help-26

[fleet walrus]

If you go over 4 it counts as blue

[cinder sequoia]

well the probability of getting a red at some point in the 4 rolls for scenario 2 is 1 minus the probability of getting all blues

which is 1 - (1/2)^4, assuming the rolls are independent

[fleet walrus]

Which is 93.75 percent

[cinder sequoia]

right

[fleet walrus]

Which is why I chose that for the first scenario

[cinder sequoia]

so what do you mean by "is it safe to treat the second scenario as the first?" technically the probabilities of getting red are the same

but whether you want that to be a higher probability for a single roll or a lower probability across multiple rolls depends on the purpose

[fleet walrus]

Basically, if I didn’t know that scenario 2 and 1 were performed differently would it matter?

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[fleet walrus]

<@&286206848099549185>

[lucid bridge]

yes u can treat them the same, theres no difference

@fleet walrus

[fleet walrus]

Thank you so much 🙂

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[spare fiber]

uh, I dont know understand what this question part b wants me to do

[spare fiber]

if I still have to use trigo sub at the end, why dont I use it at the beginning

thx

[finite storm]

to show symmetry, maybe?

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[spare fiber]

across the y axis?

[finite storm]

ye

i showed

${\int_{-a}^{a} \frac{1}{1+x^2} \dd x = 2\int_{0}^{a} \frac{1}{1+x^2} \dd x}$

[thorny flame]

k

[finite storm]

but itstrue for any even fx anyway

[spare fiber]

yes i think so

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[hallow viper]

.open

How would I calculate the level of fuel in this tank for a given role and pitch angle ?

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[grand pawn]

I need to find the surface area of the Blue triangle. Could anyone help me?

[inner wren]

What is given?

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[grand pawn]

.reopen

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✅

[grand pawn]

<@&286206848099549185>

[warm turtle]

are these the only given ?

[grand pawn]

yes

[wooden osprey]

interesting problem

[fallen vector]

no dots ?

[grand pawn]

Nope

[warm turtle]

has a ratio between the lengths of the sides been given?

[grand pawn]

nop

[fallen vector]

you just see there is a triangle square

=> 2 edges = 90

[grand pawn]

Santa faces a difficult task. Unfortunately, the 3 departments of the Christmas factory have fallen out, as the gift distribution center has less space than the gift shopping department. Kann-nix's section, who is the boss of the "mess, mistakes and rejects" department, clearly has the most space. However, Santa can't knock down load-bearing walls, so he thought he could work out the area of the gift distribution center and give them the chance to divide up this space well. Santa has printed out the cross-section, but due to all the things standing around he could only measure one side, namely 8m as drawn. He also knows that one of the triangles is at a right angle. Can you help Santa calculate the area of the blue section?

---

this is translated from another language so the translation might lack understandibility xD

The thing is, i tried to find out the Hypothenuse of the tops triangle because then I would know the Hypothenuse of the blue one but I am missing values to plug in

[fallen vector]

so triangle square is 2 edges = 90

[grand pawn]

and how do I go off of that?

[fallen vector]

the sum of the 3 angle in triangle is 180

the sum the 2 angles is 90 if it is a he triangles is at a right angle

know it

Pythagoras: 8^2 + ?^2 = a^2

i give you some hint

this is the first hint'

[grand pawn]

Hmmm, okok, I will try it later and contact you if I need more help

thx ^^

[fallen vector]

area of triangle: 8 x h : 2

...

draw a height of blue triangle

properties of height is

[wooden osprey]

not enough information

[fallen vector]

height is perpendicular to v

hypotenuse

here

[wooden osprey]

again,not enough information

if you shorten/lengthen OB, the properties are still retained, but area of AOD might change

[fallen vector]

thanks

draw OE is a height of area of AOD triangle

[wooden osprey]

không đủ thông tin. đừng có ráng giải nữa

[fallen vector]

kk

i think use sin cos tan

to solv this

[wooden osprey]

brother

well if we also have additional information we might be able to solve it

but as of now, i don't think so

[inner wren]

Is it known that ABCD is a square?

[wooden osprey]

looks like so

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[grand pawn]

.reopen

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[grand pawn]

yes

But do ya'll have the answer or should I just print it out and check the angles and solve it this way?

[warm turtle]

the figure might not necessarily be drawn to scale

oh wait, but is abcd proven to be a square?

[grand pawn]

Well, I am pretty sure but this seems hopeless xD

[warm turtle]

oh omg sorry i didnt read thiss

[grand pawn]

Let me check again

[warm turtle]

if it is there is a higher chance that it is actually drawn at least near to scale

because everything is enclosed in the square

[grand pawn]

In the instructions there isnt anything that says that it is a square but as far as I know my math teacher, he wouldnt give us a rectangle

[grand pawn]

would that make it easier to solve mathematically?

[warm turtle]

there shouldn't be distortions in the angles or the lengths because if that were the case the square would not be drawn to scale, which does not seem to be the case

[warm turtle]

i dont know 😭😭😭

[grand pawn]

hahahaha okok

[warm turtle]

i feel like there is not enough given information

[warm turtle]

there may be a chance that the measuring of the printed figure may work hahahaha i dont guarantee

[grand pawn]

🤷 well then back to the good old paper, pen and protractor 😂

[grand pawn]

yea, but better trying that than not trying at all

[warm turtle]

yess

better than nothing !

[grand pawn]

still thanks

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[normal otter]

idk how to do this 😭

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[round pagoda]

pls explain why 3k+2 was used

if question says clearly multiple of 3 OR one more multiple of 3

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[round pagoda]

pls

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[dusk hedge]

hi

[dusk hedge]

can I ask for help on circuit related questions here?

I need to find Vout

I have no idea where to start doing KVL / KCL

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[young lagoon]

did that once

was told i was weird

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[winged wigeon]

Can anyone help? If i show that A1B1 and A2B2 are parallel, its done, but i have no idea on how to do that. I was trying to prove that A1B1A2B2 was a trapezoid, but idk

[winged wigeon]

I know that the lines are not parallel in the image, but it is just a sketch

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[sonic escarp]

A1B1C1 and A2B2C2 are obviously similar. this should give you the first step.

[winged wigeon]

I dont see why its obvious

[sonic escarp]

what can you say about the angles at P? concrete: angle B1PA1 an angle B2PA2.

[winged wigeon]

Oh i see

Thanks

[winged wigeon]

Ooooooo i got it

Thanks!!!

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[dusk hedge]

lol

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[frozen solar]

can someone check this for me?

[frozen solar]

and this one too pls

[raven field]

vertex is correct the mistake is in the coefficient

[frozen solar]

would the coefficient be 2 then?

wait the coefficient for both is wrong?

[raven field]

Then it would be open to the right so no

[frozen solar]

-2 my bad

[raven field]

(h,k) is correct
It’s just the c value

[frozen solar]

but the c value would just be the distance from the vertex to the foci right?

[raven field]

Yep

[frozen solar]

in this case wouldn’t it just be -1?

[raven field]

Yes

[raven field]

This is correct

[frozen solar]

can i get this one checked too when you have a chance?

[raven field]

And this too

[frozen solar]

yay ty

[raven field]

Yea this is correct

Man it’s been a long time since I did conic sections

[frozen solar]

lmfao yeah this is mac 1146 i think

integrated algebra

does this look correct?

whops i took a picture of the wrong one

okay there

and this one

[raven field]

This is not C

[frozen solar]

ah i see now, it’s 0,4 lol, it’s supposed to be a right?

[raven field]

It’s supposed to be D

[frozen solar]

oh shoot why, i thought the coordinates were 0,4 and 0,-4

also last question

my bad for all the overwhelming questions

[raven field]

These numbers are toooooo big

[frozen solar]

i should simplify the bottom?

okay i fixed it

i put 25^2 and 15^2

that was the most i could simplify it

[raven field]

Man I’m a little rusty but this should be C

[frozen solar]

oh so it was c? thank god lol

[raven field]

Ngl my brain got confused a lot I redid it 3 times

[frozen solar]

do not fret lol, tysm i appreciate your help

[raven field]

Correct

[raven field]

Correct

[raven field]

And this is correct

[frozen solar]

YAYYY

thank you so much

[raven field]

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[radiant sky]

Hi I need help my tutor left

[topaz sinew]

Please don't occupy multiple help channels.

[radiant sky]

He left one me

[valid pulsar]

do you need help for all the questions?

or just the general concept

[radiant sky]

Yeah but sorry I just started eating can you please wait

Sorry

I just started

It will be quick

[valid pulsar]

yes

[radiant sky]

I’ll ping you when I’m back

[valid pulsar]

ok

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@radiant sky Has your question been resolved?

[radiant sky]

No

@valid pulsar has

Il back

Sorry

I tried to find

Yo

[valid pulsar]

sorry

yes

im here

[radiant sky]

Ok

I need help with all

[valid pulsar]

alright

[radiant sky]

And general convey

Concept

[valid pulsar]

so for the top left picture, set x as the hours of babysitting done

[radiant sky]

I don’t understand

How to find variable

[valid pulsar]

by shery

oh

the variable is

basically the unknown value the question is asking for

[radiant sky]

Yeah

[valid pulsar]

in the first question it asks how many hours she worked

so set the amount of hours she worked as a variable

[radiant sky]

So x? Is 8.25

[valid pulsar]

no

[radiant sky]

8.25x

[valid pulsar]

yes

+15

[radiant sky]

But why wouldn’t it be 15 or 64.59

50

[valid pulsar]

the equation in that question is 8.25x+15=64.5

[valid pulsar]

because 15 is the one time charge, there is only one time that she adds that value

[valid pulsar]

64.59 is the total value or the amount she earns

[radiant sky]

Ohh

so 8.25x +15=64.50

[valid pulsar]

yess

[radiant sky]

Im solve it

[valid pulsar]

ok

[radiant sky]

Rq

Wait

How do you divide

8.25x=49.5

Cause we’re getting x alone

Divide by 8.25?

[valid pulsar]

yes

[radiant sky]

Ohh

Ok

One sec then

[valid pulsar]

49.5/8.25

[radiant sky]

X=6

[valid pulsar]

yup

[radiant sky]

Do you see where I’m having trouble

Like finding

The variable

[valid pulsar]

yeah from the question

[radiant sky]

Yeah

So the next one I finished awhile ago

[valid pulsar]

the friday one?

[radiant sky]

No cam one

[valid pulsar]

yeah that one

ok

and it equals to 4?

[radiant sky]

Yeah

[valid pulsar]

alr

[radiant sky]

So we’re on Ayden one

[valid pulsar]

ok

do yk what the x is

[radiant sky]

35

35d

35x

Frick

[valid pulsar]

lol not quite

[radiant sky]

What

[valid pulsar]

35 is the flat rate every month

but the question is asking for the amount of additional classes he took

[radiant sky]

Oh so 12x

[valid pulsar]

yes

[radiant sky]

Ok finding the variable is rellay hard so after the questions can you teach me how to find them

In question

Please

[valid pulsar]

I can give you some other questions and you can tell me the variable

without solving it

[radiant sky]

That’s fine

After tho

Alright so it’s 12x

35+12x=143?

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[valid pulsar]

alr give me 10 min after though i need to eat

.reopen

[radiant sky]

That’s fine

Please ping me

After tho

So I can come back

[valid pulsar]

yeah after I help you with the origenal questions

[radiant sky]

Ok

[valid pulsar]

yes

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[sacred shale]

Wojtek had 30 two-zloty coins and 48 five-zloty coins. He exchanged half of the five-zloty coins for two-zloty coins. The amount from exchanging the five-zloty coins was equivalent to the value he received in two-zloty coins.

Calculate how many two-zloty coins Wojtek now has in total. Show your calculations.

[sacred shale]

30 - 2 zł
48 - 5 zł

Half 5 zł

24

Divide by 2

12 - 2 zł

So it’s gonna be ?

42 - 2 zł
24 - 5 zł

[rigid ivy]

Oh "half of the five-zloty coins" misled me for a moment. I was asking myself "How do you exchange 2.5 coins?? "

[rigid ivy]

Forgive my ignorance on your currency, but is zł your currency symbol, or are you doing z as a variable, and then ł alone is your currency symbol?

[sacred shale]

zł is the currency symbol like $ or €

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[royal sky]

why is A not?

[vernal matrix]

What's the subspace test say?

(also hiii )

[royal sky]

hii

scalar mult and adition

[vernal matrix]

Well, for example, is (a + b)^2 the same thing as a^2 + b^2?

[royal sky]

ahhhhhhh

okoko

uh

lets say theres

v=(a, a^2,a^3) u=(b,b^2,b^3)

addition

wouldnt it be

u+v=(a+b, a^2+b^2,...)

[vernal matrix]

Yep, you then would want, by the definition of the set, the second entry to be (a + b)^2, the third to be (a + b)^3 and so on

But, in general, it isn't

[royal sky]

ah

yeah cuz final form has to saty (z,z^2,z^3...)

okokok

ytytytytytytytyttyty

tytytt

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[royal sky]

also im not saying im learing matrix

dimensions

vector space to dimensions in one night

but

[vernal matrix]

anything is possible they say

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[empty fractal]

I’m doing #4, this should be easy but honestly I’ve been throwing up and sick all day and my brain isn’t processing anything. I’d prefer if someone helped me out quick instead of an entire math lesson

[steady escarp]

So there are 2 qtys involved: set fee per month, and fee for the number of movies per month

They probably want u to recognise which of these is represented by the slope

And which is represented by the intercept

[empty fractal]

yeah

so y is the total cost that depends on how many movies are streamed

but how do I write that

[empty fractal]

if that’s right

like uh if I’m right idk

[steady escarp]

Y is the total cost yes

But what is ur answer to 3?

[empty fractal]

omg sorry I didn’t seee this

I’m like floating half asleep

uh y=3x+22

[steady escarp]

What is 3 there called?

[empty fractal]

slope

[steady escarp]

And which of these 2 qtys does 3 represent

[empty fractal]

number of movies

?

[steady escarp]

I don't want u to guess

[steady escarp]

Why do u think it is this

[steady escarp]

Also what does x here represent?

[empty fractal]

I have no idea

my brain isn’t working

[steady escarp]

What does x mean?

[empty fractal]

Number of movies streamed

[steady escarp]

Yes

So to find the total cost u multiply x by 3

Number of movies streamed * 3

So what must 3 mean?

Fee for what?

[empty fractal]

uh the movie

the movies

wait I had it I forgot hang on

the cost of the movies you streamed which is 3 pppperrrrr movie?

[steady escarp]

Yes

3 per movie * number of movies = total cost for all the movies

But there's another cost involved that is always charged, on top of this

[empty fractal]

yea

Uhhhh

[empty fractal]

the flat fee of the service per month?

[steady escarp]

Yeah

So it's total cost for all the movies + flat fee

(3 per movie * num of movies) + flat fee

So what is the slope, and what does it represent and what is the intercept and what does that represent

[empty fractal]

the slope represents the number of movies x 3 per movie and then the y intercept represents the flat fee of the service per month?

okay ty

.close

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[west canopy]

unit vectors are (3 1) for x axis and (1 2) for y axis
vector is equal (1 4) after transformation
find the original vector

[west canopy]

why doesnt this work?

[steady escarp]

It shud work

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[steady escarp]

Why r u saying it is wrong?

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[high hare]

can someone tell me the answer of this i need to verify

[coarse tusk]

what's your answer

[high hare]

-2/5

[coarse tusk]

that doesn't sound right

,w limit as x tends to 0 of (sqrt(1+sin^2(x))-cosx)/(sqrt(cosx)-sqrt(cos3x))

[thorny flame]

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Requested by kheer257

[coarse tusk]

yeah that seems reasonable

!show

[topaz sinew]

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

[high hare]

damn i tried every math solver and it didnt work

alr how do i close?

.close

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[livid badger]

hey just a quick question, can anyone remind me why x^2 can go into the square root resulting in square root of x^4? i learned how this worked back in like gr 10 but now i forgot and then it came back to haunt me again so just wondering again how it works again lol

[lethal anchor]

it is just that [
x^2 = \sqrt{x^4}= x^{4/2}
]

[thorny flame]

Aero

[livid badger]

but how does it become x^4?

[lethal anchor]

well, squaring and taking the square root are inverse operations. So: [
x^2 = \sqrt{(x^2)^2}
]
what's happening here is that we took both the square root and square simultaneously. So the equality holds

[thorny flame]

Aero

[lethal anchor]

and $(x^2)^2 = x^4$

[thorny flame]

Aero

[livid badger]

ahh okay i see

thank you so much!

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[distant spoke]

could someone help and teach me with these problems for my homework? i'm a bit confused with what to do

[topaz sinew]

@distant spoke Has your question been resolved?

[distant spoke]

<@&286206848099549185>

[fallen vector]

what is m ?

i only study like BAC, ABC,...

[distant spoke]

m is the arc

[fallen vector]

this one is find the angle ?

[distant spoke]

yea the measure

you find the measure

my bad, m is not arc, its measure

so what is the measure of m<1, m<2, m<3, etc.

[fallen vector]

AMH = 70

[distant spoke]

wait

is m<1 = CA or CH?

[fallen vector]

measure of arc 1

[distant spoke]

yeah

[fallen vector]

let me find

[distant spoke]

okay ty

[fallen vector]

find the

mAC

mAc = 180 degress

[distant spoke]

yeah

so 180/2

= 90*

?

[fallen vector]

the first one

[distant spoke]

okay so m<1 = 90*

[fallen vector]

mCH = 110

Notice

[distant spoke]

okay

what is mCH?

[fallen vector]

m<4 is 70 degress

[distant spoke]

m<1?

wait can we do this in order?

and elaborate me

[fallen vector]

let me caculate

[distant spoke]

m<1 = mCH?
mCH = 110*
mCH = 110/2
mCH = 55

is that right

[fallen vector]

The sum of the angles subtended by arcs in a circle equals the total arc measurement.
𝐶
𝐴
CA being the diameter implies that it subtends a right angle (
9
0
∘
90
∘
) at any point on the circle.

[distant spoke]

okay

.close

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[topaz sinew]

@pearl valve Has your question been resolved?

[pearl valve]

<@&286206848099549185>

If it isn't too much trouble.

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@pearl valve Has your question been resolved?

[smoky sparrow]

I'm confused

you want to disprove $\lim_{x \to -3/2} (1 - 4x) = 6$ right

[thorny flame]

south

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[surreal mural]

is that rank(B) = n?

[strong sable]

r

Also m≠n

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@strong sable Has your question been resolved?

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@strong sable Has your question been resolved?

[sharp wolf]

rank of the product (ab) is always less than or equal to the lowest rank out of its constituents (a, b)

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[sharp wolf]

so if m>r, rank(ab) is r

else m

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[halcyon owl]

can how to find AB guy

[halcyon owl]

help me =))

<@&286206848099549185>

guys its so simple =))

.solved

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[hollow star]

how do i solve this one algebraically

[sharp wolf]

get the coefficients of x outside

[hollow star]

huh

[sharp wolf]

let me rephrase

make the bases same

[hollow star]

yeah to what number

7x or 9x

[sharp wolf]

make it to the base 10 to make it simpler

[hollow star]

blud ts does NOT work

<@&286206848099549185> this guy is no help

[sharp wolf]

$\frac{\log(6\sqrt{21})}{\log(9x)} = \frac{\log(14\sqrt{3})}{\log(7x)}$

[thorny flame]

cps

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[hollow star]

ts did not help help help

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[devout flower]

The teacher has given us a challenge question, and I would like some help with completing it. My problem is with the number c) of the question. To solve it, I figured that making myself a sort of formula for Gn would be helpful, but I can’t seem to figure out what it would be.
Unfortunately, this homework is in French, and I’m not confident in my ability to translate it without losing the nuance of the question.

[devout flower]

This is about as far as I got for making the Gn formula

[hollow star]

answer is 15

[topaz sinew]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[hollow star]

my bad gang

[dense lily]

Wait

! nosols makes more sense

[topaz sinew]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[hollow star]

nuh

i know beter

[topaz sinew]

@devout flower Has your question been resolved?

[devout flower]

?

How did you come to that conclusion though?

[devout flower]

! nosols?

[topaz sinew]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dense lily]

That's a factoid

Unrelated to your question though, sorry.

[devout flower]

Oh, alright! Thank you,

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[jolly storm]

Hi, I don’t understand my mistake in pt c

[neon iron]

hi there

[jolly storm]

Hi!

[neon iron]

I have a feeling your examiner is wrong here

,w plot 1/(e^(x^2) + 3)

[thorny flame]

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Requested by alison_burg3r5

[pseudo horizon]

they're saying it's 1/4

[neon iron]

The answer should be ebtween 0 and 1/4

[neon iron]

oh okay lol :)

I thought they were saying that its between 0 and 4

[jolly storm]

So how should I find the domain?

[pseudo horizon]

maybe but i think they're just correcting the denominator

[neon iron]

No I think you are correct here. They are indeed correcting the denominator upon closer inspection

[pseudo horizon]

you wrote, as x goes to infinity y goes to 0
as x goes to 0 y goes to??

[neon iron]

ok so we know that for a function, the derivative at the $x_{\text{min}}$ of that function is 0. Let's differentiate $h(x)$

[thorny flame]

Edmund Cloudsley

[jolly storm]

Um I’m so sorry but I don’t know calculus yet

[thorny flame]

Edmund Cloudsley

[neon iron]

and then we set this derivative to 0 to find x_min

and x_max

and subsquently find the highest and lowest value of h(x)

remember that the second derivative of h(x) must be < 0 at the maximum and > 0 at the minimum

[neon iron]

oh okay lol. Then I guess you would have to rely on intuition

[jolly storm]

Is there a way to solve the domain without derivatives? Even if it’s longer?

[neon iron]

Course there is!

okay so you found out the lower value quite easily

as 0

and you had issues with the higher value

[jolly storm]

Yea

[pseudo horizon]

i think you just plugged in 0 and said e^0=0 instead of 1

[thorny flame]

Edmund Cloudsley

Edmund Cloudsley

[jolly storm]

I think what I thought was that the inside of a log couldn’t be 0 or negative, so I just said x couldn’t be 1/3 or less

For h^-1(x)

Is anyone there?

[pseudo horizon]

didn't you mix up the ln and sqrt in part (b)?

[jolly storm]

I’m not sure, part b was marked as fine so I haven’t really double checked it

[pseudo horizon]

i think you did

[jolly storm]

Wait I’ll redo my work

I’m getting the same inverse function

At least, I don’t know how to find the inverse in a way other than this

[pseudo horizon]

you can't take the sqrt like that

[jolly storm]

Oh ok, could u explain why? Also, what should I do instead?

[pseudo horizon]

because:

$e^{y^2}=e^{(y^2)}$

[thorny flame]

Axe

[pseudo horizon]

instead you should take the ln()

[jolly storm]

Oh like ln(stuff) = y^2 and then square root it after?

[pseudo horizon]

yeah

[jolly storm]

Got it, can u explain why the other way doesn’t work?

[pseudo horizon]

because you're trying to act like it's:

$\left(e^y\right)^2$

[thorny flame]

Axe

[jolly storm]

Don’t they both equal e^(2y) tho? Also, does that mean when dealing with exponents, I should always treat them as a term before considering their base?

[pseudo horizon]

no, they're different

yeah if there aren't parentheses you have to consider the exponent as a term before considering its base

[jolly storm]

Ok got it

So then with the square root function, how do I find the max?

[pseudo horizon]

the radicand must be >= 0

so you can solve:

$\ln(1/x-3)\geq 0$

[thorny flame]

Axe

[jolly storm]

Am I meant to solve the range for the above?

[pseudo horizon]

no you're looking for the domain of h^-1

did you find h^-1?

sorry, typo

[jolly storm]

This?

[pseudo horizon]

yeah

it will be defined when?

sqrt of a negative is undefined

[jolly storm]

So ln has to be +

[pseudo horizon]

or 0

[jolly storm]

Oh yea or 0

But how does ln = 0?

Oh wait when it equals 1 right?

[pseudo horizon]

yes

[jolly storm]

Like the inside

So when x is 1/3, ln is 0?

Oh wait no

[pseudo horizon]

when x is 1/3, ln is undefined

[jolly storm]

1/x = 4

[pseudo horizon]

yes

[jolly storm]

So x = 1/4!

Wait but if x goes above that, then ln inside becomes 0 or negative which just isn’t possible right?

Which is why x is less than or equal to 1/4

[pseudo horizon]

yeah

ln becomes negative

[jolly storm]

And if x is 0 or negative, then ln is undefined again

I think I got it!

Thank you!

[pseudo horizon]

you're welcome

[jolly storm]

How do I end this chat?

[pseudo horizon]

.close

[jolly storm]

Got it thank you!

.close

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[tepid tree]

Can I js get confirmation if this is correct or not pls

[peak glen]

yes

it's correct

[tepid tree]

Okay tysm

[rigid ivy]

Are you following order of operations correctly? Should I inerpret your math as
$$((5\div 1000)\times60)\times60$$

[thorny flame]

SWR

[tepid tree]

Yes

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[rigid ivy]

May I suggest a more complete way to write your unit conversions? It's a little more tedious, but it really helps you make sure that you converted your units correctly

Once it's all crossed out, you are left with km/hr

This is how I always did it and it always helped me

But it's just a suggestion

[tepid tree]

I will save this

Thank u very much

[rigid ivy]

Awesome. Hope it helps. And good luck on your sciences

[tepid tree]

<3

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[junior lake]

Help i got stuck in this step i dont know what to do with the numbers below

[acoustic pecan]

try simplify 3^(2x), or split it rather

or am i speaking the nonsense this time, indeed i am

you want to end up with something in the form u^2+pu+q=0
with p and q being constants

so how could we go about that

[junior lake]

Well

Chat gpt says this is the next step to simplify it

But idk how we got there

[acoustic pecan]

that saddens me

[stiff lava]

factor something out

[acoustic pecan]

more just subtract 108
then solve the quadratic however you please

[junior lake]

I just got it , thank you guys

.close

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[finite mural]

Help

[finite mural]

Si la soluci´on de un sistema de ecuaciones lineales ho-
mog´eneo es ´unica, la matriz de coeficientes del sistema es
invertible

[cedar wagon]

Unique solution only

Unique solution => invertible

[finite mural]

I have to say wheter is false or true

but this says is false

[rigid ivy]

you translated from spanish to spanish

[cedar wagon]

Oh

[river crown]

Hey azon, Im Adding and Subtracting Rational Expressions With Unlike Denominators and im a little confused could you help me out

[cedar wagon]

And i read it in fr

[rigid ivy]

I read it in es

[finite mural]

Bilinguals

[acoustic pecan]

#❓how-to-get-help

[finite mural]

I had one like that specifically the a in which by a theorem showing that the columsn are linearly independant is the same as beign invertible so the only chacne y is the hypothesis of the theorem that are square matrix is not true

[cedar wagon]

Sure i was translating

[finite mural]

But i not find simething like that here

[finite mural]

Acutally it may be that no? Imso dumb

I asked gemini and reasoned but got the same answer that whenever it has only one solutions ( and it is a square matrix is does have to be that way)

Thanks

.close

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[glass lantern]

Let n be a positive natural number. Prove that f(x) = x^n is continuous at every point of the domain

[thorny flame]

levi

[glass lantern]

∀ε > 0, ∃δ > 0 such that |x - p| < δ ⇒ |x^n - p^n| < ε

Let, say, n = 2.
Then |x² - p²| < ε ⇒ −ε < x² − p² < ε ⇒ −ε < (x − p)(x + p) < ε

(1) x + p is positive

−ε < (x − p)(x + p) < ε ⇒ −ε/(x + p) < x − p < ε/(x + p)

Then δ ≤ ε/(x + p)

oh

what happens if x = −p?

[pseudo horizon]

doesn't delta-epsilon say something like 0<|x-p|<delta?

[glass lantern]

Yes

[pseudo horizon]

oh wait

[glass lantern]

I kind of forgot about that

[pseudo horizon]

but it doesn't help in the way i thought

i missed a negative

[glass lantern]

I think it's different for continuity though

[pseudo horizon]

ok

[raven sparrow]

For those the usual trick is to do some sort of estimation on delta and pick the smallest that works.

The issue is that you get $|x^2 - p^2| = |x-p| \cdpt |x+p| < \delta |x+p| < \epsilon$ and you somehow need $|x+p|$ in terms of $\delta$.

[pseudo horizon]

if x=-p,

[thorny flame]

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

[pseudo horizon]

then (x+p)(x-p)=0

and -epsilon < 0 < epsilon

this will be true for any delta

[glass lantern]

I'm very confused, but let's keep going

Just let me try something

[raven sparrow]

=> is \implies or \Rightarrow

[glass lantern]

lmao thank you

$\forall\epsilon > 0, \exists\delta > 0 | -\delta < x - p < \delta \implies -\epsilon < x^n - p^n < \epsilon$

[thorny flame]

levi

[glass lantern]

seen worse

Still, n can be any natural positive number

Even if we solve for n = 2, there's still a lot to cover

[pseudo horizon]

you might be able to factor out x-p from x^n-p^n for any natural number

[glass lantern]

https://www.youtube.com/watch?v=b5MmJKGXbKw

[pseudo horizon]

lol

well it's gonna work out, because p is a zero of x^n-p^n

it's gonna be a sum

[glass lantern]

A zero?

Like a root?

[pseudo horizon]

yeah

[glass lantern]

What theorem is that?

[pseudo horizon]

$x^n-p^n=\left(x-p\right)\sum_{i=0}^{n-1}p^{i}x^{n-1-i}$

[thorny flame]

Axe

[pseudo horizon]

i used polynomial long division

[glass lantern]

is it just my adhd or does that symbol look like majin vegeta

[glass lantern]

someone just mentioned that

i'll look into it, thank you

[pseudo horizon]

sure

[glass lantern]

It's much simpler than that, actually

|xⁿ - pⁿ| < ε
−ε < xⁿ - pⁿ < ε
−ε + pⁿ < xⁿ < ε + pⁿ

Assume n is odd. Then:

−ε + pⁿ < xⁿ < ε + pⁿ
√(−ε + pⁿ) < x < √(ε + pⁿ),
where √ is the nth root

If n is even and p = 0, then
−ε + pⁿ < xⁿ < ε + pⁿ
−ε < xⁿ < ε
|x| < √ε
−√ε < x < √ε

why did i choose maths

[pseudo horizon]

well it seems like you're getting there but i don't think you finished the n odd case

[glass lantern]

Yeah

It’s kind of pointless anyway because I found the solution on the internet

If I learned anything from that is that I should take roots to get rid of exponents

And just isolate that xⁿ to define an interval

If n is even and p > 0, then
−ε + pⁿ < xⁿ < ε + pⁿ
|x| < √(ε + pⁿ)
–√(ε + pⁿ) < x < √(ε + pⁿ)

[pseudo horizon]

first step seems sus

[glass lantern]

How so?

[pseudo horizon]

how did you get that?

remember the implication needs to work in reverse

[glass lantern]

|xⁿ – pⁿ| < ε
–ε < xⁿ – pⁿ < ε
–ε + pⁿ < xⁿ < ε + pⁿ

[pseudo horizon]

yeah up to there seems fine

[glass lantern]

Let p > 0

–ε + pⁿ < xⁿ < ε + pⁿ
√(–ε + pⁿ) < |x| < √(ε + pⁿ)
||x|| < √(ε + pⁿ)

I think

Since |x| ≥ 0, then there exists √|x|, for all x

[pseudo horizon]

umm

what if x is -2

and n is 2

[glass lantern]

Whatever, because n is even

[pseudo horizon]

sorry not x i mean p

[glass lantern]

That’s assuming p > 0

I don’t really know what to do if p < 0 lmao

[pseudo horizon]

oh

[glass lantern]

oh

But n is even

So it’s whatever either way

|xⁿ| < ε + pⁿ

n > 0, so xⁿ ≥ 0, so |xⁿ| = xⁿ and √|xⁿ| = √xⁿ = |x|

|xⁿ| < ε + pⁿ
√|xⁿ| < √(ε + pⁿ)
|x| < √(ε + pⁿ)

Not sure what that means anyway

Oh

Since ε > and pⁿ > 0, then ε + pⁿ > 0 and, more importantly, √(ε + pⁿ) > 0

[pseudo horizon]

|x| < √(ε + pⁿ) implies –ε + pⁿ < xⁿ ??

[glass lantern]

|x| < √(ε + pⁿ) implies –√(ε + pⁿ) < x < √(ε + pⁿ)

And since √(ε + pⁿ) is always defined, then there exists –√(ε + pⁿ)!

So x is within that interval

[pseudo horizon]

ok i might not be able to help lol

i think i am lost

[glass lantern]

lmao that's just calculus

i mean who doesnt get confused

[glass lantern]

b > 0
|a| < b ⇔ –b < a < b

Ex.:

|5| < 6 ⇔ –6 < 5 < 6
|–5| < 6 ⇔ –6 < –5 < 6

It makes sense. In the number line, if we think of the absolute value of a number as the distance from that number to 0, then

|–5| < 6 means 6 is further away from 0 than –5 is

Therefore, –6 should also be further away from 0 than –5, because –6 has the same distance from 0 than 6

[pseudo horizon]

my brain is fried, you must be right

sorry

[glass lantern]

When you realise it, you’ll be like OMG

[pseudo horizon]

let x=1/2, p=1, e=1/2, n=2
|1/2|<sqrt(1/2+1^2) =>? -1/2+1^2<(1/2)^2
1/2<sqrt(3/2) =>? 1/2<1/4

this fails

[topaz sinew]

@glass lantern Has your question been resolved?

[glass lantern]

|½| < √(½ + 1)
–√(½ + 1) < ½ < √(½ + 1)
–√(3/2) < ½ < √(3/2)

[pseudo horizon]

nothing wrong with what you've written there

[glass lantern]

lmao

[glass lantern]

I don’t get what you did

[pseudo horizon]

ok we're looking for delta
we have this condition
−ε + pⁿ < xⁿ < ε + pⁿ
and you went to this
|x| < √(ε + pⁿ)

[glass lantern]

Go on

[pseudo horizon]

if we are to end up with a condition like |x-p|< something...

that condition must imply the previous statement

which must imply the previous

yes?

[glass lantern]

Indeed

[pseudo horizon]

so |x| < √(ε + pⁿ) must imply −ε + pⁿ < xⁿ < ε + pⁿ

[glass lantern]

Sure

[pseudo horizon]

just remove one restriction, and
|x| < √(ε + pⁿ) must imply −ε + pⁿ < xⁿ

[glass lantern]

Yes

[pseudo horizon]

that fails with the given values

x=1/2, p=1, e=1/2, n=2

[glass lantern]

So it’s |½| < √(½ + 1²), right?

[pseudo horizon]

yes

that is a false statement, yes?

sorry

it's true

[glass lantern]

It’s true, in fact

[pseudo horizon]

lol

yes it's true

but what about −ε + pⁿ < xⁿ

[glass lantern]

|½| < √(½ + 1)
–√(½ + 1) < ½ < √(½ + 1)
–√(½ + 1) < ½

I think there’s the problem

It’s not like equality where you can just put ² and be done with it

[pseudo horizon]

you can't square both sides of an inequality because one or both of the sides might be negative which might flip the inequality

[glass lantern]

Square is a nice verb

[pseudo horizon]

[glass lantern]

Yes, which is obviously –√(½ + 1) that is negative

[pseudo horizon]

and besides you'd be missing a negative before the epsilon

[glass lantern]

lmao

always a missed sign

now i'm confused

that property is true though

trust me bro

[pseudo horizon]

what property?

this one?
|a|<b => -b<a<b

that is true

[glass lantern]

oh

then what are we discussing

that’s exactly the property i applied

[pseudo horizon]

[glass lantern]

[pseudo horizon]

this is all true and valid, it just doesn't imply −ε + pⁿ < xⁿ

[glass lantern]

I mean, it depends on what values of epsilon and x you choose, no?

And p as well

[pseudo horizon]

yes

[glass lantern]

I should've went with Stewart

[pseudo horizon]

sorry i have to eat dinner now

[glass lantern]

k tysm for your time and help though

[raven sparrow]

I think Axe had correct doubts about your argument.
This is from the even case, but I think the same thing applies to the others.

−ε + pⁿ < xⁿ < ε + pⁿ
|x| < √(ε + pⁿ) <---------- this doesn't follow from the above, it's not strict enough
–√(ε + pⁿ) < x < √(ε + pⁿ)

since it would mean that the inequality you start with is
−ε - pⁿ < xⁿ < ε + pⁿ, when it's not...

The problem is that if you have the case, say, n=2, p=4 and ε = 1, then your last inequality says that you should have |x| < sqrt(17) for this to work. But clearly if you take like x=0 this can't be the case, since |0^2 - 4^2| = 16 > 1 = ε.

For any n, for the p=0 case, you can take the nth root on both sides of a chain of inequality like this, but in general this doesn't work

[pseudo horizon]

delta can be in terms of p but not x right?

[raven sparrow]

Indeed.

delta restricts x to some distance of p, so it can't depend on x.

[pseudo horizon]

then the whole factoring thing from earlier is irrelevant

[raven sparrow]

Not exactly

For the general case, the standard trick is to assume that delta is at most some number, like 1 for instance.
Then you can go about the factoring argument and use that to get your bound on delta and take delta to be whichever is smallest between 1 and that, because x will be at least bounded above by p + 1.

The rationale behind that is that if epsilon is really big, you allow so much error in the output that delta = 1 is sufficient. If it's smaller then you can ensure a better bound.

[pseudo horizon]

ah, that seems promising

[raven sparrow]

In practice, it's helpful because in a term like $(x-p)(x+p)$, you want a bound on $x+p$ to be able to say that $$|x-p| < \frac{\epsilon}{\text{something that doesn't have any x in it}}.$$

If $x$ has an upper bound, say $p+1$, then this factor is bounded above and you can argue that
$|x-p| |x+p| < |x-p| |2p + 1| < \epsilon$
and you have an upper bound on delta :

[thorny flame]

Azyrashacorki

[raven sparrow]

This works for n=2 in this case, but the same argument can be adapted to arbitrary n, since the remaining factor you get after the factoring thing is always bounded above when x is bounded above.

[pseudo horizon]

where did you get 2p+1 though

oh i see

[raven sparrow]

It's not particularly evident why you'd do that though, but it's a trick used enough across different epsilon-delta proofs that one should have it in their arsenal.

Another maybe easier approach to this specific problem would be to show that the product of continuous functions is continuous, and then only prove that f(x) = x is continuous. It follows directly that x^n is continuous.

[topaz sinew]

@glass lantern Has your question been resolved?

[pseudo horizon]

okay

$\left|x-p\right| < -\left|p\right|+\left(\epsilon+p^n\right)^{1/n}
\implies
-\epsilon+p^n < x^n < \epsilon+p^n$

[thorny flame]

Axe

[pseudo horizon]

that might work for even n

[topaz sinew]

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[lucid cedar]

What happened.

My answer and symbolab's answer do not match

[wooden osprey]

what's your answer

[lucid cedar]

-Ln|x|-Ln|x+1|

[lucid cedar]

first, I factor -(x^2-x) as -x*(x+1)

[river valve]

that looks right from partial fractions

have you tried simplifying the symbolab answer

[lucid cedar]

Ah, how?

I do not know how to do that.

[river valve]

use the log rule, log(a) - log(b) = log(a/b)

[lucid cedar]

I mean

I end up with -Ln |x|/Ln|x+1|

[river valve]

and if you apply that rule again thats the same as your answer

[lucid cedar]

How?

[river valve]

log(a) - log(b) = log(a/b) ==> $-ln|x| / ln|x+1| = -ln|x| - ln|x+1|$

[thorny flame]

Adum

[wooden osprey]

$\log a-\log b = \log\frac a b$

[lucid cedar]

but, how can I apply it again? I already have -Ln|x|/Ln|x+1|

[golden mesa]

so apply log laws to this too, and it should be the same answer

[river valve]

i misread this, i think you applied the rule wrong it should be $-ln\frac{|x|}{|x+1|}$ at this step

[thorny flame]

Adum

[lucid cedar]

oh, what?

How?

[river valve]

$\ln|2x+2| - \ln 2|x| \to \ln \frac{|x+1|}{|x|} \to \ln |x+1| - \ln |x|$

[thorny flame]

Adum

[river valve]

now subtract the extra $2 \ln|x+1|$, and its your answer

[thorny flame]

Adum

[topaz sinew]

@lucid cedar Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[viscid wren]

here is the function i(x)=kx²+kx+15. one of the abscissas at the origin is 1.5. Find the value of k and the other abscissa

[viscid wren]

I tried but I just can't

[finite storm]

one of the abscissas at the origin is 1.5?

[viscid wren]

yes

[finite storm]

isnt origin (0,0)

[viscid wren]

i translated my exo in french so maybe it doesn't mean the same

[pseudo horizon]

do you mean one of the roots / zeroes?

[viscid wren]

yes

[pseudo horizon]

then i(1.5)=0

[viscid wren]

yes

(1.5, 0)

[finite storm]

oh makes sense

[pseudo horizon]

you can plug this into the polynomial and solve for k

[viscid wren]

okay

[pseudo horizon]

i(x)=kx²+kx+15
try plugging in x=1.5 and i(x)=0

[viscid wren]

do k is -4?

so*

[pseudo horizon]

that seems right

[viscid wren]

okay thanks!

[pseudo horizon]

you're welcome

[viscid wren]

.close

[topaz sinew]

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• Show your work, and if possible, explain where you are stuck.
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[rocky pumice]

im sure i did something wrong here

[sweet shard]

,w factor 2k^2 + 6k +4

[thorny flame]

Results provided by WolframAlpha

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

Requested by rie.mann

[sweet shard]

Nope. Just gotta keep going

[rocky pumice]

it would be wrong if i didnt factor it?

[sweet shard]

Probably

[red star]

You have to factor for it to represent the question

[sweet shard]

2n(n+1) becomes this when you substitute n+1 for n

[rocky pumice]

okay cool cool

i understand 😄

thanks

.close

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[mortal shadow]

is this correct? for $E_{-2}$ i keep getting s[1 0 0] + t[-1 0 0] and my friend is too

[thorny flame]

sunsick