help-10

[strange tinsel]

$178^2+273^2=106213$. Subtract from both sides, knowing $235^2=55225$, and you get $-50988=-2 \cdot 278 \cdot 273 \cdot \cos B$. Then further simplify.

[warm shale]

Rallph82

[obtuse pebble]

@upper grotto Has your question been resolved?

[strange tinsel]

Generally it's easier to work with all constant terms combined.

[obtuse pebble]

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[floral canopy]

Hey can someone help me solve this equation:
x''+x'+x=0

[floral canopy]

Btw the function is x(t)

Can I just say that x(t) = e^sx and solve for s?

[gilded needle]

sounds like a viable approach, did you try it?

[floral canopy]

Yes

But then there is root of -3

[gilded needle]

yes

[floral canopy]

You get e^-1+-√-3/2

[gilded needle]

$s = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$

[warm shale]

OurBelovedBungo

[floral canopy]

Yes

That's what I got

But then you get i sin(t)

[gilded needle]

so $e^{-x/2}e^{ix\sqrt{3}/2}$ and $e^{-x/2}e^{-ix\sqrt{3}/2}$

[floral canopy]

Wait what

[gilded needle]

sry bad typesetting

[floral canopy]

Yea right

Ok

[gilded needle]

damn it

[floral canopy]

So then you get i *sin(t)

I got it dw

But is I*sin(t) a solution?

[warm shale]

OurBelovedBungo

[floral canopy]

Yes

[gilded needle]

there finally

lol

[floral canopy]

Lol

[gilded needle]

and for the ones with the imaginary exponents you can use euler's formula

$e^{i\theta} = \cos(\theta) + i\sin(\theta)$

[warm shale]

OurBelovedBungo

[gilded needle]

damn i cannot type tonight

[obtuse pebble]

@floral canopy Has your question been resolved?

[obtuse pebble]

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[wanton dagger]

Does this mean (-4x+5y)-(3x+2y)

[fallen mantle]

no

try to analyse the meaning of subtracted from

[wanton dagger]

So -4x+5y-3x+2y

[fallen mantle]

no

[wanton dagger]

Oh wait my bad

The other way

So (3x+2y)-(-4x+5y)

Thanks

.close

[obtuse pebble]

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[fallen mantle]

welcome

[obtuse pebble]

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[timid silo]

x |x-a| = 3 has one real solution, find the least real number of a

[stark ether]

what have you tried?

[timid silo]

so case 1, x-a=3/x which is x^2-ax-3=0 and case 2, x-a=-3/x which is x^2-ax+3=0. Since one solution, b^2-4ac = 0. For case 1 I get a = +-sqrt -12 which is invalid and for case 2 I get a=+-sqrt 12

[gloomy valve]

sounds good

[stark ether]

Yeah

[timid silo]

shouldnt the lowest value be - sqrt 12? the answer says there are no lowest value

[gloomy valve]

plug in both

into the original equation

and try solving it

for one of them you'll get two solutions

[timid yacht]

for sqrt(12) you will have 2 solution

and for any bigger number, youll get 1

so there is no lowest value

[timid silo]

where did you plug it in?

[timid yacht]

wdym

[timid silo]

how did you get two solutions?

[timid yacht]

divided the abs into 2 case

like always

or by looking at the graph

try to draw x and abs(x-a) on the same plane

its x*abs(x-sqrt(12))

a normal parabola

but its reflected after the 'a' value

[timid silo]

so a=sqrt 12 gives x = sqrt 3 or - sqrt 3

[timid yacht]

no

[timid yacht]

this is the EQN and as you can see its equal to 3 in 2 points

[timid silo]

yeah

[timid yacht]

but when a grows, the parabola's max will go up too

oh, then its reverse, it doesnt have a lowest value, cuz its -inf

[obtuse pebble]

@timid silo Has your question been resolved?

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[jade cape]

Hi how come this is the asnwer pekase?

Im confused on how to do it

[obtuse pebble]

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[obtuse pebble]

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[sacred root]

How am I supposed to find the gof of this

So looking at this ik f(x)=x+1 g(x)=3x

So gof(x)=3x+3

[obtuse pebble]

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[flat zephyr]

!open

[flat zephyr]

Help

[pine sail]

Read this.

[flat zephyr]

Help me

I dont know everything

So hard

I need help

Pls

What do i do

Pls anyone

[flat zephyr]

Can u help me

[flat zephyr]

.

[pine sail]

I sure can.

What have you tried so far ?

[flat zephyr]

Nothing

I dont understand anything

This is so hard

[pine sail]

Surface area.

[flat zephyr]

Ya

[pine sail]

You have to calculate the area of all the faces of the prism.

That would be the area of the pyramid's surface.

[flat zephyr]

Yep

[pine sail]

Pyramid

It's a triangular pyramid.

Can you not calculate the area of a triangle ?

Do you know any formula for that?

[flat zephyr]

Well i dont understand this

Im still new to grade 7 for 3 days

My friend got to grade 7 up to 3 months ago

[pine sail]

How old are you?

[flat zephyr]

13

[pine sail]

I see

Well do you know how to calculate the area of a triangle?

[flat zephyr]

Hmmm i didnt learn nothing last 3 days. Its still greetings for new students

My friend is forcing me tho

[pine sail]

I mean that doesn't answer my question.

Do you or do you not?

[flat zephyr]

I dont know

[pine sail]

Okay.

[flat zephyr]

I do not know

[pine sail]

Can you see this figure has 4 total triangles, the area of which you have to figure.

[flat zephyr]

Hmmmmm

[pine sail]

Okay now

You need to learn what area of a triangle is.
The formula is 1/2b*h but you need a proper understanding of what b and h here imply.

So I suggest look videos regarding that.

[flat zephyr]

Where

Youtube?

[pine sail]

Sure that should work.

Just write area of triangles and then if it helps your grade also.

[flat zephyr]

Ya

I apologize

My friend have been forcing me

[pine sail]

What for? It's alright.

[pine sail]

Forcing you to do maths?

[flat zephyr]

Yes eventho i hate it

[pine sail]

Lol

Then don't do it.

[flat zephyr]

And forcing me to ask on public servers

Lel

I rather play games

War thunder and valorant

💀

[pine sail]

And why are you doing it for them?

[flat zephyr]

Cus im a good friend

He helps me make cookies too

[pine sail]

That's not the definition of a good friend though.

[flat zephyr]

We both help each other

Ngl

[pine sail]

Why can't they come here and seek help.

Although that's none of my business.

[flat zephyr]

Because he was a lazy monkey

😂

[pine sail]

Now that you know you need to know the formula for area I suppose that's all you can do.

[flat zephyr]

I just need the answer

He told me ' tell me the answer rn. I dont need the formula or anything i just need the answer'

srry tho

He forced me

I apologize

Im gonna learn that first

@pine sail thanks

[pine sail]

I can't give you the answer.

I mean I can, but I won't.

And to be fair, no one here will.

[flat zephyr]

ya

thats bad tho

im gonna close this

see ya bro

thx for ur help

cheers

.close

[obtuse pebble]

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[obtuse pebble]

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[toxic tinsel]

when im solving an inequality like this, why do i need to multiply both sides by the square of the denominator rather than just the denominator itself, like is this just a rule that I should know

[glossy ibex]

cus square of it is always positive

[toxic tinsel]

so even though i was right in that particular questtion i couuld have got the wrong answer by not squaring

[glossy ibex]

probably yeah

[toxic tinsel]

is there a rule for what you can and cannot do to both sides of the inequality like do you have to flip it in some examples if you multiply both sides by somthing

[glossy ibex]

Yes!

[toxic tinsel]

i remember my techer saying somthing like this but cant remember exactly what he said

[glossy ibex]

if you multiply by a negative number you have to flip the ineq sign

[toxic tinsel]

that is the only time you have to flip it?

is there like an easily understanable reason as to why thats the case or should i just blindly remember it

[glossy ibex]

for example

3 < 4

-1*3 > -1*4

[toxic tinsel]

ahhh

makes sense

just isnt as clear as that when using unknown values

thank you

I really appreciate it 🙂

[glossy ibex]

[toxic tinsel]

how do I close this

.close

[obtuse pebble]

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[open panther]

hey im stuck on a bit of trig (im only 14 btw) i know all the formulas but i just dont know when to use them, as in in what situations i would use certain formulas (cosine, sine rule, pythagoras, sohcahtoa)

[robust sleet]

cosine is for non right angle triangles, sine and sohcahtoa for right angle triangles and pythagoras is to find the missing side if two sides are given

[open panther]

... you just done better to help me understand in one minute what my maths teacher coudnt do in 4 weeks

tysm

[high lily]

consider what you have and what each formula and identity uses

[open panther]

ty

wait how do i close this

. close

[high lily]

no space

[open panther]

oh ok

tysm btw all

.close

[obtuse pebble]

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[gilded needle]

"cosine is for non right angle triangles"? I guess the advice you get here is worth what you pay for it...

[toxic tinsel]

(the joke is you don't pay for it)

[obtuse pebble]

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[rough pollen]

There's a right triangle ABC with the leg AC=10cm, BC=15cm. Inside the triangle is a square with one of the vertex on C, and the opposite vertex on the hypotenuse AB. Find the area of the square.

[rough pollen]

ive found the hypotenuse which is 5√ 13

[dense imp]

with problems like this, it usually helps to draw yourself a diagram, see if you can find "hidden" triangles etc that you can use, have you done that already?

[rough pollen]

i have drawn a diagram, and i see a couple of other triangles but they're not helping me very much ig

[dense imp]

so we know what the area of the whole large triangle is, right? (10 x 15 / 2) what could we subtract from that to get the area of the square?

[rough pollen]

the two areas of the two other triangles?

[dense imp]

yep

easiest way to set this up i think would be:

[rough pollen]

area of the square + area of the two triangles= the whole triangle?

[dense imp]

yes, exactly

[rough pollen]

its ok i can do it by myself

thanks for the help

.close

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[royal solar]

my friend, you’ve been mislead

@rough pollen

[rough pollen]

what why

[obtuse pebble]

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[dense imp]

$x^2 + \frac{x(15-x)}{2} + \frac{x(10-x)}{2} = \frac{10 * 15}{2}$

[warm shale]

Sooshon

[dense imp]

this simplifies very simply to 25x = 150 and x = 6, so the square's area is 36

[royal solar]

do you know similar shapes

(10-x)/x = x/(15-x)
so x^2-25x +150 = x^2

so yeah, I got the same answer

[dense imp]

not sure why you think this

[royal solar]

apologies,

I initially didn’t know why you did areas, but hey

anything that works ig

@dense imp .close it so that others can use this

[dense imp]

.close

[obtuse pebble]

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[royal solar]

and once again, my fault man

[open panther]

wait is that not right

[dense imp]

someone responded to you earlier: "cosine is for non right angle triangles, sine and sohcahtoa for right angle triangles and pythagoras is to find the missing side if two sides are given"

[open panther]

is that wrong?

idk

[dense imp]

well that is NOT correct, because sine, cosine, tangent and pythagorean theorem ALL apply only to right triangles

they are right in saying you use pythagorean theorem to find the third side of a right triangle if two are given

[open panther]

so then how do you do calculate a side for not right anglefd triangle

[dense imp]

do you have a specific example problem in mind?

[cedar lichen]

Maybe they were referring to the law of cosines?

[dense imp]

there is this formula, Herod's formula which relates the 3 sides of a triangle, s = (a+b+c) / 2

[obtuse pebble]

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[dense imp]

if you need the area without knowing angles , but in general if you know two sides of a triangle but no angle there are infinite possibilities for the length of the third side

[open panther]

im in s3 (im 14) i havent done like advanced trig i've only started the basic (cosine, sine, sohcahtoa)

[dense imp]

to answer your original question...

[open panther]

yes

[dense imp]

all these things are concerning right triangles only, which is what you will usually work with

you can think of pythagorean theorem as relating all 3 sides of a right triangle, without worrying about angles

sin, cos, tan relate two sides of a right triangle, together with one of the angles

so theres always kind of 3 elements involved in each relationship

.close

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[worldly falcon]

how do i find the steady state vector of

[worldly falcon]

0.8 0.5

0.2 0.5

(matrix)

i did M-I_2

which gives

-0.2 0.5

0.2 -0.5

which gives

-0.2 0.5

0 0

[obtuse pebble]

@worldly falcon Has your question been resolved?

[worldly falcon]

.close

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[deft slate]

There is something wrong with this proof: I believe it is that she assumes the contrapositive of "if x is not in L(M'), then x is in L(M)" is "if x is in L(M'), then x is not in L(M)"

[deft slate]

At the same time, idk what is the contrapositive of that - bc honestly that's what I would've wrote myself

How does contrapositivity work in set theory?

[warm canopy]

$P \implies Q$ is the same as $\neg Q \implies \neg P$

[warm shale]

iCaird

[deft slate]

OH ITS DEMORGANS (???)

my god i completely forgot about that man

[warm canopy]

not de morgan

thats for $\wedge$ and $\vee$

[warm shale]

iCaird

[deft slate]

oh it doesn't apply to sets

but propositions

so then the real contrapositive of "if x is not in L(M'), then x is in L(M)" is "if x is not in L(M), then x is in L(M')"

(?)

I think I understand but can someone doublecheck im understanding this contrapositivity concept right?

the negation of "x is in L(M)" is "x is not in L(M)" ?

and similarly, the negation of "x is not in L(M')" is "x is in L(M)" ?

[warm canopy]

this is the correct contrapositive

[deft slate]

awesome thank you so much!!

.close

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[trim island]

why is the differential of ln3x = 1/x
if you integrate that again you get lnx

[warm canopy]

$\ln(3x) = \ln(3) + \ln(x)$

[warm shale]

iCaird

[trim island]

ln3 is just a constant

[warm canopy]

note that they differ by a constant $\ln(3)$, when you integrate its unique up to adding a constant

[trim island]

damnnnnnnnnnnnnnnnnnnnnnnn

[warm shale]

iCaird

[warm canopy]

yup you got it

[trim island]

wtaf

okay

makes sense

thanks

[warm canopy]

no worries

[trim island]

.close

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[half marsh]

Hi! I was asked to calculate sin(arcos(1/5) without a calculator. My teacher did this and said we used Pythagoras but I’m not sure what he did. Could someone help me ?

[tardy epoch]

Which part don't you understand

Start top to bottom

Actually I guess start with the drawing

[timid silo]

$sin^2(a) = 1-cos^2(a)$

[warm shale]

Jester

[half marsh]

first of all I don't understand how A^2 = sin^2(arcos(1/5), is Pythagoras, what triangle are we working on, and shouldn't there be a sum = something ? instead of just one value equals the other

[timid silo]

so A is the answer u are looking for

$A^2 = (sin(arccos(1/5)))^2$

[warm shale]

Jester

[half marsh]

i don't think i understand this part either

[timid silo]

A is answer

[half marsh]

but why is A placed where it is on the drawing then

[timid silo]

because A sin of arccos(1/5) which is alpha on the drawing

[half marsh]

oh okay i think i get it

[tardy epoch]

maybe it's helpful if you draw the triangle yourself.

[half marsh]

is it the one with A, p(A) and O ?

[tardy epoch]

nah start from scratch

i mean it will be

but if you're just copying the drawing then that defeats the purpose

you're given $\arccos(\tfrac{1}{5})$. Draw a triangle with that as an angle

[warm shale]

riemann

[half marsh]

it's not that I don't want to I just don't know how. For example I don't know how I'm supposed to know roughly where arccos(1/5) is located

[timid silo]

like this

[tardy epoch]

looks good

Pythagorean theorem tells you $A^2 + (1/5)^2 = 1^2 = 1$

[warm shale]

riemann

[tardy epoch]

Using the same triangle, you can now calculate the sin(arccos(1/5))

[half marsh]

I see, so since the circle has a radius of length 1, I just cut the other side and get 1/5. That's how I get arccos(1/5) right ?

[half marsh]

okay thank you i will try this now

[tardy epoch]

drawing always helps

[obtuse pebble]

@half marsh Has your question been resolved?

[obtuse pebble]

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[worn ocean]

This is the question I have, but my answer is 4! x 2!

which is incorrect

My logic is if we make a bundle of 2 boys, then we're left with 4 dudes, which is then 4! and then the bundle can be done in 2! ways

which would give us 4! x 2!

[timid silo]

5x4x3x2 x 4C1

[haughty coyote]

Take the first 4 boys. Give each of them a different room. Then give the final boy his room.

[worn ocean]

This is incorrect too ;-;

[timid silo]

no way

[worn ocean]

It is

[timid silo]

4x5!

[worn ocean]

yes thats incorrect

[worn ocean]

4! x 4?

[timid silo]

ah

[haughty coyote]

Is it ?

[worn ocean]

no? ;-;

[haughty coyote]

It was actually a question

But normally yes

[worn ocean]

i think it is 4! x 4 if I do what you suggest here ;-;

[worn ocean]

the answer is 5C2 x 4! but i dont get why

[timid silo]

bundle 2 kids then give them ttheir rooms

bundle 2 kids in a box and treat them as 1 kid

[worn ocean]

thats exactly what i did tho

[timid silo]

5c2?

[worn ocean]

My logic is if we make a bundle of 2 boys, then we're left with 4 dudes, which is then 4! and then the bundle can be done in 2! ways

[timid silo]

no?

bundle is 5c2

choose 2 kids from 5

[worn ocean]

there arent 5 entities

when u bundle, there's only 4 entities

[timid silo]

after bundling that is

[worn ocean]

yes

[timid silo]

so 5c2

take 2 kids and bundle them

[worn ocean]

5c2 is just choosing 2 kids from 5

[timid silo]

yeah

[worn ocean]

ohhhhhhh

you are saying the ways of making the bundle

[timid silo]

make them roommates

[haughty coyote]

That didn't account for everything because it forced boy n°5 to have a roommate

[worn ocean]

no it didnt, i put in x4

so he could go into any room no?

[worn ocean]

i get it, now

[timid silo]

but he'll have a roommate regardless

[worn ocean]

choose 2 and then 4!

[haughty coyote]

But it didn't account for something like boys 2 and 3 being together

[worn ocean]

yeah

[timid silo]

3 boys is false

[worn ocean]

he meant number 2 and number 3

[timid silo]

ah

[worn ocean]

anyways, thanks again

.close

[obtuse pebble]

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[cursive robin]

How to do this? Algebra I

[warm canopy]

Would be great if we had 2^(stuff) on both sides

See if you can make the right hand side like that

[cursive robin]

2^x+3

[urban patrol]

No

[cursive robin]

heh? Then what i it

*is

[urban patrol]

2^3x

when you stack exponents you multiply

2^x+3 is 2^x*2^3

[warm canopy]

$\left(a^b\right)^c = a^{bc}$

[warm shale]

iCaird

[urban patrol]

^

[cursive robin]

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[stiff bear]

Hey I am having trouble with this question I was able to find all of the sides and angle values it’s just the tower height I am confused on any help on how I can solve this?

[stiff bear]

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[hushed cairn]

For a definite integral, why do you take the function with the top number subtracted by the function of the bottom?

[hushed cairn]

Also this isn't related to the question but how would I solve something like this?

[urban patrol]

just solve the sigmas individually

[hushed cairn]

but they want me to write it as a single sigma equation

sorry I should've posted what they wanted me to do as well

<@&286206848099549185>

[obtuse pebble]

@hushed cairn Has your question been resolved?

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[lunar laurel]

I need some help solving this

[lunar laurel]

I already did this:

I don't know what or how I can factorize this any further

in the paper its v instead of x

[gilded needle]

are you allowed to use l'hopital's rule?

[lunar laurel]

im not

[gilded needle]

you can rewrite both the numerator and denominator in terms of sine as follows:

cos(2x+pi/2) = -sin(2x)

and

sin(6x+pi) = -sin(6x)

then, do you know the limit of sin(x)/x ?

[lunar laurel]

1 right?

[gilded needle]

yes

[lunar laurel]

why is it that i can do this?

[gilded needle]

those are trig identities

you could derive them for example from the formulas for cos(a+b) and sin(a+b)

so after doing that your fraction becomes $\frac{\sin(2x)}{\sin(6x)}$

[warm shale]

OurBelovedBungo

[gilded needle]

which you can very cleverly rewrite as

$\frac{6x}{\sin(6x)} \cdot \frac{\sin(2x)}{2x} \cdot \frac{2}{6}$

[warm shale]

OurBelovedBungo

[lunar laurel]

I dont understand how but wouldn't that still be 0?

Im confused

[gilded needle]

well what's the limit of the first of those three fractions?

6x / sin(6x)

(let y = 6x)

[lunar laurel]

is it 1? is the rule reversible?

I thought it was only when sine was at numerator

[lunar laurel]

What is this called btw? I had never seen this

Like the rule

[obtuse pebble]

@lunar laurel Has your question been resolved?

[gilded needle]

yeah, if the limit of f(x) is L then the limit of 1/f(x) is 1/L (as long as L is nonzero)

that's in general, not just for f(x) = sin(x)/x

[gilded needle]

maybe look up "algebra of limits", I've seen that term

it's the set of rules that say lim ( f(x) + g(x) ) = lim f(x) + lim g(x) and lim ( f(x)g(x) ) = (lim f(x)) (lim g(x)) and so forth

so you can use the 1/f(x) rule to conclude that lim (x / sin(x)) = 1 / (lim (sin(x) / x)) = 1/1 = 1

and similarly if we replace x with 2x or 6x like in your problem

[lunar laurel]

Ill look it up to see if I understand better, thank you so much!

.close

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[lunar laurel]

If limit x-0 sen(-2x) / x is -2, how is sen(-2x) / -x equal to 2?

[gilded needle]

factor out a -1 from the denominator in the second one?

$\frac{\sin(-2x)}{-x} = -\frac{\sin(-2x)}{x}$

[warm shale]

OurBelovedBungo

[lunar laurel]

Ah thanks!

.close

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[jade cape]

Hi guys

Can I Get some help on this topic please?

Im not to sure on how to work it out

Would help a lot thank you.

[obtuse pebble]

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[craggy pagoda]

how do u determine what sign goes here

hes saying

to visualize whats happening outside the intercepts

idk what that means

[tardy epoch]

Plug in any point in the given range

0 is probably the most convenient

[craggy pagoda]

so

-2 < 0 < 3

which would be a true statement ?

[tardy epoch]

I mean plug it into the quadratic

[craggy pagoda]

how abt for something like this tho

it would be -3>/0

which isnt right so im not sure

im just kinda confused bc in the vid my teachers saying to visualize it

[minor cypress]

It should be < right?

[craggy pagoda]

no i think thats right

[minor cypress]

See x< -3 implies x+3<0

similarly 2x-1>0

Ive forgotten the equal sign but just assume it's there

Now if you multiply the two you get your LHS

On your RHS you are multiplying something -ve with something +ve

So the result should be -ve

[craggy pagoda]

still kinda confused man

could u try to explain what he means by this

i’m trying to understand what he means by visualizing it

[minor cypress]

Oh wait I'm so dumb

Ignore whatever i said above

you do know that the roots for the final quadratic are -3,0.5 right?

[craggy pagoda]

yes

[minor cypress]

and it's upward U shaped

[craggy pagoda]

yea

[minor cypress]

Yeah so for x < -3 and >0.5 what is the value for the parabola

Note that these are it's roots

[craggy pagoda]

3 and -1/2?

[minor cypress]

No

See where the parabola cuts the x axis

[craggy pagoda]

ya the x ints

[minor cypress]

Those are the roots

-3 and 1/2

[craggy pagoda]

kk

oh yea sorry i misunderstood the question

how do u determine

which sign to use from there

[minor cypress]

yeah so you decrease x from -3 and increase x from 1/2

What is the sign(+ve or -ve) of the quadratic function

[craggy pagoda]

wait

what does

ve evens tand for

stand

sorry ive never heard of that

[minor cypress]

+ve - positive

[craggy pagoda]

OH

[minor cypress]

-ve - negative

[craggy pagoda]

wdym by

decrease x from -3

[minor cypress]

Go left from -3

I'm talking graphically

[craggy pagoda]

ok yea

-4

[minor cypress]

Any arbitrary number i mean

-4 is included in it

What is the sign for the quadratic

Is it < or > 0

[craggy pagoda]

?

which one do u look at

the -3 or the 0.5?

[minor cypress]

Either

Should be the same sign anyway

[craggy pagoda]

-3 > 0?

woudlnt that be a false statement or w/e

[minor cypress]

not about the value of x

The value of y

You need to compare what value you get after putting x in the original quadratic

[obtuse pebble]

@craggy pagoda Has your question been resolved?

[obtuse pebble]

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[jagged edge]

Do you guys know what I did wrong here?

[jagged edge]

It’s suppose to be 35 = w

But I’m getting 35 = w^2 for some reason

[acoustic bronze]

you messed up on the step of going from the volume to the derivative of the volume

[jagged edge]

Wdym

Which line did I mess up

[versed turret]

Missing a w in 210w

[jagged edge]

OHHHG

[obtuse pebble]

@jagged edge Has your question been resolved?

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[opaque vessel]

.close

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[magic comet]

Could anyone help me with part 3? I actually don't understand how the hint is used...

[obtuse pebble]

@magic comet Has your question been resolved?

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[versed cave]

Hi!

[versed cave]

Why do I get different results if I do implicit differentiation of x² + y² = c?
Shouldn't d/dx (y) be 0?

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[wary fable]

how do i solve for A with only vertex form

[wary fable]

.close

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[feral swallow]

Is the answer for e) y=9/20 - 23/80?

[royal basin]

are you sure you didn't miss anything there

did you maybe mean y = (9/20)x - 23/80?

[feral swallow]

Yeah

[royal basin]

your y-int looks off

[feral swallow]

I know there is an x but im making sure if the y int is corrrect

Okay let me try again

13/80 = 9/20 + c
In the form of
Y= mx + c

Yeah i got the same and

Answer

[timid silo]

$\frac {9}{20} x - \frac {23}{80}$

[warm shale]

CatHashira

[feral swallow]

Yepp

[royal basin]

9/20 + c?

not 9/40 + c?

the point of contact is x=1/2 after all

[feral swallow]

Well what i did was i used dy/dx for the y equation which made it from

$\frac {1}{4} x² + \frac {1}{5} x$

[warm shale]

X̵͒̅

[feral swallow]

Tooo

$\frac {1}{2} x + \frac {1}{5}$

[warm shale]

X̵͒̅

[feral swallow]

Then i subsituted x with 1/2

Am i correct until now?

[royal basin]

yes you are correct however i have identified your mistake

[royal basin]

.

[feral swallow]

Yeah?

[royal basin]

y = mx + c

you forgor to plug in x = 1/2 into this

and mistakenly put x = 1

[feral swallow]

OHHH YEAH

Yes yes u are right

9/20 × 1/2 which is 9/40

And now we solve for c right?

[royal basin]

yes

[feral swallow]

$\frac {13}{80} - \frac {9}{40}$

[warm shale]

X̵͒̅

[feral swallow]

This

Alright i got it

Thank you so much ann have a good day

I have a question for u if u dont mind, did u do igcses and took further maths?

[royal basin]

uh no

that sounds like a UK thing, which i am not from

[feral swallow]

Okkkaayy

Thanks again and yes its uk

[timid silo]

A bit off topic but are you an actual mathematician as written in your bio?

[royal basin]

i have a bachelor's in applied math, does that count

[wanton dagger]

I think so.

[timid silo]

Ya almost ig

bachelor is almost?

I am guessing from google to be honest

[wanton dagger]

From what I have found on google, it should count

[timid silo]

i will study to become a pure mathematician

[wanton dagger]

Nice, good luck

[dense imp]

I'd say when you have your own Erdős number there would be no more question about it, but that's a high standard : )

[wanton dagger]

What’s an ‘Erdos number’

[dense imp]

https://en.wikipedia.org/wiki/Erdős_number

[timid silo]

aah i hear about this

[wanton dagger]

Hmm I see

Wow

[timid silo]

my erdos number must also be high 😎

[timid silo]

I see

[wanton dagger]

Lol

[timid silo]

ah but i dont have an official number

That will be given to the real talents in math ig lol

[wanton dagger]

Surely

[timid silo]

Imagine if someone's number is -1

Or if someone's number is sqrt(-1)

[wanton dagger]

That would be insane

How do you even get an erdos number though?

[timid silo]

That person has broke all human expectations

[wanton dagger]

Yep

[timid silo]

When you reach a certain degree in maths maybe

idk tho tbh

[wanton dagger]

Mmm.. makes sense though

[timid silo]

We should switch to #discussion

[wanton dagger]

Uh ok

[dense imp]

you would have to coauthor a paper with someone who already has an Erdos number 😄

[royal basin]

fun fact

i have an erdos number

[timid silo]

you phd?

[royal basin]

no, i'm not a phd

[timid silo]

how

[earnest elk]

Wow thats so Cool

[timid silo]

Hey I need help with a question

Omg only 268k people have erdos number

[royal basin]

@timid silo this channel is occupied, please move

anyway

[timid silo]

It's rare ig

[royal basin]

i wrote a research paper in year 2 with my algebra prof as my advisor

[timid silo]

damn nice

[wanton dagger]

Offff yr 2!

[royal basin]

and summer of that year he asked me if i was ok with him sending it off to be published under mine and his names

and i said "hell yes sign me up"

[timid silo]

can u send link

[royal basin]

no, that would doxx me

and reveal my legal name which i would rather not

[timid silo]

what was the research on

[royal basin]

group theory

proving that a certain group is isomorphic to the free group on two generators

anyway i looked up my algebra prof and found that he has an erdos number of 5, making my erdos number 6

[timid silo]

wow

[earnest elk]

Ann, are you researching on something atm?

[timid silo]

Damn

[royal basin]

ostensibly yes

bashing my head against the wall

but yes i do have a research problem before me

[timid silo]

I thought ansh was best here but now I think its ann

[wanton dagger]

Then an erdos no. would be enough to call yourself a mathematician

[dense imp]

bravo @royal basin : )

now let's publish something together so I can get an Erdos number 🙂

[timid silo]

Level 6 Mathematician

[timid silo]

Dealing be like

[earnest elk]

Lol

[obtuse pebble]

@feral swallow Has your question been resolved?

[obtuse pebble]

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[light ridge]

Can someone give me hints?

[light ridge]

please

<@&286206848099549185>

[obtuse pebble]

@light ridge Has your question been resolved?

[light ridge]

.close

[obtuse pebble]

@light ridge Has your question been resolved?

[eternal bloom]

To prove linear dependence, show that some vector in the set can be written as a linear combination of the others.
To prove linear independence of {v1, v2, v3}, show that if x1v1 + x2v2 + x3v3 = 0 for some scalars x1,x2,x3, then we have x1=x2=x3=0.
To prove that the set is a basis of V, prove that it spans V and it is linearly independent.
To find the dimension, write down the number of elements in a basis.

[obtuse pebble]

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[fallen hollow]

Hello guys

[fallen hollow]

I need help with percentage

Anybody up to help

[meager siren]

[earnest elk]

whats your question?

[fallen hollow]

How do I find

15 is what percent of 300

And

30% of 450

[arctic iron]

% = 1/100

[fallen hollow]

What

@arctic iron

[earnest elk]

https://www.youtube.com/watch?v=WYWPuG-8U5Q

[arctic iron]

Hm

X% of Y is X/100 • Y

So 10% of 20 is simply
10/100 • 20

Which gives you 20/10 = 2

[fallen hollow]

30% of 450 is 30/100 x 450?

[arctic iron]

Apply that logic here

Yes precisely

[fallen hollow]

The answer is 135?

[arctic iron]

Then for the first question you have to rewrite this

[arctic iron]

Yes

[fallen hollow]

Let's goo

[arctic iron]

So have you attempted to rewrite it

[fallen hollow]

Alright so

[arctic iron]

This time around you don't know what X is

[fallen hollow]

15% of 300 is 15/100 x 300

?

[arctic iron]

No, wrong interpretation

"15 is what percent of 300"

That's your question

[fallen hollow]

Yes

[arctic iron]

X% of Y is X/100 • Y = Z

[fallen hollow]

So we don't know what x is

We don't put anything there?

[arctic iron]

Actually we don't know what X is

[fallen hollow]

Then

[arctic iron]

We find it

[fallen hollow]

X% of 300 is x/100 x 300?

[arctic iron]

Yes but results to 15

[fallen hollow]

And that's not the answer?

[arctic iron]

Yes

[fallen hollow]

Is 45 the answer?

[arctic iron]

I doubt that

Does 45% of 300 give you 15?

You need to understand the question well, they're different from the other one

"15 is what percent of 300"

What percent of 300 gives you 15?

That's the question

[fallen hollow]

Ohh

But what I'm asking is

How do I solve that

[arctic iron]

X% of Y is X/100 • Y = Z

Do you know how to change the subject of a formula

[fallen hollow]

What do I put at x place??

[arctic iron]

You put X, because you want to know X

[fallen hollow]

Oh-

[arctic iron]

So when we're done solving you get
X= a value

What do you think Y is from the question

[fallen hollow]

300?

[arctic iron]

Yes.

Then what do you think Z is

[fallen hollow]

Wow I'm confused

I get what you're saying

But I don't understand how to use that

[arctic iron]

You only have to substitute the values

[fallen hollow]

That's what I'm not getting

Cause you confused me with the x thing

Cause we know y is 300

But idk what to put at x

And then how to solve it

[arctic iron]

What percent of 100 gives you 20?
X/100 • 100 = 20
X = 20
So 20% of 100 gives us 20

[arctic iron]

We can't put anything at X because that's exactly what we're finding, we only know Z and Y then we try to find X

[fallen hollow]

Ok ok wait

Is 5 the answer?

[arctic iron]

Yes

[fallen hollow]

Let's gooo

Thank you

[arctic iron]

You're welcome

Would you be able to solve a question if they asked you
20% of what gives you 10?

[fallen hollow]

@arctic iron gives you?

[arctic iron]

Yes

20% of what is 10

[obtuse pebble]

@fallen hollow Has your question been resolved?

[obtuse pebble]

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[mental sun]

I don't quite follow how they can deduce from this that I becomes 0 when t is infinite

[mental sun]

nvm I'm dumb

,close

.close

[obtuse pebble]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[wanton kraken]

"Given the vectors AM=3a and AN=ka find for which value of k the points M and N are matching"

[wanton kraken]

help

I know vectors and stuff, but those kind of problems I don't know how to solve

I just finished learning about the rule of subtracting vectors

is it 3a=ka ?

then 3=k?

cuz you subtract a from both sides if I have to make it an equation?

I've looked everywhere on the internet

[brave bramble]

3a = ka
Which does imply k = 3.
However no we're not subtracting a to get this result

I assume a is itself a vector?
Then the only way these can match is if their scalar multiple is the same

[wanton kraken]

oh yeah my bad I meant divide both sides

[brave bramble]

So that's still a no-good, haha. Can't divide by vectors, such an operation makes no sense

[wanton kraken]

oh well dunno

but I see like

3a if a is a vector

is multiplied three times

so ofc k has to be 3 as well

[brave bramble]

Let's say
3x + 2 = ax + b
Regardless of x.
Then you'd have to conclude that a = 3, b = 2

We're kinda doing that

[wanton kraken]

oh

I see it

but then it gets weirder

I have another problem from the like

[brave bramble]

But you said it perfectly. If 3 times some vector is equal to k times that same vector, then k = 3

[wanton kraken]

MN=(3k+2)a and NQ=(4-k)a
MN, NQ, and a are vectors

[brave bramble]

Then 3k + 2 = 4 - k

[wanton kraken]

oh so those problems are from the same nature

I was just confused, because

it was never explained

from KhanAcademy, internet as a whole, teacher, textbook

and then all of the sudden I get those problems

thanks a lot

[brave bramble]

Oh wait, mb. MN and NQ don't have the same setup

[wanton kraken]

yea

M to N then from N to Q

I guess it's the triangular rule of adding vectors

[brave bramble]

Okay so I used the fact that "M and N were matching" from the last question

[wanton kraken]

has something to do with it?

[brave bramble]

Are M and Q matching for this question?

[wanton kraken]

yeah, the problems states to find for which value of k M and Q are matching

[brave bramble]

Okay. So MN and NQ are vectors going in the opposite direction

That means,
(3k + 2) = -(4 - k)

[wanton kraken]

aren't they something like this?

[wanton kraken]

yeah I get to the correct answer this way, but how did you understand that they are in the opposite direction?

[obtuse pebble]

@wanton kraken Has your question been resolved?

[wanton kraken]

@brave bramble there?

[obtuse pebble]

@wanton kraken Has your question been resolved?

[wanton kraken]

<@&286206848099549185> MN=(3k+2)a and NQ=(4-k)a
MN, NQ, and a are vectors
How do I deduct that the vectors are in the opposite directions?

The problem states

"Find for which value of K points M and Q are matching"

[brave bramble]

MN is a vector going from M to N.
NM is a vector going from N to M.
They are the same vector, but opposite direction.

[wanton kraken]

ohh damn it it's so obvious

thank you

.close

[obtuse pebble]

Channel closed

Closed by @wanton kraken

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dense pier]

Two sets are given - A and B. The intersection of the sets is

[dense pier]

i've solved the set B but im not sure what should i do with A

[copper latch]

Whats that < > notation though

Is it () or []

[dense pier]

[]

[copper latch]

Okay for the intersection maybe you look where would the graph cross

Also if it helps
sin(π-x) = cos (x)

Then try to draw the graph,

Also do note that you should take in the consideration of domain while plotting the graph

[dense pier]

so instead of 2sin(x-π) i get -2cos(x)?

[copper latch]

Yes correct

So simply put, you want -2cos(x)= x

But keep in mind the domain

Ohh shoot

Wait

I’m sorry

[dense pier]

huh

[copper latch]

sin(π-x) = sin (x)

Not the cos

I’m dumb

[dense pier]

me too becase i have this formula in a front of me and i still wrote cosx haha

[copper latch]

So you want -2sinx = x

For that domain

Btw bro can you find domain

[dense pier]

i mean

from 0 to π

isnt it

like π/3, π/2

[copper latch]

No its not correct

Because take a look at the condition of second set B

|x-1| >=1

[dense pier]

ah so it is from -inf to 0, including 0 & from 2 to + inf including 2

[copper latch]

Yea correct

[timid silo]

A language I can't understand

[copper latch]

So ur domain for the function would be..?

[dense pier]

(∞; 0] & [2; +∞)

[copper latch]

Yea but 2sin(x-pi) is restricted to (0,π)

And remember we want to find their intersection

[timid silo]

Ig u want

$-2 sin(x)=x$

for $0<x<\pi$

and
x greater or equal 2

[copper latch]

Ig u want

$-2 sin(x)=x$

for $0<x<\pi$

and $x \geq2$

[warm shale]

Deep.

Pi Creature

[copper latch]

Its not correct @timid silo

[timid silo]

Ig u want

$-2 sin(x)=x$

for $0<x<\pi$

and $x \geq2$

Ty

[warm shale]

Pi Creature

[timid silo]

The latex or the math?

[copper latch]

Math

[timid silo]

Oh lemme cross check it

[copper latch]

$-2 sin(x)=x$

for $0\leq x\leq\pi$

and $x \geq2$ U {0}

0 is in the domain

[timid silo]

Ya it's in absolute bars

[warm shale]

Deep.

[timid silo]

Ya that's good

[copper latch]

Idk how to write union in latex

[timid silo]

Btw I am Frustrated Cat if you didn't know

[timid silo]

Same

[copper latch]

Ya i saw

[dense pier]

lmao

[timid silo]

So @dense pier what did you thought?

Now

[dense pier]

well it starts to make some sense to me

although i dont really get substitution of a variable

like if we take x=2

-2sin(x)=2?

[copper latch]

You substitute both in lhs and rhs

[dense pier]

i feel like i'm so close to figuring it out but at the same time i feel kinda stupid cuz i dont figure it yet

[dense pier]

so -2sin(2)=2?

how does that make sense