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[high ether]

Write it where

Ahhhh

[minor needle]

so we put it there deliberately

[high ether]

Thank you very very much

[gray isle]

you can't quite write 5x + 5y - xy as a product either,
if you were on the other hand asked to find the integer solutions to the equation
then it'd be worth looking up something like Simon's favourite factoring trick

[minor needle]

and now we have to stay with sense

[high ether]

Its way easier than i thought

Thanks again

.close

[lone heart]

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[high ether]

Yeah i was asked this

[gray isle]

...

i asked multiple times whether factorise equation was the original question
and it turns out it was indeed not

[lone heart]

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[lilac thicket]

hey everyone , I would appreciate your help "debugging" my solution, image follows

[lilac thicket]

idk where I did something wrong. The solution should not be correct, as the values in our solution sheet differ from mine

nevermind

.close

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[high ether]

How to show that (gcd(a,b))^2=gcd(a^2,b^2)

[wind cloak]

Consider their prime factorization

[modern sedge]

is this even true?

Cause I dont think so

[high ether]

I am not sure

[modern sedge]

its true iff a and b are coprime I think

[high ether]

But i feel it is

[modern sedge]

Counter example: gcd(2,4)=2 gcd(4,16)=4

[high ether]

Sry i am meant the gcd ^2

[modern sedge]

Alright

then do what NEONPerseus said

[high ether]

Yeah... Okay

Thanks

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[stable warren]

I need help with this step of this question:

[stable warren]

How is √40 = to 2√10

Please help

[gray isle]

consider sqrt(ab) =sqrt(a)sqrt(b) for a,b >= 0

and identify perfect square factors of 40

[stable warren]

what are perfect square factors

plz

@gray isle

[gray isle]

factors that are perfect squares

[stable warren]

bro there are no perfect squares of 40

:/

[gray isle]

can you list the factors of 40

[stable warren]

1, 2, 4, 5, 8, 10, 20, 40

@gray isle

[gray isle]

dw about the 1 since it has no impact on simplification,
is 2 a perfect square?

[stable warren]

um

2 is a surd??

[gray isle]

2 itself is an integer

i'm not asking you to take square roots yet

only to identify whether something is a perfect square, (square of an integer or not)

[stable warren]

no 2 is not a perfect square then

only 4 is

:/

[modern sedge]

So you can write 40 as 4x10, right?

[lone heart]

@stable warren Has your question been resolved?

[stable warren]

yes

but but he wrote it as 2√10

that is my question

[gray isle]

after writing 40 as 4 * 10

consider sqrt(ab) =sqrt(a)sqrt(b) for a,b >= 0

[stable warren]

$sqrt(ab) =sqrt(a)sqrt(b) for a,b >= 0$

[ocean seal]

TechySkills

[stable warren]

brekh

$√ab = √(a) . √(b)$

[ocean seal]

TechySkills

[stable warren]

wait

so

$√4 = √2^2 = 2$

[ocean seal]

TechySkills

[stable warren]

$2√10$

Now i understand

[ocean seal]

TechySkills

[lone heart]

@stable warren Has your question been resolved?

[stable warren]

.close

[lone heart]

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[steady salmon]

if i have 2 parallel vector how can i find the distance vector between them

[balmy light]

are your vectors 2D or 3D?

[steady salmon]

3d

it is parallel sorry

$u=i+2j+3k
\n v=3i+4j+5k$

[ocean seal]

1667
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[balmy light]

Take the dot product of A and B. Since they are perpendicular, their dot product will be zero:
A · B = 0

[steady salmon]

no i need the red vector

[balmy light]

could you dm me?

[steady salmon]

no

[balmy light]

|u - v| = sqrt((u - v) · (u - v))
u - v = (i + 2j + 3k) - (3i + 4j + 5k)
= -2i - 2j - 2k
(u - v) · (u - v) = (-2i - 2j - 2k) · (-2i - 2j - 2k)
= 4i^2 + 4j^2 + 4k^2 + 8ij + 8ik + 8jk
= 4 + 8ij + 8ik + 8jk
(u - v) · (u - v) = 4 + 0 + 0 + 0
= 4
|u - v| = sqrt((u - v) · (u - v))
= sqrt(4)
= 2

feels wrong, not sure

[sharp girder]

Aren't vectors freely movable in space? You should be given lines corresponding to them to find the dist.

[steady salmon]

seems complicated lemme send the question

[sharp girder]

..

[steady salmon]

i thought that i can find the dist vector betweem them then find equation

[sharp girder]

Lol why did you need dist. For this

Look for line you need a point and a direction

[steady salmon]

i dont know how to find with perpendicular

[sharp girder]

As your direction is perpendicular to the vectors just take their cross product

The cross product of two vectors represents a vector perpendicular to those two vectors

[sharp girder]

Then write the equation of line in standard form

[steady salmon]

yeah yeah now get it

i am learning the formulas forgeting the theory part thats bad, now get it man

thanks

[sharp girder]

Its fine keep practicing 👍

[steady salmon]

appreciate

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[solid swan]

Why is value taken to be 2 in poison and not rounded up or down

[lone heart]

@solid swan Has your question been resolved?

[mellow grail]

Could you type it..?

[lone heart]

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[frosty kayak]

why is physics impossible to learn?

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@frosty kayak Has your question been resolved?

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[desert marlin]

how to do 2.4

i have AC from 2.2

how do i workout 2.4

[pseudo ice]

Work out BP, and do the dot product of that and AC, setting it to zero?

[stable drift]

For 2.4 you need the following:

1. Find out AC. This can be done as you know AC=OC-OA.
2. You know OP and OB. find out BP (in terms of alpha beta and gamma)

3. Their dot producr is zero

As cos(90)=0, so I think you will be able to establish relation between alpha beta and gamma

[pseudo ice]

[desert marlin]

i get (alpha-6,beta-4, gamma+1) for BP

i have 3 unknowns how do i do dot product

[pseudo ice]

What was your AC you had?

[desert marlin]

-8,0,4

[pseudo ice]

Multiply each coordinate and add them together

[stable drift]

-8alpha-48+4gamma+4=0. Divide both sides by 4 and you get the answer

[desert marlin]

i whould multipluy -8*(alpha-6)+...

[pseudo ice]

Like that, yep

[stable drift]

Its gamma-1 i think

[desert marlin]

no because gamma--1

[stable drift]

Oh yeah, you're right.

[desert marlin]

Is that right?

[stable drift]

Precisely.

[desert marlin]

I have 2alpha-gamma=11

But I can't use that at this point

[pseudo ice]

Did you show that?

[desert marlin]

Oh nvm

[stable drift]

?

What do you need that for?

[desert marlin]

I misunderstood question

[stable drift]

I see.

[desert marlin]

OH

GOt it

Lol

Dam I was overcomplicated it

THANK YOU SO MUCH

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[balmy light]

Let S be the set of all positive integers whose base 10 (decimal) representation consists of one or more 1s followed by one or more 0s. For example, 11100 and 10000 and 10 are elements of S, but 111 and 111101 are not. Determine the number of elements of S that consist of at most 100 digits and are divisible by 2020.

I just need help starting this, haven't been more lost than this

[lone heart]

@balmy light Has your question been resolved?

[prime badge]

i kinda got it

like if you divide 11110000 by 20 you get 555500
and that has to be divisible by 101
but it's usually not
only when it's like 550055..0055

hm

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[ivory igloo]

wait

.reopen

oof

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[balmy light]

waitt

[balmy light]

.reopen

@ivory igloo @prime badge any luck?

i barely understand whats going on

[ivory igloo]

wait im sending

1 min

[balmy light]

yes no worries

[prime badge]

my thing solves it, but it's maybe not the intended way

[balmy light]

i mean probably just share the method since atm i plan on skipping this question

[prime badge]

the "method" is that all such numbers are A × 2020 where A is 55, 5500, 550055, 55005500, ...

so it's approximately 100/2

[balmy light]

didnt they say the element of S has to be in form of 11100...

[prime badge]

you multiply by 2020

[balmy light]

yeah but i dont think thats what the question is asking

[prime badge]

what do you mean

[balmy light]

lets wait for GameSwitch and see what he has to say

[balmy light]

the first part, the base 10 thing

[prime badge]

the question is asking that, it's just my solution is relying on an unproven claim

[ivory igloo]

2020 is a 20 multiple. So, the elements are 20 multiples. The only way to achieve this is by setting the final two digits as 0 (since its a 100 multiple now)
So, now we need to find at most 98 digit numbers which are 2020/20 = 101 multiples
Here, instead of factorizing and using divisibility rules, we can think about the format of 101 multiples.
Let us take a number, for example, 111.
Now, 111 * 101 is just 111 * (100+1) = 111 * 100 + 111 * 1 =

11100
+ 111
-----
11211
-----

We can visualize it as shifting the number two digits left and adding it to the original number. Using this thought, we can understand that numbers like 11, 110011, 11001100, ... when multiplied by 101, do not give any numbers with digits other than 1. For example:

110011 * 101 =
11001100
+ 110011
--------
11111111
--------

So, the key idea is to multiply numbers of form 11001100... with 101 to get the numbers we need
for 11, we will get 1 repeated 4 times; for 11001100, we get 1 repeated 8 times
So, the numbers desired are 111.. (4 multiple) number of times, and some number of 0s added to them
All the possibilities are (excluding the two final 0s):
1 (four times) and 0 (0 to 94 times)
1 (eight times) and 0 (0 to 90 times)
.
.
.
1 (96 times) and 0 (0 to 2 times)
So, number of elements are 94+93+...+2 = (96-1) + (92-1) +... + (4-1) = (24+23+...+1) * 4 - 24 = 300 * 4 - 24 = 1200 - 24 = 1176

done

[prime badge]

oh right, i forgot that step, you can add more zeroes

[balmy light]

give me a min, lemme read this and text back

[ivory igloo]

the answer's verification 👇

i accidentally edited it to 1152, now edited it again, so check again if you read it as 1152 already (:

(i mean the answer)

[balmy light]

waittt

@ivory igloo can i please DM you?

[ivory igloo]

yes

sure

[prime badge]

94 numbers that look like 00055000, 90 numbers that look like 0000055005500000 etc.

not saying 1176 is wrong

actually 95, 91...

yay

[balmy light]

@prime badge if i may ask, how did you arrive to 4n-1?

[prime badge]

it's made up

i arrived at 95, 91 same way as gs did

it has to be 4n to go with a step 4 and it has to be − 1 to reach 95 instead of 96

[balmy light]

let me read all this once again with a clear mind

ill get back to both of you once i start understand whats going on here

[prime badge]

we did exactly the same thing but i think neither of us can explain it

they said a number becomes like 1110000 when multipled by 101 only if it starts like 110011 and then zeroes

it just visually makes sense

[balmy light]

yeahh i mean i kinda understood the explanation of 101 and all, its just when the possibilities came in, i dont know what the possibilities are of and its a bit confusing there

[prime badge]

so i'm multiplying by 2020

that makes a number 4 digits longer

so my longest 55005500... number is 96 digits long

[balmy light]

one min, so 55005500. is a element of S?

[prime badge]

no

[balmy light]

its the product of those numbers that is a element of S (some number * 2020) = S

[prime badge]

it's an element of s divided by 2020

yes

[balmy light]

okay okay i see your approach, you are getting the multiples by just multiplying the 5 and 0 to get 1100 combinations

[prime badge]

i'm not sure what you mean

[balmy light]

can we talk on DM?

[prime badge]

sure

[lone heart]

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[hollow hawk]

Hello on this problem I got .2 but I’m not sure if I did it right

[hollow hawk]

“A new restaurant is trying to attract customers and has decided to run a promotion. During their grand opening celebration, everyone who stops by will be given one ticket which contains an area that can be scratched off to reveal a single prize.
Consider a case where 50% of the tickets reveal a discount, 20% of the tickets reveal a free beverage, 15% of the tickets reveal a free appetizer, 10% of the tickets reveal a free dessert, and the remaining tickets reveal a free entree.
What is the probability that the individual who receives the first ticket will win either a free appetizer or free entree (assume that this individual is able to pick one of the tickets at random)?”

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@hollow hawk Has your question been resolved?

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[wind hawk]

Is there anything fundamentally different about a function mapping whose image is [0, 1] but codomain is the reals, and the same function but with a codomain and image of [0, 1]?

[mortal trellis]

well one is surjective and the other isnt

[lone heart]

@wind hawk Has your question been resolved?

[wind hawk]

Lol yep that’s a good point

Dunno why I didn’t think of that

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[wintry birch]

Hi how do I solve sin 3x = sin 2x + 5 sin x

[wintry birch]

I tried to use simpson so I get 2 cos 5x/2 * sin x/2 - 5 sin x = 0

[tacit arch]

Simpson?

[wintry birch]

Yea the formula of sin x - sin y = ...

The formulas of Simpson?

[trim wagon]

nvr heard of "simpson"

[wintry birch]

Wait

[trim wagon]

[wintry birch]

This

[trim wagon]

why not just use sin(3x) formula

and sin(2x)

[wintry birch]

What is the formula of 3x?

I don't think we have learned that one

[trim wagon]

sin(3x) = sin(2x+x)

[wintry birch]

Ooooh

Let me try that

[wintry birch]

So I am trying the 2x+x way bc we haven't learned that formula but then I get sin 2x * cos x + cos 2x * sin x and idk how to simplify it

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[karmic osprey]

https://media.discordapp.net/attachments/373217754666500097/1079146683079864381/image.png

[karmic osprey]

to find k(1)

i use the quotient rule

but the thing is g(1) looks like its near 1.25

but i dont know if i should use that number because its not sitting at one of those points

because to find f'(1) i was going to use the point (0,0) and (2,0.5)

but im not sure if theyre the right numbers to use

because i estimated

for 0.5

[marsh rapids]

The slope of that line is 4/3. That gives you g(1)

[karmic osprey]

how did u find the slope

[marsh rapids]

Look at the endpoints to figure it the precise equation of the line

[karmic osprey]

what do u mean by endpoints? can u take a screenshot and label them?

[marsh rapids]

The non-differenriable points

Endpoints of the middle segment of g's graph

[late ridge]

By any chance. Do you know what type of equations or format is called and could help to explain it to me

[karmic osprey]

wym?

[marsh rapids]

You're learning derivatives. You know how to find the equation of a line. Cmon

It goes from (-1.5, -2) to (1.5, 2)

[karmic osprey]

o

sry

i just dont know vocab words

like endpoints

😅

[marsh rapids]

Point at the end

Sometimes English is actually a well made language

[karmic osprey]

when u said endpoint i thought u meant like

i thought thats what u meant by endpoint

😅

but thanx

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[candid torrent]

ummmmm

i can try to help

i think you should start be subbing 1/2 * x into cos(2theta)

yeah

so that would give you 2cos^2 (1/2 * x) - 1

consider that cos^2(theta) = (cos(2theta) + 1)/2

sub 1/2 x into theta

and you can solve

that's fine

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[sharp sonnet]

how would I solve that

I tried using the pwoer rule it didn't work

[lone harness]

close

[lone harness]

you differentiated the first term, check the other terms again

[sharp sonnet]

so would it be 77x^76+4

[lone harness]

not quite

what does -x differentiate to?

[sharp sonnet]

would it jsut be x

so 77x^76+x+4?

[lone harness]

no

[sharp sonnet]

how would you do the power rule on x

[lone harness]

(-1)x differentiates to -1

[lone harness]

x¹ differentiates to 1×x⁰= 1

and then you can use the fact that d/dx kf(x) = kf'(x)

[sharp sonnet]

wait what is that for

[lone harness]

it's just a general rule

[sharp sonnet]

so my final answer would not be 77x^76-1+4

[lone harness]

no because 4 differentiates to 0 remember

a constants derivative is 0

[sharp sonnet]

ahhh cuz there is no exponent so its 0 and 0x4 is 0

[lone harness]

this is just a rule

I think maybe do some recap on differentiation

[sharp sonnet]

yea no we are just starting this topic and we were assignned this homework we started this yesterday

[lone harness]

okay makes sense

so what's the final answer

[sharp sonnet]

77x^76-1 ???

alright so it was correctthat makes sense

but one more question

why wouldn't that be correct

is ti becuase it is only asking for the t

[limpid spade]

Yes

e^6 is a constant

[sharp sonnet]

alright thank you

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[grand violet]

LES GO

[grand violet]

I GOT NUMBER 0

ok anyway

why can't I ping helpers in forum channel

[wary stream]

Pretty sure it does ping helpers

[grand violet]

it said it doesnt

it said failed

[mortal trellis]

discord forums arent great

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[vapid steppe]

yo

[vapid steppe]

if something can be explicitly defined with no ± or whatever does it mean it’s a function

[last ether]

Not necessarily

Like

Take y = {x for all R, 15 when x = 0}

Like that isn't a function

Because at x = 0, y = 0 and 15

A function is when an input has only one output

[vapid steppe]

isn’t that a piecewise function then

[last ether]

Well no I didn't restrict the top one

[vapid steppe]

i mean like if it has a collection of x terms and collection of y terms and you can solve for y and it won’t have ± then is it function

[last ether]

Supposedly yeah

[vapid steppe]

alr

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[main oracle]

what do I do when I have degree 50, cant find it on the unit circle

[lone heart]

@main oracle Has your question been resolved?

[main oracle]

<@&286206848099549185>

sorry to ping but i was just looking at the bot rules

[minor needle]

what's the problem?

[main oracle]

Im just not sure how to simplify this expression down

[minor needle]

yeah I mean, don't you know formula?

[main oracle]

I tried using the half angle formulas but I have been confused due to the fact that I cant find 50 degrees on the unit circle

[main oracle]

would it help if I showed you my work?>

sorry, im just really lost on these new concpets

[minor needle]

yes please,

np

[main oracle]

thgis is as far as i wen t

i just seriously dont know what to do from her e

[minor needle]

first one is true

it's just sin(x/2)

where x = 50 deg

or theta, whatever

[main oracle]

ohh wait would it be sin(25)? because its just trying to simplify it down to sin(theta/2)?

[minor needle]

yeah, it's sin(25)

[main oracle]

oh my gosh thank you i was stuck on that for 2 hours

you are the best

i was just overthinking it

.close

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[soft sedge]

How to do this

[tacit arch]

what's the quesiton

[soft sedge]

find the value of x and y

[tacit arch]

can you screenshot the whole page

[last ether]

There's two variables, one equation

[wary stream]

I see 2 variables, two equations

[soft sedge]

how do i do it

idk how to do 3

[wary stream]

Could be wrong but relate each part with the other side

[last ether]

Oh they want you to compare terms

[wary stream]

Recall that $\frac{1}{2}\sqrt{6} = 0+ \frac{1}{2}\sqrt{6}$

[soft sedge]

huh

[wary stream]

Bot's dead still

[soft sedge]

oh

[wary stream]

That's what I was trying to say

[soft sedge]

ye

[wary stream]

You can relate the whole number terms on the left with the right side

And the sqrt terms on both sides

[soft sedge]

x-2y=0

what is 1/2 root 6

idk how to do that part

[wary stream]

That's equal the other stuff

[soft sedge]

but how do u simplify 1/2 root 6

[wary stream]

You can square both sides to get rid of the radical

[soft sedge]

ok

so it would be 1.5

x+y=1.5

[wary stream]

Yes

[soft sedge]

y=0.5

x=1

?

[wary stream]

You can plug in the values to check

[soft sedge]

yup i checked

im correct

ty

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[acoustic crow]

Integration by substitution:
I have the following function with substitution t = sqrt(x). I need to write the integrable function that depends on the variable t after the substitution t=sqrt(x).

As far as I've understood, I need to get rid of the variable x. I've tried to solve t and dt, but can't get rid of x

[minor needle]

show your work

if possible

[acoustic crow]

well

dt/dx= 1/(2*sqrt(x))
and therefore dt=1/(2*sqrt(x))*dx
==> dx=2*sqrt(x)*dt```

so I cant really use that dx if I still have variable x in there

[minor needle]

from the last line

we can deduce

dx = 2t dt

[acoustic crow]

Ok yeah, it's definitely too late for me. That was too simple. Thank you very much

.close

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[leaden aspen]

Hi I'm solving this curve fitting LA problem

I'm just a bit confused on why the derivative solved for the last equation has the constants 1, 2 and 3 added to the terms

Can anyone explain?

[minor needle]

what's y

[leaden aspen]

oh shit

because of the exponents coming down i see now

I should actually take the derivative next time before I get all crazy huh ? haha

[minor needle]

yea, just power rule

[fervent bane]

can I send my problem right here?

[minor needle]

open your own channel, #❓how-to-get-help

[fervent bane]

ok

thanks

[leaden aspen]

Thank you much !

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[raven field]

where does the 75 come from?

[minor needle]

180 - 105

[raven field]

oh ok

thx

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[alpine sable]

Hello all, I have a quick question. I am given three sides of a triangle; 10, 15, and 5 radical 13. How would I figure out if 5 radical 13 is bigger than 15? That's mainly what I'm unsure about

[alpine sable]

oh wait

that's simple

.close

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[halcyon egret]

how do i find the period ?

[minor needle]

you know how to find period of the function sin(ax)?

[last ether]

I can't tell if the argument is pi/3 or pi/3 (x-1/4)

[lone heart]

@halcyon egret Has your question been resolved?

[surreal meadow]

the latter

[last ether]

#❓how-to-get-help

Delete your post.

[young ore]

wdym

oh my bad

[last ether]

This is not your channel

[young ore]

how to find my channel

[lone heart]

Please read #❓how-to-get-help

[young ore]

ok thanks

[lone heart]

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[candid bronze]

help

[pine kettle]

no

[candid bronze]

So

Here is the equation

[(2)^5] x 67

I put it in my ti84

[strange stag]

Hi everyone I help with this problem

[candid bronze]

it didnt work

[lone heart]

Please read #❓how-to-get-help

[sonic ridge]

channel is occupied

[wary stream]

What does it show?

[pine kettle]

can u show us what u put into ur calculator?

[candid bronze]

syntax

error

[pine kettle]

can you show what you put into your calculator

please

[candid bronze]

The equation I showed above

[(2)^5] x 67

[wary stream]

Even with the []?

[candid bronze]

Yeah

[pine kettle]

try using parenthesis instead

[candid bronze]

Okay

[wary stream]

Because I don't think ti84 uses those as parentheses equivalents

[candid bronze]

Alright it worked. tysm

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[next osprey]

I’m stuck on the first of 3/x^2

Should I bring it up

And I forgot how

[surreal meadow]

use linearity

then express the 3/x^2 as 3x^-2

[next osprey]

I don’t think I’ve learned what liberality is

[surreal meadow]

it means you can split the integral at the + or - signs

[next osprey]

I haven’t learned that. I only know the FTC theorem (Fundamental Theorem of Calculus)

This is what I did for the previous question

[surreal meadow]

never seen this?

[next osprey]

Nope

[surreal meadow]

then how did you evaluate the first question?

[gray isle]

well you unknowingly intuitively applied that in the previous question

[hardy tiger]

How to print 3 spaces tab in python?
I know /t does indent but thats 4 spaces and i need it to be exactly 3 spaces

[next osprey]

Hmm

I’m just lost on the 3/x^2

[surreal meadow]

write it as 3x^-2

[next osprey]

Bc I know the 1 is 1x/1

Ok

Bc negatives go down right?

🤦‍♂️

I just realized

Should be like that correct?

[gray isle]

no

[next osprey]

Rip

[gray isle]

1 isn't 1x^(-2)

[next osprey]

Is it just 1^-2

[gray isle]

why even bother with an exponent

for that

1 is 1

[next osprey]

That’s true

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[frigid zinc]

In the football game, the quarterback throws the ball at an angle of 45 degrees relative to the ground and at a speed of 30 meters per second. The team player who is 15 meters ahead of the center back runs at a speed of 5 meters per second. The player sees the ball and begins to accelerate at a constant acceleration. What is the acceleration needed by the player so that he can catch the ball exactly at the height at which it was thrown?
Is my working correct?

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@frigid zinc Has your question been resolved?

[frigid zinc]

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[urban kindle]

how do I find I0 and I1 using x with this tabe

[urban kindle]

this is a Modified Bessel functions of the first
and second kinds*

here is how the book did it

[lone heart]

@urban kindle Has your question been resolved?

[urban kindle]

<@&286206848099549185> please

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@urban kindle Has your question been resolved?

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[wet veldt]

explain how question b is solved

[naive valley]

no solution = the equations are inconsistent, right?

[vale wigeon]

have you been able to solve part a?

these subproblems don't really have any drastically different ideas in them

[wet veldt]

i didnt attempt a since i was only tasked with b and c

o wait i solved it

.close

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[civic wharf]

need help with decimals

[serene junco]

#❓how-to-get-help

[civic wharf]

suggestions needed

[tacit arch]

The suggestion is to read that channel if you want help

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[alpine sable]

hello Im not sure what the difference is when using domain is (−π/2, π/2) compared with (−π, π)

how does either one restrict the answers?

could someone explain

[velvet cliff]

the trig functions are periodic

for example sin and cos repeat every 2 pi radians

[alpine sable]

ok I understand that part

[velvet cliff]

without a domain there would be an infinite amount of solutions

[alpine sable]

but I don't understand what is negative 2 pi radians?

and negative pi

what do they mean by those

[velvet cliff]

it's just the other direction

the x axis is radians

so it's to the left of the y axis

[alpine sable]

but how does the negative affect the answer

because usually the domain starts from 0

which makes sense for me

[velvet cliff]

it doesn't affect it much

still acts the same

[alpine sable]

Do you know the restrictions of a trig function

that make it monotonic

i don't know what u mean by montonic

im in highschool

I mean sorry one to one function

no sorry

I still don't understand

im hs too 😂

💀

maybe my school not as good then

Is this trig identities?

no

[merry depot]

because they are periodic, tan(x) = 0 when x = 0, pi, 2pi, 3pi, 4pi,..... By restricting the domain to [-pi/2, pi/2] the extra solutions are unnecessary.

pi, 2pi, 3pi,.... are not in the domain, so the only solution to tan(x) = 0 on the domain [-pi/2, pi/2] is x=0

[alpine sable]

hmm

[merry depot]

but on [-pi,pi] the solutions are -pi, 0, pi

[alpine sable]

but are there any solutions with negative

[velvet cliff]

don't be afraid of negative solutions

[alpine sable]

because if not why don't they just say the domain starts from 0

[velvet cliff]

you can just find them by subtracting the period

[merry depot]

yes, there are negative solutions'

[alpine sable]

hmm

ok thank you

I'll try them now on my own and will see if there are any more issues

.close

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[idle spindle]

shouldnt it be 2.7 since in the 2 numbers added 0.31 only has 2 sig figs

[idle spindle]

got this from organic chem tutor lol

am confused

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[patent frost]

Hello! Im helping my sister w her physics homework but I don't know either, physics is the end of me. Thanks!

[gusty gorge]

you should see if you can bring your sister here, because it's easier to fill in gaps in her understanding than play a game of telephone pertaining to only these problems

but if you really want to know

the first one can be done by drawing a diagram and using some formulae to convert the vector to polar form

the magnitude is given by the pythagorean theorem

and the direction (north of east) is given by atan(10/20)

and the second one is basically the same thing

[patent frost]

We both did not understand anything, what's the final answer? But thanks for clearing it up a bit though.

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@patent frost Has your question been resolved?

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@patent frost Has your question been resolved?

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[brisk radish]

What will be the answer?
Can someone explain the steps as well please

[gusty gorge]

ye^x

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[cerulean turtle]

I am wondering how i would go about figuring out how this question works 'The first, second and fourth terms of a particular arithmetic sequence
are the second, fourth and sixth terms of a geometric sequence.
Find the common ratio(s) for the geometric sequence.'

[gusty gorge]

maybe you should think of it as "the second, fourth, and sixth terms of a geometric sequence are the first, second, and fourth terms of a particular arithmetic sequence"

and then you can put some variables to it

maybe say r is the ratio and v is the second term of the geometric sequence

[cerulean turtle]

i think that helps, im still not too sure though

[gusty gorge]

so write out some equations

the fourth term of the geometric sequence is what in terms of r and v?

[cerulean turtle]

2(vr)?

[gusty gorge]

no

I suggest you review what a geometric sequence is

[cerulean turtle]

ive been doing that for the last 30 mins and its not really clicking

i understand how a geometric sequence works, just not this one

[gusty gorge]

but maybe it would help you answer this question

the fourth term of the geometric sequence is what in terms of r and v?

[cerulean turtle]

ill have a look and get back to you, thanks

[gusty gorge]

also not sure if that's all of the question

like this common ratio will in general depend on the parameters of the arithmetic sequence

[cerulean turtle]

well its the last question in a practice test, thats all there is

and is it either (vr)r or vr⌃2

[gusty gorge]

those are both the same, so either will work

[cerulean turtle]

i thought so

[gusty gorge]

but if you're looking for specific sequences, you won't find anything

[cerulean turtle]

so how could i solve it?

[gusty gorge]

like the question, as given, only has so much of an answer

[brisk radish]

Can you explain

[gusty gorge]

for example, if you have an arithmetic sequence 2, 4, 6, 8, 10, ..., and you pick the geometric sequence to be sqrt 2, 2, 2 sqrt 2, 4, 4 sqrt 2, 8

it satisfies

but then again, the geometric sequence 0,0,0,0,0,0,0,0,0 and the arithmetic sequence 0,0,0,0,0,0 also satisfy it

[cerulean turtle]

yeah that makes sense, is there a way to find it other than trial and error?

[gusty gorge]

the issue is not that you can't find it

the issue is that there isn't a particular unique pair of sequences that have this property

[cerulean turtle]

so there isnt a set way to find it?

[gusty gorge]

if there isn't even a unique answer

how would there be a way to find the answer?

the only thing you can give is either a characterization of all possible answers or just a single answer

[cerulean turtle]

i think that makes sense, it said to find the common ratio(s)

so i assume i would write a few and talk about that a bit saying how there could be a lot more or something like that?

[gusty gorge]

so you can find the ratio from the beginning and increment of the arithmetic sequence

but not sure why that would be particularly useful

[cerulean turtle]

yeah alright

that helps a lot with how i approach these questions

unless theres anything else i hope you have a great day!

[pseudo ice]

Do they say anywhere that the geometric sequence needs to be non constant (or imply it anywhere)?

[cerulean turtle]

no it doesnt, ill take a snip of the question

[pseudo ice]

Yea then as above, if the geometric sequence was not constant then you’d have a set of answers, but if not then it’s a bit

[cerulean turtle]

yeah, i dont think ill be getting anywhere near that kind of stuff for a while

[pseudo ice]

Hmmm, well you could make the assumption that it’s constant to begin with then try and solve that for the ratios, just so you have an idea of how to do it?

[cerulean turtle]

i really wouldnt know where to start, other than just guessing

[gusty gorge]

well if the geometric sequence is constant then isn't it just 1

lol

[cerulean turtle]

i think i found some answers and its 0, +/-sqrt2 and +/-1

you guys have a great day!

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[dense lily]

how do I find the height here?

i have to find the surface area, but i dont get how to find the height

is the height 95

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[solar trail]

Bezout’s Theorem: Polynomial P(x) is divisible by binomial (x – a) if and only if P(a) = 0.
Factor Theorem: A polynomial P(x) divided by Q(x) results in R(x) with zero remainders if and only if Q(x) is a factor of P(x).

what's the diff between the two? aren't they essentially the same or am i missing smt?

[solar trail]

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[fluid rampart]

claim

[fluid rampart]

how does cos^2(x)/sin(x) equal cot(x)cos(x)

[long axle]

What have u tried

[fluid rampart]

well i know that the two are equal, im just not sure how to get from cos^2(x)/sin(x) to cot(x)cos(x). I haven't tried anything yet because I don't understand what to do

[long axle]

Hmm, alright, are u allowed to manipulate either side?

Or just one

[fluid rampart]

yeah

either side is fine

[long axle]

Ok, so let’s start with cot(x)*cos(x)

Try rewriting cot in terms of cos and sin

[fluid rampart]

cos/sin

[long axle]

So what’s the right side now

[fluid rampart]

cos/sin*cos/1

i see

[long axle]

U understand?

[fluid rampart]

yep. Thanks for your help 🙂

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[hasty bluff]

Hi. I keep getting an answer that is too high to this question and I'm hoping someone can explain how I'm getting it wrong. The question: A go-kart racing track is made up of two circular arcs connected by two straights. The circular arcs are determined by circles of radii 32 metres and 60 metres and their centres are 68 metres apart. Calculate the length of the racing track to the nearest tenth of metre

[hasty bluff]

I found calculated the major arc length for each circle and their sum is more than the solution (without the straight lengths added) so I know I'm doing something wrong.

I found the angle in radians of the triangles that I constructed from the two radii (sides 32 and 60) and the distance between the two circle centres (side 68). The angle is 1.0808 radian. I doubled it then subtracted it from 2 pi to get the major angle then multiplied it by the radius of 32 to get the arc length of 131.89

I found the remaining angle of the triangle from the two radii and the distance between the centres by subtracting 1.0808 and pi/2 from pi. Doubled it and subtracted it from 2pi, then multiplied that angle by the 60 radius to get the arc length of 318.20. Add the two arc lengths together and I have more than the 436.7 metre answer that the textbook tells me I'm meant to have. 😦

@fluid rampart

@long axle

@gusty gorge

<@&286206848099549185>

<@&286206848099549185>

Hi, thanks so much for getting in touch. I was wondering if I'd done something wrong

<@&286206848099549185>

Oh, ok. Thanks. That's reassuring. I got really paranoid because after I'd done it I read somewhere that you're not meant to ping users directly. Whoops . . .

<@&286206848099549185>

@lost rampart

[gray isle]

stop abusing pings

[lost rampart]

Send problem

Nvm

[hasty bluff]

sorry. I didn't mean to abuse anyone. my bad

thank you noboru

A go-kart track is made up of two circular arcs connected by two straights. The circular arcs are determined by circles of radii 32 metres and 60 metres and their centres are 68 metres apart. Calculate the length of the race track to the nearest tenth of a metre

[lost rampart]

Lemme try solving

[hasty bluff]

thank you!

[lost rampart]

903.7 metres

This is what I got

[hasty bluff]

textbook says 436.7 metres

i got 450.09

[lost rampart]

Js a sec

[mellow grail]

Do you have a figure?

I recommend drawing one

[hasty bluff]

do you mean a diagram?

[mellow grail]

Yeah

[hasty bluff]

Oh yeah I'll try figure out how to post it. I'm not great at this carry on sorry but I'll try my best

[mellow grail]

It doesnt need to be perfect, just draw a rough diagram

[hasty bluff]

it's the posting bit I'm worried about. This is my first time on discord

[mellow grail]

Ah

[mellow grail]

Are you on PC/Laptop?

[exotic belfry]

i think there should be more information, otherwise their are different solutions possible.

[hasty bluff]

I'm on a laptop and I tried to drag and drop a photo

[mellow grail]

Nice

[hasty bluff]

wait - that worked??

ok that was easier than i expected

[mellow grail]

yeah 🙂

anyways, we are asked the perimeter of the race track right?

[hasty bluff]

yeah. and I'm meant to get 436.7

[exotic belfry]

where are the two straights which should connect the acrs?

[hasty bluff]

but i'm already too much and i haven't thought about any straights

[mellow grail]

arent the straights those radii?

[hasty bluff]

yeah, I tried another tactic for that but it was worse

honestly, i have no clue. I thought maybe the straights were the rectangle sided I tried in that diagram

[exotic belfry]

by the reading of the question i thought about something like this:

[mellow grail]

Hmm...That does make more sense

[hasty bluff]

yeah, definitely. Except that that circles are so large they're going to need to over lap

they're 68 metres apart and their radii are 32 and 60. So I figure they're gonna have to overlap

[mellow grail]

Well the circles arent part of the track. Its the straights + piR + pir, where R is 60 and r is 32 we have to calculate

even tho they overlap, it shouldnt make a significant difference to the perimeter

[exotic belfry]

but: if my interpretation is right, the given data are not enough to get a unique solution.eg. are the half circles or not? should the lines connect the arcs tangential? etc. etc.

[mellow grail]

I figured that the lines are tangential, that is a pretty common question

[hasty bluff]

it's from a high school textbook

[mellow grail]

Hmm...Then the lines being tangent is probably true

[hasty bluff]

and the ones before this one I was doing the overlap thing to find segment areas and stuff like that. but I can't get to their magical 436.7

[mellow grail]

this is the figure I think of the question

[hasty bluff]

that's awesome

[mellow grail]

only part of the circles circumference is used

[mellow grail]

Paint 3D on Windows 10 🙂

[hasty bluff]

i'm loving what you're doing, thank you

my brain can't get past the idea that they're only 68 metres apart

[mellow grail]

The circles arent part of the diagram, they are only needed to calculate the arc lengths

Unfree channel buddy

[exotic belfry]

you assume that the line from one center to the other is horizontal, but what if one of the straight lines are horizontal? I think there are different solutions.

[glad dew]

Where can i send my question?

[mellow grail]

#help-7｜zen1thxyz

[mellow grail]

Horizontal isnt possible IF the lines are tangent

and if the textbook gives a definite answers, either its wrong, or we have to consider the lines are tangential to the arcs

[exotic belfry]

i mean this:

[mellow grail]

Ah..I thought you wer saying both horizontal

[hasty bluff]

So the Pythagoras thing that i was trying is closer to what i'm meant to be doing? how would I figure out the angle? I was actually planning on adding the whole lot together. the arc lengths from my first diagram and the rectangle length (around 62 metres) from my other attempt that looked more like yours but it's too much by a long shot

[exotic belfry]

i said: "one of the straight lines is horizontal"

[mellow grail]

Sorry, I must've misread it

[mellow grail]

I think the 'two tangents are drawn from an external point of the circle, then they are of equal lengths' plays some role, but I dont think its applicable anyways

[exotic belfry]

different positions of the arcs and the assupmption of tangential straight lines gives different "open angels of the arcs" which gives different length of the arcs.

[mellow grail]

What about considering AOPD? its a trapezium with 3 sides known

or the one adjacent to it

[hasty bluff]

that's clever

[mellow grail]

Wait I got it

Its pythagoras

[exotic belfry]

yes, of course it is ok for me. my point is: there are also other solutions with the given data.

[mellow grail]

Nah it doesnt help much I think

[mellow grail]

I couldnt agree more with that

[hasty bluff]

argh

[mellow grail]

But with the assumptions that either straight isnt horizontal, and they are tangential to the arc, I think i have an answer

[hasty bluff]

please let it be 436.7

[mellow grail]

Lemme give a diagram

Okay

So:
AQ is a perpendicular to DP

[hasty bluff]

yup

[mellow grail]

We know it is parallel and equal to OP, as it is originating from the same horizontal line and the also ending on the same horizontal line

(the diameters are the straight lines)

[hasty bluff]

i did try pythag, it's on my second photo

[exotic belfry]

if you assume that CD and AB are diameters then your lines AD and CB are not tangential to the arcs.

[mellow grail]

Well u applied it wrong (I think)

[mellow grail]

Why so?

[hasty bluff]

It wouldn't surprise me at all to hear I applied it wrong. I'm forever making the weirdest, smallest mistakes

[exotic belfry]

look at the slopes of the line and the arc in point C. if DC is a diameter, then the slope of the arc is 0, the slope of the line is not 0.

[mellow grail]

...

[mellow grail]

Can you verify the books solution? With certain assumptions?

[hasty bluff]

yes, please?

I like the pythag plan. It makes sense to me

[mellow grail]

Could you send a photo of the question in the text book?

[exotic belfry]

i would prefer the suggestion of @mellow grail. trying to find additional assumptions to get a given solution can end up in mission impossible.

[hasty bluff]

It's from a high school text book for 16 year olds

[mellow grail]

Could you send PDFs? I cant open HEICs on my laptop

[hasty bluff]

oh, sorry.

i think my file size is enormous. are you able to see the pdfs i'm posting or are they too large?

[mellow grail]

Able to see

[mellow grail]

I meant the question, not the cover page of the textbook....

the page on which the question is

[hasty bluff]

sorry, i'm losing track of which photo is which

[mellow grail]

aight

its the question

[hasty bluff]

not enough info to answer it?

[mellow grail]

Ig..I am making another diagram, tryna make sense of it

[hasty bluff]

my hero. thank you so much for this

[mellow grail]

Let just this be the figure

[hasty bluff]

works for me

[mellow grail]

What was the books answer again?

Mines 435.9 apprx

[hasty bluff]

436.7!!! i think you've nailed it!!!

what the heck did you do??

[mellow grail]

Hmm..Could be different cus I used calculator pi, not 22/7 or 3.14

[mellow grail]

Simple

pytha

[hasty bluff]

hey, it's the closest i've heard and i've been working on this stupid thing forever

[mellow grail]

You alr know the concept, you just applying it wrong

One sec

[hasty bluff]

thank you

[mellow grail]

Aight

Now, calculate pythagoras for both of the triangles

(64 diameter, 32 radius)

[hasty bluff]

wait, i had 68 as the hypotenuse

[mellow grail]

Thats where you were wrong

the reason the length is 68 is because the distance between the centres is 68

[hasty bluff]

yes, centre to centre is 68

[mellow grail]

well when you construct a rectangle with arms parallel to the centre to centre segment (which for the sake of simplicity is a horizontal line), we have a rectangle

its lengths will be the bases

as for the other side, it is R - r

[hasty bluff]

your diagram makes it so clear. i can see everything you're describing.

[mellow grail]

marking the centres I hope you understand why the other side (the smaller one) is R - r i.e. 60 - 32 = 28

[mellow grail]

ty 🙂

[hasty bluff]

but i'm confused as to why it wouldn't look more like my original overlapping diagram. the gap between the centres is so small for circles of their sizes.

[mellow grail]

Thats where you are wrong

the circles ARENT part of the diagram

only the arc are

[hasty bluff]

argh!!!

[mellow grail]

the arc that is outside i.e. the go kart track, is there. The circles are mentioned, just to help you calculate its length

circles Arent part of the diagram

[hasty bluff]

ok, but how am i meant to figure out the arc length if i don't know the angle size? your diagram makes it look like i only need a half circumference but every other question before this one i was mucking around with radian

[mellow grail]

Well, since the arc is made by a circle, its length must be a multiple pi times r

[hasty bluff]

fair

[exotic belfry]

the assumptions for this solution are obviously:

1. the arcs are half circles (which is not given)
2. the straight lines are not tangential to the arcs

[mellow grail]

yeah

its the closest solution

I think if she does it on paper, he should get an exact answer

[hasty bluff]

thank you!! I'm a girl btw

[mellow grail]

Oops, sorry

[hasty bluff]

you're right!! it is what the textbook is expecting!!! you've cracked it for me!!!

[mellow grail]

Np

[hasty bluff]

This is fantastic! thank you, so, so much for figuring this out for me. I'd wasted so much time on this one dumb question

[mellow grail]

It happens sometimes

[hasty bluff]

thank you. super appreciate it

[exotic belfry]

with such an assumption, the solution would be 10 m higher. thats the annoying thing, when given problems arent unique.

[hasty bluff]

super annoying question. but so satisfying to finally get to their stupid magic number

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[knotty inlet]

how do you go from top to bottom, instead of guessing that it may be sin(A-B) where B=pi/6

[whole shell]

so therefore

in this case

a is theta

b is keep it as b

[knotty inlet]

yh i get that

[mellow grail]

Question, what is r?

[knotty inlet]

but how do you notice that

[mellow grail]

Radius?

[whole shell]

you can do it

everysingle time

you see

sin(x) + cos(x)

of some form

it doesnt need to be sin either

[knotty inlet]

ah ok, so when you see sin(x) + cos(x) you can usually use that formula to simplify further?

[whole shell]

you can turn it into cos(theta - something)

[slender gull]

asin(x) + bcos(x) you can always do it by multiplying and dividing by sqrt(a^2+b^2)

[whole shell]

you can use any of these

[knotty inlet]

ah cos + cos, ud use cos(a+b) etx

okok ty

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[ancient pine]

hello

[ancient pine]

I have this grapf function problem