help-10

[zinc spruce]

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[static fulcrum]

Why my differentiation isn't correct?

[strange trellis]

you;ve done

v'u -u'v

you need to do u'v-v'u

so it would be

(e^x)*(x^2+3)-(e^x)(2x)

[static fulcrum]

Ah, I thought we could do any way as we do in product rule.

[strange trellis]

no, because you're subtracting

if you want to avoid that mistake just do product rule of e^x[(X^2+3)^-1]

[static fulcrum]

Okay thanks @strange trellis 👍

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[rancid sundial]

hey quick question

[rancid sundial]

just to confirm
they hv asked to find square root of iota

$$z= a + \iota b$$
$${\sqrt a+\iota b} = -i$$

[warm shale]

Ohio

[rancid sundial]

ib is also under the root

[tardy epoch]

show the original question

screenshot/picture is best

[rancid sundial]

it is just square roots of -i are

[tardy epoch]

just use demoivre

[rancid sundial]

how tho

wait

i can e^i(thetha) to 1/2

oh man
u are really reimann

thank u

[obtuse pebble]

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[open karma]

if $f(x \cdot y) = f(x)+f(y)$ then $f(x)=\log_{a}x$

[warm shale]

Arian

[open karma]

f(x) isn't 0

How to prove that?

[tardy epoch]

f has to be continuous?

[obtuse pebble]

@open karma Has your question been resolved?

[open karma]

It doesn't matter
But please say that your solution is for continuous or uncontinouos

@tardy epoch

[tardy epoch]

https://en.wikipedia.org/wiki/Cauchy's_functional_equation

prove it for x rational, then use density of rationals

https://math.stackexchange.com/a/98675

[obtuse pebble]

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[open karma]

Thank you for you answer @tardy epoch

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[verbal ocean]

hi

[verbal ocean]

can u help me study for the state test for 5th grade?

[tardy epoch]

how old are you

[trim portal]

Under||(age)||graduate

[obtuse pebble]

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[wicked current]

in the cartesian plane the two lines of equation:

[obtuse pebble]

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[midnight veldt]

how do i find the derivative of this graph?

[untold flax]

you'd need the function

is it a quadratic?

[nocturne minnow]

Not really. You can estimate the derivative by looking at the slopes of that function

[untold flax]

but you don't know the intervals

[nocturne minnow]

estimate the derivative

[untold flax]

.. still

how would you estimate with 0 intervals

it could literally be any quadratic rn

ok not any

any above x axis

[midnight veldt]

the function in the graph is f(x), its asking me that if h'(x) = f(x), then h(x) is increasing for xE

[grave ember]

You would have to assume each interval is 1 or something

[midnight veldt]

i may not be approaching it right

[untold flax]

You didn't really say what it's asking you

[midnight veldt]

its asking me this

[untold flax]

oh right

so h(x) is the the integral of f(x) dx right?

not the derivative

[midnight veldt]

yes

so i shouldnt try to derive the graph?

[untold flax]

no

you should be thinking abotu the integral

• the area under the graph

[midnight veldt]

im still confused on how to find h'(x)

[untold flax]

you don't need to find it

you just need to see when h(x) is increasing

[midnight veldt]

sorry i mean h(x)

[untold flax]

h(x) just represents the area under the graph for that point

so it's basically asking, for what x is the area bounded by f(x) and the x axis increasing

[midnight veldt]

so the answer would be (1,infinity)?

[untold flax]

hm, why 1?

[midnight veldt]

sorry, 3

[untold flax]

👍

I believe that's the correct answer

[midnight veldt]

i got it incorrect 😦

it was correct for the normal graph but not for h(x)

[untold flax]

hm I don't know why it would be incorrect

[untold flax]

wdym?

oh nvm I see what you mean

[midnight veldt]

so if f(x) is considered the derivate for h(x), would i need to find the cubic function then?

[obtuse pebble]

@midnight veldt Has your question been resolved?

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[shell mica]

I need some help please

[wooden cipher]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[shell mica]

1

[wooden cipher]

what type of series are these?

[shell mica]

uhh what u mean

?

like grade

[wooden cipher]

what type of sequence is it?

[shell mica]

I don't really...

know it

[wooden cipher]

is it arithmetic or geomtric

[shell mica]

oh

is geomtric

[wooden cipher]

i misspelled it whoops

geometric*

do you know how to find the common ratio?

[shell mica]

I don't know sorry

💀

[wooden cipher]

take the second term dvided by the first term

[shell mica]

can u example for me

[wooden cipher]

the first one

what is the second term?

[shell mica]

-4?

[wooden cipher]

the second term

[shell mica]

second one is 12

[wooden cipher]

divide that by the first term

[shell mica]

-3?

[wooden cipher]

yes

that is the common ratio

[shell mica]

oh okay

[wooden cipher]

to find the explicit formula, you would use the formula $\frac{first term}{common ratio} \left(common ratio\right)^{x}$ where x is the nth term

[warm shale]

GarlicBredFries

[shell mica]

so it gona be

-4/3 (-3)x

oh wait

I can use the bot

$\frac{-4}{-3}\left(-3|right)^{x}

oh nvm

$\frac{-4}{-3}\left(-3|right)^{x}$

[warm shale]

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

[shell mica]

like this

[wooden cipher]

yes

pretty much

then to find the 7th term, you would put x=7

[shell mica]

okay

so then we put the one we just do in

right?

[wooden cipher]

uhh what

[shell mica]

oh nvm I don't know what i am saying right now

please continue

[wooden cipher]

i gave you all the steps to do it, i think you should try the next one

[shell mica]

okay got it

$\frac{-4}{-3}\left(-27)$

[warm shale]

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

[shell mica]

$\frac{-4}{-3}\left{-27}{-3}$

[wooden cipher]

what values of x did you plug in?

[shell mica]

$\frac{-4}{-3}\left{-27}{-3}right$

[shell mica]

7

$\frac{-4}{-3}\left{-27}{-3}/right$

[wooden cipher]

can you take a picture of your work? it seems like youre having trouble with the bot

[shell mica]

I don't have my phone with me right now

I left it at school

[wooden cipher]

oh

[shell mica]

I am trying to use the bot for now

[wooden cipher]

$\frac{-4}{-3}\left(-3\right)^{x}$

[warm shale]

GarlicBredFries

[wooden cipher]

where does the 7 go

[shell mica]

at the x

[wooden cipher]

ok, so im assuming you have a calculator so you would plug in that

[shell mica]

still be -3?

[wooden cipher]

yes

the -3 doesnt change

[shell mica]

okay

got it

(-3)x change to -27

is this

?

[wooden cipher]

why did it become -27

[shell mica]

3x7 is 21

oh shit

:))

wrong number

sorry

[wooden cipher]

also not correct

[shell mica]

oh

....

[wooden cipher]

do you know what $3^3$ is?

[warm shale]

GarlicBredFries

[shell mica]

yes is 81

[wooden cipher]

_>

explain why

[shell mica]

i just multiply by 3 with 3 time

3x3=9

[frosty spoke]

,w 9*3

[warm shale]

Results provided by WolframAlpha

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

Requested by Saccharine#3780

[shell mica]

not that one

sorry

I feel like I trash at this

oh so it gona be 27

[frosty spoke]

wait what's your question again

[frosty spoke]

do you know what a sequence is

[shell mica]

geometric

[frosty spoke]

that is not a proper answer to the question

I asked you whether you knew what something is

and you said some random math word

we are social creatures

so I can only conclude that you have decided to turn your brain off and cough up random math words

[shell mica]

uhh

okay

[frosty spoke]

do you know what a sequence is

[shell mica]

I don't know

[frosty spoke]

have you tried looking it up or in your notes

[shell mica]

oh it might be geometric sequence

oh it is geometric sequence

[frosty spoke]

how are you going to say that if you don't even know what a sequence is smh

[obtuse pebble]

@shell mica Has your question been resolved?

[obtuse pebble]

@shell mica Has your question been resolved?

[obtuse pebble]

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[peak rain]

.close

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[timid silo]

is this right?

[latent walrus]

looks good

[timid silo]

thank uu

.close

[obtuse pebble]

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[novel gale]

what the fuck

[obtuse pebble]

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[covert girder]

can someone help me with this

[tardy epoch]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[covert girder]

i dont understand how to put it into a(1+r)^t

[novel gale]

this looks like vertex formula

[tardy epoch]

a = 1, solve 0.88 = 1+r

[upbeat lagoon]

suppose that just plugging some values till finding the best approximation is a good solution ?

this is rounded to the right values for a thousandth

[tardy epoch]

sure?

[upbeat lagoon]

yeah i was just playing with a and b till i got the best approximation

[tardy epoch]

is solving for r really that hard?

[upbeat lagoon]

he said in the question : " round a and r to the nearest hundredth if possible", if there was a direct solution why would he suggest this ?

[covert girder]

why is this wrong?

[hexed agate]

What happened to the exponent 4?

[covert girder]

oh

what do i do with the 4

[tardy epoch]

what even is the question

this?

[tardy epoch]

oh i didn't see the 4

use exponent rules

,tex .exp rules

[warm shale]

riemann

[tardy epoch]

power of power line

[covert girder]

then do i do -.12^4

[tardy epoch]

it takes multiple steps

you're probably skipping some

[covert girder]

which steps

[tardy epoch]

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[covert girder]

i dont know how to add the 4

[tardy epoch]

use this to re-write (0.88)^(rt)

make it (something)^t

[covert girder]

which rule do i use? im not sure

?

[obtuse pebble]

@covert girder Has your question been resolved?

[covert girder]

!close

.close

[obtuse pebble]

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[untold garnet]

how do i do 3d iii?

[obtuse pebble]

@untold garnet Has your question been resolved?

[frigid prism]

just plug in (2-x) for x

[obtuse pebble]

@untold garnet Has your question been resolved?

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[real reef]

should my next number be like

88? 89 87?

how does one solve this? where it says 8##

and 8##

this lay out is confusing me

https://www.youtube.com/watch?v=_7m0Q_m2ppg like this?

apparently wrong

nvm makes sense they had a horrible example

.close

[obtuse pebble]

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[thick gyro]

dont rest on your laurels yet

[obtuse pebble]

@waxen fossil Has your question been resolved?

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[wide phoenix]

<@&286206848099549185>

[obtuse pebble]

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.

[nocturne minnow]

And you never posted a question

[wide phoenix]

limx→0 f(x)/x =1 limx→0 g(x)/x =2 Evaluate limx→0 f(x)/g(x)

[latent walrus]

whats [f(x)/x]/[g(x)/x]

[wide phoenix]

1/2 ?

[latent walrus]

i mean algebraically, if you simplify that what do you get

[wide phoenix]

f(x)/g(x)

[latent walrus]

yeah, so its just this

[uncut jacinth]

how can you do that?

[wide phoenix]

What I mean is that limit x→0 f(x)/x =1 and the limit x→0 g(x)/x =2

[latent walrus]

i know

[wide phoenix]

evaluate limit limit x→0 f(x)/g(x)

[latent walrus]

yeah, since [f(x)/x]/[g(x)/x]=f(x)/g(x) then lim([f(x)/x]/[g(x)/x])=lim(f(x)/g(x)) as x approaches 0

[uncut jacinth]

but there is no given relationship

between f(x)/x and g(x)/x

[latent walrus]

there doesnt need to be you just want their limits, and since its the same limit for both of them, as x->0, those x's will cancel so you can say the limit is equivalent to just 1/2

$lim(\frac{f(x)}{x})=1$ and $lim(\frac{g(x)}{x})=2$
then $\frac{lim(\frac{f(x)}{x})}{lim(\frac{g(x)}{x})}=lim(\frac{\frac{f(x)}{x}}{\frac{g(x)}{x}}) = lim(\frac{f(x)}{g(x)})$

and i oof

[warm shale]

AℤØ

[latent walrus]

=1/2

the lims are x->0

[uncut jacinth]

how would we then evaluate the limit as x approaches zero for g(x)

?

[latent walrus]

you dont need to know that - you know that the limit of the thing before it is the same as it, you just use that

[uncut jacinth]

thats part a in the question

[latent walrus]

ooh

ok 1 min

[latent walrus]

not entirely sure how to show it at the moment, im tempted to say its just 2

i am pretty sure this holds for (b) though

as for (a) you probably want to get a second opinion

[uncut jacinth]

ok thanks for your help. Another quick question, I am trying to solve it using what we now know from b. If I did something like this, would I be able to distribute the limit back to the 1/x?

I don't think so

so my strategy may not work

[latent walrus]

you cant factor out the 1/x really, at least not like that

[uncut jacinth]

I don't it would be because then what would the possibilites of g(x) be? It must be a rational with asymptote at 0 and thus will approach negative/positive infinity at zero, no?

[latent walrus]

plus you would be factoring out (1/x)/(1/x)=1

[uncut jacinth]

yeah I see what u mean

[latent walrus]

why does g(x) necessarily have to have an asymptote at 0?

[uncut jacinth]

sorry not g(x) itself but g(x)/x, so if I plug 2 back in I will get 2/0 which is not = 2

so it does not satisfy

[latent walrus]

its also not necessarily true that g(x)/x has an asymptote at 0

[uncut jacinth]

common factor?

g(x) could be x(x+2) right, and then the limit would be 2?

[latent walrus]

possible yeah

in fact neither g(x) or g(x)/x can have an asymptote at 0 or this limit doesnt exist

[uncut jacinth]

so would the only equations for both be horizontal lines with a factor x?

how would I algebraically prove that?

[latent walrus]

why horizontal lines? g(x) could be x(x^40+x^39+....+x^2+2)

[uncut jacinth]

oh yeah I see

then how could I approach that problem? I am kind of lost for ideas

[latent walrus]

b seems quite straight forward. but i havent really dealt with something like a without knowing what g(x) is, if i did it would likely come down to l'hopitals rule

but you may want to try asking someone else for this since im not confident in giving a good logically backed answer for (a)

[uncut jacinth]

that is definitely beyond the scope of my class

[uncut jacinth]

ok, thanks again for your help

[obtuse pebble]

@wide phoenix Has your question been resolved?

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[teal turret]

What’d I do wrong? Am I using the wrong formula?

[teal turret]

It’s saying my answer is wrong

this is all it gives

[ashen dove]

Aint u doing the opposite?
I mean isn't the formula the reciprocal?!

[teal turret]

i tried the reciprocal and still got it wrong lol
(65.505)

[thick fog]

I've never used this particular form of the curvature formula before, but this looks fine to me. You should probably enter it as 8/65^(3/2) and not try to round unless it gives specific instructions on how it wants it.

[obtuse pebble]

@teal turret Has your question been resolved?

[teal turret]

ah ok lemme try that

bruh it was correct, cant believe i didnt think of that

thanks for that hexicle and sasuke

.close

[obtuse pebble]

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[timid silo]

How do i do this 😭😭 i hate kites

[broken rune]

simmilar thing

you could find th earea of all 4 triangles

idk if your teacher wants a diff way

but finding the area of each trianlge is one way

[timid silo]

it doesn't matter which way dw

[broken rune]

did that help?

[misty gate]

yo

first mark every pint

like this

now area of AOC=BOC

RIGHT?

hello

?

[timid silo]

SORRY

[timid silo]

yes

[misty gate]

hello

[timid silo]

hello

[misty gate]

do you need more help

[timid silo]

no thank u

.close

[obtuse pebble]

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[steel goblet]

How do solutions to linear diophantine equations work? Like i have the problem 169x-65y = 91.
I solved x = 14-5n and y = 35 - 13n, and x = 14+5n and y = -35-13n. I plugged both of these solutions into my calculator and they both seem correct, yet they look different. Why is that so?

[steel goblet]

nvm i made a typo

.close

[obtuse pebble]

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[shadow lava]

is this a good enough reason?

or would you recommend wording it differently

[latent walrus]

i believe thats good enough

actually

think you have them the wrong way around

approaching from the right the denominator would be -(x^2-9) since x^2 will be <9. whilst approaching from the left it would just be (x^2-9) since x^2>9 in that case

so the working is right besides you needing to swap the limits on the left and right of your working

[shadow lava]

do you think it would matter for marks?

[latent walrus]

yes

[shadow lava]

damn

i don't understand it

i put - for - with absolutes

• for +

but that's not always correct

[latent walrus]

wdym? you can kinda just think about it, if the magnitude of x is smaller than 3 (approaching from the right) then the value x^2-9 will be negative, so by definition of | | it would be -(x^2-9)

the reverse is true for the case of |x|>3 (approaching from the left)

[shadow lava]

oh so you go back to the absolute when thinking about this

not the factors

that makes sense now

x^2 - 9
-3 on the right side will be -2.9999
-2.9999^2 - 9 will be negative

x^2 - 9
-3 on the left side will be -3.0001
-3.0001^2 - 9 will be positive

[latent walrus]

yeah, put your brackets in though

[shadow lava]

for which part?

[latent walrus]

just looks nicer around the negatives

[shadow lava]

lol OK

ty!

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[obtuse pebble]

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[nocturne minnow]

. @timid silo Fyi don't give out answers like what you did in #help-18 with that user, dr buzz. It's not the purpose of the server to do the work for them or give out answers

[dreamy dock]

is C right?

[dark stirrup]

Looks reasonable

I've never seen notation like that before though, ever

I would've said NE, EH, or even NH

[dreamy dock]

oh

thx

I can't understand this

[fiery birch]

@dreamy dock What do you not understand?

[dreamy dock]

what the question means and how to answer it

[fiery birch]

It's asking about the marked angles (in black) and it's asking what can you conclude given the figure

[dreamy dock]

So letter D is correct

[fiery birch]

Yes

[dreamy dock]

thx

how to find the third one?

what formula should i use?

[fiery birch]

☝️Use what you chose as the option in the previous question

[dreamy dock]

I'm trying to find that out

[fiery birch]

You know that the sum is 180 degrees. Then if you know one of the angles, what could you possibly do with that information?

[dreamy dock]

180-50

130

I don't know what this question means

[fiery birch]

Good

[dreamy dock]

is it A?

[fiery birch]

Yes

[dreamy dock]

Thanks so much!

.close

[obtuse pebble]

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[frigid prism]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[dire sigil]

My teacher said to find the relationship between the points to find the one that's the vertex (1, ~4)

[frigid prism]

!work

!show

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[dire sigil]

1

[dire sigil]

I dont know how to start

[frigid prism]

alright so

we have three unknowns: y, m, and b

you can set up 3 equations

and solve

[dire sigil]

it's a parabola

do you mean y = mx + b like linear

or y = a(p - x)^2 + q

[frigid prism]

i was thinking y = mx^2 + b

but right you need the p

so 4 equations 4 unknowns

[frigid prism]

how do you know the vertex is 1, -4

[dire sigil]

oh ok i get it thanks

.close

[obtuse pebble]

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[pure kraken]

can someone tell me the answer to this

[nocturne minnow]

No, we're not here to give out answers

[trim portal]

Think about what fraction of the circle does the arc occupies

[nocturne minnow]

Also, from that partial screenshot, it looks like it's a step by step process

Step one says to find the circumference of the figure

Just follow the 7 steps

[obtuse pebble]

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[hidden slate]

hello

[hidden slate]

i am struggling with question 11b

I dont have a clue in the world

@brisk matrix i need you back

please come back

please

[obtuse pebble]

@hidden slate Has your question been resolved?

[hidden slate]

where did my teacher go

<@&286206848099549185>

[karmic garden]

Gl

[hidden slate]

ok i solved it

closed.

.close

[obtuse pebble]

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[obtuse pebble]

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[proper forum]

I fail to understand how the term underlined in purple = 0

[obtuse pebble]

@proper forum Has your question been resolved?

[hexed agate]

First, have you seen the line with the parenthesis with "because"?

[proper forum]

I couldn't understand its implication in the given step.

[obtuse pebble]

@proper forum Has your question been resolved?

[obtuse pebble]

@proper forum Has your question been resolved?

[lapis marsh]

Hello everyone 👋

[obtuse pebble]

@proper forum Has your question been resolved?

[obtuse pebble]

@proper forum Has your question been resolved?

[tardy condor]

@proper forum

do @ruby fulcrum

pour demande de l'aide

[proper forum]

They had to call Detective Conan himself for this one.. hmm.

[tardy condor]

???

[proper forum]

You've already done that for me now; Thank you.

ah wait lmao

<@&286206848099549185> Je demande votre aide. 😠

[tardy condor]

...

Don't understand

But I understand that you can understand french

[proper forum]

I thought you had pinged helpers for me already but you had pinged @ruby fulcrum instead.

[proper forum]

Only un peu 🥴

[restive swift]

Could it be that the purple term is just not correct? Iirc, that term should just be twice the dot product of (u+v)/2 and (u x v)

Then Dot product being linear should give meaning to the explanation in []

[proper forum]

Ahh right. My bad!

[eager shadow]

Je compris fr tu veux l aide a quoi

Mon fr est tres mauvais mais je compris

[proper forum]

The answer to my query was so simple; I'm surprised that this thread took this long to receive a legible response.

[restive swift]

Kinda am as well

[proper forum]

Regardless; thank you for your time and assistance all.

.close

[obtuse pebble]

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[sand wren]

hello

[sand wren]

how did he move

from the first line to the next

[radiant palm]

q = 2p+2

[sand wren]

This is what i got tho

[obtuse pebble]

@sand wren Has your question been resolved?

[sand wren]

<@&286206848099549185>

[fathom flicker]

Off-topic and occupying someone else's help channel <@&268886789983436800>

[obtuse pebble]

@sand wren Has your question been resolved?

[obtuse pebble]

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[trim mauve]

Can you guys choose a number between 1-2

[viral blade]

√π

[warm shale]

bee

[viral blade]

↑

[trim mauve]

Huh

Choose

1 or 2

Wuickly

1 or 2

[rocky goblet]

you asked for numbers between 1 and 2, we gave you numbers that are between 1 and 2

[fading quest]

$\ket{\text{decision}}=α\ket{1}+β\ket{2}$

[trim mauve]

I said 1 or 2

[warm shale]

Duh Hello

[rocky goblet]

you said "between 1-2"

[trim mauve]

I changed it

1 or 2

Pls

😭

I need it wuickly

[fathom flicker]

7

[trim mauve]

1 or 2

[fathom flicker]

ohhhh

8

[hexed agate]

@hexed hearth

[trim mauve]

Pls

[fading quest]

my decision is a superposition between the two, you need to measure it to find out what state the decision is in

[fathom flicker]

^

[rocky goblet]

{1, 2} is a set, and isn't empty
therefore, by the axiom of choice, there exists a choice function on that set
use the choice function on the set {1, 2}, and you get either 1 or 2, so there you go

[trim mauve]

¯_(ツ)_/¯

[hexed agate]

A "fat" creature asking for a random number who doesn't even understand what "between" means.

[obtuse pebble]

@trim mauve Has your question been resolved?

[obtuse pebble]

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[rotund vector]

can someone help me out

[rotund vector]

is 1.06x10^-19 the same as 9.97x10^-20?

[hexed agate]

How could it be same, Mr Lex Fridman?

[tardy epoch]

,w does 1.06x10^-19 equal 9.97x10^-20

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[nocturne minnow]

Love how it says "always true".

[fierce lagoon]

Can 5 = 7

Always*

For all x

[rigid lintel]

lol

[fierce lagoon]

In R

[hexed agate]

It is true in the context that both are real numbers for example

[rigid lintel]

the 10^x part is only concerning the decimals

and 1.06 is not 9.97

[drifting badger]

They are approximately the same, though. So if you for example got the two numbers from two different methods of solving something, the difference might have been caused by rounding errors.

[rotund vector]

I just did a physics problem and got the first value although the second value is the correct answer

Just not sure where I went wrong

[obtuse pebble]

Show your work, and if possible, explain where you are stuck.

[fierce lagoon]

Well they're really small numbers

[rigid lintel]

rounding error

[hexed agate]

Lex from mathematics to physics

[rotund vector]

I'll send my working hold on

[fierce lagoon]

Any intermediate rounding can result with such differences

[hexed agate]

What went wrong bro

[nocturne minnow]

Always round at the end

[tardy epoch]

yea never square brackets the end

(at the end)

[nocturne minnow]

Technically 1.06x10^-19 is equal to 9.97x10^-20 if you round them to zero since they are really really small numbers

[fierce lagoon]

They're really small

Like

Not even micro

They're like ubermicromicromicromicro

[rotund vector]

So how would I get the second value is there a rounding error anywhere in my calculations?

[drifting badger]

I would say they are average. Maybe even too big. Those numbers are more than big enough to satisfy anyone

[fierce lagoon]

Yeah my bad I live off of 9.9 * 10^(-32) calories a day

[rocky goblet]

i don't think talking about the size of these numbers is actually helpful or relevant

[rotund vector]

Cause if I made an error like this in my assessment I might get the question wrong

[tardy epoch]

don't substitute numbers until the very end

leave everything as variables

[nocturne minnow]

Also right here was rounded wrong too

It would have been 4.3e-7

[rotund vector]

So when it is 5 I round up?

[nocturne minnow]

That's how rounding works

[rocky goblet]

well in this situation you shouldn't really be rounding it at all, you should leave that as 4.25 x 10^-7 and then round at the very end of the calculation

[rotund vector]

Oh ok

[nocturne minnow]

Do you have the solution to that question? Because using the values you have, does not result in that

[obtuse pebble]

@rotund vector Has your question been resolved?

[rotund vector]

here is the answer

[tardy epoch]

it would have been a lot faster if you just showed this in the beginning

[tardy epoch]

instead of asking this question

you just xy'd a lot of people

https://xyproblem.info/

[obtuse pebble]

@rotund vector Has your question been resolved?

[obtuse pebble]

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[violet ingot]

I'm very lost on how I'm supposed to find f'(x) here

[daring rock]

Just express f'(x) in terms of p(x) and q(x)

Without equations for p and q, you can't really be more specific than that

[violet ingot]

How do I tell which point on the graph that is?

[daring rock]

You don't even need the graph for this part

All you need to know is f(x) = p(x)*q(x)

Given that, how can you express f'(x)

[violet ingot]

With what numbers ?

Nvm

I'm dumb

Just realized

[daring rock]

you got it?

[violet ingot]

Eh

Kinda?

Don't know yet

[daring rock]

since f is the product of two functions

You just need the product rule

Do you know the product rule?

[violet ingot]

Yes

[daring rock]

So if

f(x) = p(x)*q(x)

then the product rule says

f'(x) = p(x)q'(x) + p'(x)q(x)

[violet ingot]

Then I take the derivative of q'(x) multiply it by x and do the opposite for the other side and them add?

[daring rock]

hm? no

[daring rock]

this is it

[daring rock]

part a is right, do you see how we got that?

[violet ingot]

Substituting all the x in p(x) × q(x) with x?

[daring rock]

part a was just the product rule

but then for f'(0) you should replace all the x's with 0

in part b

[daring rock]

and make sure your q' stays q'

the q in the first term should be q'

But then you can just find p(0), p'(0) etc from the graph

[violet ingot]

How do I get p'(0) from the graph?

I think it's done like how lim functions are but I'm not sure

[daring rock]

p'(0) is the slope of the graph at x=0

[violet ingot]

-2?

[obtuse pebble]

@violet ingot Has your question been resolved?

[obtuse pebble]

@violet ingot Has your question been resolved?

[obtuse pebble]

@violet ingot Has your question been resolved?

[obtuse pebble]

@violet ingot Has your question been resolved?

[obtuse pebble]

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[vapid olive]

can anyone help me with part c of this question?

[vapid olive]

i posted it like 2 days ago but left it cause it got too late

[rich tapir]

write out S - S_n

what do you have

you just have to solve for n first and we know n is an integer

[obtuse pebble]

@vapid olive Has your question been resolved?

[obtuse pebble]

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[royal basin]

what's your question

[winged aurora]

If i have a/b, a/b is an integer and k|a then k|{a/b} ,is it wrong to say this ?

P|p! So can't we just say p|pcr , why is this proof required

We know pcr is alwas an integer , pcr = pq + 0 ,so p|pcr

[royal basin]

of course it is wrong. 7 divides 42 but 7 doesn't divide (42/14)

[winged aurora]

My bad ..

I was thinking about it in the wrong way my bad .

Thanks for the help

.close

[obtuse pebble]

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[frosty brook]

best way to do this?

[frosty brook]

.close

[obtuse pebble]

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[glacial horizon]

Question 13. why would C be incorrect? I mean a vector would technically be collinear with itself right

[elfin burrow]

,rccw

[elfin burrow]

lol yeah, the wording asks only for a sufficient condition for colinearity

whereas i think it probably means to ask for a necessary condition

[glacial horizon]

but if i go by the question literallt , C technically isn’t wrong right

[royal basin]

something soothing division by zero

[royal basin]

indeed it isn't but good luck explaining that to 99 exam boards lmao

[obtuse pebble]

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[obtuse pebble]

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[humble zinc]

How can we solve it without guessing?

[obtuse pebble]

@humble zinc Has your question been resolved?

[gilded needle]

the strict inequality in (a) seems suspicious, what if f is a constant function?

[humble zinc]

f is increasing is a given

[gilded needle]

does that mean strictly increasing?

[humble zinc]

Yes

It's mentioned in the first line

[gilded needle]

first line does not say "strictly", it just says increasing

[humble zinc]

I don't know, this question is from a calc 2 course

[gilded needle]

well b and d are clearly true, and if c is true then e is true

[wary vigil]

$\left( \frac{b - a}{n} \right) [f(b) - f(a)] = R_n - L_n$

[warm shale]

Katharine

[gilded needle]

can you show c?

if so, then the only possible one that's false is a

[wary vigil]

you can just fill that into all the equations

that have it

i mean apart from c because that is the same equation

[gilded needle]

also clearly either a or e is false

[wary vigil]

if you just fill in c into a and e

you will see which is false

[gilded needle]

(assuming c is true, of course)

[wary vigil]

it is but yes

[humble zinc]

Thank you ❤️ for both of you, gonna review what you said later

Both of you have the role 'advanced' this question is clearly difficult 😭

[gilded needle]

eliminating some of the answers is easy... you know (b) is true since f is increasing, and (d) is just a rewrite of the second inequality in (b), so both of those are true

(a) and (e) have the same LHS and RHS but one has < and the other has >=, so one is true and the other is false

so basic logic narrows it down quite a bit

[obtuse pebble]

@humble zinc Has your question been resolved?

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[distant schooner]

Let f:R->R, f(x) = ax+b, where a and b are non-zero constants. Given that f(1) =2 and f^-1(1) = 3, find the values of a and b

[distant schooner]

so i have formed

f(1) = a+b=2

and

f^-1(1)=a+b=3

and also concluded that a and b cannot equal to 0

however i cant simultaneous equation to find out the value of a or b since they both cancel out

so how else do i find it

[misty iron]

Hmm, your inverse looks a bit wrong.

[distant schooner]

and i think

i did one of it wrong

yeah

[misty iron]

Anyways instead of calculating inverse. Use it the other way. f(3)=1

[distant schooner]

oh

is that like a formula?

[misty iron]

So you don't need to calculate the inverse.

[distant schooner]

alright

lemme try that

f(1) = a+b+2
f(3) = 3a+b=1

lemme calculate that now

for a i got -1/2

[misty iron]

No, it's not a formula. Just an observation.
f^-1 output is the input for f.

[distant schooner]

yeah i noticed that too

and for b i got 5

lemme just check the answers

hm

i got b wrong

[misty iron]

Substitute the value of a in a+b=2.

[distant schooner]

ok i got it

before

i did

-1/2 + b = 2

and i just multiplied 2 over

without timsing the

b

xD

.close

[obtuse pebble]

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[timid silo]

Is this the correct way how I solved it?

<@&286206848099549185>

.close

[obtuse pebble]

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[weary belfry]

Can someone help me with this?

[weary belfry]

im pretty sure its the second choice but thats not my problem

I dont know what to do after i get cos(theta) in the denominator

nevermind i just forgot to sub in the x^3

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[obtuse pebble]

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[livid patrol]

can someone help me with this

[radiant palm]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[livid patrol]

1

[radiant palm]

If one bead is taken at random. Can you calculate the probability that it is red?

[livid patrol]

7/11

[radiant palm]

Now, say you picked out the bead and it was red. You take out one more bead. What is the probability that it is white?

[livid patrol]

4/10

[radiant palm]

Great.

Now, go back to the original situation. All 11 beads in the bag. If you pick out two beads at random, what is the probability that the first one is red and the second one is white?

[livid patrol]

the first one is red would be 7/11 and the second is white would be 4/10 am i wrong?

[radiant palm]

That is correct. What about the probability that BOTH happen?

[livid patrol]

what do you mean

[radiant palm]

Okay, lets do a simpler example

Lets say I toss a coin two times. What is the probability that I get heads both times?

[livid patrol]

2/4

[radiant palm]

Why do you think it is 2/4?

[livid patrol]

because if you toss it once its 1/2 and twice its another 1/2 so 2/4 no?

i got to go now ill try and figure it out for myself

i think i got it

th

thanks for the help

[radiant palm]

This is incorrect

[livid patrol]

i know

is the answer 1/4

[radiant palm]

Yes

[livid patrol]

for the tossing coin

ok i understand now

thank you for your help

[obtuse pebble]

@livid patrol Has your question been resolved?

[silver zenith]

If a polynomial function is getting divided by (x-2)^6 and remaining is x^2-5x+1, what is remanining from divison with (x-2)?

also, @livid patrol you should mark your question as resolved

[obtuse pebble]

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[silver zenith]

uhm

If a polynomial function is getting divided by (x-2)^6 and remaining is x^2-5x+1, what is remanining from division with (x-2)?

[obtuse pebble]

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[maiden nova]

Caio is a craftsman and will make a sequence of five dolls, so that the smallest is 2cm tall.

How can we get the height of the biggest doll knowing that one is 3/4 of the height of the other?

[royal shard]

suppose two dolls next to each other
the left one is a mystery, the right one has the sice 2 cm
we know that since the right one is 3/4 of the left one that:
right = left * 3/4
rearranging, we get:
left = right * 4/3

now suppose we have three dolls
let's label them a b and c

applying our just gathered knowledge:
b = c * 4/3

we can do the same for a:
a = b * 4/3
therefore
a = c * 4/3 * 4/3 = c * (4/3)^2

so if we know the one on the right, and want the nth doll to the right, we multiply 2cm by (4/3)^n

in your task, we have 5 dolls
looking at the smallest, we have to go 4 dolls to get to the one we want

so we multiply by (4/3)^4

[maiden nova]

Ok thank you

.close

[obtuse pebble]

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[frail gorge]

hi could anyone help me with this question

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[frail gorge]

I dont know where to begin

[sage geode]

You can recall that the average speed is the same thing is the total distance divided by the time of the entire journey

[misty iron]

#❓how-to-get-help

[sage geode]

Post your question in an available help channel

[quiet lily]

I need help with 21

[frail gorge]

yes i do

[sage geode]

So, what distance has been travelled from 0 to T and from T to 2T?

[quiet lily]

@sage geode

Please help me with 21

[sage geode]

.

[frail gorge]

@sage geode how do i find the value of V

[sage geode]

Can you answer this first?

Cuz that leads to the value of V

[frail gorge]

k

from T to 2T is (2T -T) = T

brb

[sage geode]

That's the time between T to 2T, not the distance travelled

[sage geode]

Alright

[royal shard]

we have been given the average speed
try to find a formula for it
between 0 and T the average speed is just V
between T and 2T the average speed is what?
once you have a formula for the complete average speed, set that equal to the 37.5 and you will get V

for the next task, try to get a formula for the function from T to 2T
i did that with integrating, but you can also just think about it
the function will have a constant that we don't know
we can find it though by setting some stuff

[obtuse pebble]

@frail gorge Has your question been resolved?

[frail gorge]

@sage geode i cant find the distance

[sage geode]

Generally the distance travelled is usually the area that the graph makes

[frail gorge]

yes

ik that

[sage geode]

Here you can cut the area into a rectangle and a right triangle

[frail gorge]

ok

[sage geode]

The rectangle has sides T and V, meaning its area is VT

[frail gorge]

yes

[sage geode]

And the right triangle has legs V and T (because 2T - T = T like you said)

So its area is VT/2

[frail gorge]

yes

[sage geode]

Adding these two areas gets us 3VT/2

Okay, now we need to divide this by the time of the entire journey

In order to get the average velocity

[frail gorge]

yes

[sage geode]

Dividing 3VT/2 by 2T yields 3V/4

But it's given that the average speed is 37.5m/s

So 3V/4 = 37.5

Can you solve for V from now on?

[frail gorge]

oh yes thanks so much

[sage geode]

You can calculate the value of T using the fact that the deceleration between T and 2T is 2.5m/s^2

[frail gorge]

thanks

[obtuse pebble]

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[subtle wolf]

How to find "a"?

[tawny fog]

Divide by 5 in both sides

[subtle wolf]

Thanks bro!

.close

[obtuse pebble]

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[bleak isle]

What is 2+2-3x3

[fiery birch]

Use PEMDAS

[bleak isle]

....

I was just testing

[fiery birch]

Testing what?

[obtuse pebble]

@bleak isle Has your question been resolved?

[obtuse pebble]

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[static patio]

Given a binary relation $R$ over $Z_n$, decide if there are 8 relations
R1, R2, R3, R4, R5, R6, R7, R8 such that R is their union and all relations satisfy:
$\forall x \forall y \forall z (Ri(x,y) \wedge Ri(x,z) \implies Ri(y,z))$.
Decomposed relations do not have to be disjoint and also can be empty.

[warm shale]

Michal

[static patio]

can someone give me an idea how to solve this with help of backtracking?

[obtuse pebble]

@static patio Has your question been resolved?

[obtuse pebble]

@static patio Has your question been resolved?

[obtuse pebble]

@static patio Has your question been resolved?

[obtuse pebble]

@static patio Has your question been resolved?

[frigid prism]

What math is this

[static patio]

It's from introduction to math logic class

[frigid prism]

I would ask it in advanced

[frosty spoke]

it's probably writing an algorithm that involves backtracking

[frigid prism]

that was my first guess but it seems like more math than cs

[obtuse pebble]

@static patio Has your question been resolved?

[obtuse pebble]

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[obsidian wagon]

I have math/programming problem, I have variable that increases every second in logarithmic speed.
Now I need another variable that calculates the opposite. I need to flip the function 🤔

[median dome]

when you flip across the line y=x, you get the inverse of the original function

so you would get an exponential function

but your graph doesn't quite look like a logarithmic growth so

[obsidian wagon]

its hand drawing

[median dome]

essentially you can just replace all the y's of the original equation with x's and vice versa

to flip across the line y=x

[obsidian wagon]

ok, but what i only have as in input is a number

[median dome]

what is your original function?

[obsidian wagon]

nothing else, so i need a "fillped" value in the specific time

like, its not exactly the math, i have a variable in javascript that adds some random number increasingly over the time

that number becomes bigger over time