help-17

[urban zinc]

alright, weird anyway, thanks

.close

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[mental cradle]

If at t=0 I have 700 mCi of substance A, and the mCi of the substance A i 5% smaller than the mci of a substance B.
Then is it B=A+5%=735, or is it A=B-5*B/100 -> B=736.84....?

[mental cradle]

I think its the latter, but i wanted to make sure

please ping me when you answer

thank you

<@&286206848099549185>

guys ignore the mci and t=0 its just a precentage problem..

[vocal sleet]

@mental cradle Has your question been resolved?

[mental cradle]

<@&286206848099549185>

[atomic summit]

b=a/0.95

[mental cradle]

.close

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[keen dagger]

how did they separate ln3 from x/2lnx

[keen dagger]

i did y = u^x/2

ln y = x/2 ln u

x/2 * ln 3x

[flat whale]

$y = \lim a\cdot b^c = e^{\lim \log(a\cdot b^c )} = e^{\lim \log(a) + c\log(b)}$

[twin meteor]

riemann

[flat whale]

identify a,b,c

[flat whale]

this is incorrect

$3 (x)^{x/2} \neq (3 x)^{x/2}$

[twin meteor]

riemann

[keen dagger]

oh is it because they dont share the same exponents

[flat whale]

yea sure

[keen dagger]

thats the only way i see it making sense for me

[keen dagger]

so if im taking the ln of something and there's a coefficient in front of that something, i have to take the ln of the coeff as well?

[flat whale]

,tex .log rules

[twin meteor]

riemann

[flat whale]

product and then power

[keen dagger]

gotcha

.close

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[pulsar star]

I need help with this question. I don't know where to start.

[pulsar star]

so what i have thought about is looking for the area of the garden first

but i dont know if that is going to work

[atomic jasper]

try setting one side as x and another side as y

then calculate the area in terms of those x and y

aswell as calculate the fencing in terms of x and y

@pulsar star

[pulsar star]

yes

so let x be the length and let y be the width

and make a function off those two sides to calculate area?

[atomic jasper]

yes

whats the area

(of the vegetable part, not the whole garden)

@pulsar star

[pulsar star]

14W−W2

[atomic jasper]

what?

whats W?

[pulsar star]

sorry i accidentally typed in my notes on the chat

[atomic jasper]

dont bother opening a help channel if youre not going to be active

[pulsar star]

no im active im just stuck and dont know what to do

[atomic jasper]

i told oyu what to do

write the area as a function of x and y

and the same for the perimeter

but for some reason you are ignoring my requests

[pulsar star]

(2x + 1)(2y + 2)

would that be the function for area

[atomic jasper]

not quite

there is a 1ft and 2ft border on both sides

[pulsar star]

(2x + 4)(2y + 2)

is that right?

[atomic jasper]

you swapped x and y, but sure

now, what is the perimeter?

[pulsar star]

2(2x + 4) + 2(2y + 2)

sorry im just not good at these types of word problems so it takes a while for me to think

[atomic jasper]

4x and 4y?

[pulsar star]

2(x + 2) + 2(y + 1)

[atomic jasper]

border thing again

[pulsar star]

yeah i just realized that

[atomic jasper]

can you fix it?

[pulsar star]

(2x + 4) + (2y + 2) = 40

[atomic jasper]

still same border issue

[pulsar star]

(2x + 4)(2y + 2) x 2 = 40

[atomic jasper]

now youre just guessing

it helps to make a drawing

the perimeter is 2 * (x+2) + 2 * (y+4) = 40

and the area is x*y

[pulsar star]

i am an idiot that helps a lot

thanks

ok

[atomic jasper]

whenever you are doing such problems yourself, make a drawing

almost always

[pulsar star]

knowing that do i just try to find x and y so i can put in the area function?

or since were maximizing

[atomic jasper]

you first need to find y in terms of x from this

(or vice versa ofc)

[pulsar star]

y = A/x

[atomic jasper]

$2x+4+2y+8=40$

[twin meteor]

Bonk

[atomic jasper]

$y=?$

[twin meteor]

Bonk

[pulsar star]

oh so since maximizing we are expanding to make a quadratic function

[atomic jasper]

this is just the perimeter first

[pulsar star]

ok i think i can do this part

[atomic jasper]

because right now, we have two unknowns with x and y

but we can represent y in terms of x so make it just one unknown

are you able to do that?

[pulsar star]

y = -(x - 17)?

[atomic jasper]

im not sure how oyu got a negative number

[atomic jasper]

!show

[vocal sleet]

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

[atomic jasper]

first of all, thats the wrong equation

its 2x+4+2y+8=40

[pulsar star]

oh shoot

[atomic jasper]

actually, the rest does look correct

and if you change it to the right numbers

you get y=-(x-14)

now, fill that into area = x*y

[pulsar star]

so A = x*-(x-14)

[atomic jasper]

yup

now how do you maximise the area?

[pulsar star]

do i expand the function into -x^2 +14

[atomic jasper]

-x^2+14x, yes you could do that

how do you maximise?

[pulsar star]

complete the square find the biggest possible x value from the function

[atomic jasper]

nothing with the derivative?

[pulsar star]

uh the problem is i have never been taught derivatives from grade 12 precalculus and college precalculus

[atomic jasper]

youre grade 12?

[pulsar star]

im a freshman college student first year and i dont remember learning derivatives at all

[atomic jasper]

okay

you need to learn those

[pulsar star]

yeah im going to look into it

[atomic jasper]

,w graph -x^2+14x

[twin meteor]

Results provided by WolframAlpha

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Requested by .bonk__

[atomic jasper]

to find the maximum of this

you take the derivative of -x^2+14x

which is -2x+14

and set it equal to 0

[pulsar star]

x = 7

[atomic jasper]

yup

and now whats y?

y=-(x-14)

[pulsar star]

7

[atomic jasper]

and now what is the maximum area?

[pulsar star]

49 ft^2?

[atomic jasper]

perfect

now, are we done?

[pulsar star]

i guess so thank you very much or until the next question

[atomic jasper]

!done

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[atomic hill]

Hello, i have a question so say you are proving a result such as any prime number + x is a compositie number, could you use the logic that prime numbers are all odd except for two and just state that like 2n + 1 (prime) + x

[heavy yoke]

but there's exactly one even prime number

[atomic hill]

yeha thats why i split it into cases no

my logic is like 2n + (1+x) where x+n is an even number

so then its ognna be even

with the gase of 2 it works as well

[heavy yoke]

what exactly is x?

[atomic hill]

uh its for an assignment so i just wanted to ask a general question

[reef agate]

!original?

[vocal sleet]

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

[atomic hill]

ok x = 7

[heavy yoke]

you can also just use the general rule of adding parity

[atomic hill]

whats that

[heavy yoke]

odd + odd = even, even + even = even, even + odd = odd

[atomic hill]

ok so thats what i was thinking

id split it up into cases tho right cause one for everything but two

and one for 2

[heavy yoke]

if you haven't covered these yet, you can prove any of them using the same sort of logic as above

[atomic hill]

no i know

[heavy yoke]

yes, you'd have to prove the case for 2 separately

[atomic hill]

alright thx

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[hexed onyx]

Can someone check these

<@&286206848099549185>

[stable fox]

😱

[hexed onyx]

not all the adding at the end ofc just the start

[hexed onyx]

Is that a yes 🤞

[stable fox]

Bro i just started learning integrals, sorry

[hexed onyx]

Oh

Anyone else

[stable fox]

@vast shale mind helping this lad?

[vocal sleet]

Please do not ping individual helpers unprompted.

[stable fox]

Oh mb

[deft oar]

checking work is really tedious 😫

do you have the OG questions?

https://tenor.com/view/megamind-gif-26070434

[deft oar]

first of all, where did 2pi come from. What are you even trying to solve?

[hexed onyx]

It’s part of the shell method formula

[deft oar]

like the OG OG question

[hexed onyx]

Yea I sent the wrong pic mb

[deft oar]

something feels wrong for your q2

[hexed onyx]

Do you know what?

[deft oar]

why do you have a r(x)?

and why are you multiplying it into the integral?

[hexed onyx]

Bro do you know how to do this stuff😭

Ur trolling me

it’s part of the formula

Just like 2 pi is

.close

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[deft oar]

WHAT?

[hexed onyx]

What

😭

Go to 5

[deft oar]

I need to you to come back here!

[deft oar]

aight

[vocal sleet]

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[stoic sierra]

ahhh I'm an idiot

[stoic sierra]

I'm pretty sure y = 1 should not be my solution

Ok, I'm 1/00 sure

Cause y(0)=1 is my condition I'm given

Ooohhh

I forgot to add c and divide that e term there

!done

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[stoic sierra]

.close

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[stoic sierra]

.reopen

[vocal sleet]

✅

[stoic sierra]

What da heck? What's this D_x?

What's D?

I understand that's the derivative wrt x, but of what?

.close

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[native willow]

vieta's formulae

can't find a pattern here

i've worked out the sum of roots, product pairs of roots and product of roots

[twin meteor]

kanyecasino

kanyecasino

kanyecasino

[native willow]

this server should have a typst bot

latex sucks

that was exhausting

[gentle nimbus]

something to do with comparing the expanded form of (x-alpha)(x-beta)(x-gamma) with the given cubic equation

i feel that would yield information about alpha beta and gamma which you could use to construct a cubic equation for the given roots

[native willow]

Elaborate?

[gentle nimbus]

solving the cubic could also be a viable method

you can factor out x-1 since all the coefficients add up to 0

[native willow]

Yeah no I'm explicitly not trying to solve the cubic

It's either through vieta's or substitution

[gentle nimbus]

it might be difficult since you need to get the value of the roots to construct the second equation

[native willow]

I asked deepseek for this question

It also gave me a hint

The issue is that deepseek is down right now

So I can't ask it to solve this

I asked it to give me this question ages ago

[gentle nimbus]

if that's the case, here's a little more help

i just hate LaTeX so i always use desmos

don't judge

[native willow]

No I know this one bro

This is how vieta's formulae are deduced

[native willow]

This is what I'm doing when I'm adding the roots

But to be fair I don't know if my new question is degree 3

Hmm

[gentle nimbus]

it should be

since there are three roots

[native willow]

Also true

[native willow]

So this is -b/a

[gentle nimbus]

for the product pairs and product, you have the value of abg and a+b+g

[native willow]

abg you say

https://tenor.com/view/gangster-dance-taeuvre-druski-white-shirt-dance-gif-9377375949202477378

[gentle nimbus]

too lazy to type of the greek letters

[native willow]

I'm making a joke

[gentle nimbus]

i don't get it :<

[native willow]

abg also means something else

Search it up

Well anyways let's keep this maths related

[gentle nimbus]

main issue is finding the values of the square roots and square root pairs

not sure how you're going to solve it without actually finding out the values of the roots

[native willow]

I don't know if the hint is helpful

What does (root(alpha) + root(beta))^2 isn't helpful since it still has a radical

(root(alpha) + root(beta))^4 ?

Not sure if that has a radical

[gentle nimbus]

it will

[native willow]

What about ^3?

[gentle nimbus]

just think it through the binomial expansion

there must be a root alpha ^ 1 and root beta ^1 term in there for any expansion

[native willow]

Oh lmao

Yes I get that

How is the hint from deepseek helpful then?

[gentle nimbus]

probably isn't

me thinks

[native willow]

Hmm

Where are you from in the world?

Which country?

[gentle nimbus]

some where

i hope

[native willow]

That's aura

Anyways this question is difficult

[gentle nimbus]

it is if you limit yourself to specifically not finding out the value of the roots

there just seems like no way to eliminate the radicals

and since we don't know their values (we choose to stay ignorant about it) we cannot substitute their values either

[native willow]

It does work out apparently chatgpt is answering it but mobile latex sucks so I can't tell

[vocal sleet]

@native willow Has your question been resolved?

[vocal sleet]

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[thorn spindle]

when should i use shell method instead of disk/washer

[hazy nimbus]

Hello

I need help

[signal pendant]

When there is radial symmetry

[hearty jungle]

!occupied

[vocal sleet]

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

[thorn spindle]

what does that mean

[signal pendant]

Do problems and you will see what that means and when to use it

[thorn spindle]

alright

.close

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[drifting yoke]

my like is so cooked Im right now awake at a time when i should be sleeping snd im sleep at a time when the day i wake up is the same day

[signal pendant]

Such is life

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[stable oar]

So I've gotten to this question and solve in by first replacing the value by

k/9 * k-1/9-1 (second time) = 7/12

by multiplying them together I got (k^2 - k)/81 = 7/12

Now doing crisscross I got 12k^2 - 12k = 637

After sending the 637 to the other side I now got: 12k^2 - 12k - 637 = 0

The next was supposedly me dividing every value by 12 but I got stuck at 637

[paper depot]

you are having trouble dividing 637 by 12?

[stable oar]

Nope

[paper depot]

hang on

[stable oar]

B shouldnt be a decimal number

I got B as a decimal 😭

[paper depot]

why not?

the problem does not say

[leaden ingot]

what is a and b

[paper depot]

ok wait hold on

how did 637 even show up

i think you messed something up

wait also

the 81 is sus as well

[stable oar]

Did I got mixed up why doing multiplication?

Oh I see it now

[paper depot]

$\frac k9 \cdot \frac{k-1}{9-1} = \frac{k^2-k}{72}$

[twin meteor]

ann.in.a.teacup

[stable oar]

Tysm 🙏

This solved it

504 can be divided for 12

I just messed up between 8 and 9

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[fathom venture]

hi

[vast shale]

hello, whats your question?

[fathom venture]

,rotate

[fathom venture]

I'm supposed to find the largest range and the domain

and i'm pretty sure any odd multiples of pi/2 aren't even included in the domain of tanx

but idk.

So thats why im asking

[fathom venture]

And if they arent included, as im supposing, would i write it down like this?

also the range of tanx is all the reals right???

[lavish sluice]

Yeah

For both the questions

[fathom venture]

alr

thanks

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[fathom venture]

factorization bc of

difference of squares

[next sparrow]

can u like solve it on paper or a note to show me with details

[fathom venture]

well to be honest i have no idea why the square root is ther

but whenever you have the form

x^2 - a^2

then youd be able to factorize it by just

(x - a)(x + a)

[next sparrow]

ik how to solve others but am stuck on this one

some ppl said its multiplied by 1/2

[fathom venture]

ahhh

thats not x^2

thats why

okay

[leaden ingot]

$x-9=\left( \sqrt{x} \right)^{2}-3^{2}$

[fathom venture]

look

[twin meteor]

TargetVN

[fathom venture]

Yup

i misread the regular x as x^2

[next sparrow]

ty

[vocal sleet]

@next sparrow Has your question been resolved?

[vocal sleet]

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[foggy sphinx]

ik this is lame
the rule says ∫ f'(x)/f(x) dx=ln∣f(x)∣+c right ?
ok then if 1/xlnx dx
the derivative should be lnx+1 right ? using ((u'v+v'u)) because x is multiplied with lnx
but if you integrate it letting lnx be u
it gives ln∣lnx∣+c
which means in way or another that ∫ 1/xlnx dx = ln∣lnx∣+c so the derivative of xlnx is lnx ????????????????????????

[leaden ingot]

"∫ 1/xlnx dx = ln∣lnx∣+c" this is correct
"the derivative of xlnx is lnx" this one is irrelevant

the derivative of xlnx is lnx+1

[foggy sphinx]

yes yes this is what is confusing me

[twin horizon]

$\frac{1}{xlnx}$?

[twin meteor]

David

[foggy sphinx]

shouldnt the lnx+1 be in the numinator ??
why is the numenator is 1 ?

its like im admitting that the derivative is 1 ?

[leaden ingot]

$\int\frac{1}{x\ln x},dx=\int\frac{\ln x+1-\ln x}{x\ln x},dx$

I mean

you can do that ig

[foggy sphinx]

yes yes clear

is that legal in terms of math ?

[twin meteor]

TargetVN

[leaden ingot]

it is legal

[foggy sphinx]

thx

.close

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[grizzled vault]

If a taxicab charges x cents for the first 1/9 mile and x/5 cents for each additional 1/9 mile or fraction thereof, what is the charge, in cents, for a ride of y miles, where y is a whole number?

[grizzled vault]

ambiguous question right? Lol

y is a whole number which means y could be 0?

idk why use whole numbers if the definition is ambiguous

x + (9xy - x)/5 would seemingly be the answer

if we have naturals (excluding 0)

but if we allow 0 then meh we have a piecewise function as an answer ig

[reef agate]

I think you colud assume y > 0 because no one wil pay a taxi to ride 0 distance

[versed bane]

just take whichever interpretation is more… natural

[paper depot]

first off i can't help but remark what a weird pricing scheme this driver's got of charging by the ninth of a mile

but also, presumably y ≥ 1 since y=0 would mean you are not riding at all?

[grizzled vault]

i guess

yeah okay fair thanks everyone

i just thought uh like if it's a math question

wouldn't it just be better to clarify

what "set" you're working over

[reef agate]

[natural numbers](#help-17 message) can include or exclude 0

[versed bane]

in reality math is full of ambiguity, but most of the ambiguity doesn’t matter :P

like here

you could just define it piecewise if you wanted to include 0

[grizzled vault]

yeah 😭

[grizzled vault]

yeah naturals aren't that bad

whole numbers historically used to be integers too iirc lol

and now they're naturals (?) seemingly

so like it's the worse word choice imo

should've just went with natural number 😭

but yeah if we're going full context-based then

y>=1

anyway okay

thanks again everyone

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[tight delta]

hello

[tight delta]

how does 3800+3800*1.8% can be rewritten as 3800(1+0.018)?

shouldnt it be 3800(1*0.018?)

[flat whale]

a(b+c) = ab + ac not abc

,calc 3800(1*0.018)

[twin meteor]

Result:

68.4

[flat whale]

,calc 3800(1+0.018)

[twin meteor]

Result:

3868.4

[flat whale]

,calc 3800+3800*0.018

[twin meteor]

Result:

3868.4

[vocal sleet]

@tight delta Has your question been resolved?

[tight delta]

still waiting...thx

[flat whale]

for what?

you got 2 explanations already

[tight delta]

that's not an explanation, that's just a screenshot

i'll wait thanks for others to hppefully chime in

[flat whale]

maybe read the screenshot ?

[vocal sleet]

@tight delta Has your question been resolved?

[vocal sleet]

@tight delta Has your question been resolved?

[civic otter]

Have you read riemann answer?

[civic otter]

Here

If you're not still satisfied, you should at least explain why you think it's multiplication instead of addition

[tight delta]

@civic otter yea i figured it out....

thanks though

[civic otter]

Alright 👍

[tight delta]

@civic otter just to explain....the question was "growth was 1.8%" which basically means 1.018 or 1+0.018 when converted to a decimal

that's it basically

[flat whale]

you didn't give the question at all

[tight delta]

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[cunning yew]

$\frac{1}{k+1} \leq \int_{k}^{k+1} \frac{dt}{t} \leq \frac{1}{k}$
how can i square this inequation ?

[twin meteor]

<rajel />

[cunning yew]

i mean the only part that isnt clear to me is getting the square root of the integrale

[bitter pilot]

if the expressions are non-negative and the inequality is true to begin with then you can square

[cunning yew]

how can i square the integral

[bitter pilot]

compute it first

[cunning yew]

should i initiate it to the previous primitive and then get them back to the integral ?

[bitter pilot]

what?

[cunning yew]

$$\left( \int_{k}^{k+1} \frac{dt}{t}\right)^2$$

[twin meteor]

<rajel />

[cunning yew]

how to do this

[bitter pilot]

compute the integral first

[cunning yew]

wdym by compute it ?

[bitter pilot]

calculate it

[cunning yew]

its clearly $\ln(k+1)-\ln(k)$

[twin meteor]

<rajel />

[bitter pilot]

abs value actually

[cunning yew]

they are postiive

k is integer

[bitter pilot]

so k > 1

[cunning yew]

now i should do $(\ln(k+1)-\ln(k))^2$ ?

[twin meteor]

<rajel />

[bitter pilot]

yes

[cunning yew]

$\ln(k+1)^2 -2\ln(k+1) \cdot \ln(k) + \ln(k)^2$

what now ?

@bitter pilot

[bitter pilot]

-2ln(k+1)ln(k)

[cunning yew]

yeah

[bitter pilot]

then you have what you wanted

[twin meteor]

<rajel />

[cunning yew]

no i want the integral symbol to stay

[bitter pilot]

what?

[cunning yew]

i want to find it without computing it

like having an expression inside the integral

instead of finding the value directly

like im expecting it to be: $\int \frac{dt}{t^2}$ or something like this

[bitter pilot]

no that's wrong

[twin meteor]

<rajel />

[bitter pilot]

you could do the following

[bitter pilot]

since we know this is the value, differentiate it

[cunning yew]

differentiate it ?

so ur saying to go the other way around

[cunning yew]

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[lost otter]

angle subtended at the centre is double the angle subtended at the circumference by the same arc

ang MOJ (obtuse) = 2 * 80
ang LOJ (obtuse) = 2 * 126

then work out x

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[atomic jasper]

<@&268886789983436800>

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[grizzled frost]

need help with this accounting assignment

.close

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[hearty delta]

Whats an example of a function thats injective but not bijective, or one thats surjective but not bijective

[bitter pilot]

What have u tried

[hearty delta]

Im just trying to understand the definitions

injective means each distinct x has a distinct y (x being inputs and y being outputs)

and surjective means each y has an distinct x

[bitter pilot]

y = x²
then y = 4 has x = 2 or x = -2 so surjective means that a y has at least one x

[hearty delta]

Ok so, this has two imputs for one output which means its not surjective

[bitter pilot]

not injective was correct

[hearty delta]

its not both right

[bitter pilot]

injectivity is basically uniquness

[bitter pilot]

the example i made f(x) = x² where f : R -> [0, oo) is surjective

[flat whale]

oo

[bitter pilot]

because for every y value you can find at least one x value that maps to it

[hearty delta]

Ok injective means for every unique input u have an unique output

so this is not injective

surjective means every output has only. 1 input

ohhh there is no only

it means there just is at least one input

[bitter pilot]

If you defined f : R -> R then it's neither surjective nor injective. For example, surjectivity fails because take for example y = -4 there doesnt exist one x-value that maps to it.
Injectivity also fails because both x = -2 and x = 2 map to 4

And if you define f : [0,oo) -> R then it's only injective but not surjective

each x value maps onto a unique y value

but not every y value is being mapped, e.g. the negative numbers

[hearty delta]

ok

[bitter pilot]

and f : [0,oo) -> [0,oo) would be bijective

can u think why

[hearty delta]

Because u can get every pos number as an output with a distinct input

[heavy yoke]

"for every unique input there is a unique output" is just a function. an injective function is the other way around, for every unique output there is a unique corresponding input

[hearty delta]

as there are no more negitive cases

Ok just to clarify

Injective : Distinct inputs ahve distinct outputs

Ie, two imputs do not map to the same output

[heavy yoke]

yes

[hearty delta]

Surjective: For every output there is at least 1 input

[bitter pilot]

yes

[hearty delta]

key being at least, which is why its diffterent from injective

and bijective means both are true

[bitter pilot]

yes

[hearty delta]

ok bet

That makes snese

sense*

I was missing the at least portion

so in ur example

y=x^2 on R -> R i not injective because u can have two inputs for the same output

[bitter pilot]

yes

[hearty delta]

Its not surjective becase the output can never be negative

but if we did y = x^2 on (0,inf) -> (0,inf) it would be both

[bitter pilot]

basically yea

[hearty delta]

as not there is only one out for each input, and every output has an input

Is another way of saying serjective that the range = codomain

[heavy yoke]

yes. or if you did R -> [0,inf) it would be surjective but not injective, or on [0,inf) -> R it would be injective but not surjective

[bitter pilot]

would add on that because for every negative number y in the codomain, there doesnt exist at least one corresponding element x of the domain

[hearty delta]

ok that adds up

Because from R -> [0,inf), multiple inputs give u the same output

and on [0,inf) -> R there is no way to get the negative numbers

ok bet, thanks for helping!

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[wraith hornet]

hi guys, ive been staring at this for a hot minute and i still do not have a singular clue of what exactly is going on

[wraith hornet]

bit of help would be appreciated

[odd helm]

68=7*10-2 yes?

[wraith hornet]

yes

[odd helm]

68 mod 7 = 7*10 mod 7 - 2 mod 7 = - 2 mod 7

ok?

[wraith hornet]

yes

[odd helm]

cube it

gets you -8

tell me what the next line you don't get is

[wraith hornet]

third from the bottom

[odd helm]

after the arrow?

[wraith hornet]

yes

[odd helm]

105 = 6*17+3

68^105 = 68^(6*17) * 68^3

[wraith hornet]

ah ok yeah i think i get it now

thanks

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[glad glacier]

hello im studying for a test and im wondering what would i do to solve these questions, i kinda know how to do inequalities but really only one step ones. im tryna practice two step then get to multip step. here's my question:

[quiet echo]

Try to solve the equation first.

As in, make the > sign an = sign.

Then, you (at least for this level of inequalities) can test to see if it is true for x>root or x<root.

[glad glacier]

oh i got it

i got less than 33/4

[reef agate]

,w solve for x, -4x + 35 > 2

[twin meteor]

Results provided by WolframAlpha

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Requested by .king.leo.

[reef agate]

✅

[glad glacier]

yup

now im tryna do 8x+2>34

i moved the constant to the right and step 2, i have 8x > 34 - 2 right now

34/8 lets see if im right

[reef agate]

Wait what

What happened to the - 2

[glad glacier]

subtracted 34

no wait

34 to 32

32/8 is what i meant to sayt

,w solve for x, 8x + 2 > 34

[twin meteor]

Results provided by WolframAlpha

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Requested by im_not_blake

[glad glacier]

hm?

oh yeah i got it right

LETS GOO I UNDERSTAND IT NOW

[reef agate]

!done?

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[glad glacier]

yeah

.close

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[hazy spindle]

I can already see them multiply by a negative number (since they are solving it as if it is an equation) and failing the test miserably.

[reef agate]

Eh theyre suppsode to test

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[echo plaza]

is this false?

[echo plaza]

i dont think its right bc we go to 2n

which is ofc greater than n

so im pretty sure some remainders will repeat

or actually it might be true from testing values

it gives all the odd remainders

no

[atomic jasper]

try for small values of n

[echo plaza]

well yeah, all odd except 0

[echo plaza]

yeah i have

and it does seem to work

but idk if im convinced

it seems to work though

[echo plaza]

is it true?

[atomic jasper]

i think its like filling up the holes

[echo plaza]

yeah

[atomic jasper]

like the first pass when youre still below n

[echo plaza]

first evens, then odds

[atomic jasper]

you fill all the evens

but showing it, im not sure

[echo plaza]

ty

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[cedar bone]

Certainly numbers <=2i are all distinct residues, the even residues specifically, with i=floor(n/2), and in fact 2i=-1 mod n. Then, 2(i+1)=2i+2=1 mod n, and the 2(i+a) for 1 < a < n-i are between 0 and n, and are all distinct. They're also all distinct from the numbers <2i since those are even and these are odd.

[echo plaza]

ty!

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[jagged cargo]

is there any tangible difference to using square bracket for matrix over parentheses?

[jagged cargo]

i have seen some people saying that square bracket is for physicians and engineer, while the latter is for mathematicians

[heavy yoke]

purely a style choice

[atomic jasper]

its the same

just preference

[jagged cargo]

alright thanks

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@worthy ferry Has your question been resolved?

[worthy ferry]

<@&286206848099549185>

[steep delta]

😟

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[worthy ferry]

.close

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[shadow surge]

Is there away to limit this function and transform it to another place?

[shadow surge]

And also shrink it

[jagged cargo]

for limit, you can put conditions

[flat whale]

e.g. { 0 < x < 1, 0 < y < 1 }

[jagged cargo]

to shift it around, apply horizontal and vertical shift

that is, adding constants to each x and y term

[shadow surge]

Ye but it doesnt work with this function

[jagged cargo]

probably because it's too complex

[flat whale]

show what you tried

[shadow surge]

Is there a way to move it around?

[shadow surge]

Wait actually it does now but i think it was lagging yesterday

[flat whale]

[jagged cargo]

f(x - a, y - b) shifts the function a unit rightwards, b unit upwards

[jagged cargo]

you can see desmos is even complaining the function is too complex to display all the details

it will take some time, but it should work

[shadow surge]

Where do we put this in?

[jagged cargo]

let's say whatever equation you have is x^2 + y^2 = 9

[jagged cargo]

to shift the function a unit towards the right, you minus each x term by a

(x - a)^2 + y^2 = 9

to shift the function b unit upwards, minus each y term by b

(x - a)^2 + (y - b)^2 = 9

do this with all x and all y

[shadow surge]

Oh ok

It workksksm

Thankk uu

[jagged cargo]

!done

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.close

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[civic otter]

!stAtus

[vocal sleet]

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

@hushed loom Has your question been resolved?

[hushed loom]

<@&286206848099549185>

[hollow sundial]

read above pls

(what step)

[vocal sleet]

@hushed loom Has your question been resolved?

[hushed loom]

the whole ques

step 2

ig

[halcyon ice]

Have you split it into multiple integrals to get rid of the absolute value?
(If not, do you know how to do that in the first place?)

[hushed loom]

i do

[halcyon ice]

Then do it here

itll get rid of the absolute value and just turn it into a pretty easy integral

[hushed loom]

yeah did

[halcyon ice]

so what does it look like now then?

(checking if you did it correctly)\

[hushed loom]

yeah yeah done now

thanks , took some to do the calc

but did it

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[mild trench]

.

i need help

[mild trench]

on part c

it doesnt make sense

i keep getting

153.4

but

the markscheme answer is

110

ive done everything

[outer warren]

show work

[mild trench]

i manage to work it out now its fine!

.solved

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[twilit needle]

So my first question already, the limit only is for 1/x?

[vast shale]

no

its for all

[twilit needle]

aaah alright

[vast shale]

because the top bound of the integral is x

and the integral is with respect of t

that's how you kno

w

[twilit needle]

hmmm okay oaky

i cant use FTC 1 ?

u only use that for derivatives right?

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@twilit needle Has your question been resolved?

[twilit needle]

<@&286206848099549185>

[atomic summit]

Do the integral

Then solve whatever’s left

[twilit needle]

💀

that scary thing?

alright ;-;

[atomic summit]

The integral is scary?

[twilit needle]

yes definitely for me at least

[atomic summit]

Okay.

Is it because it’s integrating a trig function?

[twilit needle]

and the power 1/t

we are only allowed to use sin cos tan and their inverses :/

and we havent done a lot of integrals so far yet

i think i have done like 30 or 40 in my life

[atomic summit]

Have you done Taylor series?

[twilit needle]

ye we had that last week

[atomic summit]

So this is a Taylor expansion

For a small t approximate tan2t

Which will just be 2t

[twilit needle]

hmm okay

so we have (1-2t)^1/t

is that this one with (1+x)^k ?

[atomic summit]

Does that look familiar?

[twilit needle]

not really yet

is this close to a maclaurin series?

[atomic summit]

(1+1/x)^x

The limit of this as x->inf is e

So I won’t bother to prove that

You can take that as a lemma

So when we inverse the variable inside the parentheses and the exponent we get e^-x

So what you end up with is e^-2 as t->0

So what you end up with is

$ \lim_{x \to 0} \frac{1}{x} \int_0^x e^{-2} dx$

Which is just e^-2

[twilit needle]

Hmmm

alright ill look into this

[atomic summit]

Easy

[twilit needle]

aaah okay i think i understand

thank you 🙂

[twilit needle]

no 💀

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[vast shale]

How to get the limit as x stands to infinity: ln(e^2x -e^x +1)-1-2x+1

[leaden ingot]

$\lim_{x \to \infty } \ln(e^{2x} -e^x +1)-1-2x+1$

This?

[vast shale]

X is above with 2

[twin meteor]

TargetVN

[vast shale]

Yes

I'm proving that the equation 2x-1 is a o.a

[leaden ingot]

Hint: $\ln(e^{2x} -e^x +1)=\ln\left( e^{2x}\cdot (1-\frac{1}{e^{x}}+\frac{1}{e^{2x}}) \right)$

[twin meteor]

TargetVN

[leaden ingot]

Can you continue from here?

[leaden ingot]

also -1+1 = 0, so the -1 and +1 cancel each other

[vast shale]

I got inf(1-0+0)-2x

So inf-2inf?

[leaden ingot]

$\ln(e^{2x} -e^x +1)=\ln\left( e^{2x}\cdot (1-\frac{1}{e^{x}}+\frac{1}{e^{2x}}) \right)=2x+\ln\left( 1-\frac{1}{e^{x}}+\frac{1}{e^{2x}}\right)$

[twin meteor]

TargetVN

[leaden ingot]

split them out

[vast shale]

Ohhhhh

Thank you

So ln(e^2x) is 2x

Thanks

[leaden ingot]

np

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[vast shale]

.close

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[twilit needle]

I think that I made a mistake

[twilit needle]

ill show my work

[rough patrol]

Don’t understand first line logic

[twilit needle]

isnt that IBP?

[rough patrol]

Oh

Shouldn’t it be minus

[twilit needle]

sin(t-s) = -cos(t-s)?

[rough patrol]

Oh wait double negative

[twilit needle]

yes

[vast shale]

what's the correct answer

[twilit needle]

uhm 1/2(e^t-cos(t)-sin(t))

[vast shale]

Alr, I'm getting just the negative of that

[rough patrol]

The calculus part is right

You just plugged in the bounds wrong

[twilit needle]

Hm?

[rough patrol]

You plugged in 0 and t wrong for s

[twilit needle]

i did?

[vast shale]

yeah

[twilit needle]

wait let me check

aaah yes

thanks guys

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[pallid stream]

in part (a) why does the bounds has to be like that for the area to come out as positive...? mine first come out as negative and i dont get why? and will i lose points if i wrote the bounds like the negtaive one and stick the absolute sign at the end to make it as an area?

[lone linden]

$\int^b_a f(x) \dd{x}=-\int^{a}_{b} f(x) \dd{x}$

[twin meteor]

Civil Service Pigeon

[lone linden]

Putting the bounds in the “reverse” order means that you’re considering the area in a clockwise direction rather than a counterclockwise one

Which has an analogous problem to when you’re finding areas in the Cartesian plane and work from right to left instead of left to right

[pallid stream]

i originally tried to trace the area by going from a smaller theta to a larger theta but since that would make the upper limit < lower limit that means I will need to stick the negative sign? so i should put the larger theta on top? trace from larger to smaller? to make it positive?

[lone linden]

i originally tried to trace the area by going from a smaller theta to a larger theta but since that would make the upper limit < lower limit

uh

[pallid stream]

what? whats wrong?

[lone linden]

How are you getting upper < lower?

Lower limit is where you start

Upper is where you end

You’re starting from a smaller angle, so that should be your lower limit

And the larger angle that you end at should be your lower limit

[pallid stream]

i was originally doing pi/6 to 5pi/6 and 5pi/6 to 13pi/6

[lone linden]

So then your integrals are from pi/6 to 5pi/6

And 5pi/6 to 13pi/6

[pallid stream]

thats how i thought its gonna be like smaller to larger is easy to think

[lone linden]

(a to b means a lower bound of a and an upper bound of b)

[pallid stream]

yea

now i know how to do it right

thanks!

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[vast shale]

hello,, if we suppose that (an) is a strictly positive number sequence. Let Sn be the sum of ak from 0 to n and we suppose that anSn -> 1. Show that the series of an diverges and that an -->0. I don't understand how to to do that pleas help me

[river arch]

We should probably start by writing all that we know down

[twin meteor]

dyxn

[vast shale]

thx for lateX didnt know how to write it

[river arch]

Let me try some stuff out

[vast shale]

the only thing I got so far is Sn*an>an^2>0

[potent beacon]

I guess a good way to start would be to assume Sa_n converges

[vast shale]

Sn or an?

[potent beacon]

$\sum_{n=1}^{\infty}a_n$

[twin meteor]

NotAPortalGuy

[vast shale]

ok thx

[potent beacon]

So Sn

[river arch]

Yeah I was able to build a proof pretty easily using contradiction for this

But I can't find a good proof for a_n --> 0

[potent beacon]

thats easy too

If this sum diverges

It must diverge to infinity

But

I think

Hm

Oh yes

It follows

[twin meteor]

dyxn

[potent beacon]

Compilcated it

If S_n converges a_n must converge to zero

[river arch]

True

[potent beacon]

So lim (an)(sn)=(liman)(lim sn)=0(lim sn)=0

[vast shale]

thx I think its good

[river arch]

But it still doesn't prove that a_n --> 0

[potent beacon]

Yes

[vast shale]

yes but we made one step

[potent beacon]

But when we know Sn diverges

And to infinity

We can prove it directly from the definition of the limit

[river arch]

If S_n diverges to infinity you can't say anything about a_n though

[potent beacon]

I mean not directly

But you can from this

[river arch]

oh yeah

[vast shale]

ok I have it I guess

[river arch]

Yeah you can probably just do it with limit rules

[potent beacon]

Here?

a_n could be divergent

So they dont work

[river arch]

No I meant with the epsilon-N thing

mb not the algebraic limit theorem

[vast shale]

if Sn diverges, let A be a stricly positive real number, then it exists N, like for every n>N; an|Sn|>Aan then an(sum of ak)>Aan, which is possible only if an converges to 0

is it good or not?

[river arch]

I don't get what you're doing

[potent beacon]

Fr

[vast shale]

why?

[river arch]

Isn't that basically what I did but you've switched some words around

[vast shale]

yes a bit but now we know that Sn diverges

so the only assumption is that an converges towards l different from 0. I can do it more cleanly by separating in the case an diverges and the case an converges

and if its different than 0, then it's contradictory

[potent beacon]

an(sum of ak)>Aan, which is possible only if an converges to 0

I dont get this sentence at all

Or rather part

[vast shale]

well an(sum of ak) is an Sn so it converges to 1, but the inequality is true for every A

so the only case in which its possible is that A*an <1 for every A, so an must go to 0

idk if its clear, perhaps my understanding is wrong but to me it makes sense

[potent beacon]

I guess that makes sense

Yes

Definitely

[vast shale]

ok thx!

[vocal sleet]

@vast shale Has your question been resolved?

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[heady atlas]

Guys help me solve 3

[heady atlas]

Anyone ?

[sly mountain]

if you consider zooming in i might help you

[heady atlas]

The picture is pixelated ?

Wait

I can’t zoom in cuz I got the picture from my friend he sent it to me