help-4

[random hull]

Come to dm so i provide you i have plenty of it

[craggy rock]

Thanks🎀

[vale dock]

@random hull Has your question been resolved?

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[midnight pier]

Hello! Why does n turn into 2pi/n in the limit? Thank you!

[frozen ledge]

supposed to be 2pi/n —> 0

[heady pawn]

typo

[frozen ledge]

and then apply lim x -> 0 sinx / x

[midnight pier]

Ah! I see. Thank you again.

.close

[vale dock]

Channel closed

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[icy ingot]

would you mind giving me some tips for integrals for you seem to be a very open person in math :3?

[turbid valve]

Saran, this is not the place to ask that - for one this is not an open channel, and for two that's an open-ended question best served in a non-help channel such as #calculus

[icy ingot]

ah i apologize for that! Will make sure it doesn't repeat again

[turbid valve]

"it doesn't happen* again" 👍

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[fervent vault]

(c) 4p / (q - r), if p=6, q=5, and r=2

[final adder]

do you want to solve for the expression given p,q,r?

[fervent vault]

yes

[wicked zenith]

what haave you tried

[fervent vault]

what do you mean

[red tulip]

Have u tried to solve it

[final adder]

show us your thought process

[fervent vault]

yes

Let's look at the expression and explore a wrong way someone might evaluate it, and then understand why it's incorrect.

Given expression:

$$
\frac{4p}{q - r}
$$

Given values:

• $p = 6$
• $q = 5$
• $r = 2$

---

Step-by-step incorrect thought process:

1. Substitute values incorrectly:
   Someone might plug in the values like this:

   $$
   \frac{4 \cdot p}{q} - r
   $$

   ❌ Mistake: They inserted the values into the wrong expression by changing the structure. They treated the denominator as just $q$ and subtracted $r$ afterward.

   Resulting expression:

   $$
   \frac{4 \cdot 6}{5} - 2 = \frac{24}{5} - 2 = 4.8 - 2 = 2.8
   $$

2. Alternative mistake – wrong order of operations:
   Another mistake might be evaluating $q - r$ incorrectly:

   $$
   q - r = 2 - 5 = -3 \quad \text{(wrong order: used r - q)}
   $$

   Then:

   $$
   \frac{4 \cdot 6}{-3} = \frac{24}{-3} = -8
   $$

   ❌ Mistake: Subtracting in the wrong order (r - q instead of q - r).

---

Correct way:

$$
\frac{4 \cdot 6}{5 - 2} = \frac{24}{3} = 8
$$

So, the correct answer is 8, and the incorrect methods usually come from:

• Misplacing parentheses or rearranging the formula,
• Applying wrong order of subtraction in the denominator,
• Incorrectly applying the order of operations (PEMDAS/BODMAS).

Let me know if you want help with spotting more of these kinds of errors!

[red tulip]

You place the values of p q and r in the corresponding places and follow bimdas

[rocky lotus]

Spiritless30
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[oak crane]

!nogpt

[vale dock]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[wicked zenith]

so, what have you tried?

[fervent vault]

sorry

for using chatgpt

i think it was correct

[red tulip]

Its fine but chatgpt doesnt rly know what you are thinking

[fervent vault]

wanna be my friend homie

[wicked zenith]

are you going to try it yourself?

[fervent vault]

i got it

first you solve bracket
5-2=3
then multiply
4*6=24 24/3=8

[final adder]

yes, that's correct

[wicked zenith]

correct

[fervent vault]

thank you

[stiff lily]

4-2=3
you mean 5-2?

[fervent vault]

yes

[vale dock]

@fervent vault Has your question been resolved?

[fervent vault]

yes

[vale dock]

Channel closed

Closed by @fervent vault

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[fervent vault]

.close

[fossil pelican]

Find the solutions to the equation x^n + y^n = z^n for any integer value of n > 2.

[tidal terrace]

[final adder]

any restrictions on x,y,z?

because uh

https://en.wikipedia.org/wiki/Fermat's_Last_Theorem

[brittle epoch]

are x, y, z integers?

[fossil pelican]

"In Fermat's equation, x, y, and z represent positive integers. It's about whole numbers, not fractions or decimals."

[final adder]

so like are x,y,z integers or not

[opal pendant]

huh this is pretty cool

whats the proof

[brittle epoch]

!occupied

[vale dock]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[final adder]

google andrew wiles proof of flt lol

[wraith heart]

don't troll

.close

[vale dock]

Channel closed

Closed by @wraith heart

Use .reopen if this was a mistake.

[opal pendant]

i think imma leave it..way beyond my scope

i thought it would be simple

[vale dock]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[opal pendant]

.close

[vale dock]

Channel closed

Closed by @opal pendant

Use .reopen if this was a mistake.

[vale dock]

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Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[wraith heart]

200 - 2 * 5 is not zero and 220 - 2 * 15 is also not zero

show the original question

[wraith heart]

then do this and show the rest of your work

you probably have a mistake earlier

you're leaving out a lot of details so it's impossible to find your error until you show the original question

¯_(ツ)_/¯

need original question

!vol

[vale dock]

Helpers are just people volunteering their time to help you. Be polite and patient.

[wraith heart]

i asked ehre

[wraith heart]

and here

doesn't seem rude

yes after your insult here

imagine insulting someone and then begging for help afterwards

[vale dock]

Channel closed

Closed by @prisma flint

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[halcyon lintel]

i odnt know what im doing wrong

this problem should be straight forward but im not getting the expected answer

basically

you just subtract 0.18 from 54.5

that gives the length of the two sides of the triangle

and the base is 49.6

to find the area, you can use the Hero's formula

this should be it right?

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fwww.chilimath.com%2Fwp-content%2Fuploads%2F2022%2F08%2FThe-Herons-Formula.png&f=1&nofb=1&ipt=f1ac67af7897a016819e8c6fef8fa1c5d90afe574bb3500ad9d0061c423d76e8

that's the formula i used to calc the area of the triangle

[next vortex]

that sounds right to me

[halcyon lintel]

yea but my answer is not matching the MC options

[fresh furnace]

Can you show your work

[halcyon lintel]

k one sec

let me share the MC options as well

[fresh furnace]

I'm getting A

[halcyon lintel]

ok what am i doing worng?

[fresh furnace]

Uh

Your side lengths seem wrong

[halcyon lintel]

sorry wrong \question

[fresh furnace]

Lol

[halcyon lintel]

my answre is close

[fresh furnace]

Where did you get 53.82

54.50 - 0.18 = 54.32

[halcyon lintel]

54 - 0.18

ohhh its 54.5

thats probabl it

[fresh furnace]

Rip

[tight kettle]

i think it's right man

[halcyon lintel]

ok i think that was my mistake....ok theres another quesatoin thats tripping me up

similar

[fresh furnace]

Ah

[halcyon lintel]

when they say area of the facade, its basically area of traingle + trapezoid right?

minus the area of square

[tight kettle]

make another square here

[halcyon lintel]

what do you mean by that

make another square where?

[fresh furnace]

Yes

[tight kettle]

man i will show you later

[fresh furnace]

900 square inches isn't 75 square feet

[tight kettle]

i am now busy with my physics

[fresh furnace]

1 square foot is 144 square inches

[tight kettle]

hey guys do you know something about reflection of lights?

[fresh furnace]

Open a different help channel
If you need help

[halcyon lintel]

im converting inches to feet right?

[tight kettle]

there are so many shits maths sorry

[fresh furnace]

You're converting square inches to square ft

[tight kettle]

ok sorry

[halcyon lintel]

square inches?

[fresh furnace]

Yes

The area of the door is 900 square inches right

[halcyon lintel]

wait

lets do it one step at a time

is my triangle area correct?

and area of trapezoid?

[fresh furnace]

Everything looks good

Except the door

[tight kettle]

i hav e some math problems?

[halcyon lintel]

so its actually 900 square inches right?

[fresh furnace]

Open a new help channel

[fresh furnace]

Yes

[halcyon lintel]

so im actually converting 900 square inches to square feet?

[fresh furnace]

Yes

[tight kettle]

sorry for everything actually i am totally new here

[halcyon lintel]

so to do the conversion?

[fresh furnace]

Mmm np post your question in #help-22 or something

If you have any

[fresh furnace]

Yeah

[halcyon lintel]

6.25

[fresh furnace]

If you want to convert square inches to square feet, you need to divide by 12 twice

Yes

[halcyon lintel]

i used an online calculator to do that

but why do you divide by 12 twice

[fresh furnace]

Because

The thing is 30 inches by 30 inches right

[halcyon lintel]

900

square inc

[fresh furnace]

If you want to convert that to feet, its 30/12 by 30/12 ft

[tight kettle]

yeah

[fresh furnace]

(each side length is 30/12 ft)

[halcyon lintel]

@tight kettle dude get out of here

yea cause its 12inches in a foot

[tight kettle]

dude i just wanna know where are you from?

[fresh furnace]

So the area in square ft is (30/12)×(30/12)

[halcyon lintel]

how do you remove this guy

[fresh furnace]

I cant

Not a mod

[tight kettle]

remove me quickly guys .

[fresh furnace]

But do you get it
Up to here

[tight kettle]

i don't know why you guys are so angry?

[halcyon lintel]

yea i get it.... 12 inches in 1 foot

just ignore him...he;ll eventually leave

[fresh furnace]

So the area is (30^2)/(12^2)

[tight kettle]

i just want to know about yourselves and those maths

[fresh furnace]

Go to #discussion

This isnt the right channel for that

[tight kettle]

i know

[halcyon lintel]

ignore him...he has no life

[tight kettle]

i have a cancer already

after someday i will leave this world. sorry please forgive me

[tight kettle]

you are right

[halcyon lintel]

how do you block?

[fresh furnace]

Think about it this way

If you have one square foot, its like having a square that with side length 1ft

Which is equivalent to having side length 12 inches

[halcyon lintel]

ok

[fresh furnace]

12 inches on each side, so there are 12^2 square inches

[halcyon lintel]

ok so if its cubic

then you divide by 12 3 times right?>

[fresh furnace]

Yes

[halcyon lintel]

ok i get it i think

lets move on

so then

i just need to subtrct 6.25 from the area of the triangle and trapezoid

[fresh furnace]

Yes

[halcyon lintel]

891

[fresh furnace]

Is that one of the answers

[halcyon lintel]

sure is lol...

so that was my mistake

[fresh furnace]

Seems like it

[halcyon lintel]

do you have time for a few more geo questions?

ive got a test on thurs and need to nail this

[fresh furnace]

I don't but someone else might

[halcyon lintel]

ok thanks

if anyone can help me with Q4 i'll be here. thanks

[next vortex]

try to find the length and width of the centre rectangle

[halcyon lintel]

yea that was my plan

[grand jetty]

1. the area of the top rectangle is given as well as its length so we can find its height (we call this a)
2. we can find the height of the bottom rectangle using the same method (we will call this b)
3. by taking a and b from the length of the whole object we can get the heigth of the middle rectangle (we will call this C)
4. knowing the bottom rectangle is equally split, we assume that 6/3 is equal to the width (we will call this D)
5. we find the area of the central rectangle with CxD
6. we add the areas of all the rectangles together

[halcyon lintel]

oh ok so

to get the height of the the top rect

its A = b x h right?

so its basically written as 28 = 7 x h

tthen we can just solve for h to find the height?

[grand jetty]

yup

[halcyon lintel]

ok let m try to work it out real quick. appreciate the help

wait

h = 4? lol

[grand jetty]

do the whole problem

dont worry about every step

[halcyon lintel]

ok i just have a question

the area is cm^2

but the length is 7cm

[grand jetty]

yes

the length and the height are both in cm

so in terms of units we have:
cm x cm = cm ^2

you can also apply this for volume to get its unit too

[halcyon lintel]

ok

let me try to work oiut the rest of hte problem

ok

so area of that middle rect is 16cm^2

so total area of that shape is 16 + 8 + 4

28cm

[grand jetty]

why 8 and 4?

[halcyon lintel]

agree?

cause those were the areas of the 2 other rectangle shapes

[west cloud]

you sure?

[grand jetty]

read the question again

[west cloud]

check the problem text again

[halcyon lintel]

yea so

28 + 18 + 16

[grand jetty]

yup

[halcyon lintel]

62

total area

agree?

[grand jetty]

good job

[halcyon lintel]

amazing

i cant believe im doing it

A 😁

is my visual representation correct?

[west cloud]

seems right

[vale dock]

@halcyon lintel Has your question been resolved?

[halcyon lintel]

thanks

[west cloud]

if you're done with this channel, you can .close it

[vale dock]

Channel closed

Closed due to timeout

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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.

[halcyon lintel]

.close

[halcyon lintel]

hello there

so im trying to solve this problem

my initial reaction is to try to solve the missing side of the upper most trapezoid shape

so to get the height of the trapezoid, you just do 113-58-32 = 32mm

h = 32, length_a = 120

wat nevermind.

i think my steps is incorrect

[golden cloak]

Where is the 115 coming from?

[halcyon lintel]

sorry, should be 113

[golden cloak]

And it’s not -58mm bc the line doesn’t go in the right direction

You have to use pythagoras

[halcyon lintel]

but it shows 58mm poiinting to the blue dotted line

isnt that the height

oh wait

i see what you're saying

thats the height of the triangle shape

not the actual height of the diagram

i think thats what you mean

[golden cloak]

Ye

[halcyon lintel]

yea so i first use py. theorem to solve for the height of that triangle

[golden cloak]

Idk, didn’t know how to say it

[halcyon lintel]

ok so first step is to solve the height of that triangle

cool

step 2

this is where im still kinda lost

my first instinct is to solve the missing side on the top trapezoid shape

the line that is parallel to the 120mm

the blue dotted line

[golden cloak]

Yes and you know the height of the trapezoid

[halcyon lintel]

yea

[golden cloak]

So you could solve for the triangle with length of 64mm (I’ll make a picture)

[halcyon lintel]

oh ok

so then i can use py.theorem again to solve for that missing triangle base?

[golden cloak]

Ye

Then x is what?

[halcyon lintel]

wait lol

120-58+ the base of the triangle

[golden cloak]

Yes

So you just have to do two times pythagoras

[halcyon lintel]

so its basically using py.theorem on both of the triangles?

yea

[golden cloak]

Yes

[halcyon lintel]

yea makes sense....appreciate it. thanks 🫶

[golden cloak]

Okay, you could post your result so I could check, if mine is the same

[halcyon lintel]

oh, you want to see my work?

warning. its gonna take some time lol

[golden cloak]

Yes

I’ll wait

[halcyon lintel]

k one sec

thats the height of the trapezoid shape on the top

now to find base of that triangle

[golden cloak]

That can’t be: 110-80 is not 75

[halcyon lintel]

the 75 is just the length of that missing triangle

this one

oh wait

x cant be longer than the hypotenuse

no wait

im confused

[golden cloak]

But 113-52-32=29, not 75

[halcyon lintel]

51,84 is the length of that missing side from that triangle

[golden cloak]

That’s right

[halcyon lintel]

75 is the height of that trapezoid shape on the top

[golden cloak]

No

You made a calculation error

[halcyon lintel]

29.16

[golden cloak]

Yes

That’s right

[halcyon lintel]

how tf did i get 75

[golden cloak]

Idk

Ok, so now the second triangle

[halcyon lintel]

ok let me work it out

2mins

agree?

something is not right

[golden cloak]

No

You forgot the square under the root

[orchid fjord]

When is there going to be a math lecture

[golden cloak]

?

[halcyon lintel]

which square?

[halcyon lintel]

honestly

i keep making these small stupid errors and it keeps costing me

[golden cloak]

Looks good

Do you still have an open question?

[halcyon lintel]

yea seems to be good

yea i have lots of questions

but i dont want to take anymore of your time

appreciate it. 🫶

[golden cloak]

Ok, then could you close the channel please?

[halcyon lintel]

sure

.done

[golden cloak]

.close

[orchid fjord]

.close

[golden cloak]

You have to write it

[halcyon lintel]

@golden cloak hey sorry

if you could help with one other question, i just realized i got one of my other quesetions wrong and not sure why

if you cant help its ok but ill post it if anyone can take a look

this was my solution but it appears to be incorrect

[ebon kernel]

@everyone

[orchid fjord]

Hello everyone when is the next math lesson?

@everyone

Does anyone know?

[normal hollow]

6th aug,2026

U r a year too early

Go back to sleep

[orchid fjord]

Oh…

Sorry

[halcyon lintel]

im still here if anyone can help thanks

[normal hollow]

Procedure seems right

Idk abt the calculation

Calculation seems right too

[normal hollow]

Why do u think so

[halcyon lintel]

im trying to upload a screenshot that shows my answer is incorrect but library wifi is slow af

sorry

[normal hollow]

np i'll wait

[halcyon lintel]

there you go

finally

[normal hollow]

Plug in 63.98

Cuz that's the actual value u got here

[normal hollow]

Here

Exceeding 64 would be wrong

[halcyon lintel]

63.98?

[indigo temple]

that is the closest number to your calculation

[halcyon lintel]

doesnt it say to round to nearest dollar

?

find to the nearest dollar

ok i think i get what it means

thanks

.done

[normal hollow]

That is ur calculation

[halcyon lintel]

yea

[normal hollow]

Yes if u had 64 ,u would be rounding it to integer

It didnt ask for rounding it to an integer form

[halcyon lintel]

i guess they wanted to the nearest dollar without going over?...

i dont know

i dont know actually

they just trying to make it confusing for you

all good though

[normal hollow]

It says rounding it to nearest value as in rounding it to lower decimals i suppose

[halcyon lintel]

yea im just gonna move on

thanks anyway

[normal hollow]

Alr if done then .close or .solved

[halcyon lintel]

.closed

[normal hollow]

.close

[vale dock]

@halcyon lintel Has your question been resolved?

Channel closed

Closed by @halcyon lintel

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.

[rare sleet]

How to indenify what I'm solving for . Cos,Sin, Tan

[unborn knot]

wdym

u have example?

[normal hollow]

Do u have a specific question

[rare sleet]

Wiat a sec

Question 3

[unborn knot]

SOH CAH TOA

learn it imo

[fickle rose]

sin is opposite/hypotenuse, cos is adjacent/hypotenuse and tan is opposite/adjacent

[rare sleet]

"If it says how much height did you gain" what is it, is it tan,cos or sin?

[normal hollow]

"The side opposite to the angle is opposite "

[fickle rose]

draw it out

it will help

[normal hollow]

Depends on angle

[rare sleet]

Ok

[unborn knot]

its probably to the horizontal

so adj

[normal hollow]

It's vertical,no?

Height

[unborn knot]

youd slip and fall if it was 20° to the vertical

[unborn knot]

far too steep no

[rare sleet]

That how?

[normal hollow]

Draw the figure

[rare sleet]

Idk how to do that...

[unborn knot]

^^

[unborn knot]

try

if u fail retry

best way to learn

[normal hollow]

Draw a path going upwards

[normal hollow]

To vertical?

[unborn knot]

yh defo 20 to the horizontal

[normal hollow]

Not to the norizon?

I meant the angle is formed with horizon

[unborn knot]

ye

thays what i said

[normal hollow]

But the vertical as in opposite is the height

[unborn knot]

ye

opposite to angle

[unborn knot]

so tan

[normal hollow]

Yea

[unborn knot]

or wait

[normal hollow]

Yes so put in the angle

[unborn knot]

ye

[rare sleet]

I don't have protracter

[unborn knot]

sin

im buggin

[normal hollow]

The moutain is 20° with horizontal

[unborn knot]

just label it

[normal hollow]

Js put in

Foesnt need to be exact

[unborn knot]

its not scale

[unborn knot]

yeah that works

but ur giy is walking backwards

[rare sleet]

Uh

[unborn knot]

now solve

[rare sleet]

Ok

[unborn knot]

j personal preference ur fine dw lol

[fickle rose]

you have an angle, and two sides, what are those two sides?

the possibilities: adjacent, opposite, and hypotenuse

[rare sleet]

This good?

I do it correctly?

[fickle rose]

not quite, correct that sin is opp/hyp, but you plugged 100 into the opposite, 100m is not on the opposide side of the angle

[rare sleet]

100 is under the line right?

Which is hypi

Hypo

[loud heron]

yes

[rare sleet]

Oks

I got the answer right though rightM

?

But wrong place

[loud heron]

in SOH
it means
sin = Opposite/hypotenuse

[rare sleet]

Its 34.2m

[fickle rose]

yes because you also messed up your algebra

[rare sleet]

Right?

[rare sleet]

Sad

[fickle rose]

technically right

[rare sleet]

Yes

[fickle rose]

but try to find your error

the two mistakes cancelling out here was concincide and may not work in general

[rare sleet]

Ok

Question 4

The shadow is 5 m

Tan = opp/ adjac

?

[loud heron]

nah

its sin

because you have the hypotenuse

and are working out the opposite

Ohh wait

mb

[rare sleet]

I am right?

[loud heron]

yes

[rare sleet]

😁

[loud heron]

i got confused

[loud heron]

because you labelled the 5m on the hypotenuse

[rare sleet]

So 5m is vertical?

[loud heron]

no

[rare sleet]

And hypo is opp?

[loud heron]

5m is horizontal

[loud heron]

no you are trying to work out opposite

you have adjacent and the the angle

[rare sleet]

This how?

[loud heron]

yeah

thats good

[rare sleet]

Opp = cos

So

Cos= ad/hypo

?

[loud heron]

you have adjacent and opposite

so uh

i wouldnt use cos

[rare sleet]

Tan

[loud heron]

yeah well done

[rare sleet]

Good?

[loud heron]

yeah

[rare sleet]

Yipee

[loud heron]

thats the right answer

[loud heron]

✅ ✅

[rare sleet]

Now I can rest in peace

[loud heron]

well done vro

[vale dock]

@rare sleet Has your question been resolved?

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[halcyon lintel]

hello there

im completely stuck on this question

[heady pawn]

where are u stuck

[halcyon lintel]

the tangent line should be parallel to QP?

where the line just touches the circle?

[heady pawn]

?

you're trying to find OP

[halcyon lintel]

yea but what does the QP is the tangent to the circle mean?

is that what they mean?

[heady pawn]

not exactly.

[jade ivy]

The line QP only touches the circle at Q

[heady pawn]

draw a line from Q to P, then it's indeed tangent to the circle

and perpendicular to OP

[halcyon lintel]

???

[heady pawn]

P is not the midpoint of the circle?

[halcyon lintel]

Liek that??

image still uploading

there you go

like that??

[heady pawn]

yes thats tangent to the circle.

[halcyon lintel]

which point is O?

i only see QP

[heady pawn]

remake the image exactly.

[halcyon lintel]

its actually QO

[heady pawn]

then draw a single line FROM POINT Q TO POINT P

and nothing else

[halcyon lintel]

not QP....my bad

ok

one sec

where is P?

ionly see Q and O

am i supposed to solve for P or something?>

[pale kettle]

Can you not see P in the picture?

[halcyon lintel]

no

i see only Q and O

my eyes are bad

is the point in the middle of the circle supposed to be a P?

[halcyon lintel]

oh wtf....its al l the way out there?

i thought it was somewhere near the circle

[pale kettle]

I would ask your teachers for a larger print, this sounds like a genuine issue you could have with your education

[halcyon lintel]

There are many issues but I won't go into it atm

Ok so the tangent line would be from q to p points?

[pale kettle]

Yep

[pale kettle]

Correct

[halcyon lintel]

angle QOP is 32.8 deg

[pale kettle]

Mhm, so draw that on

[halcyon lintel]

igree?

[pale kettle]

Yep

[halcyon lintel]

ok so

oh ok...so now i need to find distance to OP

[pale kettle]

Mhm

[halcyon lintel]

is this a trig problem?

[pale kettle]

Yep

[halcyon lintel]

so basically, this is whats given

my first instinct is to try to solve for the radius

caues that would give me a second length

of the triangle

and i dont see how the angle given helps me

is angle OQP supposed to be a right angle or something?

[manic hatch]

Pretty sure you can use this rule

Where x = 148/2

You know that the arc is 32.8

[halcyon lintel]

honestly, this question is supposed to be strictly related to circles

so im going to assume that i need to use formula for area or circumference of a circle but i dont see how any of these formulas apply to this scenario

[pale kettle]

This question revolves around you noticing a right angled triangle

[halcyon lintel]

so OQP is indeed a 90deg angle?

[pale kettle]

Yes

As QP is perpendicular to the radius

[halcyon lintel]

and we're just supposed to know that?

[pale kettle]

And we are using the radius as one of the lengths of the triangle

And perpendicular lines are, by definition, at right angles to eachother

[halcyon lintel]

this looks so wrong

[manic hatch]

It's a property you have to memorize

[manic hatch]

Whenever you connect a radius to a tangent line, it creates a right angle

No need to use the theorem of cosines here

[halcyon lintel]

so a tangent line will always have a 90deg angle?

[manic hatch]

Only if you connect a radius to it

QP^2 = PE * PO

This is another property you have to know

[halcyon lintel]

so only the end of the radius that connects to the edge of the circle right?

where it will form the 90deg angle?

[manic hatch]

What do you mean?

[halcyon lintel]

let me draw

so the 90 deg angle will be from A to B

[manic hatch]

This is actually wrong

[halcyon lintel]

never from C to B

[manic hatch]

You have to extend this to a secant

[halcyon lintel]

what?

[manic hatch]

So this picture that we have

[halcyon lintel]

is my answer correct? the one i shared earlier

25 units

[manic hatch]

We have to extend it like this

[manic hatch]

Perhaps, I'm just showing you an easier way

[manic hatch]

From this. QP^2 = PE * PK

[halcyon lintel]

is this a shorcut?

[manic hatch]

It's a helpful theorem

[halcyon lintel]

so line AB in my drawing is not a tangent line?

[manic hatch]

Yeah it is

We just needed a secant line

For the theorem

[halcyon lintel]

oh ok but back to my original question

the 90dg will always be from A to B

never from C to B?

[manic hatch]

This is how the angles are being made

Every time you connect a radius to the tangent line

It creates a 90 degree angle

[halcyon lintel]

ohhh ok that makes more sense

yea so its the radius on the edge of the circle that forms the 90deg angle

thanks for clarificatoin

hopefully my finall answer is correct

appreciate it

🫶

[manic hatch]

It's also helpful to know that if they make a 90 degree angle on 1 line, they are parallel to eachother

It's not drawn correctly, but you get the idea

[halcyon lintel]

yea got it....

i was just expecting this answer to be less about trig and more about using circle formulas

but i was wrong

thanks

.closed

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[royal basalt]

Just checking to see if my homework is good my work is the last image

[distant galleon]

,w sin(x)=-0.23, 0 \leq x < 2pi

[rocky lotus]

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[distant galleon]

Looks like you may have rounded early

[distant galleon]

This is fine

[distant galleon]

,w (180/pi)arctan(1/5)

[rocky lotus]

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[distant galleon]

This is fine to the nearest tenth

[distant galleon]

,w 4sin(pi/2 * x)=3, 0<x<10

[rocky lotus]

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[distant galleon]

Yeah this one looks fine

[distant galleon]

,w 3sin(2w)+5sin(w)=0, 0 \leq w<2pi

[rocky lotus]

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[distant galleon]

The rounding on this is off

[royal basalt]

2.56

Thank you

[distant galleon]

,w sin(165 deg)=(sqrt 6-sqrt 2)/4 true or false

[rocky lotus]

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[distant galleon]

Checks out

[royal basalt]

Changed 2.55 to 2.56 and rounded correctly everything else solid

Thank you for the help!

Was there anything else I missed?

[distant galleon]

Unless I mentioned

No?

[royal basalt]

Yea just double checking my work

Should be finished in a minute

.end

[distant galleon]

!done

[vale dock]

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

[royal basalt]

.close

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[pine prairie]

ok my adventures into PDEs continue

[pine prairie]

I want to solve this problem

$u_t + (x^2-1)u_x = 0$ with the boundary condition $u(0, x) = e^{-x^2}$

[rocky lotus]

frosst

[pine prairie]

following Oliver's intro to pdes book, we find c and 𝛃 accordingly

$c(x) = x^2 - 1$, $\beta = \int \frac{1}{c(x)},dx$

[rocky lotus]

frosst

[vale dock]

@pine prairie Has your question been resolved?

[pine prairie]

oh i forgot about this

do the integration we get $\beta(x) = \frac{1}{2}\ln\left|\frac{x+1}{x-1}\right|=t+k$

[rocky lotus]

frosst

[pine prairie]

and then im a bit lost

chatgpt says you let $t = 0$ and $x(0) = \xi$ but idk why you'd do that

[rocky lotus]

frosst

[wild linden]

Generally, how we approached this sort of thing in engineering was we assumed u(t, x) = T(t) X(x). So du/dx = TX', du/dt = T'X, so we have TX' (x^2 - 1) + T'X = 0. We can isolate. T'/T = (1 - x^2) X'/X. The left hand side is solely a function of t, and the right hand side is solely a function of x, so the only way that these can both be true is if both sides are constant. So T'/T = C, and (1 - x^2) X'/X = C, and we just have two ODEs now.

However, I was never really happy about the first step where we assumed the form of u(t, x) was decomposible like this.

[pine prairie]

I don’t think that’s right

The solution is $u(t,x) = \exp(-\left(\frac{1+x+(x+1)e^{-2t}}{1+x-(x+1)e^{-2t}}\right)^2)$

[wild linden]

does that satisfy the boundary condition? It looks like if t = 0 you get exp(-(x+1)^2), should the numerator have an (x-1)exp(-2t) instead?

[rocky lotus]

frosst

[pine prairie]

Oh it’s + on the bottom

[wild linden]

wouldn't the + on the bottom give you a div by 0?

[vale dock]

@pine prairie Has your question been resolved?

[pine prairie]

Errr let me check

[pine prairie]

Oh you’re right

[earnest ingot]

I think you get x-1 instead of x+1

[pine prairie]

It’s x-1 for both

Both numerator and denominator are x + 1 +- (x-1)e^-2t

[pine prairie]

It said to do this

[pine prairie]

Then sub x = xi and t = 0 into here to solve for k in terms of xi

[pine prairie]

Then plug it back into this equation for an equation of x, t and xi

[distant pulsar]

oliver gives you this

[pine prairie]

Solve for xi then plug it into f from here

[pine prairie]

Okay but also the next line in here says this formulation isn’t particularly enlightening

[pine prairie]

But I also tried to graph this for other examples above and those worked

So I’m not sure what isn’t enlightening about it

[pine prairie]

,rccw

[pine prairie]

I think Oliver is wrong and it is very enlightening

[pine prairie]

.close

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[ivory valley]

I need some help understanding this proof

[ivory valley]

I understood the first line, basically the big rectangle within [x_(j-1), x_j] is the same as thinner rectangles within summed up, but how does the upper inequality happen?

I made a drawing and it seems to make sense

[hot tide]

any more questions?

the gist is finer partition => better riemann sums

[ivory valley]

I am not sure but I think I will move on

.solved

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[glass kelp]

,rccw

[rough talon]

trick is to draw lines trough the center of the circles

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@tardy hamlet Has your question been resolved?

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[halcyon plank]

Need help with the approach

[tidal swift]

what did you call me and exactly me for

[halcyon plank]

I needed help

[ebon glade]

helps to know all small primes

[halcyon plank]

with combinatorics

[halcyon plank]

thats the issue

idk how to deal with big primes

there must be some technique or approach I'm unaware of

[ebon glade]

start with the small primes anyway

that should already be quite restrictive

[earnest anchor]

btw

[halcyon plank]

But when Ill have to reach bigger primes what do I do

[earnest anchor]

what is the different between combinatorics

and permutations?

[ebon glade]

worry about that later

[halcyon plank]

its combinations and permutations and as a whole they are called combinatorics

permutations are arrangements

[tawny moon]

combinatorics include permutations. if you mean combination vs permutation, combination doesn't care about order

[earnest anchor]

ohh

[earnest anchor]

what is combinatorics means though?

[tawny moon]

the study of combinations and permutations

[earnest anchor]

OHHH

[halcyon plank]

Basically whole things is about arrangements and combinations

[earnest anchor]

so its like

a parent branch

or is there something different?

[tawny moon]

parent branch of what exactly

[earnest anchor]

cuz like Trigonometry is litterly combinations of both geometry and algebra but its pretty diff

[halcyon plank]

like number of ways you can pair your sneakers with a cool jacket or stuff

[earnest anchor]

Both Combinations and permutations

[earnest anchor]

yes factorial

[tidal swift]

i think you might want to open a new help channel for this

there's quite a bit to talk about

[halcyon plank]

yes

[earnest anchor]

Really?

arent the definitions of combinations should be pretty simple

[halcyon plank]

you can talk abt it in general or discussion channel

[tidal swift]

this channel belongs to brischkl at the moment

[earnest anchor]

That is true, sorry for my rudeness

[tidal swift]

judging from your curiosity (which is all good!), i sense a lot of questions from you

[earnest anchor]

[tidal swift]

so best to open a new channel so we can answer your questions without bothering OP

[earnest anchor]

oke okee imma made one

thanks

sorry @halcyon plank

[halcyon plank]

NP

btw

I need help someone

[ebon glade]

well what have you done in the meantime

have you looked at the small primes

[lost marlin]

What can a prime number end with ?

[halcyon plank]

they just don't end with even

[halcyon plank]

jwait i'm computing them

@ebon glade I can't solve it

[lost marlin]

Hmmm what about 1 digit primes

[halcyon plank]

2 ways

[ebon glade]

can you list all 1 digit primes

[halcyon plank]

2,3,5,7

[ebon glade]

so what color do those digits need to have

[halcyon plank]

green or red

since they can't be blue

[ebon glade]

good

what about all primes < 20

[halcyon plank]

when one of them is blue its 2 ways and when one of neither of them is blue its 3 ways

total 5 ways (probably)

[lost marlin]

What can we say about the colour of 2

[halcyon plank]

2 ways

[lost marlin]

All evens have to contain at least 1 non green colour

So what does that tell

[halcyon plank]

1 possible way for 2 then

[ebon glade]

lets think about 22

[halcyon plank]

it isn't a prime tho

[ebon glade]

but it is composite

[halcyon plank]

yes so

2 ways

[lost marlin]

?

[lost marlin]

You said this earlier

[halcyon plank]

because 2 is a prime

and an even number

[lost marlin]

So how many colours can 2 take

[halcyon plank]

wiat 2 can take 2 colors

there is no such restriction on even

[lost marlin]

Also I recommend using RGB for colours

Instead of numbers

[ebon glade]

2 can be red or green

now what about 22

[halcyon plank]

it can be red or blue

[ebon glade]

22 needs to have a digit that is not green

[halcyon plank]

yes

[ebon glade]

so what color can 2 have

[halcyon plank]

ummm

it can have red

[ebon glade]

it has to be red

now what about 3

[halcyon plank]

it can have either red or green

@ebon glade pardon me for late reply

[ebon glade]

lets think about 33

[halcyon plank]

so 33 can have either red or green too

it can have 5 ways in total

when one of them is blue

[ebon glade]

I have no clue how you arrived at 5

the digit 3 cant be blue

but also it cant be green

cause its contained in 33

[lost marlin]

Wait I think you are misinterpreting the question

[halcyon plank]

when you exclude blue you get RG,RR,GG and with one of them blue BG,RG

[lost marlin]

We don't have to find the ways in which each number can be represented

[halcyon plank]

total 5

[ebon glade]

all 3s have the same color

[lost marlin]

We have to find the ways to colour just the digita

[halcyon plank]

will it be so??

[ebon glade]

you cant have the first 3 be red and the second one be green

[halcyon plank]

ohh

then there'll be 3 ways

wiat

2 ways

[lost marlin]

No need to count the ways of each number

We need to eliminate possible colours of digits

[halcyon plank]

yes'

but I'm still confused and unable to solve the problem

[ebon glade]

the digit 3 cant be blue cause 3 is prime and cant have a blue digit

the digit 3 cant be green cause 33 is composite and cant have a green digit

so what color is 3

[halcyon plank]

Red

so it can have only red colour

must*

[ebon glade]

yes

what about 5

[halcyon plank]

5 is prime so RG

[ebon glade]

lets think about 55

[halcyon plank]

It'll be only red

[ebon glade]

so for which digits do we know the color now

[halcyon plank]

single digits and multiples of 11

[ebon glade]

do you know what the word digit means

[halcyon plank]

yes

oh sorry

2,,3,5

[ebon glade]

what other digit is also easy

[halcyon plank]

7 I guess?

[ebon glade]

yes

so 2,3,5 and 7 are all red

[halcyon plank]

It'll be same thing

yes

[ebon glade]

and we realized along the way that numbers consisting of only one digit are interesting

so what about 11

[halcyon plank]

11 is prime and it's digit is 1 so if we consider 1 as prime we get RG

and it isn't composite

[ebon glade]

what about 111

[halcyon plank]

111 is divisible by 3 so it's composite and it can have R

[ebon glade]

so 1 is red

[halcyon plank]

yep

[ebon glade]

so now we know that all numbers that end in 1,2,3,5,7 are fine

cause no matter whether they are prime or composite, they contain a red digit

[halcyon plank]

yep

[ebon glade]

what about numbers ending in 4

[halcyon plank]

they are composite so for them it'll be (RB)

[ebon glade]

so 4 is either red or blue and no matter what, every number ending in 4 is fine

what about 6

[halcyon plank]

same for 6

[ebon glade]

what digits are left?

[halcyon plank]

we are left with 8 and 9

[ebon glade]

one more

one of those is easy

[halcyon plank]

0?

[ebon glade]

for the other ones you'll have to think a bit more

gtg

[halcyon plank]

kk

[lost marlin]

8,9 are slightly difficult

[vale dock]

@halcyon plank Has your question been resolved?

[ebon glade]

8 is easy

9 seems to be very difficult actually

unless I'm missing something

well I suppose 4,6,8,9 as a combination is very difficult

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[frozen laurel]

Help

[next vortex]

what's your question?

[frozen laurel]

If α and β are the distinct roots of the equation x² - 3¼ x + 3½ = 0, then find the value of α⁹⁶(α¹²-1) + β⁹⁶(β¹²-1) and I did it using Newton's formula and also playing with sum and product of roots but my friend said complex numbers could be used to solve this question but I don't understand how

[red tulip]

the descriminant of the quadratic is less than 0

so the roots are on the complex plane

if the two roots are in the complex plane then they are conjugates of each other

[frozen laurel]

That much I know