help-27

[hollow walrus]

thank you guys though for coming to help

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[devout plover]

🤔

[surreal night]

You're right

[devout plover]

yay

[devout plover]

this would be like

a/b = d/c

right

[surreal night]

I think so

[devout plover]

ok

.cloSe

woops

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[oak jasper]

Hello! Im currently doing equations with parentheses: decimals and fractions. I hate fractions and don’t understand them so could someone break down as much as possible on where do I start and how do I go about doing equations with fractions?

[thorny raft]

So the main thing you need to remember about solving equations is that you want to get the variable by itself

so think about, for example, 3x = 6

how could you get x by itself here?

you divide both sides by 3

but you could also write it as (1/3) * 3x = (1/3) * 6

so you multiply both sides by 1/3 to get x by yourself

so now say you have (1/3)x = 6

to get rid of this now you just multiply both sides by 3

[proud acorn]

man I'm slow

[thorny raft]

it's basically just using the rule of reciprocals (something) * (1/something) = 1

[proud acorn]

its like you are speaking a whole other language 😭

[thorny raft]

what are you confused on?

[oak jasper]

Me too😭

[proud acorn]

can you check dms

its private kinda

[thorny raft]

lol why does it need to be a dm

we're talking about fractions

[oak jasper]

Okay I do think I get it now after figuring it out myself a little but im just stuck on where to go after taking everything out of parentheses

This is where I am now

[thorny raft]

ok so this is an equation that has a variable on both sides

[proud acorn]

nah its abt something else

lol

[thorny raft]

there's multiple ways to approach this but I like to get the variable on one side first since it's simpler

[thorny raft]

I'm doing this rn

[thorny raft]

do you think you can figure out how to get the r to one side or do you need some help?

[oak jasper]

I will need a help 😅 I know I need to cancel out something but idk which one

[thorny raft]

no problem

so let's think about a different example

if we have 3x = 2x + 1 how could we get the x to one side

[oak jasper]

Canceling out the 1 ?

Oh

Wait

Canceling the 2x

[thorny raft]

exactly

then you'd be left with x = 1

so with (1/2)r - 3 = 12 - (9/2)r it's the same idea

[oak jasper]

So I should cancel out the 9/2r ?

[thorny raft]

yes

[oak jasper]

Do I minus that from the 1/2r or the 3?

[thorny raft]

well let's think, you have -(9/2)r and to cancel that you need to add (9/2)r to both sides
So we would have (9/2)r + (1/2)r - 3 = 12 - (9/2)r + (9/2)r
and then (9/2)r + (1/2)r - 3 = 12
next is just combining like terms

[oak jasper]

Ah I see

Okay

I basically did the whole problem

But I came out with a decimal…? For the R=

[thorny raft]

the answer will be a whole number

what are your steps right now?

[oak jasper]

So would it be 15/1 divided by 10/4 ?

[thorny raft]

where are you getting your 10/4?

[thorny raft]

ah

(9/2) + (1/2) is not (10/4)

remember you only add the numerator

[oak jasper]

Okay so it would just be 10?

[thorny raft]

(10/2)

[oak jasper]

Right so then id divide 15/1 and 10/2

[thorny raft]

yes but I would recommend evaluating 10/2 first

[oak jasper]

Meaning… ?

[thorny raft]

well what is 10/2

[oak jasper]

5

[thorny raft]

yes, so then you'd have 15/5

[oak jasper]

Omgg

Okay

Thank you so much 🫶🏾🫶🏾🫶🏾

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[zenith tulip]

What's the logic of this? I'm completely lost. The answer is 288. I couldn't do anything, I only understood it increments and decreases by one

[zenith tulip]

that long line is just ........

[dense lynx]

combine the stuff inside the parenthesis

like do (1 + 1/2) = 3/2 and (1 + 1/3) = 4/3 etc

[zenith tulip]

isnt there any practical way

since it goes till 47

[dense lynx]

just do it for the first few terms

you will see what happens and why this is very easy to evaluate

[zenith tulip]

ok, let me try

[zenith tulip]

so i got 48/47 for the upper side

and 11/12 bottom

i get 576/517 when multiplicating them

i dont know how to reach 288

[dense lynx]

1 + 1/47 = 48/47
1 - 1/12 = 11/12

it looks you took the last term for both the numerator and denominator and disregarded everything else

[zenith tulip]

oh

theres also

multiplication

between them...

but i mean

multiplicating everything between 1 and 47 will give me huge numbers

feels like i dont understand it enough

my book doesnt literally explain the logic behind that

[devout snow]

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[unborn atlas]

41b

[unborn atlas]

How do I construct the integral?

I think the shell method works well here

[devout snow]

@unborn atlas Has your question been resolved?

[winter torrent]

well if you've done 41a it's very simple

[devout snow]

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[tawdry venture]

how do i integrate this

[heavy current]

hello Flappie

[surreal night]

Hi higher

[surreal night]

!show

[devout snow]

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

[tawdry venture]

yes

i literally ahve no idea how to start it

[heavy current]

we need to find $\int 3e^{-3x}dx$?

[woven radish]

higher!

[heavy current]

what do you think the first thing you should do is?

[surreal night]

Try looking at the derivative of e^ax

[devout snow]

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[raven moss]

regarding line integrals solved numerically vs. analytically

[raven moss]

$\int 25x^{2}-8x^{3}dx$

[woven radish]

Stormerz

[raven moss]

integrated from 2 to 3 gives 28.33

but numerically this gives me 27.41

what did i do wrong?

[surreal night]

,w int 25x^2-8x^3dx from 2 to 3

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[surreal night]

what happens when you let n go bigger?

[raven moss]

it still converges to 27.41

[last burrow]

(25/3)x^3-2x^4

[raven moss]

this is the idea

sum all the l's

when delta_x and delta_y approaches 0

[devout snow]

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@raven moss Has your question been resolved?

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[real whale]

Hey, I have to resolve an inequality and I got x-1 < 2 sqrt(x+1). How I can find x?

[feral agate]

Squaring both sides, along with checking the domain of the square root term

[real whale]

Okay so we don't have to separate at 0

I mean separate in 2 cases where x-1<0 and a second case where x-1>0

[feral agate]

That would have been necessary had the inequality been reversed

$x-1 <2\sqrt{x+1}$

[woven radish]

kheerii

[feral agate]

If x-1<0, then all values of x (that are in the domain) are valid

Since the left hand side is negative and the right hand side is necessarily positive (as the square root function is always positive)

If x-1>0 then we actually need to solve the inequality

$x+1\ge 0$ is also a necessary condition, due to the domain of the square root function

[woven radish]

kheerii

[real whale]

Effectively

Okay I understand thank you!!

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[somber marsh]

I need help with q4 a and b, 5, 6b and 7 a b

[restive river]

How can I make a channel

Is it possible?

[somber marsh]

u cant make one u have to go to the unoccupied help and dm in those ones what u need help with

[restive river]

There's in no unoccupied

[somber marsh]

Integration and using calculus

[somber marsh]

go to how to get help

[devout snow]

@somber marsh Has your question been resolved?

[devout snow]

@somber marsh Has your question been resolved?

[devout snow]

@somber marsh Has your question been resolved?

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[sweet bobcat]

Hiya so i got a question
is like computing definite integrals the exact area

[sweet bobcat]

or is it just an estimation

when we like do a question like this

is it the like, exact area

[wicked turtle]

for nontrivial curves, it defines the area under the curve
but it does agree with the geometric area in cases where the geometric area is known

[sweet bobcat]

what's a non trivial curve?

[wicked turtle]

well for example $\int_0^\pi \sin(x),dx$
you don't have a separate geometric definition of the area under this curve

[woven radish]

Bungo

[wicked turtle]

(or choose some more complicated function if you like)

[sweet bobcat]

so in a function like this ^

its just defining the area under the curve?

instead of actually finding it

[wicked turtle]

well that one goes negative so you also have to remember that the integral is computing the signed area

(area of the part above the x axis minus the area of the part below)

[sweet bobcat]

this area right?

[wicked turtle]

yea, the rectangles are an approximation

[sweet bobcat]

all those like rectangles

[wicked turtle]

that's what a riemann sum is

the integral is what you get in the limit

[sweet bobcat]

but im not solving for the rectangles, im just finding the actual area

[wicked turtle]

yea, the integral is the area more or less by definition
it's defined for that purpose

[sweet bobcat]

So then this is the exact area, as a defined integral is the area under a curve?

[wicked turtle]

yes

it's not an approximation

[sweet bobcat]

uh what's the exact definition of a definite integral(could you share a link?)

[wicked turtle]

the most intuitive one is given in terms of upper and lower sums, which are approximations from outside and from inside using rectangles, lemme dig up the definition

basically says that if you make the rectangles thinner and thinner, the upper and lower approximations become closer and closer to each other, and they converge to the same limit

that limit is by definition the definite integral

[sweet bobcat]

slrighty so this was for a math assigment just like tell me if this sounds like smart

[surreal night]

<@&268886789983436800>

[sweet bobcat]

?

go off dude

hey flappie

[wicked turtle]

go away

[sweet bobcat]

I think this is the exact area, in our graph, as the rectangles get more thinner, both the upper and lower approximations become closer and closer, and eventually converge on the definite integral.

[sweet bobcat]

is this the right way to think about it?

[wicked turtle]

yea that's basically it

[sweet bobcat]

ik i just basically copied off your thing

is there anything you think i should add?

[wicked turtle]

like i said, the function in your specific example takes on both positive and negative values, so you may want to at least mention that the definite integral computes a signed area

but aside from that it's fine

[sweet bobcat]

alrighty!

[sweet bobcat]

thank you, mind if you proofread my final response?

[sweet bobcat]

actually one last doubt

whats a signed area?

[wicked turtle]

it's the area of the part of the function above the x axis, minus the area of the part below the x axis

[sweet bobcat]

Oh ok , so like the integral of the lower bound minus the upper bound?

[wicked turtle]

like for example:

,w plot sin(x) from x=0 to x=2*pi

[woven radish]

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[wicked turtle]

this guy has integral zero

the part above the x axis has the same area as the part below

they cancel each other out

[sweet bobcat]

oh ok, but in this case we dont have that

we can't like, fold it in right?

[wicked turtle]

right

more generally you just take (area of the part above the x axis) - (area of the part below the x axis)
and that's the integral

[sweet bobcat]

oh alrighty

[wicked turtle]

(you don't have to do this manually, that's what the integral computes automatically)

if you want to treat the part below the x axis as positive area as well, then that's equivalent to integrating the absolute value of f(x) instead of just f(x)

[sweet bobcat]

I think this is the exact area, in our graph, as the rectangles get more thinner, both the upper and lower approximations become closer and closer, and eventually converge on the definite integral. Which computes over a signed area.

so i just added ethe last part

is this like mathicametically correct or whaterver

[wicked turtle]

i would get rid of the word "over"
it computes the signed area

[sweet bobcat]

oh ok

[wicked turtle]

but that's all correct

[sweet bobcat]

alrighty!

thanks

also one more thingy about this question, the estimated area changes, but the actual area doesnt change, this is due to there being like less or more rectangles right

which increase or decrease the estimation?

[wicked turtle]

yes

as you make the rectangles thinner, in general the upper sum gets smaller and the lower sum gets bigger
because you can fit the rectangles more accurately

eventually in the limit, the two estimates approach the true area (if the function is integrable)

[sweet bobcat]

alrighty sounds good! thanks so much dude ❤️

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[prisma quail]

I don't know why my answer to this problem is incorrect: The base of a solid is the region between the parabolas $x = y^2$ and $2y^2 = 3 - x$. Find the volume of the solid if the cross-sections perpendicular to the $x$-axis are equilateral triangles. I got $\frac{\sqrt{3}}{4}\int_{-1}^{1}\left(-3y^{2}+3\right)^{2}dy = \frac{12\sqrt{3}}{5}$

[woven radish]

Suzzzzzzz

[mild sorrel]

,w time usa

[woven radish]

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[mild sorrel]

,w time new york

[woven radish]

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[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[pure spruce]

can someone explain this anymore

no one ?

[eager nova]

Your channel is here #help-12 you have been asked something there

Close this channel

[pure spruce]

ooh sorry

[eager nova]

.close

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[obtuse wind]

Hi, why is this false? I know it's supposed to use the chain rule but I don't know why

[timber pebble]

chain rule

well, okay, not really what you asked

it matters what f(x), yea

i mean, say f(x) = 4

then the derivative is 0!

so, it seems like our answer should involve f'(x) somehow

otherwise i see no immediately good way to produce this 0

why sully

[safe jasper]

unexpected factorial?

[mighty knoll]

0 factorial

[timber pebble]

the symbol is used here to emphasize a remark

[mighty knoll]

I'm meming obv OP understood it was 0

[obtuse wind]

this is what is answer key says

[timber pebble]

its true

[obtuse wind]

i just don't get why do that instead of the power rule

[timber pebble]

how do you know the power rule is applicable

say f(x) was ... sin(x)

[safe jasper]

you can view f(x)^1000 as a composition of two functions, g(x) = x^1000 and f(x) = f(x), then f(x)^1000 is g(f(x)). by the chain rule, the derivative of g(f(x)) is g'(f(x)) * f'(x)

when calculating g'(f(x)), this is where the power rule comes in, since g is a power function

but since two functions are composed, the chain rule must also be included

[timber pebble]

i dont know like okay you can totally use the definition to see how the chain rule should give you this answer

[timber pebble]

but this is more concerning because you should try to come up with an example that breaks it

[obtuse wind]

OH and the power rule is when it doesn't have f(x) in the parenthesis?

[timber pebble]

the power rule is when f(x) = x

and infact, you can use the chain rule here, too

(its just that f'(x) = 1)

[safe jasper]

what if f(x) = 2? if you believe that you should only be able to use the power rule, then d/dx (2^1000) would be 1000 * 2^999. but 2^1000 is a constant, clearly its derivative is 0

[timber pebble]

yea, this is my reasoning as to why we should expect the answer to have f'(x) in some form or another

because otherwise it's very hard to make a constant function give the answer we would expect

[safe jasper]

there isn't really a rule like "we only do power rule when we don't see f(x)". you do chain rule when you see a composition of functions, you do power rule when something is raised to a power.

in this example, you have a composition of functions, you have u^1000 but instead of u, you've composed that with f(x)

so because there's something being raised to a power, your intuition about using power rule is correct

but because f(x) has been composed inside another function, you must also take the derivative of the "inner" function

[obtuse wind]

i'm lost again 😭 How come these use the chain rule?

[timber pebble]

you did it correctly

[timber pebble]

its the exact same thing as before

i mean, literally

$\dv x \qty( f(x)^n ) = n \qty( f(x) )^{n-1} f'(x)$

[woven radish]

jan Niku

[timber pebble]

here's our rule

can you see how these were applied above?

[obtuse wind]

no 😭

[timber pebble]

$\dv x \qty[ (5x+3)^4 ]$

[woven radish]

jan Niku

[timber pebble]

whats f(x) and whats n?

[obtuse wind]

but it looks like this 😭

[timber pebble]

no it doesnt

[timber pebble]

it looks like this

[obtuse wind]

.close

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[restive river]

Hi, can someone help?

[wicked turtle]

what do you think is the right answer?

[restive river]

The thing is, I’m not sure how to solve it at all

[wicked turtle]

probably you have some theorem involving parallel lines and interior angles?

[restive river]

J know it’s part of the triangle sum theorem

[crisp zealot]

extend the line segment PQ to an actual line

and use the fact that it intersects line n and line m with the same angles (since n and m are parallel)

[restive river]

0.0

[devout snow]

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[devout snow]

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[next nacelle]

Can someone explain to me what I did wrong? The original problem is factoring the polynomial

x^4 - 9x^2

[surreal night]

you didnt write the second term as a square

[safe jasper]

sqrt(9x^2) != 9x

[next nacelle]

Wait so is it
A = x^2
B= 9x^2

I thought that you exclude the exponent. Like in A, I only wrong the x^2 @surreal night

@safe jasper

[surreal night]

no, its x^2 and 3x

[next nacelle]

I meant the b term

In the formula

[surreal night]

$\left(x^2\right)^2=x^4$ and $(3x)^2=9x^2$

[woven radish]

Flappie

[next nacelle]

OHhh I see

[surreal night]

:D

[next nacelle]

I see now

Thank you

[surreal night]

i figured youd see it after i wrote it out

you were close though

only a factor of 3

[next nacelle]

Haha makes sense

Ty

.close

[devout snow]

Channel closed

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[next nacelle]

.reopen

[devout snow]

✅

[next nacelle]

wait

so would the final answer be (x^2 + 3x) (x^2 - 3x)?

@surreal night

because i entered that in, and it said it was wrong 😐

[fierce heath]

you can still factor x

[surreal night]

Indeed it is

[next nacelle]

i see so i would get:

(x) (x+3)(x-3)

[fierce heath]

x^2

one x from x^2+3x

one x from x^2-3x

[next nacelle]

wait what

im factoring out an x^2?

[fierce heath]

x(x+3) * x(x-3)

[next nacelle]

OHH

youre right

[fierce heath]

you couldve gotten straight to that btw

[next nacelle]

but im confused why i would multiple the 2 x's

[fierce heath]

if you factored x^2 from this

x^2(x^2-9)

becomes

x²(x+3)(x-3)

[next nacelle]

in a previous problem where i factored a polynomial, i got

4x^2 + 24 x - 9x - 54
= 4x(x+6) - 9(x+6)
= (4x - 9) (x+6) <-- in this example, i added the 2 things i factored out

so if the problem is

x(x+3)x(x-3)

would it be x^2 bc im multiplying the 2 factors?

[fierce heath]

yea

[next nacelle]

ok thanks and last question

this makes sense, because since the original problem had a highest degree of 4

my final factored form should have 4 x's right

so x^2 is 0 , 0

[fierce heath]

yes

[next nacelle]

x+ 3 = 0 / x = -3

x - 3 = 0 / x = 3

[fierce heath]

seems good to me

[next nacelle]

if i put anything squared = 0

doesnt the answer always have to be + or -?

so if it was 4x^2 instead of x^2

the answer would be +2x, -2x

[fierce heath]

if youre factoring by difference of two squares yes

like 9 - 4x^2

becomes (3+2x)(3-2x)

[next nacelle]

ok thank you

.close

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[devout snow]

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[edgy path]

Honestly i have no idea what to do.

[humble gyro]

Do you know the definition of tan x?

You can replace tan x with sin x / cos x and then add the two terms

[edgy path]

so just use the identity and solve?

Like this?

[urban harbor]

how did sin^2x / cos x become sin^2x * cos x on the 2nd line?

cos suddenly jumped to numerator

[edgy path]

my mistake

what would it be?

[urban harbor]

well it's just sin^2x / cos x, pretty much what you have on the first line, now add up the two terms by coming up with a common denominator

what do you need to multiply the first term by (cos x) to have a common denominator?

[edgy path]

cosx? im not sure

[urban harbor]

multiply by cos x/ cos x yes

[edgy path]

so, then am i left with sin^2x = 1 ?

[urban harbor]

how?

[edgy path]

becuase the cosx cancel?

[urban harbor]

can you show your work youre doing?

[edgy path]

wait, can we rewind

[urban harbor]

$\cos x\times\f{\cos x}{\cos x}+\frac{\sin^2 x}{\cos x}$

[woven radish]

∫oosh (lemonsaurus appreciator)

[urban harbor]

we're multiplying the left term by cos x / cos x to get the same denominator right, with this so far?

so we have the common denominator cos x

[edgy path]

yup

[urban harbor]

so we have

$\frac{\cos^2 x+\sin^2 x}{\cos x}$

[woven radish]

∫oosh (lemonsaurus appreciator)

[urban harbor]

what do you notice now?

[edgy path]

can cancel out cosx?

[urban harbor]

what do you mean cancel out

[edgy path]

like get rid of 1 cosx

[urban harbor]

so cancel out with denominator?

[edgy path]

yes

[urban harbor]

are you saying if you have $\frac{A+B}{A}$ you can cancel out the A's?

[woven radish]

∫oosh (lemonsaurus appreciator)

[edgy path]

yea

[urban harbor]

then you have a fundamental misunderstanding of how fractions work

let's try an example:

$\frac{2+3}{2}=3?$

[woven radish]

∫oosh (lemonsaurus appreciator)

[edgy path]

ok, so what would i do instead?

[urban harbor]

well the most important thing is to go back and work on your fundamental algebra skills, because if you don't fix the holes in your understanding then it's useless to continue to learn new concepts, you will just get tripped up in problems by algebra and you won't have fun doing it.

to keep solving this, we would use the trig identity sin^2 x + cos^x = 1 which makes the numerator become 1 and we simply end up with 1 / cos x which is what we wanted

$\frac{\cos^2 x+\sin^2 x}{\cos x}=\frac{1}{\cos x}$

[woven radish]

∫oosh (lemonsaurus appreciator)

[edgy path]

Wait so multiplying by cosx doesnt work then?

[urban harbor]

at what step?

we DID multiply by cos x / cos x or are you referring to something else?

[edgy path]

at this step can i not just multiply by cos x

[urban harbor]

multiply what by cos x

both sides?

[edgy path]

yes

[urban harbor]

sure, then we just end up with 1 = 1 instead of 1 / cos x = 1 / cos x, kinda same thing and doesn't make a difference

but iirc, don't these trig proofs kind of want you to work with just 1 side of the equality and make it become the 2nd side without messing with the 2nd side?

whether you cancel out the denominators or not doesn't matter that much, the key thing is using the cos^2 + sin^2 = 1 identity to transform the numerator

gotta go, hope that was mildly helpful, gl

[edgy path]

thank you

.close

[urban harbor]

just go to khan academy algebra stuff and review the fundamentals because thats what tripping you up

[devout snow]

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[steel sun]

what does the underlined sentence mean

[steel sun]

what is the floor of a logaritm

[winter patrol]

door you know what the floor function is

they've also made a mistake, that should say
$$2 < \log_{10} 400 < 3$$

[woven radish]

ℝαμΩℕωⅤ

[steel sun]

ye thats why its highlighted

[steel sun]

jo

n

no

[steel sun]

whats a floor func

[winter patrol]

rounds a number down to the nearest integer

floor(2.6) would be 2

[steel sun]

isnt 3 the nearest?

[winter patrol]

round down

[devout snow]

@steel sun Has your question been resolved?

[devout snow]

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[thin geode]

How do i solve 2d

[thin geode]

6^x = 0.25 ^2x-1

I get x lg 6 = (2x-1) lg 0.25 but i dont know how to continue

[tender cobalt]

you treat lg 6 and lg 0.25 as numbers then continue to solve for x

think about if you had x 999 = (2x - 1) 888

you dont need to calculate 2 * 888 to know that (2x - 1) 888 is 2 * 888 * x - 888

that way you can write an answer without needing to "combine" or "solve" that 888 with anything

same thing applies with lg 6 and lg 0.25

(2x - 1) lg 0.25 is 2x lg 0.25 - lg 0.25

you keep doing this until you solve for x

[thin geode]

But the logs have different numbers one has 6 and one 0.25

[tender cobalt]

what does that have to do with what I said

[thin geode]

You said to multiply (2x-1) with lg 0.25. But Ive done that but i dont know how to progress because lg6 is doffent

[tender cobalt]

you know that lg 6 - lg 0.25 is just lg 6 - lg 0.25 right

again you dont need to combine the numbers with anything

you need to know how you usually solve a problem like this

say you have 999x = 2 * 888x - 888

how would you solve for x

[thin geode]

Isnt it lg 6/0.25?

[tender cobalt]

1. you dont have to do that

2. combining numbers is never necessary until the very end where it goes into the calculator

so you can feasibly leave lg 6 - lg 0.25 just like that, you dont gain anything by calculating it to be lg 24

the calculator calculates all of it by the end

[thin geode]

ok so i separated the variable and the constants so its now lg 0.25 = 2x lg 0.25 - x lg 6

What i do next?

[tender cobalt]

again, if you have 888 = 2 * 888x - 999x, how would you solve it

[thin geode]

solve 2 multiplier and divide with a number

[tender cobalt]

what

[thin geode]

solve 2 x 888 -999 and divide with whatever number that makes

[tender cobalt]

you know x is a variable

[thin geode]

i meant multiplication sign

[tender cobalt]

I wrote $888=2\cdot888x-999x$

[woven radish]

mtt

[tender cobalt]

that is what I mean with the * and the x

[thin geode]

But we cant do that here

[tender cobalt]

cant?

[thin geode]

the logs are different

[tender cobalt]

let me repeat

combining numbers is never necessary until the very end where it goes into the calculator

so you can feasibly leave 2 lg 0.25 - lg 6 just like that, you dont gain anything by calculating it

[thin geode]

No no its not just constant its 2x lg 0.25 and xlg 6 where x is variable

[tender cobalt]

that didnt seem to phase you here

what said that you had to calculate 2 * 888 - 999?

theres xes in there

[thin geode]

hey just give me the next step

[tender cobalt]

you told me the next step

I am in the process of convincing you to do the same step to the logs

you told me its not possible

Im telling you to write it down

write it down privseb

you wrote down 2 * 888x - 999x to be (2 * 888 - 999)x

now how can you write down (2 lg 0.25) x - (lg 6) x?

[thin geode]

i dont understand what you are meaning

x ( 2lg 0.25 -lg 6)

[tender cobalt]

there you go

now as you said, what is the next step?

as a hint, you already said this step

[thin geode]

Divide the logs with the lhs

[tender cobalt]

write it out

no calculating

WRITE it out

[thin geode]

Hey dont be rude about it

[tender cobalt]

I am giving you the same amount of repetition as I do with anyone else

its just a reminder, its not personal

no calculating, write it out

[thin geode]

then id recommend you to stop with that, because that attitude is not nice when approaching someone who just doesnt understand

Ya ill write it out

Lg 0.25 /( 2lg0.25 -lg 6)

[tender cobalt]

youre forgetting parentheses, can you see where

yep

so x = lg 0.25 / (2 lg 0.25 - lg 6)

the calculator from here can calculate x

[tender cobalt]

with that out of the way, can you elaborate on this
I wasnt deliberating being rude

[thin geode]

maybe you wasnt, but it sure did come off like that

[tender cobalt]

no no we're being productive by considering this

you see if I am, I find out something I did wrong

its not usual to see these kinds of things until after its all over, so its hard to see in the moment

[thin geode]

ya math is not easy, thats why you need patience. it doesnt help by starting to use allcaps. just be more chill

[tender cobalt]

thats not what caps means

you ever read the EULA

they have these important sections that are in all caps

theyre not meant to say the agreement is yelling at you

they mean this part of the agreement is important and you need to pay special attention to it

[tender cobalt]

same thing here, you seeing the caps as yelling is a misinterpretation

but then there needs to be a less ambiguous way to see it if thats the case

[thin geode]

why didnt u just write it normally?

[tender cobalt]

because generally if you dont get the first "write it out", repeating it again isnt going to help

the "write" part is more important than the others

so somehow that word needs special emphasis

italics wont work, its condescending
bolding will be yelling
underlining wont work, its too flashy but its second-best

capitals is the lightest version

thats what I was thinking at the time

[thin geode]

whetever it was personal or not, i thought it was a bit rude. Just see it as a was to better approach someone. im not going to argue anymore.

[tender cobalt]

youre leaving without giving me an alternative? (meant to mean without ultimately leaving an alternative that works for the both of us)

also me explaining what I did and you asking good questions isnt an argument

we arent disagreeing on anything

[thin geode]

rn i dont even know what you are talking about. just move on, if you wanna take my advice or not thats up to you, im not trying to persuade you to do what i think is good.

[tender cobalt]

you didnt know what was happening the whole time??

you didnt tell me until now, damn

you said what you said like you did, nvm Ill close this
sorry about that

[thin geode]

yk now your really being rude, but whatever

[tender cobalt]

I still dont get how

[thin geode]

just have a good day man, thanks for the help. take it how you want.

👍

[tender cobalt]

youre not having a good day though

.close good luck on the rest of it regardless

[devout snow]

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[primal ferry]

when taking the first differential of both sides, why does d(s^2)/dt become 2s ds/dt

[safe jasper]

by the chain rule, since s is a function of t, then differentiating it requires also multiplying by its derivative wrt t

d/dt ((s(t))^2)

[primal ferry]

see i know chain rule as du/dv = du/dt * dt/dv

so here it would be uhhh

wait

lemme do this myself

ds^2/dt = ds^2/ds * ds/dt?

[safe jasper]

yes

[primal ferry]

and ds^2/ds is 2s

alright gotcha thanks

.close

[devout snow]

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[primal ferry]

.reopen

[devout snow]

✅

[primal ferry]

uhh i got stuck 😭

chain rule again?

wait no just quotient rule?

v is s and u is at+b

du/dt is at+b

ds/dt is v(elocity) which we have as at+b/s

now what

the a on the lhs is acceleration

[primal ferry]

this is the question

im so confused

[urban harbor]

first line correct, can you explain how you progressed from there?

[primal ferry]

divided bs by 2 and made v the subject

then differentiated both sides to get an equation for acceleration

[urban harbor]

yes sounds good, a bit hard for me to read some of your writing

so you have $sv = at+b$

[woven radish]

∫oosh (lemonsaurus appreciator)

[primal ferry]

fair enough

[primal ferry]

yep

[urban harbor]

i actually would recommend against dividing by s now, just differentiate both sides implicitly again and use product rule on the left since both s and v are functions of t, should be less fugly than messing around with fractions and quotient rule

[primal ferry]

hmm okay wait let me try that

[urban harbor]

so you'd have $s'v + v's = \f{d}{dt}(at+b)$

[woven radish]

∫oosh (lemonsaurus appreciator)

[urban harbor]

s' = v and v' = a

[primal ferry]

ok but where do i go from here. s' is v so we get v^2 and as

yes

but this a is not the constant a on the rhs

v^2 + As = a

A = (a-v^2)/s

[urban harbor]

hmm

since the answer key seems to want a relative to s, maybe now go back to the expressionw we got for v and plug that in for v

[primal ferry]

the question is asking for A's proportionality to s if im comprehending this right

[urban harbor]

yep, exactly

so we had sv = at+b, guess we can write v as you did originally v = (at+b)/s and plug that in for v maybe?

[primal ferry]

we had v = (at+b)/s

[urban harbor]

yes, exactly

[primal ferry]

lemme go do that

[urban harbor]

see what that simplifies to

[primal ferry]

wait i forgot to reverse the signs

[urban harbor]

that doesn't seem quite right

how did a become sa

[primal ferry]

i forgot to put s^2

[urban harbor]

but where did the s^2 come from

if you multiply top and bottom by s^2, shouldnt you now have a s^3 in the denom

[primal ferry]

wait what

let me do all steps

[urban harbor]

$\frac{a-\frac{a^2t^2+2abt+b^2}{s^2}}{s}$

[woven radish]

∫oosh (lemonsaurus appreciator)

[urban harbor]

can multiply the whole big fraction by s^2 / s^2 (thats one wya to look at it it anyway)

[primal ferry]

youre right

so it would be option a

a is directly prop to s^-3

[urban harbor]

hm but theres also as^2 in the numerator

[primal ferry]

that's what's throwing me off

[urban harbor]

the work seems correct as far as i can tell, unless im missing something, but none of the answers seem to match up really

maybe someone else can verify, sorry i can't help further

[primal ferry]

the problem is there's no answer key

😭

thanks for the help tho

[urban harbor]

i tried checking work with chatgpt just to see if its result would differ and it gets same thing (disclaimer i dont recommend using it except as a means to check something youve already worked out yourself)

[primal ferry]

what if it has something to do with those acceleration equations

im still in hs i havent done that chapter of kinematics

[urban harbor]

im inclined to think the problem might just be badly written at this point

i'd ask teacher for clarification

[primal ferry]

it's from a sample exam paper

oh well i'll leave this thread open for a few hours

in case someone figures it out

[devout snow]

@primal ferry Has your question been resolved?

[urban harbor]

@primal ferry ah, so you can use s^2 = at^2+2bt+c and plug that in for the s^2 in the numerator

also btw there's a negative typo in your last image and just combining the exponents in denominator

so you'll get: $\frac{(at^2+2bt+c)a-a^2t^2-2abt-b^2}{s^3}$ then you can cancel some stuff

[woven radish]

∫oosh (lemonsaurus appreciator)

[urban harbor]

and youre left in numerator with just a constant so now it works 🙂

[primal ferry]

.close

[devout snow]

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[primal ferry]

tysm

[urban harbor]

.close

[devout snow]

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[lost crag]

given

how do i prove this?

[stone stump]

.

[lost crag]

ty

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[subtle elk]

Please, help me with the proof of this statement:

[devout snow]

@subtle elk Has your question been resolved?

[gritty terrace]

Ooh I didn’t know this result was true

Seems pretty cool

[devout snow]

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[subtle elk]

.reopen

[devout snow]

✅

[subtle elk]

<@&286206848099549185>

[devout snow]

@subtle elk Has your question been resolved?

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[short hare]

Ill give an example first and say where im confused

[short hare]

Y=x^5

Y’=5x^4

Y’’=20x^3

Y’’’=60x^2

The question states

[short hare]

If x^2 was x^n-3

What is the lowest possible value for n that makes d^3y/dx^3 not zero

The answer is 3

Okay

But

Wont 2 make it x^-1

Which is not zero.

So wth?

[frozen stump]

uhh

idk why they gve you like x^2 or anything

wait let me think what to say first

[short hare]

Take your time

Ill give the original question to avoid confusion

And i think i figured it out

[frozen stump]

ok

[short hare]

Y=x^n

N is positive

X is not zero

[short hare]

.

[frozen stump]

wait so the x^2 isnt or is real

[short hare]

Its an example when n=5

[frozen stump]

ah ok

[short hare]

Yeah

[frozen stump]

you have this

[short hare]

Y’’’=n(n-1)(n-2)y^n-3

It cant be 2

[frozen stump]

yea i just ofcused on the pwers for this

let me write it out

[frozen stump]

this part of the q or you just saying

[short hare]

Im just saying

[frozen stump]

im pre sure its asking for just the 3rd derivative to equal 0 right

[short hare]

Trying to fully conprehend it

Comp

[frozen stump]

if n is 1 or 2 then other derivatives would be 0

wait

lmao

i fucked up

[short hare]

Its asking what values for n that’d make it anything but zero

Yeah i mean

[frozen stump]

sorry this

[short hare]

Its correct

[short hare]

Mhm

Lowest possible value is indeed 2

3*

Bc 2 and 1 would make n negative

And anything above 3 is bigger than 3

And thus the answer is 3

Ty sho

[frozen stump]

1 and 2 would make it 0

not negative

cause the bracket thing stays forever

[short hare]

Yeah i was talking aby the example 🥲

[frozen stump]

fucking weird lmao

[frozen stump]

ah soz lmao

[short hare]

Yes okay

[frozen stump]

yo derivative man

come check my channel pls

[short hare]

Who

[frozen stump]

real easy q i highkey just forgot

[frozen stump]

you

[short hare]

Oh ok

[devout snow]

@short hare Has your question been resolved?

[short hare]

I didnt close this?

[devout snow]

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[surreal night]

you should just give it some time :)

[devout snow]

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[ancient valve]

I need help understand eigenvalues and eigenvectors. How come there is X2 = s "is a free variable", this doesnt make sense to me. Also, without the echelon matrix aftr the subtraction of eigenvalues from A, i can still get x1 = -1/3X2 so is it really necessary. What about normalisation of X1 ^2 + X2^2 =1 ?

[devout snow]

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[ancient valve]

close

.close

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[devout snow]

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[solid osprey]

find $y-x$ if x and y are real numbers and
$$7^x-16y=0$$
$$4^x-49y=0$$

[woven radish]

Skill_Issue

[solid osprey]

idk where to start

[winter torrent]

solve for y

[solid osprey]

how

[arctic field]

solve for x

[solid osprey]

what

😭

wair

[arctic field]

,, \f {7^x} {4^x} = \f {16} {49}

[solid osprey]

$$7^{x+2}-16y\cdot49=0$$
$$4^{x+2}-16\cdot49y=0$$
$$7^{x+2}-4^{x+2}=0$$

[woven radish]

Skill_Issue

[arctic field]

what did you even do

[arctic field]

just work from here

[mild sorrel]

7 = 4?

[devout snow]

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[opaque haven]

hey so i have a problem. given that z=e^(2πi/7), i need to find 1+z+z^2+...+z^5 and 1+zbar+zbar^2+...+zbar^6...anyone have any ideas?

[opaque haven]

as zbar i mean the conjugate of z

i assume theres some neat way to solve this but i just cant see it

[wicked turtle]

these are geometric partial sums, there's a formula for that

[opaque haven]

ah i could use the formula for complex numbers too huh?

[wicked turtle]

yea it works for any complex number except 1

[opaque haven]

okay thank you, ill try again and if im stumped again ill come back

ill leave the channel open just in case

the common ratio is one though

doesnt that leave a denominator of 0?

[safe jasper]

the common ratio isn't 1

1 + z + z^2 + ...

what's each term being multiplied by to get to the next?

[opaque haven]

right, z

im stupid

thanks

.close

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[vernal flame]

Is this the right way to work this question out?

[royal pivot]

What do you need to find?

@vella?

@vernal flame

[vernal flame]

how high is the tower

[devout snow]

@vernal flame Has your question been resolved?

[gentle granite]

The process seems to be correct, but when you input the values of the sides in the final Pythagoras theorem, FB= h/tan12 and similarly FA=h/tan10

That will cause an error to be made

Once you find the height, you can find FA and FB which will help you find the direction of motion of the car

[devout snow]

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[solid osprey]

ABCD is a square with sidelengths 4, D is rotated 60° clockwise with the centerpoint A, where the new D is called D', A and B and C are mirrored from D' to create A' B' C' respectively, find the area of A'B'C'D'

[solid osprey]

they want it in some sort of a+bsqrt3 form

im pretty sure A'B'C'D' can be made into A'B'D' and B'C'D' triangles, where A'B'D'=ABD' and B'C'D'=BCD'

BAD'=30 deg

ABD'=AD'B=75

using some trig you can get the area of it

idk how to get the B'C'D'

[devout snow]

@solid osprey Has your question been resolved?

[dire forge]

yup exactly

to get BCD', use the properties of angles formed by isoceles triangles to figure out all the remaining angles in this diagram and then you should be able to calculate BCD' given one side and two angles

(DD' is length 4 bc ADD' is an equilateral triangle)

[solid osprey]

oh is BCD'=CBD'=15?

ok ty

.close

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[dire forge]

np!

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[zealous pasture]

when dividing a rectangular into 4 triangles, are the area of triangles equal or not?

[zealous pasture]

uhm

[dire forge]

The top and bottom triangles are congruent, and the left and right triangles are congruent. But all of them don't have to be equal.

Wait actually

[zealous pasture]

yea i know, but im asking if the areas of all 4 triangles are the same or not

[dire forge]

Ignore what I just said, they are all equal. If you name the height of the rectangle h and the width of the rectangle w

Then the top triangle has area 1/2 (w) (h/2), and the left triangle has area 1/2 (h) (w/2)

Which are equal

So they are all equal, my bad.

[zealous pasture]

alright thanks!!

.close

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[dire forge]

you're welcome!

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[restive river]

@devout snow help

[restive river]

@restive river

@devout snow help

?

[spring salmon]

what u need help with

[restive river]

How do I create channel

[spring salmon]

u have this channel

this is your channel

[restive river]

So that I can ask help for my math question

Oh

I need help with these 2

😅

If sin(A + B) = 1 and 2sin(A - B) = 1 where 0 < A B < π/2 then what is (tan A) / (tan B) equal to?

[devout snow]

@restive river Has your question been resolved?

[manic condor]

!status

[devout snow]

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

[restive river]

1

[restive river]

1

[manic condor]

Okay.

Well, do you know where sine takes the value of 1?

[restive river]

Yes

Sin 90

[manic condor]

Yes.

Which is same as sin(pi/2) .

Now, if both A and B are less than pi/2, can you find an upper bound for A+B?

[restive river]

If Sin ( A + B ) = Sin π/2 , A+ B = π/2

Isit?

Sin sin get cancel, right?

[manic condor]

Yes. Only because they have given that A and B lie between 0 and Pi/2.

It won't happen always. It's important that you understand that.

[restive river]

😯

[manic condor]

Otherwise, Sin(5Pi/2) is also 1.

[restive river]

How would have you solved this if in the question it was given B lies between 0 and pi/3?

[manic condor]

But since sum of A and B can't reach even Pi, all other options than sum being pi/2 are ruled out.

[manic condor]

Nothing about the A?
Anyway, most probably there is a way to manipulate expresssions directly into what you need but in our case, it's simpler this way.

[restive river]

You get that by doing this right?
Sin ( 4π + π / 2 ) = sin ( 2π + π )

But sin 2π is 0?

[manic condor]

Umm... No. 2π + π/2.

[restive river]

🤔

[manic condor]

Sin(5π/2) = sin( 2π + π/2) = sin(π/2)

[restive river]

Oh, you seperate the 2

[manic condor]

You have made a mistake while adding. 4π + π/2 = (9π)/2

[restive river]

Sorry I'm rlly not good in maths 😅

[manic condor]

Well, as long as you actually try. If you don't understand any part, ask me to elaborate.

[restive river]

Ty, let me try solving it now , if I need help I'll let you know

Tysm ❤️

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[tender cobalt]

are you stuck solving it

[restive river]

Yes 🥲

[tender cobalt]

can you show your work so far

[restive river]

Oki I sec

I got a + b = π/2

Couldn't solve after that

[tender cobalt]

since sin(pi/2) = 1,

we say that a + b = that pi/2

now remember that we got this from sin(a + b) = 1

lets go over to 2 sin(a - b) = 1

ok?

[restive river]

Ok

[tender cobalt]

this would mean sin(a - b) = 1/2

[restive river]

How 1/2

Ohhh 2 went that side

Got it

[tender cobalt]

yes

[primal ferry]

👍

[tender cobalt]

now so you keep track of whats happening,

what angles have a sine of 1/2?

sin(what angles) = 1/2?

[restive river]

30

[primal ferry]

angleS

[tender cobalt]

are there any others?

[restive river]

Idk

[primal ferry]

do you know what quadrants are

[restive river]

Sin 30 = 1/2

[tender cobalt]

wait nvm

[restive river]

Yes

[tender cobalt]

it says 0 < A, B < pi/2

so both A and B are forced to be acute

[primal ferry]

oh wait yeah

[tender cobalt]

sin(30) = 1/2 is the only one youll need to consider for now

now in degrees, we know 0 < A < 90 and 0 < B < 90

[restive river]

😯

[tender cobalt]

we also know sin(A - B) = 1/2

[tender cobalt]

in the problem, it says $0<A,B<\frac\pi2$

[woven radish]

mtt

[tender cobalt]

now do you know what pi/2 radians is in degrees?

[restive river]

Yes

[restive river]

90°

[tender cobalt]

yep

so $0<A,B<90$

[woven radish]

mtt

[tender cobalt]

this is short for $0<A<90$ and $0<B<90$

[woven radish]

mtt

[restive river]

And a and b both are small angles than 90

[tender cobalt]

yes