help-27

[granite anchor]

no its not log9 18 its just log 18

[lament kraken]

is just log 18

[granite anchor]

yeah

if there is no number on the log its assumed to be 10 and u dont need to put it

[stray spade]

also not log9 2 just log2

[granite anchor]

yeah

[lament kraken]

power rule: says log_a(b^c) = c log_a(b)

[granite anchor]

yeah

[languid iris]

the expand part

its lowk confusing

[granite anchor]

so its like

imagine u have for example

log(16)

u can make it

log (4^2)

no even better

log(2^4)

and u move the exponent outside

4log2

[lament kraken]

go to definiton of log and its intutive

[granite anchor]

so whatever power u have just take it outside to multiply it

[languid iris]

ahh

alright

[solar goblet]

this very depends on what subject you are doing

mathematicians assume log as base 10, physicists assume log as base e, computer scientists assume log as base 2

[granite anchor]

wouldnt log base e just be written as ln

[stray spade]

wolfram alpha writes log base e as log

[granite anchor]

oh wow i wasnt even aware of that

ok on the bottom that's what it tells him so he prolly dont gotta worry about those yet

[languid iris]

What yg thinking

[stray spade]

bring down the exponent

the exponent x - 1

[languid iris]

so x-1log2 6

[stray spade]

yeah

but use parenthesis to not make it ambiguous

[languid iris]

(x-1)log2 6

thx bro appreciate it

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[wheat basin]

whenever i try to solve this,

i get like the biggest denominators

i got x/15/4 = 18/5

[solid hull]

Do you have any examples of how you would solve it?

[wheat basin]

can u just see if thats correct or not

[solid hull]

Alright

[solid hull]

That's what I got off the top of my head

[wheat basin]

alr alr ty

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[devout snow]

Please don't occupy multiple help channels.

[gleaming nexus]

.close

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[calm patio]

can someone quickly explain

why the bounds are -4 and 4

[restive river]

10-6

[calm patio]

but if the radius on the left side and right is 5

and the radius of the drill bit on the left side and right is 3

why not 5-3

[restive river]

gimme a sec

[devout snow]

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[restive river]

https://www.youtube.com/watch?v=MwACWGGDgNo

watch this video

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[elder goblet]

given that y=x-1 the straight line and curve equation is (y-1)^2=4(x+1) which is like Y^2=4aX

here I need to find the area of this shaded area by integration

so I am looking for some ideas which makes integration easier

[rare merlin]

you can integrate in respect to y

[long pasture]

you mean like shifting the whole graph down by 1 unit and right by one unit?

[lean nova]

It’s probably ideal to integrate with respect to x first here as a function of y and set your outer bounds as constants (y) so dxdy

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✅

[elder goblet]

Someone wrote such formula as a shortest way

I never saw it

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[grand river]

I'm trying to derive the solution for an underdamped second order differential equation

[grand river]

But I end up with a complex sine term at the end, while in the book it isn't complex

I started from this. But instead of $\alpha$ it's $-\alpha$, and $\beta$ is $w_d$\

[woven radish]

johnseymour20

[grand river]

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[long pasture]

hmmm, I have worked a bit on this question, it seems minus one sqrt term on both sides will work, further simplify to something nice like in a conic section

[dense lynx]

geometric interpretation

this equation describes ellipse with foci (0, -1) and (1, 0)

you can intuitively justify x^2 + y^2 is maximized along the ellipse minor axis, and problem is not hard to solve from here

(side note: this question could also be framed as |z + i| + |z - 1| = 8 to find max |z|)

[devout snow]

@urban otter Has your question been resolved?

[urban otter]

Thwt was the actual question 💀

I just framed it into coordinate geometry

[dense lynx]

yeah I guessed

anyway still just think about it geometrically,
basically all complex number questions that look like that should be done this way

[urban otter]

ahhh right it represents an ellipse

I suck at conic sections lol

Thanks a lot :prayge:

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[subtle shell]

i know i have to use points in hte graph and ha and properties to find an equation but not sure how

[restive river]

Lagrange interpolation

[subtle shell]

how do i do that?

[restive river]

https://youtu.be/bzp_q7NDdd4?si=0T7UKXDmrXkwC_jv

[subtle shell]

i dont think we were taught this in class (at least not yet), but it makes sense to use this to find equations. during class my teacher said to use the formula y=ab^(c(x-h))+k

and to find points on the function and ha to find possible values for a, b, c, h and k

I found the points (-4, -3), (-5,-1) and ha: y= -4

[subtle shell]

heres an example of what my teacher showed us

can i ping helpers?

ill wait like 10 minutes first

<@&286206848099549185>

i means it 5AM so makes sense they arent as many helpers avaible now , thank you edmund for helping me

.close

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[subtle shell]

.reopen

[devout snow]

✅

[subtle shell]

thx

im still stuck 💀

i dont really understand lagrange interpolation

all i know if that for y=ab^(c(x-h))+k, the value for b will be 2 since it goes down 2 units from each point and the value for k will be -4 since the ha is y=-4

im lost on the rest

<@&286206848099549185>

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[untold pivot]

Is anyone familiar with what $\rho(D) \equiv 0 \pmod u$ could mean here?

[woven radish]

somethingwrong

[sand dove]

Rho(D) is a multiple of u

[untold pivot]

but isn't u an integer while \rho(D) is the image of D

[sand dove]

In cardinal probably

In cardinal

[untold pivot]

that means the number of elements in the set?

[sand dove]

Yes

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[wary ginkgo]

J'ai du mal à déduire la question b

[coral violet]

english?

[wary ginkgo]

Wiat
I will translate the questions for you

[coral violet]

alright!

[wary ginkgo]

1 Show that Vx€ [1, +inf] ; rac(x-1)/x <= 1/2

2 (a) Show that: (V(a;b) R'xR'); a+b=0 <=> a=0 and b=0.

2 (b)Deduce that: (V(x;y)€ R²): rac(x²+1) + rac(x²+1) = 2 <=> x =0 and y = 0

[wary ginkgo]

I didn't know how to answer 2(b)

[coral violet]

can you give me 5 mins?

alright..

so listen..

to do this, you need to consider the minimum value of rac(x^2 + 1)

Its 1 right?

and the minimum lf rac(y^2 + 1)

its 1 again

notice how 2 is the sum of their minimum values?

if we take other values for two terms and add it, then the sum is always greater than 2

so, that means they MUST be their minimum

And their minimum occurs at only 1 value for both x and y

and it is (0,0)

if you need more info about how it occurs at only one value, you can ask me out!

[devout snow]

@wary ginkgo Has your question been resolved?

[wary ginkgo]

Yes please

[coral violet]

alright

so listen

[wary ginkgo]

♥️

[coral violet]

the root fuction is a strictly increasing fuction right?

[wary ginkgo]

Yes

Always positive

[coral violet]

mhm…

so if x^2 is varying from 0 to infinite, the value of the overall function keeps on increasing…

[wary ginkgo]

Yes

[coral violet]

so we can agree on its minimum occuring at zero?

[wary ginkgo]

Yes

[coral violet]

and if you put zero, the value of the function is 1

[wary ginkgo]

[coral violet]

So the minimum of the function is 1 right?

[wary ginkgo]

Yes

Ok I get it now

I have another question.

1 Show that (Vn€N); n(n+1) is even.

2 Show that the number A=n(n²+5) is even for any natural integer n.

@coral violet

[coral violet]

yes…

[wary ginkgo]

[coral violet]

alright… can we get into dm?

[wary ginkgo]

Yes

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[potent kelp]

tex: \left(\frac{{14}}{{121}}\right)^{-3}\cdot\left(\frac{{11}^{{1}}}{{22}^2}\right)^{{2}}\cdot\frac{{7}^{4}}{{11}^3}

question is "Present the expression in the most reduced form of multiplying primes by powers" and i dont understand it can anyone explain it to me please?

[dapper fable]

we can factorize each term into a product of primes, for example 121 = 11x11

[potent kelp]

i dont really understand the question it self

whats multiplying primes by powers

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@potent kelp Has your question been resolved?

[cyan surge]

14 = 2x7
121 = 11x11= 11^2
like so..

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[carmine cedar]

The last question

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@carmine cedar Has your question been resolved?

[glossy dew]

the match type?

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[raven leaf]

Can someone walk me through part b please

[devout snow]

@raven leaf Has your question been resolved?

[twilit comet]

i mean just do it for each side individually?

not too hard

[raven leaf]

should this not be zero

[twilit comet]

what is [-eps, eps]^2 supposed to mean

don't use that notation

[faint gorge]

it's the square region of the cube

[twilit comet]

no ik

i'm just saying it's not very good notation

ik vector calculus man 😭

[raven leaf]

ok why are you nitpicking on notation which is completely clear here rather than the actual problem man

[twilit comet]

okay be civil

i'm only trying to help

[raven leaf]

I don think this was very civil but I ignored it.

[twilit comet]

pointing out something which might confuse other people isn't wrong

[twilit comet]

care to elaborate?

[raven leaf]

Your tone suggests "this is easy, why is this even a question"

I don't think that's very polite, do you?

[twilit comet]

i apologize if that's what you thought i meant

and no, it certainly is a common question

[faint gorge]

I am not completely sure, but maybe you are supposed to do f(x) = f(n)

[twilit comet]

but i still don't think this was an appropriate response

[twilit comet]

no, you compute it for each individual side and then add them up

[twilit comet]

and yes, it should be zero

[faint gorge]

It should be (-1,0,0)

[raven leaf]

i get this by cartesian formula

[arctic field]

yeah if you do it like this you will just get 0

which isnt the answer you're looking for

the key thing to note is that z is not just z

z is the z-coordinate of the point at which you're integrating

[twilit comet]

so you need to actually consider it in the second and third integrals

it's a constant in the first (although different for both heights)

[raven leaf]

and this is incorrect because I'm treating z constant?

[twilit comet]

yes

fyi split the top integral into 2

[raven leaf]

then why does cancelling the cross products make it incorrect?

[twilit comet]

one for z = eps, the other for z = -eps

[arctic field]

its incorrect because you didnt plug in the value of z for each side

[raven leaf]

aaahhhh

[arctic field]

z at the point of writing down your integral is no longer z

[raven leaf]

understood

so epsilon i think

[arctic field]

there exists a value of z to each side here

[raven leaf]

then -eps

which combines them rather than cancelling

[arctic field]

wait thats the wrong integrand

[raven leaf]

don't worry

[raven leaf]

i gotcha

[arctic field]

this is the one

[raven leaf]

ya

understood

[arctic field]

i never worry

[raven leaf]

yabois a potato

[raven leaf]

cold

[arctic field]

tis in the name

[raven leaf]

awesome, thank you

[faint gorge]

So if you did a triple integral there wouldn't be this issue in the first place?

[arctic field]

you know discrimson?

[arctic field]

well you'd be integrating z inside of the integral

so you have to deal with it there

[raven leaf]

.close

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[hard musk]

wow

[unborn cedar]

how do i do b

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[restive river]

hello

[restive river]

.close

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[bronze bay]

Hello I am trying to solve this angle between 2 planes question and I am not really able to find it

[bronze bay]

I've located the two normal vectors of the planes used the dot product rule and found the cosine

my answer rounded to 1 d.p was 57.8

[north roost]

Can u show ur full calculation

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[granite verge]

i did 3/4 * 9/10 but i dont seem certain on this answer

could someone explain how i might go about this question using the "P(X)" please

[tall knoll]

what's the "P(X)"

[granite verge]

the probability

[tall knoll]

alright let's first simplify some notation

let's let C be the event that the machine is set up Correctly, and R be the event that the parts are of the Required precision

can you translate the info given in the problem using appropriate probability notation?

[granite verge]

C= 3/4
R = 9/10 * 3/10?

[tall knoll]

i want probability notation, not just numbers

P(C) = 3/4 is fine

[granite verge]

P(C) = 3/4
P(R) = 27/40

[tall knoll]

your second equation is incorrect

where is that coming from?

[granite verge]

90% if C happens
30% if C doesnt ?

[tall knoll]

you should be able to construct 3 equations from the given info

P(C) = 3/4 is one of them

[granite verge]

Oh

P(R) = 9/10
P(R) = 3/10

[tall knoll]

P(R) can't be both 9/10 and 3/10

look at the sentences once more

If the machine is set up correctly, 90% of the parts are of the required precision

alternatively, 90% of the parts are of the required precision given the machine is set up correctly

[granite verge]

so if C then 90% of parts are of the required precision

[tall knoll]

and how do you handle "if C" or "given C" with probability notation?

[granite verge]

that line

P(C|R)

i think

[tall knoll]

that's the probability that the machine is set up correctly given the parts are of the required precision

[granite verge]

the only other thing i can think of would be
P(C'R)

apart from this i am lost, i only ever learnt probability using tree diagrams. teacher taught a bit about probability notation but only to write something like "P(X)=1/4"

[tall knoll]

perhaps it would be best if you expanded on what you know about "the line"

what goes on either side of this line

[granite verge]

ok, if possible could you suggest me a video that could teach me the fundamentals of probability notation and how to use them?

[tall knoll]

https://www.khanacademy.org/math/statistics-probability/probability-library/conditional-probability-independence/v/calculating-conditional-probability

[granite verge]

thanks!

.close

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[restive river]

i suck at math. like im really bad. i i need urgent help…i have to hand my assignments tomorrow. and its just too much. The 193 km long country road from A to B was straightened and improved into a 183 km long new road. By how many percent did the driving time from A to B decrease when the speed increased by 10 km/h and it was 80 km/h before the increase

[lusty sapphire]

Break it down into pieces.

1. Overall, the problem is asking you to find the percent change in driving time.
2. Use what you know about percent changes. You're going to need to know the original driving time and the new driving time.
3. Consider how you can get driving time. You need to know the distance traveled, and at what speed.
4. You know the distances, so you need to know the speeds.
5. You know the original speed.
6. Find the new speed.

Then work backwards to get to your answer.

[restive river]

so am i like supposed to figure out the original ones driving time percent? or whatever its called

[lusty sapphire]

You're supposed to find the difference between new and old triving times, and express that difference as a percent

[restive river]

its the distance and

the increase of speeding limit

so 10km and 10km/h

but how do i make it percent?

[lusty sapphire]

not quite, no

[restive river]

…oh

[lusty sapphire]

you're not trying to find the difference between 10km and 10km/h. Fundamentally, that makes no sense. We're trying to find the difference in travel time. Like, for example, if originally the drive was 2 hours, but now its only 1 hour. You need to find the percent difference.

[restive river]

ohhhh

so

first i have to figure the time on the original one

and then the new one

okay okay

hold on

the first one is 2h and 41 minutes

the second one is 2h

how do i turn the difference to percent

[lusty sapphire]

Close but a little off

[restive river]

…

[lusty sapphire]

2.41 hours does not equal 2 hours and 41 minutes

[restive river]

oh…

true…

[lusty sapphire]

This is slightly off too, but within significant figures, so I won't complain

[restive river]

is this close enough tho…

[lusty sapphire]

no

[restive river]

…

okay so

2.41

uhm

okay i do not get it

so 193km / 80

and its 80/h

2.41

[lusty sapphire]

$%$ change $=100%\times\frac{\text{new}-\text{original}}{\text{original}}$

[woven radish]

SWR

[restive river]

oh

so how is it not 2h and 41 minutes

is it because hour has 60 minutes?

whys math so hard…

[lusty sapphire]

The same reason 2.60 hours would not be 2 hours and 60 minutes

[restive river]

how do i turn 2.41 to hours

[lusty sapphire]

It's demands a lot. But its utility is beyond compare.

[lusty sapphire]

You don't need to. 2.41 hours is perfectly fine.

[restive river]

oh

[lusty sapphire]

anyway I gtg. You should be good from here.

[restive river]

thank you!

A group of 6 people completed their assignment in three weeks while working 40 hours a week. How many people would be needed for the job if the contract is done in two weeks and only 36 hours are worked per week.

<@&286206848099549185>

🥲

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[frank crag]

im trying to solve for the derivative, so far im here

[frank crag]

now here

and then i got stuck

i need to somehow reach the answer of 1/(x^2+1) ^ 3/2

[upper schooner]

Actually, retracting the , you're forgetting a - sign for the second term here

[frank crag]

oh right

[upper schooner]

Cause you should have brought down -1/2

[upper schooner]

(it just makes you get a - sign for the second term here though, so you can have that one )

[frank crag]

confused

[upper schooner]

About why you bring down a -1/2?

[frank crag]

what do you mean by "bring down"

like move the -1/2 into the denominator?

[upper schooner]

As in, you know the whole "x^n differentiates to nx^{n - 1}" rule

[frank crag]

yes

[upper schooner]

I'm saying that here (taking into account chain rule of course) that your "n" is -1/2, so you're multiplying by -1/2 (rather than +1/2)

[frank crag]

ohhh

you're talking about the negative

?

[upper schooner]

Yep

[frank crag]

okay, so if i were to add that

it would be

[upper schooner]

Yep, what I meant

[frank crag]

how to go to

as a final answer

[upper schooner]

Well, you're aware that $(x^2 + 1)^{-3/2}$ is the same thing as $\frac1{(x^2 + 1)^{3/2}}$, right?

[woven radish]

@upper schooner

[frank crag]

yes

i just dont know how to get rid of the top part

and make it a 1

[upper schooner]

Well, how about we rewrite what we have a little bit first
[
\frac1{(x^2 + 1)^{1/2}} - \frac{x^2}{(x^2 + 1)^{3/2}}
]

[woven radish]

@upper schooner

[frank crag]

ok i get that

now get it under a common denominator?

[upper schooner]

Yep

[frank crag]

would factoring work

like factor out the

[upper schooner]

Well, I mean, you may want to factor something else instead to make your life nicer

[frank crag]

like this?

[upper schooner]

Wait hang on a second-

My eyes

The (x^2 + 1) should not be squared (because -1/2 + 3/2 is 1)

[frank crag]

oh right

wait

im confused on factoring

so its -1/2 - (-3/2)

anyways

so simplifying is 1/ (x^2+1)^3/2

[upper schooner]

Yep, that's pretty much it

[upper schooner]

(are you still confused, or is it alright now?)

[frank crag]

oh yeah

why factor out the

and not the other part

its probably a dumb question

[upper schooner]

As that means that you'd be left with things that only have positive powers after you've factored, so they're a bit easier to deal with (and if you have some foresight, you'd see that you'd instantly get the (x^2 + 1) out, and that the -x^2 would cancel for you)

[upper schooner]

It's relatively alright to deal with this, but a bit more pain to deal with $\frac1{(x^2 + 1)^{1/2}} \qty( 1 - \frac{x^2}{x^2 + 1} )$

[woven radish]

@upper schooner

[upper schooner]

(well, not really that much more pain, to be fair, it's not the worst to deal with(!))

[frank crag]

oh i see

thanks for your help

👍

!

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[lost cargo]

https://gyazo.com/a4f06e95180833c48dcf18ddd10976f5 what would you consider for a?, ik its l(x) - f(a) + f'(a)(x-a), but how do u find a?

[lost cargo]

and my approximation the way i did it was -2/3 which is no where near the real value

unless im just overthinking this

<@&286206848099549185>

<@&286206848099549185>

[polar ridge]

ok so

i imagine maybe 1/e

so the anser is -1

[lost cargo]

yes how did u get that approximation tho

i know the answer since i can just put it in a calculator

like what should i use for a

i used 1 for a

but idk if thats right

[polar ridge]

well... i jsut thought like an engineer

π = 3 = e

[devout snow]

@lost cargo Has your question been resolved?

[devout snow]

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[storm solar]

i need help

[storm solar]

im doing simultaneous linear equations

my qquestion is

[median pilot]

I assume we're trying to find what values of x this holds for. What have you tried so far?

[storm solar]

i solved it but i just wanted to check if the answer is correct

my answer for this was

[neon wadi]

yes

[median pilot]

^

let me double check. also want to simplify the fraction

[storm solar]

if i simplify then answer is x is grreater than 2.5

-2.5

[median pilot]

The only thing I'm seeing different is the greater than vs less than sign. Did it get flipped at some point?

[storm solar]

yes

i had to flip it so that i could divide -10 with 4

[median pilot]

I see. You only flip the inequality sign when you divide by a negative number.

So, for example, if you divide by negative 4 rather than positive 4.

so:
4x < -10
x < -10/4

[storm solar]

no in linear equations when there is some number with x and you need to find only x then you flip it and the 4 goes to the other side

not if its negative or positive

[median pilot]

hmmm lets test it. you're saying x is greater than -2.5.

what happens if you try x = 0 (0 is greater than -2.5).

[storm solar]

there is no number with x so x = 0

[median pilot]

7(x+2) < 3x + 4 [evaluated at x = 0, in other words plug 0 in for x]
7(0+2)< 3*0 + 4
7(2) < 0 + 4
14 < 4

but 14 isn't less than 4, so there must be a mistake somewhere.

[storm solar]

then idk

u can ask someone else

[neon wadi]

o yeah i didn’t notice he flipped that

[median pilot]

I'm just showing a way to test out your answer.

But the issue is just where you flipped the sign. You only flip the sign if you multiply or divide by a negative number. If you multiply or divide by a positive number, you do not flip the sign.

[storm solar]

ok ty for the clarification

also -10 is a negative number so my answer is correct?

[median pilot]

Not quite, but you're close.

The rule only cares about the number you are dividing (or multiplying) both sides of the equation by. So the -10 there doesn't matter. The only thing you consider when deciding whether to flip the sign or not is the sign on the 4

[storm solar]

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[brisk raptor]

i don't know how to do this, i honestly don't even know where to start

[tight prawn]

have you learned what the inner product is?

because that is what part a is asking you for.

[brisk raptor]

there was one lecture but i spaced out a lil and dont understand the internet

[faint gorge]

the inner product is distributive use that

[devout snow]

@brisk raptor Has your question been resolved?

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[mellow willow]

I’m not sure where I went wrong

[mellow willow]

Because I have to get 10 I. That final pet of the box

[junior chasm]

why is it just 7

before 10

[mellow willow]

What do you mean?

[mellow willow]

Oh wait

It is meant to be x

[junior chasm]

just x?

[mellow willow]

Just forgot to add the variable

7x

[junior chasm]

it is correct

[mellow willow]

But I can’t get it into factored form

Because -1•-3 does not equal 10

Im really confused

Because I have to put it in factored form

But I cant figure out the box

Ok I did it again but I still can’t find a solution

[obsidian nebula]

are you factorising x^3 - 4c^2 -7x + 10 and x-1

*-4x

[mellow willow]

This I’m not really sure

I also tried with long devision and can’t do it

My teacher might of made a typo

He does it a lot

But I have a quiz

[obsidian nebula]

yes this bit is correct

[obsidian nebula]

xd

[mellow willow]

I just want it to be confirmed

Bevause it is almost 3 am

I’m tired 😭

So like is this even possible

[obsidian nebula]

This should be

[mellow willow]

I’ve been stuck on this problem for like a hour

[obsidian nebula]

if you put 1 into this polynomial it guves 0, so by factor theorem x-1 is a factor

[mellow willow]

Yea we it is given to us that (x+3) and (x-1) are factors In trying to find the third one

[obsidian nebula]

wait I think I know whu

[torn vessel]

-4-(-1)

[obsidian nebula]

here should be -3x^2 as you are subtracting negative x^2

[mellow willow]

Wdym

[obsidian nebula]

this

[mellow willow]

So it equals -3

[torn vessel]

you have $-4x^2-(-x^2)=-3x^2$

[woven radish]

Vѳrtєx-

[torn vessel]

not -5x^2

[obsidian nebula]

yes

[mellow willow]

What about the box tho

I’m really confused

[obsidian nebula]

kinda same thing here

remainder at that is $-3x^2-7x$

[woven radish]

VjFaU1dsb3hjRVZhTW1SUFVqQXhibGRY

[mellow willow]

Wdym

Wait I’m confused

What long division was wrong

[obsidian nebula]

and then you subtract $-3x^2+3x$

[woven radish]

VjFaU1dsb3hjRVZhTW1SUFVqQXhibGRY

[mellow willow]

I still don’t understand where I went wrong with the box

[obsidian nebula]

$-7x-(+3x)=-10x$

[woven radish]

VjFaU1dsb3hjRVZhTW1SUFVqQXhibGRY

[mellow willow]

I know I’m repeating myself but I’m very confused

[obsidian nebula]

Im not exactly sure how the box works
Ive inly learnt the long div relly

[obsidian nebula]

don't worry about it

[obsidian nebula]

how does the box work

[warped jasper]

in the box method youre just multiplying the 2 numbers of the row and the column onto chosen box right?

like x^2 • -1

= 1x^2

[mellow willow]

I think

[warped jasper]

top number multiplied by side number?

[mellow willow]

I got some variables in that

I haven’t gotton any sleep this week because I’ve had three quizzes and a test

So I’ve been going to bed at the earliest 3 am since Sunday

[warped jasper]

its alright that makes sense

but do you see where you went wrong in the box?

[mellow willow]

I’m trying it again

[mellow willow]

Nope 😭

[warped jasper]

you wrote different numbers this time

[mellow willow]

Wait

[torn vessel]

x^2 times -1 is what?

[mellow willow]

I need sleep 😭

[torn vessel]

real

[warped jasper]

you should prioritise sleep rather than revision

if ur cooked ur cooked just revise earlier next time

[mellow willow]

I’m trying to make the equation x^3+-4x^2+-7x+10 inside the box

[mellow willow]

I mean I got home at 7pm and started hoemwork at 9

[mellow willow]

So the numbers on top of the box change

[warped jasper]

i think the box method only works when you have the long quadratic and youre trying to find a root

like you already have the numbers on top of the box and you’re trying to find the numbers on the side

[mellow willow]

I did it for another EWUATION and it worked

[warped jasper]

maybe it was just coincidence or you had the numbers on the side and solved for the top?

[mellow willow]

I just realized that is wrong

I need one more variable

Omg 😭

Should I not do the box

[warped jasper]

use long division or factor theorem first for questions like that

i say use the box when you get given
ax^2+bx+c and you have to find a b and c. and you know what one of the variables/root already is

so you have the numbers on the side and ur trying to figure out the numbers on the too

top

also in the box you did before you were so close but you multiplied 3 and 1 and got 4

ihave to fo now

[devout snow]

@mellow willow Has your question been resolved?

[obsidian nebula]

yea the box isn't a general method I guess

[devout snow]

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[lone stirrup]

Could someone explain me, why do we ^2 both whole sides?

[lone stirrup]

My idea was that we have 4 terms.

so x+2 becomes x-2

x-2 = root(5x-4)

^2 both sides you'd get

x^2-4 = 5x-4

But that's where it goes wrong

[restive river]

(x-2)^2 is not equal to x^2 - 4

[restive river]

to eliminate the square root

[lone stirrup]

So basically to eliminate the square root we have to put parentheses around all of the left and right side?

[devout snow]

@lone stirrup Has your question been resolved?

[devout snow]

@lone stirrup Has your question been resolved?

[plush atlas]

Is this normal solving for x?

[plush atlas]

If you remember surd rules, for example root(a) × root (a) = a, which is basically just (root(a))^2

So yeah to eliminate square root, you have to multiply the term by itself which in other words is just squaring the term

But what you do on one side of the equation you must do on the other side so you also have to square the entire other side

Which is (x-2)^2, so just expand and solve for x from there

[lone stirrup]

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[devout snow]

Please don't occupy multiple help channels.

[loud hemlock]

Can someone help me
Find x and y please

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[twilit comet]

use the fact that the cointerior angles of parallel lines add to 180

i.e., x - y + 145 = 180 and x + y + 125 = 180

[stable wraith]

@loud hemlock close #help-12 before opening a new one

[loud hemlock]

Sorry

I got y=45 is that right?

[stable wraith]

and for x?

[loud hemlock]

I got x=45 and y=10

Is that right?

@stable wraith

[stable wraith]

yup

[loud hemlock]

Ty

What can I do to solve this one

5x-3y+6x+9y=180?

Could I also do ×+3y+6x+9y=180?

So
X=12 and y=8?

@stable wraith

Is x=12 and y=8

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[wraith cloud]

Check if function is odd or even: sin(2x)
I'm quite bad at dealing with sin and cos in equations and I don't know how to do this.

[fossil locust]

the definition is that f(x) = f(-x) for any even function

can you check this for f(x) = sin(2x)?

[wraith cloud]

So at first I wanted to do -sin(2x)=sin(-2x) but at that point I'm already stating it's an odd function.
So I wanted to go with just sin(-2x) and try to get it to the state where I have a*sin(2x) where a = -1 v a = 1 at the end.

But I don't know what to do with sin(-2x)

[fossil locust]

yeah that's the right identity to use

so starting from $f(-x)$, $\sin(-2x) = -\sin(2x) \ne \sin(2x)$

[woven radish]

south's secret twin brother

[fossil locust]

so it's not an even function

[wraith cloud]

Don't I have to prove it some how rather than just state it?

[fossil locust]

you've shown that $f(-x) \ne f(x)$

[woven radish]

south's secret twin brother

[fossil locust]

I think you're overthinking the problem

[wraith cloud]

Ok, thanks

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[forest palm]

kinda stuck after this step, what do i do?

[coral violet]

is the answer 2?

For small values… sinx = x

or sinx/x -> 1

[forest palm]

wont there be an issue if our x/y both are 1 we get sin(k(2))/k?

[coral violet]

But, it approaches 1, not exactly 1

[forest palm]

hmm yeah you are right, so how do i advance from the step im currently on?

[coral violet]

so

well

here

[forest palm]

how did you get rid of (x^2 + y^2)

[coral violet]

Well…

Substituting it isnt an issue

here

[forest palm]

are you sure that you can divide like that when its inside a sin?

[coral violet]

Well, yes

since, as x->0

sinx / x = 1

here imagine k->0

so K(x^2 + y^2) ->0

[forest palm]

Yeah i see it now

thank you!

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[forest palm]

cant tell if i made right sub here since i dont see any advancing possible

[devout snow]

@forest palm Has your question been resolved?

[forest palm]

<@&286206848099549185> ?

[orchid sierra]

There are a couple of things you can do

since this isn't indeterminant form (0^0, 0/0, inf^inf, etc) you can get the right answer with direct substitution. that would be 9^(9/0) -> 9^inf ~~ inf.

but there is a more rigorous way

use the fact that

$a^b = e^{\ln{(a^b)}} = e^{b\ln{(a)}}$

[woven radish]

Melvin Eugene Punymier

[orchid sierra]

when a and b are both functions (continuous at the limit), we can use the Composition Limit Law to bring the limit inside the exponential function (because the exponential function--outer function--is continuous). Then we use the Product Law.

You can see a summary of those laws here https://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html#Composition_Law

I got the general idea from this paper https://sites.science.oregonstate.edu/math/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/power_limits.html#:~:text=All of the different forms,how this method can work.

so then this will be (this will take a sec to write)

$\lim_{(x,y) \to (0,0)} \left(9 + 2x^2 + 2y^2\right)^{\frac{9}{x^2 + y^2 + 2 x^2 y^2}}$

ugh latex is breaking hang on

[woven radish]

Melvin Eugene Punymier

[orchid sierra]

$\exp\left(\lim_{(x,y) \to (0,0)} {\frac{9}{x^2 + y^2 + 2 x^2 y^2}} \lim_{(x,y) \to (0,0)} \ln{\left(9+2x^2 + 2y^2\right)}\right)$

[woven radish]

Melvin Eugene Punymier

[orchid sierra]

there we go 🙂

[forest palm]

seems i need to do reading on this 😄

[orchid sierra]

meh, barely.

[forest palm]

i didn't follow how we got the 9 there but everything else was clear 🫡

[orchid sierra]

you know that "exp" is "e^(something)", right?

isn't the 9 in your limit twice?

looks right to me, I copied it the same way you wrote it?

[forest palm]

thats a 1

[orchid sierra]

you've made it this far in math classes writing your 1s like 9s???

[forest palm]

yes 😭

[orchid sierra]

or 7s

[forest palm]

But yeah so everything else makes sense then

[orchid sierra]

oh good!

yeah and it really wouldn't matter what number that is for this limit anyway

[forest palm]

so this is out e^bln(a)

from this formula

[orchid sierra]

e is the outer function

the product b*ln(a) is the inner function

[forest palm]

which we can solve by using product law

[orchid sierra]

the inner function is the product of two functions

[orchid sierra]

yup

the outer function and inner function had to be continuous to use composition law lim f[g(x)] = f[lim g(x)]

and the factors f(x) and g(x) need to be continous at the limit for lim f(x)*g(x) = lim f(x)* lim g(x) to work too

which they usually are...this being math homework.

[forest palm]

wow, everything went in my brain and acctaully stayed. nicely explained 🫡

[orchid sierra]

😊

damn, I'm good at 630 am

[forest palm]

ill return the favor by writing my 1s better in the future

[orchid sierra]

🫡

[forest palm]

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[devout snow]

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[random spindle]

Can I cancel units cubed with units squared? I'm trying to find the weight of brass required in kilograms (density is 8.73g/cm3) from an area of 3508cm2

[random spindle]

When I cancelled the units out, I got a total of 30,624.84g, 30.62kg, but for the project I want that seems wildly high

nvm, its late and I made a really dumb error

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[plush harness]

can anyone explain how 2nd to 3rd transition

[misty crest]

they subtracted 1

$1 = \frac{x+2}{x+2}$

[woven radish]

knief

[misty crest]

$\frac{x|x|}{x+2} - 1 = \frac{x|x|}{x+2} - \frac{x+2}{x+2}$

[woven radish]

knief

[plush harness]

oh thats dumb from me srry

[misty crest]

then distribute the negative

[plush harness]

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[plush harness]

anyone help me do this.....i dont know where to even start

ping me plz as im not here

[devout snow]

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[plush harness]

anyone??

[plush harness]

plz ping me

[olive snow]

Try to get rid of abs value by spliting the interval of solving

[plush harness]

wdym?

im sorry but i am very dumb

[olive snow]

Rewrite as a piecewise function your expression

[fathom flint]

squaring would work but it'd be lengthy asf

[plush harness]

could you explain in a bit more detail

[fathom flint]

my first thought was to square both sides to get rid of modulus

But since the question is big we have to apply a different method

[plush harness]

any tips on how to

[fathom flint]

i suggest don't try it because it take alot of time

takes

[plush harness]

if you could show me the soln in like a paper thatd be great but maybe like a general walkthrough if you dont have the time

[plush harness]

is it really that hard?

[fathom flint]

not hard just takes alot of time

[plush harness]

so whats the approach i should take

[devout snow]

@plush harness Has your question been resolved?

[devout snow]

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[unborn atlas]

I don’t get the intuition behind how a change in power is all it took to take it from converging to diverging, both graphs approach 0 at infinity, I know it algebraically adds up, but how can one be finite and one be infinite?

[twilit comet]

because the rates at which they approach 0 are different

not entirely sure what you're confused about here

[unborn atlas]

Sure, I agree, but how does that make such a significant difference, the rate at which graphs approach 0 can be different yet both can still converge

[twilit comet]

it just does lol

[unborn atlas]

How?

[zealous sinew]

This also confuses me intuitively

It makes sense mathematically, but not mentally, you know?

Anyways, I tried asking Meta AI about explaining it to me, and this is what it said:

The intuition behind convergence/divergence of series like ∑1/x^n (p-series) lies in the rate at which terms decrease.
Convergence (p > 1)
For ∑1/x^(p+1), terms decrease rapidly (e.g., 1/x^2, 1/x^3). The sum of these terms is finite because:
Each term is smaller than the previous.
The decrease accelerates.
Think of it like a snowball rolling down a hill, slowing down rapidly due to friction. The snowball's kinetic energy (term value) decreases quickly.
Divergence (p ≤ 1)
For ∑1/x (or ∑1/x^p with p ≤ 1), terms decrease slowly (e.g., 1/x, 1/√x). The sum is infinite because:
Terms decrease, but slowly.
The harmonic series (p = 1) is a border case.
Imagine a snowball rolling down a hill with negligible friction. The snowball's kinetic energy decreases slowly, and it travels far.
Critical threshold (p = 1)
The harmonic series (∑1/x) is a critical case:
1/x + 1/2 + 1/3 + ... = ∞
This series diverges, but slowly.
Why the power matters
Increasing the power (p) accelerates term decrease:
1/x^2 decreases faster than 1/x
1/x^3 decreases faster than 1/x^2
This acceleration affects convergence:
∑1/x^2 converges (∑1/n^2 = π^2/6)
∑1/x diverges (∑1/n = ∞)
The power changes the rate at which terms approach 0, influencing convergence.
Intuitive analogy
Consider filling a bucket with water:
Convergent series (∑1/x^2): Water flows rapidly, filling the bucket to a finite level.
Divergent series (∑1/x): Water flows slowly, overflowing the bucket.
The power (p) controls the flow rate.

[supple knot]

!nogpt

[devout snow]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[zealous sinew]

oh hectic, I thought the analogy might help. whoops, I'll refrain from using the tool on here in future then; my apologies

[unborn atlas]

do you have any intuition by chance

[supple knot]

what exactly are you confused by in this sentence

[unborn atlas]

Well graphically they both appear similar, and both approach 0 as x approaches infinity, I don’t see why the rate at which they approach 0 determines whether they converge or not if they both “reach” that point eventually

[supple knot]

what is "they"

[unborn atlas]

Graphs

1/x and 1/x^2

they approach 0 at different rates

it’s obvious when looking at 1/x^p when p< 0 but p=1 itself makes no sense to me

[supple knot]

yes they do approach zero at different rates.

what is "it's" in this sentence

[unborn atlas]

So does 1/x^2 and 1/x^3 though, I don’t see it as grounds to determine convergence

[unborn atlas]

It is obvious that 1/x^-2 doesn’t converge

[supple knot]

right

[unborn atlas]

But I can’t draw a parallel similar to that seen in p=1

[supple knot]

the simplest way is just to use power rule for integrals

have you learned that yet

[unborn atlas]

Yes

I think?

[supple knot]

for $p \neq 1$, then we have $\int \frac{1}{x^p} dx = \frac{x^{-p+1}}{{-p+1}} + C$

[woven radish]

riemann

[supple knot]

when $p=1$, what's the antiderivative?

[woven radish]

riemann

[unborn atlas]

1/0?

[supple knot]

no

do you know the derivative of $\log(x)$?

[woven radish]

riemann

[unborn atlas]

Yes

[supple knot]

what is it

[unborn atlas]

$\frac{ln|x|}{x*ln(b)}$?

[woven radish]

Nathan

[supple knot]

$\log$ in math means natural log

[woven radish]

riemann

[unborn atlas]

?

[supple knot]

fix this answer

[unborn atlas]

log is based on 10 by default no?

[supple knot]

no

[unborn atlas]

ln isn’t log

[supple knot]

also wrong

[unborn atlas]

??

ln assumes log_e

log itself is assumed base 10

[supple knot]

not in higher math but sure

[unborn atlas]

wait I’m being stupid

$\frac{1}{xln(a)}$

[woven radish]

Nathan

[unborn atlas]

@supple knot

?

[unborn atlas]

This is just lnx

[supple knot]

yes

[supple knot]

so why is this confusing you

[unborn atlas]

I thought you were asking me to plug in 1 to what you determined next it it I thought it was a big reveal or something

[unborn atlas]

Because it doesn’t really answer my questions

Both approach 0

Meaning it’s essentially a bounded area

[supple knot]

wrong

[unborn atlas]

I guess ln(x) approaches infinity sure, but I’m saying graphically it makes no sense to me

[supple knot]

$\lim_{b \to \y} \int_1^b \frac{dx}{x} = \lim_{b \to \y} \log(b) = + \y$

[woven radish]

riemann

[supple knot]

,w plot log(x)

[woven radish]

Results provided by WolframAlpha

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

Requested by rie.mann

[supple knot]

this logic is your main flaw

[unborn atlas]

The antiderivative doesn’t approach 0 yeah I know

I know the other one does, but why must it

Again I understand it algabraicly I’m not asking to be told that lnx approaches infinity, I’m saying graphically 1/x and 1/x^2 both approach 0 why can’t they both have finite areas intuitively

[supple knot]

have you seen the divergence of the harmonic series?

and how that compares to the integral of 1/x

[devout snow]

@unborn atlas Has your question been resolved?

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[nova basin]

would this be...
a) False (cusp at x = -1)
b) True
c) 2 (x = -1, 0)
d) 2 (x = -1, 0)
e) Down
f) False (2nd derivative is negative because the function is concave down at x = 1)
?

[devout snow]

@nova basin Has your question been resolved?

[devout snow]

@nova basin Has your question been resolved?

[wise storm]

Yeah that all looks correct, do you have any specific doubts about any of your answers you want to discuss more maybe?

[nova basin]

the graders said c was incorrect i guess? not sure why

[wise storm]

I guess it depends on what their definition of critical number is, I'm kind of rusty myself I forget if it means the derivative is 0 or if that also includes when the derivative is undefined or not

Paul's online notes seems to agree that it's when it's undefined so I think you're right. So that matches since at -1, 0 are the only points in that interval (-2,2) where that's satisfied

https://tutorial.math.lamar.edu/classes/calci/criticalpoints.aspx

it might be possible that some people only call it a critical point when the derivative is 0 and not when it's undefined and they're using that convention? You'd have to ask your teacher on that before/after class or at office hours I think and see if you can get points back

or email them maybe

[nova basin]

hmmm oke i will thx

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[nova basin]

i just realized x = 1.5

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[dapper crane]

Hello , this might be a stupid question , but ; when f(x) = x^2 , we have the completed square (x+2)^2 - 5 , to translate the graph we sub x+2 , therefore f(x+2) = f(x+2)^2 , why do we subtract 2 to get back to the original (f(x) = x^2) , I understand its to translate the graph horizontally by -2 , but how does this get you back to the origin ? It gets you further away

[devout snow]

@dapper crane Has your question been resolved?

[trail eagle]

g(x) = f(x+2) moves f(x) 2 units to the left.

If you want to get back, then notice that g(x-2) = f((x-2)+2) = f(x), which is equivalent to translating g 2 units to the right, i.e undoing the translation you did to f in the first place

[dapper crane]

Maybe I incorrectly worded my 'question' , I'm trying to understand why ,by subtracting 2 from f(x+2) gets you a different x^2 curve compared to an x^2 curve that passes through the origin .

[trail eagle]

You mean why is f(x+2)-2 != f(x)?

[dapper crane]

yes

i understand 'mathematically ' it isnt , but graph wise i dont

[trail eagle]

f(x+2) is translated 2 to the left

Then you subtract 2 from it, that means you translate it 2 down

So the total graph has been moved 2 to the left and 2 down

Not back to its original state

[dapper crane]

were subtracting 2 to get back to f(x) no ?

[dapper crane]

sorry i dont understand what you mean by 2 down as well ? I thought its just 2 left

[trail eagle]

When you subtract inside the function, the translation obtained is horizontal

[dapper crane]

yep

[trail eagle]

When you add/subtract outside, it’s vertical

Think of x^2.
Clearly (x+2)^2 is not the same as x^2 + 2

[dapper crane]

ye thats true

[trail eagle]

The former has been translated 2 units to the left, the latter 2 units up

[dapper crane]

ah ok ,

but i still dont get why our aim is to get to f(x)

like , f(x) = x^2 , sub x+2 --> f(x+2) = (x+2)^2 , then subtract 2 (hence translate left by 2) to get to f(x) = x^2

but if we subtract 2 it results in f(x) = x^2 which is just a parabolic curve that crosses through origin

[trail eagle]

What is your original problem

[dapper crane]

no problem im just trying to understand , (ifyou need to know its x^2 + 4x - 1 -> (x+2)^2 - 4 -1 -> (x+2)^2 - 5 -> a curve thats minimum point is 2,-5 )

[trail eagle]

There is no “need” to get f(x)

[dapper crane]

isnt that the theory to draw the curve

[trail eagle]

And I don’t see what you’re trying to do here. If anything you would start with f(x)=x^2 and then sub in x+2 so as to get

f(x+2) = (x+2)^2
And then subtract 5 to get

g(x) = f(x+2) - 5 = (x+2)^2 - 5

Then g represents your function f that was translated left by 2 and down by 5

So if you want to draw the curve that’s helpful because x^2 is easy to draw and your function (x+2)^2 - 5 is just that but translated

[dapper crane]

so we use f(x) = (x+2 )^2 to help draw the curve

it isnt the theory to understand why we shift to the left by 2 ?

[trail eagle]

If you have a function g and a reference function f then if you can write g(x) = f(x-h) + k, that means that g is actually f translated h units to the right and k units up

In other words the origin has moved to (h,k)

[dapper crane]

sorry I dont understand this g stuff ? is it f(x) ? but like a superior ?

[dapper crane]

ohhhh

[dapper crane]

this is making a bit more sense

so does g(x) represent the origin ?

[trail eagle]

g is just another function that is a translated version of f

So if you take the gtaph of f, and move the whole axes to (h,k), the new graph is the graph of g

[dapper crane]

so g(x) = f(x+2)^2 - 5

[trail eagle]

Yes

So that is equivalent to sending the origin to (-2,-5)

[dapper crane]

ahhhhh ok , because in videos / images it show a new curve has been translated on the same axis they use

but in reality its just, the parabolic curve hasn't moved, the axis has moved instead ?

[trail eagle]

I think you would benefit with playing a bit with desmos

I used to be able to share a desmos graph idk how to anymore

But yes the idea is that the curve has moved

You’ve moved its whole axes

[dapper crane]

the 'idea' or reality ?

like is it just to explain it more easily , instead of having to teach the whole g(x) method or whatever in order to understand that the origin has moved

[trail eagle]

I mean f is still the same, g is a curve whose graph is the same as f but translated

[dapper crane]

?

[trail eagle]

f is still defined just like you did in the beginning. It's g that is a representation of the graph of f that has been translated

[dapper crane]

making a little bit more sense ...

so g(x) = f(x+2)^2 - 5 is an 'imaginary' curve ?

and f(x) is still there we can see it

[trail eagle]

There is a reference curve, the one you're comparing to. And a resultant curve, that is the application of the transformation you're doing to it, such as translating by h to the right and k up

They still exist, just one is a translation of the other

x^2 is a reference function

x^2 + 1 can be seen as x^2 translated 1 upwards

You could also think of x^2 + 1 as a reference function

And then x^2 is x^2 + 1 translated 1 downwards

[dapper crane]

reference = f(x) , the resultant is g(x) , g(x) is an application of transformation on f(x)

[trail eagle]

Yes.

https://www.desmos.com/calculator/zsqetgqvkm