help-36

[tired walrus]

maybe we could read it together to find out how much of an issue it is.

[strong flax]

hmm

ok

i dont have it in hand

i guess i could get it

[tired walrus]

yes, go get it.

[strong flax]

well

i know its either 32k or 36k

[proper dagger]

per how long?

a month? a year?

[strong flax]

why dont we just look for both cases

3 years

36 months 32k

or 36 months 36k

[tired walrus]

ok so 32,000 miles per 3 years.

[strong flax]

yea

[proper dagger]

so ~10,667mi/yr

[strong flax]

this will just be alot easier ngl

[tired walrus]

in the meantime, im seeing 8 breaks per week in your current schedule.

[strong flax]

if i just lived on campus

[tired walrus]

so assuming you drive to and from home on each one of them (generous)

that'll be 9 * 2 * 8 = 144 extra miles per week

with an academic year 39 weeks long (give or take) that comes out to

,calc 144 * 39

[soft zealot]

Result:

5616

[tired walrus]

about 5,500 miles driven extra

per annum

ok so i guess if you drove back and forth every single break this would eat like half of your distance limit.

[strong flax]

i see

[tired walrus]

if you stay behind for the shorter breaks between your bio and chem classes, which are only 1.5 hours long and so conceivably could be spent on campus without rotting too much, all this driving will only eat 25% of your limit.

do you have anywhere else to drive regularly besides uni?

[strong flax]

i guess work

where i work its kind of close actually

[proper dagger]

how far?

and i'm guessing you work only on weekends?

[strong flax]

7 minute drive

2 miles

[tired walrus]

how many days a week?

[strong flax]

i actually work for my mom and dad

so i can just chill out there

in the business and theyll let me chill out and relax

[tired walrus]

ok great

wait actually

[strong flax]

maybe

i think i just need to bite the bullet

i mean autumn quarter is only like 11 weeks

then my suffering ends

its either that or i can just drop every single class and create a different schedule

[strong flax]

home to uni is 9 miles 15 minutes

uni to work is 7 minutes 2 miles

so that must mean home to work is 7 miles

[tired walrus]

that so?

do you drive past your workplace on your way to uni?

[strong flax]

yes

[tired walrus]

right ok

so you can drive to work during your multi-hour breaks and chill out there instead of going all the way home

that way your extra gas costs from that become even lower

[strong flax]

yea

yea

[tired walrus]

so it is even less of an issue

[strong flax]

ok i see

[tired walrus]

so like it's up to you to decide obviously

but we've looked into your concerns about time, fuel costs and car-lease distance limits

and by the looks of it, you won't really bump up against any of them, and you definitely won't have to rot on campus for 12 hours at a time

[strong flax]

hmm ok

thx ann

[final saddle]

@strong flax Has your question been resolved?

[final saddle]

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.

[final saddle]

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.

[whole halo]

<@&268886789983436800> @patent spire

[final saddle]

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.

[boreal smelt]

how do I solve this?

[terse crypt]

gp?

[lime crest]

geometric series

[terse crypt]

r = -tan^2 x

i think

[lime crest]

(since tan^2x < 1 in the given interval)

who the hell writes 1 + 2 + ...infinity though

never seen thst before

[terse crypt]

good question lol

is ... infty even a legit notation 😭

[boreal smelt]

yeahhh

how did I not see it😭😭😭

$\int^{\pi/2}_{0}\frac{dx}{(a^2sin^2x + b^2cos^2x)^2}$

[soft zealot]

Prathmesh

[boreal smelt]

I tried by multiplying sec^2x in both numerator as well as denominator

and then I substituted t=tanx

this is what I got $\frac{1}{a}[\int^{\infty}_{0}\frac{dt}{(t^2 +b^2)^2}]$

[soft zealot]

Prathmesh

[boreal smelt]

can anybody please help me with this integration

a and b are constants

@rustic sequoia

[delicate sorrel]

Hnm

[boreal smelt]

?

.close

[final saddle]

Channel closed

Closed by @boreal smelt

Use .reopen if this was a mistake.

[stuck juniper]

@boreal smelt do u know feynmann?

Very cool way to do this

I just worked it out

[boreal smelt]

yea

[stuck juniper]

And got pi/4ab(1/a-1/b)

Is this right

[boreal smelt]

just a sec

[stuck juniper]

(1/b^2 -1/a^2)

[boreal smelt]

yea

you are correct

[stuck juniper]

Cool

[boreal smelt]

your answer is right

how did you do this?

I wanna know

[stuck juniper]

Feynmann

[boreal smelt]

is it a method?

[stuck juniper]

First i took function of a

I got a result

Then function of b

Got a result

Added em up

Got the answer

[boreal smelt]

can you show?

cuz i didn't understand

[stuck juniper]

Yes

@boreal smelt

The first thing i got by simple integration

[boreal smelt]

ok so the first thing by simple integration

[stuck juniper]

<@&268886789983436800>

[boreal smelt]

please explain how you did the second step

[final saddle]

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.

[stuck juniper]

U can differentiate

[stuck juniper]

Wrt to a variable

Other the integrating variable

[tired walrus]

with respect to to

[supple mantle]

When the question is an answer

[boreal smelt]

ohh partial differentiation?

[stuck juniper]

im sleep deprived

[stuck juniper]

This is a classic jee question

[opaque ember]

jeopardy

[boreal smelt]

yeaa I'm solving it for jee

[stuck juniper]

This is definitely one step above advanc

I got the idea cause i had a similar question on my test

[boreal smelt]

this question is in my dpp

[stuck juniper]

Understandable

I think u could by parts as well

But cooler method

🥸

[stuck juniper]

@boreal smelt this is much more of a doable thing in the exam

Unless ur air 1

[boreal smelt]

I am trying so hard to understand what you did 😭😭

[boreal smelt]

by parts?

[stuck juniper]

Mhm

[stuck juniper]

Differentiated wrt to a once

And b once

Added them

Sin^2+ cos ^2 is 1

[boreal smelt]

but how did the integrand disappear ?

[stuck juniper]

I just didnt write it

dx is still there

Where ever it needs to be

Solving habits

[fast smelt]

you can differentiate with respect to whatever variable you want

are you referring to the fundamental theorem of calculus?

[boreal smelt]

and how did you write it directly as pi/2a^2b?

[stuck juniper]

Differentiate pi/2ab

Wrt a

[boreal smelt]

wouldnt it come with a negative sign?

[stuck juniper]

It will cancel with the one of lhs

[boreal smelt]

ohhhhh I finally got it mannnn

[fast smelt]

d/da{ pi / 2ab } = d/da{ pi/2b a^(-1) } = pi/2b -1 a^(-2) = -pi / (2 a^2 b)

[boreal smelt]

thanksssss

[stuck juniper]

U could byparts

[boreal smelt]

i well understand it now how you did it

[stuck juniper]

But even that requires manipulation

[boreal smelt]

btw can we introduce partial differentiation in any integration? Like does it affect the integration?

[stuck juniper]

Should be continous

[boreal smelt]

alright

[stuck juniper]

Just watch some vids on it

[boreal smelt]

can you recommend some?

[stuck juniper]

Anything is fine

Or even ask ur coaching sir

My sir taught it

[boreal smelt]

ohh

okayyy

thanks!

[stuck juniper]

Also if want the by parts help

Just multiply snd divide with the derivative of a^2sin^2x+b^2cos^2x

Then by parts

.close

[final saddle]

Channel closed

Closed by @stuck juniper

Use .reopen if this was a mistake.

[final saddle]

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.

[fleet hinge]

Hello, first time here.
This is what I am struggling with.

[thin cloud]

,rccw

[fleet hinge]

So far i have tried making a figure of sorts.

[supple mantle]

Which question are you stuck with again?

[fleet hinge]

Here @supple mantle

I am sorry for not specifying.

[supple mantle]

Ah

[fleet hinge]

It is from the chapter quadratic equations. A basic introduction of sorts to it where only the formula is taught.

[supple mantle]

Implicit differentiation

[supple mantle]

Oh...hmm

[fleet hinge]

This is the figure i've made so far.

[supple mantle]

Well

[fleet hinge]

I.... haven't studied calculus.

[supple mantle]

Do you know what the shape you made is?

[fleet hinge]

It is a triangle.

[supple mantle]

Which type of triangle?

[fleet hinge]

Right angled, since the tower must be standing perpendicular

[supple mantle]

Exactly

[fleet hinge]

I tried to go that way, got 20√5

As the line that connects the initial points of the drone and the car

At least as far as the question considers.

[supple mantle]

What's the rest of the question?

[fleet hinge]

At what distance from the base of the tower will the drone intercept the car

[supple mantle]

Did you take trigonometry?

[fleet hinge]

A drone is hovering at the top of a vertical tower that is 20 meters high. A car is moving in a straight line toward the base of a tower. At a certain moment, the car is 80 meters away from the base. At the same moment, the drone begins to descend in a straight line to intercept the car. Assuming the drone and the car are moving at the same constant speed, at what distance from the base of the tower will the drone intercept the car.

[fleet hinge]

Again. The basics.

Specific angles like 0, 30, 45,60 and 90

[supple mantle]

So just using pythagoras

Well we don't really need to use much anyway

[fleet hinge]

ahh, its still bound to quadratic equations as the worksheet in its entirety is based on it.

[supple mantle]

The triangle's base is 4 lots its height

[fleet hinge]

Yes

[supple mantle]

That's the slope of the straight line the drone will travel in

[fleet hinge]

So, like the graph.

[supple mantle]

Yea

[fleet hinge]

x/4x?

[supple mantle]

Mhm

[fleet hinge]

okk.

But neither of them are still, the drone must constantly change its path to intercept the car somewhere, right?

And they both have a constant speed

[supple mantle]

Well as the car gets closer to the tower, the drone's direction will become more vertically downwards

[fleet hinge]

Yes.

[supple mantle]

Wait

[fleet hinge]

But, we aren't taught how to connect graphs with quadratic equations

[fleet hinge]

ok

[supple mantle]

$(80-vt)^2 + 20^2 = (20\sqrt{17}-vt)^2$

[soft zealot]

VulcanOne

[fleet hinge]

v is the change in time? or velocity?

[supple mantle]

Velocity

t is time

[fleet hinge]

Why the 20²?

[supple mantle]

From Pythagoras basically, we constructed this equation, but we kept in mind the constantly changing distance for the car and the drone

The only distance that doesn't change is the tower's height

[fleet hinge]

So we add it to the change in the car's movement?

[supple mantle]

Wait hmm

$(80-vt)^2 + (20-v_y t)^2 = (20\sqrt{17}-vt)^2$

[soft zealot]

VulcanOne

[fleet hinge]

I don't really get it but thanks a lot. I will try to figure it out.

[supple mantle]

v_y is the vertical speed the drone moves in, v is the speed that both the car and the drone move in

$-40v_y t +v_y ^2 t^2 = 40(\sqrt{17} -4) vt$

[soft zealot]

VulcanOne

[supple mantle]

That's the most I can come up with

Atm

[final saddle]

@fleet hinge Has your question been resolved?

[final saddle]

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.

[tardy hamlet]

YOLO just need to ask if I’m going the right way

[tired walrus]

$\frac{2}{5} \cdot 10 \neq 8$ so i am afraid even your first line is mistaken

[soft zealot]

Ann

[tired walrus]

that and the letter x seems to have vanished somewhere?

dunno what's up w that.

[final saddle]

@tardy hamlet Has your question been resolved?

[south harbor]

Can some one help W 20

[craggy plank]

claim your own channel, check out #❓how-to-get-help

[south harbor]

I can't find that

[craggy plank]

you mean the channels? Check out channels and roles

and click on "follow category"

[dusky shadow]

,rccw

[final saddle]

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.

[lyric moat]

Was this solved right?

[rocky tusk]

this is illegible

[lyric moat]

Handwriting or the steps?

[rocky tusk]

it looks like you wrote y/y

[rocky tusk]

handwriting

[lyric moat]

Y prime/y

[rocky tusk]

where is y’ coming from?

this is what i see

$y = e^{2x + 3}$

[lyric moat]

The question was asking to differentiate y=e^(2x+3)

[soft zealot]

knief

[lyric moat]

Yes this is it

[rocky tusk]

so why are you doing y’/y?

logarithmic differentiation?

just use the chain rule

[lyric moat]

If I apply chain chain rule it would be 1/y * y' right?

[rocky tusk]

🤔

[lyric moat]

I am tweaking aren't I

[rocky tusk]

left side is y’, right side is 2e^{2x + 3}

multiply by the derivative of 2x + 3, which is just 2

[lyric moat]

FUCK

YOU ARE RIGHT

thanks bro

[final saddle]

@lyric moat Has your question been resolved?

Channel closed

Closed by @lyric moat

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.

[hollow plover]

can anyone help me solve this economics questionn

[craggy plank]

just post it here

[night raft]

!da2a

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[ripe jewel]

!da2a

[final saddle]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[ripe jewel]

damn MxRgd's lightnign fingers beat mine

[hollow plover]

not sure how to solve for the change in producer revenue

[final saddle]

@hollow plover Has your question been resolved?

Channel closed

Closed by @hollow plover

Use .reopen if this was a mistake.

[final saddle]

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.

[tall kindle]

I have no answer key but I just wanted to double check that the answer is A and not B right?

[kindred flicker]

When speak spanish 😭💔

[void ivy]

No sorry it's A

[loud sundial]

Yes

[tall kindle]

but it's two positive roots and one negative no?

[loud sundial]

The given graph is shifted down by some amount. If we consider shifting the graph upward until the turning point below the x-axis is a double root, then the root with multiplicity 2 is positive.

[void ivy]

yes

[loud sundial]

yes

[coarse turtle]

Option A

[tall kindle]

ty

.close

[final saddle]

Channel closed

Closed by @tall kindle

Use .reopen if this was a mistake.

[loud sundial]

[final saddle]

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.

[quasi plume]

how do you do these?

[fallen vigil]

what is taylor series

[quasi plume]

P_n(x) = bla bla bla

huh

what's the first step with problems like this?

lets just choose 49 for example. take the limit I assume

[rocky tusk]

rewrite cos and similar functions as taylor series

[quasi plume]

okay

[rocky tusk]

and do the algebra

[quasi plume]

oh that's it?

[rocky tusk]

yes

[quasi plume]

I shouldnt try to evaluate the limit first?

[rocky tusk]

the whole point is to do it with taylor series

try the first one and it should be clear

cos(x) = 1 - x^2/2 + x^4/24 - x^6/720 + …

[quasi plume]

okay

so I did it and got

x^4/24 - x^6/720 + …

[rocky tusk]

/x^4

[quasi plume]

right

[rocky tusk]

so what does that give you

[quasi plume]

lim x to 0 (x^4/24 - x^6/720 + …) / x^4

[rocky tusk]

no but more simplification

divide each term in the numerator by x^4

[quasi plume]

okay

lim x to 0 (1/24 - …) ?

[rocky tusk]

yes

which is

[quasi plume]

1/24?

lemme write out the other term also

lim x to 0 (1/24 - x^2/720 + …) ?

oh

x to 0 i'm dumb

I thought it was inf

okay bettt this is light

[rocky tusk]

yep

[quasi plume]

big ups knief, thanks

.close

[final saddle]

Channel closed

Closed by @quasi plume

Use .reopen if this was a mistake.

[final saddle]

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.

[rigid nacelle]

Increasing(Why does this mean above the X-axis?): (0, 2) U (4,6) U (8, inf)
Decreasing: (2,4) U (6,8)

Local Max: 2
Local Min: 4, 8

Concave Down: (0, 3)
Concave Up: (3,6) U (6, inf)

Inflection: 3

[rigid nacelle]

Is this right and how would I sketch it based on the results?

[rocky tusk]

you can prove that f is increasing when f' > 0 and f' > 0 means f' is above the x axis

[rigid nacelle]

Ah, ok

[rocky tusk]

you can think of f' > 0 as f has positive slope

so its increasing

[rocky tusk]

what about x = 6?

you sort of just ignored it

local max/min can occur where f is not differentiable

[rigid nacelle]

On which?

[rocky tusk]

not just at points where f' = 0

for example, |x| has a local minimum at x = 0

[rigid nacelle]

I thought if it crosses the X-axis then it's will be an local max/min

but 6 tping between the X-axis also counts?

[rocky tusk]

although |x| is not differentiable there

[rocky tusk]

thats one possible scenario

but it doesn't exhaust all possible situations in which f has a local max/min

[rocky tusk]

consider this example

,w graph |x|

[soft zealot]

Results provided by WolframAlpha

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

Requested by kniefy

[rocky tusk]

clearly not differentiable at 0 since the slope from the left doesn't agree with the slope from the right of 0

however, it is a local minimum and in fact a global minimum

[rigid nacelle]

Why though?

Is it because at 0, it changes directions? (goes from decreasing to increasing)

but I thought that was an infleciton

[rocky tusk]

do you know the formal definition of the derivative

using the limit?

[rigid nacelle]

Nope

[rocky tusk]

$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

[soft zealot]

knief

[rocky tusk]

never seen this?

[rigid nacelle]

I thought it was

(f(x+h) - f(x))/h

[rocky tusk]

its an equivalent definition

this gives the derivative at a point

[rigid nacelle]

Ah

[rocky tusk]

if you'd like, just substitute h = x - a then as x -> a, h -> 0 and f(x) = f(h + a) and so it becomes f(a + h) - f(a)/h

[rocky tusk]

we say that f is differentiable at x = a if this limit exists

now you should hopefully remember that if the limit from the left doesn't agree with or equal the limit from the right then the limit doens't exist yes?

[rigid nacelle]

Yeah

[rocky tusk]

so in our example with |x| the difference quotient (just the f(x) - f(a)/x - a stuff) becomes 1 on the right but - 1 on the left of zero

clearly 1 isn't equal to -1 so the limit doesn't exist

and thus f is not differentiable at 0

but its clear that |x| has a minimum at 0

[rigid nacelle]

So you can also count 6 as a local min?

[rocky tusk]

why?

how did you determine its a min and not a max

[rigid nacelle]

It went from -1 to 1

neg to pos = min
pos to neg = max

[rocky tusk]

in the original?

[rocky tusk]

here?

[rigid nacelle]

|x|

[rocky tusk]

or in |x|?

6 is not a local min for |x|

[rigid nacelle]

Going left to right, it starts -1 then stop 1

[rigid nacelle]

I meant 0 lol

It's why 0 is the local min

[rocky tusk]

yea 0 is a local min because there is an interval around 0 so that all points in the interval are above or equal to f(0)

and in fact its a global minimum

meaning its the smallest y value for the entire domain

[rigid nacelle]

For 6, it's local max then?

[rocky tusk]

yes

increasing before 6, decreasing after 6

[rigid nacelle]

This si much easier to understand lol

but yeah I get it, thanks

Last questin, how would I sketch it?

[rocky tusk]

f'(0) is not zero though?

do they mean f(0) = 0

[rocky tusk]

because clearly f'(0) = 3

[rigid nacelle]

This is the entire question

[rocky tusk]

i think they mean f(0) = 0

[rigid nacelle]

I don't know, I am just a humbly confused student

[rocky tusk]

assume they mean f(0) = 0

[rigid nacelle]

Sure

[rocky tusk]

then just do interval work

you could just see that this looks like cosine

and deduce that f must be a sine wave from 0 to 6

[rigid nacelle]

Is just doing sin to cos enough?

[rocky tusk]

idk what you mean

[rigid nacelle]

Is it enough to just to sine wave from 0 to 6

[rocky tusk]

but what about after 6?

[rigid nacelle]

THat's my question lol

[rocky tusk]

looks almost like a parabola so it doesn't really matter that much so long as you capture the increasing/decreasing behavior and concavity

[rigid nacelle]

.close

[final saddle]

Channel closed

Closed by @rigid nacelle

Use .reopen if this was a mistake.

[final saddle]

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.

[spiral dock]

i need help with this question

[spiral dock]

1. g(t) = 4cos^3(1-2t)

idk wat to do next

2. cos(1-2t) -> -sin(1-2t)

[grave axle]

it's a chain rule question imo

first find the derivative of cos⁴() then derivative of cos() and then the (1-2t)

[grave axle]

yeah now find the derivative of (1-2t)

and multiply the three of them together

(1) (2) and (3)

[spiral dock]

-2 right

[supple mantle]

Right

[grave axle]

yes

[spiral dock]

then i multiply it all together/

[grave axle]

yeah try it

[spiral dock]

hm okay

[supple mantle]

Keep in mind you need to apply Power rule in the beginning

[spiral dock]

4cos^3(1-2t)-sin(1-2t)(-2)

[supple mantle]

Yep

Correct

[spiral dock]

-8cos^3(1-2t)-2sin(1-2t)

ermm i dont think i did that right

[supple mantle]

g'(t) = 8cos^3(1-2t)sin(1-2t)

[spiral dock]

oh

how isnt the 8 infront of cos a negative

[supple mantle]

Because of the -2 and -sin

The negatives cancel

[spiral dock]

okay can u help me w the next question

or do i ahjve to start a new channel

[supple mantle]

You can send it here

[spiral dock]

for this i start with the product rule which is (f)(g') + (f')(g)

[supple mantle]

Mhm

[spiral dock]

f' = 2x g' = -5cscx

[supple mantle]

Hmm

[spiral dock]

ngl i dont know how g' becomes that

[supple mantle]

Check g' again

[spiral dock]

yeahh i knew it

cot5x

[supple mantle]

What's the derivative of cot(x)

[spiral dock]

-csc

[supple mantle]

Only?

[spiral dock]

ya

o wait

-csc^2

:D

[supple mantle]

Yep

Now what's the derivative of cot(5x)?

[spiral dock]

5-csc^2

[supple mantle]

-5csc^2(???)

[spiral dock]

x?

oh 5x

-5csc^2(5x)

[supple mantle]

Nope

You forgot that you used chain rule

[spiral dock]

(x^2)(-5csc^2(5x)) + (2x)(cot5x)

oh

[supple mantle]

Correct

[spiral dock]

-5x^2csc^2(5x) + 2cot(5x)

final answeR ^

[supple mantle]

Ye

Correct

[spiral dock]

yess

im lowk understanding it now ty

first we do the derivatrive of the squareroot to get rid of it right

so it would be

1/2(r^2 + r / r^2)^-1/2

[supple mantle]

Yeah

[spiral dock]

then i can cancel out both the r^2 in the numerator and denominator right

[supple mantle]

1/[2 * \sqrt{(r^2 + r)/r^2}]

[spiral dock]

can u show that written if possible

nvm i see it

wait nvm i dont

[supple mantle]

Hm

[spiral dock]

so if the r^2 cancels out

would the remaining r be r/1

so sqrt 1 + r

[supple mantle]

$\frac{1}{2\sqrt{\frac{r^2 + r}{r^2}}}$

[soft zealot]

VulcanOne

[supple mantle]

It'll be (1+r)/r

$\frac{1}{2\sqrt{\frac{r+1}{r}}}$

[soft zealot]

VulcanOne

[spiral dock]

i dont get it lol

[supple mantle]

That's the derivative of the root

[spiral dock]

i thought we got rid of the root

[supple mantle]

We simplified what was in the root

[spiral dock]

1/2(r^2 + r/ r^2)^-1/2

[supple mantle]

We can't get rid of the root here

[spiral dock]

ohh

[spiral dock]

wats the next step

finding the derivative of it all?

[supple mantle]

You can further simplify the fraction and make it $1 + \frac 1r$

[soft zealot]

VulcanOne

[supple mantle]

And take the derivative of that and multiply it with the whole thing

[final saddle]

@spiral dock Has your question been resolved?

Channel closed

Closed by @spiral dock

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.

[quasi plume]

can someone help with #65 please

[drowsy epoch]

which part

[quasi plume]

all of it

basically

[drowsy epoch]

[quasi plume]

I managed to figure out a I think

ln(1+x) = series 1^k * (-1)^(k+1) / k which is just (-1)^(k+1) / k

part b is kinda cooked

so for that I tried

idk what the point of the -ln(1/2) is

but I did - series x^k / k = - - series 1/2^(k)*k

answer is correct, I'm sure my methodology is wrong though

I got the extra - from the -ln(1-x)

and then I took x=-1/2

and moved the negative outside bc I felt like it

therefore you get series 1/2^(k)*k

[coral thunder]

Part (a) uses ln(1+x) rather than ln(1-x)

[quasi plume]

yeah

it's straight forward

part b is the weird one bc there's random negatives present now

I wrote the series for "b" as series (-x)^k / k

but idk if that's right

[coral thunder]

For part (c) you can write $\ln \frac{1+x}{1-x} = \ln(1+x) - \ln(1-x)$ and subtracts the corresponding terms of each series.

[soft zealot]

Hardy

[coral thunder]

And yes, your series for part (b) looks fine.

[quasi plume]

yeah the series looks fine bc I looked at the answer key, but is the work right?

I did the work after knowing the answer so I molded it as I saw fit

[quasi plume]

is the logic of the double negatives here canceling correct?

[coral thunder]

Yes

[quasi plume]

okay cool

so part c now

do you mean for me to plug in these?

[coral thunder]

Right

[coral thunder]

Well, they aren't equal to one another. The series you want is given by subtracting one from the other.

[quasi plume]

yeah for sure, that pic was from a formula sheet. sorry should've been more clear

okay so we end up getting

x-(x^(2)/2)+(x^(3)/3) - (-x - (-x)^(3)/2 ...)

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

[soft zealot]

Paul04

[quasi plume]

seem right?

wait

$\ln \frac{1+x}{1-x} = \ln(1+x) - \ln(1-x)$ is the same as $\ln \frac{1+x}{1-x} = \ln(1+x) +- \ln(1-x)$

[soft zealot]

Paul04

[quasi plume]

so I can just directly do $$x - \frac{x^2}{2} + \frac{x^3}{3} - \dots + \left(x - \frac{(x)^2}{2} + \frac{(x)^3}{3} - \dots \right)$$

[soft zealot]

Paul04

[quasi plume]

right?

[coral thunder]

Right

[quasi plume]

okay so the resulting series is just gonna be

plus

[coral thunder]

That's right

[quasi plume]

denominators are the same so we can just add and get series [(-1)^(k+1)*2x^k]/k

[coral thunder]

The numerators aren't the same

[quasi plume]

oh

right

[(-1)^(k+1)*x^k]+[x^k]/k

I guess u can factor out x^k now if that helps

[coral thunder]

Yep

[quasi plume]

[ x^k * ((-1)^(k+1) + 1) ] / k

[coral thunder]

Yep. You could leave it like that, or check behaviour for odd & even k.

[quasi plume]

oh cool, that wasn't too bad

what does part d mean?

[coral thunder]

In part (e) they probably want you to only evaluate terms with nonzero coefficients.

[quasi plume]

okay

[coral thunder]

Oh, you want $\ln 2$ and have a series for $\ln \frac{1+x}{1-x}$. So you want $x$ such that $\frac{1+x}{1-x} = 2$.

[soft zealot]

Hardy

[quasi plume]

okay cool

[coral thunder]

Have to go sorry, hope that's helped and maybe someone else can help more if needed.

[quasi plume]

sounds good, thanks so much!

.close

[final saddle]

Channel closed

Closed by @quasi plume

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.

[timber plume]

I need help with 11. b)

[timber plume]

we were taught

that we cannot deal with

-1 exponents

because the denominator would be 0

if we tried integrating it

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

[vital crag]

you don't need to

if f(x) is even, then the integral of -1 to 1 of f(x) can be expressed in terms of the integral from 0 to 1 of f(x)

[tired walrus]

was that the exact wording...?

your teacher said, verbatim, "we cannot deal with -1 exponents"?

that aside though there is in fact no need to break your back with any integration at all here, as riemann said

[tired walrus]

ok but this rule is entirely inapplicable here.

[proper dagger]

um, that's a quadratic in the denom anyway

[tired walrus]

that's precisely why the rule is inapplicable.

[proper dagger]

yea pointing it out to OP

[tired walrus]

yeah i guess when i say it there's like a debuff to comprehensibility or something

[proper dagger]

i didn't mean it that way, sorry if it came across as such

[timber plume]

wdym?

[tired walrus]

consider this

you've shown that 1/(x^2+1) is an even function

that means its graph is symmetric across the y-axis

what does that let you say about the green and blue areas as shown in my pic?

[timber plume]

same area

[tired walrus]

ok, do you see how to express $\int_{-1}^1 f(x) \dd{x}$ in terms of $\int_0^1 f(x) \dd{x}$ now?

[soft zealot]

Ann

[timber plume]

but

don't we need to add a constant

of 2

before the integral

[tired walrus]

adding that constant is your job. you're not trying to say i omitted it by mistake, are you?

[pliant shore]

woah there Ann

[timber plume]

I really thought

you missed a 2

[pliant shore]

yes, and can you explain in your own words why

[tired walrus]

i didn't miss shit, i said "express <this> in terms of <that>"

i didn't put any equals sign or do the expressing

that was deliberate because i wanted YOU to do it

[vital crag]

the last time ann left out a factor of 2, it was last century

[timber plume]

lmao

[tired walrus]

nah don't exaggerate

[timber plume]

I know this but how does this relate to my op?

[tired walrus]

your op literally \textbf{asks for} $\int_{-1}^1 f(x) \dd{x}$.

[soft zealot]

Ann

[timber plume]

yeah but then how will we integrate the integrand?

[tired walrus]

... you don't, not in this question.

the whole POINT was:
\begin{itemize}
\item they give you the value of $\int_0^1 f(x) \dd{x}$ for free
\item they ask for the value of $\int_{-1}^1 f(x)\dd{x}$
\item you need to find a link between the two and then use it
\end{itemize}

[soft zealot]

Ann

[vital crag]

Re-read the question

[timber plume]

ohh got it

pie/2

[tired walrus]

pi, not pie.

do not ever spell it as "pie".

[timber plume]

pi/2

[timber plume]

my bad 😭

thanks

.close

[final saddle]

Channel closed

Closed by @timber plume

Use .reopen if this was a mistake.

[final saddle]

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.

[jagged flare]

$A,B,C,D,E,F$ are on a circle, $P,Q,R$ are the intersections of $AD$ and $BE$, $AD$ and $CF$, $BE$ and $CF$. If $AP=21$, $CQ=13$, $ER=35$, and $PQR$ is an equilateral triangle, then find $BP+DQ+FR$

[soft zealot]

ihave<skissue>

[jagged flare]

ignore the 60

im confused, how do you do this? im guessing maybe poap? but how do you pull it

[robust mulch]

honestly my first guess is to let the equilateral triangle have side length 0

so you have three lines that intersect with 60deg angles

kinda stuck from there

worst case you can coordbash

just a guess but it may be 59

69*

theres probably some weird property about equilateral triangles and circles

[final saddle]

@jagged flare Has your question been resolved?

[coral thunder]

One thing that occurs to me is that if you take a line parallel to AD, then the length of AP and DQ change by the same amount.

Likewise BP and QC

[obtuse epoch]

there's a symmetry

[coral thunder]

So you can choose the parallel line with Q=D=C, and then do the same for A=B=P

[obtuse epoch]

something like this ig AP+CQ+ER=BP+DQ+FR.

[coral thunder]

Yes, those transformations preserve (AP+CQ+ER)-(BP+DQ+FR).

And the final result has ER=FR

But there's probably something nicer and more rigorous.

[jagged flare]

what?

[obtuse epoch]

yeah and AP , CQ , ER are given

21+13+35=69 deg

[coral thunder]

If you imagine "sliding" one line of the equilateral triangle in and out to get different intersection points.

If you move the line outward, then AP and QD get longer by the same amount as each other, and BP and CQ shorter by the same amount as each other.

ER and FR don't change.

So the diagram with the larger triangle has the same value for (AP+CQ+ER)-(BP+DQ+FR).

[obtuse epoch]

if you rotate it then P becomes Q and Q becomes R and 1 more

[jagged flare]

does the expanding/contraction cancel eachother out?

[vital surge]

a = AP, b = BP, c = CQ, d = DQ, e = ER, f = FR

power of a point:
(s + d)a = (s + e)b
(s + a)d = (s + f)c
(s + b)e = (s + c)f

rearrange:
(a - b)s = be - ad
(d - c)s = cf - ad
(e - f)s = cf - be

(1) - (2) + (3):
a + c + e = b + d + f = 69

[jagged flare]

oh

i see, thank you guys!

.solved

[final saddle]

Channel closed

Closed by @jagged flare

Use .reopen if this was a mistake.

[coral thunder]

Oh yes that's a nicer one 🙂

[final saddle]

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.

[safe cradle]

16*55 please

[tired walrus]

<@&268886789983436800> troll?

[safe cradle]

I got 28

[tired walrus]

your phone definitely has a calculator

[fallen geyser]

Don't use help channels for troll questions, thank you.

.close

[final saddle]

Channel closed

Closed by @fallen geyser

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.

[little finch]

Find the coefficient of x^10 in the expansion of (1 + x^2)^20

[proper dagger]

oh boy. looks like a long day of binomial expansion with pascal's triangle?

[little finch]

yes

[little finch]

yes

[desert mantle]

(not like you need to compute the entire triangle)

(and you could just compute the binomial coefficient with the factorial def)

(or maybe its even enough to just write down the binomial coefficient itself)

[proper dagger]

then that's out of my league

[ivory bear]

use binomial theorem directly

[lucid nymph]

you can use binomial general term

[ivory bear]

more better

[little finch]

ncr one?

[ivory bear]

in the end my and @lucid nymph 's approaches are same

[lucid nymph]

$T_{r+1}=\binom{n}{r}x^{n-r}y^r$

[soft zealot]

ImOakley

[little finch]

k

[proper dagger]

huh. looks remarkably like Bernoulli trials

[tired walrus]

there is a connection there but that's best kept out of this help channel

[little finch]

x to the power 2r so r = 5 ?

[tired walrus]

r=5 indeed

[proper dagger]

fair

[little finch]

ok got it 20 c 5

Hey I've got another one

[sour zealot]

ok!

[little finch]

Q4

[desert mantle]

,rotate

[desert mantle]

wrong pic?

[little finch]

naah the question below

[lucid nymph]

you mean the sum to 20 thing?

[little finch]

yes

[desert mantle]

thats an expression

[lucid nymph]

whats the question

[desert mantle]

not a question

[little finch]

I hereby apologize.

help tho ;-;

[sour zealot]

what do you need help with ;-;

[lucid nymph]

im confused are u supposed to express the sum?

[tired walrus]

!xy

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[little finch]

i showed the picture

[lucid nymph]

that cant be the original question

[little finch]

that is

ig they want the sum

[lucid nymph]

are you able to take a screenshot of something?

[little finch]

i took pic from my nb

[tired walrus]

we do not understand WTF the question is asking

[lucid nymph]

where is the question from

how was it given to you

[bold turtle]

of the original question, not of your WO

[little finch]

its asking the sum

i suppose

[desert mantle]

pull out a factor of 2

[late bolt]

Is it this?

[little finch]

i got this solved

[little finch]

that becomes( 2 to the power 20-r )/ 2 ?

use the general term then?

[desert mantle]

2*2^(20-r)

[little finch]

oh yeah

ok got it thanx

.close

[final saddle]

Channel closed

Closed by @little finch

Use .reopen if this was a mistake.

[little finch]

.reopen

[final saddle]

✅

[solid notch]

do you have a question?

[proper dagger]

if you have, please send it

[little finch]

yes

wait a minute im taking picture

The empty space has the summation symbol

[final saddle]

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

[sour zealot]

could you show your work?

[little finch]

word?

[proper dagger]

work

[little finch]

its quite messy i did it in rough i rewrote it as 1/n(1+k/n+(k/n)^2
ig its riemann summ of 0,1

[tired walrus]

for the future, btw:

!1q

[final saddle]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[tired walrus]

and also ^ for exponents

[little finch]

howd you make that on keyboard

[tired walrus]

shift 6

[little finch]

ok

ok thx

so will i recieve help tho?

[proper dagger]

i don't see a question

[little finch]

this

the empty space is summation idk y it glitched

[little finch]

see it now?

[proper dagger]

yes

[little finch]

ok

[little finch]

.

sed

[little finch]

...

.close

[final saddle]

Channel closed

Closed by @little finch

Use .reopen if this was a mistake.

[final saddle]

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.

[past cargo]

not math

but can someone help

[solid notch]

give input on what you think

[past cargo]

i think it will affect momentum because mass is changed

[flint saffron]

Hello i have to do a decomposition in fourier's series of Bcos(wt)cos(phit) so i said that it is B/2*(cos((w+phi)*t+cos((w-phi)*t)) ( idk if it is correct but it seems logical, fourier's decomposition is transforming an expression to the sum of cosins and sin, however it asks to draw the graph of fourier. I looked it up online and it would seem that i have to draw triangles but I do not get it. May someone explain it to me ? I am french so the question I have of my book is french but i do not know the method.

[night raft]

!occupied

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[final saddle]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[solid notch]

sniped

[thin cloud]

that was fast

[proper dagger]

i barely typed

[flint saffron]

sorry

[night raft]

lol

[night raft]

no worries

[meager mirage]

That kind of reasoning is very wrong

[solid notch]

think bigger

[past cargo]

is momentum not mass x velocity

[night raft]

yes

[past cargo]

mass increases

[meager mirage]

Velocity can decrease

[past cargo]

oh true

[lucid nymph]

conservation of momentum is one of the fundamental laws

[meager mirage]

It only changes if net external force / thrust is applied

[past cargo]

so does it do nothing then, because they act perpendicular to each other. like its moving horizontal and that thing is being dropped vertically

[lucid nymph]

no

the mass of the truck increases

so the velocity decreases

conservation of momentum applies in each direction

so the fact that they are perpendicular doesnt matter

[past cargo]

ok so no change of momentum then?

[lucid nymph]

impossible

[meager mirage]

No

[night raft]

you're confusing it with moments, when you say perpendicular distance

[meager mirage]

Use energy

[lucid nymph]

this isnt a question about energy

[past cargo]

if mass increases and velocity decreases then its the same

[meager mirage]

Mgh = 1/2(m+M)v^2

[past cargo]

wdym

[solid notch]

?

the cart is already at the runaway

read the question again

[meager mirage]

Take M as mass of truck, m as mass of collected water

[meager mirage]

Oh sorry

[solid notch]

he asked 2.4 , your suggestion was prol relevant for 2.1, 2.2

[meager mirage]

Mb

[past cargo]

so what happens then

[solid notch]

yeah pricely

pricisely

[meager mirage]

Yeah velocity decreases

[past cargo]

what about for the collected water

[lucid nymph]

thats what were talking about

the water adds mass to the truck so the truck has to slow down

[solid notch]

we were talking about the system no? if he solely is looking at the water , it did have a direction change, hence a change in momentum?

[past cargo]

yes we discussed effect on the truck

now what about the water

[lucid nymph]

the vertical movement isnt relevant

[past cargo]

well read the question

[lucid nymph]

no you dont understand

momentum is conserved in each direction

the rain has no horizontal momentum

[past cargo]

does it not convert

[lucid nymph]

and in the vertical technically momentum is still conserved its just it comes in the form of going into the whole earth and giving it a change in velocity so tiny its literally negligible in every sense

[past cargo]

it falls and hit the truck, the tuck is moving so the water also moves in the truck

so vertical to horizontal

[lucid nymph]

?

[past cargo]

so a increase in momentum

[lucid nymph]

like i said momentum is conserved in each direction

you cant bring momentum from the vertical to the horizontal

if two balls hit off each other they still retain the exact same speeds in the directions perpendicular to where their centers align

because its only the direction they impact along that is affected

[past cargo]

yh i see

thanks for the help

.close

[final saddle]

Channel closed

Closed by @past cargo

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.

[ember escarp]

how to get the rational equation of 1/x = 2/3x-1?

[terse crypt]

Wdym ‘get’

[ember escarp]

oh what i mean is what is the rational equation

sorry, wrong sentence

[terse crypt]

Do u mean

‘How do I solve this rational equation?’

[ember escarp]

yes

[delicate sorrel]

[terse crypt]

Eliminate the denominators first

And note where x are undefined

[ember escarp]

what I have so far is I found the lcd

then i tried multiplying both sides

and from there, im lost

[delicate sorrel]

$/fract{1}{x}/ = /frac{2}{3x-1}$

[delicate sorrel]

You can just inverse the two sides and work normaly

[coarse mantle]

multiply both sides by x

[ember escarp]

so like x(1/x = 2/3x-1)x?

[coarse mantle]

you'll get $2x = 3x-1$

[soft zealot]

Bakoles

[severe canyon]

I have a hunch that we are solving a different expression

!original

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[delicate sorrel]

Just inverse the both sides and work normaly

You can its allowed

[ember escarp]

can u show me how?

[delicate sorrel]

I mean by inverse like 1/x become x

And 2/3x+1 becomes 3x+1/ 2

[ember escarp]

aren't we supposed to look for the lcd?

[delicate sorrel]

Youll have x= 3x + 1 /2

[delicate sorrel]

Huh?

[coarse mantle]

$\frac{1}{x} = \frac{2}{3x-1} => x = \frac{3x-1}{2}$

[soft zealot]

Bakoles

[ember escarp]

cuz i know u find the lcd then multiply it both sides then solve and check the solution

I understand the instructions, its just that I can't comprehend on how to solve it

or the steps

[coarse mantle]

look, get rid of the fraction 1/x

how?

multiply it by x

you get $x\cdot\frac{1}{x} = x\cdot\frac{2}{3x-1}$

[soft zealot]

Bakoles

[coarse mantle]

do you see it now?

[ember escarp]

mhmm

[coarse mantle]

can you multiply them?

what do you get

[ember escarp]

wouldn't be just the same cuz x = 1?

[coarse mantle]

i mean yes, x = 1

wait

what was your question?

didn't you need to find x?

[ember escarp]

yes

[coarse mantle]

okay great you found it

x = 1

[ember escarp]

oh

i have another one tho, its much different

x-2/2x+4 = 1/5

[coarse mantle]

did you try anything so far

[ember escarp]

still the same

stuck in multiplying both sides

thats what im having a hard time

[coarse mantle]

use the butterfly technique

[ember escarp]

does it apply to any equation?

[coarse mantle]

see this example:
if a/b = c/d
then ad = cb

$5(x-2) = 1(2x+4)$

[soft zealot]

Bakoles

[coarse mantle]

can you solve it now?