help-26

[twin linden]

yes

Substituting OA = a and AC = (4/5)(b - a)

[fossil moss]

I did AC= a+(4/5)(b-a)

[twin linden]

yea

OC = a + (4/5)(b - a)

OC = (1/5)a + (4/5)b

[fossil moss]

Thats what I got for AC, not OC lol

But yh

[twin linden]

OC is (1/5)a + (4/5)b.

[fossil moss]

Yeah

So my answers rn

Are

AB= b-a
AC= (4/5)(b-a)
CB= (4/5)b+a
OC= (1/5)a +(4/5)b

[twin linden]

Remember, vectors represent both magnitude (length) and direction.
So, a and b not only tell you the distance from the origin O to points A and B, but also the direction.  
Sources and related content

[twin linden]

yes

[fossil moss]

Yhh

[fossil moss]

yeah ik lol im just being braindead

[twin linden]

anymore questioN?

[fossil moss]

I don't think so, ty

[twin linden]

!done

[topaz sinew]

If you are done with this channel, please mark your problem as solved by typing .close

[fossil moss]

.close

[topaz sinew]

Channel closed

Closed by @fossil moss

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[sweet vigil]

Discrete math btw,

Why did we put S----> t

at the sstart

and i dont understand the steps below

[craggy haven]

um, well, s -> t is the second premise

[sweet vigil]

Ik but why didnt we put -r ----> s at the start

why did we put the second premise at the start

[craggy haven]

because that's what we were using first

[topaz sinew]

@sweet vigil Has your question been resolved?

Channel closed

Closed by @sweet vigil

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[toxic stirrup]

how would i do this?

[sweet shard]

You can factor A^2 + I

[toxic stirrup]

How?

I thought you can’t factor a sum of squarws

U there?

[sick escarp]

use imaginary numbers

[toxic stirrup]

.close

[topaz sinew]

Channel closed

Closed by @toxic stirrup

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[stoic badger]

Stuck with all of this

Uni economics assignment

[hollow drum]

,rotate

[topaz sinew]

@stoic badger Has your question been resolved?

[topaz sinew]

@stoic badger Has your question been resolved?

[topaz sinew]

@stoic badger Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[glossy ruin]

Surely there's an easier way

[glossy ruin]

Idk if this is even right?

[neon iron]

why did u do F'(x)/F(x) ?

[vernal matrix]

Seems you changed the function midway

[vernal matrix]

Derivative of ln(f(x)) wrt x

[neon iron]

what ?

i mean ok but why ?

[vernal matrix]

Other than this, seems alright (and easily fixed, change the 4s to 1s within those logs)

Presumably they want the derivative of f?

[glossy ruin]

my thought process was: ln(fx)' = 1/f(x) * f'(x)

[neon iron]

ok

[glossy ruin]

cuz I didnt know how to do it any other way

which is why I'm asking if there's an easier way or if I'm even correct cuz this is quite the process

[neon iron]

u could use the dervivative of a composition of functions

(gof)' = f' * g'of

are u french ?

[glossy ruin]

I see, but how would I go about it here

I speak both

up to u if you wanna use french

[neon iron]

no it was to know if u were in terminale

[glossy ruin]

Oh nah nah

I'm undergraduate

1st year

[neon iron]

ok

[neon iron]

and actually i thiink it would be the same with it

so idk another way

[glossy ruin]

yeah, probably. welp tis ok

.close

[topaz sinew]

Channel closed

Closed by @glossy ruin

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[wet steeple]

If (1^2)cm = 1cm, then why when we convert the centimeters to millimeters 1cm = 10mm, when we do the same operation the result is 100mm which is 10cm?
Is it that 1^2 = 10?

[wet steeple]

Like, how does this work? Because as far as i know 1^2 = 1*1 = 1

[brazen cobalt]

I don't understand

What is the question?

[worthy storm]

probably what you're looking for is:
(1 cm)^2 = (10 mm)^2 = 100 mm^2

[wet steeple]

why is 1*1cm = 1 but 10*10mm = 100 if 10mm = 1cm

[rigid ivy]

$1^2$ cm and $1$ cm$^2$ are two different things. You probably meant the latter

[thorny flame]

SWR

[brazen cobalt]

Units

[wet steeple]

Yeah, the first one

not cm^2

but n^2

[wet steeple]

Indeed

[brazen cobalt]

Ok give me a second

Okay

1cm =10mm

You are wondering why 1cm²=100(mm)²

Correct?

[wet steeple]

Not quite, i am wondering why
1²cm = 100²mm

i am going linear

[brazen cobalt]

100² as in 100×100?

[wet steeple]

wait i just got entangled

1²cm = 10²mm

[brazen cobalt]

It's not true

[wet steeple]

Yea i just wrote it wrong, sorry

[brazen cobalt]

1²cm≠10²mm

Where do you get that from

[wet steeple]

due to 1cm = 10mm

[brazen cobalt]

Yeah

Ok so what operation are you doing to get 1²cm=10²mm

I'll assume you are squaring the equation 1cm=10mm

[loud oasis]

if you square both sides of the equation
1 cm = 10 mm
you get
(1 cm)² = (10 mm)²
1² cm² = 10² mm²
the units square too

[brazen cobalt]

Exactly

[wet steeple]

Ooohhh

[brazen cobalt]

Treat units like variables

Similarly

1L=?

Litre

[wet steeple]

1000ml

[brazen cobalt]

In volume

[wet steeple]

1000ml^3

[brazen cobalt]

No 😭

Litre is a unit of volume

[wet steeple]

Oh wait

1000^3 ml^3

[brazen cobalt]

Milli litre is just thousanth of a litre

Noooo

1L =1000mL

Milli stands for thousandth

Ok

Try this

How many millimetres in a metre

[wet steeple]

1000mm

[brazen cobalt]

Yes

There are many units of volume

One is Litre

1L=1000cm³

Remember volume is some sort of side×side×side

So cm³

Ok

Now can you tell me how many Litres does 1m³ of water have

[wet steeple]

1L?

[brazen cobalt]

Try again

Ok you know the relationship between 1metre and 1cm

And there is a relationship between cm and litre

So let's find it?

1m=100cm

[wet steeple]

yes

[brazen cobalt]

Ok now

Cube both sides

[wet steeple]

1m^3 = 100cm^3

[brazen cobalt]

100³

So that's 10⁶

Correct?

Or are you stuck

[wet steeple]

I am a bit stuck on the 1m^3 part

would it be 1m^3 = 100^3m^3?

[brazen cobalt]

1m=100cm

[brazen cobalt]

Yess

[wet steeple]

Ooohh

I understand

[brazen cobalt]

And 100³=10⁶

[wet steeple]

What about 1?

[brazen cobalt]

Let it be

Lol

We have to eliminate cm

[wet steeple]

Oh

[brazen cobalt]

Don't bother about metre side

Ok so what do we have now

[wet steeple]

1m^3 = 100^3cm^3

[brazen cobalt]

Or 1m³=1000×1000cm³

We can write that?

[wet steeple]

Wait wait wait

yes

[torpid sparrow]

[brazen cobalt]

So now what is 1000cm³

1L

[wet steeple]

Or 1m^3

[brazen cobalt]

so 1m³=1000L from 1m³= 1000 × (1000cm³)

[wet steeple]

Oh

Well i think i understand now

So due to 1000cm^3 = 1L, 1000cm^3 = 1m^3, thus 1000L = 1m

[brazen cobalt]

1000000cm³=1m³

That's 6 zeroes

Or 100³cm³

[wet steeple]

So lemme get this straight , 1L is the 1000th of 1m^3

[brazen cobalt]

Yes

[wet steeple]

Ooohh

anyways i gtg now

[brazen cobalt]

👍

[wet steeple]

seeya!

.close

[topaz sinew]

Channel closed

Closed by @wet steeple

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[supple merlin]

Solve the equation cos 3𝑥 =1/2 Answer exactly in radians.
what do I start with?

[candid salmon]

cos^-1 on both sides

[brazen cobalt]

cos(3x)=cos(pi/3)

Now when you remove cosine

3x=2n(pi)±pi/3

[supple merlin]

Should there be 2 solutions for it?

[brazen cobalt]

No

Infinitely many

Cos(60°)=cos(420°)

Please check the question

[brazen cobalt]

Where n is any integer

[supple merlin]

what about the rule cosθ= 1/2 ⇒θ= π/3 or θ= 5π/3
​

[brazen cobalt]

???

Cos(theta) never = 2

[supple merlin]

typo

mybad

[brazen cobalt]

Yeah exactly

You see you are getting 2 results

But theta =-5pi/3 is also a correct solution

So is theta =-pi/3

Please check the question, they would have given a particular interval

[supple merlin]

Just solve it and answer in Rads

[brazen cobalt]

Then just go for pi/3

Rads

You sure they have not provided anything else?

[supple merlin]

yeah 100%

[brazen cobalt]

Textbook?

Could you send a picture

[supple merlin]

It's Swedish though

[brazen cobalt]

Ok so 3x=pi/3

X=pi/9

But tell them that there are infinitely many solutions

[supple merlin]

what about this?

[brazen cobalt]

That's the complete solution

Accounting for all possible values of x

[supple merlin]

do you solve 3x = ∏/3 + 2n∏

x = ∏/9 + 2n∏/3

[brazen cobalt]

±

Yeah

[supple merlin]

do you need ± if there are infinite solutions?

[brazen cobalt]

Yes

Not really

Haha

If n is integer, not really

Because n can be negative also, yes?

So + is also enough

[supple merlin]

could be anything, everything on the table

[brazen cobalt]

A + or - should satisfy to get all possible values so yes

Anything works

[supple merlin]

gotcha

.close

[topaz sinew]

Channel closed

Closed by @supple merlin

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[torn geyser]

5/12 is used as a/b for division point

[torn geyser]

i cannot find x and i have test tmrw

im going crazy how do i start

i found out the point xp,yp as 540, 180

but idk how to aplly the forula

formula

how do i use pythagoras?

no?

[mint crescent]

Assuming the two lines are perpendicular, you can find the equation of this

[mint crescent]

And then the intercept follows directly

If you’re not given that they’re perpendicular, then there’s not enough information to solve

[topaz sinew]

@torn geyser Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[scenic flare]

We were told to show f:R²->R with f(v) = v1+v2² in the two euclidean metric spaces (R², d1) and (R, d2) is continuous, but I can't prove it (via epsilon-delta)

[scenic flare]

It's trivially continuous by different arguments, but I can't determine a delta for eps-delta

[craggy haven]

helps to remember that the distances you're talking about are well under 1

so squaring makes things get smaller

[scenic flare]

hm but how can I relate the two distances
|f(v)-f(w)| = |v1+v2²-w1-w2²|
|v-w| = √((v1-w1)²+(v2-w2)²)

v2 I think otherwise yes, but then it becomes
< delta + delta * (delta + 2|v2|)

But don't get to a point where it's independent of w by bounding |v-w| < delta

[cursive patrol]

|v2^2 - w2^2| = |v2 - w2| |v2 + w2|

[scenic flare]

Yeah, I've been there at the start too, then I used delta ≤ 1

which meant |v1-w1| ≤ 1

and |v2-w2| ≤ 1

but then I still can't get of w in |v2+w2|

[cursive patrol]

|v2 - w2| < delta, so |v2 + w2| leq |v2 - w2| + 2|w2| < delta + 2|w2|

[scenic flare]

huh my messages disappeared

nice

in short: then I'm at < delta + delta * (delta + 2|v2|)

here I wasn't sure if one can find a delta

[cursive patrol]

oh i am just blind, lol

[scenic flare]

my main issue is that I either stumble into not eliminating w or not being able to determine a delta, so I should rather focus on getting better at finding a formula for delta :]

[cursive patrol]

should this be v2 or w2?

[scenic flare]

Does this approach work generally, trying:
delta + delta * (delta + 2|v2|) < eps
and solving for delta

[scenic flare]

v2, v is given

Or no I should limit delta again right

e.g. delta < 2|v2|

[cursive patrol]

so v is your arbitrary point not w

[scenic flare]

ys

then
... < delta + delta * 4|v2| = delta * (1+4|v2|)

delta < eps / (1+|v2|)

yay

why didn't I limit delta again before

[scenic flare]

thx for guiding 😄
high lin alg no issue but when encountering basic ana problems I feel like I'm stuck in the 1st semester haha

[cursive patrol]

i prob would have done delta^2 leq delta there, similar to what Hayley was getting at

I.e. < delta (2 + 2|v2|)

[scenic flare]

I'll remember repeating limiting

[cursive patrol]

since the issue here is if v2 is 0

and even if not 0 then i don’t know if that works

oh well

[cursive patrol]

🎉

[scenic flare]

ah ryt yeah that case can be specifically excluded with delta < 1 if v2=0

.close

[topaz sinew]

Channel closed

Closed by @scenic flare

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[old plover]

How do I find the area of the blue circle here?

<@&286206848099549185>

[brazen remnant]

draw in the segments connecting the centers of the circles

[old plover]

I got 9pi for the area, but I’m not sure if that’s right @brazen remnant

[unreal tinsel]

show steps

[brazen remnant]

i don't think that's correct

[old plover]

How come?

[brazen remnant]

try to use this:

hint: ||use the fact that its a square||

[unreal tinsel]

the top of the smaller circles are not the same height as the bottom of the bigger circle, thats why your step of doing 8 - 2 to find the diameter of the bigger circle is incorrect

[old plover]

Yeah, I realize that now

[old plover]

How would I approach it? Like this? 2^2 + (2+r)^2 = 4?

[brazen remnant]

what does the length of the red segment need to be?

[old plover]

8

[old plover]

So my equation would work, wouldn’t it?

[brazen remnant]

no?

[old plover]

Why not?

Half of 8’s 4

[brazen remnant]

where did you get your equation from

[brazen remnant]

now what is the length of the green segment

[old plover]

4

That’s where I got my equation from

To solve for r

[brazen remnant]

no, it's not 4

if the whole thing is 8 and one part is r, what is the other part?

[old plover]

Idk

[brazen remnant]

if i have a stick of length 8 inches and i cut off r inches, how long is the remaining piece?

[old plover]

8-r

[brazen remnant]

yes

do you see how this corresponds to finding the length of the green segment?

[old plover]

Yes

[brazen remnant]

so you now have a right triangle with legs of length 2 and 8-r and a hypotenus of length r+2

[old plover]

Yeah, I see it now

Thx

[topaz sinew]

@old plover Has your question been resolved?

Channel closed

Closed by @old plover

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[merry bridge]

Hey, I'm not sure where to go from here

[earnest herald]

@merry bridge

[merry bridge]

sup

[earnest herald]

after you reach that point where you put m as dy/dx

you can put the point 12,3 in place of x and y

to find x0 and y0

[merry bridge]

for the actual slope?

[earnest herald]

you will need another equation

which will be by satisfying the point in the actual curve

[earnest herald]

yess dy/dx as you found out -x/4y

[merry bridge]

so -12/4(3)?

[earnest herald]

you have 2 equations and 2 unknowns you should be able to solve remember you'll get 2 answers

[merry bridge]

but that's -1 which seemingly misses

the tangent line isn't at (12, 3), it crosses (12, 3)

[earnest herald]

that is why you put 12,3 in place of x and y

not x0 y0

[merry bridge]

oh

[earnest herald]

you can also put x0 y0 in curve to get another equation

[merry bridge]

like so?

[earnest herald]

then try solving simon-tenously (idk how to spell)

[merry bridge]

I can't put it in the -x/4y because that's the slope, and the slope can't be -1

[earnest herald]

there is an easier method that requires you to use the direct formula... you study it in the ellipse chapter in coordinate geometry

the slope is also -x0/4y0

its not -x/4y

[topaz sinew]

@merry bridge Has your question been resolved?

[merry bridge]

I'm not sure where to go from here:

Or more accurately, here

[earnest herald]

you are doing it wrong sir

you have to put 12,3 in place of x y (or x1 y1 in your equation)

not x0 y0

x0 and y0 is a point on the curve for which dy/dx is equal to -x0/4y0

[merry bridge]

that doesn't make sense? why would it be in the place of x and y? Normally when we find the equations of tangent points, we put the points where the tangent line must pass through on x_0 and y_0, then we set the equation of the point-slope form equal to the original equation

[earnest herald]

yes you are correct when the point is on the curve

in this question the point lies outside of the curve

[merry bridge]

no, even when the point isnt on the curve.

[earnest herald]

so your method is wrong

no no

that is not how it works

i suggest you try revising your notebook or textbook you refer to

or try solving a solved problem for same

[merry bridge]

This is how both the textbook and my instructor have taught it.

[earnest herald]

i am sorry but that's not how i was taught

i did the question with my method and i got the same answer

your instructor taught you correctly but its different

[topaz sinew]

@merry bridge Has your question been resolved?

[merry bridge]

yeah I figured it out myself

.close

[topaz sinew]

Channel closed

Closed by @merry bridge

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[lunar oriole]

hi so if i have to make an equation in the form of y=-(x+a)(x+b) and i have the coordinates 5,0 and 3,0 to sub in how would i do that

[lunar oriole]

and why would you do it

[brazen cobalt]

Providing the original question can give more context

[smoky sparrow]

oh wait I misread

[smoky sparrow]

well by definition, (5, 0) and (3, 0) are roots of the quadratic

[lunar oriole]

okay

[smoky sparrow]

and the roots of y = -(x - a)(x - b) are a and b

[lunar oriole]

so where do you sub it in

i'm so confused

[brazen cobalt]

In the same equation

[lunar oriole]

okay

so then what do you do after

[smoky sparrow]

basically you need the zero product property actually

[brazen cobalt]

You should get a linear equation in 2 variables a and b

[smoky sparrow]

don't think about subbing in directly cause it won't work

or it will be longwinded

[lunar oriole]

hold on i'm so confused so

[smoky sparrow]

so if y = 0, -(x + a)(x + b) = 0

zero product propertly implies x + a = 0 or x + b = 0

[lunar oriole]

oh yeah

got it

yeah

that's fine

but like

for

[smoky sparrow]

so now you can sub in x, say 5 + a = 0 and 3 + b = 0

[lunar oriole]

a and b

OH

wouldn't it be -3 and -5

[smoky sparrow]

yep exactly

[lunar oriole]

bc it's negative

yeah

thought so

okay got it

[smoky sparrow]

so it's y = -(x - 5)(x - 3) bingo

[lunar oriole]

oh got it

i think idk i'm so confused on why we have to solve for x when we already have it

ngl

[smoky sparrow]

yeah like you want to solve for a and b ofc

[lunar oriole]

OH

so we found b

and a

[smoky sparrow]

yeah like there is a small thing we glossed over

say we can sub in x = 5 into x + b = 0 instead

that's not an issue

and x = 3 into x + a = 0 instead

so you get -(x - 3)(x - 5) then

but that's the same, cause the order of multiplication doesn't matter

[lunar oriole]

so then x is five and x is 3

but tf

i'm so confused

but wait nvm

we found a and b

okay so it's not as complicate as it looks

[smoky sparrow]

more like we can choose which x value to use

and both of them give the y value as 0

[lunar oriole]

so if we get two points with x coordinates we just sub it in

[smoky sparrow]

yes

sub the 2 different x values into the 2 linear equations

[lunar oriole]

what if u get a y coordinate

[smoky sparrow]

oh yeah if you have two different points in general

say 5 = -(2 + a)(2 + b)
and 10 = -(3 + a)(3 + b)

now we're talking, that's fairly spicy actually

(if you have two points with the same x-coordinate but different y-coordinates, that's not a function by definition)

[topaz sinew]

@lunar oriole Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[smoky sparrow]

@sonic ibex

[smoky sparrow]

next time ask in an open help channel

[sonic ibex]

was it not open?

[smoky sparrow]

you can open one yourself

yeah it was closed

[sonic ibex]

i though if you typed on a close it open

does it not?

[smoky sparrow]

it doesn't

[sonic ibex]

it always worked for me liek that

btw is ther ea simpler way for the equations/

[smoky sparrow]

yes I was about to show you the way

can you post the original system of equations pls

[sonic ibex]

k

x-3y=5 and 2x-2y=y^2-5

[smoky sparrow]

okay so you want to find (y - 2)^2 = y^2 - 4y + 4

y^2 = 2x - 2y + 5

so y^2 - 4y = 2x - 6y + 5

= 2(x - 3y) + 5

so there you go, y^2 = 2 * 5 + 5 + 4

[sonic ibex]

huh

[smoky sparrow]

which part did you not get?

[sonic ibex]

most of it

from so

[sonic ibex]

where did you get the = part

[smoky sparrow]

just expand (y - 2)(y - 2)

[sonic ibex]

k

how you solve for y tho

[smoky sparrow]

so you just rearrange for y^2 like I did

and subtract -4y

the trick is you don't need to find what y is!!

[sonic ibex]

still dont get it

[smoky sparrow]

ok now which part don't you get

[sonic ibex]

why you dont need to solve for y and expading it

[smoky sparrow]

cause there's a trick in the question

such that the numbers work out

that solving for y is not the most efficient way

[sonic ibex]

.close

[topaz sinew]

@smoky sparrow Has your question been resolved?

[whole geode]

.close

[topaz sinew]

Channel closed

Closed by @whole geode

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[neon iron]

Hello

[neon iron]

I need help

Should I just write Inf(4A-2B)= Inf(4A)-Inf(2B) , release the two and switch inf to sup?

[strong hollow]

Sry wrong channel

[haughty adder]

First, maybe think about why -B is a bounded below set, when B is a bounded above set

[neon iron]

🤔

Should I just write Inf(4A-2B)= Inf(4A)-Inf(2B) , release the two and switch inf to sup? (My assumption doesn’t work) ?

[haughty adder]

Forget inf and sup, and recall the definition of being bounded below / above

[neon iron]

Okay this may sound dumb, but for every x, x<s (s is the upper bound). Multiplying by negative will flip the inequality -x>-s boom lower bound

[haughty adder]

It's better to say that you subtract x and s from both sides of the equation rather than multiplying by (-1) to get -x > -s imo, but yeah, that's it essentially

[neon iron]

Ah

[haughty adder]

Now, try to show that 4 A + 2 (-B) is bounded below

You already know that A and (-B) are bounded below

[neon iron]

I mean the thing is, it’s common sense but I am struggling to translate it to math

[haughty adder]

You already wrote what it means to be bounded below

All elements a of A are bounded below by some constant c, so a >= c. Same thing holds for elements -b of -B with another constant d, so -b >= d.

What can you say about elements of 4 A + 2 (-B) then, which can be written as 4 a - 2 b, where a is in A and b is in B?

[neon iron]

4 a - 2 b >= 4c +2d , where 4c +2d equals to some constant k?

[haughty adder]

Yeah, that's it

[neon iron]

Massive W

Thank you 🙌🏻

.close

[topaz sinew]

Channel closed

Closed by @mystic wasp

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[rocky prawn]

The question asks to find a double intégral of 3x^2 with the limits of the D region. The D region is defined by the inequilaties from the first picture, graphically is just an intersection marked in red from the second picture. I tried to make limits from the given inequalities, personally I found describing the shape in limits rather difficult so I skipped any circle coordinates. Unfortunately, I don't have any corrections, so I would like to check what are you thinking about this question.

[rocky prawn]

I do have a suspicion that the big fraud moment might be coming from the first integral's limits.

feel free to ping me :3

Here's the clarification what needs to be calculated (if the word description is bizzare)

[topaz sinew]

@rocky prawn Has your question been resolved?

[honest canyon]

huh i got 0

did i miss something

[rocky prawn]

what did you do?

[topaz sinew]

@rocky prawn Has your question been resolved?

[topaz sinew]

@rocky prawn Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[velvet spear]

Can someone help me understanding this introduction rule in Natural Dedction?

[velvet spear]

Does it mean, assume that A is true, from there we can derive B

I.e, B is a logical consequence of A

So, anytime I have formula B, I can say that B is true because A is true?

[topaz sinew]

@velvet spear Has your question been resolved?

[topaz sinew]

@velvet spear Has your question been resolved?

[velvet spear]

.close

[topaz sinew]

Channel closed

Closed by @velvet spear

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[finite mango]

I'm somehow missing a 4 in the denominator

[topaz sinew]

@finite mango Has your question been resolved?

[sweet shard]

Where did you get your perimeter equation from. 24 = 2y+...

[topaz sinew]

@finite mango Has your question been resolved?

[finite mango]

.close

[topaz sinew]

Channel closed

Closed by @finite mango

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[pliant horizon]

Hey im trying to simplify this into q^k to calculate the limit but im kinda stuck rn. Anyone got an idea how i can simplify this furter or if 've made a mistake already

[whole geode]

It looks OK to me, but you might be able to make some progress if you let 4^k = 2^(2k) and complete the square.

[loud oasis]

rather than combining them into one sum, it would probably be simpler to calculate the two sums separately

[whole geode]

but yeah, separating the sums is probably better.

[pliant horizon]

ooh it probably is lmao ty

.close

[topaz sinew]

Channel closed

Closed by @pliant horizon

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[topaz sinew]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[vast maple]

i know how to find a tangent and normal line to the curve written in canonical form but not in general and i have no idea how to go to canonical in this example because of xy

[glad terrace]

Treat y like a function

So for example (2y)'=2y'

And just calculate y'

[vast maple]

but i cant isolate y here

i got y^2, xy and y

i tried implicit differentiation but it always gave wrong answer

[glad terrace]

y^2

(f(x)^2)'=2f(x)*f(x)'

So (y^2)' will be 2y*y'

[vast maple]

i get y' = (2x - y + 1)/(x + 2y - 1) in the result

[glad terrace]

Maybe

Idk

[vast maple]

i have tried this way but when i plug in 0.6 and 1.2 i get wrong coeffitients

[whole geode]

So if you did implicit differentiation, then you have the slope of the curve at any point on the curve

[glad terrace]

Maybe you could show?

[whole geode]

If you then use the coordinate (x, y) on the curve you have the point slope form of the line

And finally to find the normal, you just use the fact that the normal is perpendicular to the tangent

[vast maple]

wait you are right

sorry i had some misunderstanding

sorry

thank you

is this appliable to any curve?

[whole geode]

Yes

[glad terrace]

Yes

[vast maple]

in class we studied specific formulas for each curve and didnt use implicit differentiation

[whole geode]

Provided it's not misbehaving in some way

[glad terrace]

Um

Like what?

[whole geode]

Like infinite slopes or kinks in the curve or so on

[glad terrace]

I mean you have to derive those formulas somehow

[vast maple]

this for parabola

and other for others

.close

[topaz sinew]

Channel closed

Closed by @vast maple

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[shrewd shore]

What am I missing here for (b):

[shrewd shore]

The answer is 50pi

radius is 10 (each block is 5)

(pi(10)^2 ) / 2

[half carbon]

its negative lol

[shrewd shore]

HECK

Thank you, damn it.

.close

[topaz sinew]

Channel closed

Closed by @shrewd shore

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[echo dew]

hi, quick question

[echo dew]

can i sketch 1/x given a function f(x)

i know i can graph 1/f(x) but is 1/x something that can be done ?

[agile harness]

!xy

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[echo dew]

.close

[topaz sinew]

Channel closed

Closed by @echo dew

Use .reopen if this was a mistake.

[craggy haven]

oh ok

[cedar wagon]

Job done

xd

[half carbon]

success!

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[strong sable]

If A is a positive semidefinite matrix, then is the null space of X^T A X the same as AX?

[topaz sinew]

@strong sable Has your question been resolved?

[topaz sinew]

@strong sable Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[vocal dew]

why am i getting -189 as my determinant and not 189?

[craggy haven]

,w Det[{{-1, 6, -2}, {6, -2, -1}, {-2, 3, 5}}]

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by kitten.in.a.teacup

[craggy haven]

[vocal dew]

but if i were to manually find the determinant?

[craggy haven]

didn't you do that?

with the minors

[vocal dew]

oh wait lol mb yea i did and im not sure why the textbook says that its 189

[craggy haven]

misprint

[vocal dew]

probably, tysm tho!

.close

[topaz sinew]

Channel closed

Closed by @vocal dew

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[wintry siren]

-sigh-

even after simplifying the root id still be wrong

idk where i went wrong or if i even did this right ;-;

[wintry siren]

answer!!

[sterile finch]

,rotate

[wintry siren]

here is it possible to rotate now?

[sterile finch]

,rotate

[sterile finch]

Give me a moment.

[wintry siren]

nwrs! im greatful for your help ❤️

[sterile finch]

You used the diameter instead of the radius.

[wintry siren]

oh my god.

OKOK WILL TRY AGAIN!!

wait would large cyl be (x + 1)^2(pi)(x)?

[sterile finch]

Yes.

[wintry siren]

and small be x^2(pi)(x)?

[sterile finch]

Yes.

[wintry siren]

okok!!

[sterile finch]

Doh. I'm wrong.

The diameter of the outer radius is 2x + 2 + 2.

[wintry siren]

idk where im going wrong

[wintry siren]

oh?

OH

NOW I GET IT

okok i got it now!!

tysm kookiemon!!

❤️

[sterile finch]

yw

[wintry siren]

.close

[topaz sinew]

Channel closed

Closed by @wintry siren

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[paper ether]

hey can't find exemple of a surjective group homomorphism $f : G \to H$ such that $f(Z(G)) \subsetneq Z(H)$. Can someone help me pls ?

[thorny flame]

mid

[topaz sinew]

@paper ether Has your question been resolved?

[opal vault]

sorry I misunderstood your problem

lemme rethink

[paper ether]

i think that i should take H abelian group so that H = Z(H) but have no idea how to find G and the morphism

[opal vault]

take the signature homomorphism on S3

you know it's surjective onto {-1,1}

and yet identity is the only element of Z(S3)

@paper ether does that help

[paper ether]

yep i'll try this

thank you it seems to work

.close

[topaz sinew]

Channel closed

Closed by @paper ether

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[glad root]

for the 2nd problem i don't know how to involve theta in the equation

when the system isn't at an angle, the equation for mass I got was m=(kx^2)/v^2

[tranquil jetty]

just add the gravitational potential energy of the mass into the original equation you derived, so:
1/2 (mv^2)=1/2 (kx^2)+mgh

[glad root]

where is theta going to be used?

[tranquil jetty]

I don't think it is needed

[glad root]

its part of the answer though

[tranquil jetty]

I cant figure how it is necessary. It is only necessary to calculate the acceleration, and forces, so f=mgsintheta

[glad root]

try to avoid force equations

i don't think theres any force equation in the answer

[tranquil jetty]

yeah i dont think so to

[glad root]

i think the angle somehow affects the potential energy but idk how

[tranquil jetty]

If we are to calculate the potential energy, it will simplify to mgh? E=fd= mgsintheta*h(1/sintheta)=mgh? Is it going to affect the potential energy in the spring?

[glad root]

i have the answer to the solution but idk how to get there do you want me to send it

[tranquil jetty]

ok

[glad root]

they didn't even simplify it and set it =m but it doesn't really matter

im not sure why they add the horizontal distance

[tranquil jetty]

OH when you extended it it will go further back than h

that additional height is calculated with x*sintheta

[glad root]

isn't xsintheta the distance it moves along the track

?

oh

[tranquil jetty]

No it is the height that is gained by the block when you pulled it back by x

[glad root]

so they're using the x like the "hypotenuse" right and sin of the angle is the additional height right

[tranquil jetty]

yeah basically using geometry to calculate the distance

*height

[glad root]

i thought they were doing like the force of gravity and doing sin Fg which would've been the distance traveled along the track

ok that makes a lot more sense now

[tranquil jetty]

yeah its simpler than that

[glad root]

Just had the problem of fully understanding the question lol

alr tysm

[tranquil jetty]

no problem

[glad root]

.close

[topaz sinew]

Channel closed

Closed by @glad root

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[dawn ridge]

How do I do this??

[dawn ridge]

My teacher isn’t giving any pre-lesson instructions. This if from Algebra 2

If someone can give me an example of how to complete this, I can do the rest on my own

[rigid ivy]

Are you familiar with the equation of a parabola in vertex form?

[dawn ridge]

Nope

Every time I ask my teacher for help, he says the same thing but with different words and doesn’t explain anything

[rigid ivy]

poopoo shared it just above

[dawn ridge]

lovely username

[rigid ivy]

The basic graph of a parabola is described by the equation y=x²

[dawn ridge]

okay so it’ll be y = a (x - 3)^2 + 2

[rigid ivy]

for which problem?

[dawn ridge]

1. p(3,2)

[dawn ridge]

is this the “equation of the parabola”

And how do I tell if the graph opens up or down

because they’re positives, it’d open up right?

[unique quarry]

yes

[dawn ridge]

yes to thats the equation of the parabola, yes to it opens up, or yes to all of the above?

[unique quarry]

both

[dawn ridge]

okay awesome

[unique quarry]

if x^2 is positive it opens up

[dawn ridge]

so i just apply the same knowledge to the rest of the questions?

[unique quarry]

and vice versa

yes

[dawn ridge]

awesome thank you. as for 9-17, how do i turn those graphs into an equation?

[unique quarry]

same thing

find the vertex

see if it opens up or down

[dawn ridge]

uhh

yee okay

tyty

[rigid ivy]

This is incorrect

[dawn ridge]

whart

how do i find the correct answer then

[rigid ivy]

it would help to first learn about the vertex form of a parabola. I can give you a crash course, but reading your math book will help you so much more

[dawn ridge]

we dont have math books really

[rigid ivy]

wild

[dawn ridge]

yeah my teacher is ass. not much i can do about it

[rigid ivy]

idk how you're expected to do this then without the knowledge or an ability to learn it

[dawn ridge]

he kinda expects us to already know

he just gives us our paper, tells us how one problem is solved, and expects us to do the rest, but i havent been able to be in class so ive been doing it online

i havent had the example

[rigid ivy]

ouch

okay well, then yeah you need to learn about parabolas and vertex form

the basic form of a parabola is simply y=x². By basic form, I mean that the shape of its graph will be the overall characteristic of what a parabola looks like, but it does not describe every parabola. You can think of it like y=x describing the basic form of a line, but y=x is not the only line. You can translate lines, scale them, what have you

[dawn ridge]

Aight

So what formula can I use to solve these stupid questions

[rigid ivy]

First, it's important to describe some properties of parabolas

The vertex of a parabola is defined as the turning point. Visually, it's the lowest point in a parabola that increases upward, or its the highest point in a parabola decreasing downward

,w plot y=x^2

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by spelwerdzrite

[dawn ridge]

Yeah I know the vertex

[rigid ivy]

for y=x², the lowest point of the parabola is at (0, 0), so the vertex of the parabola y=x² is simply (0, 0)

[dawn ridge]

Yee

[rigid ivy]

are you familiar with translations of graphs?

like, what h and k do?

[dawn ridge]

Partially

[rigid ivy]

So let y=f(x) be some arbitrary function. It doesn't matter what f(x) is, just draw in your head whatever graph you like

[dawn ridge]

Okay

[rigid ivy]

imagine now you wanted to graph y=f(x)+1

[dawn ridge]

okay

[rigid ivy]

idk if makes visual sense to you that basically the whole graph would move up now by 1 unit

example:

,w plot y=x^2+1

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by spelwerdzrite

[dawn ridge]

Yeah I gotchu

[rigid ivy]

Note how the graph translated upward compared to the original graph here

there are vertical translations, because we move the graph vertically.

Next, we care about horizontal translations

These are a bit harder to describe conceptually

[dawn ridge]

So we change the X?

[rigid ivy]

yes, but do you know how?

[dawn ridge]

Nope- Im a bit dense

[rigid ivy]

no worries, it's hard to visualize at first

So let y=f(x) be the graph of some function.

Let's say that we want y=g(x) be some graph the looks exactly like the graph of y=f(x), but moved 1 to the right

[dawn ridge]

it’d be y=g(1) right?

[rigid ivy]

So if (a, b) is some point in the graph of f, then we're gonna (a+1, b) to be a point in the graph of g. Does that make sense at all?

[dawn ridge]

kinda yea

im sorry you have to educate me lmao

math isnt my strong suit if it isnt obvious

[rigid ivy]

no worries

[rigid ivy]

But essentially, this is what we want to do

Something that gets overlooked often, but is very important to know: for the function y=f(x), the graph of f is basically a collection of points in the plane. Arbitrarily, every point can be described as just (x, y). But the function is given as y=f(x). So every point on the graph of is just (x, f(x)).

To reiterate, the graph of a function f is just every point (x, f(x))

(for which x is a valid input for f)

[dawn ridge]

okay i think i understand

[rigid ivy]

awesome

So, the graph of f is every point (x, f(x)), and like I shared earlier, the graph of g (which is f translated to the right by 1 unit) needs to be (x+1, f(x))

okay and here's where we get a little creative. We're going to make a new variable. Let's just call it z. The name doesn't matter. We're going to define z simply as z=x+1

If we solve for x, then x=z-1

we replace x in (x+1, f(x)) with z-1 and get (z, f(z-1))

So, the graph of g is just every point (z, f(z-1)) where z-1 is any valid input for g.

[dawn ridge]

what would that look like?

[rigid ivy]

soon

[rigid ivy]

but remember that z was just an arbitrary variable name. It could still be x. That is, the graph of g is just every point (x, f(x-1)). By the definition of a point, we have y=f(x-1). But, but the definition of "graph of g", we also have y=g(x). Therefore, we get g(x)=f(x-1) (since they were both equal to y, they're equal to each other)

[dawn ridge]

uh

yes

[rigid ivy]

In summary, to move a function 1 unit to the right, it becomes f(x-1). In general, if we wanted to move some arbitrary k to the right, the translated function is f(x-k)

[rigid ivy]

where did I lose you?

[dawn ridge]

okay that makes sense

[rigid ivy]

example:

,w plot y=(x-1)²

[thorny flame]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by spelwerdzrite

[rigid ivy]

this is just the graph of y=x² moved 1 to the right

[dawn ridge]

okay

that makes sense

[rigid ivy]

This is known as a horizontal translation

Okay, now this is where it gets important: we can mix horizontal and vertical translations.

Let's say we wanted to move a parabola 2 to the right, what would the equation be?

[dawn ridge]

y = (x - 2)^2?

[rigid ivy]

yes

that is exactly right

good work

[dawn ridge]

phew thank you ;-;

[rigid ivy]

Now, let's say wanted to move this 3 units down. What would you do?

[dawn ridge]

y = x^2 - 3?

[rigid ivy]

close. I meant how would you translated the function that you already translated 2 to the right? (Or, how would you move y=x² 2 to the right and 3 down)?

[dawn ridge]

OH

y = (x -2)^2 - 3?

[rigid ivy]

yes. Good work

[dawn ridge]

Okay yeah this makes sense

[rigid ivy]

Now here's the important bit. We're gonna taklk about vertices

Remember that the vertex of y=x² was (0, 0)

[dawn ridge]

yeah

[rigid ivy]

(0, 0) was the lowest point on the graph. But we moved the graph 2 to the right and 3 down, so the vertex will also move 2 to the right and 3 down. So what must the coordinate of the vertex of the translated parabola be?

[dawn ridge]

(2, 3)?

[rigid ivy]

exactly right

[dawn ridge]

Or well, because we moved it down, it’d be -3?

[rigid ivy]

oh

haha

you are right

[dawn ridge]

did we mean to move it up?

[rigid ivy]

nope

I asked you to move it down

[dawn ridge]

So 1. would be y = (x - 2)^2 +3? and the graph would open up?

[rigid ivy]

the vertex would be (2, -3), not (2, 3) I was mistaken earlier

[dawn ridge]

okay

[rigid ivy]

yes

[dawn ridge]

YAY im actually learning 😭

[rigid ivy]

So, having found that the vertex of y=(x-2)²-3 is (2, -3), what do you suppose is the vertex of the general parabola y=(x-h)²+k?

[dawn ridge]

(h, k)?

[rigid ivy]

yes

congrats

[dawn ridge]

Awesome!! Thank you so much

[rigid ivy]

That's exactly why this is called the vertex form of the parabola

the only thing we did not discuss was a, the scale factor

but I assume you know how to use scale factor?

[dawn ridge]

So in that case, problem 9 would be y = (x - 0)^2 + 4?

[dawn ridge]

uhh maybe?

[rigid ivy]

correct

[dawn ridge]

I know what certain things look like, but im completely blank when it comes to the names

“things” being equations or formulas or whatever

What does scale factor look like?

[rigid ivy]

I can explain in like an hour. I need to run to the pharmacy

[dawn ridge]

All good! I actually think I can figure out Scale Factor on my own. Thank you so much!!!

I think I can solve the rest of the problems on my own now. You were a huge help so thank you

[rigid ivy]

[dawn ridge]

im gonna close this so someone else can use this channel

if i need more help ill reopen another channel

thank you!!

.close

[topaz sinew]

Channel closed

Closed by @dawn ridge

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[lusty quest]

how do i get 80?

[celest creek]

120+320 is 440 right

[lusty quest]

yes

[celest creek]

which is 360 + 80

cos 360 + 80 = cos 80

[lusty quest]

thank you!

.close

[topaz sinew]

Channel closed

Closed by @lusty quest

Use .reopen if this was a mistake.

[lusty quest]

.reopen

[topaz sinew]

✅

[lusty quest]

now i got 160

[celest creek]

okay?

[lusty quest]

how do i get 160 when z/w

[celest creek]

wait

do you know Euler's form of complex numbers?

r(cosx + i sinx) = r e^ix

[lusty quest]

ol

ok

[celest creek]

what's z and w btw

z = 5(cos 120 + i sin 120)

and w = 3(cos 320 + i sin 320)

right

[lusty quest]

yes

[sweet shard]

Oh that's a ÷ symbol, not+

[celest creek]

so z is 5(e^i(120))

and w is 3(e^i(320))

so z/w is (5/3)* e^i(120-320)

if that makes sense

[lusty quest]

how do i got 160?

[celest creek]

-200 + 360 is 160

[lusty quest]

oh got it

appreciate it

[celest creek]

we can add and subtract 360 for trig angles remember that

[lusty quest]

thankyou!

.close

[topaz sinew]

Channel closed

Closed by @lusty quest

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• 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.

[silent frigate]

so when i try to prove this limit of the sequnece using defitnion of limit

why doesnt it work when L = 1.1?

like limit iis 1, but when i put in L = 1.1, it seems to work as well

as well as plugging in 0.9

[raven sparrow]

It doesn't work.
When epsilon is smaller than 0.1

1. you get that any N works since your choice of N just requires that N is bigger than some negative number
2. the last step in the inequality is not true.
   0.1 + 5/N > 0.1 for any N,
   so in particular if epsilon is less than 0.1, you can't have
   0.1 + 5/N < epsilon.

[topaz sinew]

@silent frigate Has your question been resolved?

[silent frigate]

oh wait i see hmm.

[silent frigate]

why would we get any N works since N is bigger than some negative number?

like when epsilon is smaller than 0.1?

[raven sparrow]

You say that given any epsilon, we can just let $N > \frac{5}{\epsilon - 0.1}$.

[thorny flame]

Azyrashacorki

[silent frigate]

yes

[raven sparrow]

But if epsilon is less that 0.1 this is negative

[silent frigate]

yes

[raven sparrow]

So then any N works

Because any N is bigger than a negative number

[silent frigate]

yes, but isnt that okay?

because all that is reqwuired is that there exists some N

[raven sparrow]

Well it's a hint that something is wrong with your choice of N.
If any N works and your sequence converges, then that means that the sequence is constant.

[silent frigate]

hmm i see

[silent frigate]

is it because of this then?, like the second parrt

[raven sparrow]

Yes, essentially.

The difference between the terms and 1.1 gets close to 0.1, not 0

[silent frigate]

oh i see. Becuase you found an epsilon that doesnt work for this, since if epsilon < 0.1 then 0.1 + 5/n < 0.1, but at the same time is greater than 0.1

[raven sparrow]

Yes that's the idea

[silent frigate]

i see. But what about for 0.9 then?

because my methdology was this

in order to find N

i did $|1 - 5/n - 0.9|$ < $\epsilon$

[raven sparrow]

It's the same idea. The difference will get close to -0.1

[thorny flame]

LXDL