help-10

[rare oak]

Anyone know how to do this question, I found the bottom was 30 but I am confused.

[obtuse pebble]

@rare oak Has your question been resolved?

[obtuse pebble]

@rare oak Has your question been resolved?

[hardy widget]

(c) This is the same as the circumference of an enlarged version of the original figure, with a rectangle of dimensions 84.39 by 77.88. The distance is thus $\boxed{84.39(2)+77.88\pi}$ \ \ (d) Find the diameter of the semicircle in terms of the number of lanes, and find the length ran using the logic from (c).

[warm shale]

messy circle creation

[rare oak]

So C is 413.4?

[hardy widget]

yeah

[rare oak]

ok Thank!!!

You are a life saver!

.close

[obtuse pebble]

Channel closed

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[rare oak]

.reopen

[obtuse pebble]

✅

[rare oak]

ummm

[fierce lagoon]

What

[rare oak]

sorry

did I do it wrong?

It says its wrong and I followed the previous persons steps

im confused

[fierce lagoon]

Yeah your first answer

It's off by a bit

[rare oak]

ah

[hardy widget]

oh wait nearest meter

[fierce lagoon]

You would assume the person is running along the far end of the first track

[rare oak]

did I round it wrong?

[fierce lagoon]

Nearest meter

[hardy widget]

actually this too

[fierce lagoon]

No decimals

[hardy widget]

imma just leave it to you lol

[fierce lagoon]

But also like

Your answer is quite off

[rare oak]

so

413?

[fierce lagoon]

You assume the person is running along the far end of the first

[rare oak]

yes

[fierce lagoon]

No because 413 is also wrong, idk how you got that

[rare oak]

ok I get it now I messed up part c

[rare oak]

I got 413 from messy circle , he showed me how to do it and that made more sense

[fierce lagoon]

They goofed up because 73 + 2(1.22) = 75.44

[rare oak]

ah

[fierce lagoon]

84.39(2) + 75.44π

[rare oak]

405.8 so thats the answer

Thanks for the help!!

[fierce lagoon]

Nearest meter bud

[rare oak]

oh

406

Thanks for the help!!!! 🙂

[fierce lagoon]

Np

[fierce lagoon]

Are you done with the channel?

[obtuse pebble]

@rare oak Has your question been resolved?

[obtuse pebble]

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[fierce lagoon]

What's the domain of sqrt(1-x^3)

You sure?

Plug in x=3

See if you get a real answer

Wdym FI?

Here's the thing. Your function has a domain restriction because of the $\sqrt{1-x^3}$. The domain of that is $x \leq 1$, so the domain of the function that you're taking the limit of is ALSO $x \leq 1$.

[warm shale]

Umbraleviathan

[fierce lagoon]

Because of the domain restriction, as x approaches infinity you're gonna get complex numbers that don't exist on the normal XY plane

@wild matrix

[fierce lagoon]

Read from here

You can't

[obtuse pebble]

@wild matrix Has your question been resolved?

[fierce lagoon]

No it doesn't change anything

[obtuse pebble]

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[teal breach]

How would I find the limit for this problem?

[fierce lagoon]

Combine the fractions or use L'Hopital

[teal breach]

Thanks!

[obtuse pebble]

@teal breach Has your question been resolved?

[obtuse pebble]

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[crimson lion]

im a little stuck

[crimson lion]

1 over -7x-5 = -1

how would I go about solving this?

[halcyon plover]

You might want to learn some algebra to do that

[ruby path]

$\frac{1}{-7x - 5} = -1$

[warm shale]

NEONPerseus

[crimson lion]

ive gotten this far, just figuring out this last bit

[halcyon plover]

Can you show the whole process?

[crimson lion]

ignore the little top part that says u = 5x+6 , its from another problem

obviously

[halcyon plover]

Thats correct, just continue

[crimson lion]

right, so hwo though?

hence the first question i had

$\frac{1}{-7x - 5} = -1$

[warm shale]

Biblicalwizardx

[crimson lion]

like, whats the next step?

[halcyon plover]

Find the x

Have you done some problem like this before?

[crimson lion]

yes, but im not great at fractions

:/

[halcyon plover]

Ok for simple, what fraction make -1

[crimson lion]

like, multipy both sides by -7x-5?

[halcyon plover]

Yes, its true

[crimson lion]

so that'd be...

[crimson lion]

@halcyon plover -1/-1

?

[halcyon plover]

-1/-1 equal 1

[crimson lion]

so the answer is 1?

[halcyon plover]

Just follow this

[halcyon plover]

No

[crimson lion]

if i multiply both sides by -7x-5, wouldn't that cancel them out in eachside?

[halcyon plover]

It cancels the left one, not exactly cancel, its like (1/2)*2

[crimson lion]

= 1

right, so its 1 = 1 / -7x-5

[halcyon plover]

The right side is wrong

[crimson lion]

which right side?

[halcyon plover]

This one

[crimson lion]

so up until here im good?

[halcyon plover]

Yeah, and dont forget the other x too

[crimson lion]

exactly, sorry, but how does $\frac{1}{-7x - 5} = -1 (-7x-5)$

[warm shale]

Biblicalwizardx

[crimson lion]

on each side?

dont they cancel out

[halcyon plover]

You have to multiply both side

[crimson lion]

?

[halcyon plover]

$\ (frac{1}{-7x - 5})(-7x-5) = -1 (-7x-5)$

[warm shale]

Rise

[crimson lion]

so it'd be, 1 = -1(-7x-5)

then solve for x

$(-1)frac{1}{-7x - 5} = -1$

[warm shale]

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

[halcyon plover]

Yes

[crimson lion]

so x = -4/7?

[halcyon plover]

Yes

And dont forget the other one

[crimson lion]

and the other is -16/21?

[halcyon plover]

Yes

[crimson lion]

omg thank you u r a saviour!

[halcyon plover]

Np

[crimson lion]

p^4-26p^2?

would equal

26p^2?

actually 25p^2 because p - 26p = 25p

and 4-2 = 2

right?

.close

[obtuse pebble]

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[vestal burrow]

hi! im trying to find x in this question

the background of this question has to do with triangles but this is the part im stuck on

ive tried using elimination and substitution but it always ends on a dead end

im also not very sure on what you can "cancel out"

[desert sinew]

Row 2: $y^2 = (x + 5)^2 - 12^2$

Use this new row to subtract row 1, and y-terms should be gone.

[vestal burrow]

how did you get (x+2)^2

did u mean 5?

ill try that

ok but after u subtract x^2 from x^2 u get 0

[warm shale]

jimmy1234

[desert sinew]

Typo sorry.

[vestal burrow]

nw but even if u use that one and subtract from the first equation dont the variables just cancel themselves out?

[desert sinew]

I think the new Row 2 will have like x^2 + 10x + …, and substracting Row 1 should still leave you an 10x term to solve for 🙂

[vestal burrow]

omg i just realized my mistake

im so dumb

TYSM JIMMY

.close

[obtuse pebble]

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[timid silo]

how will you find the total surface area of this shape

[timid silo]

can you help@on it

[nocturne minnow]

Find the area of the sector, then multiply by the height
Oops

[pine sail]

That's volume.

[timid silo]

to find the total surface area

you have to take out shapes

two sectors

don’t know what the yellow shape is

they have said in the question you have to find the total surface area of the slice of cheese

so i think you don’t have to consider other areas

does that make ssense

[timid silo]

can you explain

[pine sail]

Sure I can.

So you find the areas of all the figures that make sense.

First, the sector on the top.

[timid silo]

thanks and they said in the question that you have to find the area just of the slice of cheese one

[pine sail]

I mean, that means the area of the entire thing lol.

[timid silo]

oh even though they said that you still have to find all the areas individualli oh

[pine sail]

yes.

So the two sectors are fairly easy to compute.

(top and bottom)

Right?

[timid silo]

Right

[pine sail]

OAB is where I'm getting at.

[timid silo]

like they are obvious

[pine sail]

Other than that, you have the rectangle with side lengths 5 * 6

Two of them...

Yes?

[timid silo]

the bigger ones

the green

but if you see doesn’t it look like a different kind of a rectangle

because of the top line it’s not straight a bit

length into width

[pine sail]

It's 3 dimensional, cut them some slack for that.

[timid silo]

oh no

now it looks like a rectangle

[pine sail]

Imagine a pizza slice bruh.

[pine sail]

Okay so the area of the rectangle in red like I mentioned, what would that be?

[timid silo]

my teacher taught me with this name

the formula is length into width so 6 into 5

[pine sail]

Yes.

And you have two of these rectangles too..

One behind.

Also.

[timid silo]

6 looks like the length

[pine sail]

That doesn't matter lol.

[timid silo]

i kept that in mind

therefore

[pine sail]

So let's recall, we have the area of 2 sectors and 2 rectangles.

Now we have the right most curved area remaining.

[timid silo]

it looks like a rectangle

[pine sail]

It's not.

AB is not straight.

It's curved.

[timid silo]

can you tell that how@do@you figure that out

[pine sail]

Yes.

Can you find the length of AB?

[timid silo]

sometimes the shapes are like very complex so@how can you get it

[pine sail]

AB is an arc.

[timid silo]

i have found it

[pine sail]

Always break the figures down into simpler fundamental figures.

[timid silo]

in part a

[pine sail]

Good what would it be?

[timid silo]

right ill practice that for later shapes

it is 39.26990817 cm

[pine sail]

,calc 25/180pi6

[timid silo]

oops

[warm shale]

Result:

2.6179938779915

[pine sail]

Wait what

Am I doing

Wait I did right

Yeah, what did you do?

[timid silo]

for arc lengths you’ll use 25/360 * 2 pie (6) square

[pine sail]

6 square?

No, it's just 6.

[timid silo]

oh no

i confused this a bit

2 pie are

[pine sail]

Yeah.

[timid silo]

because you are finding out the arc length

[pine sail]

Yes.

[timid silo]

it would be this one

[pine sail]

Yes.

Now you have arc length

[timid silo]

four areas and one arc length

[pine sail]

The area of that curved shape would be arc length* height. . .

So that's your fifth area.

For total surface area, add them up.

[timid silo]

can i ask you one question that the bigger rectangle that you highlighted will that be at the back too?

[pine sail]

Yes, that's why 2 rectangles.

[timid silo]

okay also the teacher said you can take out shapes if you want to got for easy working

the confusing thing here was that of the fifth shapes that was the arc length one

[pine sail]

2 sectors.
2 Rectangles.
And one curved surface area of the cylindrical segment.

[timid silo]

that’s only one

[pine sail]

That's the area of a pizza slice.

[pine sail]

Yes, now you know.

[timid silo]

maybe the shape wouldn’t make sense if you said that curbed surface would be at the left side too

it’s only one hence

[pine sail]

If you can think of a shape, it always makes sense.

[timid silo]

how do you get better at math

in case if you forget things so early

[pine sail]

Someone tell that to me.

[timid silo]

haha

thanks for your help

[pine sail]

You're welcome.

[timid silo]

here i close the channel

.close the channel

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[wraith bane]

Why does 1 satisfy the conditions of a prime? If a prime is a number only divisivle by 1 and itself, isn't 1 divisible by both 1 and itself despite the fact that itself is indeed 1? Is the primacy of 1 mathematically refutable?

[obtuse pebble]

Please don't occupy multiple help channels.

[wraith bane]

Apologies. I had a new question lol.

[nocturne minnow]

You still don't open multiple channels so you should close the other channel you have

[wraith bane]

If I have a second question unrelated to the first, then why "don't" I?

[nocturne minnow]

Why don't you open multiple channels? Is that what you are asking?

[wraith bane]

First one is closed. Any further issues or are am I free to go?

[nocturne minnow]

You're good now, it was mainly because there's limited amount of channels and many people here could have multiple questions and if people opened multiple channels each,.. one people wouldn't know which one to help you in because you're in multiple channels and two because the server would run out of channels

[wraith bane]

Okay.

[opal trench]

1 is not a prime, which is really just by definition/convention

take the fundamental theorem of arithmetic, which says every natural number greater than 1 has a unique factorization of powers of primes. it's not as fun if 1 is prime, bc then it's not unique anymore. eg. 6 can be written as 2 x 3, or 2 x 3 x 1, or 2 x 3 x 1 x 1 x 1... not unique anymore. so maybe it's easier to just say 1 isn't prime rather than exclude 1 as a factor in that theorem

actually iirc the greeks didnt think 1 was a number at all lmao

[wraith bane]

So, they say that every number can be made from primes in one and only one way. But what about 3? Without one, three cannot be made from two primes but itself.

[wraith bane]

I kinda feel that lmao

[opal trench]

3

[wraith bane]

1 + 1 + 1 = 3
2 + 1 = 3
1 + 2 = 3
but that is all

[opal trench]

we're talking about factors here, not sums

[wraith bane]

But every prime composes itself. Naturally.

[opal trench]

yes!

[wraith bane]

No, you were talking about factors and I was talking about sums. Lol.

[opal trench]

this is what it means to be prime.

prime numbers are all about factors

[wraith bane]

What is a factor, again? Numbers that can be multipled to make another number?

[opal trench]

yep, those would be factors of a number

[wraith bane]

I am trying to understand the mindset of those individuals who consider primes to be sacred.

And still failing, unfortunately.

[opal trench]

I mean they are just numbers

but it's neat that there are numbers with that prime property, and there are infinitely many of them

lots of math involves the distribution of these primes among all the others, and you can do pretty neat stuff with all that

(disclaimer i am not a number theorist)

[wraith bane]

Funny, I just began reading an article about this same thing.

[opal trench]

the riemann zeta function is well known example of something related to primes, obv a lot more advanced than what we were touching on, but it's famous (maybe in part bc of the small cash prize associated w it)

[obtuse pebble]

@wraith bane Has your question been resolved?

[wraith bane]

Just read about that as well.

[obtuse pebble]

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[jovial plaza]

How do i extract the square root of a quadratic equation

[timid silo]

please elaborate with you question. What do you mean by "extract"

do you mean completing the square?

https://www.mathsisfun.com/algebra/completing-square.html

[obtuse pebble]

@jovial plaza Has your question been resolved?

[obtuse pebble]

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[pallid cloak]

if theres a root number in stuff like cost, should i find the root or just leave it as it is?

[pallid cloak]

like if they ask "find the cost of laying grass in 105 root 50 m^2 at the rate of some $ per m", the cost would have a root

[kindred oasis]

No, keep it as 105√50, don't approximate

[pallid cloak]

alright tysm

.close

[obtuse pebble]

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[timid silo]

I need help with the second one

I know how to calculate the equation of planes with a line and point but idk how with a line and vector. Are vectors just points?

Can I treat them the same

[viral blade]

I think so

There's not much else they could mean

Oh wait hang on

[timid silo]

I was thinking I could subtract the vector by the initial points of the direction vector but idk what that would give.

[viral blade]

They probably mean it as a point

There's a chance they mean "if the plane contains x, it contains x+the vector"

[timid silo]

In my other homework it said point

Which is why I’m confused that it says vector here

[viral blade]

We can wait for another opinion

[timid silo]

Alright

[viral blade]

I'm voting point but not too confident

[obtuse pebble]

@timid silo Has your question been resolved?

[plain stag]

,rotate ccw

[merry peak]

nice uh

username

@timid silo

[timid silo]

Thanks

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

i need help andrea

[timid silo]

from this question i got -9 and -2

when i write the equation by adding those 2 replacing -11x

does it matter which one i put first

-9x or -2x

[restive ridge]

no

[timid silo]

(6x)^2 -9x^2-2x^2+3
or
(6x)^2 -2x^2-9x^2+3

ok so i shouldnt care which one i put first

just do whatever

[restive ridge]

addition is commutative

[timid silo]

wym

wont i factor from here

[restive ridge]

a+b=b+a

[timid silo]

what do i do next

oh i got it

.close

[obtuse pebble]

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[coral junco]

a friend sent me this proof he was doing, i thought it was correct at first but after i re read it, it seems to have an error, is this step valid? we are supposing true (n k) not (n+1 k) so in theory we shouldn't be able to replace it, right?

[obtuse pebble]

@coral junco Has your question been resolved?

[coral junco]

ok i think it is wrong, i'm gonna redo it and send it to him

[obtuse pebble]

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[pallid cloak]

If AB, BC, AD, and AE are parallel to a line l, then show that A, B, C, D and E are collinear points.

[obtuse pebble]

Please don't occupy multiple help channels.

[pallid cloak]

.close

[obtuse pebble]

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[brisk arrow]

Any angles that I missed?

Can I find any other angle in those triangles

[obtuse pebble]

@brisk arrow Has your question been resolved?

[brisk arrow]

<@&286206848099549185>

[obtuse pebble]

@brisk arrow Has your question been resolved?

[obtuse pebble]

@brisk arrow Has your question been resolved?

[viral blade]

No

All remaining angles can be stated in terms of x

And we can find multiple values of x that all work

If you squash the top of the triangle down the point on the right just gets closer, that's all

[brisk arrow]

got it

thank you so much, for checked those out for me

.close

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[mossy ibex]

What is the difference between a field, a ring, and a group?

[desert sinew]

You may want to check other lecture notes on web or wikipedia. They must already make very clean explainations on these jargons 🙂

[civic zealot]

a group is a set closed under an operation like addition or multiplication (with a bunch of other properties)
a ring is a group under addition, that's also closed under multiplication
a field is a ring that's commutative under multiplication with multiplicative inverses and an identity

[mossy ibex]

Thank you both

Do the operations of fields and rings have to be addition and multiplication

Or are the operations just called addition and multiplication

https://www-users.cse.umn.edu/~brubaker/docs/152/152groups.pdf

Based on this page it sounds like there can be other operations

[kind hawk]

they are just called addition and multiplication

because they "behave" like addition and multiplication behave for integers

[obtuse pebble]

@mossy ibex Has your question been resolved?

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[bronze moth]

hey every1, this year i got unlucky and got the math teacher who isnt pationated about math and doesnt know how to explain things properly so im already confused and lost my way in the forest, can someone explain everything to me in detail pls. thxxxxxx

[sweet crystal]

..

[quiet wadi]

You sound like you're looking for a dedicated tutor. That's not what this is for

This is for answering individual questions

[tawny fog]

What's the question

[bronze moth]

yeh im downloading the things into the channel, but its in dutch so ill need to translate

[stable rain]

how to leave the forest

[quiet wadi]

Downhill until you find water. Then follow it upstream.

[stable rain]

LMAO SENDING ME TO THE MOUNTAINS!!

[quiet wadi]

You got out of the forest

[bronze moth]

these are the exercises that i dont understand

[bronze moth]

😉

there we go

i dont understand a single damn thing

i know its very much

[sweet crystal]

Translate that first q

[bronze moth]

oke oke

1s

[bronze moth]

without using a calculator give the exact value of...

[sweet crystal]

ok

120 is in which quadrant

[bronze moth]

2nd

[sweet crystal]

is cos -ve or +ve in 2nd quadrant

[bronze moth]

-ve +ve ?

whats that

[sweet crystal]

negative

positive

[bronze moth]

ahhhh

negative sorry

[sweet crystal]

yes

does cos change to sin when an odd multiple of 90 is added to the angle?

[bronze moth]

no

it doesnt

[sweet crystal]

it does

[bronze moth]

hmmm

im sorry

[sweet crystal]

so cos120 = cos(90+30) which is -sin30. do you follow?

[bronze moth]

yes yes

im here

[sweet crystal]

and what is sin30

[bronze moth]

-cos 120

ohhh nooo

im sorry

[sweet crystal]

well yeah but im asking for the value, thats what the q is asking you to do

[bronze moth]

sin 30

= 1/2

right?

[sweet crystal]

yes

[bronze moth]

yehh

[sweet crystal]

so your answer would be -1/2

[bronze moth]

why -?

[sweet crystal]

.

[bronze moth]

ah yes yes

[sweet crystal]

you yourself said negative

[bronze moth]

yeh

i get it

nice nice

[sweet crystal]

ok i hope its clear, now try the other parts

[bronze moth]

i got the 1st 1

[bronze moth]

where did you get these formula

cus i recognize them from somewhere previous year

but dont really have a paper or something with them

[sweet crystal]

when its 120

the x coordinate will be negative

that is, cos will be negative

since its the x coordinate

[bronze moth]

so cos120=-sin 30

[sweet crystal]

yes

[bronze moth]

and is that a formula or

how are those equal to each other

man im just asking questions because i dont understand it. sorry

[tidal hawk]

issokk!

yesh those are formulas

there’s more

if u want to know why there are equal u can watch a vid on YouTube! it explains pretty well

https://youtu.be/FCQ9xOJxmgU

[obtuse pebble]

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[mellow igloo]

Hi everyone, would like to know if it is possible to use the method of differences to find the general formula for the summation of an arithmetic sequence.

[mellow igloo]

Would appreciate any assistance with this, as I have tried and failed to generate the formula so far.

Thank you

[obtuse pebble]

@mellow igloo Has your question been resolved?

[obtuse pebble]

@mellow igloo Has your question been resolved?

[obtuse pebble]

@mellow igloo Has your question been resolved?

[worn swan]

idk about that, but there must be some yt videos about deriving that formula

[viral blade]

You can try

But it's definitely way harder than other methods

[obtuse pebble]

@mellow igloo Has your question been resolved?

[obtuse pebble]

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[heavy lake]

How am I supposed to find a counter example for 9.?.

[heavy lake]

Im confused on what this paper is asking me

[modern solstice]

replace x and y

[heavy lake]

For 9?

[modern solstice]

yeah

[heavy lake]

How do I know when to replace x and y

If there is an even power exponent doesn’t that automatically make the function not invertible

I have x = y^5(2y+3)(3y-5)

[modern solstice]

How?

[heavy lake]

By replacing x and y

[modern solstice]

oh damn you factored already

[heavy lake]

I think?

[modern solstice]

multiple y-values for 1 x-value => not inversible (not a function)

[heavy lake]

Oh so then how do you find a counter example

[modern solstice]

lets say x=0

[heavy lake]

Ok

[modern solstice]

0= y^5(2y+3)(3y-5)

now, y can be 0, or -3/2 or 5/3

see?

[heavy lake]

Ohhh

Can I ask

One more question

[modern solstice]

yh

[heavy lake]

So for the rest of these questions should I replace x and y

Or if I see just one exponent that is even

I should rather just go and find a counter example

[modern solstice]

Always change x and y

then it'll be easier to find a counter example

[heavy lake]

Okay thanks for your help

[modern solstice]

gl and good work

[heavy lake]

👍

Test

.close

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[quartz lantern]

Can you change the signs of an equation without flipping it

[timid silo]

without flipping it?

[quartz lantern]

Like moving all the variables to the other side

[timid silo]

you can just multiply -1

[quartz lantern]

oh

Ok thank you

.close

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[novel crystal]

Can somebody help me with this problem?

[novel crystal]

Number 5 is the one i"m having trouble with

[south inlet]

take a pic

[novel crystal]

I sent the pdf

its the last problem

[south inlet]

im not tryna install that

[novel crystal]

oh you cant view it

[south inlet]

no

u can?

[novel crystal]

I'll take a pic and send it

[south inlet]

ok so we need to find the equation of the company a and b

u know how to do that?

[novel crystal]

I'm looking at the book right now

equation of the line ?

[south inlet]

yeah

[novel crystal]

yeah, this is an inequality though, does that matter?

[south inlet]

no

[novel crystal]

im just stumped

[south inlet]

basically you have to find the equation of the two lines

lets say equation of company A is A, equation of company B is B

[novel crystal]

ok

[south inlet]

we just do A < B

and then solve for x

[novel crystal]

sweet, i'll try it

I'm confused how to find the slope of the line without any coordinates

[south inlet]

i think just by the graph

choose points that are on intersections of the x and y axis

because they are easiest to tell what the coordinates are

[novel crystal]

(50, 32.5)?

[south inlet]

thats a point for both

but there is also the points on x = 0

those are easy

[obtuse pebble]

@novel crystal Has your question been resolved?

[obtuse pebble]

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[vestal cape]

can someone explain this

[vestal cape]

why does |x-2| get put there

formal definition of a limit

[zenith raft]

post the definition and then we'll see where |x-2| comes from

[vestal cape]

what do you mean

[zenith raft]

like a screenshot of the definition of a limit

[vestal cape]

well this is the whole answer

gimme a sec

[zenith raft]

do you see where |x-2| could come from?

[vestal cape]

well I understand that you factor x^2 - 4 into

x -4 and x + 4

2*

not 4 sorry

but I don't get how it ended up being multiplied to the ϵ

[zenith raft]

I can't tell what's confusing you

[vestal cape]

so we have |x^2 - 4| < ϵ

and then we turn that into

|x-2 * x +2| < ϵ

I get that

but then

a 5 comes out of nowhere?

and the |x -2| pops up too

this part

I thought it was supposed to be in terms of ϵ

but it just decided to dip

[zenith raft]

how you think to do this might be tricky but in order to accept this part of the answer you just need to agree that if |x-2| < 1 then |x + 2| < 5

I know it looks like it comes out of nowhere

[vestal cape]

but it's not |x + 2| < 5 is it

it's 5 times |x + 2|

[zenith raft]

if |x + 2| < 5 then |x + 2||x-2| < 5|x-2|

(for x not equal to 2)

[vestal cape]

ok

and what does that prove

cause it just ends there

[zenith raft]

that's just scratch work

[vestal cape]

hm?

[zenith raft]

the proof outline continues below

and presents more of the argument

the point of that is just to show that with an appropriate delta, if |x - 2| < 1, then |x^2 - 4| < 5|x - 2|

5|x - 2| is easier to work with

[trail musk]

It might help to look at it like this

[vestal cape]

sorry but why 5

[zenith raft]

because... it works

[vestal cape]

so if I am doing a question by myself how am I supposed to find that

[zenith raft]

welp

it's tricky but you can try to follow the example problems as templates

[vestal cape]

😭 🔫

ig I will have to keep practicing until I get it

ok actually I found another method

oough kinda hard to read sry

where does the 5 come from in this one

just the sqrt 25?

also this doesn't have absolute values

<@&286206848099549185>

[viral blade]

Well it was originally 75

Or wait hang on I'm misinterpreting your question

They subtracted 5 from all sides

Because they wanted the center to be x-5, not x

[vestal cape]

why's that

just to get x - c?

couldn't the value of c just be 0

[viral blade]

the value of c has to be 5

because we want the limit as x approaches 5

[vestal cape]

oo I see I see

[viral blade]

we want to find δ such that |x-c| < δ etc

[vestal cape]

so that 5 has nothing to do with sqrt 25

just a coincidence

[viral blade]

Indeed

[vestal cape]

🙏

I see

[viral blade]

Well

Maybe not entirely coincidence

[vestal cape]

🤨

[viral blade]

Think about it

We're proving basically that the limit as x->5 of x² is 25

It's not just a coincidence here that 25 = 5²

[vestal cape]

oh

yea ig that makes sense

[viral blade]

But yeah it's not too important

[vestal cape]

and then after that?

[viral blade]

We found a range to put x-c in

based on ε

[vestal cape]

yes

[viral blade]

so that F(x) is always within ε of L

the problem is the range isn't of the form |x-c|<δ

[vestal cape]

because it's in the terms of ε?

[viral blade]

Sorta

Your final δ is actually gonna end up being in terms of ε if that makes sense

But try converting the range in the third to last box into the form |x-5|<δ

You'll end up running into a certain problem

[vestal cape]

?

[viral blade]

Yeah that one

[vestal cape]

ok gimme a sec to think

I don't get it

|x - 5| < δ

δ = E/5?

idk what to do w the stuff on the side

the sqrt stuff

[viral blade]

could you do it if I said -3 < x-5 < 3 instead

just as an example problem

[vestal cape]

uhhh no

prob not

[viral blade]

|x-5| < 3

[vestal cape]

oh

and then you 3δ?

[viral blade]

well δ=3 here

That's what I mean put it in that form

see

|x-5| < δ

δ=3

[vestal cape]

ok

[vestal cape]

so then doesn't the rest of it play out like

sorry if I am a little slow rn it is kinda late

[viral blade]

don't worry

[vestal cape]

ok so what is the significance of δ = 3 in the example

[viral blade]

Just a random value I chose

[vestal cape]

yes but what does that change

[viral blade]

To show how to go from inequalities on both sides to one side

So in general the process is

-δ < x-5 < δ

which is the same as |x-5|<δ

Forget the 3 here, this is true for any δ

[vestal cape]

huh?

so then does it make any difference

if there is stuff on the sides or not?

[viral blade]

What do you mean by that?

[vestal cape]

ohw ait

if δ = 3 or whatever

doesn't need to be put in terms of E?

[viral blade]

well

δ=3 is in terms of ε

again that was just an example and not the value you should actually use

[vestal cape]

ok so are we going back to the original q then

[viral blade]

yeah

[vestal cape]

so δ = -5 sqrt 25 + e/3?

[viral blade]

Well, that's the unfortunate part

If that's δ

The thing on the left isn't -δ

They're not negatives of each other

So that's good thinking, but we run into a problem

[vestal cape]

does it have to be +5 instead of -5?

[viral blade]

It's like if I asked -3 < |x-5| < 4

[viral blade]

The bounds are correct as is

We haven't made any mistake in getting to them

They're sadly just not useful as is

[vestal cape]

yes but for me to do the δ = -5 sqrt 25 + e/3

the -5 has to be +5 on one of the sides

so that they are true opposites right

is that why it doesn't work that way in this question

I am not questioning -5ing everything

[viral blade]

Ah ok

Yeah in order for them to be negatives you would need the left side to be like 5 - sqrt (25 + ε/3)

[vestal cape]

ok

[viral blade]

in the meantime

Something important to note

We have a range, that includes 0, but is not centered about 0

On that range, we know |f(x)-L| < ε

That's all the work that's been done so far

Now we need to find a range that is centered about 0, for which |f(x)-L|<ε

Just by visualizing the problem, do you think that's possible?

[vestal cape]

how would I visualize this in my head

like I understand the relations of δ and E

[viral blade]

Don't worry about the δ,ε part for now

We have a range that contains 0 but isn't centered about it

Every point in that range satisfies an important property (the ε thing)

Can we find a range that's centered about 0 (any range!) such that every point in it also satisfies the important property

[vestal cape]

wdym centered about 0

[viral blade]

the range is -a < x < a for some a

[vestal cape]

actually let me help you understand my situation a little bit

I'm not sure what is going w prof but we are using calc early transcendals textbook

we went from section 1.5 to 2.9

[viral blade]

ah hmm

[vestal cape]

like in one day to the next

[viral blade]

what class is this?

[vestal cape]

so I have 0 clue

[vestal cape]

calc for math and stats 1

[viral blade]

Is it 300 level?

[vestal cape]

this be the cover

[vestal cape]

worth full course marks yea

[viral blade]

never heard of it

[viral blade]

ah as in like a 3rd year of college course

is what I mean

[vestal cape]

no

see my uni we do each course is worth x amount

like most courses are worth 3.0

I thought that's what you meant

nah it's university year 1

[viral blade]

ah

[vestal cape]

it's the first calc course I can take

[viral blade]

😨

[vestal cape]

so went from grade 12 to this

[vestal cape]

uh oh

[viral blade]

Calc 1 is a normal first year course

Rigorously proving limits of quadratics was more of a 3rd year thing though

[vestal cape]

🤷‍♂️

[viral blade]

Definitely a difficult topic

[vestal cape]

2 days ago we were talking about inverse functions

[viral blade]

Have you taken courses on logic?

[vestal cape]

and function notations? or smth like that

the day after that

we just went to a completely different unit

[vestal cape]

no this is my first semester

it is like my 3rd day

I haven't done any courses

[viral blade]

ah all right

[viral blade]

that's brutal ooof

[vestal cape]

I couldn't really understand his speaking or writing so now I am self studying

😂

[viral blade]

ah

3 days and already getting the full college experience

[vestal cape]

👍

not like this is a class I can drop either

it is literally the core of my major

there is calc for math and stats 2 next sem

so if I mess up foundations I gonna have a tough time

[viral blade]

I would tentatively suggest that unless you see otherwise there's a chance you don't need to rigorously prove limits

The way to solve (finite) limits for most functions is to just plug in the value

[vestal cape]

like lim x ->2 plug in 2.0001 kinda thing?

[viral blade]

only when you get 0/0 or something as a result do you need to dive deeper

[viral blade]

Just plug in 2 lol

[vestal cape]

oh what

[viral blade]

It'd work for this problem

It's not a proof

But you'll usually notice when the method doesn't work because you'll get 0/0 and then you'll have to work harder

Like lim x→0 of x/x

[vestal cape]

I think we do have to prove limits like that though sadly

like as shown in the textbook

all this stuff

[viral blade]

ah that's brutal

[vestal cape]

yea

maybe I'll try again tmr lol

[viral blade]

Fair enough

It's a really hard concept so don't worry if it takes a while

[vestal cape]

but yea this is more or less the kinda qs he assigned

[vestal cape]

what time zone are you in

is it est by chance

[viral blade]

central

it's 10:17 pm here

[vestal cape]

ok an hour behind

nice

do you mind if I ping you again tmr

won't be at like 5am or anythin

[viral blade]

Not at all go ahead

Although you might get in trouble you can tell them you have my permission

I'll be on around 10:40 est probably lol

9:40 central

am

[vestal cape]

alright 👍

thanks bro you a legend

[viral blade]

no problem, good luck with your class

[vestal cape]

thanks

.close

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[rough epoch]

Am I allowed to do this with limits?

[zenith raft]

no

[wild swallow]

this is not a limits problem

you cant do that with sin in general

[zenith raft]

that's just coincidence

that it's true

[rough epoch]

Ah so this is a coincidence

.close

[obtuse pebble]

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[obtuse pebble]

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[brave meadow]

.close

[obtuse pebble]

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[restive sapphire]

if cot(a) = 4/5, find sec a

[restive sapphire]

so cot a = 4/5

id go hyp = sqroot(4^2 + 5^2) = c^2

do i j solve from there

[desert sinew]

Yes. You may want to draw a right-angled triangle to figure things out.

[restive sapphire]

ok

[desert sinew]

Notice that you may have to consider all 4 quadrant 🙂

[restive sapphire]

wdym

[desert sinew]

Say if A is in quadrant 1 and 3, cot is positive, but cos can be positive or negative there.

[restive sapphire]

oh ok

so i got sq root 41

So what do i do after i got √41

<@&286206848099549185>

[obtuse pebble]

@restive sapphire Has your question been resolved?

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[obtuse pebble]

Please don't occupy multiple help channels.

[hushed zenith]

So I know 4 is a factor

and then I did quadratic to get complex roots

But it's just +/- 5?

[royal basin]

,w solve x^3 - 4x^2 + 25x - 100 = 0

[warm shale]

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Requested by Ann#0413

[hushed zenith]

Oo

[royal basin]

did you get x^2 - 25 or x^2 + 25?

[hushed zenith]

I'm a tad rusty so maybe I just needed to include the i

I used quadratic formula

So I used synthetic to get (x-4)

Which leaves you with x^{2}+0x+25

Using quadratic, you get +-sqrt{100}/2

ohhh wait

It's sqrt{-100}

My bad

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[obtuse pebble]

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[lean tinsel]

How would you solve \sqrt(3)tanx=1 for 0<=x<=2\pi ?

[lean tinsel]

Tan x is equal to 1/root 3

but i'm not sure how the rest of the 0<=x<=2\pi comes into it

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[royal basin]

have you made any progress?

this problem yields quite easily to applying the definition of linear independence and the definition of kernel

[lone blade]

so i know if v is in the kernel and we let v be a linear combination, that combination is 0

[royal basin]

nyeh? what?

[timid silo]

btw for this cant u just factor 25 from 25x-100 so 25(x-4)

and x^3-4x^2 factor x^2 so x^2(x-4)+25(x-4)=(x^2+25)(x-4) then zero product rule

[royal basin]

@lone blade would you like me to give you the start of a proof for you to continue

[obtuse pebble]

@lone blade Has your question been resolved?

[lone blade]

Ya

[royal basin]

We wish to show {v, w} is linearly independent.
To this end, let a, b ∈ R be scalars such that av + bw = 0.
...

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@lone blade Has your question been resolved?

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[sacred root]

Hello

[sacred root]

How do I do number 6

Wo I opened up in the formula : (a+b)(a^2-ab+b^2)

Then

What should I do

[obtuse pebble]

@sacred root Has your question been resolved?

[sacred root]

No

<@&286206848099549185>

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[tidal hawk]

ohh u don’t need to use that formula

[tidal hawk]

can use triple angle formula instead

[sacred root]

Ye I figured it out

Now

[tidal hawk]

ohh okie good job (:

[sacred root]

Thx

[stable rain]

another instance of

helping someone inadvertently starting a new channel

@tidal hawk

[tidal hawk]

ooops HAHAH sryyy

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[inland pine]

so i just started studying matrices and this question came up
can someone please tell me how to do this question

[stable rain]

use induction i guess

[royal basin]

there is a good geometric visualization for this

[inland pine]

ooh what is it

[royal basin]

A is the matrix for a clockwise rotation by alpha

[wild swallow]

still have to do algebraic proof :/

which sucks

[stable rain]

yea normally geometric is just to visualise

[inland pine]

oh cool
but like how to prove that

[wild swallow]

what circle said

try using induction + angle sum formulae

[inland pine]

i haven't studied induction yet so can someone teach me

[stable rain]

@wild swallow uve a template right

[wild swallow]

ugh

im on a phone

[stable rain]

LOL

[wild swallow]

#help-7｜zen1thxyz message

here

but obv replace the thing want to prove

[inland pine]

i didn't understand anything when i read that 🥲

[royal basin]

well first observe that your formula works for n=1

which should be obvious

then from the formula for A^n deduce that it works for A^(n+1) too

i.e. verify that $$\bmqty{\cos(n\alpha) & \sin(n\alpha) \ -\sin(n\alpha) & \cos(n\alpha)} \bmqty{\cos(\alpha) & \sin(\alpha) \ -\sin(\alpha) & \cos(\alpha)} = \bmqty{\cos((n+1)\alpha) & \sin((n+1)\alpha) \ -\sin((n+1)\alpha) & \cos((n+1)\alpha)}$$

[wild swallow]

that line break

[warm shale]

Ann

[wild swallow]

better

[royal basin]

i can fix my own tex if needed

[wild swallow]

yes indeed

[inland pine]

ok i put n=2 and got the RHS