1}+A)^{-1}A= X
]

[lean otter]

that looks fine

,close

[obtuse plover]

Use .close to close a channel

[worthy hemlock]

.close

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[vale hollow]

What does it mean for a function f to be less than a function g, also written as f < g ?

[split ether]

It means that for all x the inequality f(x) < g(x) is true

[vale hollow]

ah so for all x the function f will yield a y value less than the y value given by g for the same x?

[upbeat swan]

Yup

[split ether]

Yup

[upbeat swan]

Jinx

[lean otter]

Jinx

[vale hollow]

ty

[split ether]

Really important to note that there are many functions f and g such that neither of f < g, f > g or f = g is true

[vale hollow]

ye

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[dull crane]

Find an equation for the tangent to the curve at the given point. $y=x^2-3, (-2,1)$

[flat frigate]

happy cat

[dull crane]

$\frac{dy}{dx}=2x$

[flat frigate]

happy cat

[dull crane]

What do I do after this step?

[primal granite]

use the formula f(a)+f'(a)(x-a) to find the tangent line. with a being the x value of the point given

[dull crane]

Oh ok

$-2-4(x+2)$

[flat frigate]

had2grind4this

[dull crane]

wait one second

f(a) = -2

[primal granite]

f(a) doesnt equal -2

the other stuff is correct though

[dull crane]

oh it's the regular function of f

whools

[primal granite]

ye lol

[dull crane]

f(-2) = (-2)^2-3 = 1

f'(x) = 2x -> f'(-2) = 2(-2) = -4

$1-4(x+2)$

[flat frigate]

had2grind4this

[primal granite]

correct, now put that into standard form and you should be done :D

[dull crane]

$-4x-8$

[flat frigate]

had2grind4this

[primal granite]

i think you might have missed smthing

[dull crane]

what did you mean by standard form

[primal granite]

oh dont forget about the one

[dull crane]

OH in my head i multiplied 1(-4) 💀

[primal granite]

standard form as in ax+b

[primal granite]

oh lmao

[dull crane]

-3x-6

[primal granite]

maybe we should do this step by step

first thing to do is distribute the -4

what would that get us?

[dull crane]

oh

because its being multiplied by the backrets

brackets

i see

$-4x-7$

[flat frigate]

had2grind4this

[primal granite]

yep

[dull crane]

Thanks

[primal granite]

that is the tangent line

[dull crane]

😂

[primal granite]

np :D

[dull crane]

.close

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[lean otter]

Quick question

since the triangles are similar

aka triangle edf is a smaller scaled version of triangle abc

is it safe to assume that cos f is 4/5

as the answer

since it will have smaller side measurments

[lilac patio]

yeah think so

[proper crypt]

yes

[lean otter]

okay thanks

.close

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[versed lance]

I am so lost, I just need the solution as my tutor is unavailable for the next few days

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[versed lance]

.close

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[light hatch]

i dont understand the english behind this question

[lilac patio]

me neither

[light hatch]

like

[pulsar pecan]

have you played cards before?

[light hatch]

i'd understand it if it saids 5 cards are dealt from a deck of 52 player cards, how many possible combinations can contain jack, queen and king of hearts

[light hatch]

yes sir

[pulsar pecan]

this is what it says

[light hatch]

huh????????//

the language is messed up in the question itself

[safe radish]

@light hatch Has your question been resolved?

[safe radish]

@light hatch Has your question been resolved?

[spare kraken]

The number of ways to select the jack, queen, and king of hearts is 1 since there is only one set of these three cards in the deck.
The number of ways to select the remaining two cards from the remaining 49 cards is given by the combination formula:

$C(49, 2) = 49! / (2! * (49-2)!) = (49 * 48) / (2 * 1) = 1,176$

[flat frigate]

purrthagorascat

[spare kraken]

hope it helps @light hatch

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@light hatch Has your question been resolved?

[light hatch]

o

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[obtuse igloo]

wat did i do wrong?

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Channel closed

Closed due to the original message being deleted

[broken yew]

.reopen

just testing

[subtle juniper]

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[lunar edge]

I was here yesterday asking about this

[lunar edge]

Why is this not the answer that I’ve picked? I don’t see how it’s continuous

It says that when x=1 the right side of the limit is -5

[cunning shard]

show the function

oh it’s written in answer

[lunar edge]

Yee

[cunning shard]

What you think is a continuous function?

[lunar edge]

A function is continuous is f(x) = f(a) as x approaches a

So if I sub a in for x it equals a function that is connected throughout, in layman’s

[cunning shard]

A function is continuous if $\lim_{x \to a} f(x)= f(a)$ for every a in the domain?

[flat frigate]

dotdoc.

[cunning shard]

Is that what you mean?

[lunar edge]

If it doesn’t go to infinity

[cunning shard]

what do you mean?

which one going infinity?

[lunar edge]

F(a)

[cunning shard]

technically there is no number as infinity

[lunar edge]

Omg

Are you able to help me or

[cunning shard]

i’m trying to

[lunar edge]

Why is this the answer

That is literally what im asking

[cunning shard]

This is the definition of continuity, one other way to see continuity is that you should be able to draw the function without taking pen up

[lunar edge]

Yes

I understand that

[cunning shard]

now you see yourself

[lunar edge]

Now why is -5 not the right side

It says

When x=1

The right side is -5

The answer says it is

The answer says it exists at 3

[cunning shard]

What’s the limit of the piecewise function?

[lunar edge]

Why’d you delete that

[cunning shard]

Do this

[lunar edge]

The limit is what I got wrong

[cunning shard]

and see of it’s continuous

[lunar edge]

I don’t understand why it is the way it is

[cunning shard]

That’s what i was trying to do

[lunar edge]

Why is the right side of the function not -5

[cunning shard]

^

[lunar edge]

I know

I appreciate it

But I just want to know

Why is the right side of the function not -5

[cunning shard]

Which function is determined by the right side of x=1?

[lunar edge]

The right side

[cunning shard]

write that here

[lunar edge]

Lim x -> 1 = -5

[cunning shard]

if i ask you to evaluate f(1)

Where do you evaluate?

forget about limits

[lunar edge]

On right side

[cunning shard]

type it here

[lunar edge]

It’s just -5

[cunning shard]

how about right side of 1?

or left side of 1?

[lunar edge]

Left side is not an issue

The right side says when x = 1 it’s -5

[cunning shard]

just post a work of yours

[lunar edge]

I don’t know what that means

Why isn’t the right side -5

Can you answer this

[cunning shard]

The function is discontinuous anyways

[lunar edge]

So you can’t answer this

[cunning shard]

limiting value and f(1) aren’t same

[lunar edge]

Okay

So when x = 1, which it does in the limit

Why isn’t the right side -5

[cunning shard]

dude the function is discontinuous and your question is wrong

That’s the reason

idk how to be more helpful

[lunar edge]

Then you shouldn’t answer questions if you can’t obviously answer

<@&286206848099549185>

[cunning shard]

that’s literally the answer

[lunar edge]

You are not,

Answering MY question as to why

I know it’s wrong

Obviously

[cunning shard]

A function is continuous if $\lim_{x \to a} f(x)= f(a)$ for every a in the domain

[lunar edge]

I’m asking for a reason as to why the right side in my answer is incorrect, not where it is

[flat frigate]

dotdoc.

[spare kraken]

hi

[cunning shard]

Because it doesn’t hold the condition for continuity

[lunar edge]

I know

[cunning shard]

that’s simple as that

[lunar edge]

I said that

[spare kraken]

@lunar edge what is the question?

[lunar edge]

And my answer is wrong

[random cobalt]

this is getting heated

[cunning shard]

pls help him

[lunar edge]

In my pins, why is the right side not -5

[lunar edge]

I appreciate your attempt, and your time, thank you.

[random cobalt]

da**

anyway

i think whats confusing u is that the limit of x is not always equal to f(x)

when calculating limits, it might help to ignore what f(x) is and focus on what it looks like its gonna be

[lunar edge]

So where does the right side of -5 fit into all this

[random cobalt]

can you show me the original function plz

[lunar edge]

-x^2-2x+6

And then -5

[random cobalt]

i cant see the full function

can u plz show me

[lunar edge]

That’s the whole thing

[lunar edge]

It just says suppose f(x)

[spare kraken]

it's on the answer

[random cobalt]

thx

so

[random cobalt]

mb

now

[distant widget]

its a limit saying that it tends towards 1, not that it equals 1, so you cant use the part of the function for x = 1

[random cobalt]

is the condition for x =/ 1 continuous

[random cobalt]

exactly

[cunning shard]

$\lim_{x \to 1+} = -(1)^2-2(1) + 6 = \lim_{x \to 1-} = -(1)^2-2(1) + 6 = 3$

[random cobalt]

thank u

[random cobalt]

thank u too

see?

[lunar edge]

So then the part that’s the left side just carries over

[flat frigate]

dotdoc.

[cunning shard]

Does this help?

[lunar edge]

No sadly

[cunning shard]

Notice both cases you plug in 1

[lunar edge]

Because it still is equaling 1

[random cobalt]

i see yall got this

imma head out

ping if needed

[lunar edge]

Yz has it I think

[cunning shard]

you should read about limits please

[lunar edge]

I did thanks

[cunning shard]

you need plug 1 for 1- or 1+

[lunar edge]

I know

Please

[cunning shard]

even if 1 is not available, you check what happens naturally if one was there

[lunar edge]

Yes, I know

You’re not helping

I appreciate your attempt

But you can move on now, thank you

@distant widget

[lunar edge]

Did you see this? Is this correct?

[cunning shard]

I get where you are confused

there is where

It doesn’t say when x=1 right side of limit is -5

[lunar edge]

Yeah how

So wait

[distant widget]

could you explain what you mean by this

[cunning shard]

The question just gives a function. Functions can also be piecewise

[distant widget]

when the others say you can substitute 1, they are right

and the answer there proves that for you

[lunar edge]

Since it doesn’t exactly equal 1, it doesn’t exactly equal -5

[cunning shard]

The notation of curly braces only talks about how function is defined nothing about limits

[lunar edge]

Doc thanks yz has it

[cunning shard]

you use that definition to evaluate limits

[distant widget]

that piecewise function isnt really a good example but yes, an x value that tends towards 1 such as 0.99999 but never actually is 1, would approach the value of 3

[lunar edge]

So the right side of this is basically irrelevant because it will never exactly equal 1

[distant widget]

when x is actually equal to 1, it has a different value which is - 5

[lunar edge]

Okay

So in that case when one side is completely irrelevant, the only side that exists is the side that holds through the whole function then

Which is why both sides are 3 right

[distant widget]

by side do you mean the x->1- and x->1+

[lunar edge]

Yeah

[distant widget]

normally for the left side you would look for something slightly smaller than 1 such as 0.99999, and for the right side something slightly greater such as 1.00001

[lunar edge]

Yeah

[distant widget]

you can substitute both numbers into the equation and find that they tend towards 3

[lunar edge]

Yeah

[distant widget]

you need both of these sides to tend towards the same thing

[lunar edge]

That part I get

Yeah

And since x isn’t exactly 1

The right isn’t -5 right ?

[distant widget]

yes you treat it the same as how you done the left side

[lunar edge]

And if I only have one side that exists that side is correct for the whole function right?

[distant widget]

its a strange example you have, normally they would do this for when the numerator of some fraction would be zero for some x

[lunar edge]

Yeah that I can do

Yeah it is weird

These were the only problems I got wrong on this test

3 out of 20

[distant widget]

the answer is there, if both limits are not the same, then x->1 would not exist

[lunar edge]

And I just wanted to double check my thoughts

[lunar edge]

So that’s a yes then

Because both sides of the piecewise are not the same but one isn’t really relevant

[distant widget]

you cant say its correct for the whole function because the limit doesnt exist in the first place

[lunar edge]

Because x doesn’t exactly equal 1

[lunar edge]

The answer says it does exist though

It says it exists at 3

[distant widget]

i meant for a case where it doesnt exist

[lunar edge]

For both sides

Yeah

[distant widget]

so for example x->1- is equal to 20 and x->1+ is equal to 50

[lunar edge]

What I’m asking though is if one side exists and the other side isn’t in play, the side that does exist is actually relevant for both sides

[distant widget]

they arent equal so x->1 doesnt exist

[lunar edge]

That’s what this answer is saying

[distant widget]

the limit wouldnt be the same on both sides though, so you cant say a limit exists even if it does on one side

[lunar edge]

So then how is the answer for this 3 and not “does not exist” if the side is relevant when x = 1 is -5

The left side has to carry over

The left side has to be the whole function

Otherwise this answer is wrong and I was correct, which I don’t think is the case

The listed right side is -5 when x = 1, but because x doesn’t exactly equal 1 here, the right side is not in play

[distant widget]

-(1.0001)^2-2(1.0001)+6 = 2.99959999

this is for the right side, this approaches 3

[lunar edge]

Therefor, I have to assume that the left side, which is true, Carries over to the whole function

[lunar edge]

Yes, but it says suppose the right side is -5 when x = 1

So my axiom on this has to be true

[distant widget]

but x isnt equal to 1 here

[lunar edge]

Exactly

And that’s the same side as the left

So the answer is “yes” then

[distant widget]

both limits would be 3

[lunar edge]

Yes

If what I’m saying is true

That if one side is irrelevant

The side which is becomes the whole function on both sides

[distant widget]

what do you mean by irrelevant

[lunar edge]

Yes, right?

[lunar edge]

Irrelevant like the right side becoming -5 when x = 1, which it doesn’t, so it is irrelevant

You can just say it’s true

That’s all I was looking for mainly, now that you helped me earlier with seeing it that way

That aspect was just never explained, so I had missed an assumption and just directly plugged 1 into it and got -5

[distant widget]

we are interested in seeing when it approaches 1 not when it is equal to 1, so i wouldnt call either side irrelevant

you need to know both to decide whether a limit truly exists

[lunar edge]

Yeah but that’s frankly beyond the scope

Yes

But for the sake of knowing past that

[distant widget]

also if you substitute 1 into the first expression of the piecewise function and not the entire piecewise function, you will get 3

[lunar edge]

Yes

I got that

And then I saw x = 1 is -5

Put that in

Saw them different

“Does not exist because different”

Answer wrong

“Why why why”

[distant widget]

i see why you got confused

[lunar edge]

Yeah

A little gotcha in there

[distant widget]

if it said lim x=1 instead then your thinking would be fine

[lunar edge]

Yeah then it would be correct

[distant widget]

but it said x->1

[lunar edge]

But then it’d just be algebra basically

Yeah

I just didn’t know that if one said was in play and the other wasn’t effectively, the function could still exist as one side

And you helped me with that

That stumped me from last night so I greatly appreciate this

[distant widget]

i feel like you have the wrong idea, both sides here do exist and so the final limit also exists

[lunar edge]

Yes

I get that

But one side is not pertinent to the limit

They could have well as added

“500000 x = 1”

And it would still be the same

[distant widget]

are you saying that if one side existed and had some value say 1, and the other side did not exist - then the final limit is 1

[lunar edge]

No

I’m saying that the right side is quite literally out of the problem

Because x never reaches 1

[cunning shard]

says who?

[lunar edge]

So it’s not important to my solving and understanding of what the limit is

[lunar edge]

What

[cunning shard]

the function does reaches 1

[distant widget]

i understand now what you mean, but you can also decrease towards 1

[lunar edge]

Says the problem lmfao

[lunar edge]

Sure

[cunning shard]

That’s why it’s piecewise

[lunar edge]

This is dangerously wrong

[distant widget]

so then you would use the right side

[cunning shard]

It does rearches one

[lunar edge]

No

[cunning shard]

at f(1)= -5

[lunar edge]

Because the right side says -5 when x = 1

Not when x approaches from that side

It’s not listed and I have no information to assume that

[lunar edge]

Please leave

[cunning shard]

Mr. know it all cannot be helped to he honest

[lunar edge]

You’re just saying wrong things and it’s confusing people

[cunning shard]

just go take a book, figure it out for you

[distant widget]

we can approach one by going from 0, 0.1,..., 0.999 and by 10, 5, 3, 1.00001

[lunar edge]

What you’re saying isnt factual is antithetical to solving the problem

[distant widget]

think of a number line

[lunar edge]

I understand this

But there’s a level of assumption you have to make because that isn’t listed as part of the piecewise

[cunning shard]

what is a piecewise function to you?

It’s a normal function

[lunar edge]

Technically the right side is still -5 when x = 1

[lunar edge]

I’m asking you again to please leave

Second time

[cunning shard]

It’s a public help platform

You may leave sir

[lunar edge]

This is my help channel currently

[distant widget]

we arent interested in x = 1 so you dont have to consider that

[lunar edge]

Yeah that’s my point

[distant widget]

at x = 1.00001 the value approaches 3 rather than 5

[lunar edge]

“In play” “irrelevant “ etc yeah

[lunar edge]

Yeah I understand that

I think we’re talking in circles and semantics

You helped me big time though

Thank you thank you thank you

Seriously

I was stumped on that for a bit

.close

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[tall bough]

does anyone know what this symbol is

[light shoal]

it means surjection in many contexts, not sure about this particular context though

[tall bough]

ohhhh

its makes sense now

hihi i finally can continue reading

.close

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[lean otter]

can someone explain to me the inverse of g(x)=−3(x+6) is this? I'm confused why we didn't disturbed the -3

[brazen helm]

you mean g^(-1) right

[lean otter]

yes

[brazen helm]

well you know how to find it right?

g is the inverse of f if f(g(x)) = x

so you have to solve x= -3(y+6) and solve for y

[lean otter]

okay

[brazen helm]

got it?

[lean otter]

nope

[brazen helm]

you have to isolate y in the equation

[lean otter]

I know but what should I do with the -3?

[brazen helm]

you divide by -3

It stays

notice you will indeed get -1/3 x

and then the 6 goes right over to the other side

[lean otter]

why not distributing?

[brazen helm]

you don’t have to, it will come out to be the same thing

try it for yourself

[lean otter]

but I want to distrubite and I got wrong answer

[brazen helm]

can you show me your work

[lean otter]

y = (x + 18) / -3

[brazen helm]

Why!

?

[lean otter]

why what?

okay thank u for your time I have to go @brazen helm

.close

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[dusk vapor]

how to solve this, no clue how to begin

[toxic stratus]

did you even attempt part (a)

[dusk vapor]

no hint

snow

frick u

im trynan do b without doing a

[toxic stratus]

well do part (a)

[dusk vapor]

i did it

its not that deep

[toxic stratus]

you're gonna need it

[dusk vapor]

no what if theres a way without a

[toxic stratus]

yeah brute force calculate the whole thing

[dusk vapor]

(x^2+ai)(x^2-ai)

[toxic stratus]

no

bruh

[dusk vapor]

are u trolling...

[toxic stratus]

factorise it without complex numbers

[dusk vapor]

...

ur trolling the hint said use complex numbers

[toxic stratus]

yeah but the answer shouldnt have any complex numbers in it

oh part (a) says linear factors

well do part (a), but pretend it said quadratic factors with real coefficients

treating a as a real number

[dusk vapor]

question 6 is so f uped snow

snow Ou is confused of why this isn't over 100

28+64 is 92

@kind flax

where did he make the mistake

WHERE DA MISTAKE

well he found out

its 28+96

cool

.close

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[fallow dock]

Is this right?

[split ether]

cos(180 - x) =/= cos(x)

[twilit harness]

no

[split ether]

Instead apply cos(360 - x) = cos(x)

[fallow dock]

Is the answer then cos190?

360-170

[split ether]

Yup

[fallow dock]

Thank you

.close

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[fallow dock]

.reopen

[safe radish]

✅

[fallow dock]

How do I solve this?

This is what I have so far

[split ether]

That doesn't really look like a right triangle where the hypotenuse is 5cm long

[fallow dock]

Oh right triangle

[split ether]

Right, label any of the other vertices as A

[split ether]

Right, so we are given that cosA = 0.75

Meaning that the ratio between the side adj to A and hypotenuse is 0.75

So adj/5 = 0.75 and adj = 3.75

Right, let's now find the opposite side

You could either calculate sinA using the Pythagorean identity and solve for it the same way

Or just use the Pythagorean theorem

[fallow dock]

.close

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[potent osprey]

Anyone know whether this has a closed form?

[potent osprey]

As for context, it's my result for part (b) of this question:

[safe radish]

@potent osprey Has your question been resolved?

[ruby sequoia]

is that the fractional part?

[potent osprey]

No, should be the end result

Wolfram gives this for n=100

https://www.wolframalpha.com/input?i=sum+for+k+%3D+0+to+infinity+k+*+((1-2^(-k))^100-(1-2^(-k%2B1))^100)

[ruby sequoia]

no i meant if {x} denotes the fractional part of x

[potent osprey]

Oh no its just brackets

parentheses

[ruby sequoia]

ohhh

i think you could make this a telescoping series

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[potent osprey]

I also thought about that but the problem is that you cant really cancel $k(1-2^{-k})^n$ with $-(k+1)(1-2^{-k})^n$

.reopen

[flat frigate]

_daili

[safe radish]

✅

[potent osprey]

And $\sum_{k\geq 1} - (1-2^{-k})^n$ diverges so also cant split sum as far as I know

[flat frigate]

_daili

[ruby sequoia]

$k(1-2^{-k})^{n}-(k-1)(1-2^{-(k-1)})^{n}-(1-2^{-(k-1)})^{n}$
how about writing it this way

[flat frigate]

lag8280

[ruby sequoia]

then using this, you could find a closed form for the partial sum

then take the limit

i think it should work

[potent osprey]

That would leave $k(1-2^{-k})^n + \sum_{l=1}^{k-1} - (1-2^{-l})^n$ for the partial sum I think

[flat frigate]

_daili

[ruby sequoia]

yup i believe so

the sum to the right can be simplified

by using the binomial expansion

then swapping the double sums

[potent osprey]

Let me try

If I didn't make a mistake, I currently have $-\sum_{m=0}^n (-1)^m \binom{n}{m} (2^{-mk} - 1)$

[flat frigate]

_daili

[ruby sequoia]

hmm now we've got to take the limit as k goes to infinity

[potent osprey]

Yea, I feel like I did something wrong

[ruby sequoia]

hmm idk i feel like this is correct

[potent osprey]

But $k(1-2^{-k})^n$ goes to infinity as $k$ goes to infinity right?

[flat frigate]

_daili

[ruby sequoia]

so does this sum

theyll cancel out

also, you can simplify it even further i believe

[potent osprey]

wait why would this second sum not go to a finite value?

I mean I know that it should diverge but am having difficulty seeing how the current sum diverges

[ruby sequoia]

hmm

[potent osprey]

thats why I probably made a mistake somewhere

let me check my work

[ruby sequoia]

notice the m=0 terms

[potent osprey]

ah right

oh i did make a mistake btw

I guess the problem is also equivalent to $\lim_{k\to\infty} \sum_{m=0}^k (1-2^{-k})^n - (1-2^{-m})^n$

[flat frigate]

_daili

[potent osprey]

Honestly, it seems like this way will not lead to a closed form, I don't even know if a closed form exists (though I assume it should since they ask for stdev in the next part)

I found this OEIS sequence for the numerators:

A158466

so yeah since they don't give an explicit formula there, i'll just assume it does not exist

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[uneven pulsar]

is this just essentially asking me to plug in 3 for x everywhere possible?

[latent heath]

Yes , that's right

You have to plug the 3

[safe radish]

@uneven pulsar Has your question been resolved?

[uneven pulsar]

oops, sorry for not seeing this

so it would just be 14 for both?

[latent heath]

Yes hence it prove it's parallel

[uneven pulsar]

Ohh simple

Then for this it would be the same process?

[latent heath]

It's the same has before this time they given it in degrees

If you looked at j and l they have the same degree

So for m and k has to have same degree

Hence it's prove it parallelogram

[uneven pulsar]

Oh ok

Thx

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[hardy lion]

this is a related rates problem, what are you knowns and what are your unknowns?

[safe radish]

@white field Has your question been resolved?

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[languid stratus]

I don't really get this problem at all. This is my first time working with the other trig functions aside from sin/cos/tan, but I think I understand them fine. I just don't understand how I'm supposed to do this. The video they provided didn't really help.

[upbeat pewter]

sin is negative and tan is positive

So the point shall lie on Q3

[languid stratus]

Is that the bottom left one? I've been out of high school for a while and they didn't really explain the whole quadrants thing on this review.

But I think I get it, so sin(theta) is just entirely negative. The video they showed had an example with a negative fraction for tan(), and their explanation didn't make much sense.

This does help though

thank you very much

.close

[safe radish]

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[humble heron]

I need someone to look over my questions to determine if I made any mistakes

[humble heron]

<@&286206848099549185>

<@&286206848099549185>

[safe radish]

@humble heron Has your question been resolved?

[full thicket]

1st page looks fine
2nd page, the slope of #4 should be negative, I can't read #5, #6 the x intercept is wrong, can't read #8
3rd Page, 2nd graph is wrong(look at the parenthesis)
4th Page, 10a is wrong(look at the parenthesis), I can't read the answer for 11

Might have made a mistake somewhere, its a little blurry

[humble heron]

I just saw how blurry it is. Does this help?

[full thicket]

Yeah ok, thats much better

Now I can see 18b is correct, little unsure what the number at the end wa at first

[humble heron]

ah ok do you mean 10b?

[full thicket]

oh yeah

my b

[humble heron]

all good! what did i do wrong with 10a?

[full thicket]

So, when you moved the two across the equal sign you added it to the 7 that was inside the parenthesis. Before you're able to add it to the other constant, you would have to distribute the -5/3 to both terms.

[humble heron]

Ohh ok that makes more sense.

[full thicket]

ok good, I don't know how to use the formatting thing, so im trying to get by the best I can without it

[humble heron]

All good. Is 7, 8, and 11 ok after sending a better pic?

[full thicket]

11 is good, just make sure you write the x in teh equation

let me recheck 7 and 8

[humble heron]

Kk

[full thicket]

7 is good, 8 I am getting x=20

[humble heron]

Hm ok I’ll def re do that one and see where I went wrong. And 5 is ok?

[full thicket]

That one is wrong, there is only one vertical line that passes through that point.

[humble heron]

Right, sorry I’m still confused where I went wrong for 5.

[full thicket]

So for 5, you are given the fact that it is a vertical line, if I were to say that any number plugged into x=c would hold that point, it would be false. If I had the line x=0, it will never hold the point(-5,2)

It will have points like (0,0),(0,1),(0,100),(0,-5)

but not (-5,2)

[humble heron]

Ohh ok I see

[full thicket]

what do you think it is

[humble heron]

So given what you said, would it be the points in between - 5-2 ?? Or am I so far off lol.

[full thicket]

A little off. Start at a basic vertical line. x=0 has point (0,2). x=-1 has point(-1,2).

[humble heron]

Hm ok

[full thicket]

Here's the pattern. The green point is the point you need to include. The rest are vertical lines. You can see the equations for each on the side.

[humble heron]

Ah ok I’ll do that question again. Tysm!

.close

[safe radish]

Channel closed

Closed by @humble heron

Use .reopen if this was a mistake.

[full thicket]

np

[humble heron]

.reopen

[safe radish]

✅

[humble heron]

Wait so for 4 I just put a negative before the fraction?

And I’m confused with 9b, 6, and 10a

<@&286206848099549185>

[full thicket]

yes for 4, the line is pointed downwards from left to right, so the slope is negative.

For 6, the sign for the the 18 switched, im going to guess its just a writting error

9b, its the same issue you had in the other problem. When moving the 2 from the left to the right, you can't add it to the 4 in the parenthesis until you remove the parenthesis.

10a, is the same problem as with 9b

Lemme write something down rq

[humble heron]

Kk

And those are the only mistakes out of the 4 pages?

[full thicket]

im pretty sure

The +2 has to be placed outside of the parenthesis and the 2 and 4 can't be added together until you distribute the -1/4 and remove the parenthesis

This was the problem for both 9b and 10a

[humble heron]

Ohh ok I see

Is that 1/4 for in the brackets or 1/6

[full thicket]

1/4 sorry, i have bad handwriting

[humble heron]

All good!!

How did you get rid of the 1/4 in the brackets?

[full thicket]

are you talking about how I went from 4(-1/4) to -1

[humble heron]

Wait nvm I see now

[full thicket]

k

[humble heron]

So I need to fix 9b, 6, 10a, 5, 8,6 and then the 2nd graph? The graph one I’m a bit confused on

[full thicket]

The graph is the one I showed in the work

[humble heron]

Ah right my b

[full thicket]

u good. for 10a, its the same issue with the parenthesis

[humble heron]

How in 10a do you distribute the -5 over 3?

[full thicket]

multiply both the x and the 7 by -5/3

[humble heron]

I keep getting a decimal

[full thicket]

don't evaluate -5/3 with a calculator, just leave it in fractional form

and do the operations from there

[humble heron]

Wdym?

[full thicket]

So, just do fraction multiplication and addition. Don't turn the fraction into a decimal.

It's not a very pretty answer either way

[humble heron]

Oh ok I"ll try that. One second.

Wait so would the answer for 10a is y= - 5/3 x + 7/3 -2?

[full thicket]

It shouldn't be 7/3. 7*(-5/3) means you have to multiple 5 and 7 together

then put that over 3

[humble heron]

so 35 over 3?

[full thicket]

yeah, so now you got y=-5/3 x - 35/3 + 2

just combine the -35/3 and the 2

[humble heron]

thats a decimal 13.666666 etc,

unless im doing it wrong

[full thicket]

they are different signs, the number should be getting smaller

[humble heron]

i got y= - 5/3x - 33/3

[full thicket]

sorry, forgot to mention, you need to make the 2 have the same denominator if you add them together it

[humble heron]

Im so confused

the answer wouldnt change

[full thicket]

See how the 2 was changed to 6/3 so that both fractions have the same denominator

This lets me add the two numbers together to get the final answer

Since 6 divided by 3 is 2, I can replace the 2 in the problem with 6/3

[humble heron]

y= -5/3x - 29/3 ?

[full thicket]

yeah

[humble heron]

ohhh ok

[full thicket]

Sorry for confusing you, I'm not great at teaching others

[humble heron]

im so sorry im still confused on 5

[full thicket]

Is that the vertical line one?

[humble heron]

yes

[full thicket]

Ok. So we need to make a vertical line that has the point (-5,2). The form for a vertical line is x=c where c is some number.

[humble heron]

How do I do that?

[full thicket]

Well, there's only one vertical line that can have a -5 as the x value

[humble heron]

so Is x= -5

[full thicket]

yeah

[humble heron]

is that the equation?

[full thicket]

yup

If you're ever asked that kind of question, its as simple as taking the x part of the coordinate

and putting it in the equation

[humble heron]

Oh ok sounds good. My final question is just making sure 12 is correct

oh and 6 because i re did it and i dont see anything wrong with it

[full thicket]

12 is a little off. its got the right intercept, but its not perpendicular

small math error, 20 divided by 4 is 5 not 6 (this is for problem 12)

And for number 6, look at the sign of the 18 on the x intercept. It switched randomly

[humble heron]

Ok i fixed all of that. For the 9b I got y= - 1/4 x -1 is that correct now that I changed it?

[full thicket]

Seems so

[humble heron]

okay great tysm for your help I know that was a lot.

[full thicket]

No problem. It helps me out too.

[humble heron]

🙂

.close

[safe radish]

Channel closed

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• Ask your math question in a clear, concise manner.
• 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.

[vapid anvil]

Why is this wrong?

[stoic dune]

Is it wrong? That's quite the algebra lol

[plucky elk]

What is your last step doing

sin(x)/x goes to 1 if x goes to 0

Is your x going to zero as h goes to zero

[vapid anvil]

It's not

How would I do this then

Wait

I think I know