help-38

[jovial sluice]

Helo

<@&286206848099549185>

@fervent thorn

@warm kernel

@trim jolt

#❓how-to-get-help

@trim jolt

@warm kernel

<@&268886789983436800>

Bruuuuuhhh

[minor shadow]

id start with messing around with some angles

lets denote m(XAD) = x and m(YAB) = y

since X-A-Y, it means DAB = 180-(x+y)

now since Z is a circle it means DAB is an angle on the circle

which is half of arc DCB

same logic for C

which would mean C = (360-BCD)/2

slap that all together and youd get

$C = \frac{360-2(180-(x+y))}{2}$

[solid kiln]

Kel.plush

[minor shadow]

i think i fucked up somewhere

hold on

[jovial sluice]

I don't know even what is going on

I need to write a prove

[minor shadow]

dunno how you write proofs i dont know ur school system

but this is the general idea

[jovial sluice]

Okay

Thx

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[stoic saddle]

7(e^5-e^3)

you understand it

just made a mistake I think

your constant applies to the integral you get

[wide mural]

oh for real?

[granite cove]

A blunder 😂

[stoic saddle]

[wide mural]

😭

my fault og

[stoic saddle]

was it correct?

[wide mural]

checking rn

[stoic saddle]

most important part of that q is getting the correct values of e^n I think

[wide mural]

would be 7e^5+7e^2 yeah'

[wraith hinge]

you put a plus instead of a minus here too

[wide mural]

incorrect @wraith hinge i tried that

[wraith hinge]

wdym

[stoic saddle]

no

it is correct

I think ;-;

[stoic saddle]

you forgot to apply the constant properly

[wide mural]

unless i wrote it wrong again

[wraith hinge]

7()

u can factor out the constant

[wide mural]

u right

u right

my fault

alr appreciate it gentlemen

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[tawdry iris]

im confused bc i never learned any shapes besides bell curve, bimodal, pos/neg skew, and symmetrical lol.. could someone please help me with examples of the other types of histograms, or if these arent even histogram types?

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[quasi sorrel]

you are on a scale in an elevator and you are rising up towards your destination. The elevator stops suddenly causing you to feel lighter. The elevator slows to an acceleration of -3 m/s^2. you have a mass of 70 kg. what is your weight force on the scale?

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[tacit echo]

How can we prove that $a \in \mathbb{R} \implies \frac{1}{a} \in \mathbb{R}$?
Its a Real Analysis question, btw.

[solid kiln]

n11

[valid moss]

because R is a field

[tacit echo]

Hmm

which property we use

[valid moss]

should be $a \in \bR \setminus \Set{ 0 }$ though

[solid kiln]

i believe in mathemagic

[valid moss]

existence of inverses

[tacit echo]

Yes true

[real sinew]

41 rooks are placed on a 10 × 10-chessboard. Prove that there are five rooks among these 41 that
do not attack each other (i.e., no two of them are in the same horizontal or vertical line).

[valid moss]

go away

[real sinew]

Why

[valid moss]

are you literate

[real sinew]

Yes

[valid moss]

go read #❓how-to-get-help then

[tacit echo]

I'm kinda confused...
We are frequently using a set P for all positive real numbers, to prove these identities, if it helps explain?
I'm not sure how I can prove it...
Best I could think of was this: a is in P, 1 is in P, so 1/a is also in P (R being closed under multiplication). Does it feel right?

[valid moss]

1/a is just yet another notation for a^(-1) (by definition)

a^(-1) here being the inverse of a

which exists and is in R because R is a field

[tacit echo]

true true

[tacit echo]

I see

Alright, thanks a lot!

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[wraith hinge]

I don’t get how to prove the relations for symmetric reflexive and transitivité

write out the definitions

I mean reflexive is like (x,x)

Symmetry is (x,y) implies (y,x)

so in terms of maximum elements...?

That’s the part that confuses me

The relation

X and Y are related when the have the same max element

Like the max element of {1,2,7} and {2,7#

so are X and X related?

Is 7

[wraith hinge]

Yes

X has the same max element as itself

yeah exactly

And vice versa Y

now if X and Y have the same max element, do Y and X have the same max?

I don’t know how to justify that besides words

[wraith hinge]

Yes ofc

words are fine here

Like in my proof or just rn

words in your proof

Okay and then for transitivity like

If you define z as the same max element as X and Y it’s pretty easy to explain ?

[wraith hinge]

Can I define transitivity similarly

yes

Also is the equivalent class of 1 just itself

no, theres one more set in that class

Equivalent class of 1 are all the sets with 1 as it’s max element

Does it have something to do with the empty set

no

find another set that has maximum element 1

I genuinely cannot think of one???

0<1

Oh shoot

Rigut

Would I use Cardinality of the power set to find the total amount of distinct classes

how many possible maximums are there?

I don’t know

how many elements in S?

how many of these elements can be maximums?

10 elements

Well actually 11 cause the empty set

?

Can’t the empty set be a distinct element

OH rip

and im asking for elements in S

not A

2^10??

no

10 was correct

Just S is 10

the empty set is in P(S), not in S

What does 10 tell us

come up with a set such that each of these are maximums

Like where every element is the maximum???

where 1 is a maximum, 2 is a max, 3 is a max etc

That’s going to take forever

a single set for each maximum

you already have {1} has max 1

find a set that has max 2

The answer is 10

yeah

TYSM

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[midnight lantern]

$$6x

[midnight lantern]

$$6x^2+13x=5$$

[solid kiln]

_sami

[midnight lantern]

i turned it to
$$6x^2+13x-5=0$$

[solid kiln]

_sami

[midnight lantern]

now i need to factorise but im stuck

[jolly basin]

two numbers multiple to ac and add to b

what is ac in this case

[midnight lantern]

wait i think i figured it out holup

i have a different method than the one yall seem to use

[jolly basin]

[midnight lantern]

so i cant rlly explain how im thinking

[jolly basin]

there are a lot of ways to factor

[midnight lantern]

ok what if the first one is (2x ? 1)(3x ? 5)

[jolly basin]

mmmmm

smth is wrong

ur number are correct

their placements are not

[midnight lantern]

really

let me think ab this

oh shi i see

what do i do

help plz

[near spade]

try using quadratic formula

[midnight lantern]

my current part is supposed to solve this without quadratic formula

just null factor law n shi

I DID IT

$$(3x-1)(2x+5)$$

[solid kiln]

_sami

[jolly basin]

yes

that's right

[midnight lantern]

ok now what is x

i got this

1/3 or -5/2

right?

i dont need it in decimals, fractions are accepted

[jolly basin]

yeah

[midnight lantern]

YESSIRRRRR

thank u :)

[jolly basin]

it's and btw

[midnight lantern]

oh fr?

cant x only be 1 value

[jolly basin]

zero product property

[midnight lantern]

otherwise itd be both x and y?

[jolly basin]

either value keeps the equation true

nvm

it's a wording thing

[midnight lantern]

lol ill js accept that its and

[jolly basin]

either is fine

[midnight lantern]

oh bet

well thanks, and have a good day :)

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@jolly mountain Has your question been resolved?

[jolly mountain]

<@&286206848099549185>

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[mint fern]

I have no idea what 23°40’ means can anyone help?

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@mint fern Has your question been resolved?

[wise cipher]

It's an angle, but in minutes.

Like you've seen a clock, right?

[mint fern]

Yes

[wise cipher]

It's just 23 + 40/60

Like 23 is the degrees

And between 23.01 and 23.09, they just made it 1 = 60

It's just a convention that mathematics, mostly astronomers use for angles

Since 60 has so many divisible numbers

it can be either divided to 2, 3, 4, 5, 6, 10, 12, and so on

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[coral raptor]

Hello, so I’m having a hard time figuring out how to find horizontal and veritical asymptote of a function that’s NOT rational

[coral raptor]

I’ve gave my shot at the problem

Oups

And basically I have to find the horizontal asymptote of

x+sqrt(x^2-2x)

I tried with limits

And I tried turning it into a rational expression?

But I’m not sure where to go afterwards

<@&286206848099549185>

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@coral raptor Has your question been resolved?

[coral raptor]

<@&286206848099549185> ;-;

[slender shard]

Did you try with lim as x-> -infinity?

[coral raptor]

Yess I did

In the picture above

But I’m kinda stuck after that

Cause I used lhopitals

And then the bottom is more complex

[slender shard]

Try multiplying by the whole fraction conjugate.

$\lim_{x\to -\infty} (x+\sqrt{x^2 -2x}) \cdot \frac{x - \sqrt{x^2 -2x}}{x - \sqrt{x^2 -2x}}$

[solid kiln]

Kookiemon

[slender shard]

See if that helps. I dislike this type of hand waving but it should work.

Hmm, it looks like you did that.

$2\cdot \lim_{x\to -\infty} \frac{x}{x-\sqrt{x^{2}-2x}}$

[solid kiln]

Kookiemon

[slender shard]

$2\cdot \lim_{x\to -\infty} \frac{\frac{1}{x}}{\frac{1}{x}}\cdot \frac{x}{x-\sqrt{x^{2}-2x}}$

[solid kiln]

Kookiemon

[slender shard]

$2\cdot \lim_{x\to -\infty} \frac{1}{1 - \frac{\sqrt{x^{2}-2x}}{x}}$

[solid kiln]

Kookiemon

[slender shard]

$2\cdot \lim_{x\to -\infty} \frac{1}{1 - \sqrt{\frac{x^{2}-2x}{x^{2}}}}$

[solid kiln]

Kookiemon

[slender shard]

$2\cdot \lim_{x\to -\infty} \frac{1}{1 - \sqrt{ 1 - \frac{2}{x}}}$

[solid kiln]

Kookiemon

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@coral raptor Has your question been resolved?

[normal grail]

Hi

[restive drift]

@frail current come back fren

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[languid pier]

alr the 1/ bit may be wrong but if its = to f(x) why is it wrong? i did the 1/ bc it was 1/x

[zinc ginkgo]

Chain rule

[languid pier]

oh

is the 1/ right?

[zinc ginkgo]

There's a couple things wrong with your answer

[languid pier]

mk so no 1/

[zinc ginkgo]

Try again and send your answer here

[languid pier]

mk

[frail current]

Fren?

[languid pier]

but wait I thought h'(x)=f(x) but thats if that 1/x up there is x, why would there be chain rule? also what do i do with the 1/x?

[random sky]

u use chain rule with ftc part 1

[languid pier]

let me just...

[restive drift]

friend. You’re unbanned from the server

[languid pier]

wat dat

[frail current]

ill rejoin

lol i cant it wont let me

[zinc ginkgo]

@restive drift @frail current talk somewhere else please

[random sky]

you said it here

so just replace t with 1/x, then multiply it by the derivative of 1/x

[languid pier]

ok

-1/x^2 right

[random sky]

yep

[languid pier]

it was wrong

[random sky]

why the 3 in the denominator?

[languid pier]

bc i forgot to put it on the numerator

duh

[random sky]

remove the 3 lol

[random sky]

?

bro where are you getting 3 from

[languid pier]

cahin rule?

[random sky]

noo youre not supposed to use the chain rule for the integram

just the 1/x

[zinc ginkgo]

There is no f' on the right

[random sky]

^

[random sky]

ok

[languid pier]

https://tenor.com/view/po-dies-dies-of-death-teletubbies-gif-23885910

[random sky]

and why is arctan3/x on the denominator

lmao

[languid pier]

im done with this shit

ugh

idk why i cant think the same as I used to

[random sky]

just put arctan(3/x) on top

[languid pier]

maybe i just havnt been sleeping enough

[random sky]

-arctan(3/x) / x^2

[languid pier]

but i cant sleep bc i have to much work

[random sky]

whys man having a mental breakdowwn

[languid pier]

college 👍

[random sky]

college 👍

[languid pier]

college 👍

ill be back

.close

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[languid pier]

hello I am back

[languid pier]

is it bc of the x on the bottom

[frigid pine]

kinda confused how you got that answer in the first place

[languid pier]

i replaces u with the top one and multiplied it by the dirivitive

[frigid pine]

umm

i dont understand what that means

ill just tell u how to do it

[languid pier]

ok

[frigid pine]

remember the derivative of arctan?

[languid pier]

i/x^2+1

[frigid pine]

oh wait

i just saw its g'(x) i cant read

[languid pier]

yes

[frigid pine]

okay so your answer is wrong due to 2 reasons

[zinc ginkgo]

At least 2

[languid pier]

;n;

[frigid pine]

(1) you didn't put the (6x) in parenthesis in 6x^2

[languid pier]

fair

[frigid pine]

(2) You didn't consider the bottom bound

[languid pier]

you mean ur mom?

im tired, continue

[frigid pine]

okay so the bottom bound

literally do the same thing for the top bound which you just did (with the parenthesis this time again tho)

and just subtract it

[languid pier]

wym

[frigid pine]

so you have the thing in the box

and then you add parenthesis around the (6x)

and then you have that

and you make a copy of it where you change all the 6s to 4s

[languid pier]

mk so i have parenthasees

seeeeees

[frigid pine]

and you subtract that from the original 6s

[languid pier]

why4?

[frigid pine]

because bottom bound

[languid pier]

but y

[frigid pine]

you are adding all the extra area coming on the right side

but you need to subtract all the extra area disappearing from the left side

[languid pier]

mk

but, how u get 4

do I always use 4

no matter the equasion?

[frigid pine]

you get 4 from 4x

the same way that you got 6 for the upper bound

[languid pier]

ohhhhhhhhhhh

oh ok

[frigid pine]

you get 4 for the lower bound

[languid pier]

so do the whole thing over again

but with 4x

and subtract it

[frigid pine]

yep

[languid pier]

wow

ty

ill be back

.close

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[dense agate]

I need helps.

[dense agate]

plssss

<@&286206848099549185>

i need help anyone?

[languid pier]

bro what

1 dont ping helpers for 15

2 say what the problem is

[random sky]

"1 dont ping helpers for 15
2 say what the problem is" -🤓

[languid pier]

,_,

[languid pier]

whats up

[dense agate]

ik this is not math but I can’t find the difference of this.

[languid pier]

I hate this

[dense agate]

loll

🥲

[languid pier]

go away

[random sky]

weather is like short term whereas climate is long term

does that help?

[dense agate]

yeah ik that but look at the little sticky notes.

I don’t understand that.

[random sky]

all of them?

the hurricane and blizzard r good

average rainfall is climate

raining is good

cold weather is good

georgia seasons is climate

[dense agate]

bro i swear thank u sm

[random sky]

the ones wrong in climate are both orange, blue and the bottom right pink (bird)

[dense agate]

It came like that.

I did not put that lol.

my teacher put that.

[random sky]

which ones?

[dense agate]

all of them.

[random sky]

lmfao what

[dense agate]

she wanted me to put it in the correct spot though.

[random sky]

ok

[dense agate]

cuz ofc she ain’t giving me the answers

[random sky]

then some of them might be wrong

[dense agate]

yk strict teachers

[random sky]

ig

[dense agate]

what do i do with these?!!

[dense agate]

look

6 wrong I think.

which are in the climate.

[random sky]

6 in each?

[dense agate]

No.

[random sky]

this should be good

[dense agate]

which ones are wrong?

[random sky]

ignore that

wait

yeah

circled ones

r wrong

[wraith hinge]

Climate refers to what the weather is like over a long period of time

What it is usually like in a specific area

Weather is a temporary thing

So you just look at each and think whether or not it is long term

[random sky]

weather or not

[dense agate]

(Corrections)

now this?

[random sky]

no

dude

[dense agate]

wymmmm?

[random sky]

put the joe one in climate

joe skiing

[random sky]

should be good

[dense agate]

alr tysmmm

[random sky]

np

[dense agate]

this is very helpful.

what grade r u in btw?

[random sky]

cegep1

[dense agate]

:00

then i must get a 100.

🤣

[random sky]

maybe but i might be wrong lol

take information from strangers with a grain of salt

[dense agate]

loll

alright tysmmm

this was so helpful

<@&286206848099549185>

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[trim jolt]

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[dense agate]

Is this right?

[waxen egret]

no

[dense agate]

why?

[waxen egret]

orange occupies a greater portion than black

[dense agate]

ohh ok ty

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[wraith hinge]

how to find the perimeter of AED if ED and BC are parallel?

[waxen egret]

what's p?

[wraith hinge]

perimeter

[waxen egret]

hmm

[wraith hinge]

oops sorry you have to find perimeter of AED not area

[waxen egret]

ok

[wraith hinge]

.

[paper swan]

well the problem is theres not enough information

We dont know the scale of the smaller triangle at all

in fact it could be larger

[wraith hinge]

bruh these problems are killing me 😭

[paper swan]

take a pic of the original question

there is something missing

[wraith hinge]

it's written in foreign language

[paper swan]

spanish by chance?

[wraith hinge]

Georgia

n

[paper swan]

georgian?

[wraith hinge]

yes

[paper swan]

well dont waste your time with this question

its impossible to solve without the correct information

[wraith hinge]

ah ok

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[wraith hinge]

sin 2x/2x and 3x/cos3x is 1 ....

[wraith hinge]

can anyone explain how ??

[stoic iron]

They're not

[wraith hinge]

I'm going to assume you are talking about their limits as they approach 0

But the second one wouldn't be even true in this case

You need to elaborate

[wraith hinge]

i forgot to mention that sry

its limit x tending to 0

$\lim_{x \to 0} \frac{\sin(2x)}{2x} = 1, \lim_{x \to 0} \frac{3x}{\cos(3x)} = 0$

[solid kiln]

♡LexQa♡

[wraith hinge]

is it some kind of property or ..

One day I will know how to write more than one line in latex

[wraith hinge]

The first one is indeed a representation of the common limit of $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ the geometric proof for that takes a while though so u might as well look that up

[solid kiln]

♡LexQa♡

[wraith hinge]

Second one is just

Not an undefined limit to begin with

Just substitute 0 in

[wraith hinge]

k

so does that mean 2x/sin2x is also 1?

yeah because you can always say that $2x = u, \lim_{u \to 0} \frac{\sin(u)}{u}$

[solid kiln]

♡LexQa♡

[wraith hinge]

Whatever approaches 0 is still going to approach 0 regardless of what constant you multiply it with

[wraith hinge]

one more thing is this limited only to 0 or ..

I mean yeah

If x is approaching other stuff then it is going to give you different stuff as a result

ok

in my view i think they multiplied numerator and denominator by (cx/2)^2 but how did that sin x became sinx/x (marked part)??

[wraith hinge]

They just multiplied the thing with x/x

[wraith hinge]

but it is only on denominator ... for it to be equal shoudn't they also multiply numerator with x/x?

It doesnt matter here

Because x/x is 1

Multiply that in the numerator too if u want

It's just 1

[wraith hinge]

lol i always forgot these kind of things

thx btw

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[tough wedge]

How does this become 1/sqr10, if we derive U (sqr(PM)), while P is 2 and M is 5? I get 5/2sqr(10)

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<@&286206848099549185>

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[fickle ledge]

how do I find sum from k=1 to infinite of 1/(2k-1)^2

[fickle ledge]

if 1/k^2=pi^2/6

[delicate bobcat]

assume sum 1/k^2 = 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ... = pi^2/6
then sum 1/(2k)^2 = 1/2^2 + 1/4^2 + 1/6^2 + ... = 1/2^2 * pi^2/6 = pi^2/24
think about the difference between these twos

because your sum is 1/1^2 + 1/3^2 + 1/5^2 + 1/7^2 + ... (generally next odd numbers in the denominator)

[fickle ledge]

right, but why is the second series a fourth of the first

[delicate bobcat]

because it gives even numbers in the denominator which we want to get rid of

[fickle ledge]

yea ik

but why 1/2^2 * pi^2/6

[delicate bobcat]

because we have sum of 1/(2k)^2

1/2^2 is constant

[fickle ledge]

oh its 1/4*k^2

[delicate bobcat]

ye

[fickle ledge]

alr I get it

ty

I got it

[solid kiln]

Mehdi_Moulati

$\sum_{k=1}^{\inf} \frac{1}{k²} = \sum_{k=1}^{\inf} \frac{1}{(2k)²} +  \sum_{k=1}^{\inf} \frac{1}{(2k-1)²}$

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[stable drift]

Can someone please explain 12 a)

I understand what all of that information is

But when it comes to actually graphing it I get confused where to plot the points

The x axis points confuse me

Like I understand in terms of the y axis you add 3 from 5 to get the max point and to get the min point you subtract 3 from 5

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[wraith hinge]

I really dont get how to solve for F 😦

[small pivot]

Why do the weird stuff with the indicies

Isn't is just 100F=16(480-20F)

[wraith hinge]

It worked

Thanks so much

[small pivot]

Nw

Don't overcomplicate stuff lol

[wraith hinge]

Altough i dont know what indicies means lol

[small pivot]

Like u did to the 0.5 and stuff

[wraith hinge]

I wouldnt know how to calculate with ^-0,5 thats why i wrote it the other way 😦

[small pivot]

Yeh just try not to get powers involved unless u have to

Makes stuff more complicated

[wraith hinge]

You notice the bottom version is the one i modified to get the top one?

I basically got rid of them to make it easier

Maybe you read it the other way

[small pivot]

Oh yh nvm

[wraith hinge]

lol

[small pivot]

I didn't see the arrow lol

[wraith hinge]

😄 alright then

[small pivot]

Nvm ignore me then lol

[wraith hinge]

May i ask how you identified the next step though? Is it always a priority to get rid of the numbers below the -------- ?

[small pivot]

Yeh I always try to

I just timesed both sides by the bottom

To get rid of fractions

[wraith hinge]

Do you know if there is like a general scheme one would use when solving for something?

Like, first get rid of of all ...
Second do ...
If there is ... do ..
And so on

[small pivot]

Idk not rlly

[wraith hinge]

I was wondering if there is like a scheme you can follow to identify what the next step would be

[small pivot]

Maybe there is but I don't know of any

Acc no scrap that

[wraith hinge]

So you just use your general math knowledge ?

I never heard of bidmas lol

[small pivot]

U sorta do it automatically after a while

I rlly can't think of the right order

U can try Google maybe or smthn

Sorry I just have no clue rn

[wraith hinge]

Thank you still!

[wraith hinge]

May i lastly ask if you used some source for practice to get used to it?

Like a collection of practice questions somewhere ?

[small pivot]

Um good question

I did it a while ago NGL lol

[wraith hinge]

I really dont have much experience and get stuck with lots of calculations because of it :/

[small pivot]

U want practice for just solving equations?

[wraith hinge]

I think so yes

Problem solving with math rules

[small pivot]

https://corbettmaths.com/wp-content/uploads/2022/06/Advanced-Changing-the-Subject.pdf

This is like changing the subject of a formula

Which is kinda the same thing but harder ig

Cause u need to manipulate the equations

[wraith hinge]

Thank you! I will have a look at it 👍🏻😊

[small pivot]

Nw

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[warm sierra]

If let’s my prof said this is true

Since the denominator value is lambda and the coefficient bfr the identity matrix is 1 therefore it’s just K=Lambda

If we were to say 2 * identity matrix - 1/4 * A would the lambda value be 2*4=8=lambda?

Assuming that the determinant of A is 0, not invertible.

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[warm sierra]

Bruh

.close

<@&286206848099549185>

Previously asked but no reply and channel closed

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[mental frigate]

hm

[mental frigate]

A computer that was purchased for $6000 depreciates at a rate of 13.4% per year. An equation modeling this is, f(x)=6000(0.866)x, where x is the number of years.
a) What is the computer worth after 2 years?

b) When will the value of the computer be $2530?

[warm sierra]

A)4392$ since each year ur depreciating 13.4%

6000$ * 0.134=804/year
804 * 2(for 2 years)=1608
6000-1608=4392

Can someone also confirm

[night zodiac]

First, don't do the work for people or give out answers

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[mental frigate]

[mental frigate]

I know the answer

[mental frigate]

Can you see if mine correct

[night zodiac]

Post your work

[mental frigate]

I was going to

[warm sierra]

Ah ok first timer Ty

[mental frigate]

Had to walk the dog :-:

After two years, the computer will be worth 6000 * .866 * .866 = 6000 * .866^2 = 4499.736.
Using the following formula, you can find how long the computer will take to equal 2530.
2530 = 6000 * .866 ^ n
In this case, n represents the number of years. divide both sides of the equation by 6000
to get: 2530/6000 = .866 ^ n
Calculate the log of both sides of the equation to get, log(2530/6000) = log(.866 ^ n)
by log rules, this becomes
log(2530/6000) = n * log(.866) solve for n to get n = log(2530/6000) / log(.866) = 6.002209934.
confirm by replacing n in the original equation to get: 2530 = 6000 * .866 ^ 6.002209934.
this becomes 2530 = 2530, confirming that the value of n is correct.
The solution is that the value of the computer will be 2530 in 6.002209934 years.

[mental frigate]

correct?

looks better in my doc

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[woven valley]

We have to solve for r

[woven valley]

I’m a little hung up on where to start first, and also how to handle the r cube

I’m guessing to divide the 4/3 and pi but I really am not getting the fluency of how that’s supposed to look like

[zinc ginkgo]

Dividing by a fraction is the same as multiplying by its reciprocal

x = 2/3

[woven valley]

Reciprocal meaning the denominator right

[zinc ginkgo]

Then (3/2) x = (2/3)(3/2) = 1

Reciprocal of a/b is b/a

[woven valley]

Meaning the fractions are just flipped

Right

That word messes me up

[zinc ginkgo]

Say it enough so it becomes part of you

[woven valley]

So that is correct though? The reciprocal is just flipping the fractions upside down?

[zinc ginkgo]

I wouldn't call it that, but if you want to go ahead

[zinc ginkgo]

As long as you memorize this

[woven valley]

Yeah, I feel like I understand what it represents switching the numerator and denominator, but that word I feel like I don’t understand why that’s the word

But anyway

So I tackle the fraction first? Or the exponent?

[zinc ginkgo]

This is right

[woven valley]

So I dont try and get rid of the exponent first? I’m just unaware of what to clear order wise

[zinc ginkgo]

You can try both ways

One is fewer steps

[zinc ginkgo]

You divided by pi on the right, but multiplied on the left

You have to be consistent to keep the equation the same

[woven valley]

Oh damn brainfart

I’m guessing here but would it look like this?

Because that looks super strange to me

[zinc ginkgo]

Right

[woven valley]

That’s right!?

And then I’m assuming I cube root off r?

[zinc ginkgo]

Yes cube root both sides

[woven valley]

Would I have mattered which I started with? Would I still end up with the same answer or could I have ended up with numbers outside of the cube root? It kind of throws me off

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[humble sierra]

trying to find the answer to D

i feel like i've tried every possible combination but can't find it.

[lament jewel]

what did u try

did u round

[humble sierra]

yes i did

answer b/answer c and rounded

i did answer b/answer b+answer c

not sure what else to try

[lament jewel]

er

i think its

(c-b) / c

[humble sierra]

here's the begnning btgw

i'll try that

PLEASE

PLEASE PRETEND I NEVER ASKED

i rounded so terribly

thanks for your help !!1

[lament jewel]

lol

wot

gj solving it

💕

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[round sedge]

I’m confused on how I prove it’s an isos

[long basin]

The base angles are congruent

[round sedge]

Omfg

I just realized that’s a reason

I’m stupid as hell 😭

[long basin]

It happens

[round sedge]

Thank you mr leviathan

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[clear thistle]

Can anyone helo me out

Idk what's its saying??

Maybe it's telling me this??

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<@&286206848099549185>

Man

<@&286206848099549185> can you pretty pls help me?? It's been like 8 hours and I'm not finding any idea what the f is this question saying.

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[wraith hinge]

what is this " ln x " ?

[wraith hinge]

what your problem??

y = e^x
x = ln y

i am just learning formulas

[wraith hinge]

can you tell me what does "ln" mean?

[wraith hinge]

read

e is neper number

ln is log
e

so it's lust d/dx(log x) = 1/x ??

yes

ok

but in e

log
e

not log
10

understand?

[kindred pier]

"natural log"

[wraith hinge]

k it is only applicable for natural logs is that it?

[wraith hinge]

k

k ??

https://en.wikipedia.org/wiki/E_(mathematical_constant)

read it

[wraith hinge]

i meant ok....... i want to learn maths faster for entrance exam you got any tips to memorize every formulas?

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[dusky frost]

Yello

[shadow glacier]

yello

[dusky frost]

Is this right?

Im practicing inverse variations

[balmy basin]

looks like it

[dusky frost]

Thank you

Question

If x is followed by directly varies or inversely varies then after that is y

Does that mean like that x = yk

Like directly or inversely is an sign of =

[balmy basin]

yes, you can treat it as "=k"

[dusky frost]

Oh nice

I always though it was something lime that but never asked cher about it

Thanks for clearing it up for me

Oh also is the dif between direct and inverse that inverse decreases y when x increases or vice versa?

[balmy basin]

yep

[dusky frost]

Thank you

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[cursive halo]

Hello! I've given this info (on the pic) and i need to slove the line middle point and length.

[cursive halo]

and i dont know where to start

i missed my class where they talked about that

ough and ive given this aswell

[shadow glacier]

ru sharing the answers too

[cursive halo]

yep

[quaint gazelle]

find the line starting (A) and ending (B) points then calculate the middle point using (A+B)/2

and then do lenght using the distance between A and B

[cursive halo]

(3,5;6;3) and the lenght is square root of 21

and how do i find starting line A?

[quaint gazelle]

t=0 and t=1 are the values for the points

[cursive halo]

But if t=0 everything is 0 right?

[shadow glacier]

instersection of all three

equations

[cursive halo]

and now if i change x all of the others have to equal to each other right?

[cursive halo]

how do i do that.... im sorry for asking too much

[shadow glacier]

well draw three lines

by putiing

x=0 and find t

then t=0

and find x

you will get two points

make them

lemme solve it

hold my beer🍺

[cursive halo]

holding 🍺😀

[shadow glacier]

ok

here

is the steps

find t in terms of x

from 1st equation

t=x-3

now put this in both 2nd and 3rd equation

now form lines

of 2nd and 3rd equations

try

[cursive halo]

like this?

[shadow glacier]

yup

form a line

of 2nd equation

and now think

the range of t

as you know t

must be in some range

thats 0,1

[cursive halo]

?

second one is Z not 2

or am i doing something wrong now?

@shadow glacier

now when i replace X in the equation before i get 0,1 which is t = y+1/2+3 = z-1/4 + 3?

[shadow glacier]

nice

try forming

lines

with these equations

[cursive halo]

how do i form lines doe? 👀

emmmm

im confused now

@shadow glacier sorry for mentioning again

im confused as hell

@quaint gazelle am i doing something wrong?

yeah ive gave up

<@&286206848099549185> if you can guide me it would be nice

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[zinc ginkgo]

pain

can you pick JUST ONE and work on it on your own and then show your attempt

where are your steps?

you figure out the limits from this

draw a picture

where's your normal vector?

$F\cdot n dS$

[solid kiln]

riemann

[zinc ginkgo]

what's n and dS there?

https://tutorial.math.lamar.edu/Classes/CalcIII/ParametricSurfaces.aspx
https://tutorial.math.lamar.edu/Classes/CalcIII/SurfaceIntegrals.aspx
https://tutorial.math.lamar.edu/Classes/CalcIII/SurfIntVectorField.aspx

if you're supposed to use divergence theorem, then
https://tutorial.math.lamar.edu/Classes/CalcIII/DivergenceTheorem.aspx

don't ping random mods for math help

yes you have to figure out the volume from your equation

[zinc ginkgo]

.

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@wraith hinge Has your question been resolved?

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@wraith hinge Has your question been resolved?

[zinc ginkgo]

try both ways and see which is easier

There are worked examples in the links if you read them

Work through it yourself then ask when you get stuck

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@wraith hinge Has your question been resolved?

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[last flume]

How do I graph this?

[vocal hazel]

make a straight line from the origin, in the direction of 2pi/3

so in the direction of 120 degrees from the positive x axis

[willow crane]

i need help bruh

[fathom ivy]

#❓how-to-get-help

[last flume]

Oh thats it? I thought i needed to find R lol

[vocal hazel]

yep, if r is not specified, we can assume it can be any value, thus we only need to restrict theta

[last flume]

Oh ok got it, thank u

[fathom ivy]

hard to solve for r when you know nothing about it

so you only have the implied r >= 0

#❓how-to-get-help

[night zodiac]

#❓how-to-get-help

[last flume]

thanks 😺

.close

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[wraith hinge]

Can someone explain why it’s wrong

[long basin]

$3x^4 ≠ (3x)^4$

[solid kiln]

Umbraleviathan

[wraith hinge]

oh ok

thank you for helpiong

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[trail pike]

can someone help me past the dy/dx part

i cant solve for a and b

[wraith hinge]

you're given two things, $y(1) = 4$ and $\frac{\dd{y}}{\dd{x}}\left(1\right) = -\frac{3}{2}$

[solid kiln]

45

[trail pike]

yea

[wraith hinge]

compute y(1) first

[trail pike]

a/(b+1)^(1/2) = 4

[wraith hinge]

yes

[trail pike]

my dy/dx is

[wraith hinge]

a^2 = 16(b+1) in other words

[trail pike]

wait how

[wraith hinge]

what happens if you square this

[trail pike]

if u square it

isnt it now different

than what we had originally

?

[trail pike]

u still there

[compact sigil]

The equation for a curve that passes through the point (1,4) is $y=4$. The gradient at this point is given by $-3/2$, so the equation of the curve can be written as $y = -3/2 (x-1) + 4$.

The given equation of the curve is $y=\frac{a}{\sqrt{bx+1}}$. Setting this equal to the equation for the curve that passes through (1,4) and has gradient -3/2 at this point, we have

[\frac{a}{\sqrt{bx+1}} = -\frac{3}{2}(x-1)+4]

Rearranging and squaring both sides, we have

[a^2 = \left(-\frac{3}{2}(x-1)+4\right)^2 (bx+1)]

Evaluating this equation at the point (1,4), we have

[a^2 = \left(-\frac{3}{2}(1-1)+4\right)^2 (1+1)]

This simplifies to $a^2 = 16$, so $a=\pm 4$. However, since the original equation has a square root, we must have $a \ge 0$, so $a=4$.

Substituting this value of $a$ into the equation of the curve and simplifying, we have

[4 = \frac{4}{\sqrt{bx+1}} = \frac{2}{\sqrt{bx+1}} \cdot \frac{2}{\sqrt{bx+1}} = \frac{1}{bx+1}]

We can solve for $b$ by squaring both sides, which gives

[16 = \frac{1}{(bx+1)^2} \implies 16(bx+1)^2 = 1]

Again, we can evaluate this equation at the point (1,4) to find $b$. This gives us

[16(1+1)^2 = 1 \implies 36b = 1 \implies b = \frac{1}{36}]

Thus, the value of $b$ is $\frac{1}{36}$. The curve y=a/((bx+1)^(1/2)) that passes through the point (1,4) and has gradient -3/2 at this point is given by

[y = \frac{4}{\sqrt{\frac{1}{36}x+1}}.]

[solid kiln]

MeatBagFrank

LaTeX source sent via direct message.
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.61 [\frac{a}{\sqrt{bx+1}}
                            = -\frac{3}{2}(x-1)+4]
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```

[compact sigil]

@trail pike does this help? if not, let me know what you have trouble with.

[trail pike]

yeah um the y = -3/2(x-1)+4

im not sure where that came from

[compact sigil]

The equation $y = -\frac{3}{2}(x-1)+4$ is the equation of a line that passes through the point (1,4) and has a slope of -3/2. This equation is derived using the point-slope form of a line, which is given by $y-y_1 = m(x-x_1)$, where $(x_1,y_1)$ is the point through which the line passes and $m$ is the slope of the line. In this case, the point through which the line passes is (1,4) and the slope is -3/2, so the equation of the line is $y-4 = -\frac{3}{2}(x-1)$. To put this equation into slope-intercept form, we can solve for $y$ to get $y = -\frac{3}{2}(x-1)+4$.

The equation $y = -\frac{3}{2}(x-1)+4$ is used in the given equation $\frac{a}{\sqrt{bx+1}} = -\frac{3}{2}(x-1)+4$ because this equation defines the points on the curve where the line intersects the curve. The values of $a$ and $b$ are determined by solving this equation for specific values of $x$ and $y$.

[solid kiln]

MeatBagFrank

[compact sigil]

@trail pike

[trail pike]

yea

sorry im reading it atm

one sec

wow i never thought about it that way

from what i understand in txtbooks

that point slope form is only used to find tangent for a point on a curve

[trail pike]

also

how can u just square both sides like that

doesnt the product become different

[compact sigil]

Squaring both sides of an equation is a mathematical operation that allows you to eliminate the square root on one side of the equation. In the given equation, we squared both sides in order to eliminate the square root on the left side and simplify the equation.

Here's how we did it:

First, we started with the equation $\frac{a}{\sqrt{bx+1}} = -\frac{3}{2}(x-1)+4$.

Then, we squared both sides of the equation to get

$$\left(\frac{a}{\sqrt{bx+1}}\right)^2 = \left(-\frac{3}{2}(x-1)+4\right)^2.$$

The left side of the equation is the square of the original equation, which means that we have squared the fraction $\frac{a}{\sqrt{bx+1}}$ and obtained

$$\frac{a^2}{\left(\sqrt{bx+1}\right)^2} = \frac{a^2}{bx+1}.$$

The right side of the equation is the square of the line's equation, which means that we have squared the expression $-\frac{3}{2}(x-1)+4$ and obtained

$$\left(-\frac{3}{2}(x-1)+4\right)^2 = \frac{9}{4}(x-1)^2 + 8(x-1) + 16.$$

Since the left side of the equation is equal to the right side, we can set the two expressions equal to each other and obtain the simplified equation

$$\frac{a^2}{bx+1} = \frac{9}{4}(x-1)^2 + 8(x-1) + 16.$$

This equation is easier to solve than the original equation because it does not have a square root on the left side. By solving this equation, we can find the values of $a$ and $b$ that satisfy the original equation.

[solid kiln]

MeatBagFrank

[trail pike]

but if u look at it from a graphical point of view

the graph would have changed

[compact sigil]

how so?

[trail pike]

like if u took 3x^2 - 3x - 9 = (4x + 1) ^ (1/2)

squaring both sides

then taking either side of the eq and comparing it bfr and after it was squared