help-10

[scenic fjord]

Actually I got the form already

[jolly ginkgo]

just notation used fornormal integration

[scenic fjord]

Just that I don’t know why f(x) and g(y)

And also

Why I got an extra x

🙃

[jolly ginkgo]

dw you haven't done the previous part right too

I'll tell you

[scenic fjord]

Which parts I’m wrong

[jolly ginkgo]

when you integrate partially the constant factor is actually function of x

let me write it down

[scenic fjord]

Can you point out where?

The first part isn’t should be function of y

Why it’s not A(y)

But it’s A(x)

[warm shale]

fäf

[scenic fjord]

Oh, if we don’t have x part, we need to write constant this way?

[jolly ginkgo]

yes

[scenic fjord]

Then why not A(x) x?

Isn’t we do integration again?

[jolly ginkgo]

yeah let me make that clear too

[scenic fjord]

I mean why not like this

[jolly ginkgo]

yeah i know

you are integrating A(x)

not a constant to get x

integration of A(x) gives us f(x)

[scenic fjord]

Why integrate A(x) get f(x)

[jolly ginkgo]

that's just a notation they used

[scenic fjord]

And A(x) is not a constant?

[scenic fjord]

I didn’t learn this before

[jolly ginkgo]

no it's not

[jolly ginkgo]

there's nothing to learn

[scenic fjord]

Why it’s not a constant? Because we haven’t consider x case?

[warm shale]

fäf

[jolly ginkgo]

we integrated partially with respect to y, right? so there will be a function of x

[scenic fjord]

Oh, different notation is it? Because it’s u(x,y), not just u(x)

[jolly ginkgo]

as i showed here

[scenic fjord]

Anyway, for Yc, am I getting the correct answer?

Because answer given us sin(-2y) instead of sin(2y)

I think mine is correct but just in case there anything I didn’t wrongly

[jolly ginkgo]

yeah yours may be correct

[scenic fjord]

Then the answer wrong I think

[jolly ginkgo]

let me check

,w y"+6y'+13y=0

[warm shale]

Results provided by WolframAlpha

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

Requested by yes sam#5028

[jolly ginkgo]

yeah you are right

[scenic fjord]

Oh, can check like this

Just realised

Thanks for your help

[jolly ginkgo]

you're welcome dear ❤️

[scenic fjord]

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[obtuse pebble]

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[oblique thunder]

Im lost. How do i solve this

[oblique thunder]

Here's the clear equation, sorry

In my attempt, i used e^x as my u but when i finally apply the formula the 2+ is left out

do i need to separate it as 2 over e^x + e^x

[shy vigil]

i might be wrong but

can't we just

$\begin{align*} \int \left( \frac{2+e^x}{e^x} \right) dx &= \int \frac{2}{e^x} dx + \int dx \
&= 2 \int e^{-x} dx + \int dx
\end{align*}$

wow. ok

[warm shale]

n11
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[shy vigil]

fine

just

int 2/e^x + int 1

[oblique thunder]

bahaha ive ran out of brain juice to work this one out

[shy vigil]

-2e^(-x) + x + C

is that the answer by any chance

[oblique thunder]

it says so on my book, yes

[shy vigil]

oh wow

so do you see what happened there?

[oblique thunder]

nope. it's an exercise and the book only shows the final answer

[shy vigil]

It's like $\frac{a+b}{c}$ right? I just wrote it like $\frac ac + \frac bc$ and then integrated separately

[warm shale]

n11

[shy vigil]

godddd

ok

[oblique thunder]

well, i know the integrate of 1, it's x. can you guide me through the 2 over e^x

[shy vigil]

Of course!!

[warm shale]

n11

[oblique thunder]

may i ask how did the negative got in front of the 2e^-x

[steep flax]

Substitute $u=-x \implies du=-dx \implies dx = -du$:
$$ -\int e^{u}, du $$
Integrate
$$ -e^u + C $$
Back-substitute $u=-x$
$$ -e^{-x} + C $$

[warm shale]

waris

[steep flax]

Remember to multiply by the 2 in front of it.

[oblique thunder]

thank you @steep flax @shy vigil it's really confusing when i get to scramble diff ways of solving a problem and overcomplicating v simple like this one. i really appreciate it

.close

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[scenic fjord]

Maybi know how we integral this? Not just sub in formula? But I got 2pi h instead of just pi h

Thought that we are using this

[jolly ginkgo]

yes

[scenic fjord]

But why isn’t it has 2

[jolly ginkgo]

2?

[scenic fjord]

I did this

Because lamda is the whole part behind negative and before x^2

[jolly ginkgo]

yeah

so what are you trying to say?

[scenic fjord]

Why answer no 2

[scenic fjord]

Can you see the difference for mine and the solution?

[jolly ginkgo]

there's an extra 2

your handwriting is nice

@scenic fjord shouldn't the 2 go when you square

[scenic fjord]

huh? Like how?

[jolly ginkgo]

what is realtion of $\phi_0$ and $1$

[warm shale]

fäf

[scenic fjord]

?

[jolly ginkgo]

what is the relation of cri and 1

[scenic fjord]

I don’t know 🌚

[jolly ginkgo]

$\pai_0$ and $1$

[warm shale]

fäf
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[scenic fjord]

I just wanna know why there isn’t 2 inside

Like I just sub the formula

And it’s wrong

[jolly ginkgo]

yoou need to give some essential info in order to get a response

[jolly ginkgo]

tell me the formula

that's what I'm asking for

[scenic fjord]

This is the formula

I shown

[jolly ginkgo]

not that one

the formula after you get value of $\psi_0$

[warm shale]

fäf

[scenic fjord]

I don’t have any formula after getting this

That’s what I got in my textbook

Or you mean

The formula before I do normalisation?

[jolly ginkgo]

How did you get from that step to the next step?

there is no psi in next step

[scenic fjord]

What? I’m doing normalisation

That’s why I only send the integral

Because that the part only related to the question I’m asking

[jolly ginkgo]

how did you do that?

we do normalization by $\int_{-\infty}^{\infty} |\psi_0|^2 dx = 1$ right

[warm shale]

fäf

[jolly ginkgo]

then obviously the 2 won't stay in denominator of power of e

it gets cancelled out by square

[jolly ginkgo]

this was what I was asking for whole time

[scenic fjord]

This is normalisation, textbook show me

I just don’t know why don’t have 2

That’s my main question

Erm , internet got problem just now

[scenic fjord]

Actually I don’t really und about normalisation

[scenic fjord]

May I know why it’s cancelled out?

[jolly ginkgo]

tell me what you get by this

[scenic fjord]

Something like this

[jolly ginkgo]

No

[scenic fjord]

That’s why I said I don’t und normalisation

[scenic fjord]

Oh okay, maybe it’s the same

Maybe is something like this

I think I got it

[obtuse pebble]

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[brisk peak]

Just wanted to know what this relationship is - is it a commonly used formula in structure material studies ? I am using it to solve a euler-bernoulli equation about buckling of a tower.

[obtuse pebble]

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[obtuse pebble]

@brisk peak Has your question been resolved?

[obtuse pebble]

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[autumn kiln]

Can someone help pls

[glass dagger]

What is the question?

to figure out m?

[autumn kiln]

Yes

[glass dagger]

Do you know about derivatives and how they give you the slope of the tangent line?

And do you know that two lines are perpendicular precisely when the product of their slope coefficients is -1?

[autumn kiln]

Yeah

[glass dagger]

Let's define f(x) = 25 - 3x^2 and g(x) = mx - 3 for some m. If the graphs of these functions meet at a point (x0, y0) such that they're normal to each other, then what sort of equations can you form?

[autumn kiln]

Simultaneous?

25-3x^2 = mx -3

[glass dagger]

If x is the point at which we have this normal intersection then yes, that's one equation that must hold: the graphs must intersect

but what else? it's not enough that they intersect, they have to also be normal to each other

[autumn kiln]

6mx=1

[glass dagger]

exactly

Now you have two equations and two unknowns: x (the point of intersection) and m (the slope you're trying to find)

You might even find several solutions

[autumn kiln]

great I’ll solve them now👍

I’m not sure if I’ve done this right

[glass dagger]

what did you do here?

[autumn kiln]

I replaced mx with 1/6 from 6mx=1

[glass dagger]

oh yeah true

seems right to me

visually it looks very convincing so well done

[autumn kiln]

I got srt18/6srt167 but that’s not the same as the answer in the mark scheme

they said the answer is root(1/334)

Is that the same thing

[glass dagger]

yeah that's the same thing because 36 * 167 / 18 = 334

[autumn kiln]

Ohh I see

Thanks for helping me!

[glass dagger]

no problem! :)

[autumn kiln]

.close

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[past walrus]

can someone explain to me how its -27x^2?

like how is it not -9x^2?

Because like if they did the prblm like this shouldn't 18x^3 be 54x^3?

[obtuse pebble]

@past walrus Has your question been resolved?

[past walrus]

<@&286206848099549185>

[chilly lotus]

Seems like an error. Coz it is supposed to be 9x^2

[timid silo]

yeah the teacher made an error

[obtuse pebble]

@past walrus Has your question been resolved?

[obtuse pebble]

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[wide hearth]

I am super lost on how to do th8s

This

[tardy epoch]

the picture of your screen is unreadable

i suggest screenshot and upload

[wide hearth]

Ok

is this better @tardy epoch

[obtuse pebble]

@wide hearth Has your question been resolved?

[wide hearth]

<@&286206848099549185>

[obtuse pebble]

@wide hearth Has your question been resolved?

[wide hearth]

<@&286206848099549185> can I have help please

[glacial obsidian]

You should find the average value of the unpaid balance in the 30 days

For example on day 1 the unpaid balance was 742.11

On day 15 it was 742.11 + 44.62

Et cetera

So you find the average value of the unpaid balances on days 1, 2, 3,... until 30

[obtuse pebble]

@wide hearth Has your question been resolved?

[wide hearth]

Help please

[obtuse pebble]

@wide hearth Has your question been resolved?

[wide hearth]

<@&286206848099549185>

[tardy epoch]

.close

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[spice zenith]

Help me I don’t know how to do this

[spice zenith]

.close

[obtuse pebble]

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[inner spindle]

How do i find 27 and 14 on this

[inner spindle]

<@&286206848099549185>

[amber wedge]

Mf what am I looking at.

[inner spindle]

is the photo blurry

[grizzled shore]

Don’t immediately ping helpers

and show the question

[amber wedge]

Choc / total number of cookies.

[inner spindle]

Is this clearer

[grizzled shore]

Don’t cut out the words at the top

[amber wedge]

What class is this.

Who would want someone to do this kind of trig

[inner spindle]

Geometry kap im a freshman

its not trig just a matter of finding angles to find the rest of the other angles

[amber wedge]

What does kap stand for

[inner spindle]

idk its like an advanced class

[amber wedge]

What year

[inner spindle]

so we get harder questions

[inner spindle]

im freshman

[amber wedge]

Ah

I’d rather go through chemo then do that kind of shit

[inner spindle]

do u know how to find 27 and 14 tho?

@amber wedge

@grizzled shore

<@&286206848099549185> pls i wanna go to sleep 😭

[timid silo]

how can I help you ?

[inner spindle]

how to find 14 and 27

[timid silo]

you mean angle MFE and angle MEF, correct ?

[inner spindle]

Yes sir

[timid silo]

alright

for 14 its simple

you got the measure of the angles 15 and 16, right ?

[inner spindle]

oh my god how did i miss that

yeah i see it

i get 110 for 14 and then 65 for 27

@timid silo thank you i love u

[timid silo]

It's normal you just said need to sleep 😂

[timid silo]

no problem happy to help

[inner spindle]

thanks im gonna sleep now

i got a test over this tmr lol

[timid silo]

sweet dreams

[inner spindle]

thanks bro

[timid silo]

any time

[obtuse pebble]

@inner spindle Has your question been resolved?

[obtuse pebble]

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[lofty hare]

Could someon help with (c)?

[obtuse pebble]

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[autumn kiln]

Can someone help me with part c please

I got this as E but I think it’s wrong

[obtuse pebble]

@autumn kiln Has your question been resolved?

[autumn kiln]

<@&286206848099549185>

[obtuse pebble]

@autumn kiln Has your question been resolved?

[autumn kiln]

<@&286206848099549185>

[obtuse pebble]

@autumn kiln Has your question been resolved?

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[celest helm]

This isn't a question about a math problem more of a question about math

[celest helm]

So im trying to learn anti differential and integrals

can you recommend me a place like a website or videos of it

and i also hope its not questioning me with some basic integrals

I need the hard ones .-.

[median yacht]

Are you starting with integrals?

[celest helm]

not really, i just need to practice on it cause im failing my calc classes

i could solve the basic one's but my prof is always one step ahead of my class and gives very hard questions

and hehe i hope if ur recommending me a site or service its free

since hehe broke college student

[grizzled shore]

i think professor leonard has really really good videos on calc if you want to try them out

https://www.youtube.com/watch?v=fYyARMqiaag&list=PLF797E961509B4EB5

each episode is titled with the topic so you can just kinda skip around to what you need help with i suppose

[celest helm]

ohh thankyouu ❤️

[grizzled shore]

i learned calculus off this man too

absolutely jcaked dude as well

[celest helm]

I'll just leave the chat open after 10min if ever other people would want to recommend

[grizzled shore]

👀

[celest helm]

im not into guys... 😶

HAHAHHAHAHAHHAHA

[grizzled shore]

neither,. but is he pretty jacked

[celest helm]

.close

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[ivory cargo]

I need help showing the following series limit

[obtuse pebble]

@ivory cargo Has your question been resolved?

[fluid snow]

Does n go to infinity?

[ivory cargo]

yes i assume so

but it it like sectionsums

its a series S_n of sectionssums

[fluid snow]

Then this has been asked on Math StackExchange before.
https://math.stackexchange.com/questions/2986785/how-to-find-a-limit-of-this-sequence-lim-limits-n-to-infty-sum-limits-k?noredirect=1
https://math.stackexchange.com/questions/482003/how-to-evaluate-the-sum-frac1-sqrtn-frac1-sqrt2n-cdots-frac1?noredirect=1
https://math.stackexchange.com/questions/2440224/what-is-the-value-of-lim-limits-n-to-infty-left-sum-limits-i-1n-frac?noredirect=1

[ivory cargo]

okay thanks

[rich token]

Can some one teach me how to do differentiation and Integration ?

with Example

Plz

[obtuse pebble]

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[modern ravine]

Can anyone tell me where the first line comes from?

Why is b/a = 2a/b?

This is from an exercise that I don't have because it was from a test. It's something about proving the ratio of the measurements of the DIN A4 paper size standard

Which are these, with the following ratio

Which is also the sqrt of 2

[obtuse pebble]

@modern ravine Has your question been resolved?

[obtuse pebble]

@modern ravine Has your question been resolved?

[knotty gulch]

paper has a property that if you stack them together along the long edge, the new long edge (2a) to the new short edge(b) is equal to the ratio of the long to short

basically $\frac{b}{a}=\frac{2a}{b}$

[warm shale]

[code{RED}]

[knotty gulch]

that ratio you have is very close to

,w \sqrt{2}

[warm shale]

Results provided by WolframAlpha

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Requested by [code{RED}]#8104

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[scenic fjord]

May I know why it’s a linear superposition and how it works out from first step to second step?

[haughty coyote]

Split/distribute things appropriately to make the eigenstates appear using tbe double angle formula

[scenic fjord]

May I know which formula?

[haughty coyote]

Very first one

[scenic fjord]

But there’s a 1+cos theta

[haughty coyote]

I said distribute

[scenic fjord]

Let me try is this what you mean

[scenic fjord]

Like this right?

[haughty coyote]

Looks good

[scenic fjord]

Okay, thanks for your help

.close

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[obtuse pebble]

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[fierce vale]

Find a,b,c are the zeroes of the polynomial P where c = a+b

[ruby path]

Use Vieta's formulae

[fierce vale]

$a+b+c=4 \Rightarrow c=2$

[warm shale]

adamchebil33

[ruby path]

Now you can just long divide

[fierce vale]

long divide what?

x-2?

[ruby path]

(x - 2) is a factor

Yes

[fierce vale]

I managed to get a system of equations instead

[ruby path]

I guess that works too

¯_(ツ)_/¯

ab + (a + b)^2 = 6
ab(a + b) = 4

This?

I'd rather do the long division

[warm shale]

adamchebil33

[ruby path]

What

[warm shale]

adamchebil33

[ruby path]

$$ab + bc + ca = 6$$
$$ab + c(a + b) = 6$$
$$ab + (a + b)^2 = 6$$

[warm shale]

NEONPerseus

[fierce vale]

yeah that's what I did

but c = 2

so ab = 6-4

[ruby path]

Oh right

Mb you're good

[wooden cipher]

This looks like ((x-1)^4-1)/x, idk if that helps

[fierce vale]

I got 2 values for b though

I got quadratic equation when I solved the system

[ruby path]

One of them would be for a and the other for b

Since the equations you get are symmetric

[fierce vale]

what does that mean?

[ruby path]

You can replace a with b and b with a, and still have the same equation

I think symmetric is the wrong word

Cyclic I believe

[wooden cipher]

Symmetric is fine

[fierce vale]

how do i prove that though

[ruby path]

Just replace a with b and b with a

Or do it again with a and realize you get the same thing

[wooden cipher]

For two variables, cyclic should also be fine i think

[ruby path]

I'd prefer cyclic

I think the group theorists have that word trademarked

[fierce vale]

oh ok

but how do i solve it using long division?

[ruby path]

Do you not know how to long divide polynomials?

[fierce vale]

ik

[ruby path]

Yeah then just do P(x)/(x - 2)

[fierce vale]

oh ok got you

then i get quadratic

and solve for a and b

[ruby path]

Yes

[fierce vale]

it's way simpler that way

thank you

[ruby path]

,w divide (x^3 - 4x^2 + 6x - 4)/(x - 2)

[wooden cipher]

Ooh you can use this to your advantage by using diff of squares

No vieta needed

[ruby path]

I think you'll get a perfect square

That's nice

[warm shale]

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Requested by NEONPerseus#2128

[ruby path]

Smart man

[fierce vale]

?? how

,w develop (x-1)^4

[warm shale]

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Upgrade to WolframAlpha Pro!

Requested by adamchebil33#3772

[wooden cipher]

$\frac{(x-1)^4-1}{x} \newline
\frac{(x-1)^2-1)((x-1)^2+1)}{x}\newline
\frac{(x^2-2x)(x^2-2x-2)}{x} \newline
(x-2)(x^2-2x-2)$

[warm shale]

GarlicBredFries

[fierce vale]

oh ok got you

how did uk that it's (x-1)^4-1

[wooden cipher]

Hanging paren in line 2 😔

[fierce vale]

do you memorise (a-b)^4?

[wooden cipher]

Instinct

It looks like the fourth row of pascals triangle

Practice more and you recognize it

[fierce vale]

hopefully one day i'll be able to recognize it 😅

ty guys bye

.close

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[brittle drift]

can someone help me solve 26(ii)-

[tardy epoch]

try drawing the triangle that represent the given equation

[brittle drift]

then?

[tardy epoch]

show your drawing

[brittle drift]

ah

actually ty for this but imma just show it to my teacher🏃‍♂️😭

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[hexed apex]

What I've done so far is equate the area to the perimeter and try to simplify, but I got stuck as you can see in bottom left

[obtuse pebble]

@hexed apex Has your question been resolved?

[obtuse pebble]

@hexed apex Has your question been resolved?

[obtuse pebble]

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[marsh maple]

There is triangle ABC. we have X, Y, Z points on the sides of triangle, such as AZ CY and BX are concurrent. In triangle XYZ we have D, E, F such as XF ZE YD are concurrent and we have to proof that BF CD and AE are concurrent

[marsh maple]

<@&286206848099549185>

[timber island]

!15mins

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

@marsh maple Has your question been resolved?

[obtuse pebble]

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[velvet briar]

graph the ellipse. identify the coordinates of vertices and foci. x^2/9 + y^2/16 =1

[wooden cipher]

cool problem

whats your question

[velvet briar]

i need explantion on the steps like i understand but not quite

like how can i solve it

also whats c exactly?

[wooden cipher]

things are labeled differently, what is the name for "c"

[velvet briar]

c^2 = a^2 -b^2 whats c in the formula? where can i graph it?

[wooden cipher]

ok, lets start with this: how is the equation of an ellipse usually written?

[velvet briar]

x squared / a squared + y squared / b squared

thats horizontal

the vertical is the otehr way

[wooden cipher]

do you think the ellipse is vertical or horizontal?

[velvet briar]

vertical

[wooden cipher]

ok, so what is your a and b?

[velvet briar]

a is 16, b is 9

[wooden cipher]

a^2 is 16, b^2 is 9

so that means

[velvet briar]

oh yeah root

ok a is 4 b is 3

[wooden cipher]

so you can find c

[velvet briar]

okay yeah but what is c called do you know?

[wooden cipher]

i dont

[velvet briar]

okay thank you

[wooden cipher]

youre welcome

[obtuse pebble]

@velvet briar Has your question been resolved?

[obtuse pebble]

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[hallow hatch]

could anyone please help me with this? im not sure what to do

[hallow hatch]

anyone? <@&286206848099549185>

ill ask again later.

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[lean spear]

not too sure what kind of answers they are expecting here

[ancient ridge]

i think in the first parenthese for the limit i think they want you to put in the equation you would use

so like for the first one with the limit approaching 1 from the left, what equation would you use?

[lean spear]

I see

[ancient ridge]

yeah you got it

good job

[lean spear]

functions are continuous when they = 0

[ancient ridge]

yeah exactly

do you have any other questions?

[lean spear]

ya, are they expecting me to plug in a number for x here?

is a function discontinuous when it has parameters like the -1 <= x < 1

[ancient ridge]

not necessarily

so with this you might have to test a couple values to make sure each equation links with each other

[lean spear]

< and > are open circles so are those discontinuity points?

[ancient ridge]

they could be, but it isn't necessarily an indicator

just because they could be, doesn't mean they always are

do you know how you would test continuity with this?

[lean spear]

umm, maybe I plug in a number less than -1 for the 1st function

[ancient ridge]

no you don't have to do that

so you see how the bounds of the first function is x < -1

and how the second equation has the bounds -1 <= x < 1

[lean spear]

yes

[ancient ridge]

these two functions are "suppose" to connect at x = -1

but we aren't sure if they are connected

so we should put in x = -1 into both equation to see if they have the same y value

does that make sense?

[lean spear]

ok ill do that

[ancient ridge]

if they both have the matching y values, that means we know it continuous

[lean spear]

so 6(-1)^2 gives me 6, and 6(-1) = -6

[ancient ridge]

okay, so do you know what that means?

[lean spear]

they do not = each other

meaning discontinuity

[ancient ridge]

yeah so there is going to be a jump discontinuity at x = -1

okay, so applying the same idea

what do you think we should do next?

[lean spear]

plug in x = 1 into the 6(x) and 6/x. giving me 6 and 6

[ancient ridge]

yeah

good job

so what does that mean

[lean spear]

so they are continuous

[ancient ridge]

yes

you got it

[lean spear]

so now I need to visualize the direction it is continuous from

[ancient ridge]

okay, so you already calculated some of the points

i would focus on the points that you've already calculated, and use that as a guide

[lean spear]

so it looks like the correct graph is the one on the right because there is a "jump" from 6 to -6 and there is an open circle for the 6 because of the <

[ancient ridge]

yeah you're correct

[lean spear]

meaning continuous from the left

ope

thats wrong

[ancient ridge]

the graph is correct, but the continuity reasoning is incorrect

[ancient ridge]

your arrows are pointing rightward meaning that it should be?

[lean spear]

cont. from left would mean its coming from the left and going to the right or no?

[ancient ridge]

oh right my bad

my brain went to mush

[lean spear]

maybe the answer is neither because of the jump on the same x value

[ancient ridge]

yeah exactly

my bad i thought we were looking at 1, but we should be looking at x = -1

so yes it would be neither because there is a jump discontinuity

[lean spear]

wow

that was wrong too

[ancient ridge]

wait really?

[lean spear]

yes

[ancient ridge]

we are analyzing x = -1 right?

[lean spear]

yes

[ancient ridge]

or is it a different value?

oh

hold on lemme stare at the graph again

[ancient ridge]

oh that is wonky wording

so then it has to be the right

i think it's because as the limit approaches infinity it is still continuous

[lean spear]

Ok so for the first function I plug in 2 and I get 4. I set the second function = to 4 and plug in 2 for x and that gives me 3 = 4a - 2b

and then I set that 4a - 2b = to 4x - a + b. plug in 3 for x to have 12 - a + b = 4a - 2b

that becomes 12 + 3b = 3a

[ancient ridge]

okay so this first step is correct

the next part is incorrect

you see how the 2nd and 3rd equation connects at x = 3

[lean spear]

ahh ok, I plug in 3 for the 2nd and 3rd equation

[ancient ridge]

yeah that's right

and then what do you think we do next with those two equation?

[lean spear]

set them = to each other and solve for a and b

[ancient ridge]

yeah

let me know what you get

[lean spear]

I'm at 10a = 12 + 4b

[ancient ridge]

okay good

so you know how earlier we had 4a-2b = 3

when x = 2

[lean spear]

yes

[ancient ridge]

you can use that equation and the one you just got to make a system of equations to solve for a and b

does that make sense?

[lean spear]

yes but am not too sure how that will be set up

[ancient ridge]

instead of 12 it should be 11

hold on lemme show you

[lean spear]

yes I understand why

just saw it

[ancient ridge]

\begin{cases} $10a - 4b = 11\ 4a - 2b = 3$ \end{cases}

[warm shale]

tkumi
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[ancient ridge]

do you know how to solve this?

[lean spear]

I used to,

[ancient ridge]

okay do you know where to begin with this?

[lean spear]

get variables by themselves

like 10a = 11 + 4b

[ancient ridge]

yeah you could do it via substitution

so get a by itself with that

and then plug it into the other equation

[lean spear]

is there another way?

[ancient ridge]

yeah

so it's called elimination method

do you know how to do that?

[lean spear]

I've probably done it before but don't remember it right now

[ancient ridge]

elimination might be easier than substuiting

[lean spear]

ya I agree

[ancient ridge]

okay so with elimination we want to get at either a and b to be matching, so we can get rid of it

so with this i'm going to focus on getting rid of b since i can find a common LCM between 4 and 2

so to get the second equation to get 4b, i would multiply the whole second equation by 2

[lean spear]

one sec

what do we do to the 1st equation?

[ancient ridge]

we keep it as is

[lean spear]

ok

[ancient ridge]

does that make sense?

[lean spear]

almost

I know what ur trying to get at

[ancient ridge]

ah

[lean spear]

but unsure how it looks

[ancient ridge]

hold on

it should be something like this \begin{cases} $10a - 4b = 11\ 8a - 4b = 6$ \end{cases}

[warm shale]

tkumi
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[ancient ridge]

do you see how i got here?

[lean spear]

yes we multiplied by 2 to make the -4b a common number between the 2 equations

[ancient ridge]

do you know what should happen next?

[lean spear]

I'm thinking get one of the variables by itself

not sure what it will be

[ancient ridge]

okay so with elimination method we are able to get one variable by itself by subtracting from top to bottom

[lean spear]

oooooo

[ancient ridge]

so i would go $10a - 8a$

[warm shale]

tkumi

[ancient ridge]

then $-4b + 4b$

[warm shale]

tkumi

[lean spear]

2a = 5

[ancient ridge]

then $11-6$

[warm shale]

tkumi

[ancient ridge]

right

[lean spear]

so a = 2.5

[ancient ridge]

so then a would equal to 5/2

yeah

[ancient ridge]

so then we can take 5/2 and plug it in to either of those equation

[lean spear]

now that I know one variable

yes

[ancient ridge]

either of those equation should work

lemme know when you got b

[lean spear]

14 = 4b

14/4 = b

[ancient ridge]

yeah

you can reduce it 7/2

[lean spear]

or 3.5

[ancient ridge]

yeah

[lean spear]

wow

[ancient ridge]

so you got your a and b

does that make sense?

i think something that needs to be reviewed again is system of equations

[lean spear]

yes thank you so much

[ancient ridge]

but besides that i think you got the whole limit thing down

good job

[lean spear]

how many other methods are there?

elimination seems like it will end up being the easiest

[ancient ridge]

for system of equations i know the two commons methods are

elimination and substitution

both are great methods

however at certain times sometimes elimination is easier to work with

or vice versa

its good to know both either to double check your work or have an easier method of solving them

[lean spear]

yes, will practice this some more

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[obtuse pebble]

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[obtuse pebble]

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[sonic jungle]

Calvin's car is guaranteed for 30 days or 4,500 miles (whichever comes first). He bought the car on July 7 with 38,472 miles. On August 7, the odometer reads 42,950. Is the guarantee still in effect?

[sonic jungle]

Bruh

[obtuse pebble]

@sonic jungle Has your question been resolved?

[meager sierra]

I'm pretty sure the guarantee is no longer in effect. Calvin drove the car for 4,478 miles in the 30 days since he purchased it. Since the guarantee expires after either 30 days or 4,500 miles, whichever comes first, and since 4,478 is less than 4,500, the guarantee has expired.

[obtuse pebble]

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[neat heart]

some graph theory help

[neat heart]

"let u and v be two vertices of degree 3 in a graph G of order 6, if the degrees of the vertices of the graph G - {u,v} are 2,1,1,0 then what is the size of G

im not even really sure how to start this problem, ive been trying to find a video that would explain what its asking but all the videos im seeing just go over basic graph theory

[obtuse pebble]

@neat heart Has your question been resolved?

[frosty spoke]

hmm well you could solve it by drawing out an example graph that does it

I think you could probably get away with just adding the degrees together and dividing by 2

[obtuse pebble]

@neat heart Has your question been resolved?

[neat heart]

which would be 5

because

u and v have degrees of 3

the remaining vertices have added degrees of 4

because 2+1+1+0 = 4

[frosty spoke]

I don't think that works

[neat heart]

so then 3 + 3 + 4 is 10

[frosty spoke]

when you add u and v, you're changing the degrees of the other vertices too

you need to be a little smarter about it I think

[neat heart]

uhh

then how would i approach this

[frosty spoke]

try drawing an example graph out

[neat heart]

im not really sure how to draw it out

im confused on how that doesnt work though

cus it says if the degrees of the vertices of the graph G - {u,v} is 2,1,1,0

which is 4,

the degrees added up would be 10

[neat heart]

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[polar nexus]

how do I translate this to logical expressions
The reciprocal of every positive number is also positive.

[polar nexus]

is this correct ∀x((1/x)>0)?

[frosty spoke]

no

[timid silo]

for all x: x > 0 => (1/x > 0)

[polar nexus]

so ∀x(x>0 → 1/x>0)

thanks!

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[polar nexus]

.reopen

[obtuse pebble]

✅

[polar nexus]

hi again
how do I do that for this
There is no smallest number.

∃x ∀y (x < y)?

[smoky slate]

The other way around, for every x there's a y that's smaller

[polar nexus]

∃x ∀y (x > y)?

ooh then inverse it

~∃x ∀y (x > y)

[smoky slate]

Either negate it or reverse the inequality, but not both

[polar nexus]

I dont know how to negate inequalities
so
∃x ∀y (x > y)?

im confused

[smoky slate]

That says there's an x that's larger than any number

~∃x ∀y (x < y) means there's no x smaller than any y

[polar nexus]

what suppose to represen the smallest number

x or y

[smoky slate]

It'd be x

[polar nexus]

I dont know the de morgan law for inequalities

do they flip like the ~(pvq) = ~p^~q

[grizzled shore]

shouldn't you do the forall first

[polar nexus]

s0 ∀x ∃y (x > y)

[smoky slate]

Then, ∀x ∃y (y < x)
For all x there's a smaller y, ie no smallest number

[grizzled shore]

cos you want to say no matter what x, there's always a number y that's smaller

the other way round sounds a bit weird

[polar nexus]

ooh that makes more senes

thanks

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[onyx lodge]

How do I do this?

do I set up an inequality? I tried that but nothing got me to "therefore x must be an integer"

idk what do to

[grizzled shore]

show that x is an integer => 0.141x + 0.859x = x

and then show show that 0.141x + 0.859x = x => x is an integer

[onyx lodge]

wait do u mean floor(0.141x) + ceil(0.859x)?

but how do i show either of those

[grizzled shore]

oh yeah that

hmm, have you done much with floor/ceil functions?

[onyx lodge]

not really

im revising for finals

i just know what floor and ceil mean

and like floor(x)<x ceil(x)>x but thats about it

[wild swallow]

that is not quite right

floor(1) is not < 1

[grizzled shore]

im not sure how you can do this step without manipulating floor/ceil functions in some way, maybe someone else has a better idea

[wild swallow]

you need inequalities on what the values of floor and ceil can be

in particular, one side will be
[
\floor x \le x \textqq{and} \ceil x \ge x
]

[onyx lodge]

oops <= => i mean

[wild swallow]

can you bound it from the other side?

[
{?} < \floor x \le x
]

[wild swallow]

fill in the ?

[onyx lodge]

x-1?

[wild swallow]

yes

what about ceil

[onyx lodge]

x+1

[wild swallow]

good

so we can apply it to the question

what are the bounds for each of the terms

floor(0.141x) and ceil(0.859x)

[onyx lodge]

0.141x-1 < floor <= 0.141x
0.859x <= ceil < 0.859x + 1

[wild swallow]

should keep a strict inequality for the left one

[onyx lodge]

oh right

[wild swallow]

[ 0.141x - 1 < \floor {0.141x} \le 0.141x ]

[wild swallow]

and the other one?

okay good

so what can you say about their sum

[onyx lodge]

x-1 < floor + ceil < x+1

[wild swallow]

yep

but x is integer

what can you say about the value in the middle then

[onyx lodge]

integer

wait but why

[wild swallow]

and how many integers exist between x-1 and x+1

[onyx lodge]

cant it be non integer

oh nvm

[wild swallow]

can floor + ceil be non-integer?

[onyx lodge]

cuz floor and ceil is int

ok so its x

[wild swallow]

exactly

[onyx lodge]

ahh

[wild swallow]

yep

[onyx lodge]

ah so its x

what about the other way then

[wild swallow]

well i think the other way is almost obvious

[onyx lodge]

hmm ok i try it then

thanks 🙂

[wild swallow]

x = floor(blah) + ceil(blah)

[wild swallow]

same logic as here

[obtuse pebble]

@onyx lodge Has your question been resolved?

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[lucid girder]

Can someone check my answer to this epsilon proof?

[obtuse pebble]

@lucid girder Has your question been resolved?

[deft gust]

It looks good to me. Your choice of N works. I would prefer to say "let N be a natural number greater than cbrt(1/2ε)" because I think N has to be a natural number, but I guess it depends on what you use as your definition of convergence of a sequence.

[lucid girder]

Awesome thanks a lot

What about this one?

[deft gust]

Yes tht's correct

[lucid girder]

Awesome thanks heaps

[lucid girder]

.close

[obtuse pebble]

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[lusty sentinel]

The rectangle in the figure is divided into 8 regions. The areas of three of the regions are 2, 3 and 20 as indicated in the figure. Find the area of the shaded region.

[trail musk]

M appears to be the midpoint of AB

[royal basin]

appears to be != is

[trail musk]

Yeah but seems like not enough info

Unless I'm missing something real obvious

[lusty sentinel]

Mmm

[trail musk]

If you're willing to take M as the midpoint of AB then this simplifies really nicely. Unsure if you're meant to take that as a given though

[lusty sentinel]

Mhm

I think we should take like

The triangle above M and N

[obtuse pebble]

@lusty sentinel Has your question been resolved?

[lusty sentinel]

Nope

<@&286206848099549185>

<@&286206848099549185>

[timid silo]

this is easy AF

dm me i will teach you just Pythogoreoustheoram

[royal basin]

you're either lying or being pretentious

also how exactly do you propose the pythagorean theorem be used here

[timid silo]

coz i had done this before

plus some properites will also be used

[lusty sentinel]

Okok

[royal basin]

may i ask that you share an outline of your solution

[lusty sentinel]

Mmm

Nope...

[royal basin]

was asking @timid silo, who claims this problem is "easy AF"

[lusty sentinel]

I just dm-ed him but he hasn't tell me how to😂

@royal basin Do u know how to solve it?

[royal basin]

not immediately, no.

[lusty sentinel]

Ye he's lying lol

He says he's from IIT

[timid silo]

you really think thiss lol

[timid silo]

tbh i m not here to prove anyone ..... i already proved

[royal basin]

@timid silo do you or do you not have a solution for this problem

[timid silo]

https://www.youtube.com/watch?v=N4x8ex7rODg

this will help you

to get solution

[royal basin]

did you not claim the pythagorean theorem could be used here somehow...?

[lusty sentinel]

I got it

Thank u

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[timid silo]

I got [
a_n = 2^{n-2} + 4a_{n-2} + a_{n-1}
]
for $a_n$ possible solutions

[timid silo]

wait i completely messed that up i thought that said two consecutive 0's

.close

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[strange plover]

how do I dolse this

[slender nebula]

Integral from a to b + integral from b to c= integral from a to c

So 2to4 + 4to7 = 2to7

Thus 4to7= 2to7 - 2to4

=3-5

[strange plover]

2

[slender nebula]

Nope

[strange plover]

-2

[slender nebula]

Yep

[strange plover]

thank youu

[jolly ginkgo]

[strange plover]

is the particle slowing down?

[jolly ginkgo]

What do you think?

[strange plover]

yes

its correct?

[jolly ginkgo]

You don't have to ask it everytime, try getting answer yourself

[strange plover]

I think is -2 and 5

because in both are changing from a decreasing to increasing

[jolly ginkgo]

Okay

[strange plover]

i was wrong

it was 8

[jolly ginkgo]

It was asking for minimum of g(x) and the graph was for f

[jolly ginkgo]

This one is pretty easy one

[strange plover]

for you

[jolly ginkgo]

This only needs basic knowledge, what does differentiation of f shows and what does double differentiation of f shows

[strange plover]

how do i solve this?

this is the washers formula, right?

[jolly ginkgo]

Yes you have to use it

[strange plover]

its the first one?

the second to the right

[jolly ginkgo]

First of all what is the rotation along?

[jolly ginkgo]

Idk by second one you mean the right one or down one

[strange plover]

top right

[jolly ginkgo]

No

First check what it is rotating along

Whenever you are rotating along x=c the integration will be in terms of y

[strange plover]

its B?

its D

?

.close

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[gentle cipher]

Find the missing number in the parentheses in the sequence below:
0, 1, 4, 15, 56, ( )

[royal basin]

malicious compliance with underspecification says 19

[gentle cipher]

Ok, thanks

.close

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[empty hull]

lmao

[obtuse pebble]

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[spiral wraith]

is there a shorter way to solve this... I feel like I'm doing it wrong

[high lily]

don't bother expanding

[spiral wraith]

oh i forgot to say it asked for dy/dx in terms of x

[high lily]

and you don't need to expand for that

(x^2+2x+7)^3 can be left as is

there's no requirement to expand that and its more ideal to leave it as it is in that form

[spiral wraith]

when it asks for dy/du does that mean y' over u'

my teacher never went over the notation

[high lily]

dash notation gets a little ambiguous with multiple varibales

[spiral wraith]

I see

[high lily]

derivative of y wrt u

[spiral wraith]

with respect to?

[high lily]

yes

[spiral wraith]

so 4u^3 ?

[high lily]

that's what you're supposed to have for dy/du,
and not dy/dx

[spiral wraith]

I see

[high lily]

dy/dx = dy/du * du/dx
from chain rule

[spiral wraith]

and dy/dx is y' ? if we were to use dash notation?

[high lily]

yes

[spiral wraith]

I see

?