help-33

[still temple]

for instance at the 3 mark of the graph there is a single dot above it

which means that there was one game in which the difference in goals scored between the us team and the other tam is 3

[swift beacon]

ohhhhhh okay, so as the numbers go up is the difference of scores between the teams? the 3 is 3 points difference

[still temple]

yeah exactly

if a dot is at 1 then that means that one game, the differenec was 1

and since there are 6 dots over 1 that means that in the entire set of games represented, there were 6 games where the difference was 1

[swift beacon]

okay thank you that helped. sorry sometimes i feel dumb when i ask questions like these but having someone just explaine it helps a lot. i can finish this on my own i believe. im sorry haha wee can close this

[still temple]

u good bruh

[marsh citrus]

@swift beacon Has your question been resolved?

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[marsh citrus]

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[normal bough]

How to solve number 6?

[viral dome]

alright

[late geode]

what's the slope of
y = x+4

[viral dome]

so start by finding slope of the new line

[normal bough]

x

[late geode]

no

[normal bough]

1/1 ?

x/1 ?

[late geode]

stop guessing

random things

the slope of an equation in slope intercept form
y = mx + b
is m

what is your m in
y = x + 4

[normal bough]

y = _x + 4

I'd just say it's the x.

[bright jay]

m is a coefficient

What's the coefficient of x?

[normal bough]

1

[bright jay]

Then the slope of y = x + 4 is?

[normal bough]

slope would equal 1

[bright jay]

If a line is perpendicular to another, what does that mean?

[normal bough]

90 degrees?

its at an angle of 90 to

the other line

[bright jay]

What does it mean for the slope?

[normal bough]

Not sure

Going down? aka negative?

[bright jay]

You don't have to reply to every one of my messages

[normal bough]

eh

[bright jay]

https://youtube.com/watch?v=Sc35n_Y93-o

#❓how-to-get-help

[normal bough]

still confused.

[late geode]

do you know the relationship between the slopes of perpendicular lines?

if I gave you the slopes of two lines, would you be able to tell me if they would be perpendicular to each other without doing any graphing?

[normal bough]

possibly

[late geode]

lines with slopes of: 2 and -3

would lines with those slopes be perpendicular to each other,
why/ why not?

[normal bough]

perpendicular to each other

[late geode]

why/ why not?
how did you reach that conclusion

[normal bough]

does it count if you graphed it in ur head?

[late geode]

no graphical approach at all

[normal bough]

ok.

.

[normal bough]

nvm that

[normal bough]

no idea.

[late geode]

did you watch the video that was linked?

[normal bough]

yes

didnt make sense

[late geode]

google slopes of perpendicular lines

[normal bough]

oh.

huh

[normal bough]

the 2 becomes -2,

-3 becomes +3

[late geode]

no

what are you reading...

[normal bough]

¯_(ツ)_/¯

ima just skip that for now.

.close

[marsh citrus]

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[shadow wadi]

(4x+ 3y) / 2x+ y = 2 + y/(2x+y)

subsitute x= 2y/7

divide

add and subtract y in 4x + 3y

(4x+ 2y + y) / (2x+y)

[(4x+2y) / (2x+y)] + [y/(2x+y)]

{ 2(2x+y) / (2x+y) } + [y/(2x+y)]

2 + y/(2x+y)

got it? @still temple

[marsh citrus]

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[surreal marsh]

Need help with this multivariable calculus problem. Currently doing linear approximations

so I am guessing I use the equation of a cylinder and take the partial derivatives?

[worn nimbus]

by amount of metal do they mean volume?

or mass

[surreal marsh]

Im guessing the material that makes the can

okay now im confused

so I guess i use surface area??????

or volume

[wind wadi]

The amount of mental will be the difference of the volumes of two cylinders, an outer one and an inner one (which has dimensions smaller by 0.05 and 0.4 respectively)

[surreal marsh]

i don think this is integration

[wind wadi]

You can write down the difference and model the small quantities as dr (radial direction) and dh (along the can's height). Then, for example, if you get products of these small quantities (dr^2 or dr * dh), you can omit them in your approximation

[surreal marsh]

yeah idk this is so confusing. I found the formula online... don't know how it works. Im just gonna ask my professor

thank you for the help though

.close

[marsh citrus]

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[vapid sun]

Please could somebody help me finish this integration

[vapid sun]

I'm unsure what to do from this point

[vapid sun]

<@&286206848099549185>

[wet zenith]

I cant read

[vapid sun]

Here you go

I had a feeling people couldn't read it

[wet zenith]

Well write the numerator as derivative of the stuff in the denominator times some constant plus some other constant maybe?

But theres probably a simpler way xd

[vapid sun]

You mean to integrate it in terms of ln?

[wet zenith]

Do we get ln?

[vapid sun]

No we won't be able to

[wet zenith]

Whatever we get, we get first try that imo

[tight furnace]

Substitution

Our numerator is close to a du

[vapid sun]

Would I do a trig substitution?

[tight furnace]

for a u that works well in the denominator

Not trig

[vapid sun]

Or just the denom

[tight furnace]

more direct

[vapid sun]

Ok

[tight furnace]

What's du then

[vapid sun]

So du is 4-2x dx

[tight furnace]

ah or wait hang on

yeah I was being dumb

[wet zenith]

If you can integrate linear/ root quad and just root quad in denom, u should definitely try this

[vapid sun]

I'm sorry man I don't understand what you mean

[tight furnace]

This might still be it tho

[vapid sun]

Probably because I'm tired @wet zenith

[tight furnace]

If u try idk

[wet zenith]

The stuff inside the square root

In the denominator

[vapid sun]

Yeah

[wet zenith]

Wait no numerator is expressed as constant( derivative of stuff inside sq root in denom) + some other constant

Im also sleepy mb

[vapid sun]

lol

The chapter is trig substitutions

[wet zenith]

U get the constants by comparing coefficients

[wet zenith]

Yea but who actually uses trig subs💀

People just memorize the general formulas

[vapid sun]

lol

I haven't been taught your method yet

What year are you in?

[wet zenith]

12

[vapid sun]

I'm in year 13 lol

[wet zenith]

U can try it

[vapid sun]

You from the UK?

[wet zenith]

No

[vapid sun]

Ah

Well I don't know how your method would work

That's what I would get

And then the numerator has another constant added to it

[wet zenith]

x= a(-2x+4) +b

a and b are constants

[vapid sun]

ok

[wet zenith]

Compare coefficients on both sides now

[vapid sun]

What do I do after that

[wet zenith]

To find a and b

[vapid sun]

so a is -1/2 and b is 2?

[wet zenith]

Right

Now break the original question into two parts

[vapid sun]

Wdym

[wet zenith]

Use this

And the values of a and b u got

[vapid sun]

Oh

[wet zenith]

Dont multiply a

But just seperate the question

[vapid sun]

Do you mean just substitute X in?

As in sub in the -2.... Etc in for x

[wet zenith]

Yea

[vapid sun]

Only on the numerator?

[wet zenith]

Yea

[vapid sun]

Right done

[wet zenith]

(m+n)/c= m/c +n/c use this

As in ur numerator is in addition of two parts

Break them and write them seperately

[vapid sun]

Ok

Done

[wet zenith]

Can you show what u have

[wet zenith]

Ok just put an integration symbol on the right part as well, we solve these two seperately

[vapid sun]

Ok done that

[wet zenith]

So for the left part, use a substitution of like t= the stuff inside the sq root in denom

Find dt

[vapid sun]

Done all that

And I'm left with this

Forgot to put a sign

One sec

[wet zenith]

Yea now u should know how to solve this part

[vapid sun]

Ok

[wet zenith]

In order to solve the 2nd integration ull have to complete the square inside the sq root in the denom

[vapid sun]

Done that

[wet zenith]

Show what u got

[vapid sun]

For the second part

[wet zenith]

Nice

[vapid sun]

For the first part I got the wrong answer i think

[wet zenith]

What did u get

Second part is correct tho

[vapid sun]

How old are you if you don't mind me asking?

[wet zenith]

17💀

[vapid sun]

Same lol

I'm surprised I've never learned this

Because I'm doing further maths which is meant to be the most advanced maths at a level

And I've never once been taught that method you showed

[wet zenith]

Im doing basic maths

🥸

[vapid sun]

You've learned how to integrate to arcsin etc in basic maths?

[wet zenith]

Well tbh i havent finished this chapter yet

Theres a problem related to this chapter, in my channel, if u want to have a look at it

[vapid sun]

Yeah sure

.close

[marsh citrus]

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[marsh citrus]

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[vestal mist]

How many ways are there to park two different buses and two different cars on seven consecutive parking spots if each bus needs two consecutive parking spots to park? Assume that the spaces are along a one-way street, and that all the vehicles are pointed in the direction of traffic.

[vestal mist]

can i get a hint

no solutions pls

[upper briar]

i think

cuz the buses are always being used

u can treat it as buses takin up 1 space

n total 5 spaces

[vestal mist]

thats actually a good point

but will it always work?

[upper briar]

um

in this case yea?

cuz u r always using 2 buses

if u r using a variable num of buses then prob no

aka if u can choose to use 2 or 3 or 4 buses

[vestal mist]

gotcha

thanks!

@upper briar is the answer 5C4

[upper briar]

idk

[vestal mist]

well its like how many ways to order 4 vehicles upon 5 spots

i sorta want someone to confirm lol

bc im not so sure

[marsh citrus]

@vestal mist Has your question been resolved?

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[cunning blaze]

My friend needs help with this. I have no clue how to do it and neither does he😅

[main idol]

,rotate

[main idol]

f(x+h) means wherever you see an x, replace it with x+h

e.g. g(x) = 4x^2

g(x+h) = 4(x+h)^2

[cunning blaze]

So I just plug it in

Simplify it

And bam?

I got an overly long answer though

[main idol]

then you probably have an algebra mistake

or it's just a long answer

[cunning blaze]

?

[main idol]

that's only part of the question

[cunning blaze]

Help me with the other part😭

[main idol]

idk what you're stuck on

[cunning blaze]

Is that right?

[main idol]

for this?

[cunning blaze]

Yes

It wants the difference quotient

I don’t get it am I right

[main idol]

looks like you left out some of the terms

[marsh citrus]

@cunning blaze Has your question been resolved?

[marsh citrus]

@cunning blaze Has your question been resolved?

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[viscid plinth]

could someone check my answer please

[marsh citrus]

@viscid plinth Has your question been resolved?

[viscid plinth]

okay wait i fixed some errors

<@&286206848099549185>

[silk sable]

Looks good to me.

[silk sable]

I took a slightly different route though.

[viscid plinth]

ahh i see, thats similar to one tht someone else showed me as well, thank you very much!

[silk sable]

yw

[viscid plinth]

.close

[marsh citrus]

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[primal sparrow]

Hello. I am having issues with this problem here. Attached are the problem itself (the png) and the work I have done so far towards solving the problem (the pdf)

[main frigate]

draw a force diagram with all the forces on the plane

[primal sparrow]

That is what I did in the pdf

That was the whole point of me drawing the coordinate plane thing with lines

[main frigate]

oh

might want to ask the physics discord #old-network

[primal sparrow]

Did not know about the physics discord, thanks

.close

[marsh citrus]

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[vapid glen]

quick q when determining degree its just the highest right and in factor its just the exponents

[vapid glen]

so if its like (x+1)^2(x-1)(x-4)

the degree would be 4 even right

[main frigate]

yeah

[vapid glen]

ok ty

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[split kestrel]

could someone help with this pelase

[marsh citrus]

@split kestrel Has your question been resolved?

[marsh citrus]

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[zealous cove]

https://discord.com/channels/268882317391429632/1027030518061076490 hope some people would see this...

[marsh citrus]

@zealous cove Has your question been resolved?

[zealous cove]

<@&286206848099549185>

[marsh citrus]

@zealous cove Has your question been resolved?

[zealous cove]

.close

[marsh citrus]

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[dim plover]

how would i dpo this

[dim plover]

i know how to do m = tan Θ

but not sure how to do it backwards

[sharp vessel]

$\tan^{-1}(x)$

[elfin berry]

RedstonePlayz09

[sharp vessel]

This is the inverse tan function. Also known as arctan(x)

You can use this on your calculator to find the angle when given the value of tan.

Just to note, it's NOT this: $\frac{1}{\tan(x)}$

[elfin berry]

RedstonePlayz09

[sharp vessel]

Don't confuse it with 1/tan(x)

[desert dragon]

I want 210,000 apple.
Every 15 minutes I get 50 apples

How many minutes/hours/days is it going to take to get 210,000 apples?

[dim plover]

is ti possible to do without calculator?

[sharp vessel]

Well, unless it's a known value, not really?

You can approximate it

Why would you want to do it without a calculator?

[sharp vessel]

#❓how-to-get-help
This channel is taken

[dim plover]

just I am revising for a test

but i thought it was a non calc

but maybe its not

[sharp vessel]

You might want to ask your teacher

[dim plover]

ok yeah

so

in the calculator would it be

arctan(2)

[sharp vessel]

No

the slope is 1, recheck your work

[dim plover]

omg

yeah

[sharp vessel]

$\theta = \arctan(1)$

[elfin berry]

RedstonePlayz09

[dim plover]

ty

:)

[sharp vessel]

No problem!

[marsh citrus]

@dim plover Has your question been resolved?

[sharp vessel]

.close

[marsh citrus]

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[wet peak]

How do I solve x^2-2x-8

[cunning fiber]

that's an expression with no directive tbh

[eternal tundra]

you mean it equals 0 right?

[wet peak]

Yea

[eternal tundra]

you either use a formula

[wet peak]

I think

[eternal tundra]

or you do complete the square

i suggest complete the square

[wet peak]

Ok

I want to get the answer (x-4)(x+2)

[eternal tundra]

do you know how to do complete the square?

[wet peak]

No

[eternal tundra]

have you never done it?

[wet peak]

No

[eternal tundra]

lets take a general approach

consider this expression

we want to just get one x

this helps to achieve this

rearranging we get:

now substituting in we get

this is the general approach

[wet peak]

I dont get it

[eternal tundra]

which step?

[wet peak]

All of it

[eternal tundra]

the first picture is just the expression

we don't have to understand how to derive the second picture in order to use the formula

so let's not derive it right now

do you understand how (x+0.5a)^2 equals the right hand side?

[wet peak]

No

I don’t get anything

[eternal tundra]

(a+b)^2=(a+b)(a+b)=a^2+2ab+b^2

that should be familiar

as for the third picture, we just rearranged

for the fourth picture, we substituted it into the original equation (first picture)

[wet peak]

Yea

So can we do that for this

[eternal tundra]

yes

we have this

now consider

[wet peak]

Yea

[eternal tundra]

(x-1)^2

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

rearranging we get

(x-1)^2-1=x^2-2x

notice that the right hand side is also present in the original equation

so we substitute in

and get

[wet peak]

I’m confused how you got (x-1)^2-1

[eternal tundra]

btw:
how do we get the (x-1)^2 you may wonder.
step one: make sure the coefficient of x^2 is 1
step two: divide the coefficient of x by 2
in this example we have -2x, so we divide -2 by 2
thus we get -1

maybe this helps:
x^2+bx+c=0
(x+0.5b)^2=x^2+bx+(0.5b)^2

[wet peak]

Na I still don’t get it

Idk why

I gotta go

.close

[marsh citrus]

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[eternal tundra]

pls don't occupy multiply channels

[crude marlin]

How are 11 and 14 alternate interior angles?

[crude marlin]

Srry

It was a mistake

But could someone pls help? I’m rlly stuck

[eternal tundra]

,rotate

[still temple]

Can someone help me please

[eternal tundra]

@still temple #❓how-to-get-help

[warm galleon]

Do u know how to multiply the parenthesis

Do u know what (m - 4)^2means?

[lean flame]

What do you know about alt int angles

[crude marlin]

I mean, they are interior

Only one these are

14 is exterior, while 11 is interior

@lean flame

So how is it possible that they’re alternate interior angles

[lean flame]

They are interior if you are looking from the perspective of the lines r and p

[crude marlin]

What’s r?

There’s no r

Also, how am I supposed to know what the transversals are?

[lean flame]

r or n

i can't see properly

[crude marlin]

But I still don’t understand how they’re alternate interior @lean flame

I’m so lost

[lean flame]

You know this right? c and f are alt int angles agree?

[crude marlin]

Yes

That makes sense

But this doesn’t

@lean flame

[lean flame]

Look at this picture careful, let the L line be the transversal

[crude marlin]

Wait, I might get it

Is it bc u kinda have to, like, look at the bigger picture?

[lean flame]

Do you see the similarity?

f is literally 11 and c literally 14

[crude marlin]

Ohhhh

I was looking at it differently

So how do I know whether something’s a transversal? @lean flame

[lean flame]

In geometry, a transversal is a line that passes through two lines in the same plane at two distinct points. Transversals play a role in establishing whether two or more other lines in the Euclidean plane are parallel. Wikipedia

So in this case every line is a transversal

[crude marlin]

Oh, ok

[marsh citrus]

@crude marlin Has your question been resolved?

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[crude marlin]

Tysm btw

[marsh citrus]

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[crude marlin]

Oh and also @lean flame, what’s the difference between the line segment notation and this notation?

[crude marlin]

In numbers 9 and 10 for example

[lean flame]

What is "this notation"?

[crude marlin]

Nvm, it’s line notation I think

[late geode]

double arrow indicates the line that extends infinitely in both directions

look up line vs segment vs ray

[marsh citrus]

@crude marlin Has your question been resolved?

[crude marlin]

Ty

Also

Do u mind helping me with #28?

,rotate

[crude marlin]

@late geode

[marsh citrus]

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[crude marlin]

I don’t understand how I’m supposed to know

[marsh citrus]

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[silent surge]

Find the equation of a parabola with vertex on the line y = -3x, axis parallel to the y-axis and and passing through (-7, 13) and (5, 1)

Help, please.. Tell me what to do with it or instead, give the solution

[silent surge]

Nevermind, I think I got it XD

.close

[marsh citrus]

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[hushed dagger]

Let a = (1,2,1) and b = (1,2,-2) be two vectors in space. Decide the two vectors p and q so that a = p + q.
p should be parallel to b, q should be perpendicular to b;

[hushed dagger]

I'm thinking about dot product and equation system

[echo bough]

yeah good ideas

[hushed dagger]

But I got stuck

[spark otter]

a and b do not even have the same amount of coordinates

[hushed dagger]

Oh wait srry

1sec

[echo bough]

what does p parallel to b mean?

and q perpendicular to b?

[hushed dagger]

Fixed

Parallel vectors means that one of the vectors can be multiplied by a scalar k so that it gets the same vector.

This is perpendicular

Basically the angle between the two vector should be 90

[echo bough]

i meant in terms

of vectors

[hushed dagger]

Oh

[spark otter]

Yes, and using mathematical terms and not graphs ?

[echo bough]

this but as an equation

[hushed dagger]

Dot(a,b) = 0 is perpendicular

[echo bough]

yeah ok

[spark otter]

Yes

[echo bough]

so b . p =0

[hushed dagger]

Yea

And for the parallel I did this. (x,y,z) = k*(1,2,-2)

[spark otter]

b . q = 0 more like

[spark otter]

Yes this is good

[echo bough]

yeah ok right

and then last thing is b+q=a

[spark otter]

Now, if you write q = (x,y,z), how do you write b . q = 0 in terms of x,y,z ?

[hushed dagger]

(x,y,z) . (1, 2, -2) = x + 2y - 2z

This is where I got stuck

[hushed dagger]

Don't I need to do smth before that?

[echo bough]

ok

so call x_q,y_q,z_q and x_p,y_p and z_p the cordinates

as you already calculated

x_q +2 y_q -2 z_q =0

[echo bough]

then from this

x_p = k

y_p=2k

z_p = -2k

and lastly

[hushed dagger]

Yes, I got that

[echo bough]

x_p+x_q = 1

y_p+y_q= 2

z_p+z_q=1

so now

try to do some algebra to find k

[hushed dagger]

Ok give me a min

Wait what

[hushed dagger]

Shouldn't this be -2?

[echo bough]

p+q=a

[hushed dagger]

Oh right

[marsh citrus]

@hushed dagger Has your question been resolved?

[hushed dagger]

What kind of algebra?

I got stuck

[echo bough]

hinnt

express q cordinates

in terms of k

by subbing the p cordinates

[echo bough]

then sub those values into this

you'll have an equation that only depends on k

then you find k

that gives you p

and then q

[hushed dagger]

Ok I got k = 1/3

Is that correct?

[echo bough]

idk

didnt do it

you'll check at the end

if that gives you a q

that is indeed

perpendicular to b

[hushed dagger]

Ah

[echo bough]

so now p = (1/3)b

[marsh citrus]

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[hushed dagger]

.reopen

[marsh citrus]

✅

[hushed dagger]

OK I think I got the right answer now

@echo bough Thanks for the help!

[echo bough]

no prob

[marsh citrus]

@hushed dagger Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[cobalt anchor]

Hey guys I need help on 12

I did it but I got a different answer than the key

Can you guys spot any mistakes??

Or show me another way of doing it

[true flicker]

1. differentiate left side of the eqn
   a) rewrite (2x-1)^2 as (2x-1)(2x-1)
   b) expand
   c) simplify and combine
   d) use binomial theorem
   e) differentiate using sum rule and evaluate

2. 8 is constant wrt x, therefore derivative of 8 wrt x is 0

3. set left side to be equal to right side

4. solve for y'
   a) rearrange terms
   b) factor
   c) simplify

[marsh citrus]

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[marsh citrus]

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[sinful edge]

Can someone help me solve this one

I’ve been stuck on where to start

[marsh citrus]

@sinful edge Has your question been resolved?

[sinful edge]

<@&286206848099549185>

[sinful edge]

<@&286206848099549185>

[marsh citrus]

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[still temple]

Hi is this correct?

[marsh citrus]

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[marsh citrus]

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[cursive spruce]

<@&286206848099549185>

im not sure on the method

please can u help me

[proven timber]

!15m

[marsh citrus]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[cursive spruce]

sorry

[proven timber]

Lets say each cup starts with x milliliters

How many milliliters does cup A have after pouring?

[last oriole]

x-120

[proven timber]

I was asking Cade2icy

[last oriole]

we r working together

we both stuck lol

[proven timber]

Ok then

How many ml does cup B have

[cursive spruce]

x+120

[last oriole]

x+120

[proven timber]

It's given cup B has 5 times as many ml as cup B

So how does x - 120 and x + 120 relate based on that

[last oriole]

what?

[cursive spruce]

x+120=x-120

[proven timber]

Let's call x - 120 = A, and x + 120 = B

B is 5 times as much as A

So what does that mean as an equation

[last oriole]

so would it be 5x=600

i mean

[cursive spruce]

yhh

[proven timber]

No

Let's work just in A and B

B is 5 times bigger than A

[last oriole]

okay

[cursive spruce]

yhh

[proven timber]

So what does that mean

[cursive spruce]

ax5=b

[last oriole]

B=5xA

[proven timber]

Yes, B = 5A

Now, B = x + 120 and A = x - 120

So what does B = 5A become in terms of x

[last oriole]

5x-120?

5x-600

[cursive spruce]

one sec

[last oriole]

im not sure

[cursive spruce]

jud thinking

[proven timber]

Just replace B and A with their respective values

[last oriole]

sorry?

i dont get what u mean?

[proven timber]

B = 5A

B = x + 120, A = x - 120

[cursive spruce]

x+120=5x-24

[proven timber]

Why 24?

[cursive spruce]

cus 24

x 6

=120

24x 5

=120

therews 5 x

[proven timber]

So you divided 120 by 5?

Why?

It's 5A, not A/5

[cursive spruce]

idk

srry

[last oriole]

can we go on a vc so u can explain i dont get it

[cursive spruce]

yh pls

if u dm

[proven timber]

I'm not in a place where I can go on vc right now

[cursive spruce]

oh

[last oriole]

oh

np

[cursive spruce]

dw

[proven timber]

Just type B = 5A, but replace B with x + 120 and replace A with (x - 120) (make sure to remember the parenthesis)

[cursive spruce]

x+120=5x-120

[proven timber]

Remember the parenthesis with A

5A = 5(A)

[cursive spruce]

x+120=5x-600

[proven timber]

Yes

Solve for x

[cursive spruce]

but why -600

??

[proven timber]

Distributive property

[cursive spruce]

wht??

[proven timber]

5(x - 120) = 5x - 5 * 120 = 5x - 600

a(b + c) = ab + ac

[last oriole]

x=180

[proven timber]

,w x + 120 = 5x - 600

[elfin berry]

Results provided by WolframAlpha

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

Requested by Gamer Dio#7663

[proven timber]

Yes

So of x = 180

How many ml does cup A have after pouring

[cursive spruce]

60

[proven timber]

Yes

Tada

[cursive spruce]

buts what x===

??

[last oriole]

i dont understand how u guys got to x+120= 5x-600

[proven timber]

B = 5A

[last oriole]

yes

[proven timber]

B = x + 120, A = x - 120

Thus x + 120 = 5(x - 120)

[last oriole]

so

you x a

so it would be B=5x-600

OHHHHH

i kinda understand it now

thank you

[cursive spruce]

thankyou

[marsh citrus]

@cursive spruce Has your question been resolved?

[marsh citrus]

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[fossil river]

hello

[fossil river]

bello

how do i.do this

a yone

<@&286206848099549185>

winus ru still tbere

@sullen lichen

i.neednhelo

<@&286206848099549185>

[sullen lichen]

just don't freely ping helpers (look at the green message above), and i'm not sure i comfortable to help on this since this looks awful a lot like a test of some sort of kind.

[fossil river]

we can share with class as u see there @sullen lichen

which mean its just classwork

.close

[marsh citrus]

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[fringe quarry]

Can u help me

[fringe quarry]

How to solve this

@sullen lichen

[marsh citrus]

@fringe quarry Has your question been resolved?

[agile spindle]

You have an equation like 28 + 33 - 2k = 60 - 10 - k, k is the number of students that learn both languages

@fringe quarry

[marsh citrus]

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[sudden shoal]

How would I graph this? Im just getting 1

[silk skiff]

hi

i did it on desmos

then copied it

into my book

or test

[glass perch]

Just draw a graph of ln x and then mark the area enclosed by the integral

[sudden shoal]

Where is “e” on ln(x) ?

[glass perch]

1 and e are the boundaries of the integral

[sudden shoal]

Oh e as in the value? 2.17

[glass perch]

https://images.app.goo.gl/fwZatQBPMLkGbriU6

You mark them on the coordinate axis like in the picture

Yes

[sudden shoal]

Kk got it

Thanks a lot

I forgot e had a value

.close

[marsh citrus]

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[quaint arrow]

im trying to establish the probabilities of each thing

is the probability of choosing 3 red balls 3c3/11c3?

@wind wadi and is the prob of getting 3 white balls 4c3/11c3?

wait no

.close

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[narrow breach]

Would like some help on c please

[narrow breach]

ive figured that un+1 is just un multiplied by the common difference

so i did Un multiplied by 2/3

which equals Un+1

but i dont know how to get it in that form

[hidden plaza]

Simplify this $u_{n+1} = 3(\frac{2}{3})^{n+1} -1$

[narrow breach]

so

[elfin berry]

Pure

[narrow breach]

its just 3(2/3)^n x (2/3) - 1 right?

thats the equation

$u_{n+1} = 3(\frac{2}{3})^{n}*(\frac{2}{3}) -1$

[elfin berry]

Smooth

[narrow breach]

didnt expect that to work

anyways simplifying this

so just $u_{n+1} = 2(\frac{2}{3})^{n} -1$

[hidden plaza]

$u_{n+1} = \frac{3(\frac{2}{3})^{n}\times {2} -3}{3}$

[elfin berry]

Smooth

Pure

[hidden plaza]

By multiplying the whole thing by 3/3

Now recall that $u_n = 3(\frac{2}{3})^n -1$

[elfin berry]

Pure

[narrow breach]

so then

the numerator gets replaced with Un?

6

[hidden plaza]

$2u_n -1 = 2 \cdot (3(\frac{2}{3})^n -1) -1$

[elfin berry]

Pure

[narrow breach]

$2u_n -1 = \cdot (6(\frac{2}{3})^n -2) -1$

that feels wrong

whoops

the 2 is still there

[elfin berry]

Smooth

[hidden plaza]

$2u_n -1 = 6(\frac{2}{3})^n -3$

[elfin berry]

Pure

[hidden plaza]

That’s what we have here

So replace the numerator with 2u_n -1 and we are done

[narrow breach]

so then add 1 and divide by 2?

[hidden plaza]

?

[narrow breach]

wait

oh

i see

ok yeah i get it now

thank you for your help

.close

[marsh citrus]

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[solar light]

How do I find the general solution?

I got it down to
z y x w
3 -3 -3 -9 = 18
0 -2 -1 -5 = 11
0 0 0 0 = 0

but where do I go from here?

[marsh citrus]

@solar light Has your question been resolved?

[solar light]

pls someone help so I can go to sleep this is the only problem I havent solved yet then im done

[marsh citrus]

@solar light Has your question been resolved?

[solar light]

.close

[marsh citrus]

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[river ivy]

prove that

[river ivy]

with lamda= any real number, it's a solution of

in the first part, i know that vector b and vector x are perpendicular by doing b.x

now, what about the magnitude?

[marsh citrus]

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[marsh citrus]

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[river ivy]

.reopen

[marsh citrus]

✅

[marsh citrus]

@river ivy Has your question been resolved?

[marsh citrus]

@river ivy Has your question been resolved?

[marsh citrus]

@river ivy Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[hollow dock]

Why is it called relative min and max?

[hollow dock]

Is it actually relative?

[marsh citrus]

@hollow dock Has your question been resolved?

[left hornet]

A relative maximum or minimum occurs at turning points on the curve where as the absolute minimum and maximum are the appropriate values over the entire domain of the function.

In other words the absolute minimum and maximum are bounded by the domain of the function.

[marsh citrus]

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[still temple]

Help

[upper briar]

helped

[still temple]

A

[upper briar]

OMG

THIS

u were here b4

did deep fail

[still temple]

I

Yeh

Sadly

[upper briar]

LOL

[still temple]

I need help or at least a reference

But idk why people is using circles

😭

[upper briar]

they

like me

💕

[still temple]

.close

[marsh citrus]

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[upper briar]

oh

can u make 2 circles of equal radii?

@still temple

[marsh citrus]

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[severe smelt]

Is there a good resource/method used to create a spline through known points? Not influenced by points. See image. I found this but it only shows how to do it through 4 points. Ideally I'd need it to work for any given number I plug in: https://www.particleincell.com/2012/bezier-splines/ I also found this but I don't understand some of the notation used and it's also through a limited number of points: https://youtu.be/tXrkNfHFakg Even just a pointer in the right direction would be amazing.

[severe smelt]

This is basically what I'm trying to achieve visually. I place the points and my code will use them to create the spline: https://www.youtube.com/watch?v=UyRl1Vpecko (but I only need it to work in 2D)

[marsh citrus]

@severe smelt Has your question been resolved?

[marsh citrus]

@severe smelt Has your question been resolved?

[marsh citrus]

@severe smelt Has your question been resolved?

[sharp vessel]

Try having lots of part, each representing a point.
The points will be defined by:

$P(x) = y_0\frac{x-x_1}{x_0-x_1} + y_1\frac{x-x_0}{x_1-x_0}$

[elfin berry]

RedstonePlayz09

[sharp vessel]

You plug in x values between 2 control points, and you get the y values

(Found this on the internet, try to see if it works)

[marsh citrus]

@severe smelt Has your question been resolved?

[sharp vessel]

@severe smelt So?

Did you try it?

[severe smelt]

I don't know how to implement it and it doesn't look like it'll create a spline in any way

[grave ginkgo]

that doesn't really look like a spline, it's just a linear equation

[sharp vessel]

Iterate through each control point you have. Let's call the current point P_0 and the next point P_1.
Then, create many small parts (cubes or spheres) that will represent the points on the spline.
The first cube's x value will be P_0's x value, the second cube's x value will be the last cube's x value plus a small value. Basically each cube will have a x value between P_0 and P_1.
Then, you can use the formula I sent to find the y values.

[sharp vessel]

It's to find the y value of EACH point on the spline.

Between 2 control points

As I said, this is just something I found

I'm pretty sure that's what you need

Your spline will be a polynomial, with degree corresponding to the amount of control points.

[grave ginkgo]

okay, it took me a while to find anything on the internet that resembled your answer, but what you're describing seems to be a Lagrange polynomial

[marsh citrus]

@severe smelt Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[timber tundra]

Let a and b be natural numbers.
Give all solutions to the following equation: 2^a + 1 = b^2
Hint: Rewrite the above such that it talks about multiplication (.) instead of addition (+).

[devout mauve]

what have you tried

[timber tundra]

This is all I came up with a = log2 (b^2 - 1) and b = √ (2^a + 1)

[devout mauve]

too complicated

but b^2-1 is a good step

do you know an identity which turns this into multiplying some stuff?

instead of currently adding stuff?

[timber tundra]

(b + 1) (b - 1) ??

[devout mauve]

yes

so 2^a = (b+1)(b-1)

so on the LHS we have a product of only 2s

the prime factorization of numbers is unique. so what does that tell us about the RHS

[marsh citrus]

@timber tundra Has your question been resolved?

[timber tundra]

LHS will always be even, i don't know 😅

should b be prime?

2 is a prime factor to (b + 1) (b - 1) ?

[devout mauve]

and it is the only prime factor

so what do we know about b+1 and b-1 then

[timber tundra]

(b + 1 )( b - 1) = even and we can only divide it by 2 ??

no, (b + 1 )( b - 1) divided by 2 should be even, but what should I do now?

b can only be 3 😎

how should I give all solutions? @devout mauve

[devout mauve]

why can b be only 3 all of a sudden

where did you take that from

[timber tundra]

just ignore that😂

[devout mauve]

by the uniqueness of prime factorization, we need that b+1 = 2^k for some k and b-1 = 2^m for some integer m. cause then 2^a = (b+1)(b-1) = 2^k*2^m = 2^(k+m)

so both b+1 and b-1 are powers of 2

what now

[timber tundra]

I understand so far, can you give a hint?

[marsh citrus]

@timber tundra Has your question been resolved?

[devout mauve]

b+1 and b-1 are 2 apart from each other

and they are both powers of 2

that is enough to determine what b+1 and b-1 are

[timber tundra]

so b is an odd number?

[devout mauve]

can b+1 be 64 for example?

[timber tundra]

yes. 63 + 1 = 64

or 65 - 1 = 64

but all odd numbers don't work, how should I find the right b?

[devout mauve]

well but if b+1=64, then b-1=62

but 62 is not a power of 2

both b+1 and b-1 have to be powers of 2

[timber tundra]

oh right haha

[timber tundra]

ok, is there a method to get all solutions out?

in Number Theory

[devout mauve]

there arent that many powers of 2 that have a difference of 2

this is not hard. dont overthink this

[timber tundra]

2 and 4

[devout mauve]

so b+1=4 and b-1=2

so b= what?

and then a= what?

[timber tundra]

b= 3 and a = 3

[devout mauve]

yes

and that is all solutions

[timber tundra]

hahahha omg 💀

thx😅

.close

[marsh citrus]

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[unborn echo]

How do I do this problem?

[wind wadi]

use
$$A \setminus B = A \cap B^c$$

[elfin berry]

nvx

[wind wadi]

you can also use a venn diagram to prove these

[unborn echo]

Ok

Is this $A \setminus (B \setminus C^c)$

[elfin berry]

bluefire

[unborn echo]

to the c

didn't include that part

[wind wadi]

I'd first apply this rule to B \ C^c

[unborn echo]

Ok

B intersect C?

[wind wadi]

yes

[unborn echo]

Ok

So we are left with A \ (B intersect C)?

[wind wadi]

yes

[unborn echo]

What do I do now?

[wind wadi]

you can apply the rule again

[unborn echo]

A intersect (B intersect C)^c?

[wind wadi]

yes

[unborn echo]

Ok

[wind wadi]

now what happens if you apply the complement to (B intersect C)?

[unborn echo]

So A intersect (B^c union C^c)

[wind wadi]

yes!

[unborn echo]

so A intersect A?

[wind wadi]

no

[unborn echo]

Ok

[wind wadi]

do you know the distributive property for intersect and union?

[unborn echo]

Yes

So (A intersect B^c) union (A intersect C^c)?

[wind wadi]

correct

now can you see it?

[unborn echo]

does it simplify to A union A?

[wind wadi]

the rule we used before

you can now "undo" it

on each of (A intersect B^c) and (A intersect C^c)

[unborn echo]

Oh ok

I see so when you do that

it proves that the left will equal the right

Since it is (A\B) U (A\C)

[wind wadi]

right