help-38

[winter ingot]

would u sya its easier to find x

or q?

so my question would be x = 17-q-x

?

or would q = 17-2x

be easier?

[mighty canyon]

q = 17-2x

because you want to eliminate $q$ and turn it into single variable equation

[winter ingot]

k

[solid kiln]

print("NAME")

[winter ingot]

yh

[winter ingot]

is $4.06 correct?

i mean

i got

4.9625

[winter ingot]

4.0625

it's not a whole number

do i did do something wrong?

[solemn thorn]

can you give the equation you used

1 quarter = 25c
1 dime = 10c
1 penny = 1c

just for my reference

[winter ingot]

wdym

[mighty canyon]

like, how do you get 4.0625?

[solemn thorn]

like what equation did you use to solve

you should get two

[winter ingot]

i did q = 17-2x

[solemn thorn]

seems right to me

[winter ingot]

and then did 25(17-2x)+2x=230

[mighty canyon]

that is where it goes wrong

[solemn thorn]

well i dont think pennies have the same value as dimes

[winter ingot]

shoot

my bad

;-;

[solemn thorn]

nobody perfect

[winter ingot]

a chocolate bar cost 4 times as much as a stick of candy. if 5 sticks of candy and 2 chocolate bars cost 1.82. how much does a stick of candy cost

how do i do this

i am struggling

4x+x=1.82

right?

because x = one stick of cnady

?

[solemn thorn]

one moment

yea the chocolate bar will be equal to 4x thats correct

now how much will 5 sticks of candy cost?

[winter ingot]

oh

[winter ingot]

5x?

[solemn thorn]

very nice

how about 2 chocolate bars

[winter ingot]

0.14?

[winter ingot]

is the answer?

[solemn thorn]

yes this is the price of 1 candy stick

[winter ingot]

tyty

[solemn thorn]

2 chocolate bars would be equal to 2(4x)

or 8x

[winter ingot]

done my hw now i can sleep 🙌

[solemn thorn]

then youd add them up and divide

[winter ingot]

alr i got it

[solemn thorn]

14 cents for candy sticks are a steal

[winter ingot]

real

gn

thx

thx @mighty canyon

[solemn thorn]

close the channel when youre done

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[flat wing]

can someone help me with this please

[flat wing]

would righting usin x =v cos x be right?

x for alpla

and then vsinx / ucosx

= 1/3

be right?

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[modern ferry]

i dont get how to solve this

[modern ferry]

i use v/|v| right

and times it with gradietf(A)?

so that we get 2(x, y, -z)^t/ x^2+y^2-z^2 * v/|v|?

yes i do

so we get 2/9(3,4,-4) for the first part

and 1/3(2,2,-1)?

yes but its okay i have not gotten an answer yet

[whole coral]

,w gradient ln(x^2+y^2-z^2)

Bruh

[modern ferry]

haha

[solid kiln]

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[modern ferry]

but it think i get 4/3

[solid kiln]

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[whole coral]

[probably do the previous one and press the more option]

[modern ferry]

haha yes you tried your best

but for the second part do i do f(A) + gradientf(A)?

or wait we need h too right

2ln(3) + 4/3(3)

hmm but what is h?

wait is h = (0.002, 0.002, -0.001)?

Yes okay

i think i get 2ln(3) + 0.012?

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[urban copper]

should I put everything under same fraction

[chilly remnant]

just the second part

[carmine spade]

not in this case
it looks like (and will turn out that) the second two terms can be handled as a seperate limit

actually its easier to see that you can deal with the first term on its own

[urban copper]

,w lim n to +infinity (5 + cos(n))/(3 + n^4) + sqrt(n^2 -6n) -sqrt(n^2 +11)

[solid kiln]

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[carmine spade]

||the top of the fraction is bounded and the bottom -> infinity so that term must go to 0||

[urban copper]

OH

[carmine spade]

for the first term ^

[urban copper]

yes numerator is bounded for first term

so that whole fraction is 0

[carmine spade]

yh

[urban copper]

okay

conjugate?

[carmine spade]

yep

[urban copper]

okay I did it

it was literally that trick that gave it away

that the left fraction is literally bounded as fuck

tysm everyone I will be closing

love you guys, xoxo

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[urban copper]

Let ( f : \mathbb{R} \to \mathbb{R} ) be defined by
[
f(x) = \begin{cases}
\frac{7e^x - e^{2x} - 6}{x} & \text{if } x \ne 0, \
a & \text{if } x = 0.
\end{cases}
]
Find ( a \in \mathbb{R} ) such that ( f(x) ) is continuous at ( x_0 = 0 ), and for the found value, decide using the incremental coefficient if ( f'(0) ) exists.

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

lowkey this is what incremental coefficient means

lowkey just means using the definition of derivative as limit

can someone help me now please

<@&286206848099549185>

[worldly field]

yo whats up

lemme read the question real quick

sorry, cant help you:(

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@urban copper Has your question been resolved?

[carmine spade]

we use lhopital to fine the value of a,
namely a = ||lim x->0 (7e^x - e^(2x) -6)/x||
= ||lim x->0 7e^x - 2e^(2x)||
= ||5||

then we use the definition of the derivative. if the limit exists then f'(0) exists
lim x->0 ||[ (7e^x - e^(2x) -6)/x - 5] / x = lim x-> 0 (7e^x - e^(2x) - 5x - 6)/x^2 on which we can again use l'hopital||
= ||lim x->0 (7e^x - 2e^(2x) - 5)/(2x) again we can use l'hopital||
= ||lim x->0 (7e^x - 4e^(2x))/2||
= ||3/2||
and so ||f'(0) = 3/2 exists||

@urban copper

[urban copper]

we cannot use lhopi though

its not a 0/0

or is it?

im confused how you calculated a

[carmine spade]

it is

top -> 7(1) - (1) - 6 = 0

[urban copper]

oh yeah my bad

a = 5

sure

[carmine spade]

yep nice

[urban copper]

this is the hard parat, how do I find if f'(0) exists

[carmine spade]

like you said

[urban copper]

did you used this ?

[carmine spade]

yeah

well i used the other one by its basically hte same

[urban copper]

which one

[urban copper]

do you mind we use the other one

lowkey I dont get what x -> a means in denominator!!!

x approaching a?

[carmine spade]

that should be a -

[urban copper]

ohh

[carmine spade]

you can change from this one to the other one by setting x = a + h

then the bottom becomes just h

and the top is f(a + h) - f(a)

[urban copper]

what is a here

5?

[carmine spade]

a is 0

because thats the x coordinate we want to investigate

different a from the question

[urban copper]

im so confused rn

mmm

[carmine spade]

what about

[urban copper]

what is f(x)

in terms of our piecewise

[carmine spade]

the function in the question

ah well since x->0 we know that x != 0

so its the first definition

[urban copper]

?????????????''

[carmine spade]

?

[urban copper]

can you elaborate

[carmine spade]

its just how limits work

[urban copper]

okay

[carmine spade]

think about x moving in toward 0

but never 'reaching' it

[urban copper]

true

what about f(a)

if we know a=0

[carmine spade]

yep

in part one we found what f(0) was

[urban copper]

f(a)=5

[carmine spade]

yep

[urban copper]

,, \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}

we know $f(x) = \frac{7e^x -e^{2x} -6}{x}$

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

we know f(a) = 5

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

$\lim_{x \to 0} \frac{\frac{7e^x -e^{2x} -6}{x} - 5}{x} \ = \lim_{x \to 0} \frac{\frac{7e^x -e^{2x} -6 - 5x}{x}}{x} \ = \lim_{x \to 0} \frac{7e^x -e^{2x} -6 - 5x}{x^2}$

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

0/0

7e^x -2e^(2x) -5

,w differentiate 7e^x -e^(2x) -6 -5x

[solid kiln]

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[urban copper]

okay

bottom is 2x

$\lim_{x \to 0} \frac{\frac{7e^x -e^{2x} -6}{x} - 5}{x} \ = \lim_{x \to 0} \frac{\frac{7e^x -e^{2x} -6 - 5x}{x}}{x} \ = \lim_{x \to 0} \frac{7e^x -e^{2x} -6 - 5x}{x^2} \ = \lim_{x \to 0} \frac{7e^x -2e^{2x} - 5}{2x}$

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

7e^x -4e^2x

,w differentiate 7e^x -2e^(2x) -5

[solid kiln]

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[urban copper]

$\lim_{x \to 0} \frac{\frac{7e^x -e^{2x} -6}{x} - 5}{x} \ = \lim_{x \to 0} \frac{\frac{7e^x -e^{2x} -6 - 5x}{x}}{x} \ = \lim_{x \to 0} \frac{7e^x -e^{2x} -6 - 5x}{x^2} \ = \lim_{x \to 0} \frac{7e^x -2e^{2x} - 5}{2x} \ = \lim_{x \to 0} \frac{7e^x -4e^{2x}}{2}$

[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

okay and if you take it from either left or right both are 3/2

quite interesting

[carmine spade]

nice !

[urban copper]

so this function is differentiable

both limits yields the same thing

[carmine spade]

yep

[urban copper]

we know its differentiable at x_0 = 0

and a=5

this is done.

which theorems and criteriums we used tho????????'

this feels like black magic to me . . .

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[delicate lance]

can someone expalin how to do in the most easiest explanation ever

[slim hedge]

fr cant think of a easy way to explan

you're on your own

:/

[delicate lance]

BRO 💔

[hollow frost]

Give me a sec and i may explain it to you!

Ok nevermind my bad this is outside of my knowledge, i know how to solve it but i don't think i'd be the best to explain it, hopefully someone else may answer it for you in better terms

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[delicate lance]

oh okok

[hollow frost]

Okay so it works basically on this formula

Radius times force times sin x (R x F x sinx)

Doing that and applying everything you will get the magnitude of the torque of this force.

[delicate lance]

Ok and how do u find the radius

[hollow frost]

the radius is the initial point of the wrench to it's final point

in other words 14

14 cm

see it as a circular movement

[delicate lance]

ohh

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[ionic pendant]

,rccw

[sleek flare]

How do I get 4e 23 as a decimal number

I'm using a ti 84

[ionic pendant]

well you gave that $-4E-23 = \num{-4 e-23}$

[solid kiln]

cloud

[sleek flare]

How do I get this as a normal number

I think I pressed something on my calculator

[ionic pendant]

scientific notation is the only reasonable way to express the number, since it would have 23 zeros, but you would get $-0.\foreach\x\ in{0,...,22}{0}4$

[solid kiln]

cloud

[sleek flare]

I'm doing a lown formula

How do I get this as a money number

Not scientific notation

[ionic pendant]

i think it is more likely that you entered the formula wrong, since that is not a realistic answer to get

[sleek flare]

Hmm

I got it to work thanks

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[polar quarry]

help

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[scenic roost]

i think x axis label is asking for the period

do you know how to calculate that?

i have to leave soon so ill just tell you

||3/5 tells you the period of the graph, which is always 2pi/ the value given at x, which in this case is 3/5||

||10pi/3 on the x axis||

[polar quarry]

Yes that is right

Im just trying to figure out the whole shift thing

I think i got it, thanks!

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[round galleon]

I dont really know much about calculus but is limit just another way for substitution

[pseudo grotto]

it is if the function is continuous at the limit

but in general no

[round galleon]

oh wdym by that

[round galleon]

tbh this question appeared out of my thought when i saw a yt short for lim x→∞ (xsin(1/x) so i assumed that all we have to do was substituted it but apparently was not all of the process so i assumed there has to be more than meets the eye, but uhm anyways u got vid/book reccs for understanding limits?

[pseudo grotto]

the issue is if you were to just substitute in x=0 you get sin(1/0) whicj is the problem

instead we think about it as "what happens to xsin(1/x) when x is getting extremely close to 0"

as for a resource, I would just look up a khan academy yt vid or something, there's plenty of resources that explains what limits are

[runic dew]

Khan Academy is not the best resource for limits

[round galleon]

oh ty thats actually a decent information

alright ima close then

.close

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[frail heron]

May I see your attempt?

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[full dock]

just post ur question here

[ornate thicket]

alr

.close

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[lilac cloud]

Could someone help with this question? Let me post my working out

[dusty sleet]

When you calculate the determinant, the element -2 should be a +2 (because you did λI **-**A)

[lilac cloud]

ah right

does it make a difference?

since its multiplied by 0

yeah i changed the -2 to a 2 but it doesnt change my answer

it says the correct answer is 1

how do they get a value larger than 1 as the other eigenvalue

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[ruby narwhal]

I have three sides of a triangle, and one of its angles. How can I find the other 2 angles. Keep in mind its required that I calculate everything by hand

[ruby narwhal]

i tried the law of sines, but I got the wrong answer

and now i tried cosine

but im stuck at calculatinig arccos(2.344)

this is something i cant do by hand or with the tools my professor has alowed

[old lance]

can you send the angles and side lengths?

[ruby narwhal]

this is what i am given

and i found the last side

9.7

[old lance]

did you find it using the law of cosines?

[ruby narwhal]

yes

its rounded

[old lance]

well unless it has a "clean" solution, I dont know how you'd go on to find it without a calculator

what I mean by clean is something that can be solved by inspection

[ruby narwhal]

yes

[old lance]

let me try some stuff out, give me a bit

[ruby narwhal]

got it

thank you so much!

[old lance]

how did you find cos(35) by hand?

[silk hull]

a^2=b^2+c^2-*2bc-cos35

[old lance]

yeah but he's not allowed to use a calculator

so trig functions are a bit of a problem

[wraith hinge]

pretty sure arccos(2.344) isn't even possible

[old lance]

yeah its out of its domain lol

I just noticed

[ruby narwhal]

i mentioned the professor has given us some tools

this includes mainly a table with some values

should i send images of those

[old lance]

oh can you send the values?

yeah

[ruby narwhal]

sorry for it being difficult to read

these are in degrees

[old lance]

oh that makes it much easier

you can use the law of sines

and using the table you can find an approximate to sine inverse

[ruby narwhal]

i tried that but had gotten the incorrect answer

maybe i did it wrong

[old lance]

okay i'll try it

[old lance]

yeah try it again

I just did it and it worked for me

here are my steps:

using the law of sines:
$$\frac{\sin(35)}{(10^{2}+16^{2}-2(10)(16)\cos(35))^{\frac{1}{2}}}=\frac{\sin(\theta)}{16}$$

[solid kiln]

proofAd

[old lance]

substituting in the value for cos(35) from the table

and performing long multiplication , we get:

$$\frac{\sin(35)}{\left(\frac{2348}{25}\right)^{\frac{1}{2}}}=\frac{\sin(\theta)}{16}$$

[solid kiln]

proofAd

[old lance]

you can calculate the prime fractors of 2348 and 25 to simplify this to:

$$\frac{\sin(35)}{\left(\frac{2\sqrt{587}}{5}\right)}=\frac{\sin(\theta)}{16}$$

[solid kiln]

proofAd

[old lance]

multiply both sides by 16 and take the sin inverse (also substitute the value of the table in place of sin of 35)

you'll get sin inverse of 0.9469

which rounds to sin inverse of 0.947

you can find this value on the table and look at its corresponding angle

[ruby narwhal]

ok so i had gotten that

[old lance]

you'll find it is 71 degrees

[ruby narwhal]

yes

[old lance]

so the final answer is approximately 71 degrees

[ruby narwhal]

that was my original answer

[old lance]

okay thats corrrect

[ruby narwhal]

but i thought the largest side should correspond with the largest angle

and if B angle is 71, that would mean C angle is 96

[old lance]

how did you get C to be 96?

I got C to be 74

which is fine because we're doing approximations

I calculated 180-71-35

[ruby narwhal]

ok

so i actually just stupid

[old lance]

its fine, we all make silly mistakes lol

hopefully your question is answered

[ruby narwhal]

thank you so much

i am genuinely mad at my self

[old lance]

dont be, the fact that you're checking your answer using common sense after obtainning it is great

I think many people just gloss over that

[ruby narwhal]

thank you thank you

ok

ii must get back to it

have 2 more problems

[old lance]

alright, good luck with your studies and dont forget .close

[ruby narwhal]

.close

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[unborn torrent]

Hello

[unborn torrent]

Please can someone explain what I've done wrong ? Also, why are they finding x ba and not y ba...?(Part 3a)

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[unborn torrent]

<@&286206848099549185>

[unborn torrent]

<@&286206848099549185>

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[unborn torrent]

<@&286206848099549185>

[terse bison]

which part are you struggling with

@unborn torrent

[unborn torrent]

Part a

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[unborn torrent]

<@&286206848099549185>

[tall kernel]

!15min

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[tall kernel]

once

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[jagged fiber]

can someone explain what is going on here i have my final TMROW

😭

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[fierce coyote]

hi

[fierce coyote]

can you help me make it one equation with x and y

[trim jolt]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wraith hinge]

Can you please send the entire question

[fierce coyote]

it says :show that the curve K has two tangent lines at the point P(0,0) and determine the equation of this tangent line

its 27

[fierce coyote]

i think i can go further if i have this but im not sure

are you stil here?

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[dull sandal]

i got a very basic problem with fractions (was unable to learn math for years)

struggling to expand these denominators to the very same value. isn't there any practical way to handle this?

[cyan raven]

Hey! so it would be easiest to compute the lcm of 9,3,7

lcm is the least common multiple!

[last falcon]

in this case it would be 63 right?

We can handle 3 and 9 so i focused on the 9 and 7.

[cyan raven]

Yup!

[last falcon]

Do you know how to have the same denominator when you have got 2 fractions ( starting off basic)

[dull sandal]

my book doesn't cover that but i'll take a look, hank you!

[cyan raven]

lowest common multiple is the lowest multiple shared by two or more numbers. So in this case, @last falcon used the fact that 9 and 3 both "share" a 3, where as 7 shares a 7 on its own. so we want the final denominator to share everything in the denominators!

[dull sandal]

with very basic numbers, like 2, 3, 9 - yes. but couldnt do that for entire hour

[last falcon]

It should, which textbook are you using

[storm burrow]

a multiple is just like when you add a number to itself over and over again

[dull sandal]

i mean, it does - but about 20 pages later

its a non-english math beginner book

[storm burrow]

so like 27 is a multiple of 9

[last falcon]

Oh got it

[storm burrow]

because 9 + 9 + 9 is 27

[cyan raven]

i'm sure this will be explained later on--it is more for building intuition at this stage 🙂

[last falcon]

To be honest i am doing it in a way where i be like okay 3 and 9 got a point imma take 9 and 7 now. Idk how im doing exactly but i get it right 90%

[dull sandal]

yes, you're definitely right

but then, how do i solve this without lcm if i want to follow the syllabus

[last falcon]

tbh you have to follow it because i never saw a way to do it otherwise

[storm burrow]

idk how you would do it without having a common denominator

[dull sandal]

then i'll go learn it first

[cyan raven]

so it's about asking "how do I make the denominators the same"

[dull sandal]

i was like crazy "why i cant do"

but you guys relaxed me

thanks very much!

[storm burrow]

https://www.youtube.com/watch?v=RMJs_I9Nydw

here's a great video explaining lowest common multiples

[dull sandal]

i'll watch!

im grateful

[cyan raven]

we can try to figure out what is outlined in the textbook, if you'd like?

[last falcon]

I know the lcm or lowest common multiple term sounds so mathematical especially if you’re a non English speaker who is trying to learn math from scratch but i swear it is not hard because i am doing it as well

[dull sandal]

just fractions at this section, something called "ladder operations" (book says its just called as this by students, might be a bad translation), multiplication, division etc. with fractions

it did cover basic algebra before

[dull sandal]

ik what you meant in my language

i remember that from middle school

but i will take a look from scratch

[storm burrow]

is english not your first language?

[dull sandal]

no

[last falcon]

whats your first language

[dull sandal]

turkish

[last falcon]

i was just wondering

[storm burrow]

oh wow, you speak very well for a non native speaker lol

[last falcon]

Dude

I am turkish

LMAO

[dull sandal]

cool

[dull sandal]

thank you!

[last falcon]

text me any time you want i’ll try helping. I am texting in english here to not to break rules

[dull sandal]

i'll text you, ty!

[last falcon]

you can add me if you want its up to you

[dull sandal]

so im now closing this

[last falcon]

you’re welcome!

[dull sandal]

thank you all

[last falcon]

alright see ya!

[dull sandal]

.close

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[fallow tapir]

how would I do a problem like this? i dont really know where to start and my teacher was gone during this lesson so it was never really explained.

[keen igloo]

I think you just need to give a value for x, and then solve the math and plot the point

[sweet hare]

just pick 2 points for x, e.g., 0 and 1

or 1 and 2

[keen igloo]

You can give 0 to x, it's easier

[fallow tapir]

i see, thank you both!

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[wraith hinge]

for this

[wraith hinge]

this equals

-57/16

why cant i multiply by q/2

[sweet hare]

what do you mean?

[wraith hinge]

like

[wraith hinge]

this

is equal to -57/16

why cant i just multiply the denominator (q/2)

[sweet hare]

you could

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[rapid horizon]

Hints please

Yes

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[prime forge]

so i'm a little stuck on this problem
it's an improper integral, i've splitted it in two: from 1 to 3pi/2 and from 3pi/2 to inf and i've found the limit of integrand as x approaches inf but i'm not exactly sure what to do with the part from 1 to 3pi/2

[prime forge]

i've also looked at the graph of the integrand in desmos and it looks like it's divergent because the area under the [1;3pi/2] part looks infinite but i don't understand how to prove it mathematically

<@&286206848099549185>

oh sorry i didn't read that it's for pre-university questions, that's quite advanced for that

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[zealous ravine]

why'd they do this step

[zealous ravine]

is it because if an equation has no real roots the only way to find out which way the parabola is facing is trhough the constant term?

if constant term is +ve then its upward facing and downward for -ve?

[jaunty talon]

yeah, since c>0, that would mean that the parabola would be on the upper part.

honestly use use determinant

b^2 -4 ac < 0
(4b)^2 -4(3a)c<0

since you know c is positive

you know which signs

16b^2 -12ac <0

[trim jolt]

@zealous ravine Has your question been resolved?

[zealous ravine]

Ahh this would mean a is positive for it to be possibly less than zero

which means its facing upward

?

right

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[last kraken]

actualy

[last kraken]

@warm minnow come here

someone already claimed 15 lol

Can you explain to me how you got to your answer?

[warm minnow]

Ale

[last kraken]

aight

[last kraken]

there we go

[warm minnow]

As I was saying

If radians = arc length / radius

Than number three is not true because the radius is on top and the arc length is on bottom

Is that right

[last kraken]

yes

but what's confusing you with 1 and 2?

[warm minnow]

Than number 4 would be true because the radius is on bottom and arc length is on top

[last kraken]

true

[warm minnow]

I just don’t understand it

What does it mean by dilation and factor of 10

What does factor of 10 mean

[last kraken]

So

[warm minnow]

Does it mean multiplied by 10

[last kraken]

basically

Circles are always proportional to one another

They just change in size

[warm minnow]

Because there all 360°

[last kraken]

mostly bcuz of their symmetry

so the only factors that change are the radius and their circumference

[warm minnow]

Ok

[last kraken]

and to compare one circle to another

we can compare their radius with a proportionality factor

[warm minnow]

Right

[last kraken]

so basically what the box says is "is cicle B 10 time bigger than A?"

[warm minnow]

No

[last kraken]

why?

[warm minnow]

Because 10x bigger means it’s 3 x 3 10 times

Or no

[last kraken]

Actually that would be 3^10 times bigger

[warm minnow]

Oh

So it would be true because 10x 3 is 30

[last kraken]

correct

[warm minnow]

And 10x 2 is 20

[last kraken]

correct

and now for the second box, what is it that makes you confused

[warm minnow]

It doesn’t make sense

[last kraken]

why is that

[warm minnow]

It says the ratio l/r is = to ration l/r

There the same thing so of course it would be equal

[last kraken]

actually

[warm minnow]

But I feel like there a trick in it

[last kraken]

it says is l/r = L/R

different letters mean different variables

and usually mean different values

if you look to your paper, you might notice that the variables associated with their arc lenghts and radii have a variable assigned to them

namely l, r, L and R

[warm minnow]

I see

[last kraken]

in this case they capitilized the l and the r for the B circle just bcuz it is "the big one"

[warm minnow]

L = 20 R = 30 l = 2pie r = 3

[last kraken]

correct

well

L = 20*pi and l = 2*pi

just don't forget pi, it is a number as well

[warm minnow]

Ok

[last kraken]

now that we've cleared the misunderstanding

would you mark that box?

[warm minnow]

Well it ask if there = there not the same sized circles

Because one is bigger

But they use the same formula

So when it asked if there the same are they talking about the size or the formula

[last kraken]

they are asking about the ration

that is

l/r

so the value that comes out of it

[warm minnow]

So there not the same

Because there different numbers

[last kraken]

try calculating l/r

what value do you get?

[warm minnow]

Wym

How would I calculate them

[last kraken]

it is a division

[warm minnow]

Do I do 20pie divided by 30

[last kraken]

correct

actually

let me use a different approach

do you know how to compare fractions?

[warm minnow]

No

Actually yeah

I think

[last kraken]

how would you compare fractions?

[warm minnow]

So let’s take 1/3 and 3/4

[last kraken]

alright

[warm minnow]

You would multiply 1 x 4

And 3x3

So 9 and 4

And 9 is bigge

R

[last kraken]

correct

so if you do this for l/r and L/R, to what conclusion do you get?

[warm minnow]

That the L/R is bigger

Than l/r

[last kraken]

why?

[warm minnow]

Wait

There. =

So number 2 is fire

True

[last kraken]

correct

So I guess that's it

.close

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[lapis cosmos]

help, im very stuck here

[lapis cosmos]

I set 2 sin - 1 aside and work with the tangents, adding and sustracting I get to 1 / cos - sen

[mighty canyon]

ok

[lapis cosmos]

then 2 sin - 1 transforms to - 2 cos, and so, -2cos/cos-sen

and thats where I dont know how to progress from, if its correct until there

[hybrid vine]

Are you aware of trigo formula for multiple angles?

Cos 2x = Cos^2 2x - Sin^2 2x

[lapis cosmos]

I am

[mighty canyon]

then you probably know what $2\sin^2(\theta) - 1$ will be

[solid kiln]

print("NAME")

[mighty canyon]

you probably know that you have to either get the tan term to be that or get sin term to be tan term

and if you have goal in your mind, then the simplification can be made easier

so, you are correct that you would get something that looks like $\frac{1}{\cos^2(\theta) - \sin^2(\theta)}$

[solid kiln]

print("NAME")

[lapis cosmos]

and that'd go to cos 2x, I assume

[mighty canyon]

yes, but there is an easier way to solve this

actually...

yeah, that is one way

another way will be ugly but more straightforward

so you got $\cos(2\theta)$ here and what is $2\sin^2(\theta) - 1$?

[solid kiln]

print("NAME")

[lapis cosmos]

im getting -2 cos^2

as in, sin^2-1 + sin^2 -1 = -cos^2 - cos^2

[mighty canyon]

notice that $1-2\sin^2(x) = \cos(2x)$, right?

[solid kiln]

print("NAME")

[mighty canyon]

that is double angle formula. Then, what you have is $2\sin^2(\theta) - 1 = -(-2\sin^2(\theta) + 1 )$, agree?

[solid kiln]

print("NAME")

[mighty canyon]

and you almost have it

[lapis cosmos]

how do i get from cos^2 - sin^2 to this? or is it just a different formula

[mighty canyon]

ok... so you know $\cos(2x) = \cos^2(2x) - \sin^2(2x)$, right? you can write $\cos^2(2x) = 1-\sin^2(2x)$, agree?

[solid kiln]

print("NAME")

[mighty canyon]

it is just $\sin^2(x) + \cos^2(x) = 1$ identity

[solid kiln]

print("NAME")

[mighty canyon]

move sin term to right side and done

[lapis cosmos]

done, its -1, I understand

thank you very much, I always get in disbelief that all trigonometry comes from simple formulas an think that what im missing comes from somewhere else

[hybrid vine]

||Isn't it +1? ||

[lapis cosmos]

but its always the same few things manipulated

[mighty canyon]

-1 indeed

[hybrid vine]

Oh yea it is mb

[lapis cosmos]

thanks for the help 🙂

.solved

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[past meadow]

https://cdn.discordapp.com/attachments/880616249283973140/1245951048695152710/image.png?ex=665a9dd6&is=66594c56&hm=9723d5b287a19f61f12f28c0e63f2b5fea5aae916f038ccb312b54e3d8350d09&

[past meadow]

i believe the answer is 3

[eternal token]

you would be correct. what do you need help with?

[silent oyster]

correct?

[eternal token]

am i tripping

one moment

[austere cedar]

No using ÷

[past meadow]

well its still PEMDAS

[silent oyster]

BODMAS

[past meadow]

or if / and X on same line its still left to right

not right to left

[silent oyster]

I THINK 48

[silent oyster]

? r u writing a speech?

[eternal token]

(18/(1/2*4)/)3

[past meadow]

1/2 doesnt mean .5 in this case

[silent oyster]

why did u put brackets??

[austere cedar]

Nope! Just reminding people that ÷ is unclear.

[eternal token]

it is how i am interpretation the order of operation of that image

[silent oyster]

so whats the answer?

[austere cedar]

We don't use it past basic PEDMAS lessons. It's not real math

[silent oyster]

no its wrong

[austere cedar]

There's no answer as long as ÷ is there

[silent oyster]

really?

so r u telling pedmas is useless?

[past meadow]

oh boy

i feel really lost here

[austere cedar]

If that's what you're hearing

But I'm pretty sure I only said ÷ is not a symbol that can be used

[silent oyster]

ok

[mighty canyon]

here we go again... I think the ambiguity is not on the division symbol but on the lack of bracket

[austere cedar]

I mean sure, if it was bracketed through the roof that would also fix it

[past meadow]

https://bidasci.is-ne.at/6B_9ZIUiP.png

this is someone's possible answer

[mighty canyon]

like $18 \div \frac{1}{2}$ has one meaning

[solid kiln]

print("NAME")

[past meadow]

so its left to right

not right to left

when / and X are on the same line

[mighty canyon]

it does not matter... because you can swap the multiplication order

and division is a kind of multiplication

like $ab = ba$

[solid kiln]

print("NAME")

[past meadow]

so that is 9

[mighty canyon]

so $abc = cba = cab$

[solid kiln]

print("NAME")

[mighty canyon]

it does not matter.

here, if I were to do this, I would have (18)(2)(4)/3 = 48 as in the wolframalpha

[silent oyster]

but 18 divided by 1/2 and 1/2 divided by 18 are not same

[mighty canyon]

$a\div b = \frac{a}{b}$

[solid kiln]

print("NAME")

[mighty canyon]

so $18\div \frac{1}{2} = \frac{18}{\frac{1}{2}} = 18 \times 2$

[solid kiln]

print("NAME")

[mighty canyon]

you are wrong, you need to write that in terms of multiplication first

not wrong that both are equal

but wrong to not see that you can swap the operation under the multiplication

[silent oyster]

i just said they r not equal

ok ok

[mighty canyon]

like $a \div b \times c = \frac{a}{b}c = a \frac{c}{b} = a(c\div b)$ that is my point

[solid kiln]

print("NAME")

[silent oyster]

ok

[mighty canyon]

so yes $a\div b \neq b \div a$ but it does not matter you calculate the nominator first or denominator first

[solid kiln]

print("NAME")

[silent oyster]

...

[mighty canyon]

Like, you need to understand first that addition and subtraction are the same operation under the real number. You need to understand that you can swap the order of additions. You have to understand first that multiplication and division is the same operation under the real number.

so on and so forth

[silent oyster]

thts basic math right

[mighty canyon]

yes. and that way you don't need to debate on PEMDAS BOMDAS BODMAS or whatever mnemonic people use these day

[silent oyster]

well these r introduced to just make it easy

[mighty canyon]

but also make people trippy on this kind of question

[silent oyster]

u know all the pedmas and bodmas come to the same conclusion

[past meadow]

this is so confusing

because PEMDAS goes out the window

does not apply here

[silent oyster]

its not pemdas

its pedmas

[mighty canyon]

see?

confusing

[silent oyster]

yea haha but

its because they jumble the letters

[past meadow]

ah

[silent oyster]

they do not remeber the real thing and jumble the letters and say its wrong

[mighty canyon]

that's my point, if we just remember this, it is bad

[silent oyster]

hmm

[mighty canyon]

if you understand that, however, good thing

[silent oyster]

i usually follow ur approach

but u cant explain tht to evryone right so i just go saying pedmas or bodmas

[mighty canyon]

I mean, PEDMAS PEMDAS or whatever is useful when you have multiple incompatible operation like addition and multiplication together. When it is like this is like you have $abcd$ and you can just permute it to get the same answer. Can it be applied, obviously yes. But it is misleading for this kind of problem

[solid kiln]

print("NAME")

[mighty canyon]

not that you get the wrong answer but you got my point. conceptually you lose the way to simplify things which is the needed skill for higher level mathematics.

[past meadow]

so dont always use PEDMAS?

[mighty canyon]

PEDMAS PEMDAS is there to help guide you

just not a dictate rule

like PEDMAS and PEMDAS are equivalent

[silent oyster]

ur right many get confused

[past meadow]

what i did wrong was not working left to right then

i would of had it right

if i worked left to right with / and X

[mighty canyon]

well, even if you don't, you still can get the right answer

[silent oyster]

its not left to right

or right to left

[past meadow]

well if we did 18 / 1/2

first

and that part was left

[mighty canyon]

see: $\frac{18\div \frac{1}{2} \times 4}{3} = \frac{18 \times 2 \times 4}{3} = 6 \times 8 = 48$

[solid kiln]

print("NAME")

[mighty canyon]

I don't even do left to right, I even jump out and divide by 3 before I finish my calculation

[silent oyster]

haha

i think @past meadow is confused rn

[mighty canyon]

yeah, this is the first time I guess that someone told you PEMDAS is wrong

[past meadow]

oh you basically flipped the factions

[mighty canyon]

yes.

[past meadow]

ah i see

now by that

its much clear

[mighty canyon]

the main point for this problem is to see that $\div \frac{1}{2} = \times 2$

[solid kiln]

print("NAME")

[past meadow]

yep with that understanding

it all fights into place

[mighty canyon]

basically yes

[silent oyster]

thanks @mighty canyon

[past meadow]

thank you all

oh boy, i am actually past this math

but its been a minute since i touched algebra

[silent oyster]

haha adavnced i see

[past meadow]

just terrible how we forget simple stuff

ugh!

[mighty canyon]

well, you definitely want to ditch out those PEMDAS concept when you study advanced stuff. Not like you don't need to know order of operation but hell if you follow that, your life will be too hard

[silent oyster]

haha yea i followed it but while doing operations i stoped using it idk when

[past meadow]

oh boy

i am greatful

for the help

[grim sparrow]

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[gaunt temple]

Hello

[gaunt temple]

Can somebody please proof check this?

[marble wharf]

dont write max if you dont know that a set has a max. thats the whole point behind introducing sup, because that one always exists

you have only shown that x is greater than supA and supB. not that it is equal to the bigger one of those

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@gaunt temple Has your question been resolved?

[gaunt temple]

Thank you for your reply. What about this?

x = sup(AUB)

Then x is an upperbound of AUB and x>=y for all y in AUB. Since y is in AUB, y is in A or y is in B. Hence, we have that x>=a for all a in A and x>=b for all b in B.

If x is in AUB then x = sup(AUB) = max(AUB).
Without loss of generality, suppose x is in A. Then x = max(A) = sup(A).
Now suppose x is in B. Then x = max(B) = sup(B).
Since x = max(AUB), we conclude that x = max(sup(A), sup(B)).

If x is not in AUB (i.e. AUB doesn't have a maximum), then we have three cases:

1. A has a maximum and B does not;

2. B has a maximum and A does not;

3. both A and B don't have a maximum;

4. x > max(A) = sup(A) and x >= sup(B). Since x is not in AUB it must be x = sup(B).
   Therefore sup(AUB) = sup(B).

5. Using the same logic, we get sup(AUB) = sup(A).

6. x >= sup(A) and x>= sup(B).
   Therefore sup(AUB) = max(sup(A), sup(B)).

Combining case 1) and case 2) we also get that sup(AUB) = max(sup(A), sup(B)).

[marble wharf]

dont do cases about whether maximums exist or not

only work with supremums

you still have only shown that sup(AuB) >= max(sup(A),sup(B))

[trim jolt]

@gaunt temple Has your question been resolved?

[gaunt temple]

Sorry, I don't understand why is it wrong

For example, why is this wrong? "x > max(A) = sup(A) and x >= sup(B). Since x is not in AUB it must be x = sup(B)"

Sure I'm saying that x is greater than sup(A) and sup(B) but since AUB is defined as the union of A and B, sup(AUB) must be either sup(A) or sup(B)

therefore sup(AUB) = max(sup(A), sup(B))

[marble wharf]

why cant it be something else

you arent using the property that the sup is the least upper bound

you have to use that

[marble wharf]

its not wrong. but also not helpful. the third case works anyway

[gaunt temple]

can you make an example of two nonempty sets A and B s.t. sup(AUB) is not equal to neither sup(A) nor sup(B)

[marble wharf]

well I mean of course it cant because you are asked to prove that it is

but the point is your proof doesnt show that it has to be

[gaunt temple]

so like this

x = sup(AUB)

Then x is an upperbound of AUB and x>=y for all y in AUB. Since y is in AUB, y is in A or y is in B. Hence, x is an upperbound of A and an upperbound of B.

Let epsilon be greater than 0.

x - epsilon = sup(AUB) - epsilon must be in AUB.
Hence, x - epsilon must be an element of A or an element of B.

[marble wharf]

if you have A=[0,1] and B=[0,2] and set x=3, then it also satisfies these equations

[gaunt temple]

then i'm stuck

[marble wharf]

but clearly 3 is not the supremum of AuB

ok better now with the epsilon

but x-eps does not have to be an element of AuB

but, there needs to be an element y in AuB such that x-epsilon < y

[gaunt temple]

but we said that x is sup(AUB), so we can only set x=2 here

[marble wharf]

we dont know what sup(AuB) is yet

currently x=3 also satisfies all the other equations you wrote down

what it doesnt satisfy is that it is the least upper bound of AuB. but your argument didnt say anything about having to be that

[gaunt temple]

yeah right

[gaunt temple]

okay makes sense now

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[desert osprey]

I’ve got that a must be divisible by b but I’m not sure how I’d count the number of integer divisors for each value from {1..25}

[frozen plover]

eh casework isnt too bad here

[desert osprey]

What do you mean?

Manually count each number!

?*

[frozen plover]

yeah

25 --> 5
24 --> 2, 4, 6, 8, 12
23 --> ...
22 --> 2, 11

its easy to list them all off

[desert osprey]

Is there not a more efficient way?

the solutions state this but I don’t understand it:

I can’t tell if that’s a variation of the pigeon hole principle

I don’t understand how they’ve derived that summation

[desert osprey]

Ig it wouldn’t take that long and that’s what I was tempted to do

[eager summit]

$\log_{m^b}m^a = \frac{a}{b}$

[solid kiln]

SirGareth

[eager summit]

slightly better

[desert osprey]

k=a/b where k is an integer

I get that part

[frozen plover]

yeah me neither

[acoustic kraken]

can someone help me with the twin prime conjecture

[desert osprey]

.

[desert osprey]

Do you understand how they’ve got that?

[eager summit]

,rotate

[desert osprey]

,rotate

[desert osprey]

ah

[eager summit]

@solid kiln rotate

;-;

[eager summit]

,rotate

[eager summit]

uh

idk

[desert osprey]

Pain

[eager summit]

wait its just number of multiples of j right

[desert osprey]

Yea I think so

Honestly I think exhaustion is probably the best method

I don’t understand how they’ve derived that

[eager summit]

a =/= b right

[desert osprey]

yea

[eager summit]

yeah normally number of multiples of b till a is a/b

(for example number of multiples of 3 till 30 is 30/3)

[desert osprey]

yea

[acoustic kraken]

@desert osprey after you finish with this guy can you help me

[eager summit]

however here since a =/= b you subtract the -1

[desert osprey]

It’s the other way round and there’s really free help channels

[desert osprey]

Then what’s with the floor function

[eager summit]

yeah its not necessarily an integer

suppose multiples of 3 till 31

would be floor of 31/3

[desert osprey]

Oh I see

So they’ve just chosen 25 and calculated all the factors of it

well multiples

[eager summit]

yep

[desert osprey]

that is slightly confusing but i kinda see how it works

thank you

[eager summit]

np

[desert osprey]

.close

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[torpid magnet]

?????

[runic dew]

There seems to be a mistake in the question.

[torpid magnet]

i know that its asking for an angle measure of 180

[torpid magnet]

why?

[runic dew]

456pi/6 is extremely large

converting to degrees

we get

[torpid magnet]

it asks in radians

[runic dew]

180/pi * 456pi/6

[torpid magnet]

a measure of 720 is the same as a measure of 360 despite not being on the circle

this is normal unit circle stuff but i forgot how

to do it

it just measures in rotations

or im delusional 😭 idk i hate sat math

[lucid burrow]

Oh maybe the answer options are modulo 2pi ?

[runic dew]

This 13680 degrees. I disagree with treating 360 degrees as 720 degrees.

[acoustic flint]

Do you know what the angle abc would make between 0,2pi

[shy nacelle]

Answer should be pi or any odd multiple of pi

[torpid magnet]

180

[shy nacelle]

So its c

[torpid magnet]

wait

but how did u convert it

[shy nacelle]

,calc 462/6

[torpid magnet]

idk what im supposed to do w that number