help-38

[unborn estuary]

.close

[trim jolt]

Channel closed

Closed by @unborn estuary

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[shell barn]

Find the smallest prime factor of 2^23 - 1

[shell barn]

<@&286206848099549185>

So we have that 2^23 = 1 mod p

[zinc ginkgo]

!15m

[trim jolt]

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.

[shell barn]

So that means that the order of 2 mod p is 23

And then since Zp is a cylic group under multiplication that means 23 has to divide the order of the group which is p-1

Which is what my prof told me but I don't get that

What does a cyclic group mean and why does 23 have to divide p-1

@zinc ginkgo

[zinc ginkgo]

???

!noping

[trim jolt]

Please do not ping individual helpers unprompted.

[zinc ginkgo]

.

[shell barn]

Oh mb

My prof went into something about cosets but I don't get that either

[trim jolt]

@shell barn Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[iron swallow]

Suppose we draw a simple continuous curve bounded to the x-axis so that the curve starts at a point (x1, 0) with x1 < 0, and ends on (x2, 0) with x2 > 0, and the entire curve lies above the x-axis. I want to show that for any pair of points of intersection between this curve and y = |x|, that the curve can always be horizontally translated (can be translated both left and right) in such a way that the points of intersection will become (t,-t) and (t,t) for some real t > 0, if it is not already.

So far, I was thinking, for however you choose to draw the curve in that manner, we can consider the slope between the 2 points of intersection (which we will call m) and its sign, and we note if m = 0, then the points of intersection will be in the desired form ((t,-t) and (t,t)). If m > 0 on the chosen curve, we can horizontally translate it so eventually m < 0 and since this transformation is continuous, then we can assert that m = 0 at some point in between and likewise, if m < 0, horizontally translate until where m > 0, asserting m = 0 in between.

I am not sure how to further formalise this and/or if there are errors in this proof sketch, especially with the nature of how wildly one can draw such curves. Any tips on how to move on from here, or if there are any errors so far that I may have missed in this sketch?

[tepid hamlet]

i'm confused, did you mean to say (t, -t) and (t, t) or was it supposed to be (-t, t) and (t, t) ?

i assume you meant 2nd because abs value is positive but you are suggesting there's some point of intersection with a negative y

[iron swallow]

oh yeah sorry i did mean (-t, t) since it is absolute value

[tepid hamlet]

if your curve contains the points (x1, 0) and (x2, 0) then you cannot say that "the entire curve lives above the x-axis" because those two points are ON the x axis not above it and they are part of the curve

so are you trying to say that all y values of points on the curve should be non-negative, but it is ok for them to be 0?

anyway i think i kinda get what you're trying to say and i feel like it's just some kinda strange re-telling \consequence of Rolle's Theorem \ Mean-Value Theorem or something along those lines

like basically, if one point is higher than the other, then we can move the curve horizontally and there must be some y value between 0 and that point's y value that is better suited, well that is just a consequence of continuity and the mean value theorem

@iron swallow are you still here? 😛

[trim jolt]

@iron swallow Has your question been resolved?

[iron swallow]

Yeah i meant to say nonnegative to that as well

But you got what i meant lol

[iron swallow]

Would this apply to any possible simple curve?

Like non function ones too?

That part im not necessarily certain about

[tepid hamlet]

well i think so, i'm not entirely sure on the proof but

if you could find some translation that guarantees a positive slope between the 2 points

then some translation that guarantees a negative slope

since the slope is also changing continuously as you translate

mean value theorem guarantees there is some translation for every value in between (i.e. 0)

it's kind of a classic mean value theorem set up but i'm not exactly sure on how to prove

[iron swallow]

Ahh ok i see

[iron swallow]

Oh like this is a format that one could use to prove it similar to that of the proof for the mvt?

[trim jolt]

@iron swallow Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[paper torrent]

so...this friday i have my annual exam of math, and i've done pretty less. But i'm kinda good at math but i'm still scared (like for eg on tests i get 25/30 and im in india so you can guess the educational system). They make like extra complex questions and some guides are just impossible to do. Many easy chapters are cut and this is the weightage

Number system-10
Polynomials-17
Linear equations in two variables- 7
Triangles-10
Heron's Formula-6
Surface area and volumes-9
Quadrilaterals-4
Circles-14
Statistics-3

(Its an 80 marks paper)

[trim jolt]

Please don't occupy multiple help channels.

[wraith hinge]

Your question being?

[paper torrent]

uhh soo no specific question just need to know if i can do math in 2 days..(i practice almost everyday in tuition but im not sure)

[wraith hinge]

your question is better suited for #discussion or something

these channels are meant for specific mathematics questions

not to provide opinions or such

[remote wave]

i mean i dont even know what youre asking

what kind of question is "can i do this or not"

just try

and you'll find out

[paper torrent]

ohhk thanks

so i'm confused in question number 8
i used the theorem angle subtended by the arc at the centre is double the angle subtended by it.
so <APB = reflex<AOB
2x50 = 100
so <AOB = 360-100 = 260 (but that cannot happen because according to ASP property of a triangle 260 cannot be an angle. Help?

[tepid hamlet]

I think you are misunderstanding a bit the theorem with the angle subtended by an arc at the center

that has to do with an angle with its vertex at the center (i.e. angle POQ in this figure subtending the arc PQ) vs the angle PAQ where the vertex of the angle is on the CIRCUMFERENCE of the circle. in the case of problem 8, it doesn't involve a point on the circumference so it doesn't apply

in this case angle POQ has double the measure of angle PAQ

[paper torrent]

oh yes

the circumference!

[tepid hamlet]

the most obvious way to think of problem #8 is to use the idea that tangents to the circle will make right angles with a radius meeting at the same point and consider this quadrilateral:

what is the sum of the angles of a quadrilateral (four sided shape)?

[paper torrent]

360

[tepid hamlet]

yes, so then what would be the measure of angle AOB?

[paper torrent]

OHH

230?

ohh

[tepid hamlet]

130

[paper torrent]

oh shit

yeah

130

[tepid hamlet]

and now triangle AOB is isosceles (the two radii are congruent) so you can find out the other angles in it with the obvious sum of angles in the triangle

so final answer is? : )

[paper torrent]

65?

[tepid hamlet]

how are you getting that

[paper torrent]

oh wait

no

NONONO

25?

[tepid hamlet]

yes

[paper torrent]

lol

i tried to do mental math but yeah it doesnt work for me ig

btw one more thing

[paper torrent]

how did yk that the angles were 90? (stupid ques but wtv)

[tepid hamlet]

because PA and PB are tangents

[paper torrent]

tangents? 👀

i'm just in 9th grade oml 😭

[tepid hamlet]

the problem says: "in the figure PA and PB are tangents to the circle"

[paper torrent]

oh....

they didnt teach us that but wtv

[tepid hamlet]

a tangent touches the circle at exactly that 1 point and is perpendicular to the radius to that point

[paper torrent]

urea'
kl;

[paper torrent]

ohh

okay!

tysm!

this was really helpful!

and also

what abt ques number 7?

[tepid hamlet]

https://en.wikipedia.org/wiki/Tangent_lines_to_circles
read just the first paragraph of this

its a concept that comes up quite a bit

[paper torrent]

ohh

[paper torrent]

done!

[tepid hamlet]

let me throw a look

[paper torrent]

but in the 7th ques they say that tangent PR is of 50 degree, shouldnt it be 90?

[tepid hamlet]

its 50 degrees to that chord, not to the radius

its 90 degrees to the radius given there though

[paper torrent]

so the radius <POQ is 90?

[tepid hamlet]

what?

the angle between the tangent and the radius is 90

OP is the radius

PR is the tangent

so OPR is 90

[paper torrent]

ohhk

tangents are rlly confusing

[tepid hamlet]

so that's actually what you need to use to solve the problem

[paper torrent]

ohh okay

then like radiis are equal and then isoceles tri property

and then using ASP find the value of <POQ?

[tepid hamlet]

its actually really simple, let me clarify just a moment...so the red is the tangent line right? and its intersecting the circle at a single point, im drawing a red circle around that one point yes?

now draw the radius from that one single point to the center of the circle...

[paper torrent]

yeah?

[tepid hamlet]

and so you've drawn a right angle

that's all there is to it

you have the tangent line

then you have the radius between the center and where the tangent hits the circle

that's the right angle

[tepid hamlet]

yeah so what's the measure of angle <OPQ?

[paper torrent]

OH

IT ALL MAKES

so much more sense now

[paper torrent]

40 degree

so

<POQ = 100 degree?

[tepid hamlet]

yes and because isoscles as you say, you would also know the measure of which other angle?

yeah, exactly, you got right to the conclusion

[paper torrent]

oh

[paper torrent]

<0QP

[tepid hamlet]

yeah yeah you got it all

[tepid hamlet]

lemme just do one more illustration...

consider a round ball resting on the ground, from the side it looks like a circle and the ground is touching it in a single point right? so you can think of the ground as tangent to the circle

[paper torrent]

ok

[paper torrent]

ohh

[tepid hamlet]

and if you draw the radius from that point to the center, the center is directly above that point of contact right? because it balances out, so that's kind of another way to see maybe how the radius is perpendicular to the tangent because the ground is horizontal and that line through the point of contact and the center is perfectly vertical

or if you consider your position on a spherical earth...going along the surface of the earth is what you think of as the horizontal direction wherever you might be on the globe right? and straight down is the direction to the center of the earth, which is perpendicular to moving along it

(anyway in case that helps)

[paper torrent]

ohh

umm so hey

i tried this ques

ques 10...is the ans 40?

and also question number 9 is so confusing...help?

@tepid hamlet ?

[tepid hamlet]

yes that's right

[paper torrent]

ohhkk

[tepid hamlet]

for 9 what have you tried? did you draw a figure?

[paper torrent]

i couldnt figure it out

the figure 😭

[tepid hamlet]

something like that

just two tangents to a circle, they meet with a 60 degree angle and the radius of the circle is 3

[paper torrent]

ohhh

[tepid hamlet]

so if you draw a segment from the center of the circle to the vertex of that 60 degree angle, you split it in half

then each side becomes a 30 / 60 / 90 triangle, you might know how to solve it from there

[paper torrent]

what

sorry i didnt get it 😭

[paper torrent]

this part

[tepid hamlet]

this segment bisects that 60 degree angle

[paper torrent]

oh

[tepid hamlet]

because everything is symmetric (even though my drawing is crap and it doesn't look like it)

so each side is a 30 degree angle

[paper torrent]

its so good

i could never draw that

😭

[paper torrent]

oh

[tepid hamlet]

so you have right triangles and you can also get this angle

[paper torrent]

by ASP?

oh yeah

[tepid hamlet]

30 + 90 + ? = 180?

i have no idea what ASP is

[paper torrent]

angle sum property?

[tepid hamlet]

ya that

[paper torrent]

yee

[paper torrent]

60

[tepid hamlet]

i assume you haven't learned trigonometry yet?

[paper torrent]

nope

its in 10th

basically

oh wait

[tepid hamlet]

or at least properties of 30 / 60 / 90 triangles?

[paper torrent]

im going to be in 10 this year

😭

[paper torrent]

👀

which property??

[tepid hamlet]

i'm actually not sure how to do this without that knowledge

[paper torrent]

whats that property?

oh wait

they havent taught us that yet

but i just googled it

ig i can learn new stuff idm

[tepid hamlet]

well just that the lengths of sides in a 30/60/90 triangle are always in a specific ratio and its a triangle that comes up so often that its worth memorizing basically the sides are in the ratio of 1 : sqrt(3) : 2

[paper torrent]

oh

so the length of the tangents are 3root3?

[tepid hamlet]

so the large side (hypotenuse) is always twice the length of the smallest side and the middle one is square root of 3 times the smallest side. in this case the small side is the radius (3) and the problem is asking for the length of each tangent so i guess it would be $3\sqrt{3}$

[solid kiln]

Soosh

[tepid hamlet]

ya exactly

[paper torrent]

it all makes soo much more sense now

btw question

did u do a PHD in maths or smthg?? ur maths is too good

[tepid hamlet]

no, but i kinda minored in it in college

[paper torrent]

ohh

our school teachers are weird

they dont teach us this shit

but they expect us to know all this fucking mess

[tepid hamlet]

and i've tutored math a decent bit so i at least have some decent grasp of what comes up often in some high school and college classes

[paper torrent]

ohh

*not decent

ALOT BROO

[tepid hamlet]

well i'm lucky i knew these things at least : )

[paper torrent]

wdym atleast?? yk alott

u have a bright future in math 👍

[tepid hamlet]

bad teachers is unfortunate, but glad you are finding ways to learn anyway; definitely try not to get behind in math especially because it all builds on itself and well if you get to something like calculus, there's a saying that "most people who fail calculus actually fail algebra in a calculus class" because they are basically lacking the foundation in the basics and it finally comes to bite them at that point

it's actually pretty sad what a poor foundation in basic algebra most students in calculus have these days and they struggle like crazy because they cant do basic math

[paper torrent]

is calculus that hard?

[tepid hamlet]

-end rant about today's math education- lol

[paper torrent]

ohh

[paper torrent]

lmaoo

[tepid hamlet]

actually i really believe it's very intuitive and beautiful \ fun but yeah it's hard for people who are also trying to learn the last 4 years of math they have been "getting by" in for the last 4 years

that's pretty much always the case for the people who struggle with it

[paper torrent]

ooh

damn'

[tepid hamlet]

the actual concepts intoduced in calculus can be explained more easily than how to do basic algebra and geometry in my opinion to anyone who is well versed in those things already

anyway cheers, i've got to head off and best of luck : )

[paper torrent]

okay!! thank you sososo much!

[trim jolt]

@paper torrent Has your question been resolved?

Channel closed

Closed by @paper torrent

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[lunar galleon]

how do i find the vertex of a quadratic function if i have two x intercepts and a y intercept

[lunar galleon]

ping me if answer please

[hot finch]

Use vertex formula

[lunar galleon]

wha is that

.close

[trim jolt]

Channel closed

Closed by @lunar galleon

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[sharp topaz]

Determine the volume of silver needed to make the napkin ring, in the following figure, out of solid silver. Give your answer in cubic millimeters.

[ocean gate]

progress ?

where are you stuck

[sharp topaz]

Do I have to convert it all to mm before starting? Or can I just convert, 22mm to cm? Do it "normally", so (pi-r^2xh)-(pi-r^2xh) and then convert that answer to mm^3? Also, I did try but according to my answer key they do pi(2.2)^2 but I really don't understand why 2.2 is the radius

[sharp topaz]

I did pix2^2 - pi2^2x2.2

no wait

[ocean gate]

waiting

take your time

[sharp topaz]

I did that

[ocean gate]

for this you can choose any unit as to perform calculations on

[sharp topaz]

ye makes sense

[ocean gate]

see in this step you are operating on the AREAS

but here the question is asking about volume

so do you how to calculate the volume of a cylinder ?

[sharp topaz]

oh yeah I did times 2 as well

[ocean gate]

nice thing

then ig you are doing it correctly just subtract the smaller volume from the larger one and you should get your answer

[sharp topaz]

pix2^2x2 - pi2^2x2.2 to clarify

[sharp topaz]

oh

then I guess I did something wrong along the way

lemme recalcualte the same thing I set up

[ocean gate]

sure

[sharp topaz]

there is one thing, I do not udnerstand. In my answer key they say pi(2.2)^2xd2 - pi(2)^2x2

pi(2.2)^2xd2 in that part, how is 2.2 the radaius that is supposed to be squared?

thought 2 mm = 2.2 cm and that would be the height, not radius

[ocean gate]

look at it from the top

there is a circle which has radius = 2

then from the edge of this circle, there lies another one which is 2mm far

so the radius of the second circle comes out to be ?

[sharp topaz]

omdddd

Lmfao i'm so dumb ty

I thought the line kept on going so that it was the height. No wonder, it would had been atrange if the heights were different anyway

[ocean gate]

no worries (:

[trim jolt]

@sharp topaz Has your question been resolved?

Channel closed

Closed by @sharp topaz

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[wraith hinge]

Does the additivity of line integrals hold even if its a vector field?

[dusty hound]

The integrand space doesn't matter

[wraith hinge]

okay

so ig define the parameterisations with something similar to
\begin{alignat*}{3}
C_1 &= \6{\vj r}t &&= t\vc\I -\vc\J &\q&-1\le t\le 1 \
C_2 &= \6{\vj r}t &&= \vc\I + t\vc\J &&-1\le t\le 1 \
C_3 &= \6{\vj r}t &&= -t\vc\I +\vc\J &&-1\le t\le 1 \
C_4 &= \6{\vj r}t &&= -\vc\I -t\vc\J &&-1\le t\le 1
\end{alignat*}

ok there u go

[dusty hound]

One copy paste later

[wraith hinge]

i think that makes sense

so yeah is that fine

[dusty hound]

The 4th one is really poorly copy pasted

[wraith hinge]

uh lemme check if that makes sense

ok first is fine

second is fine

third is fine

[wraith hinge]

and now fourth is fine?

[dusty hound]

Now yes

[wraith hinge]

evil copypastes i tell ya

so now uh its just integration

ok thanks bezier

.close

[trim jolt]

Channel closed

Closed by @hasty halo

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[ionic hull]

When solving these types of questions... When I move the matrix to the right, must it be change to negative? Since the answer I compared to the text book is

-4 2
-6 0

[uncut cove]

Hi, let me try to help you out!

[ionic hull]

Sure

[uncut cove]

When you moved the matrix from the LHS to the RHS you might have forgotton to add a minus sign?

since you are subtracting it both on the LHS and RHS

Does that make sense?

[ionic hull]

So must put a minus sign?

Here's the question...

[uncut cove]

Okay let me rephrase. In your first step, for simplicity's sake let's call the written out matrix A. So the question is:
3M + 4A = M
then in the next step we should get
3M - M = -4A, where you got 3M - M = 4A

since we are subtracting 4A from both sides of the equation

Just like how you would do it if it were regular variables instead of matrices

[ionic hull]

Ohhhh I see

[ionic hull]

Thank you very much for your teaching and time

😊

[uncut cove]

You're very welcome!

[ionic hull]

.close

[trim jolt]

Channel closed

Closed by @ionic hull

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[dapper basin]

Need massive help with this one

[frail heron]

!status

[trim jolt]

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

[frail heron]

Please place them in the right order and make sure not to violate the rule above

[dapper basin]

I guess that's how you do it?

[frail heron]

No, I mean order the boxes lol

I guess there is only one scenario that doesn’t violate the rule

[dapper basin]

I hate probability so sorry if I'm dumb as hell lol I'm fine with algebra though

[frail heron]

Nah, all good

[trim jolt]

@dapper basin Has your question been resolved?

[chilly bobcat]

yeah
if you see theres only one case for the condition to fulfil
CVCVCVC, with C for consonant and V for vowel

[frail heron]

Exactly

However, he has left

[chilly bobcat]

sad

[dapper basin]

So just 7!

[chilly bobcat]

No..

4! Ways to arrange the consonants

and 3! ways to arrange the vowels

[frail heron]

Vowels are confined in certain locations, so are the consonants

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[wraith hinge]

what is this find area inside the curves function and how to I approach and solve it

[wraith hinge]

I think it's something my online teacher yesterday was doing

xD he kinda rushed it so I still have no idea

oh ye and at the start you don't know the function

the problem is more like given this graph and how the area inside the graph shows every point that has an attribute that I forgot because I didn't understand

maybe we could find the problem relating to that online

[trim jolt]

@wraith hinge Has your question been resolved?

[trim jolt]

@wraith hinge Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[uneven hatch]

Du/dx = cosx

Dx= du/cosX

Integral(cosx*(sin(u-1))*u³)

[uneven hatch]

Multiplied by secant*du

So I get rid of the cos

Cos the cos cancels out, heh

So I now have u³*sin(u-1)

So I integrate using the product rule

(U⁴/4) × sin(u-1) + ((sin²(u-1))/2) × ((u²/2)-1u+c)

?

Replace u with sinx + 1

Is that it?

[digital bolt]

no absolutely not

it would be u^3 (u-1)

if u = sin(x) + 1, sin(x) = u - 1

[uneven hatch]

Oh yeah sorry

Lol I messed it up

(U⁴/4)×(u-1) + ((u²/u)/2)×(u³)?

[trim jolt]

@uneven hatch Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[marsh forum]

I would like to know how to obtain this result #bots message

[marsh forum]

Here's what I posted there

$\int \frac{x^7}{\sqrt{1+x^8}}dx$

[solid kiln]

Why am. I here

[marsh forum]

maybe $x^4= tan(u)$

[solid kiln]

Why am. I here

[proven hound]

1+x⁸=u?

[wraith hinge]

its quite easy if you know ||by parts|| method

like 1 step only

[marsh forum]

shit, wrong question, nvm

.close

[trim jolt]

Channel closed

Closed by @marsh forum

Use .reopen if this was a mistake.

[marsh forum]

thanks though

[proven hound]

Lol mp

[wraith hinge]

bruh 2nd time xd

[proven hound]

Np

[wraith hinge]

💀 take a chill pill

just give the og question

[marsh forum]

.reopen

[trim jolt]

✅

[marsh forum]

$\int \frac{x^5}{\sqrt{x^4+4}}dx$

[solid kiln]

Why am. I here

[marsh forum]

is the OG question

I was trying to solve for cases with lower powers

wasn't able to

I'm considering x^2= 2 tan(u)

which is probably the best way to solve it

which will give

$\int 4\tan^2\left(u\right)\sec\left(u\right)du$

[solid kiln]

Why am. I here

[marsh forum]

unless I've made a calculation mistake

.close

[trim jolt]

Channel closed

Closed by @marsh forum

Use .reopen if this was a mistake.

[wraith hinge]

would you get this ,i have bad handwriting

[marsh forum]

yeah, got it

tysm!

[wraith hinge]

:>

[marsh forum]

wait

[wraith hinge]

ik you wanna do it with your method only but still

[wraith hinge]

didnt solved it after root over t^4 +4

ig you can do that separately

[marsh forum]

Don't think That's easily integrable though

[wraith hinge]

gee! i suck

doesnt root over 2 tan(k) works ?

[marsh forum]

yeah, that should work

[wraith hinge]

oh lol i forgot one t

kekw nvm

its equally lenghty

i thought mine would be short but meh

[wraith hinge]

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[split bluff]

Hi, so for this problem I actually got an ans of 14,700N and wanted to verify the correctness of it
Ill explain I started with finding an equation which I did by multiplying 3 with 1000kg^3/L then multiplying it by 9.8
getting f(x) = 3000gx
then i applied trapezoidal rule
so b-a/n --> 1-0/2 --> 1/2
1/4[f(0) + 2f(0.5) + f(1)]
and that produces 14, 700N
not sure if it is correct so id like to hear someone else's reasoning, thanks!

.close

[trim jolt]

Channel closed

Closed by @split bluff

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[somber sable]

danggg can someone guide me through this

[somber sable]

we're suppose to simplify this using the laws of exponents

[hallow cliff]

There is something that can be factored out of the numerator and denominator

[somber sable]

the 2^x I suppose?

all of em

[hallow cliff]

Yep

So do it

Then see what you get

[somber sable]

then the 2^2 also factor out each other? leaving me with 2

[hallow cliff]

Check it again

Where did the - signs and + signs go?

[somber sable]

ahh danggg

so it becomes this right

[hallow cliff]

No, check the denominator once again

[somber sable]

the denominator is positive

[hallow cliff]

2^x(2^2) is not equal to 2^x+ 2^2 • 2^x

[hallow cliff]

Yes, but you didn't factor out the 2^x properly

[somber sable]

I don't get it

[hallow cliff]

What is the denominator originally?

2^x+2^2•2^x right?

[somber sable]

2^x + 2^(x+2)

[hallow cliff]

Yes

Now you rewrote it as 2^x + 2^2•2^x

Which is also correct

[somber sable]

yea

[hallow cliff]

Factor out a 2^x from this

Tell me what you get

[somber sable]

I'm not getting it

what you're pointing at

[hallow cliff]

I want you to factor out a 2^x from this expression

Im pointing at the fact that your denominator is missing a +1

[somber sable]

ain't 1×2^2 the same as 2^2?

[hallow cliff]

Its not a x, its a + there

[hallow cliff]

What exactly are you doing when you scratched out those 2^x ?

[somber sable]

I'm making it easier to remember for myself

[hallow cliff]

You are dividing the numerator and denominator by 2^x

[hallow cliff]

Your denominator was 2^x + 2^x • 2^2, so when you divide it by 2^x, you get 1+2^2

[somber sable]

ahh

[hallow cliff]

Your numerator was 2^x•2^2 - 2^x•2 and when you divide it by 2^x, you get 2^2-2(the numerator you have written)

[somber sable]

2^x • 2^2 - 2^x • 2*

[hallow cliff]

Yes

So what do you get at the end?

After simplifying everything

[somber sable]

I'm still stuck here

[hallow cliff]

You messed up the numerator again

[somber sable]

minus sign?

[hallow cliff]

Yeah

[somber sable]

I get 2/5 at the end

[hallow cliff]

You just showed me the correct numerator here 😭😭

Yes

It is 2/5

[somber sable]

this is what I get for not paying attention in class😔

[hallow cliff]

Yes but now you know

[somber sable]

thanks

[hallow cliff]

So just don't forget

[somber sable]

I wont🤞🏾

.close

[trim jolt]

Channel closed

Closed by @somber sable

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[honest sinew]

How do I solve this?

[tepid cobalt]

I believe the intention in "Step 1" is to tell what was done to go from this equation

$$40 = \frac{1}{2}mv^2$$

to this equation

$$\frac{40}{v^2} = \frac{1}{2}m$$.

[solid kiln]

OneTrackPony

[tepid cobalt]

You can see the options in the drop-downs.

[honest sinew]

I figured it out

I'm good

[dry viper]

you need another equation to solve this

[honest sinew]

Thanks

[dry viper]

if both m and v are variables

[honest sinew]

.close

[trim jolt]

Channel closed

Closed by @honest sinew

Use .reopen if this was a mistake.

[honest sinew]

Unfortunately I have to do all these

How do I solve it?

[dry viper]

4c-11=5c+7
c=-18

now , check those options , how many of them don't give c=-18

first one does

second one does as well

the third one doesnt

[honest sinew]

Got it

Quick Question on something else

[dry viper]

ye

[honest sinew]

Imagine Math is something I struggle a lot with. How should I go about dealing with the questions?

[dry viper]

see , the way you perceive math might be different

if you perceive it as a burden , a boring thing

you'll never find maths interesting

let me tell you a cool thing

Maths is the language in which God has written this universe

[honest sinew]

Interesting

[dry viper]

if you're struggling with this , I suggest you to polish your concepts

see , whenever you get an equation with one variable

try to bring the variable to either side

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[dry viper]

to an equation

[dry viper]

we can perform any operations on both sides

be it addition

subtraction

anything

[honest sinew]

What about equations with two variables?

[dry viper]

we need 2 equations for that

the no. of variables = no. of equations to solve it

and the lengthier the process

[honest sinew]

Ok

[dry viper]

want an example?

its pretty fascinating if you see

[honest sinew]

Sure

I'll share a problem I need help with if you don't mind

When you're done

[dry viper]

no issue

at

all

go for it

first lets solve your problem

I have to go in like 10 mins

[honest sinew]

Where am I even supposed to begin?

[dry viper]

how many variables do you see?

also , they are equations of kinematics

show me the question

the actual one

[honest sinew]

3 variables

[dry viper]

oh

see , you have to solve

for u

where do you see u?

[honest sinew]

In the middle. UT

[dry viper]

we can't separate t from u for the time being

lets try to make ut alone

from rest of the equation

would you mind if subtract x from both sides?

[honest sinew]

Go ahead

[dry viper]

5-x = 1/2 at^2 +ut

ut is getting lonlier

now how about I multiply both sides by 2? so that I can get rid of the 2 in the denominator of 1/2 at^2

should I?

[honest sinew]

I think so

[dry viper]

2(5-x) = at^2 +2ut

[hallow cliff]

you shouldn't be multiplying by 2 if you are solving for u

[dry viper]

10-2x = at^2 +2ut

[hallow cliff]

You will end up dividing by 2 at the end of the day to isolate the u

[dry viper]

now subtract at^2 from both sides

10-2x-at^2=2ut

[honest sinew]

Where should I go after subtracting X?

[dry viper]

now divide by 2t , both sides

u = (10-2x-at^2)/2t

[hallow cliff]

Do what he did, just don't multiply by 2 ( you can see he had to divide by 2t at the end , so the multiplication by 2 was useless)

[dry viper]

there you go

@honest sinew do as it seems easier , the 2 in the denominator won't help much if you're actually solving a kinematics problem

[hallow cliff]

Yep

The goal is to get the u on one side and the rest of the terms on the other side

[honest sinew]

I lost points trying to solve it, but I got it

.close

Oh wait he has to do it

[trim jolt]

@dry viper Has your question been resolved?

[dry viper]

.close

[trim jolt]

Channel closed

Closed by @dry viper

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[eager mortar]

9.a

And all the roman numerals

[trim jolt]

@eager mortar Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

@karmic spade Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[wraith hinge]

,rotate

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[rapid dock]

i dont understand what the question is asking

[dry viper]

its asking the inverse of f(x)

[rapid dock]

but isnt it in the question

[dry viper]

no?

[zinc ginkgo]

yes it is

[rapid dock]

oh okay

so x cant equal 4 is what its saying

so it can be any other number

?

[wraith hinge]

Yeah cuz then denominator 0

[rapid dock]

so can x = 1

[dry viper]

you need to find the inverse of the function , they specified the domain of f(x) that x=/=4

we just talked about finding inverse easily

replace x->y

and vice versa

[rapid dock]

yes but there is no y in the answers

[dry viper]

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

[dry viper]

f(x)=y ;-;

[rapid dock]

so f^-1 x can be y too

i thought it was only f(x) = y

[dry viper]

we reversed domain and range by replacing x and y

[rapid dock]

mhm

[dry viper]

so inverse becomes y

[lean prawn]

This is not f^(-1) it's just rewriting the original function with y instead of f(x)

[lean prawn]

Completely consistent with this

[rapid dock]

so-

[lean prawn]

It's just really just an aesthetic choice

but now you have an equation where you can try to solve for x

and you'll notice that x depends on y

[rapid dock]

would it be d?

[dry viper]

I was going to solve further .... I stopped

didn't even replace yet lol

[rapid dock]

it would be c or d right

[dry viper]

its b

[lean prawn]

well I guess I don't really know where the confusion stems from

[rapid dock]

why b?

[lean prawn]

I feel like it might be a good idea to review how functions work in general and what inverse functions are

and solve inverse functions of simpler functions

like the inverse function for f(x)=2x+1

[lean prawn]

and then try to apply that to this exercise, which is really the same thing but with a slightly harder equation to solve

[wraith hinge]

Write
y = f(x) {whatevr you are given in the questions}
Now make it so you get something which looks like
x = (something in y)
Now just write x instead of y and vice versa
You have your answer

[rapid dock]

but what does something in y mean

[wraith hinge]

Like

y = 2x + 1
x = (y-1)/2

LHS should be x , RHS should not contain x

[rapid dock]

why is it y-1 tho

[dry viper]

and to make it a bit aesthetic , swap x and y

[wraith hinge]

y = 2x + 1
y -1 = 2x
x = (y-1)/2

[wraith hinge]

Now as f(x) = 2x + 1
f^-1(x) = (x-1)/2

[rapid dock]

im confused on where you ger y-1 from

get*

[dry viper]

see how the domain for f is range for inverse?

and vice versa

[wraith hinge]

, i took it from the example i wrote thrice

[rapid dock]

so why exactly in my problem would it be b

[dry viper]

the question says y=x+3/x-4

[rapid dock]

yes

[dry viper]

find its inverse

right?

y(x-4)=x+3

xy-4y=x+3

xy-x=4y+3

x(y-1)=4y+3

x=4y+3/(y-1)

to give it a "function" look

like a f(x)

replace x with y and y with x

y = 4x+3/(x-1)

[rapid dock]

ohhhhhhhh

okok

im sorry i was braindumb for a sec

[dry viper]

tip: use x and y instead of s and t to make it easy

[rapid dock]

alr

[dry viper]

personal choice

[rapid dock]

y for s right

t for x

for this problem you have to like work backwards right

[dry viper]

yes

kinda

[rapid dock]

so youd replace...

y for x?

[dry viper]

you know its the inverse

you have to find the original now

just solve for the opposite variable

make it x = something

[dry viper]

from your assumption

correct is b

i g2g

[rapid dock]

ok

[rapid dock]

thank you for helping

[dry viper]

have a nice day

[rapid dock]

you too

.close

[trim jolt]

Channel closed

Closed by @rapid dock

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[regal hatch]

what happened to the 20x here?

[raw magnet]

(2x+5)^2 = 4x^2 + 20x + 25

[regal hatch]

yeah, but hwo did the 20x just disappear

[wraith hinge]

what

its saying

r is equal to both of those

seperate

u make the 2 equations

[frail heron]

2ab

[wraith hinge]

equal to each other

[raw magnet]

r = $\sqrt{4x^2 + 20x + 25} = \sqrt{(2x+5)^2 } = 2x + 5$

[wraith hinge]

or wait no

tf

[solid kiln]

nosqldb

[wraith hinge]

im not thinking this right

[raw magnet]

what my food friend @wraith hinge is trying to say

[wraith hinge]

ITS

THE SQUARE

[raw magnet]

is this

[wraith hinge]

yes

tyvm

i didnt realize that was it solved out

[regal hatch]

oh right

got it

thanks

.close

[trim jolt]

Channel closed

Closed by @regal hatch

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[delicate lance]

Why do i have to set the y value equal to the slope? What happens

I dont ge it

.close

[trim jolt]

Channel closed

Closed by @delicate lance

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[real saddle]

How can I solve for 9.?

I have a rough idea since 13pi/12 is 195 which is 15 degrees in quadrant 3

and I've solved for 15 degrees in 5. which was sqrt(6) - sqrt(2)/4 but I'm not sure what changes it being in quadrant 3 does when solving with addition/subtraction formulas

Would it just have the obvious effect of making them all negative or would it be different in otherways?

[bright quarry]

well you can use subtraction formula

using 225-30

[real saddle]

sin(225 - 30)
sin 225 cos 30 - cos 225 sin 30
-sqrt(2)/2 * sqrt(3)/2 - -sqrt(2)/2 * 1/2
-sqrt(6)/4 - -sqrt(2)/4
-sqrt(6) + sqrt(2)/4
?

i think the math checks out there

[bright quarry]

yea looks good

[real saddle]

ty!

.close

[trim jolt]

Channel closed

Closed by @real saddle

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[wraith hinge]

i dont understand why this is true

[wraith hinge]

since it doesnt specify that F is R, F can be C, so can you not have a matrix with all diagonals being i for example

then RREF(A) wouldn't be the identity matrix or what am i missing

wait

that kind of matrix would still be reducable

deez

.close

[trim jolt]

Channel closed

Closed by @alpine pumice

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[distant abyss]

Hello! Still not on the clear with transcendentals, but taking baby steps

[distant abyss]

Is this correct?

[trim jolt]

@distant abyss Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[zenith sinew]

Hey

[low gazelle]

!status

[trim jolt]

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

[zenith sinew]

3

[low gazelle]

What did you get

[zenith sinew]

for b 15 out of 20?

[low gazelle]

How did you answer a

How you answered a could lead to different answers for the rest of the problems

[zenith sinew]

waittt

i got it thanks

.close

[trim jolt]

Channel closed

Closed by @zenith sinew

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[proven nymph]

Hello

[proven nymph]

having troubles even understanding or starting this problem

[marsh forum]

Are you familiar with implicit differentiation ?

[proven nymph]

chain rule yes which i think implicit differentiation is based off of/almost the same

but not exactly familir with implicit no

[marsh forum]

Better learn that first

I suggest you look up implicit diff on Khan academy

[proven nymph]

i tried researching it but usually youre given two points which my teacher doesnt do

so im not even sure anymore

[marsh forum]

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-2/v/implicit-differentiation-1

try this

[proven nymph]

lemme take a look, thank you

[proven nymph]

ok I understand it now

watched the video

but now we have e^y on the other side, thats pretty confusing

in khan academy it was just = 1

[marsh forum]

try differentitaing $e^y$ using implicit diff

[solid kiln]

Why am. I here

[proven nymph]

not really sure how to go about that

[marsh forum]

hint:_ $\frac{d\left(e^y\right)}{dx}=e^y\left(\frac{dy}{dx}\right)$

[solid kiln]

Why am. I here
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[marsh forum]

oops

just a mo

[proven nymph]

np

[marsh forum]

$\frac{d\left(e^y\right)}{dx}=\ e^{y\ }\ \left(\frac{dy}{dx}\right)$

[solid kiln]

Why am. I here

[proven nymph]

so it would just be e^y * (dy/dx)

[marsh forum]

that would be equal to teh differential of the LHS

[proven nymph]

is it always gonna be like that when doing implicit diff to e raised to a variable?

[marsh forum]

wdym?

[proven nymph]

right here

i think

[marsh forum]

that's how implicit diff works

it seems to me you haven't fully understood implicit diff, please look at a few more videos to strengthen your concepts

Hope this doesn't come off as rude

[proven nymph]

not at all, i def dont

i dont understand how everything works but i tried solving it just now

solved the problem to (5x^4) / (e^y - 4y)

which is wrong lol lemme take another look at khan

[marsh forum]

Hmm, $5x^4 + 4y^3 \frac{dy}{dx} =e^y \frac{dy}{dx}$

[solid kiln]

Why am. I here

[marsh forum]

Is what you should get

[proven nymph]

yea then i subtracted the second term over to the right

[marsh forum]

Yup

[proven nymph]

ok then i did

dy/dx (e^y - 4y^3)

on the right side

is that not right

[marsh forum]

Sounds right

[proven nymph]

ok then

i divided the e^y - 4y^3 to both sides

[marsh forum]

Yes

[proven nymph]

and was left with the answer

[marsh forum]

That should give you the right answer

[proven nymph]

i think i made a error when i was writing this on my notebook when i solved it a few minutes ago

maybe wrote a number wrong

so the answer then should be (5x^4) / (e^y - 4y^3)

i believe

yeaa i see the error i made, i forgot the y was cubed lol

awesome thank you

.close

[trim jolt]

Channel closed

Closed by @proven nymph

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• 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.

[young ice]

Yoooo!

[young ice]

my question is

simplify the expression, write your answer using only positive exponents.

125^3 divided by 125^8/3

<@&286206848099549185>

Can someone help me

Thank ya bb!

[wraith hinge]

its 3

the answer is

3

[raw magnet]

!noans

[trim jolt]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[sick onyx]

u dont give the answer bro

[young ice]

dude really

!noans

[trim jolt]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[raw magnet]

I alr told her lol

But Fr

[wraith hinge]

😭

ok look

[young ice]

way to be a helper dude

[wraith hinge]

dividing by a fraction

[tepid hamlet]

it's also not even the right answer 😄

[wraith hinge]

is multiplying the reciprocal

[young ice]

wow

[wraith hinge]

oh

its tot he 8th

ok

[sick onyx]

yeah the answer is wrong-

[wraith hinge]

SEE SO IT DONT MATTER

[young ice]

yeah lol it’s wrong

[raw magnet]

Yeah ig so

[young ice]

don’t tell@me the answer also

that’s racist

[raw magnet]

Aight let’s move on

[wraith hinge]

....

[raw magnet]

Don’t clog this channel

Anyway

[young ice]

u giving it cuz i’m white and fat?

oh k

sry

[wraith hinge]

yes

exactly

[tepid hamlet]

by the way is it divided by $125^{\frac{8}{3}}$ or $\frac{125^8}{3}$

[solid kiln]

Soosh

[raw magnet]

Bro 😭 take a breather

[sick onyx]

125^3/125^8/3

[raw magnet]

Both of you guys

[young ice]

i’m offended

[young ice]

Kk!

[sick onyx]

u just put the denominator to the numerator

thus its exponent will become negative

thats the initial

[young ice]

waiy i don’t get it

how did u get that tool

the twist

tex it

@solid kiln t

[sick onyx]

like

what

u talking abt

[young ice]

i don’t understand what u r trying to say

[sick onyx]

do u know that

x^1

[young ice]

please demonstrate it

[sick onyx]

is also equal to

1/x^-1

[young ice]

no

i’m white

[sick onyx]

..

[young ice]

how am i supposed to know that

[sick onyx]

if u dont know that then its useless

how u gona solve it

[young ice]

now just tesolain pls

explain Now

NOW

[sick onyx]

how

u dont even know it

[wraith hinge]

ok

[sick onyx]

its a formula

[wraith hinge]

pause

[scarlet cloud]

hi

[young ice]

forget u

[wraith hinge]

listen

stop

[young ice]

u suck at helping

[young ice]

hi

pls help

this helper can’t help

[scarlet cloud]

iam brazilian.-.

[young ice]

so

wait do u speak english

[bright quarry]

who sucks at helping?

y r u being rude

[sick onyx]

me-

[bright quarry]

don’t listen to him

[wraith hinge]

he's js not trying to lsiten

[sick onyx]

I already told him x^1 = 1/x^-1

[wraith hinge]

and understand

[young ice]

@bright quarry please help me

i’m failing

[raw magnet]

He’s just trolling

[bright quarry]

what’s ur question

[young ice]

i’m just sad

[kindred pier]

Go to #discussion

[bright quarry]

makes sense

he said hey bbg before

[scarlet cloud]

I'm using the translator.-.

[young ice]

dude she said she gave me the answer cuz i’m white and fat

OK ANYWAY

JUSG HELP ME

[young ice]

one sec

look every helper pls leave ok i’ve chosen @bright quarry as my helper

[wraith hinge]

no u said i gave the answer cuz ur white and fat

i didnt even give the right answer

[young ice]

Thank you guys but no

[wraith hinge]

the right answer is 3/125^5

[young ice]

Leave

[young ice]

STOP

<@&268886789983436800>

[wraith hinge]

ooo bed time

bye all

lmao

[sick onyx]

Isnt even right

[wraith hinge]

ik

[young ice]

he is giving me the answer without explaining