help-10

[edgy saffron]

350

Doesn’t represent the length of RE

So it wouldn’t be proportionate

?

Hold on sorry

Gimme a sec

Sorry

My

Massive brain

Cannot

Understand this

Yeah

Mhm

Hold on

Idk but

It doesn’t feel right

That it’s a decimal

And repeating

Wait

Since they’re all similar

What if you did

Gcb

Since you have all the values

Alright

Gimme a moment

Cbp

Is similar to

Pbe

Scale factor is

2/3

400 x 2/3

Damn

It’s

The same shit

So

It might just be

Yeah

Unless we both happened to do it wrong and get the same number

But

Like

That’s not

Likely

Ok

Hold on

So

400x2/3

+400

Wait

No

Cause

The value of ge

Has to be

Has to be 2.5xeb

So then

Gb would be

666.6 repeating

Ikr

No

Ok

Wait

I think

Don’t take my word for it

But

I think I see a mistake

When you did

Ge=2.5be

Wait no

Nvm

Ok

.close

[obtuse pebble]

Channel closed

Closed by @edgy saffron

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[quaint finch]

Two ships sail from a port. One sails 14km due south whole the other sails 17km on a bearing of 120 degrees. How far apart are the ships, correct to one decimal place?

[quaint finch]

sending the diagram in two minutes...

[knotty gulch]

have you heard of the cosine rule?

@quaint finch

that’s probably the easiest way to do it

[quaint finch]

I'm rlly bad at trig rn but yes I know the cosine rule

but I think the diagram needs some fixing

so how do I apply the cosine rule in this application

[knotty gulch]

what does the cosine rule need

state the cosine rule for me

[quaint finch]

the cosine rule is c^2= a^2 + b^2 -2abcosC

[knotty gulch]

yes

[quaint finch]

where c is the unknown side

[knotty gulch]

yes

[quaint finch]

so is my diagram correct

[knotty gulch]

you need to work out C

[knotty gulch]

yes

[quaint finch]

but isn't it a right-angled tri

[knotty gulch]

no

cosine and sine rule is any triangle

you can search a proof online if u want

[quaint finch]

oh

but can't I do the sine ratio or cos ratio

[knotty gulch]

?

you can

i guess

do you want to find the x and y displacement, then find the distance between?

[quaint finch]

I want the distance straight away

[knotty gulch]

then the cosine rule

use it

[quaint finch]

alright

[knotty gulch]

you have all the information except C

[quaint finch]

alright I got the ans

[knotty gulch]

nice

[quaint finch]

thx for ur help

much appreciated

.close

[obtuse pebble]

Channel closed

Closed by @quaint finch

Use .reopen if this was a mistake.

[quaint finch]

.reopen

[obtuse pebble]

✅

[quaint finch]

@knotty gulch srry did u want to say something

I closed the channel too fast

[knotty gulch]

nah it’s fine

i was gonna tell you to close the channel

but then you closed the channel

[quaint finch]

oh lol

.close

[obtuse pebble]

Channel closed

Closed by @quaint finch

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[dusk fox]

What domains and ranges do I have to use to eliminate the graph in the bottom

[knotty gulch]

x>0

you know how to do the desmos right

put it in curly braces

[dusk fox]

Yea

Why is it not less than equals to 0

{0<_x} @knotty gulch

[knotty gulch]

try it, you have desmos

[dusk fox]

It looks the same

[nocturne minnow]

less than equals to 0
This is not the correct inequality that represents that
{0<= x}

[dusk fox]

Huh

[nocturne minnow]

{0<= x}
Is not how to represent less than equals to 0

Also, for you to eliminate the bottom part of the function, that means you want to keep the other stuff, the expression or inequality in the {} represents what you want to keep

[dusk fox]

Ok but why is it not {0<= x} instead of x<0

[nocturne minnow]

You understand that 0 <= x is x >= 0, right?

[dusk fox]

Yea

[nocturne minnow]

0 < x and x < 0 aren't the same

[dusk fox]

Yes

[nocturne minnow]

But yes, <= is valid too but not proper

Because y = 10/x, x can't be 0 already

So saying x >= 0, for y = 10/x, you can include 0 but it's already not going to contain 0

[dusk fox]

Ohhh

.close

[obtuse pebble]

Channel closed

Closed by @dusk fox

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[eager siren]

I need help graphing this domain

How do I show x and y cant equal each other on the graph

[forest sinew]

hmm you doubled up, i think youre still missing something

x != y and y != x are the same condition

but there is another condition you should add

[eager siren]

I did it this way because my teacher had it like this in the note, so I did incase. I was wondering same thing

[eager siren]

ok let me see

[reef grotto]

you need to find where the function isn't defined

so pretty much when the denominator is equal to 0

[eager siren]

the function is not defined when it is over 0

[forest sinew]

youre just missing a uhh

[eager siren]

yeah

[forest sinew]

i mean youre solving $x^2=y^2$

[warm shale]

jan Niku

[forest sinew]

so take the sqrt of both sides

$\pm x = \pm y$

[warm shale]

jan Niku

[forest sinew]

i mean or however you want to realize that theres some evenness happening

[reef grotto]

or you could factor it : $(x+y)(x-y)=0$

[warm shale]

Asagao 朝顔

[reef grotto]

but either way it gives you x=-y and x=y

[eager siren]

so when x=-y as well

[reef grotto]

exactly

[eager siren]

that is what i was missing?

oh ok

i get it thanks

[reef grotto]

indeed

[eager siren]

now i just need help graphing

[reef grotto]

what does y=x and y=-x look like

[eager siren]

its just a diagonal line but

arent i supposed to shade some regions

idk

our teacher didnt show how to do this

[reef grotto]

so pretty much it's defined over everything except these diagonal lines

[forest sinew]

there arent any regions

at least not excluded regions

the problem areas are just lines

they make that cross pattern, so if you start in one of the four regions of the cross, you cant go crossing into another

[reef grotto]

i could be wrong but i think you have to do like a dotted cross

[eager siren]

so would this be fine then

[reef grotto]

then shade everything outside

[eager siren]

oh ok i see

[reef grotto]

i think making a solid line instead of dotted means that the lines are included in the shaded region

[eager siren]

Like this?

[eager siren]

I remember something like that vaguely

Or was it not with the hollow dot and dot?

[reef grotto]

no i think that's for the single variable function graphs

[eager siren]

ah ok

i have not taken math course in 2 years lol everything fuzzy

is the graph i sent correct btw?

[reef grotto]

i think it just has to be a dashed line instead

[eager siren]

ok i will do that incase

it is not due today

so i can ask prof to be sure

What about this problem?

Did i do this right

Oh wait no i didnt

I pulled out y^2

Cant do like that

for domain can i just write xy^2!=1

[obtuse pebble]

@eager siren Has your question been resolved?

[eager siren]

<@&286206848099549185>

[eager siren]

https://tenor.com/view/fade-away-oooooooooooo-aga-emoji-crumble-gif-20008708

[drifting sun]

I'm not sure you want to square both sides like that

I think it's enough to just notice that the square root can't be negative in the reals as this would give you a complex number and the denominator can't be 0, since functions go from positive to negative whenever they cross a zero it's the same thing as solving for the zeros of $1-xy^2$ (the part inside the square root) and checking all sides around those zeros to make sure you don't dip into the negatives

[warm shale]

Blue Guilmon

[drifting sun]

also you can't have zero either so don't include the curve where that happens

because it's in the denominator

solutions to where $1-xy^2=0$ can be thought of as $1=xy^2$ so either the function $f(x)=\frac{1}{\sqrt{x}}$ or $f(y)=\frac{1}{y^2}$ graphing that curve would be your boundary condition here

[warm shale]

Blue Guilmon

[drifting sun]

the second one is probably easier to graph, so just turn the whole paper clockwise 90 degrees and pretend it's the function $\frac{1}{x^2}$ that should accomplish it

[warm shale]

Blue Guilmon

[drifting sun]

err sorry wait

ehh tbh just switch your x and y axis

and then pretend it's that

easier to just do it that way

I would just say that because you don't want $1-xy^2$ to be negative, notice that whenever $x<0$ it is quite literally impossible, no matter what $y$ is, to have a negative value here as you're adding something positive since you're subtracting a negative (and because y is squared whether its positive or negative doesn't matter) so everything on the left-half of the y-axis will definitely be in your domain, the rest will have a boundary along the function $y=\frac{1}{\sqrt{x}}$

[warm shale]

Blue Guilmon

[drifting sun]

on one side of that boundary you'll be fine, on the other you won't, check for the spot where you won't be (where it becomes negative) also the boundary wouldn't be included either as this corresponds to when the denominator is 0

@eager siren hope that helps

[obtuse pebble]

@eager siren Has your question been resolved?

[eager siren]

sorry I got caught up with something IRL, is it ok if I @ you tomorrow after I read this?

[obtuse pebble]

@eager siren Has your question been resolved?

[obtuse pebble]

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.

[astral aurora]

Is this possible to find it?

[timid silo]

yeah

split up that shape into regular shaped sections and it'll be easy to find

[astral aurora]

I tried to cut it

But failed

[timid silo]

show me what you tried

[astral aurora]

I cut it 20 m in 21 m

Same as 22 with 23

So area will be 22×20 =440

After this triangle is left

So 1/2×22 ?

[timid silo]

yes

then add them up

[astral aurora]

Se 440+11=451

Is this correct?

[timid silo]

makes sense to me

[astral aurora]

Check it please

Is there any direct formula for quadratrial

[timid silo]

i'm not aware of one

i just split it up into pieces

[timid silo]

yes it is correct

[astral aurora]

You got the sane answer?

[timid silo]

yes

[astral aurora]

I am thinking another idea

What if we add an image line

Both sides then find area and substrate it

[timid silo]

what?

can you show me a picture or something

[astral aurora]

@timid silo

[obtuse pebble]

@astral aurora Has your question been resolved?

[obtuse pebble]

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 sequoia]

could someone help me with this problem please

Draw a triangle with one side of 6.3 cm and adjacent angles of 50 degrees and 60 degrees. Measure the necessary dimensions and find the total surface area of ​​such a vertical prism whose height is 0.8dm and whose base is a drawn triangle

[velvet sequoia]

this is what i have so far

<@&286206848099549185>

[high lily]

the third angle can be determined from interior angle sum of a triangle

[velvet sequoia]

,

[high lily]

and then you can apply sine law to determine one of the other sides

which gives you enough info to determine the area of the triangle (using the relevant areaformula),
and subsequently the volume of the prism

[velvet sequoia]

is this logical

[high lily]

where's 8cm coming from

[velvet sequoia]

0.8dm = 8cm

[high lily]

the height/depth of the prism is 8dm,
this isn't the height of the triangle

[velvet sequoia]

ohhh

[high lily]

all you are given about that triangle is those two angles and the 6.3cm

[velvet sequoia]

ok

do i have to draw the triangle and then turn it to a prism?

@high lily

[high lily]

if you want to draw the prism sure

[velvet sequoia]

ok

this is right?

@high lily

[high lily]

no

[velvet sequoia]

then how do i do it?

[high lily]

the third angle can be determined from interior angle sum of a triangle

[velvet sequoia]

yeah

[high lily]

and then you can apply sine law to determine one of the other sides

[velvet sequoia]

but what is the relevant formula

how do i get it?

[high lily]

trig formula for area of a triangle

easily searchable

after completing the above steps

[velvet sequoia]

i dont understand

what now?

@high lily

[high lily]

how are you getting the 7.7cm and 7.1cm

[velvet sequoia]

some website

[high lily]

forget about the website

do the work yourself following the steps i've outlined

starting with the very first step

[velvet sequoia]

what is the relevant formula i dont understand

[high lily]

that comes later

[velvet sequoia]

english is not my first language

[high lily]

after you've done the first two steps

[velvet sequoia]

so i have to redo everything?

[high lily]

you didn't seem to have done anything i've said

at all

[velvet sequoia]

i am very confused

[high lily]

the third angle can be determined from interior angle sum of a triangle

[velvet sequoia]

yes i have all the angles

50deg 60deg 70deg

[high lily]

i didn't see 70° anywhere in your pic

anyway

[velvet sequoia]

its behind the gray line

[high lily]

next step

and then you can apply sine law/rule to determine one of the other sides

[velvet sequoia]

we havent learned about sine law yet

[high lily]

well then you've been given a question without the pre-req knowledge,
the law isn't that complicated,
easily searchable and isn't that difficult to apply

[velvet sequoia]

i have all the required stuff to calculate the surface area

but im confused on the formula

[high lily]

did you look up the sine law

[velvet sequoia]

yes and understood nothing

[high lily]

what were you looking at

[velvet sequoia]

https://www.mathsisfun.com/algebra/trig-sine-law.html

[high lily]

point to the specific spot you started getting confused

[velvet sequoia]

i have to go now bye

[obtuse pebble]

@velvet sequoia Has your question been resolved?

[obtuse pebble]

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.

[balmy mortar]

Find all equilaterals with each vertex on 3 concentric circles with radii 1, 2, 3. (answer(s) is up to rotation about the centre)

[balmy mortar]

Looking for a hint

Tried coord geom, but got stuck with something very messy, looking for euclidean geom solution (if there is one)

[obtuse pebble]

@balmy mortar Has your question been resolved?

[obtuse pebble]

@balmy mortar Has your question been resolved?

[obtuse pebble]

@balmy mortar Has your question been resolved?

[obtuse pebble]

@balmy mortar Has your question been resolved?

[open oasis]

i remember doing a similar question a long time ago so not clear on the details
but i think complex numbers might help

[wooden cipher]

@balmy mortar wlog pick a point on one of the triangles then rotate it 60 degrees

About that point

This is very similar to a 2012 aime problem

[timid silo]

Try parameterising the vertices, and use the circumcircle of the triangle

[obtuse pebble]

@balmy mortar Has your question been resolved?

[eager siren]

Sorry to post in someone else's channel but I didn't repond to you since I was busy yesterday. Thank you for helping me I got around to it now and understand everything. @drifting sun

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[prisma glacier]

Excuse me I need help with number

Six b

Is that squeeze theorem?

To be able to compute the limit?

The answer says it doesn’t apply

The divergence test but idk why

[desert sinew]

Hint: By taylor series, sin x > x for x in (0,1)

[obtuse pebble]

@prisma glacier Has your question been resolved?

[prisma glacier]

We haven’t learned Taylor series yet

But how would I do this problem without Taylor series

[north quarry]

what do u get when applying div test ?

[prisma glacier]

The limit as that approaches infinity

Is 0

Soo it converges

[north quarry]

how ?

[prisma glacier]

Plug in a large value for n

[north quarry]

no i meant how for this

[prisma glacier]

The divergence test does not apply

[north quarry]

right

div test gives u 0

and what do u conclude from that ?

@prisma glacier

[prisma glacier]

The divergence test

Does not apply

As the sum converges to 0

[north quarry]

the sum converges to 0 ?

[prisma glacier]

Yes

When you take the lim

Of a sub n

Of sin one over 2n

That goes to 0

[north quarry]

from what i see, it only asks you to look at the div test and decide if it applies or not

[prisma glacier]

Divergence test

Does not apply

What more work do I need to show for this problem

I’m good right

[north quarry]

well you need to justify why it doesnt apply

[prisma glacier]

Yea

It doesn’t apply

Because the lim approaches 0

And the divergence test needs it to not converge to 0

[north quarry]

yea

so its all good

[prisma glacier]

My bad my professor gets me anxious

The problems he has are crazy

[north quarry]

np

so 6b is solved ?

@prisma glacier

[prisma glacier]

Yes

Ain’t my writing

But

Hopefully it’s correct way of doing it

[north quarry]

im unable to read this ngl

[prisma glacier]

Dam

My class mate is trash

No wonder why she’s failing

I had just asked her

Hey send me what you have as work soo I can verify with the math community

Cause she’s also confused

She said she took the derivative

[north quarry]

for 6b ?

[prisma glacier]

It’s for not just

6b

[north quarry]

you arent allowed to take the derivative for 6b, it isnt a 0/0 or a inf/inf situation

[prisma glacier]

Other examples for the hw

That’s what I was saying

You can’t do L’Hopital’s rule right

[north quarry]

yes you cant for 6b

you need help for all the other letters of 6 ?

@prisma glacier

[prisma glacier]

Yes

[north quarry]

ok then

[prisma glacier]

My bad I’m talking to multiple people

You a tutor

And then my classmate lol

[north quarry]

np

so 6c

what would u do

[north quarry]

you cant take derivative

[north quarry]

ok

so back to 6c, how would you begin ?

[prisma glacier]

6c?

Hold on

Hear me out

This explanation

For 6c that’s crazy

I think I would split the expression into two

[north quarry]

how

[prisma glacier]

How ?

I would split

It as 1/n

1/sin n

I remembered seeing this problem from a MIT lecture hopefully the knowledge was retained lol

And not from so dumb caltech professor teaching at a community college lol

[north quarry]

i dont understand how u would split 6c honestly

the question asks for div test

[prisma glacier]

Oh shit

Divergence test

If we do the same thing and say the limit as n goes to inf of cos(1/(2n)) we'd get cos(1/(2*inf)) which is cos(0) which is 1 so the test here would apply.

[north quarry]

right

and what can u conclude for the sum ?

[prisma glacier]

The sum converges to 1

[north quarry]

no

[prisma glacier]

Shit

[north quarry]

the limit converges to 1

[prisma glacier]

Yes the limit or whatveee

That converges to 1

So the series must also converge

Or it may not converge

[north quarry]

its important to be precise

[prisma glacier]

It may diverge

Wait a second

[north quarry]

no

[prisma glacier]

Oh fk

Since that limit was 1

[north quarry]

as the limit goes to inf

[prisma glacier]

Divergence test apply

So this series diverges

[north quarry]

it gets closer and closer to 1

[north quarry]

yes

[prisma glacier]

That’s all I have to say?

[north quarry]

because of what ?

visualize it

[prisma glacier]

Visualize what?

The limit?

[north quarry]

remember, its a sum

[north quarry]

youre doing a sum, and the terms go to 1, youre therefore adding 1 and infinite amount of times

[north quarry]

yes basically

[prisma glacier]

Ohhhh

That basically diverges

To positive infinity

[north quarry]

yes

good

[north quarry]

now 6d

[north quarry]

yes

looks good

except technically you cant plug infinity

but the main idea is right

[prisma glacier]

Why can’t you

Oh

Just. A large number

It doesn’t matter fk it

I’ll do that on explain or maybe I’ll just write

N as 1000

It’s a very large number

So 1/2*1000

Is basically 0

[north quarry]

it really does tbh but its fine as long as u understand the idea

[prisma glacier]

Problem d

For number 6

Id prob do what my tutor said or friend

[north quarry]

right so how do u start

[prisma glacier]

Not for c

That was for c

Umm for d

I’m just going to plug in numbers

It doesn’t really matter

If n was 1000

[north quarry]

it matters

[prisma glacier]

Itd be 1000 times sin of 0

So the divergence test does not apply

[north quarry]

are you sure ?

[prisma glacier]

Sin 0 is 0

[north quarry]

ok

[prisma glacier]

Multipled by 1000

[north quarry]

no

[prisma glacier]

That’s still 0

[north quarry]

no

ur multiplying by something going to infinity

[prisma glacier]

Oh fk

Yea that’s still 0

Sin of 0

[north quarry]

and it isnt a 0

[prisma glacier]

Times infinity

[north quarry]

its going to 0

[prisma glacier]

Indeterminant ?

Infinity times 0?

[north quarry]

yes

but that doesnt give 0 immediately

this problem will prove u otherwise

after plugging, u get "inf*0"

right ?

[prisma glacier]

Yes

[north quarry]

yes

[prisma glacier]

Then what

L’Hopital’s rule?

[north quarry]

but infinity times 0 is indeterminate

[north quarry]

is it a 0/0 or inf/inf ?

[prisma glacier]

Idk

Why would that matter

[north quarry]

you just showed its a 0*inf

[prisma glacier]

Yes

[north quarry]

its the only scenarios where this rule is applicable

it matters

[prisma glacier]

Wait so

What’s my next step

I just showed an indeterminant form

[north quarry]

yes

u have the idea to use LHR

[prisma glacier]

LHR?

L’Hopital Rule?

[north quarry]

is there a way for u to get a 0/0 or a inf/inf ?

[north quarry]

yes

[prisma glacier]

Yes

I believe f(x)/1/g(x)

[north quarry]

uh what is f(x) and what is g(x) ?

[prisma glacier]

F(x)

Is the sin

And then g of x is n

This is what my other classmate in this other math discord has

[north quarry]

yes but theres a mistake

when derivating sin (1/2n)

it gives cos(1/2n)*(1/2n)'

=cos(1/2n)(-1/2(n^2))(2)

[prisma glacier]

It’s not negative?

Oh fk

So the 2 comes on top?

[north quarry]

well he forgot the *2 at the end

[prisma glacier]

It’s not right?

[north quarry]

no

(v^-1)'=-1(v^-2)(v')

[north quarry]

no

[north quarry]

yes but *v'

here, v is 2x

2x' is 2

thats why there is a *2 at the end

[prisma glacier]

Im confused

So what should the derivative be

[north quarry]

its just a power rule

but since it isnt a 1*x^n scenario

[north quarry]

-1/n^2

[prisma glacier]

Oh yea

You’re right it does simplify to

[north quarry]

so ur good for 6d ?

[prisma glacier]

Well

I’m still kind of on that

The derivative got me confused

But I get why he said it diverges

[north quarry]

ok

[prisma glacier]

I’m thinking

Just multiply and divide by one over n squared

[north quarry]

skipped e ?

or r u good with e ?

[prisma glacier]

Honestly

If you’re up tomorrow morning

I usually go to the math center on campus

And I try to do all the problems

I’m thinking

E diverges

[north quarry]

i can still help for some time rn if u need it

[prisma glacier]

It’s sort of the same problem

[north quarry]

why ?

[prisma glacier]

Since it goes to 0

Dt does not apply

[north quarry]

yes

u can also look at the highest powers of n

or use LHR

[prisma glacier]

When would I need to use squeeze theorem?

[north quarry]

u will probably not have to honestly

maybe for sin or cos functions

or when looking at the values for which x makes a sum converge maybe

[north quarry]

alright

how would u start

[prisma glacier]

I’m lost at this one

But let me think

Just plugging infinity

Again

You’d prob get indeterminant

But I would multiply and divide by one over n

To rearrange the inside of the ln

So it’d just be ln1/2

[north quarry]

ok

[prisma glacier]

So it doesn’t converge to 0

So it diverges

[north quarry]

ok

amazing

[prisma glacier]

You’re joking

[north quarry]

good job

no

[prisma glacier]

You’re not serious

It can’t be that simple

[north quarry]

well i am

it is

its the 1st test

so we good ?

[prisma glacier]

No

I need one more

[north quarry]

ok

[prisma glacier]

Please and thank you

Number 4 b

[north quarry]

np, glad i can help

ok but what is the question in 4b ?

[prisma glacier]

Follow the proof

Of the additivity property in theorem

[north quarry]

its a proof

[prisma glacier]

2A.2 to show that if c

Is all real numbers

Let me check the proof

[north quarry]

c is in R

[prisma glacier]

C is

IR

Or whatever it is

[north quarry]

yes

[prisma glacier]

All real numbers notation

[north quarry]

a constant

but theres no question for 4b tho ?

[prisma glacier]

The question is

Show that

Sum from n is zero to infinity of a sub n is S

And summation of n equal zero to infinity of b sub n is T

[north quarry]

its given in the 1st sentence

[prisma glacier]

First sentence of what

[north quarry]

i dont get what ur asking

[prisma glacier]

This one?

[north quarry]

its a proof

its not asking you to prove it

they give you the proof

[prisma glacier]

Fk for the last problem

My buddy is saying I did it wrong

ln(xyz/qrs) * (1/x)/(1/x) doesn't just apply to the inside of the log

[north quarry]

well u can look at the interior of the ln

and the highest powers are both 1, and have factors 1/2

so it simplifies to be ln(1/2)

[prisma glacier]

Yea

That’s what I got

Okay

If they’re not asking me to prove it

And they give me the proof

What are they asking for then

[north quarry]

nothing

[north quarry]

its just a given proof

[prisma glacier]

That’s what my professor

Had given us

[north quarry]

i think theres a part missing from 4b

it shows a partial sum

but doesnt ask anything

it probably asks to calculate the partial sum of Un

[prisma glacier]

Partial sum?

But there are no expressions

[north quarry]

well it says that Un is the partial sum of C*An

[prisma glacier]

“He wants you to prove that no matter what constant c that is a real number multiple by a series that you can show that it equals the constant times the summation”

[north quarry]

well yes C can be anything

hold on idk how to conventionally prove it

hmm

[prisma glacier]

My bad

Here is what my other classmate did

I’m trying to understand before I copy

Because basically all she did

Was pull a constant out

[north quarry]

well u could do

sum An = S

multiply both sides by c

and here you have it

[prisma glacier]

This is confusing

[north quarry]

ye it looks fine but ion get why such an obvious proof is required

[prisma glacier]

Yea

I don’t get any of this either

But

It’s calculus II

I don’t Get why there’s so many proofs

[north quarry]

ye i went thru that too but most obvious proofs i didnt have to do em

[prisma glacier]

Wtf

I thought proofs

[north quarry]

well maybe i wasnt paying attention in class

[prisma glacier]

We’re only in geometry

[north quarry]

no theyre everywhere

alright well it looks like its done, ill therefore leave, cya another time

[prisma glacier]

Cya

How do I close this

[north quarry]

do .close

@prisma glacier

[prisma glacier]

.close

[obtuse pebble]

Channel closed

Closed by @prisma glacier

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.

[solemn crater]

okay, so I started doing this question out of the textbook, and decided it might be better to split it into two cases. I got the even case figured out, but got snagged trying to do the odd case, so I decided to go look at the hint in the back of the book, and this is what they say:

but ... I have no idea why they want to split it up further or where those specific values are coming from (the +1/+3/+5). Anyone feel like explaining some math proofing to me ?

[brave bramble]

I mean, does it work?

[solemn crater]

Their odd case or my even case?

[brave bramble]

Oh, if they say that n is an odd number, then when you divide it by 6, the remainder can only be 1, 3, or 5

That's where they get those from

If the remainder was 2, for example, then n was actually even

[solemn crater]

oooh ok lol. That feels like an observation that I would not have gotten to on my own

this question def sounds like a lot of work all of a sudden

[brave bramble]

It's probably pretty fast once the hint is applied

[brave bramble]

n-1 / 2?

[solemn crater]

this is the lil bit of chicken scratch I got so far for the odd case. So I basically just need to copy this another 4 times, substitute those values and continue from there eh?

yeah, floor(n/2), if n is odd, = (n-1)/2, isnt it?

(this math proofing is def not my strong suite lol, but I think thats correct )

yeaah

[brave bramble]

It's true, but it might be better to consider the division by 6 version

[solemn crater]

to be honest I am not sure I understand that approach as of right now haha

I need to start differently too? Express that initial value as a quotient with 6 being the denominator?

is that the play?

[brave bramble]

(6k + 1) / 2 = 3k + 1/2
When floored:
= 3k

That is, replace n with 6k + 1 everywhere

Then 6k + 3, 6k + 5

[solemn crater]

on both sides of the equation?

[brave bramble]

Everywhere, yeah. Then we can look at k.

[solemn crater]

ahhh ok, yeah that makes sense. That does sounds like it'll work. I think as long as I can get the two sides to match, I shouldn't need to do much about k, do I ?

Actually yeah, taking that angle makes it waaay easier. Thank you so much @brave bramble I really appreciate your help!

[brave bramble]

Np. Feel free to ask if you have any other questions!

[solemn crater]

Sounds good, will do! Thanks again and have a great night! (or day depending on where you are lol)

.close

[obtuse pebble]

Channel closed

Closed by @solemn crater

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.

[opaque yew]

Could someone explain to me how to solve this?

[north quarry]

do yk the formula for cos2x ?

[opaque yew]

Don’t think so

[gleaming ridge]

Write tan in terms of sin and cos

[north quarry]

well it should be one you know, but here, its cos2x=cos^2(x)-sin^2(x)

[gleaming ridge]

Everything should simplify

[opaque yew]

Oh ok

This is the formula sheet I have btw

[north quarry]

cos2a is given in the bottom right

[opaque yew]

Ohh ok thanks

[north quarry]

👍

[opaque yew]

I’ll do it real quick could u lmk if it’s right?

[north quarry]

sure

[north quarry]

how it goin ?

[opaque yew]

Alright I just got stuck at a spot

[north quarry]

show me where ur stuck

[opaque yew]

I pretty sure I figured it out but I’ll show u what I’ve done so far

[north quarry]

sure

[north quarry]

yep looks good so far

you can separate the numerator

[opaque yew]

Ok thanks

[north quarry]

(a-b)/c = a/c - b/c

then simplify and you should be good

[opaque yew]

Like that?

[north quarry]

no

they arent common factors

[north quarry]

use this

[opaque yew]

How would I input the numbers in that?

[north quarry]

well first you can distribute the squares

so itll be 1-(sin^2(x))/(cos^2(x)) on top

and 1/(cos^2(x)) on the bottom

[opaque yew]

Like that?

[north quarry]

yes but the 1 shouldnt be on the same level as the sin

[opaque yew]

Oh yeah my bad

[north quarry]

but its j a detail

[opaque yew]

Then can I cancel out cos^2?

[north quarry]

not so quickly

u need to separate it first

[opaque yew]

Ok

[north quarry]

^^

[north quarry]

ok yes hut

but

if a is 1

b is sin^2/cos^2

and c is 1/cos^2

separate it with this

[opaque yew]

?

[north quarry]

the last part is right

the 1st should be

a/c

not a-c

[opaque yew]

Ohhh

Better?

[north quarry]

yep

now simplify

[north quarry]

u didnt simplify correctly

[opaque yew]

Oh

Oh my bad I was doing it like cross multiplication

[north quarry]

dw just retry

[opaque yew]

Good?

[north quarry]

amazing

good job

[opaque yew]

Awesome thank you for your help!

[north quarry]

no problem, glad i could help

[opaque yew]

.close

[obtuse pebble]

Channel closed

Closed by @opaque yew

Use .reopen if this was a mistake.

[obtuse pebble]

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.

[timid silo]

hey

[timid silo]

@dreamy forum

sorry to disturb

[dreamy forum]

np whats up

[timid silo]

but for the 3(a) how to write down the working method

for refence

[dreamy forum]

you mean 4a)?

[timid silo]

sorry question 3

for 4 a and b im clear

for 3 im clear on how to do it

however not sure of the working method

[dreamy forum]

Ok so the working method for 3 is to first analyze how the sequence works, what structure it follows, and then to use the knowledge gained by this analysis to solve the problem

So what we did is, we looked at the sequence 5,8,11 and analyzed, that the numbers in this sequence grow by 3 in every step of the sequence

Which means our sequence is 5,8,11,14,17,20...

[timid silo]

yes

[dreamy forum]

And as the prompt suggested, these are the amounts of sweets we put in a box in the current step

So first we put 5 in the box, than 8, than 11, than 14

[timid silo]

yes

and u minus each number placed in the box with 153

[dreamy forum]

At the same time our amount of availible sweets goes down so each time we put sweets in the box we subtract the same amount of sweets from our total amount

[timid silo]

so = 153-5+3(0)

[dreamy forum]

nah

[timid silo]

yes

[dreamy forum]

153 - 5 = 148
148 - 8 = 140
140 - 11 = 129
...

This is how you would write that down

oh

OH now I realized what this 0 is supposed to be

[timid silo]

it mutiple

[dreamy forum]

If you want to have a general formula for that, let x denote the step we're on

With x = 1 being the first step

[timid silo]

ok

[dreamy forum]

nvm that would be too complicated wait

[timid silo]

ok so just do the way u taught me

[dreamy forum]

yep

[timid silo]

this is simplier

and do it till it hits 0

[dreamy forum]

The general formula would probably look like this: $153 - \sum_{x=0}^{n-1} 5+3x$

[warm shale]

Cyrol

[timid silo]

that is complicated

[dreamy forum]

This first adds up 5 + 8 + 11... to the n-th step and than subtract it from 153

with x = 1 being the first step and n being the last

You could rewrite it:

$153 - \sum_{x=1}^{n} 5+3(x-1)$

[warm shale]

Cyrol

[dreamy forum]

damn thats gotten big

[timid silo]

i dont think i understand this

i will just do it the simple way

[dreamy forum]

You don't need to at this time

but this is how it would work

[timid silo]

ok i dont think i have learned to that extend yet

[dreamy forum]

Perhaps there is even a recursive formula for this

[timid silo]

but its good know for the future

[timid silo]

maths has a 1000 formulas man

[dreamy forum]

1000 forumlas but no explanation

[timid silo]

exactly

[dreamy forum]

This is why i despise school math and like university math

[timid silo]

im still in high school

[timid silo]

probably more simple

[dreamy forum]

definitely

[timid silo]

anyways thanks for the help

[dreamy forum]

np

[timid silo]

do u want me to close it?

[dreamy forum]

if you have nothing left to ask than yes

[timid silo]

.close

[obtuse pebble]

Channel closed

Closed by @mellow valley

Use .reopen if this was a mistake.

[timid silo]

ok

.close

[obtuse pebble]

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.

[timid silo]

Hi, is this correct?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

i mean its not a proof u just rewrote the sets in set builder notation

but i mean u can always argue that it is trivial

[timid silo]

.

also aint it {x \in, x \notin B} i mean it is the same but still

Yes

Thank you

.coowe

.close

[obtuse pebble]

Channel closed

Closed by @eternal chasm

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.

[fringe schooner]

Anyone know how to continue?? I tried substitution but cannot know how to subs the 3^x and 2^x together

[obtuse pebble]

@fringe schooner Has your question been resolved?

[obtuse pebble]

@fringe schooner Has your question been resolved?

[rustic bough]

$\frac{8^x-2^x}{6^x-3^x} = \frac{2^x(4^x-1)}{3^x(2^x-1)}$

[warm shale]

Mind Trickx

[rustic bough]

$(4^x-1) = (2^x+1)(2^x-1)$

[warm shale]

Mind Trickx

[rustic bough]

because of difference of squares

$(4^x-1) = (2^x+1)(2^x-1) = 2$

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

[warm shale]

Mind Trickx

[rustic bough]

$2 * 3^x = 4^x + 2^x$

[warm shale]

Mind Trickx

[rustic bough]

We see that the two solutions here are x = 1 and x = 0

obviously x = 0 dosn't work for the original solution

so x = 1 is the only solution

@fringe schooner hope that helped

[fringe schooner]

Soo no substitution and logarithm involved??

In the end, just put any x and that's it??

I did put x with any number and i got results, but I just wondering whether there is any legit algebraic method to find X

[rustic bough]

oh for algebraic methods to find x ill have to think on that for a while

but normally these don't have algebraic solutions at all

like you just need to plug in

logarithms don't help because there are two exponentials bieng added

you could sort of prove that x = 1 is the last solution however by using derivatives

at x = 1, take the two derivates

so 2 * ln3 * 3 on the LHS

4 * ln4 + 2 * ln2 on the RHS

as you can see here

the RHS is growing faster

at x = 1

so the lines will never cross again

giving x = 1 as the only remaining solution

hope that helped @fringe schooner

[fringe schooner]

Okay it means that if i found any problems that seems so impossibly done by algebraic method like this one, i just need to plug and play the values and testing the derivative?? Is that correct?

I got stumped by this kind of algebraic problems that should be easily solved by school students. As a uni grads, I feel so shameful cannot solve this for a long time

[obtuse pebble]

@fringe schooner Has your question been resolved?

Channel closed

Closed by @fringe schooner

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.