help-10

[blazing pine]

Where have you drawn g(x)??

[eager hawk]

Yh

[eager hawk]

I didnt draw it

[blazing pine]

Then draw it

[eager hawk]

I written it in piecewise function

[blazing pine]

Not that hard to do

[eager hawk]

Function

I am getting 8 points

Not continuous

And not derivable

[blazing pine]

Name the points

[eager hawk]

-1,0,1,(from here all cubeth roots- 2,3,4,5,6,7)

No sorry -1 wont come

[blazing pine]

Good

[blazing pine]

Good

So you get 8 points

[eager hawk]

-1 will come in nonderivable points

[blazing pine]

No.

Think again

[eager hawk]

Hmm

[blazing pine]

g will be 0 from -2 to 0

[blazing pine]

Youre thinking abt f(x) when saying this i presume

[eager hawk]

Uhh

Wait

Why is -1 non deriable wont come

?

[blazing pine]

Btw I've solved the exact same book : )

[eager hawk]

..

Lol

12th or 11th ?

[blazing pine]

I dont recall

Are you in the andheri batch

[eager hawk]

U passed 12th?

[eager hawk]

No im from other school

[blazing pine]

Yeah.

[eager hawk]

Not andheri

I think

[blazing pine]

Youre not from Sri Chaitanya?

[eager hawk]

Im from it

Idk andheri

Im nucleus batch

[blazing pine]

Which branch are you in then

Yk what probably best to take this chat in dms

For now we'll leave it at this

[eager hawk]

Wait a min

Just tell me athow many poins is itnon derivable

[blazing pine]

Bc g will be 0 from -2 to 0

[blazing pine]

My job is not to blurt out answers its to help you find them

[eager hawk]

No i mean i got the answer as 8

Is it correct

[blazing pine]

For non derivable points?

[eager hawk]

Yh

[blazing pine]

Yeah

Youre right

[eager hawk]

Thank youuu

[blazing pine]

Same for discontinuous as well

!done

!done

[obtuse pebble]

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

[eager hawk]

.close

[blazing pine]

[obtuse pebble]

Channel closed

Closed by @eager hawk

Use .reopen if this was a mistake.

[eager hawk]

Bro 13th one

[blazing pine]

Use .reopen

To open the channel again

[eager hawk]

.reopen

[obtuse pebble]

✅ Original question: #help-10 message

[eager hawk]

13th

[blazing pine]

!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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[eager hawk]

1

[blazing pine]

Start by taking everything to 1 side im both inequalities

[eager hawk]

Please can u give me handwritten solution??

U here?

[sharp goblet]

I can help

so can I see the image

@eager hawk

[obtuse pebble]

@eager hawk Has your question been resolved?

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[urban bolt]

im on part a idk if i did it right

[urban bolt]

this was the best i could come up with idk if its right

[wary badger]

why did you write an infinite sum

its a third degree taylor polynomial

not the full taylor series

[urban bolt]

was i supposed to write P sub 4?

it starts at the second term, so that confused me a little

[wary badger]

well no you use the graph for f(0) and f'(0)

the table is for the higher order derivatives

[urban bolt]

OH

[wary badger]

part a literally just asks for $f(0) + f'(0)x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3$

[warm shale]

knief

[urban bolt]

wait i got this now

i'll be back

[wary badger]

make sure to not write $\sum_{n = 2}^{\infty} f(x)$

[warm shale]

knief

[wary badger]

for starters it doesnt make sense to be summing f(x) here

[urban bolt]

replace it with P sub 4 right?

[wary badger]

but secondly even if you did it correctly with the f^n(0)/n! x^n its still wrong because we want the taylor polynomial, not the taylor series

[wary badger]

P_3

or T_3 however you label it

[urban bolt]

oops

typo

here we go

[wary badger]

how did you get -3x?

tangent line looks like y = -2x + 3 to me

hence f'(0) = -2

[urban bolt]

i kinda did f(0) of the tangent line

[wary badger]

🤔

that would be 3 though

the tangent line has the same value that the function does at that point

[urban bolt]

times (x+0)

[wary badger]

🤔

f'(0) is the slope of the line tangent to f at x = 0

the line is -2x + 3

so the slope is -2

[urban bolt]

i forgot calc 1 so easily

but yeah that makes me remember

[wary badger]

,w translate 爱猫咪的人

[warm shale]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by kniefy

[urban bolt]

cat lover 😭

[wary badger]

lol

interesting

can you write those symbols?

how long would it take to write that by hand?

[urban bolt]

anyways, part b should be a multiplication of powers right?

[urban bolt]

my chinese friend suggested it for me 😭

[wary badger]

oh i see

[wary badger]

well lets start with the taylor series for e^x

do you remember it?

[urban bolt]

1+x+x^2/2!+...+x^n/n!

[wary badger]

mhm

so we only want the second degree taylor polynomial for e^x f(x) so we can effectively disregard terms higher than degree 2 for both e^x and f(x)

so lets look at

$\left(1 + x + \frac{x^2}{2}\right)\left(3 - 2x + \frac{3}{2}x^2\right)$

[warm shale]

knief

[urban bolt]

right so i take the first 2 terms of my e^x and smash it with the first two terms from part a

or 3

[wary badger]

yea but we will end up disregarding some of these terms as well

[wary badger]

anything from this expansion that has degree > 2

so you can expand this with that in mind

you don't have to do the entire expansion

just take the terms that will end up with degree <= 2 after multiplying

[urban bolt]

i really have to foil all that?

[wary badger]

no you don't

for example the x^2/2 does not need to be multiplied by anything other than 3

the 3/2 x^2 doens't need to be multiplied by anything other than 1

etc

[urban bolt]

oh right, so thats what u meant

i got 3+x+6x^2/2!

[wary badger]

how did you get 6x^2/2!?

[urban bolt]

I FORGOT THE 2x^2

[wary badger]

ok but 2! = 2 and you get 3/2 + 3/2 - 2 = 1

[urban bolt]

wait what

[wary badger]

for the coefficient of x^2

[urban bolt]

hang on wait i think i see it now

[wary badger]

noooo

2/2 = 1

3 + x + x^2

[urban bolt]

I FORGOT TO CANCEL THE 2

minor mistakes

ok there we go

part c

its approximation

i think i can handle the integral part

should i take the first two terms?

i hear thats what engineers do

[wary badger]

no you use the third degree polynomial

you integrate that from 0 to 1

[urban bolt]

why 3 terms?

https://tenor.com/view/angai313-spongebob-sad-spongebob-crying-gif-21679826

[wary badger]

it tells you to use the taylor polynomial found in part a

[urban bolt]

so that entire thing?

[wary badger]

3 - 2x + 3/2 x^2 - 23/12 x^3

[urban bolt]

it says to approximate

so can i do h(1)=3?

[wary badger]

you approximate by using the taylor polynomial

since we dont know f

using the taylor polynomial is the approximation

$\int_0^x f(t) \dd{t} \approx \int_0^x P_3(t) \dd{t}$

[warm shale]

knief

[wary badger]

use $\approx$

[warm shale]

knief

[wary badger]

for the second equality

where you used to taylor polynomial

because it isnt =

[urban bolt]

i would get points marked off for that?

scary

[wary badger]

your final answer should be a number, why are you approximating the arithmetic

,calc 3 - 1 + 0.5 - 23/48

[warm shale]

Result:

2.0208333333333

[wary badger]

do not say approximately 3

we already approximated

by integrating P_3

[urban bolt]

so should i leave it unsimplified?

ap frqs dont need simplifying

[wary badger]

yea you can leave it as 3 - 1 ....

they dont care about simplification

[urban bolt]

part d should be a lagrange error bound question right?

actually i'll do it later

i need to get food

i havent eaten all day

ty for the help tho

.close

[obtuse pebble]

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[wary badger]

alternating series

[obtuse pebble]

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[warm shale]

Ashrafisgood

\begin{gather*}
If $a \in \mathbb{N}$, show that the number of positive integral solutions of 
$$x_1 + 2x_2 + 3x_3 + \ldots + nx_n = a$$ 
is equal to the number of nonnegative integral solutions of 
$$y_1 + 2y_2 + 3y_3 + \ldots + ny_n = a - n(n+1)/2.$$ 
(A solution of the above means a set $\{x_1, x_2, \ldots, x_n\}$ satisfying 
$x_1 + 2x_2 + 3x_3 + \ldots + nx_n = a$ etc.)
\end{gather*}
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.56 \end{gather*}
                  
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

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

[weak harness]

,,If $a \in \mathbb{N}$, show that the number of positive integral solutions of
$$x_1 + 2x_2 + 3x_3 + \ldots + nx_n = a$$
is equal to the number of nonnegative integral solutions of
$$y_1 + 2y_2 + 3y_3 + \ldots + ny_n = a - n(n+1)/2.$$
(A solution of the above means a set ${x_1, x_2, \ldots, x_n}$ satisfying
$x_1 + 2x_2 + 3x_3 + \ldots + nx_n = a$ etc.)

[warm shale]

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

[obtuse pebble]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[royal basin]

!msgdel

[obtuse pebble]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[weak harness]

ok

.close

[obtuse pebble]

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[brisk arrow]

Honey

[brisk arrow]

My loves

A to B, B to A

.close

[obtuse pebble]

Channel closed

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[atomic marsh]

[royal basin]

<@&268886789983436800> troll

[polar fossil]

you've been warned about wasting help channels several times before. take a week off.

[obtuse pebble]

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[graceful stream]

im completely lost while trying to learn induction, after checking n=1, 2, 3. do i insert n+1 anywhere? idk what to do with the ar^n-1?

[dark stirrup]

Assume $$\sum_{i=0}^{k-1}ar^i=\frac{a(r^k-1)}{r-1}$$ then use that assumption to prove $$\sum_{i=0}^{k}ar^i=\frac{a(r^{k+1}-1)}{r-1}$$

[warm shale]

SWR

[graceful stream]

thanks, i'll have a go at this

[red ice]

for sums, the technique is sub in the first n terms (use the induction hypothesis)
then add the (n + 1)th term on to that

and it always helps to write down your goal in what you want to prove

[obtuse pebble]

@graceful stream Has your question been resolved?

[obtuse pebble]

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[real trench]

i cant do D i

[real trench]

and C

[mighty geyser]

hmm, what have you tried so far

[real trench]

like c what are they asking like to use the graph? cause i can solve it with a calculator thing but idk if it ask me to draw a line

[real trench]

D i i tried drawing the line that only tocuhes 2 point 1 on the tangent and one on the other line

but idk what the equation would look like

[mighty geyser]

I don't think you need to find m

[real trench]

whats m?

[mighty geyser]

m as mentioned in (d)(i)

I don't think you have to find it

[real trench]

do i not need to draw the equation of the line?

[mighty geyser]

you just need to draw the line

but nowhere does it tell you to state the equation

[real trench]

ok

[real trench]

what about c

[mighty geyser]

for (c)(i) you might be able to check that directly from the graph

notice $y=\frac{1}{4}(x^3-6x^2+8x)$ so for (c)(i) you are really searching for when y=2

[warm shale]

Element118

[real trench]

oh

yeah i didnt see that

about cii

how do i use the graph for that

is it the line 8x+4?

[mighty geyser]

you are searching for when y=2

yeah the (c)(ii) you are looking for a different line

[real trench]

yeah

[mighty geyser]

careful, did you forget to multiply/divide by 4 somewhere

[real trench]

so

we try to make the equation on Cii same as the original one right

like by adding +8x and -8x

then the bring 8x to the other side

[mighty geyser]

yeah, try to spot where you can find $\frac{1}{4}(x^3-6x^2+8x)$

[real trench]

and we get a straight line

?

[warm shale]

Element118

[real trench]

man idk

how do i do it

[mighty geyser]

$x^3-6x^2+4=0$

but we wish to have $x^3-6x^2+8x$, so we have to add $8x$ and subtract 4 from both sides

$x^3-6x+8x=8x-4$

[warm shale]

Element118

[mighty geyser]

do you see how you can introduce $y$ into the equation @real trench

[warm shale]

Element118

[real trench]

@mighty geyser what happened to the 1/4

[mighty geyser]

you have to introduce it back

[real trench]

so we divide all by 1/4?

[mighty geyser]

*multiply by 1/4

or divide by 4

[real trench]

yeah

so (8x-4)/4

[mighty geyser]

you can see you recover an equation of a line after dividing by 4

[real trench]

we draw that line?

[mighty geyser]

$y=\frac{8x-4}{4}$

[warm shale]

Element118

[real trench]

2x-1?

[mighty geyser]

yeah you can simplify the right hand side

can you sketch that line

[dawn prairie]

Are u still stuck on this

[real trench]

nope

i found it

ty

also

i have a test about this kind of stuff tomorow graph of function and derivation thing

but

i can open book and bring my notes

do you guys have any tips on like what should i write on my notes

[mighty geyser]

probably understanding how intersections of two graphs work

the set of (x, y) pairs that lie on the intersection, satisfy both equations

[real trench]

just that?

if i like write every single way to solve every question do oyu think im good

[real trench]

for like every type of question

[mighty geyser]

maybe not just that

but that's one thing to keep in mind

you can go through your notes and past question and see what you are not very confident on, and add those to your note

[real trench]

like for this type of topic theres not much variation to the questions right? all they do is just change the function

[obtuse pebble]

@real trench Has your question been resolved?

[real trench]

like the hardest part is knowing what they are asking for

i dont get something what they mean by solve ... or like find ....

[obtuse pebble]

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[sonic prism]

hi

[sonic prism]

what is the limit boundaries of theta for that shape??

I know it spherical system theta is the angle between Radius and z axis But here R is not given so how can we know theta?

[obtuse pebble]

@sonic prism Has your question been resolved?

[hexed surge]

you don't need the radius to have theta

only the angle

[sonic prism]

angle between what and what ?

[hexed surge]

between the z axis and the line going through O and your point

I think here you need to verify stokes theorem for all R anyways

[sonic prism]

I see..

do u know how can we obtian the integral A.dl for the 2nd part of strokes theorem ?

dl= ax dx +aydy+az dz

but what about A?

[sonic prism]

hold up I got confused I used cartesian coordinates here

it should be dl= a_R dR + a_theta R dR + a_phi R sin(theta) d phi

[hexed surge]

dl does not depend on A, what you wrote is A.dl

oh

wait

Are a_R, a_theta and a_phi vectors?

[sonic prism]

unit vectors

[hexed surge]

ok

right

the you need to do the dot product of the field A and dl

[sonic prism]

alr

but

what is A though?

is it the given one in the question ?

bcz

[hexed surge]

yes

[sonic prism]

That's how my instructor solved it

and it confused me alot

[sonic prism]

I mean the answer doesn't look like they used A = a_phi sin (phi/2)

[hexed surge]

yes they did

[sonic prism]

oh

I see

interesting

replaced R with b theta with pi/2

because they're fixed along the path

I think

so phi is our vector and we dot product it with the dl of a_phi

alr alr I think I got it

thanks alot

you're a legend

bye

.close

[obtuse pebble]

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[zealous spade]

limits help plz, evaluate

[zealous spade]

i tried l'hoptial and got rid of the integral

then i forgot what to do

[fair tundra]

is ans 1/2?

[zealous spade]

yea

[fair tundra]

do yk limx->0 ln(1+x)/x =1

[zealous spade]

oh

that helps...

[fair tundra]

did you get it?

[zealous spade]

yea

[fair tundra]

glad to help

[zealous spade]

i'll search that lim up

thanks

.close

[fair tundra]

sure

[obtuse pebble]

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[last ore]

guys say we have this sequence righrt

[last ore]

How do we know that as n tends to inf

if an --> l

then a_n+1 --> l

aswell

like rigirously explaning

[opaque dome]

Unicity of the limit

[last ore]

how do i prove this?

[tame python]

what is unicity

[last ore]

no clue

[tame python]

do you mean uniqueness?

[opaque dome]

Yeah

Uniqueness

[tame python]

what is unicity??

did you make this word or did you gather it from somewhere else?

[royal basin]

it's french.

unicité

anyway, if a sequence approaches some limit then any subsequence of it approaches the same limit

and ${a_{n+1}}$ is a subsequence of ${a_n}$ which, in non-rigorous terms, you get by throwing out the 1st term.

[warm shale]

Ann

[royal basin]

but if you want to bury yourself in a mountain of epsilons and work out the full formal completely rigorous fully bureaucratized proof of $a_{n+1} \to L$ with the entire machinery of the $\ep$-$N$ definition then ig you're welcome to do that in your own spare time

[warm shale]

Ann

[last ore]

ohh so if i say i could remove finitely many terms of a sequence but the tail of the sequcne remians the same

[royal basin]

yes

fucking up a sequence at finitely many terms doesnt change its limit

this includes deletion of terms as well

[last ore]

and thats the same thing as showing this

if a_n tends to L

[royal basin]

1 is a finite number

[last ore]

i somewhat see it yh

its hard to picture how ur a_n+1 is just removing the first number

[royal basin]

a_2, a_3, a_4, a_5, ...

[last ore]

ohh wait yh

ahh i see

alr tysm

.close

[obtuse pebble]

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[tawdry sail]

is there a way to find the Value of the Angle D , i dont see which triangle to use

[tawdry sail]

i need to prove that its $D = 4i_2 -2i_1 - \pi$

[obtuse pebble]

@tawdry sail Has your question been resolved?

[obtuse pebble]

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[ember frost]

essentially, i need to convert a float number from one format to another, where format A has bias 15 and format B has bias 7

[ember frost]

the question mentions that if rounding is necessary, then i should round it towards +infty, but since we are increasing the precision bit, no rounding should be necessary right?

did the first 2 and this is what i got

essentially i just need to rewrite the exponent part and append a 0 to the precision bit

[obtuse pebble]

@ember frost Has your question been resolved?

[obtuse pebble]

@ember frost Has your question been resolved?

[past sand]

Why does your first value in format B have more bits?

[ember frost]

ah it's a typo, i meant 0111

[past sand]

Okay, that looks fine to me

[ember frost]

alright thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[iron edge]

Alright, time to move on to the next question

[iron edge]

I'm currently on the second question

Let me write down my attempt

Because the route with RL and with R1 are parallel, therefore they share the same magnitude of voltage

[rose scroll]

RL and R2, not RL and R1

[iron edge]

$\frac{R_LR_2}{R_L + R_2} + R_1$ is the total resistor in the system

[rose scroll]

yea, thats good

[iron edge]

Right, thanks for mentioning that

[iron edge]

is this statement correct?

[rose scroll]

you didnt change it... it still reads RL and R1

so no, its wrong

[iron edge]

Let me fix it

Because the route with RL and with R2 are parallel, therefore they share the same magnitude of voltage.

$\frac{R_LR_2}{R_L + R_2} + R_1$ is the total resistor in the system

[warm shale]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[rose scroll]

yep, good till here

[iron edge]

Right, let me continue

V = IR\
$DV_{cc} = I\times R_T$

[warm shale]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[iron edge]

Moving RT to the left and we'll get

$I = \frac{DV_{cc}}{R_T}$

[warm shale]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[rose scroll]

[iron edge]

Because we just want the current flowing to the upper route, therefore\
$I_{up} = I\times\frac{R_2}{R_L+R_2}$

[warm shale]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[rose scroll]

yep

[iron edge]

Still V = IR

Apply $I_{up}$ and $R_L$ and we'll get what we want $(V_{load})$

[warm shale]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[rose scroll]

yes, that works

[iron edge]

damn, first try

[rose scroll]

Or a much more simple way would be to say I*Rt is the voltage

coz voltage across both RL and R2 is the same

that way you dont need to calculate I_up

[rose scroll]

damn, 1 hour 15 min vs 15 min

5x improvement

[iron edge]

Right, so I think we need to remove RL, right?

That's the best way I come up with LOL

[rose scroll]

no lmao

like just solve for RL with the two expressions being equal

[rose scroll]

I mean sure thats one way to go about it, but your problem wants you to come up with a proper solution

especially since it carries as much weight as the parts 1 and 2 combined

[iron edge]

Right

What condition should it fullfils to make it unchanged?

I'm not sure about the objective/ what we need to show here

[rose scroll]

just V_out = V_load

[iron edge]

OH, I see wym

lemme try

$V_{out} = V_{load}$\
\
$\left(\frac{R_2}{R_1 + R_2}\right)\times DV_{cc} = V_{load}$\

[warm shale]

e4 e5 Qh5 Nc6 Bc4 Nf6 Qxf7#

[iron edge]

okay so, I need a confirmation rq

we basically put what we found at Q2 into Vload

and simplify it

right?

[rose scroll]

yep

[iron edge]

Right, that's neat

thanks again :DD

.close

[obtuse pebble]

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[rose scroll]

np

[obtuse pebble]

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[zealous orbit]

Yc= Ce^-()t + Be^()t

so how do i figure out which graph , these arent clean numbers

and no calculator given

[dark stirrup]

What can you say about the type of solution you should get?

Better question: what types of solutions are possible with a linear second order ODE?

[zealous orbit]

wdym ?

[zealous orbit]

exp (sin cos) , exp , and then exp + texp

everything looks trignometric other than A

[dark stirrup]

Or sin cos

[zealous orbit]

and possibly C

[dark stirrup]

And what kind of solution should you be getting?

[zealous orbit]

exp

[dark stirrup]

So that kind of narrows it down then

[zealous orbit]

yes i see

i think i got freaked out cuz i coudlnt actually compute the values

roots

-2 +- root 5

its A hopefully

[dark stirrup]

Yeah i hear you. Sometimes, you just need to fall back to other approaches, which is what this problem wants you to do

[dark stirrup]

I would agree

[zealous orbit]

thank you

[obtuse pebble]

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[nimble kelp]

Hello

[nimble kelp]

<@&286206848099549185>

[carmine spear]

aw cmon

15 mins bro

[full trench]

!15min

[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.

[royal basin]

post your question(s) here

[full trench]

And what’s your question?

[royal basin]

http://nohello.net/

[nimble kelp]

Like how to start answering it

[full trench]

Subtracting a negative is the same as adding a positive.

[carmine spear]

write the answer maybe?

[royal basin]

ok, let's see. do you know in general how to add/subtract signed integers?

if not: https://www.youtube.com/watch?v=jVvvUiExjes

[nimble kelp]

Yeah know that

[royal basin]

if you know this, then tell us more about what's having you stumped.

[nimble kelp]

I started with - 38 did I do right?

[royal basin]

where did -38 come from

[nimble kelp]

-25 plus (-13)

[royal basin]

right

yes, -25 + (-13) is indeed -38

keep going

[nimble kelp]

Then I got plus 28

[royal basin]

how did +28 happen

also your keyboard most definitely has a plus sign does it not

[chrome vault]

OH THIS MUST BE 7TH GRADE RIGHT?

[nimble kelp]

I'm bad at math ok?

[royal basin]

SOMEWHERE AROUND THERE, BUT MAYBE THERE'S NO NEED TO SHOUT ABOUT IT

[chrome vault]

is this 7th or nah

[nimble kelp]

It is

[royal basin]

with all due respect, does it matter

op needs help with this problem

[round trellis]

Just put the clinches away bc there is no * or / and - - = +

[chrome vault]

I'm going to give you this last warning stop trolling before i actually call mods

[royal basin]

fuck are you on about mate

[nimble kelp]

I got plus 28 from plus 38-(-2)

[royal basin]

yeah go ahead call the mods

you are not nearly close enough to me to be calling me that. get lost.

[round trellis]

And I got -6

[royal basin]

+38 - (-2) definitely isnt equal to 28, and also you had -38 so why did that suddenly flip

also please find the + key on your keyboard, it's really hard to read when you insist on spelling it out as "p-l-u-s"

<@&268886789983436800> maybe take a look at this btw

[nimble kelp]

I not on pc that's the problem and I can't find it on my phone I'm try again

Found it mb

[nimble kelp]

How is she trooling?

[royal basin]

your first step should have been to say

-25 + (-13) + 30 - (-2)
= -38 + 30 - (-2)

[carmine spear]

mans editing the text 💀

[unreal musk]

Really, you think editing the message will do anything like we didn't see it

[nimble kelp]

Ahh ok

[royal basin]

from here i would rewrite the -(-2) as +2 giving
-38 + 30 + 2

[unreal musk]

They're already timed out (though that one was before I could get to them )

[carmine spear]

totally no logs

[royal basin]

but the general philosophy of it is that you should keep your steps organized

and when you're working out an expression, you should make it clear what becomes what & keep it all as one chain of equalities

[nimble kelp]

Ok ty Ann for helping me how can I close this now?

[round trellis]

Sry wrong pinned

[royal basin]

.close

[nimble kelp]

.close

[obtuse pebble]

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[obtuse pebble]

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[hazy girder]

apologies if excessive opening and closing help channels isnt allowed, i finished with my old question and i have a new one

[hazy girder]

so from what i think i know

g'(x) = 1/ (f(x))

or was it f((g(x)))

[hazy girder]

could you please provide an example from one of the questions?

doesnt need to be from one of the questions i sent, just enough to picture the concept

[daring ravine]

finding the inverse of those functions is nasty so i think they just want you to find the x that gives a for each f(x) (definition of inverse)

f(x) = a
f^-1(a) = x
(supposedly)

[obtuse pebble]

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[normal wigeon]

hello guys i dont get what am i supposed to do in these mainly the first A(-3; 5) and B(1; -7).
Find the equation of a linear function f that passes through the points A(-3; 5) and B(1; -7).
a) Determine whether the function is increasing or decreasing.
b) Find the intercepts with the coordinate axes.
c) Find the equation of a function g that is parallel to f and passes through the point (0; 3).
d) Find the equation of a function h that is parallel to f and passes through the point (2; -10).

[silver flicker]

Can you send the original question?

[normal wigeon]

this is what she gave us

i just have a problem with this

idk where or how she got the 8 and b

cuz she told us that only the B needs to be canceled

yes i get it but i dont get the 18= a x (-2) + b and below it idk how to continue from there after this ik how to do the other questions

.close

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[mild mica]

how do u do this

I found fx and fy but like idk what to do now

[rough anvil]

I infer from the instructions "show your answer to 4 decimal places" - that they want you to evaluate those expressions down to 2 real values.

[mild mica]

I think something else is wrong, I've ignored instructions like that for past problems and ended up fine

[rough anvil]

Ah I see one other thing. When you are differentiating $sin(xy)$ with respect to $x$ you'll get $y cos(xy)$. Looks like you are going from $sin(xy)$ to $cos(y)$.

[warm shale]

Adam Henderson

[mild mica]

ohh 🤦‍♂️

yeah and wrt y is wrong also

tysm for the help

.close

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[inland juniper]

hello im stuck with this qustio please help

[mild mica]

for a you can sketch the graph and solve that way. or you can just plug in the points values into the equation and see if its true

just see if the point lines up on the graph

for b you can set the equations equal to each other and solve. or you can also graph that also and just see where they intersect

[inland juniper]

okay could you work me through (a) please

[past horizon]

if a point is on a line, that means that when you plug it into the equation of the line, you'll end up with a true statement like 0=0 as opposed to a false one like 1=0. You can also graph it like paul said

the point (3, -5) is where x = 3 and y = -5, so just replace x and y in the equation of the line with 3 and -5 respectively

[inland juniper]

okay thank you ill do that now

[past horizon]

and if you end up with something that doesn't make sense like 3=0, then it's not on the line

[inland juniper]

perfect so would part b be just substituion?

[past horizon]

yeah you could do substitution to solve for x and then plug that x value back into solve for y

[obtuse pebble]

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[twilit pendant]

Question and answer. How to proceed?

[twilit pendant]

All I could do is prove m = 2n, which is just a known property.

[magic plinth]

The question probably meant for <BCO to equal m instead of <BOC

in which case you can derrive the correct option

its impossible for m+n to necessarily equal 90 under the give conditions if <BOC was considered to be m

[twilit pendant]

Hmm... Thanks. I get m + n = 90 degree, taking <BCO = m

.close

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[next sparrow]

can someone please solve this problem and show the work? thank you <3

[carmine roost]

what can you simplify the denominator to?

[next sparrow]

I’m not sure lol

I don’t have all my trig identities memorized

[carmine roost]

its quite a common one

[next sparrow]

I’d need to google it I think

[carmine roost]

bruh i don't know latex

[next sparrow]

Oooh

lol

[next sparrow]

I need to work on memorizing and recognizing my trig identities

I knew this one but I need to get better at spotting when they can be used

[carmine roost]

i don't know a lot, but this one along with " tan(x) = sin(x)/cos(x) " is very useful

Do you think you are able to solve the integral now?

[next sparrow]

yes thank you for the help

.close

[obtuse pebble]

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[solid plover]

For this problem is it the statement even true? Requiring it to be a proper subset seems to make it kind of difficult

[solid plover]

figured it out

.close

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[lunar zenith]

This question is confusing me, part a should be (0.5 and 30) because the first point on the secant line starts there. Can someone please explain to me where my reasoning went wrong.

[vestal steppe]

Hi! When you are using the mean value theorem, note that the interval determines what the average rate of change is.

If you look back at the example where MVT was used, you may notice that the secant line is drawn from x = 0 to x = 1.

[vestal steppe]

The point is, for part a, the question is asking you to use MVT on the desired interval. That is, [-1, 3] starts from x = -1 and ends at x = 3.

That means (a, f(a)) and (b, f(b)) is based on this interval.

[lunar zenith]

Ok so I think I understand a little bit better now

I would now plug in a and b into the function provided?

One thing I’m also confused about is how does deriving play into the mean value theorem? I remember being told in class to derive the function provided.

[vestal steppe]

Yep :)

[vestal steppe]

Well,

This is the mean value theorem:
If f is continuous in [a, b] and differentiable in (a, b), there exists a c in (a, b) such that:
f'(c) = (f(b) - f(a))/(b - a).

Or in a more simplified message: If you have a continuous and differentiable function f, there will exist a point where the instantaneous rate of change is equal to the average rate of change over the interval.

[vestal steppe]

The point is, differentiation is explicitly mentioned in the theorem: it is pretty much entirely the point of what MVT is trying to say.

[lunar zenith]

Thank you so much for your help i understand now!

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[royal pilot]

preparing for amc test and the questions are rly weird? like idk how to approach this w no calculator

[topaz tusk]

,rccw

[topaz tusk]

hint: divisibility rule for 11

[royal pilot]

what does that mean?

[topaz tusk]

do you know when a number is divisible by 11

[royal pilot]

yeah

well for the first ten

is there a rule?

[topaz tusk]

like for 7?

[royal pilot]

wdym

sorry my brain is kinda fried rn

[topaz tusk]

what do you mean by this

[royal pilot]

is there a rule for when numbers are divisible by 11

[topaz tusk]

yes

[royal pilot]

oh what is it?

[topaz tusk]

you take the sum of digits in the odd positions and subtract the sum of digits in the even positions

take 5764 for example

the 1st and 3rd digits are 5 and 6

the 2nd and 4th digits are 7 and 4

so you take (5+6)-(7+4) = 0 which is divisible by 11 so 5764 is divisible by 11

[royal pilot]

ooo okay

so 202103 isnt prime cuz (3+1+0)-(0+2+2) = 0

tysm

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[dusk girder]

Hi just a short question: from line 3 to 4, is it allowed to just drop the 2, since i could pull it out of the sum and then divide by 2 since 0 is on the other side?

[tardy epoch]

yea 0/2 = 0

[dusk girder]

ok ty

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[radiant scarab]

I used the parallel axis theorem as I_required = I_c + md² where C is the intersection of two sides of the right triangle

On calculating I got an answer 11/36ml² but value of x = 2 is correct

[obtuse pebble]

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[vapid bobcat]

Can someone check this for me?

[obtuse pebble]

@vapid bobcat Has your question been resolved?

[languid spear]

It should be 507pi/8

[languid spear]

Third line from bottom

[vapid bobcat]

how?

OH I SEE NOW

[languid spear]

Yea

[vapid bobcat]

omg i misread my calculator

thanks!!

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[drowsy ridge]

Need assistance

[obtuse pebble]

Please don't occupy multiple help channels.

@drowsy ridge Has your question been resolved?

[obtuse pebble]

@drowsy ridge Has your question been resolved?

[drowsy ridge]

<@&286206848099549185>

[obtuse pebble]

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[obtuse pebble]

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[ember frost]

for c and d, i showed that c is not reflexive, but my intuition says d should be reflexive, but their set notations are the same

what's going on here?

[mortal blade]

d doesn't need to be reflexive right

Cuz there might not exist a page linking to page a at all

And then (a,a) wouldn't be in the relation

@ember frost

[ember frost]

hmm, that's true

by the way, if both relation has the same set notation, do they also share the same properties despite having different definitions for L?

@mortal blade

[mortal blade]

Yes but in this case they're not the same

Your definitions of L are different for c and d

[ember frost]

so basically no

but if they have the same set notation, i can do the exact same line of reasoning as c and it would still make sense, since logically they are the same line of reasoning

[mortal blade]

Essentially yea

I just realized I misread your question lol

The set notation is different BECAUSE L is different

You can't say the set notation is the same while having different L's

[ember frost]

yeah by that, i mean they are both expressible as "exists x such that L(x, a) and L(x, b)"

if i can do that then i can straight up conclude they have the same properties

despite different L

[mortal blade]

Not necessarily

Pathological example

Say L(x,a) is always true

For every a and for every x

This L gives you nothing

[mortal blade]

So in general no you can't do this

[ember frost]

so in this case the properties just happen to match up?

[mortal blade]

Yup

It is easy to see such patterns that hold when things are "nice"

But when doing logic, it's better to not fall into such traps

[ember frost]

alright thanks

.close

[obtuse pebble]

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[vale glade]

How to solve:
2(2p+1) - 30% of (3p-2) = 76?

[rose scroll]

can you write 30% in fraction form?

[vale glade]

30/100

[rose scroll]

yep

now rewrite the whole equation using that

[vale glade]

2(2p+1)- [(3p-2) × 30/100] = 76

@rose scroll

[rose scroll]

yep, now simplify everything and group he terms with p together

[vale glade]

4p+2 - [(3p-2) × 30/100] = 76

[vale glade]

Am i right?

[rose scroll]

yes, but you can do more

[vale glade]

Umm

How?

[rose scroll]

multiply everthing that you can, at least write it in separated form

[vale glade]

...

Eh?

[rose scroll]

your eventual aim is to be able to separate the variable p from all the constants

[vale glade]

I need an example

[rose scroll]

say you have an equation like (5-2)(x-1) = 5

[vale glade]

Hm

[rose scroll]

how do you simplify that to

[vale glade]

3x+3?

[rose scroll]

uhhh, 3x-3 = 5, but whatever, the idea is you split open the brackets by multiplying everything and separating x from the constant

you simplify 5-2 as 3, and hen multiply 3 with (x-1). Not leave it like that

I want you to do the same here

[(3p-2) × 30/100]

[vale glade]

Ehehe

I'm too dumb

[rose scroll]

doesnt mean you shouldnt try

give it a go

[vale glade]

...

Idk what to do

[rose scroll]

like I said, multiply

[vale glade]

That's like giving a man nothing and telling him to go to space for me 😅

[rose scroll]

[(3p-2) × 30/100]
you can see the multiplication symbol there in this expression right?

[vale glade]

Yuh

@rose scroll I'm so sorry my keyboard stopped working

[rose scroll]

Imagine if you solve this, you open a bluegem stattrak AK case-hardened

that should motivate you right?

[vale glade]

Y not js do it for me so i can find out what to do

[rose scroll]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[vale glade]

What's that

[rose scroll]

why are you cyan knife with an emerald karambit if you know nothing about cs skins lol

[vale glade]

I don't play cs cuz I don't have a pc 😭

I'm a big fan of valo and cs tho

[rose scroll]

ok whatever, just do the math bro

[vale glade]

....

IDK WHAT TO DO

[rose scroll]

are you telling me you do not know how to multiply?

[vale glade]

...

No

[rose scroll]

coz what you need to multiply is right there

and you know how to multiply

I fail to see a problem why you dont know how to proceed

[vale glade]

._.

In this case imma go leave this world and start a caveman life then

[rose scroll]

If you cant tell me what is puzzling you, I cant help

[vale glade]

I don't even feel like having 1 braincell atp

[vale glade]

I've never solved a sum like this before

[rose scroll]

are you up past your normal bedtime or smthn? go sleep and come back with a fresh brain

[vale glade]

I stopped studyin math from when i was in 6th

[vale glade]

9:25 am

[rose scroll]

then maybe the first step would be to start solving 6th grade maths first and then move on to this (unless this is 6th grade math, idk. Its all blurry to this unc)

[vale glade]

Should've asked chatgpt instead x_x

[rose scroll]

wont help you learn anything unless you do it on your own tbh

[vale glade]

I DO LEARN HOW TO DO SUMS IF I GET THE STEP-BY-STEP SOLUTIONS

🙏🏻

All i need is the next step bro c'mon

I'm begging you

[rose scroll]

well, like i have been telling you for past 10 messages, the first step was to convert % into fraction and 2nd step is to multiply the bracket with the said fraction

you did step 1

now its time to do step 2

[(3p-2) × 30/100]

[vale glade]

idk how to do step 2. That's it

[rose scroll]

I also asked you do you know how to multiply

[vale glade]

idk how to do step 2. That's it

Yes i do. But not in this case

[rose scroll]

we will take this example which you messed up: (5-2)(x-1)
I thought it was a typo

[vale glade]

I cando 752×14

[rose scroll]

but evidently its not a typo

[vale glade]

It's wasn't

[rose scroll]

ok

(5-2)(x-1)

you rightly said its same as 3*(x-1)

right? or was that also a shot in the dark?

[vale glade]

3x+3*

[rose scroll]

yea thats right

now what if instead of 3, its [\frac{3}{10} \times (x-1)?]

[warm shale]

βαχτϵρ10Φρ4γ

[vale glade]

3/10x -1?

[rose scroll]

no... you should do same as you did before, you multiply both the terms x as well as -1 with the constant outside the bracket

[vale glade]

3/10x - 3/10

[rose scroll]

this is same as $\frac{3 \times (x-1)}{10}$

[warm shale]

βαχτϵρ10Φρ4γ

[rose scroll]

yes

[vale glade]

• not -

[rose scroll]

well, then no

[vale glade]

Wai-

🙏🏻

Wish i was a girl, could js become a housewife

[rose scroll]

this is not a fastest finger first kinda situation, take your time to type an answer

[vale glade]

If i was an orphan i could do whatever i wanted

[rose scroll]

well, batman, you arent one, so its back to the math jail with you

[vale glade]

🙏🏻🙏🏻🙏🏻🙏🏻

[rose scroll]

we are still stuck at [(3p-2) × 30/100] which is now pretty much the same as what we did

[vale glade]

BRO IT'S NOT POSSIBLE FOR ME TO UNDERSTAND THE ANSWER WITHOUT THE NEXT STEP

I only learn from other's mistakes 🙏🏻🙏🏻🙏🏻

[rose scroll]

tough luck, I aint making mistakes for a problem this easy 💀

maybe you can start learning from your own mistakes

[vale glade]

No I didn't mean that

Give me the next step, if i can't find out what to do then stop teachin me

Is that a deal?

[rose scroll]

yea, the next step is to multiply [(3p-2) × 30/100] these

[vale glade]

Bro I don't know how to multiply this kind of sum

Wait imma ask chatgpt rq

[rose scroll]

we just went over this, which is basically the exact same as the problem

[vale glade]

0.9p - 0.6

Mtiy by 30, divide by 100

[rose scroll]

good job chat gpt

[vale glade]

@rose scroll isn't that it?

Multiply*

[rose scroll]

yes

[vale glade]

Yayyy

[rose scroll]

but its preferred to keep the numbers as fractions rather than decimals

[vale glade]

9/10p - 6/10

[rose scroll]

but whatever, you are using chatgpt, so it makes no difference what ppl do

[vale glade]

I'm sorry

[rose scroll]

4p+2 - [(3p-2) × 30/100] = 76
well, then the next step is to group together all the terms which have the variable p

[vale glade]

4p+2 - 9/10p - 6/10

= 76

4.9p +1.4 = 76

4.9p = 74.4

[rose scroll]

yep

[vale glade]

74.4 ÷ 4.9

[rose scroll]

idk, if you have chatgpt, what is up with keeping this channel open?

[vale glade]

I only did one step by chatgpt beo

[rose scroll]

really, then why do you again have decimals here? smells very chatgpt

[vale glade]

It's my freakin habit

My brain can't comprehend fractions

[rose scroll]

well, if you ever decide to do maths the people way, and not chatgpt way, its a bad habit that you should change

[vale glade]

I can always change stuff

[rose scroll]

time to practice till you can comprehend them

[vale glade]

So 744/10 ÷ 49/10

Imma try this?

7440/490?

[rose scroll]

yes

but now that you are at the end, its time to convert it into decimals

[vale glade]

Now again stuck

So i already did bruh

[rose scroll]

all you gotta do is p = decimal

and you are done

[vale glade]

74.4÷4.9?

@rose scroll

[rose scroll]

nahh, you messed up here

4 - 0.9 is not 4.9

[vale glade]

😅

[rose scroll]

ngl it seems like you really didnt use chatgpt afterwards, sorry for assuming that

[vale glade]

🙏🏻

[rose scroll]

also, 76 - 1.4 is not 74.4

[vale glade]

1.4?

Ahh 74.6

@rose scroll mb

[rose scroll]

theres also this

[vale glade]

EH

3.1

I'm so dumb

I did addition

[rose scroll]

yea, so p = 74.6/3.1

,calc 746/31

[warm shale]

Result:

24.064516129032

[vale glade]

Kk

Tyy

[obtuse pebble]

@vale glade Has your question been resolved?

[obtuse pebble]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pastel fractal]

very excellent

[carmine spear]

hm?

[fossil pendant]

hm?

[rapid osprey]

do you have a question?

[fossil pendant]

https://tenor.com/view/cute-babies-dancing-gif-2070683399384110126

[fossil pendant]

ping! do you have a question

[iron edge]

.solved